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Full text of "Applied mechanics for engineers"

.11111 



BY J. DUNCAN, Wh. Ex. 

APPLIED MECHANICS FOR BEGINNERS. Globe 
8vo. 35. 6d. 

STEAM AND OTHER ENGINES. Globe 8vo. 6s. 
MECHANICS AND HEAT. Crown 8vo. 55. 

AN INTRODUCTION TO ENGINEERING DRAW- 
ING. Crown 8vo. 45. 

BY J. DUNCAN, Wh. Ex., and 
S. G. STARLING, B.Sc. 

A TEXT-BOOK OF PHYSICS FOR THE USE OF 
STUDENTS OF SCIENCE AND ENGINEERING. 

Illustrated. Extra crown 8vo, i8s. Also in Parts : 
Dynamics, 6s. ; Heat, Light, and Sound, 7s. 6d. ; 
Magnetism and Electricity, 55. ; Heat, 45. 6d. ; Light 
and Sound, 45. 6d. ; Heat and Light, 6s. 



LONDON: MACMILLAN & CO., LTD. 



APPLIED MECHANICS FOR ENGINEERS 



MACMILLAN AND CO,. LIMITED 

LONDON BOMBAY CALCUTTA MADRAS 
MELBOURNE 

THE MACMILLAN COMPANY 

NEW YORK BOSTON CHICAGO 
DALLAS SAN FRANCISCO 

THE MACMILLAN CO. OF CANADA, LTD. 

TORONTO 



APPLIED MECHANICS 
FOR ENGINEERS 



BY 

J. DUNCAN, WH.EX., M.I.MECH.E. 

HEAD OF THE DEPARTMENT OF ENGINEERING AT THE WEST HAM MUNICIPAL COLLEGE 
AUTHOR OF 'APPLIED MECHANICS FOR BEGINNERS,' 'STEAM AND OTHER ENGINES,' 

'MECHANICS AND HEAT,' ETC. 
JOINT AUTHOR OF 'TEXT-BOOK OF PHYSICS* 



MACMILLAN AND CO. LIMITED 

ST. MARTIN'S STREET, LONDON 

1926 



COPYRIGHT 

First Edition, 1913. 
Reprinted 1920, 1922, 1926. 



PRINTED IN GREAT BRITAIN 



PREFACE 

THE author's object in writing this book has been to provide a 
practical statement of the principles of Mechanics. The arrangement 
adopted is similar to that of his Applied Mechanics for Beginners. 
Great pains have been taken to make the treatment adequate ; prin- 
ciples have been illustrated by numerous fully worked-out examples, 
and exercises for home or class work have been provided at the ends 
of the chapters. The working out of typical exercises must be done 
by every student of Mechanics, but the mere ability to solve examina- 
tion questions is not the only service the study of Applied Mechanics 
can render the Engineer. The problems met with in actual engineer- 
ing practice often differ greatly from the text-book form of exercise, 
and the student of Mechanics, in addition to a sound knowledge of 
principles, must learn to appreciate the assumptions involved and the 
consequent limitations which arise in their practical applications. 

Consequently, the student must be provided with frequent oppor- 
tunities for performing suitable experiments under workshop condi- 
tions. In the mechanical laboratory he must come into touch with 
practical problems, and there learn to test and apply his knowledge of 
principles, and in this work he should have the assistance of a teacher 
and the criticism of fellow-students. But if the whole value of such 
laboratory work is to be secured, no slip-shod working out of results 
must be tolerated. In recognition of the supreme importance of the 
experience gained in the laboratory, many suitable experiments have 
been described, and these have been arranged on p. xi to provide a 
connected course of practical work. The nature and scope of the 
apparatus available in different laboratories vary greatly, and some of 
the experiments included are given as suggestions only, so as to be 
applicable to any form of machine or instrument. 

Students using the book must have a knowledge of Algebra up to 
quadratic equations, and of Trigonometry to the simple properties of 
triangles. They should be acquainted also with about half-a dozen 

677095 



vi PREFACE 

rules of the Calculus, and these are given in Chapter I. Students 
able to integrate x n dx, and to differentiate #", sin#, and cosx, will be 
able to understand practically the whole volume. 

Though no particular examination syllabus has been followed, the 
book should be of service to students preparing for University degrees 
in Engineering, for the examinations of the Institutions of Civil 
Engineers and of Mechanical Engineers, and for the higher examina- 
tions of the Board of Education and the City and Guilds of London 
Institute. 

Exercises marked B.E. are from recent examination papers of the 
Board of Education, and are reprinted by permission of the Con- 
troller of H.M. Stationery Office; those marked I.C.E. are taken 
from recent examination papers of the Institution of Civil Engineers, 
and are reprinted by permission of the publishers, Messrs. W. Clowes 
& Sons. Exercises marked L.U. are reprinted, with permission, from 
recent examination papers for B.Sc. (Eng.) of London University. 

It is impossible to give in a book of moderate size a complete state- 
ment of all subjects of Applied Mechanics. For fuller information on 
special matters the student is referred to separate treatises ; the names 
of some of these are noted in the text, and the author takes the 
opportunity of acknowledging his own indebtedness to them, especially 
to Strength of Materials > by Sir J. A. Ewing (Cambridge University 
Press), and to Machine Design^ by Prof. W. C. Unwin (Longmans). 

Sir Richard Gregory and Mr. A. T. Simmons have read the proofs, 
and to their expert knowledge of books and book production the 
author owes a heavy debt of gratitude. Thanks are also due to Mr. 
L. Wyld, B.Sc., Assistant Lecturer at West Ham Institute, who has 
read the proofs and checked the whole of the mathematical work and 
the answers to the exercises ; it is hoped that his care has had the 
effect of reducing the number of errors to a minimum. 

The apparatus represented in Figs. 706, 707 and 715 is made by 
Mr. A. Macklow-Smith, Queen Anne's Chambers, Westminster, and 
the illustrations have been reproduced from working drawings kindly 
supplied by him. The illustration of a chain (Fig. 585) is inserted 
by permission of Messrs. Hans Renold, Ltd. The Tables of 
Logarithms and Trigonometrical Ratios are reprinted from Mr. F. 
Castle's Machine Construction and Drawing (Macmillan). 

J. DUNCAN 
WEST HAM, September, 1913. 



CONTENTS 



PART I. MATERIALS AND STRUCTURES. 

CHAPTER I. 
Introductory Principles - p. I 

CHAPTER II. 

Forces Acting at a Point. Parallelogram and Triangle of Forces. 
Analytical and Graphical Conditions of Equilibrium for any System 
of Concurrent Forces. Polygon of Forces - p. 19 

CHAPTER III. 

Parallel Forces. Principle of Moments. Resultant and Centre of 
Parallel Forces. Centre of Gravity. Reactions of the Supports of 
Beams. Parallel Forces not in the same Plane - - p. 40 

CHAPTER iy. 

Properties of Couples. Analytical and Graphical Conditions of Equi- 
librium of Systems of Forces in the same Plane. Link Polygon 

P- 59 

CHAPTER V. 

Simple Structures. Force Diagrams. Effects of Dead and Wind Loads 

P. 77 

CHAPTER VI. 

Simple Stresses and Strains. Cylindrical and Spherical Shells. Riveted 
Joints. Elasticity. Stresses Produced by Change in Temperature. 
^Normal, Shear and Oblique Stresses. Stress Figures - p. 93 



viii CONTENTS 



CHAPTER VII. 

Strength of Beams. Bending Moments and Shearing Forces. Neutral 
Axis. Moment of Resistance. Modulus of a Beam Section. Beams 
of Uniform Strength. Distribution of Shear Stress - - p. 131 



CHAPTER VIII. 

Deflection of Beams. Curvature. Slope. Standard Cases of Beams. 
Encastre' Beams. Points of Contraflexure. Propped Cantilevers and 
Beams. Beams of Uniform Curvature - - - - p. 163 



CHAPTER IX. 

Beams and Girders. Resilience. Gradual, Sudden and Impulsive Loads. 
Working Loads and Stresses. Wind Pressure. Travelling Loads. 
Continuous Beams. Bridge Girders. Reinforced Concrete Beams 

p. 191 

CHAPTER X. 

Columns. Euler's, Rankine's and other Formulae for Columns. Effects 
of non-Axial Loading. Arches. Suspension Bridges - p. 227 



CHAPTER XL 

Shafts. Pure Torque. Solid and Hollow Shafts. Torsional Rigidity. 
Horse-power Transmitted. Principal Stresses. Combined Bending 
and Torque. Helical Springs. Piston Rings. Coach Springs p. 251 



CHAPTER XII. 

Earth Pressure. Rankine's Theory. Wedge Theory. Retaining Walls 
and Foundations p. 279 



CHAPTER XIII. 

Experimental Work on the Strength and Elastic Properties of Materials 

p. 292 



CONTENTS ix 



PART II. MA CHINES AND H YDRA ULICS. 

CHAPTER XIV. 

Work. Energy. Power. Machines. Diagrams of Work. Indicated and 
Brake Horse-power. Dynamometers. Torsion-meters - p. 325 

CHAPTER XV. 

Friction. Laws of Friction for Dry and Lubricated Surfaces. Machine 
Bearings. Journals and Pivots. Inclined Planes. Screws. Friction 
Circle. Friction in Engine Mechanisms - - - . p. 353 

CHAPTER XVI. 

Velocity. Acceleration. Velocity and Acceleration Diagrams. Angular 
Velocity and Acceleration. Change of Velocity. Motion in a Circle. 
Simple Harmonic Motion. Relative Velocity - - - p. 381 

CHAPTER XVII. 

Inertia. Absolute Units of Force. Kinetic Energy. Momentum. Im- 
pulsive Forces. Centre of Mass. Rotational Inertia. Moments of 
Inertia. Kinetic Energy of Rotation. Energy of Rolling Wheels. 
Centrifugal Force - - p. 406 

CHAPTER XVIII. 

Angular Momentum. Gyrostatic Action. Simple Harmonic Vibrations 
and Torsional Oscillations. Simple and Compound Pendulums. 
Centres of Oscillation and Percussion. Equivalent Dynamical 
Systems - - - - - - - - - p. 430 

CHAPTER XIX. 

Link Mechanisms. Instantaneous Centres. Kinematic Chains. Inertia 
Effects in Mechanisms. Klein's Construction. Crank Effort and 
Turning Moment Diagrams. Valve Diagrams. Miscellaneous 
Mechanisms p. 455 

CHAPTER XX. 

Flywheels. Governors. Balancing of Rotating and Reciprocating 
Masses. Whirling of Shafts - - p. 494 



CONTENTS 



CHAPTER XXI. 

Transmission of Motion by Belts, Ropes, Chains, and Toothed Wheels. 

Epicyclic and other Trains of Wheels. Shaft Couplings p. 526 

CHAPTER XXII. 

Hydraulic Pressure. Centre of Pressure. Stability of Floating Bodies. 
Reservoir Walls. Hydraulic Transmission of Energy. Hydraulic 

Pressure Machines. Reciprocating Pumps - p. 564 

CHAPtER XXIII. 

Flow of Fluids. Bernoulli's Law. Flow through Orifices and Weirs. 
Flow through Pipes. Sudden Enlargements and Contractions. 

Bends and Elbows Resistance of Ships - p. 592 

CHAPTER XXIV. 

Pressures of Jets on Fixed and Moving Vanes. Hydraulic Turbines. 

Centrifugal Pumps - p. 625 

CHAPTER XXV. 

Hydraulic Experiments - p. 663 

TABLES 

Weights and Specific Gravities - - - - p. 5 

Differential Coefficients - - p. 1 1 

Integrals - p. 17 

Properties of Sections - P- 1 5 1 

Coefficients for Columns - p. 235 

Coefficients of Friction - p. 355 

Journal Friction, Oil Bath - p. 356 

Moments of Inertia - - p. 415 

Useful Constants - p. 68 1 

Coefficients of Expansion - p. 682 

Strength, etc., of Materials - p. 683 

Logarithms - p. 684 

Antilogarithms p. 686 

Trigonometrical Ratios p. 688 

ANSWERS - p. 689 

INDEX p. 710 



COURSE OF LABORATORY EXPERIMENTS 

INSTRUCTIONS FOR CARRYING OUT 
LABORATORY WORK 

General Instructions. Two Laboratory Note-books are required ; in 
one rough notes of the experiments should be made, and in the other a 
fair copy of them in ink should be entered. 

Before commencing any experiment, make sure that you understand 
what its object is, and also the construction of the apparatus and instru- 
ments employed. 

Reasonable care should be exercised in order to avoid damage to 
apparatus, and to secure fairly accurate results. 

In writing up the results, enter the notes in the following order : 

(1) The title of the experiment and the date on which it was 
performed. 

(2) Sketches and descriptions of any special apparatus or instruments 
used. 

(3) The object of the experiment. 

(4) Dimensions, weights, etc., required for working out the results ; 
from these values calculate any constants required. 

(5) Log of the experiment, entered in tabular form where possible, 
together with any remarks necessary. 

(6) Work out the results of the experiment and tabulate them where 
possible. 

(7) Plot any curves required. 

(8) Work out any general equations required. 

(9) Where possible, state any general conclusions which may be 
deduced from the results, and compare the results obtained with those 
which may be derived from theory. Account for any discrepancies. 

Notes should not be left in the rough form for several days ; it is much 
better to work out the results and enter them directly after the experiments 
have been performed. 



3rii COURSE OF LABORATORY EXPERIMENTS 



STATICS. 

1. Parallelogram of Forces - ----- p. 22 

2. Forces acting on a Pendulum - - - - p. 29 

3. Forces in a Simple Roof Truss - p. 30 

4. Forces in a Derrick Crane - - - - - - - P- 33 

5. Forces in a Wall Crane p. 34 

6. Principle of Moments - ------p. 55 

7. Reactions of a Beam - - - - - - - P- 55 

8. Centres of Gravity of Sheets p. 56 

9. Centre of Gravity of a Solid Body - - - p. 56 

10. Equilibrium of Two Equal Opposing Couples - p. 71 

11. Couples acting on a Door - P- 7 1 

12. Link Polygon p. 7 1 

13. Hanging Cord - p. 72 

14. Hanging Chain - p. 73 



STRENGTH AND ELASTICITY OF MATERIALS. 

15. Elastic Stretching of Wires - p. 292 

1 6. Tensile Tests to Rupture on Wires p. 294 

17. Torsion Tests on Wires - p. 295 

1 8. Extensions of Helical Springs - p. 296 

19. C by Maxwell's Needle - p. 297 

20. E by Torsional Oscillations of a Spring - p. 298 

21. C by Longitudinal Vibrations of a Spring p. 298 

22. E by Bending a Beam - - p- 3 

23. E by Bending a Cantilever and a Beam fixed at Both Ends p. 301 

24. Slope of a Cantilever p. 3 01 

Experiments Nos. 25 to 32 require the use of special testing machines 
and are included as suggestions. The instructions given in Chap. 
XIII. may require modification, depending on the scope and type 
of apparatus available. 

25. E for Various Materials, by use of an Extensometer - - p. 309 

26. Yield Stress and Ultimate Tensile Strength p. 311 

27. Bending Tests - - - - p. 312 

28. Shearing Tests P- 3*5 



COURSE OF LABORATORY EXPERIMENTS xiii 

29. Punching Tests - - p. 316 

30. Torsion Tests to Rupture - - - p. 317 

31. Elastic Torsion Tests - p. 319 

32. Cement, Brick and Stone Tests - - p. 320 



FRICTION AND EFFICIENCY IN MACHINES. 

33. Efficiency of a Lifting Crab - - p. 329 

34- Pulley Blocks - p. 332 

35. Weston's Differential Blocks p. 333 

36. a Wheel and Differential Axle - p. 333 

37. ,, a set of Helical Blocks - p. 333 

38. Friction of a Slider - - p. 375 

39. Angle of Sliding Friction - - p. 375 

40. Rolling Friction - - p. 375 

41. Effect on Friction of Speed of Rubbing - p. 376 

42. Friction of a Screw - p. 377 



MOTION, ENERGY, ETC. 

43. Verification of Law, F = ma - - p. 446 

44. Moment of Inertia of a Flywheel - - p. 447 

45. Centres of Oscillation and Percussion - - p. 449 

46. Centre of Percussion of a Bar - - p. 449 

47. Radius of Gyration about an Axis passing through the Mass 

Centre - p. 449 

48. Wheel rolling down an Incline - p. 450 

49. Balancing of Rotating Masses - p. 513 



HYDRAULICS. 

50. Flow through Orifices - - ' p. 665 

51. Coefficient of Velocity for a Round Orifice p. 665 

52. Flow over Gauge Notches - - p. 666 

53. Bernoulli's Law - - p. 667 

54. Venturi Meter - p. 668 

55. Critical Velocity in a Pipe p. 669 



xiv COURSE OF LABORATORY EXPERIMENTS 

56. Frictional Resistance in a Pipe - p. 669 

57. Loss of Head at Bends - - - p. 671 

58. Loss of Head at an Elbow - p. 672 

59. Sudden Enlargements and Contractions in a Pipe - - p. 672 

60. Pressure of a Jet impinging on a Plate - p. 674 

61. Horse-Power and Efficiency of a PeHon Wheel - p 675 



PART I. 
MATERIALS AND STRUCTURES. 

CHAPTER I. 

INTRODUCTORY PRINCIPLES. 

Definition of terms. Applied mechanics treats of those laws 
of force and the effects of force upon matter which apply to works 
of human art. It will suffice to define matter as anything which 
occupies space. Matter exists in many different forms, and can 
often be changed from one form to another, but man cannot create 
it, nor can he annihilate it. Any given piece of matter, occupying 
a definite space, is called a body. Force may exert push or pull 
on a body ; force may change or tend to change a body's state of 
rest or of motion. 

Statics is that part of the subject embracing all questions in 
which the forces applied to a body do not produce a disturbance 
in its state of rest or motion. When we speak of a body's motion 
we mean its motion relative to other bodies. Rest is merely a 
relative term ; no body, so far as we are aware, is actually at rest ; 
but if its position is not changing in relation to other neighbouring 
bodies, we say it is at rest. In the same way, when we speak of 
a body's motion we mean the change of position which is being 
effected relative to neighbouring bodies. Change of the state of rest 
or of motion may be secured by the application of a force or forces, 
but if the forces applied are self-equilibrating, i.e. balance among 
themselves, no change of motion will occur. Kinetics includes all 
problems in which change of motion occurs as a consequence of 
the application of forces. 

There is another division of the subject called kinematics. This 
division may be defined as the geometry of motion, and has no 
reference to the forces which may be required for the production 
D.M. A d 



MATERIALS AND STRUCTURES 



of jhe; moU 3n^ .Problems -arise in kinematics such as the curves 
described by moving points in a mechanism, and the velocities of 
these points at any instant. 

Measurement of matter. Matter is measured by the mass, or 
quantity of matter, it contains. The standard unit of mass for this 
country is the pound mass, which may be denned as the quantity of 
matter contained in a certain piece of platinum preserved in the 
Exchequer Office. A gallon of water at 62F. has a mass of 10 
pounds. In cases where a larger unit is desirable, the ton, contain- 
ing 2240 pounds, or the hundredweight, containing 112 pounds, 
may be used. Generally speaking, it is best to state results in tons 
and decimals of a ton, or in pounds and decimals of a pound. 

In countries using the metric system, the unit of mass employed 
is the gram. This may be defined as the quantity of matter con- 
tained in a cubic centimetre of pure water at the temperature of 
4C. Where a larger unit is required, the kilogram may be used, 
being a mass of 1000 grams. 

The term density refers to the mass of unit volume of a substance. 
Thus, in the British system, the density of water is about 62-5, there 
being 62-5 pounds mass in one cubic foot of water. The density of 
cast iron in the same system is about 450 pounds per cubic foot. 
The density of water in the metric system is i, and of cast iron 7-2, 
these numbers giving the mass in grams in one cubic centimetre 
of water and cast iron respectively. 

Measurement of force. Forces may be measured by com- 
parison with the weight of the unit of mass. Thus, the weight of 
the one pound mass, or that of the gram, may be taken as units of 
force, and as these depend on gravitational effort they are referred 
to as gravitational units of force. The attraction exerted by the 
earth in producing the effect known as the weight of a body varies 
in different latitudes, hence gravitational units of force have the 
disadvantage of possessing variable magnitudes. The variation can 
be disregarded in many engineering calculations, as it affects the 
result to a very small extent only. Other practical gravitational units 
of force are the weight of one ton (2240 Ib.) and the weight of 
a kilogram (1000 grams or 2-2 Ib. nearly). 

An absolute unit of force does not vary, as it is defined in relation 
to the invariable units of mass, length and time belonging to the 
system. In the British system, the absolute unit of force is called 
the poundai, and has such a magnitude that, if it acts on one pound 
mass, assumed to be perfectly free to move, for one second, it will 



INTRODUCTORY PRINCIPLES 



produce a velocity of one foot per second. The metric absolute 
unit of force is the dyne, and will produce a velocity of one centi- 
metre per second if it acts for one second on a gram mass which 
is perfectly free to move. The poundal is equal roughly to the 

weight of half-an-ounce, or, accurately, it is equal to - Ib. weight, 

o 

g being the rate at which a body falling freely increases its speed. 
For all parts of Britain g may be taken as 32-2 in feet and second 
units, or 981 in centimetre and second units. On this basis, the 

dyne will be - gram weight, or 981 dynes equal one gram weight 
nearly. * 

Newton's laws of motion. In connection with the above 
definitions, it is useful to study the laws of motion laid down by 
Newton. These laws form the basis of all principles in mechanics, 
and are three in number. 

First law. Every body continues in its state of rest or of uniform 
motion in a straight line except in so far as it is compelled by forces 
to change that state. 

Second law. Change of momentum is proportional to the applied 
force, and takes place in the direction in which the force acts. 

Third law. To every action there is always an equal and contrary 
reaction ; or, the mutual actions of any two bodies are always equal 
and oppositely directed. 

The first law expresses what is called the inertia of a body, i.e. that 
property whereby it resists any effort made to change either the 
magnitude of its velocity or the direction of its motion. In the 
second law, the term momentum may be here understood to mean 
quantity of motion, measured by the product of the body's mass 
and velocity. The law expresses the observed facts that change 
in the magnitude of the velocity of a given body is proportional 
to the force applied, and change in the direction of motion takes 
place in the line of the force. The third law also expresses 
observed facts. It is impossible to apply a single force; there 
must always be an equal opposite force. One end of a string 
cannot be pulled unless an equal opposite pull be applied to the 
other end. If the body used be free to move and an effort be 
applied, the velocity will change continuously and the inertia of 
the body provides the resistance equal and opposite to the force 
applied. 

Experimental measurement of mass and force. Masses may 
be compared by means of a common balance (Fig. i). In this 



MATERIALS AND STRUCTURES 





FIG. i. Common balance. 



appliance, a beam AB, pivoted at its centre, will become horizontal, 
or will describe small equal angles on each side of the horizontal 
when equal forces are applied at A and B. Such equal forces will 

arise when bodies C and D, having equal 
masses, are placed in the pans. This 
follows as a consequence of the fact that 
equal masses have equal weights at the 
same part of the earth's surface. Further, 
no matter at what part of the earth the 
balance is used, it will always indicate 
equal masses. It therefore follows that such a balance could not 
be used to indicate the variation of a body's weight in different 
places. 

Spring balances (Fig. 2) may be used to measure forces by 
observation of the extensions produced in a spring. As equal 
masses have equal weights, such balances will indicate 
the same scale reading for equal masses, but as it is the ^) 
weight of the body which produces the extension of 
the spring, and as it is known that the extension is 
proportional to the force applied, it follows that change 
of weight, such as would be produced by taking the 
balance to another part of the earth's surface, will be 
evidenced by a different scale reading. As has been 
already mentioned, such difference is very small. Spring 
balances are generally calibrated in a vertical position, 
as shown in Fig. 2, and will not indicate quite the same 
force when the balance is used in an inclined or inverted 
position. This is owing to zero on the scale being 
marked for the spring extension corresponding to the 
weights of the parts of the balance suspended from the 
spring, but no load on the hook or scale pan. Con- 
sequently the zero will change if the balance is used 
in any position other than that shown. 

Specific gravity. The specific gravity of a substance is the 
weight of a given volume of the substance as compared with the 
weight of an equal volume of pure water. Specific gravities are 
usually measured at a temperature of 60 Fahrenheit. 

Let V = volume of a given body in cubic feet, 



FIG. 2. Spring 
balance. 



p = specific gravity of material, 
W = weight of body in Ib. 



MATHEMATICAL FORMULAE 



Then 



Hence 



= 62-5V Ib. weight if the material is water, 
= 62-5V/> Ib. weight for the given substance. 
W 



62-5V 

This expression enables the specific gravity of a given body to be 
found roughly by first weighing it, then calculating its volume from 
the measured dimensions. 

The following table gives the weights and specific gravities of 
some common substances : 

WEIGHTS AND SPECIFIC GRAVITIES. 



Material. 


Weight of 


Weight of a sheet 
i" thick, 
i sq. foot area. 


Specific 
Gravity. 


One cub. foot. 


One cub. inch. 




Ib. 


Ib. 


Ib. 




Wrought iron - 


480 


0-28 


40 


77 


Steel 


490 


0-28 


41 


7-8 


Cast iron 


45 


0-26 


37* 


7-2 


Copper - 


550 


0-32 


4 6 


8-8 


Brass 


525 


0-30 


44 


8-4 


Gun metal 


540 


0-31 


45 


8-6 


Aluminium 


165 


0-095 


14 


2-6 


Zinc 


450 


0-26 


37* 


7-2 


Tin 


465 


0-27 


39 


7-4 


Lead 


710 


O4I 


59 


1 1-4 


Fresh water 


62-5 


0-036 





I-O 


Sea water 


64 


0-037 





1-024 



Mathematical formulae. The following mathematical notes are 
given for reference. It is assumed that the reader has studied the 
principles involved, or that he is doing so conjointly with his course 
in mechanics. It may be noted here that a knowledge of the 
elementary rules of the calculus given below is not required in 
reading the first five chapters of this book. 

MENSURATION. 
Determination of areas. 

Square, side s ; area = s 2 . 

Rectangle, adjacent sides a and b ; area = ab. 

Triangle, base b, perpendicular height h ; area = \b x h. 

Triangle, sides a, b and c. zs = a + b + c. 

Area = \ls(s - a) (s - b) (s - c). 



MATERIALS AND STRUCTURES 



Parallelogram ; area = one side x perpendicular distance from that 
side to the opposite one. 

Any irregular figure bounded by straight lines ; split it up into 
triangles, find the area of each separately and take the sum. 

Trapezoid; area = half the sum of 
the end ordinates x the base. 

A trapezoidal figure having equal 
intervals (Fig. 3) ; 

area = a ( - - + h^ + h z + h \ 

Simpson's rule for the area bounded by a curve (Fig. 4) ; take an 
odd number (say 7) of equidistant ordinates ; then 




FIG. 3. Trapezoidal figure. 



area = - (h^ + 4/i. 2 + 



+ h^ 




N G 

FIG. 4. Illustration of Simpson's rule. 

7 

Circle, radius r, diameter d\ area = 7jv 2 = - . 

4 
(Circumference = 2irr = ird.} 

Parabola, vertex at O (Fig. 5) ; area OBC = \ib. 

Cylinder, diameter d, length /; area of curved 
surface = irdl. 

Sphere, diameter d, radius r\ area of curved 
surface = ?r^ 2 = ^-rrr 2 . 

Cone\ area of curved surface = circumference 
of base x -| slant height. 

Determination of volumes. 

Cube, edge s\ volume =s 3 . 

Cylinder or prism, having its ends perpendicular g 
to its axis ; volume = area of one end x length of 
cylinder or prism. 

Sphere, radius /; volume = ^Trr 3 . 




B 

"V 




BH... ..$....*<, 

FIG. 5. Area of a paia- 
bola. 



Cone or pyramid', volume = area of base x \ perpendicular height. 



TRIGONOMETRY. 

A degree is the angle subtended at the centre of a circle by an arc 
of TrlTrth of the circumference. 



MATHEMATICAL FORMULAE 



A radian is the angle subtended at the centre of a circle by an arc 
equal to the radius of the circle. 

There are 2ir radians in a complete circle, hence 
27r radians = 360 degrees. 
|TT =270 
TT =180 
|TT = 90 

Let / be the length of arc subtended by an angle, and let r be the 
radius of the circle, both in the same units ; then angle = - radians. 

Trigonometrical ratios. In Fig. 6 let OB revolve anti-clockwise 
about O, and let it stop successively in positions OP 1} OP 2 , OP 3 , 
OP 4 ; the angles described by OB are said to be as follows : 

P^B, in the first quadrant COB. 

P 2 OB, in the second quadrant CO A. 

P 3 OB (greater than 180), in the third quadrant AOD. 

P 4 OB (greater than 270), in the fourth quadrant BOD. 
Drop perpendiculars such as PjM^ from each position of P on to 
AB. OP is always regarded as positive ; OM is positive if on the 
right and negative if on the left of O ; PM is positive if above and 
negative if below AB. 



Name of 
ratio. 


Ratio as 
written. 


Value of 
ratio. 


Algebraic sign of ratio. 


ist quad. 


and quad. 


3rd quad. 


4th quad. 


sine POM 


sin POM 


PM 
OP 


+ 


+ 


- 


- 


cosine POM - 


cos POM 


OM 
OP 


+ 


- 


- 


+ 


tangent POM - 


tan POM 


PM 
OM 


+ 


- 


+ 


- 


cosecant POM 


cosec POM 


OP 
PM 


+ 


+ 


-' 


- 


secant POM - 


sec POM 


OP 

OM 


+ 


- 


- 


+ 


cotangent POM 


cot POM 


OM 
PM 


+ 


- 


+ 


- 



The values of the ratios are not affected by the length of the 
radius OP ; taking OP to be unity, we have 
sinPOM = PM (Fig. 6), 
cos POM = OM (Fig. 6), 
tan POM = P'B or P'A, depending on the quadrant (Fig. 7). 



8 



MATERIALS AND STRUCTURES 



Figs. 6 and 7 show clearly both the sign and the varying values 
of these ratios, and enable the following table to be deduced : 





Values of the ratios for angles of 







90" 


1 80 


270 


360 


sin POM - 


O 


I 





- 1 





cos POM - 


I 





- I 





I 


tan POM - 





CO 





CO 







FIG. 6. Trigonometrical ratios. 



FIG. 7. Tangents of angles. 



The following formulae are given for reference : 

iii 



cosec A = 
tanA = 



sin A' 
sin A 



sec A 



cot A = 



cos A' 
cos A 



cot A 



tan A 
cos 2 A + sin 2 A = i . 



cos A ' sin A ' 

tan 2 A + i = sec 2 A ; cot 2 A + i = cosec 2 A. 

sin A = cos (90 - A) ; sin A = sin ( 1 80 - A), 
sin (A + B) = sin A cos B + cos A sin B. 
cos (A + B) = cos A cos B - sin A sin B. 
sin(A - B) = sin A cos B - cos A sin B. 
cos (A - B) = cos A cos B + sin A sin B. 
tan A -f- tan B 



tan(A + B) = 



i -tan A tan B 
tan A - tan B 



MATHEMATICAL FORMULAE 



If the angles of a triangle are A, B and C, and the sides opposite 
these angles are a, b and c respectively, the following relations hold : 

a = b cos C + c cos B. 
a b c 

sin A sinB sinC* 
a 2 = ft + c 1 - 2bc cos A. 

ALGEBRA. 
Solution of simple simultaneous equations. If the given equations are 

'i. (O 

* () 

then # = ^M ^-r-> 



Solution of a quadratic equation. If 

ax 2 + bx + c = o, 

then 



x 



20, 



CALCULUS. 

Differential calculus. Let AB (Fig. 8) represent the relation 
of two quantities x and y which are connected in some definite 




T O M, M a X 

FIG. 8. Graphic illustration of a differential coefficient. 

manner. Consider two points P l and P 2 on AB separated by a 
short distance PjP 2 ; then 



* 2 ; P 2 M 2 =j 2 . 
The difference between the abscissae OMj and OM 2 will be 



10 MATERIALS AND STRUCTURES 

(x 2 - XT), and may be written 8x, the symbol S signifying " the 
difference in " ; similarly with the ordinates PjMj and P 2 M 2 . Hence 

8x = x 2 - x l = MjM 2 = PjK. 

fy=y-2-yi = ?2 K >. 

The ratio of these will be 

^ = ^2-^l = P 2 K 

Sx x 2 -x l PjK* 

The value of this ratio depends on the proximity of P! and P 2 . 
If these points are taken indefinitely close together, the ratio tends 
to take a definite value which depends on the given relationship of 
x and y. This value is called a differential coefficient, and serves 
to measure the rate of growth of y with x. 

If Pj and P 2 are very close together, PjP 2 is practically a straight 
line, and we have p T/- 

|J = tanP 2 P 1 K. 

If Pj and P 2 are indefinitely close together, PjP 2 is in the direction 
of the tangent PjT drawn to touch the curve at P l ; in this case Sy 
and Sx are written dy and dx, and the final value of the ratio is 



For example, suppose a graph such as AB in Fig. 8 to have been plotted 
from the equation, y=-x i . (i) 

Then y+8y=(x + Sx? 

= X* + 2X,8x + (8x)* ........................... (2) 

Taking the difference between (2) and (i) gives 



Now (&r) 2 is the square of a quantity which ultimately becomes very 
small, and therefore becomes negligible. Hence we may write 



~" ............................................ 

Suppose, as another example, we take 

y = ax\ ........................................ (4) 

when a is a constant. It will be evident, on repeating the above process, 
that 



thus giving the rule that any constant factor appears unaltered in 
the value of the differential coefficient. 



MATHEMATICAL FORMULAE 



Take now the following equation : 

y = x* + a (6) 

The effect of the addition of a constant a to the right-hand side 
of (i) is simply to raise the graph to a higher level above OX in 
Fig. 8 ; its shape will be exactly as before, and hence the tangent at 
any point will make the same angle with OX. Therefore the 
differential coefficient will have the same value as (3), viz. 

^-2X 

dx~ 2X 



(7) 



It will also be clear that, if the equation is 



then 



dy 

-j- = 2ax. 

dx 



.(8) 
(9) 



The rule may be expressed that a constant quantity added to the 
right-hand side disappears from the differential coefficient. 

The following differential coefficients are useful; the methods of 
obtaining them may be studied in any book dealing with the calculus. 
The symbol e represents the base of the Naperian or hyperbolic 
system of logarithms, viz. 2-71828. 

DIFFERENTIAL COEFFICIENTS. 



yx n 


! = *-! 


y=ax n 


%-*#>-* 


V+ ',:; 


*-* 

dx 


y=ae bx 


~dx~ a e * 




dy _\ 


-a\o 


dy a 


y og e x 


dx x 


y a oge~ 


Tx~~x 


,=sin* 


^ = cos^- 
dx 


y a sin bx 


-=ocos&r 

dx 


,=cos* 


^=-sin* 


y=acosbx 


-2- = ab sin bx 
dx 


y = tan x 


^ = sec2;r 


y=ata.n bx 


dv 
-j- = a&sec 2 fa 




I- 




dx 



Differentiation rules. The following rules may also be stated 
here. 

If the right-hand side takes the form of the sum of a number of 



12 MATERIALS AND STRUCTURES 

terms each depending on x, then the differential coefficient is the sum 
of the differential coefficients of the terms taken separately. Thus : 
y = ax* + bx 1 



To differentiate the product of a number of factors, each of which 
depends on x, multiply the differential coefficient of each factor by 
all the other factors and take the sum. Thus : 



dy . , 

~ = zx sm x + x 2 cos x. 

dx 

To differentiate a fraction in which both numerator and denomi- 
nator depend on x, proceed thus : 

diff. coeff. of numerator x denominator 
dy _ - diff. coeff. of denominator x numerator 
dx square of denominator 

EXAMPLE. Let j/=^^. 

sin* 

The differential coefficient of the numerator is 2;r and that of the 
denominator is cos .*, hence, by the above rule : 
dy _ ix sin x - x 2 cos x 
dx sin 2 x 

Supposing we have to find the differential coefficient of 

it should be noticed that the given expression, viz. the cube of sin x, 
depends on another function of x. The rule to be followed is to 
differentiate the expression as given, viz. (sin#) 3 , the result being 
3 (sin #) 2 ; then multiply this result by the differential coefficient of 
the function on which the given expression depends, viz. sin x, for 
which the differential coefficient is cos x. 

Hence, -r- = 3 (sin x} 2 cos x 

dx 

= 3 sin 2 x cos x. 

In successive differentiation, the differential coefficient of the given 
function is taken as a new function of x and its differential coefficient 
is found ; the latter is called the second differential coefficient, and is 

written ^. The operation may be repeated as many times as may 
be necessary. 



MATHEMATICAL FORMULAE 13 



EXAMPLE. Let 



The maximum value of a given function of x may often be found 
by application of the following simple method. It will be noted that, 
in Fig. 8, at the point in AB for which y has its maximum value, 

the tangent to the curve is parallel to OX, and hence -2. for this 

point will be zero. The rule therefore is, take the differential 
coefficient and equate to zero; this will give the value of x corre- 
sponding to the maximum value of y. By inserting this value of x 
in the given equation connecting x andj^, the maximum value of y 
may be found. Thus : 

Let jj/=sin;tr, 

-^- = cos.r=o for the maximum value 
Now when cos^r = o, x is either 90 or 270, i.e. or *- radians, hence 

Maximum value of y=sin- or sin^. 
2 2 

As the numerical value of sin - is unity, it follows that the maximum 
value of y is also unity. 
As another example, take 

y = ax-x\ 

then, -j-=a-2x=0', 

/. x- for the maximum value ofy. 

Maximum value of y= = 

244 

Integral calculus. In this branch of mathematics, rules are 
formed for the addition of the indefinitely small portions into which 
a quantity may be imagined to be divided. In Fig. 9, OA and OB 
are two distances measured along the same straight line from O. 
Let these be a and b respectively, then the length of AB will be 

AB = -tf ....(i) 

The line AB might be measured also by the process of dividing it 



MATERIALS AND STRUCTURES 



up into a large number of small portions 8x l9 8x 2 , 8x s , etc. The 
total length of AB will then be 

AB = 8x t + 8x 2 + 8x B + etc. 

= b-a, from (i) (2) 

The symbol 2 or J (sigma) is used to denote the phrase "the 
algebraic sum of," and if any expression follows the symbol 2, it is 



FIG. 9. 



understood to be one only of a number of terms which are all of the 
same type. Thus, 2&r means "the algebraic sum of all terms of 
which 8x is given as a type." If we write 2*, it is to be understood 
that we are to begin taking small portions, such as Sx-^, at a distance 
a from the origin, and to finish at a distance b. Hence we may 
write (2\ fyx = b-a (3) 

In Fig. 10 is shown another example. As before, OA a and 
OB = , and the figure ABCD is constructed by making AD = a 

and BC = ^, both being perpen- 
dicular to OB. The area of the 
figure ABCD may be calculated 
by deducting the area of the 
triangle OAD from that of the 
triangle OBC. Thus : 
Area of ABCD 

= (b x \b} -(ay. \d) 
_& 2 a 2 

~ 2. 2' 

Alternatively, the area may be 
estimated by cutting the figure 
into strips, such as the one shown 
shaded. It is evident from the 
construction that its height y is 
equal to x ; let 8x be its breadth, then 

Area of the strip = x . 8x (5) 

Any similar strip will have a similar expression for its area, hence 
Total area of the strips = I^x 8x (6) 



Or 




JA | 

~4 - 

FIG. 10. 



__L 



MATHEMATICAL FORMULAE 15 

The area stated in (5) is taken as that of a rectangle, and hence 
omits a small triangle at the top of the strip. If, however, the strips 
be taken indefinitely narrow, these triangles will practically vanish, 
and the area expressed in (6) will be the area of ABCD. Hence 
from (6) and (4), p a 2 

fy.dx---- ............................ (7) 

In mathematical books, it is shown that if x is raised to a power n 
in equation (7), n having any value except - i, then the result is as 

follows : An+I _ a n+l 

^>x".dx = - ...................... (8) 

n -f i 

If n is - i, then the result may be shown to be 
, , _!:,."."- -. 2j*-'.,& = 2jf -log, ...................... (9) 

If n is zero, then #= i, and we have 

^ b a xdx = ^ b a dx = b -^ = b-a ................... (10) 

The above are examples of definite integrals, taken between given 
limits a and b the sum may be stated in an indefinite manner, 
leaving the limits to be inserted afterwards. Thus : 



It is also shown in mathematics that a constant term c should be 
added to the result. The value of c depends on the conditions of 
the problem, and can be found usually from the data. The com- 
plete solution of (n) would thus be 

- 

fm ..................... ( I2 \ 



dx 
Similarly, 2 = \og e x + <:. ..................... (13) 

oc 

If a constant factor is given on the left-hand side, it will appear 
unaltered on the right-hand side. Thus : 

x* 

a \-c. 
3 

If a number of terms be given, the result will be obtained by 
applying the rules to each term separately and then summing for 
the total. Thus : 



16 



MATERIALS AND STRUCTURES 



The rules (8), (9), (12) and (13) should be learned thoroughly. 
Some examples are given. 

EXAMPLE i. Find the area of the triangle given in Fig. 11. 
Taking a narrow strip parallel to the base and at a distance y from O, 

let the breadth of the strip be 8y and 

* its length b. 

*^ Area of the strip = b . 8y. 

KT b V 

Now ==; ; 




FIG. ii. Area of a triangle. 



.'. area of the strip = jj .y 8y. 

Any other similar strip will have a 
similar expression for its area, hence 



Total area=2"u 



_B/H_ 2 _o_ 2 \ 
~H\2 2) 



= BH 

2 

EXAMPLE 2. Find the volume of a cone of height H and radius of 
base R (Fig. 12). 

In this case take a thin slice parallel to the base ; let the radius of the 
slice be r and its thickness h . Then 

Volume of the slice = 7rr 2 . S/i. 



Now 



r 

R~H 



volume of the slice = 



Any other similar slice will have a similar 
expression for its volume, 'hence 




FIG. 12. Volume of a cone.. 



Total volume^ 



R 2 H 3 



No constant of integration need be added in either of these examples. 
Instances where a constant is necessary will occur later. 



MATHEMATICAL FORMULAE 



The following table of indefinite integrals is given here for reference. 
INTEGRALS. 



r *+! 

/ r*tt //r~ 


I sec 2 * .dx 


= tan* 


J n+i 


[-.dx = log e * 


/cosec 2 *.*/* 


= -cot* 


f-.dx = a\og e x 


lacosbx.dx 


a . , 
= T sm0* 
b 


\e x .dx =e x 


I a sin bx . dx 


= --, cos <* 



(aete.dx =-<*>* 


\a$*x*bx.dx 


<2 . 

= -T tan far 



1 cos x . dx = sin x 


/ a cosec 2 bx . dx 


= - -7 cot bx 
b 


/sin* .dx -cos* 


[ s ' m * dx 


= sec* 


J cos'*' 


/ tan x . dx = log sec x 


f COS X , 

1 . o dx 
J sin 2 * 


= -cosec* 



EXERCISES ON CHAPTER I. 

1. A masonry wall is trapezoidal in section, one face of the wall being 
vertical. Height of wall, 20 feet ; thickness at top, 4 feet ; thickness at 
base, 9 feet. The masonry weighs 150 Ib. per cubic foot. Find the 
weight of a portion of the wall i foot in length. 

2. A trapezoidal figure, having equal intervals of 10 feet each, has 
ordinates in feet as follows : o, 100, 140, 120, 80, o. Find the total area 
in square feet. 

3. Draw a parabolic curve on a base a = 60 feet ; the height y feet of 
the curve at any distance * from one end of the base is given by 



Find the area by application of Simpson's rule ; check the result by use 
of , the rule : area = 4 where b is the maximum height of the curve. 

4. Write down the differential coefficients of the following : 

(a) y 5* 3 . (d) y = sin 2 * + cos 2 *. '^ 

(b} y = 3* 2 - y* 5 . (e) y = sin 3 * + cos 3 *. 

(r) y = 2 sin*- 3 cos*. (/)j/=3tan*-cos*. 

5. In Question 3, from a point M on the base, distant 15 feet from one 
end, draw a perpendicular to cut the curve at a point P. At P draw a 
tangent to the curve cutting the base produced in a point T. Measure 

PM 
PM and MT and evaluate the ratio ~. The result gives the differential 






P,M f 






1 8 MATERIALS AND STRUCTURES 

coefficient for the curve at P ; compare this result with that obtained by 

x^ dv 

differentiation of y = 2x -- and putting x\^ in the expression for -*-. 

How do you account for the discrepancy, if any ? 

6. Take the equation y = (^-x)x. Find the value of x for which y 
attains its maximum value, and find also the maximum value of y. Check 
your result by plotting a graph from the equation. 

7. Write down the indefinite integrals of the following : 

(a) ycWx. (d) (2x* + cos x)dx. 

. dO 



8. Find the value of the following expression when R t = i2 inches and 
R 2 = 6 inches. No constant of integration is required. 

' 



9. Find the value of the following expression when 6 = 4 inches and 
H = 8 inches. No constant of integration is required. 



V 



CHAPTER II. 
FORCES ACTING AT A POINT. 

Representation of a force. Any force is specified completely 
when we are given the following particulars : (a) its magnitude, (b) its 
point of application, (c) its line of direction, (d) its sense, i.e. to state 
whether the force is pushing or pulling at the point of application. 

A straight line may be employed to represent a given force, for it 
may be drawn of any length, and so represent to a given scale the 
magnitude of the force. The end of the line shows the point of applica- 
tion, the direction of the line gives the direction, and an arrow point 
on the line will indicate the sense of the force. Thus a pull of 5 Ib. 
acting at a point O in a body (Fig. 13) at 45 to the horizontal 
would be completely represented by a line OA, 
of length 2 1" to a scale of J" to a Ib., and an 
arrow point as shown. OA is called a vector ; any 
physical quantity for which a line of direction must 
be stated in order to have a complete specifica- 
tion is called a vector quantity. Other quantities, 

such as mass and volume, into which the idea of FJG. 13. Representa- 
tion of a force, 
direction does not enter, are called scalar quantities. 

The expression "force acting at a point" must not be taken 
literally. No material is so hard that it would not be penetrated by 
even a very small force applied to it at a mathematical point. What 
is meant is that the force may be imagined to be concentrated at the 
point in question without thereby affecting the condition of the body 
as a whole. 

Forces acting in the same straight line. A body is said to 
be in equilibrium if the forces applied to it balance one another. 
Thus, if two equal and opposite pulls P, P (Fig. 14) be applied at a 
point O in a body, both in the same straight line, they will evidently 
balance one another, and the body will be in equilibrium. 

Examples of this principle occur in ties, and in struts and columns- 




20 



MATERIALS AND STRUCTURES 



Ties are those parts of a structure intended to be under pull (Fig. 15), 
struts and columns are those parts intended to be under push (Fig. 16). 
These parts remain at rest under the action of the equal and opposite 
forces applied in the same straight line. 

It is impossible for a single force to act alone. To every force there 
must be an equal and opposite force, or what is exactly equivalent to 
an equal and opposite force. The term reaction is often used to 
distinguish the resistance offered by bodies to which a given body is 



FIG. 14. Two equal 
opposite forces. 



5 00 to. 



SOOlb. 



FIG. 15. Equilibrium 
of a tie. 



500 Ib. 



^50011. 

FIG. 16. Equilibrium 
of a column. 



connected when forces are applied to the latter body. An example 
of the use of the term will be found in the reactions of the piers 
supporting a bridge girder. Loads applied to the girder are balanced 
by the reactions of the piers. 

If several forces in the same straight line act at a point, the point 
will be in equilibrium if the sum of the forces of one sense is equal to 
the sum of those of opposite sense. Calling those forces of one 
sense positive and those of opposite sense negative, the condition 
may be expressed by stating that the algebraic sum of the given 
forces must be zero. Thus, the forces P 15 P 2 , P 3 , etc. (Fig. 17), will 
balance, provided 

P 1 + P f -P t -P 4 -Pi*a 
or, 2P = o, 

the interpretation being that the algebraic sum of all the forces of 
which one only is given as a type immediately after the symbol 2 
must be equal to zero. 

Suppose in a given case it is found that the algebraic sum of the 
given forces is not zero. We may infer from this that a single force 
may be substituted for the given forces without altering the effect. 
Thus, in Fig. 18, calling forces of sense from A towards B positive, 
we have 2 + 3 + 5 _8-i=+i. 

The given forces can be replaced by a single force of i Ib. weight of 
sense from A towards B. The single force which may be substituted 



FORCES ACTING AT A POINT 



21 



for a given system of forces without altering the effect on the body is 
called the Resultant of the system. To find the resultant R of the 
system we have been considering above, we have 



The resultant R may be balanced by applying an equal opposite 
force in the same straight line, and, since R is equivalent to the given 
system of forces, the same force would also balance the given system. 





FIG. 17. Forces in the same straight line. 



FIG. 1 8. 



Any force which balances a given system of forces is called the 
equilibrant of the system. Thus, the equilibrant E of the system 
shown in Fig. 18 is a force of i Ib. weight of sense from B 
towards A. 

Two intersecting forces. To find the resultant of two intersecting 
forces, the following construction may be employed. Let P and Q 
be two pulls applied to a nail at O (Fig. 19 (a)); their joint 
tendency will be to carry the nail upwards to the right, and the 
resultant must produce exactly the same tendency. Set off, in the 
direction in which P acts, OA, to some suitable scale, equal to P, 





(a) 

FIG. 19. Resultant and equilibrant of two intersecting forces. 

and OB, to the same scale, equal to Q and in the direction in which 
Q acts. Complete the parallelogram OACB, and draw its diagonal 
OC. This diagonal will represent R completely, the magnitude being 
measured by the length of OC to the same scale. The method is 
called the parallelogram of forces. P and Q are called components 
of R. 

As R is equivalent in its effects to P and Q jointly, we may apply 
either P and Q together, or R alone, without altering the effect on the 



22 



MATERIALS AND STRUCTURES 



nail. This may be expressed by stating that the resultant may be 
substituted for the components, or vice versa. 

Substituting R for P and Q (Fig. 19 (<)), we may balance R by 
applying an equilibrant E = R as shown. Again, replacing R by P 
and Q (Fig. 1 9 (<r) ), it will be evident that P, Q and E are in 
equilibrium. 

Experimental verification. The most satisfactory proof that the 
engineering student can have of the truth of the parallelogram of 
forces is experimental. 

EXPT. i. Parallelogram of forces. In Fig. 20 is shown a board 
attached to a wall and having three pulleys A, B and C capable of 




FIG. 20. Apparatus for demonstrating the parallelogram of forces. 

being clamped to any part of the edge of the board. These pulleys 
should run very easily. Pin a sheet of drawing paper to the board. 
Clamp the pulleys A and B in any given positions. Tie two silk 
cords to a split key ring, pass a bradawl through the ring into the 
board at O, and lead the cords over the pulleys at A and B. The 
ends of the cords should have scale pans attached, in which weights 
may be placed. Thus, known forces P and Q are applied to the ring 
at O. Take care in noting these forces that the weight of the scale 
pan is added to the weight you have placed in it. Mark carefully the 
directions of P and Q on the paper, and find their resultant R by 
means of the parallelogram Qabc. Produce the line of R, and by 
means of a third cord tied to the ring apply a force E equal to R, 
bringing the cord exactly into the line of R by using the pulley C 
clamped to the proper position on the board. Note that the proper 
weight to place in the scale pan is E less the weight of the scale pan, 
so that weight and scale pan together equal E. If the method of 
construction is correct, the bradawl may be withdrawn without the 
ring altering its position. 



FORCES ACTING AT A POINT 



In general it will be found that, after the bradawl is removed, the 
ring may be made to take up positions some little distance from O. 
This is due to the friction of the pulleys and to the stiffness of the 
cords bending round the pulleys, giving forces which cannot easily 
be taken into account in the above construction. 

Notice that, before attempting to apply the parallelogram of forces, 
both given forces must be made to act either towards or from the point of 
application. Thus, given P' pushing and Q pulling at O (Fig. 21), 
the tendency will be to carry O downwards to the right. Substitute 
P = P', pulling at O for P' ; complete the parallelogram OACB, when 
OC will give the resultant R. 

It will also be noticed that any one of the forces P, Q and E 
(Fig. 19 (c)) will be equal and opposite to the resultant of the other 
two if the three forces are in equilibrium. 

Rectangular components of a force. Very frequently it becomes 
useful in a given problem to deal with the components of a given 





FIG. 2i. Parallelogram offerees applied 
to a push and a pull. 



FIG. 22. Rectangular components 
of a force. 



force instead of using the force itself. These components are 
generally taken along two lines at 90 intersecting on the line of the 
given force. Thus, given P acting at O (Fig. 22), and two lines OA 
and OB at 90 intersecting at O, and in the same plane as P. The 
components will be found by making OC equal to P, and completing 
the parallelogram of forces OBCA, which in this case is a rectangle. 
S equal to OB and T equal to OA will be the rectangular com- 
ponents of P. 

The following will be seen easily from the geometry of the figure : 



2 4 



MATERIALS AND STRUCTURES 



Also, let the angle CO A = a ; then 
OA 

^i = COS a, 

O A = OC . cos a ; 
/. T=P.cosa. 

AC . 

?^ = sin a, 



Again, 



oc 



= OC.sina, 
= P.sina. 




Triangle of forces. It will now be understood that the conditions 
which must be fulfilled in order that three forces whose lines inter- 
sect may be in equilibrium are : (a) the forces must all be in the 
same plane, i.e. uniplanar ; (b} their lines must intersect in the same 
point; (c) any one of them must be equal and opposite to the 
resultant of the other two forces. 

Condition (c) may be stated in another manner. In Fig. 23, P and 
Q have a resultant R, found by the parallelogram of forces OACB. 

A force E has been applied equal and 
opposite to R as shown; hence the 
forces E, P and Q are in equilibrium. 
The following relation evidently holds : 

R : Q : P = OC : OB : O A. 
Note the order in which the letters 
of the lines have been written ; thus, 
R is represented by OC, not by CO, 
the order being so chosen as to show 
the sense of the force. Now E is equal to R, and OA is equal to 
BC ; hence we may write 

E:Q:P = CO:OB:BC, 
OC having been altered to CO so as to give the proper sense to E. 

Expressed in words, the proportion states that the three forces in 
equilibrium axe proportional respectively to the sides of a triangle taken in 
order. The triangle OBC in Fig. 23 may be drawn anywhere on 
the paper, and is called the triangle of forces for the forces E, Q, P. 

EXAMPLE i. Given three uniplanar forces P, Q, S' (Fig. 24) acting at 
O ; test for their equilibrium. 

Using a convenient scale of force, draw ab, be and ca' parallel and pro- 
portional respectively to the forces P, Q and S'. If the given forces are in 
equilibrium, the lines so drawn will form a closed triangle. In Fig. 24, 



FIG. 23. 



FORCES ACTING AT A POINT 



it will be noticed that there is a gap aa f . S' will therefore not equilibrate 
P and Q, but may be made to do so if it is redrawn as S, parallel and 
proportional to ca, the closing line of the triangle abc. 






FIG. 24. Triangle offerees. 



FIG. 25. Triangle of forces applied to 
a push and a pull. 



EXAMPLE 2. Given two forces P and Q (Fig. 25) acting at O ; find 
their equilibrant. 

It will be observed that, in applying the triangle of forces, there is no 
necessity for first making both the given forces pushes or pulls, provided 
attention is paid to drawing the sides of the triangle in proper order. 
Thus, draw ab to represent P and be to represent Q ; then ca will repre- 
sent the equilibrant, which should now be drawn as E acting at O, parallel 
and proportional to ca and of sense shown by the order of the letters ca. 
Note carefully that the problem is not finished until E has been applied 
on the drawing acting at the proper place O. 

EXAMPLE 3. Three given forces are known to be in equilibrium 
(Fig. 26 (a) ) ; draw the triangle offerees. 

This example is given to illustrate a con- 
venient method of lettering the forces called 
Bow's Notation. This method will be found to 
simplify many of the problems which have 
to be discussed, and consists in giving letters 
to the spaces instead of to the forces. In 
Fig. 2,6(0) this plan has been carried out by 
calling the space between the 4 Ib. and the 
2 Ib. A, that between the 2 Ib. and the 3 Ib. B, 
and the remaining space C. Starting, say, 
in space A and crossing over into space B, 
a line AB (Fig. 26()) is drawn parallel and 
proportional to the force crossed, and the 
letters are so placed that their order A to B represents the sense of that 
force. Now cross from space B into space C, and draw BC to represent 
completely the force crossed. Finish the construction by crossing from 



(a) 




(b) 



tion of Bow's 
lotation. 



Applicati 
Notation 



26 



MATERIALS AND STRUCTURES 



space C into space A, when CA in Fig. 26 (b) will represent the third force 
completely. 

Examining these diagrams, it will be observed that a complete rotation 
round the point of application has been performed in Fig. 26 (#), and that 
there has been no reversal of the direction of rotation. Also that, in 
Fig. 2.6(b\ if the same order of rotation be followed out, the sides correctly 
represent the senses of the various forces. Either sense of rotation may 
be used in proceeding round the point of application, clockwise or anti- 
clockwise, but once started there must be no reversal. 

Relation of forces and angles. In Fig. 27 (a) there are three 
given forces in equilibrium, viz., P, Q and S, and in Fig. 27^) is 
shown the triangle of forces for them. From what has been said 
above, we may write 

P:Q:S = AB:BC:CA. 

It is shown in trigonometry that the sides of any triangle are pro- 
portional to the sines of the opposite angles. Hence, in Fig. 2 7 (/), 

AB : BC : CA = sin y : sin a : sin ft 
or, P : Q : S = sin y : sin a : sin ft 




Q 



A-/' / 



fa.) 




X --..C' * 

FIG. 27, Relation offerees and angles. 

It will be noticed in Fig. 2 7 (a), as shown by dotted lines, that a, ft 
y are respectively the angles between the produced directions of 
S and P, P and Q, and Q and S ; also that the angles or spaces 
denoted by A, B and C in the same figure are the supplements of 
these angles. As the sine of any angle is equal to the sine of its 
supplement, we have, in Fig. 2 7 (a), 

P : Q : S = sin C : sin A : sin B. 

We infer from this that each force is proportional to the sine of the 
angle between the other two forces. 

Any number of uniplanar forces acting at a point. The net 

effect of such a system of forces may be found by taking components 
of each force along two rectangular axes which meet in the point of 
intersection and are in the same plane as the given forces. It is best, 



FORCES ACTING AT A POINT 



27 



in order to comply with the usual trigonometrical conventions 
regarding the algebraic signs of sines and cosines, to arrange the 
forces to be either all pulls or all pushes. 

In Fig. 28, Pj, P 2 , P 3 and P 4 are the given forces acting at O, 
and OX and OY are two rectangular axes. The angles of direction 



Sifld^t 




FIG. 28. System of uniplanar forces acting at a point. 

of the forces are stated with reference to OX as a 1? a 2 , a g and a 4 . 
Taking components along OX and OY, we have : 

Components along OX, P T cos a x , P 2 cos a 2 , P 3 cos a 3 , P 4 cos a 4 . 

Components along OY, Pjsinaj, P 2 sina 2 , P 3 sina 3 , P 4 sina 4 . 
Paying attention to the algebraic signs of these, it will be observed 
that components acting along OX towards 
the right are positive, and those acting ' 
towards the left are negative ; also, of the 
components acting along OY, those acting 
upwards are positive, while those acting 
downwards are negative. Each of these 
sets of components may have a resultant, O r R x A 

or they may be in equilibrium. Suppose FIG. 29. Resultant of the system 

rr shown in Fig. 28. 

each to have a resultant, and denote 

that along OX by R x , also that along OY by R Y ; then 

P! cos a 1 + P 2 cos a 2 + P 3 cos a 3 + P 4 cos a 4 = R x , 
P! sin aj + P 2 sin a 2 + P 3 sin a 3 + P 4 sin a 4 = R Y . 
Using the abbreviated system of writing these, we have 
2Pcosa = R 




(i) 
(a) 

The system being now reduced to two forces R x and R Y acting in 
lines at 90 to each other, we have for the resultant (Fig. 29), 



(3) 



28 



MATERIALS AND STRUCTURES 



Also, 



tana 



CA = OB 
OA~OA 
Ry 



(4) 

It may so happen that either R x or R v may be zero, in which case 
the resultant of the system is a force acting along either OX or OY, 
depending upon which of the forces is zero. For equilibrium of the 
given system both R x and R Y must be zero. This condition may 

be written 2Pcosa = o, (5) 

2Psina = o; (6) 

a pair of simultaneous equations which will serve for the solution 
of any problem connected with the equilibrium of any system of 
uniplanar forces acting at a point. 

Graphical solution. A graphical solution of the same problem 
may be obtained by repeated application of the parallelogram of 
forces. Thus, given P, Q, S and T acting 
at O (Fig. 30). First find Rj of P and S, 
then R 2 of Q and T by applications of the 
parallelogram of forces. The resultant R is 
found by a third application of the parallelo- 
gram, as shown. A better solution is 
obtained by repeated application of the 
triangle of forces. 

In Fig. 31(0), four forces P, Q, S and T 
are given. To ascertain the net effect of the 
system, first find the equilibrant Ej of P and 
Q by the triangle of forces ABC (Fig. 31 (^)). E l reversed in sense 
will give R I} the resultant of P and Q, and is so shown in Fig. 31(0), 
and is represented by AC in Fig. 3i(^). Now find the equilibrant 

B 
p 






FIG. 31. Resultant by application of the triangle offerees. 

E 2 of Rj and S by means of the triangle of forces A CD (Fig. 31 

E 2 reversed gives R 2 , the resultant of R x and S, and hence the 

resultant of P, Q and S. R 2 will be represented in Fig. 3 1 (b} by 



FORCES ACTING AT A POINT 



AD. R 2 and T being the only forces remaining in Fig. 31(0), 
their resultant R will be found from the triangle of forces ADA' 
(Fig. 3i(^)), which gives their equilibrant E 3 , represented by A' A, 
and on reversal gives R. 

It will be noticed that, had the given forces been in equilibrium, 
E 3 would have been zero, and A' would have coincided with A. 
This case is shown in Fig. 32, giving a closed polygon ABCD, the 
sides of which, taken in order, represent respectively the given forces. 
We therefore infer that a given system of uniplanar forces acting at a 
point will be in equilibrium, provided a closed polygon can be drawn which 
shall have its sides respectively parallel and proportional to the given forces 
taken in order. Should the polygon not close, then the line required 
in order to close it will represent the equilibrant of the given forces, 
and, the sense being reversed, the same line will give the resultant 
of the given system. The figure ABCD (Fig. 32^)) is called the 

B 





FIG. 32. Polygon of forces. 

polygon of forces for the given forces. Note, as before, that no problem 
can be regarded as completed until R or E, as the case may require, 
is actually shown on the drawing acting at its proper place O. 

EXPT. 2. Pendulum. Fig. 33(0) shows a pendulum consisting of 
a heavy bob at A suspended by a cord attached at B and having a 
spring balance at F. Another cord is attached to A and is led 
horizontally to E, where it is fastened. A spring balance at D 
enables the pull to be read. Find the pulls T and P of the spring 
balances F and D respectively when A is at gradually increased 
distances x from the vertical. Check these by calculation as shown 
below, and plot P and x. 

Since P, W and T are respectively horizontal, vertical and along 
AB, it follows that ABC is the triangle of forces for them. Hence 



-, 

/I 

Wtana. 



(0 



MATERIALS AND STRUCTURES 



Also, 



TAB/ 



Wseca 




FIG. 33. Experiment on a pendulum. 

Measure /, also x and h, for each position of the bob, and calculate 
P and T by inserting the required quantities in (i) and (2). 
Tabulate thus : Weight of bob in Ib. = W = 
Length of AB in inches = /= 



x inches. 


h inches. 


Calculated values. 


Observed values from 
spring balances. 


Pr^W Ib. 

h 


T = ~Wlb. 
n 


Plb. 


Tib. 















The curve will resemble that shown in Fig. 34. Note how nearly 
straight it is for comparatively small values of x. 

EXPT. 3. Eoof truss. In Fig. 35 is shown a simple model of a 
roof truss consisting of two rafters made of wooden bars AB and 
BC hinged by means of a bolt at B and connected at the bottom 
by a cord AC ? wjiich takes the place of the tie-bar in the actual truss. 



FORCES ACTING AT A POINT 



Compression spring balances D and E and an ordinary spring 
balance F enable the forces in the various parts to be measured. 
C is pivoted by two pointed set screws, p LBS 
as shown in the end elevation, and a 
roller at A, also shown in end elevation, 
permits the span of the truss to be 
altered by adjusting the length of the 
cord AC. A weight W is hung from B. 

Set up the apparatus, and observe the 
push in each rafter AB and CB, and also 
the pull in the tie AC. Measure and 
note the lengths AB, BC and AC when 
the load is on. Repeat the experiment, 
using different weights and spans, being 
careful in each case to note the altered 
dimensions of the parts. Compare each 
set of readings with those found by appli- 
cation of the triangle of forces, as shown 
below. 

Make an outline drawing of the truss 
to scale (Fig. 36 (a)). If the truss is symmetrical, each rafter will 
give equal pushes, say P lb., to the joints at B, A and C. The tie 
will apply equal forces T, T at A and C. The reactions of the 
supports, Rj and R 2 , may be assumed to be vertical. Considering 



2-0 3-0 4-0 



5-0 
FT. 



FIG. 34. Graph of P and x for a 
pendulum. 




FIG. 35. Experimental roof truss. 

the forces acting at the point B, which is in equilibrium, and setting 
off ab to represent W (Fig. 36 ()), and ac and be parallel respectively 
to AB and BC, we have the triangle of forces abc for P, W and P 
acting at B. Now ca represents P acting at B, and ac may be 
taken to represent P of opposite sense acting at A. Draw cd 
parallel to AC. Then the triangle acd is the triangle of forces for 
P, Rj and T acting at A. In the same way bed is the triangle of 
forces for P, R 2 and T acting at C. Therefore, 



The results for P and T as obtained from the diagram will agree 
fairly well with those obtained from the spring balances, provided due 



MATERIALS AND STRUCTURES 



allowance be made for the effects of the weights of the various parts 
before the application of W. To do this, remove W and note the 
readings of the balances. These readings should be deducted from 




FIG. 36. Forces in a simple roof truss. 

those taken after W is applied, when the corrected results will show 
the forces in the parts due to the application of W alone. The results 
should be tabulated thus : 



Lengths in inches. 


Forces in Ib. from 
diagram. 


Corrected forces in Ib. 
from spring balances. 


AB 


BC 


AC 


P, push. 


T, pull. 


P, push. 


T, pull. 

















From your experiments, give a general statement of how P and T 
vary for the same value of W, but with increasing lengths of span. 




JL* 



FIG. 37. Unsymmetrical roof truss. 



The case of an unsymmetrical roof truss may be worked out in a 
similar manner, and is shown in Fig. 37, the lettering of which 
corresponds with that of Fig. 36, 



FORCES ACTING AT A POINT 



33 



EXPT. 4. Derrick crane. A derrick crane model is shown in 
Fig. 38, consisting of a post AB firmly fixed to a base board which 




W 



FIG. 38. Model derrick crane. 

is screwed to a table ; a jib AC has a pointed end at A bearing in a 
cup recess, a pulley at C and a compression spring balance at D. A 
tie BC supports the jib and is of adjustable length ; a spring balance 
for measuring the pull is inserted at F. The weight is supported 
by a cord led over the pulley at C and attached to one of the 
screw-eyes on the post. The inclination of the jib may be altered 
by adjusting the length of BC, and the inclinations of EC and BC 
may be changed by making use of different screw-eyes. 




FIG. 39. Forces in a derrick crane. 

Find the push in the jib and the pull in the tie for different values 
of W and different dimensions of the apparatus by observing the 
spring balances. Check the results by means of the polygon of 
forces. 

The methods are similar to those adopted for the roof truss 
(p. 30). It may be assumed that the pulley at C merely changes 
the direction of the cord without altering the force in it. Hence 



34 



MATERIALS AND STRUCTURES 



p = W (Fig. 39 (a) ). The polygon of forces is shown in Fig. 
in which w _ ^A n _ ,j 



The observed and graphical results should be compared in tabular 
form as before : 



Lengths in inches. 


Forces in Ib. from 
diagram. 


Corrected forces in Ib. 
from spring balances. 


AB 


BC 


AC 


AE 


Q, push. 


T, pull. 


Q, push. 


T, pull. 



















EXPT. 5. Wall crane. A model wall crane is shown in Fig. 40 (a). 
Its construction is similar to that of the derrick crane, and the method 
of experimenting is the same. The outline diagram is given in 





(b) 



o 



VST 



(C) 



FIG. 40. Experimental wall crane. 



Fig. 40 (b\ and the polygon of forces in Fig. 40 (c). These are lettered 
to correspond with those for the derrick crane, and will be followed 
readily. 

Forces acting at a point but not in the same plane. In Fig. 41 
is shown in outline a pair of sheer legs such as is used for moving 
heavy loads. Two legs AB and BC are jointed together at the top B, 
and are hinged at the ground at A and C so as to be capable of 
rotating as a whole about the line AC in the plan. The legs are 
supported in any given position by means of a back leg DB, which is 



FORCES ACTING AT A POINT 



35 



jointed to the other legs at B, and has its end D capable of being 
moved horizontally in the direction of the line BD in the plan. The 
load W is hung from B and produces forces T, Q, Q in the three 
legs ; these are shown acting at B. It will be noted that T and W are in 
the same vertical plane, and that the two forces Q, Q are both in the 
inclined plane which has A'B' for its trace in the elevation. As the 
legs are symmetrical, the forces Q, Q will be equal, and will have a 
resultant S, which will fall in the same vertical plane as T and W. 




FIG. 41. Forces in a pair of sheer legs. 

Draw the triangle of forces abc by making ab represent W, and be and 
ca parallel to A'B' and B'D' respectively. Then ca gives the pull T 
in the back leg, and cb gives the force S. To obtain the forces Q, Q, 
rotate the plane of ABC about the line AC, as shown, until it lies on 
the ground, when the true shape of the triangle ABC will be seen in 
the plan as ABjC. Mark off B X E to represent S, and draw the 
parallelogram of forces BjGEF, when the equal lines GBj and FB X 
will give the values of the equal forces Q, Q. 

A tripod is worked out in Fig. 42. Three poles AD, BD and CD 
are lashed together at their tops, and have their lower ends resting on 
the ground. Often the poles are equal in length, but for greater 
generality they have been taken unequal in the example chosen. To 
draw the plan (Fig. 42), first construct a plan of the triangular base 
ABC from the given distances between the feet of the poles. Con- 
struct the triangles AFC and EEC by making AF equal to the length 
of the pole AD, BE equal to that of BD, and CF and CE each equal 
to that of CD. It is clear that AFC and BEC are respectively the 



MATERIALS AND STRUCTURES 



true shapes of ADC and BDC when rotated about the lines AC and 
BC respectively, so as to lie on the ground. To find the position of 
D in the plan, draw FD and ED intersecting at D and perpendicular 
respectively to AC and BC. 

Let a weight W be hung from D, and let P, Q and S be the forces 
in the legs acting at D. P and W will be in the same vertical plane, 
and may be balanced by a third force Z applied in the same vertical 
plane and also contained by the plane of ADC. The line of Z in 
the plan will be *DG, obtained by producing BD. To obtain a true 




FIG. 42. Forces in a tripod. 

view of the forces P, W and Z, take an elevation on the ground line 
xy, which is parallel to BD ; in this elevation, B'D' is the true length 
of the pole BD. The lines of P and Z are shown by B'D' and 
G'D' in this view (Fig. 42). W will be perpendicular to xy, and 
by making D'b equal to W and drawing the parallelogram D'abc, 
the values of P and Z will be given by #D' and cD' respectively. 
To obtain Q and S, we have in the plan their lines lying on the 
ground at AF and CF, and GF will be the line of Z. Make Fe equal 
to Z and construct the parallelogram Ydef, when Q and S will be 
given by d and /F respectively. 



EXERCISES ON CHAPTER II. 

1. Two pulls are applied to a point, one of 4 Ib. and the other of 9 Ib. 
Find graphically the magnitude and direction of the resultant when the 
forces are inclined to each other at angles of (a) 30, (&) 45, (c) 120. 
Check your results by calculation. 

2. Answer Question i, supposing the 4 Ib. force to be a push. 



EXERCISES ON CHAPTER II. 



37 



3. A pull P of 5 Ib. and another force Q of unknown magnitude act 
at 90. They are balanced by a force of 7 Ib. Find the magnitude of Q. 

4. Answer Question 3, supposing P and Q . 
intersect at 45. 

5. A bent lever (Fig. 43) has its arms at ^p 
90 and is pivoted at C. AC = 1 5 inches, BC = 6 
inches. A force P of 35 Ib. is applied at A 

at 1 5 to the horizontal, and another Q is applied 
at B at 20 to the vertical. Find the magnitude 
of Q and the magnitude and direction of the 
reaction at C required to balance P and Q. 




6. A body weighing 24 Ib. is kept at rest 
on an incline which makes 40 with the hori- 
zontal by a force P which is parallel to the 

plane (Fig. 44). Assume that the reaction R F IG . 43 . 

of the plane is at 90 to its surface, and find P. 

7. Answer Question 6, supposing P to be horizontal. 

8. Four loaded bars meet at a joint as shown (Fig. 45). P and Q are 
in the same horizontal line ; T and W are in the same vertical ; S makes 




FIG. 44. 

45 with P. Given that 
and T. 




I5 tons, W=I2 tons, S = 6 tons, find Q 



9. Lines are drawn from the centre O of a hexagon to each of the 
corners A, B, C, D, E, F. Forces are applied in these lines as follows : 
From O to A, 6 Ib. ; from B to O, 2 Ib. ; from C to O, 8 Ib. ; from O to 
D, 12 Ib. ; from E to O, 7 Ib. ; from F to O, 3 Ib. Find the resultant. 

10. Two equal bars AC and BC are hinged at C (Fig. 46). A and B 
are capable of moving in guides in the straight line AB. A constant 
force P of 40 Ib. is applied at C in a direction at 90 to AB, and is 
balanced by equal forces Q, Q applied at A and B in the line AB. 
Calculate the values of Q when the angle ACB has values as follows : 
170, 172, 174, 176, 178, 179, 1 80. Plot Q and the angle ACB from 
your results. (The arrangement is called a toggle joint.) 

11. Five forces meet at a point O as shown (Fig. 47), and are in 
equilibrium. In the front elevation, P, Q and S are in the plane of the 
paper and T is at 45 to the plane of the paper ; Q makes 135 with S. 
In the side elevation T and V are in the plane of the paper. V is per- 
pendicular to the plane containing P, Q and S, and T makes 45 with V. 
Given Q = 4o tons, T = 25 tons, find P, S and V. 



MATERIALS AND STRUCTURES 



12. In a hinged structure, pieces BO and CO meet at the hinge O, and 
a force of 2 tons acts upon O in the direction AO. The angle AOB is 
115, BOC is 15 and the angle AOC is 130 ; find the forces in the two 
pieces and say whether they are struts or ties. (B.E.) 





FIG. 46. 



frunt Elevation, Side.' Elevation,. 
FIG. 47. 




13. There is a triangular roof truss ABC ; AC is horizontal, the angle 
BCA is 25 and BAC is 55 ; there is a vertical load of 5 tons at B. 
What are the compressive forces in BA and BC? What are the vertical 
supporting forces at A and C ? What is the tensile force in AC ? Find 
these answers in any way you please. (B.E.) 

14. Each of the legs of a pair of sheer legs is 45 feet long ; they are 
spread out 23 feet at their base. The length of the back stay is 60 feet. 
If a load of 40 tons is being lifted at a distance of 15 feet, measured in a 
perpendicular line from the line joining the feet of the two legs, find the 
forces in the legs and in the backstay due to this load. (It may be 
assumed that the load is simply hung from the top of the legs.) (B.E.) 

15. A tripod has the following dimensions : The apex point is O, and 
the lengths of the three legs AO, BO and CO are respectively 18-0 feet, 
17-5 feet and 16 feet. The lengths of the sides of the triangle formed by 
the feet AB, BC and CA are 9-0 feet, 9-5 feet and 10 feet respectively. 
Find graphically, or in any other way, the forces which act down each leg 
of the tripod when a load of 10 tons is suspended from it. (B.E.) 

16. If a rigid body be acted on by two non-parallel forces whose points 
of application are different and be kept at rest by a third force, how must 
this third force act, and what must be its magnitude ? A straight light 
rod xyz is pivoted freely at x, and the point y is attached to a pin -z/, 
vertically above JT, by a light cord ; xy is 3 feet, xv is 4 feet, yv is 2 feet, 
yz is 2 feet ; from z is hung a weight of 30 Ib. Find graphically the 
tension in the cord. (I.C.E.) 

17. If three non-parallel forces are in equilibrium, prove that their lines 
of action must be concurrent. A uniform plank AB has length 6 feet and 
weight 80 pounds and is inclined at 40 to the vertical. Its lower end A 
is hinged to a support, while a light chain is fastened to a ring four feet 
vertically above A and to a point on the plank five feet from A. Find 
graphically, or otherwise, the tension in the chain and the magnitude and 
direction of the action of the hinge at A. (The weight of AB may be 
concentrated at the centre of the plank.) (L.U.) 

18. Three cylinders, A, B and C, alike in all respects, are arranged 
as follows : A and B rest on a horizontal table and their curved surfaces 
touch one another. C rests on the top, its curved surface being in 
contact with both A and B. Each cylinder weighs 6 Ib. Find, by 
calculation, the mutual pressure between C and A, also what minimum 



EXERCISES ON CHAPTER II. 39 

horizontal forces must be applied to A and B, passing through their 
axes, in order to preserve equilibrium. Frictional effects are to be 
disregarded. 

19. Three similar spheres rest on a horizontal table and are in 
contact with each other. A fourth sphere, similar to the others, rests 
on the top of the three spheres. Each sphere weighs 10 Ib. Find the 
pressure communicated by the top sphere to each of the other three 
spheres. Neglect frictional effects. 



CHAPTER III. 



PARALLEL FORCES. 

Parallel forces. Confining ourselves for the present to two forces 
only, there are two cases to be considered, viz. forces of like sense 
and forces of unlike sense. To find the resultant of two parallel 
forces P and Q (Fig. 48 (a) ) of like sense, the following method may be 
employed. Let the given forces act at 90 to a rod, at the points A 
and B respectively. The equilibrium of the rod will not be dis- 



, / 




4 .S a 



3 /B 

R / 

/ * 
'k 




FIG. 48. Resultant of two parallel forces. 

turbed by the application of equal opposite forces S, S, applied in 
the line of the rod at A and B. By means of the parallelogram of 
forces A&tt, find R x of P and S acting at A ; and by means of the 
parallelogram of forces B<?/#, find R 2 of Q and S acting at B. 
Produce the lines of Rj and R 2 until they intersect at O, and let R T 
and R 2 act at O. Apply the parallelogram of forces Qhkg to find R 
of Rj and R 2 . R will clearly be the resultant of P and Q, and will 
balance P and Q if its sense be reversed. By measurement it will 
be found that R is equal to the sum of P and Q. 

The resultant of two parallel forces of unlike sense may be found 
by the same process. The construction is shown for two such 



PARALLEL FORCES 41 




forces, P and Q, in Fig. 48 (b) ; the lettering of this diagram corre- 
sponds with that of Fig. 48 (a), and may be followed without further 
explanation. If the diagram be measured, it will be found that R is 
equal to the difference of P and Q. 

Moment of a force. The moment of a force means the tendency of a 
force to turn the body on which it acts about a given axis. The moment 
of a given force depends upon (a) the magnitude of the force, and 
(b) the length of a perpendicular dropped from the axis of rotation 
on to the line of action of the force, and is therefore measured by 
taking the product of these quantities. Thus, in Fig. 49, the body 
is free to rotate about O, and a force P is acting 
on it. Draw OM at 90 to P, then 
Moment of P = P x OM. 

To state the units in which a given moment 
is measured, both the unit of force employed 
and the unit of length must be mentioned. 
Thus, in the above case, if P is in Ib. weight FIG. 49. Moment of a 
and OM in feet, the units will be Ib.-feet. Other 
units which may be used are ton-foot, ton-inch, gram-centimetre, etc. 

The sense of the moment of a force is best stated by reference to 
the direction of rotation of the hands of a clock. Thus the moment 
of a given force will be clockwise or anti-clockwise according as it 
tends to produce the same or opposite sense of rotation as that of 
the hands of a clock. 

Principle of moments. The resultant moment of two or more 
forces, all of which tend to rotate the body on which they act in the 
same sense, will be found by first calculating the moment of each 
force, and then taking the sum If some of the forces have 
moments of opposite sense, these may be designated negative, and 
the resultant moment will be found by taking the algebraic sum. 
Should the resultant moment be zero, the body will be in equilibrium 
so far as rotation is concerned. This leads to the statement that 
a body will be in equilibrium as regards rotation provided the sum of the 
clockwise moments applied to it is equal to the sum of the anti-clockwise 
moments. This is called the principle of moments. 

EXAMPLE i. A horizontal rod AB, the weight of which may be ne- 
glected, has a pivot at C (Fig. 50), and has two vertical forces P and Q 
applied at A and B respectively. Find the relation of P and Q if the 
rod is in equilibrium. 

Let AC=#, 



42 MATERIALS AND STRUCTURES 

Taking moments about C, 

clockwise moment = anti-clockwise moment, 



_ 
Q a 

It will be seen from this result that the forces are inversely propor- 
tional to the segments into which the rod is divided by the pivot. It will 
also be evident that the equilibrant of P and Q acts through C. 



AC~~ B k- ------ 6 

FIG. 50. FIG. 51. 

EXAMPLE 2. A horizontal rod BC, the weight of which may be ne- 
glected, has a pivot at C, and has two vertical forces P and Q of unlike 
sense applied at A and B respectively (Fig. 51). Find the relation of 
P and Q if the rod is balanced. 

Let AC = #, 

BC = & 

Taking moments about C, 



Q * 

Again we may say that each force is proportional to the distance of 
the other force from the pivot, and that the equilibrant of P and Q acts 
through C. 

EXAMPLE 3. A horizontal rod AB, the weight of which may be ne- 
glected, has a weight W applied at C and is 
,. . supported at A and B, the reactions P and Q 

~*~ ~ being vertical (Fig. 52). Find P and Q. 

B Let AB = ,, 



then BC = /-. 

Taking moments about B, 

Px/=W(/-) + (Q*o), 

P=(^)W (i) 

Taking moments about A, 

a = (Qx/) + (Pxo), 

W (2) 



PARALLEL FORCES 43 



It is of interest to find the sum of P and Q, using their values as found 
above ; thus (l-a\ a 



= W-ji-~+^ 

=w. 

Resultant of two parallel forces. Examining the results of these 
examples together with what has been said regarding two parallel 
forces on p. 40, we may state that the resultant of two parallel forces 
has the following properties : 

(1) The resultant is equal to the sum or diflerenee of the given forces 
according as they are of like or unlike sense. 

(2) The resultant is parallel to the given forces and acts nearer to the 
larger ; it falls between the given forces if these are of like sense and outside 
the larger force if of unlike sense. 

(3) The perpendicular distances from the line of the resultant to the 
given forces are inversely proportional to the given forces. 

We may state properties (i) and (3) algebraically : 

R=PQ, (i) 

!= (=0 

A special case. The resultant of two equal parallel forces of 
opposite sense (Fig. 53) cannot be determined from these equations. 
Here Q is equal to P, hence 

R=P-P=c 



FlG. 53. A couple. 

These results show that no single force can 

form the resultant of such given forces, and we may infer from this 
that the resultant effect is to produce rotation solely. The name 
couple is given to this system. 

Eesultant of a number of parallel forces. In Fig. 54 is shown a 
horizontal rod AB acted on by a number of parallel vertical forces 
W 15 W 2 , W 3 , P and Q. For the rod to be in equilibrium under the 
action of these forces, the following conditions must be complied with : 
(i) the forces must not produce any vertical movement, either upwards 



44 



MATERIALS AND STRUCTURES 



or downwards : (2) they must not produce any rotational movement, either 
clockwise or anti-clockwise. 

The first condition will be satisfied provided the sum of the 
upward forces is equal to that of the downward forces ; hence 



The second condition will be satisfied if, on taking moments 
about any point such as A, the sum of the clockwise moments is 
equal to that of the anti-clockwise moments, hence 
W& + W 2 * 2 4- W 3 * 3 - Pa + Qb. 

Supposing that it is found that the sum of the downward forces is 
not equal to that of the upward forces, then the rod may be equilibrated 
by application of a force E equal and opposite to the difference of 
these sums ; thus E = ( W l + W 2 + W 3 ) - (P + Q). 



1 


' W ' 1 


r W 2 


\ 


W 3 


u__ 










k *,--> 




> 


'R 




A 








IR 


* 
i 
H 
I 
U__ 


<4.. 


s 


1. 





FIG. 54. Resultant of parallel forces. 

The distance x from A at which E must be applied (Fig. 54) may 
be found by taking moments about A ; thus 

Ex = (W^ + W 2 # 2 + W 3 # 3 ) - (Pa + Q<). 

Having thus found the magnitude, position and sense of E, the 
resultant of the given forces may be found by reversing the sense of E. 
We have therefore the following rules for finding the resultant of 
a number of parallel forces P 15 P 2 , P 3 , etc. 

R = 2P, (r) 

(2) 



or, 



__ 
: 



R 



(3) 



Equation (i) will give the magnitude of R, and its position will be 
given by calculating x, obtained by taking moments about any con- 
venient point as indicated by equation (2). 



PARALLEL FORCES 



45 



EXAMPLE i. Four parallel forces act on a rod AB as shown (Fig. 55 (a)). 
Find their resultant. 

R=2P 

=2+5+7+3 

=^7 lb., of sense downward. 

Taking moments about A, we have 



feet. 



=92, 



.l 7lb.i 3/&1. 
lh ---- j --- x --- '->[ ; 

f-2-x ; f R i 

- ----- s'-- ' 



- ----- -- 

u --------- 6 - ----- ^J 

7' 



X 1 ***" 

3^ 


' > 
O 


4 

'4/6. 


" V 

^8*6. 
2tt> 

(-2-> 
y 



FIG. 55. 

EXAMPLE 2. Parallel forces act on a body as shown in Fig. 5 5 (b). Find 
their resultant. 



2)-8 
= 9 - 8=1 lb., of sense downward. 

It is convenient to take moments about a point O on the line of the 3 lb. 
force. 



(3 xo)+(4 x iJ)-(8 x 4) + (2 x 6J), 
3=-j[7 feet. 
The negative sign indicates that R falls on the left side of O. 

EXAMPLE 3. A beam of 16 feet span rests on supports at A and B 
and is loaded as shown (Fig. $6()). Find the reactions of its supports. 
Taking moments about B we have 



x 13), 



P = 2-765 tons. 

Taking moments about A we have 

Q x i6 = (2X2)+(ix5)+(fx 
Q = 1-484 tons. 



4 6 



MATERIALS AND STRUCTURES 



To check the work we have 



2765 + 1-484 = 2 + 1 +|+|, 

4-249=4-25, 

results which agree within the limits of accuracy of the answers found for 
P and Q. 



I J5- 

Y2fonj| Y^fcn 

I t"" I f"" 1 



16 



-* 



2 tons 


|3fo/u 
+5 tons 


1^4 

\4ton* 


f 


\ : ' 


B. i 


t-3' 
i 


<- in' 


^ x ' . 


i 


Af 


fQ 



FIG. 56. Reactions of the supports of beams. 

EXAMPLE 4. A beam rests on supports at A and B (Fig. 56^)), its 
ends overhanging the supports, and the beam is loaded as shown. Find 
the reactions P and Q. 

Taking moments about B, we have 



tons. 



Taking moments about A, we have 



ioQ=98-6, 

Q = 9-2 tons. 
To check the work, we have 



15 = 15. 

Graphical method of finding the reactions of a beam. The 

method will be illustrated by reference to Fig. 57, which shows a 
beam simply supported at A and B and carrying a single load W. 
Taking a base line CD projected from the drawing of the beam, set 
off CE at right angles to CD, and of length to scale to represent W. 
Join DE and project W downwards so as to intersect CD and DE in 
F and G. Taking moments about B, we have 



W* 



From the similar triangles ECD and GFD, we have 
CE = FG 
CD FD' 



PARALLEL FORCES 



47 



or 



WFG 



. 




Q 



.(2) 



FIG. 57. Beam carrying one load ; reactions found graphically. 

Hence FG represents P to the same scale that CE represents W. 
The value of Q may be found from 



Q = W-P. 

Or, by using the same construction, Q may be found by making 
DH equal to W, joining CH cutting FG in K, when FK gives the 
value of Q (Fig. 57). 




FIG. 58. Graphical solution of P for a beam carrying several loads. 

If the beam carries several loads (Fig. 58), the construction for P 
should be carried out for each load as indicated ; the total sum of the 



4 8 



MATERIALS AND STRUCTURES 



intercepts will give the value of P. Q may be found by means of a 
similar construction carried out for the other end of the beam, and 
the result may be checked by comparing the sum of P and Q with 
the sum of the given loads. 

Centre of parallel forces. Let two parallel forces P and Q act on 
a rod AB (Fig. 59). Their resultant R will divide AB in the pro- 
portion P:Q = BC:AC (i) 

Let the lines of P and Q be rotated to new positions P', Q', 
without altering the magnitudes. Through C draw DCE perpen- 
dicular to P' and Q'. Then R', the resultant of P' and Q', will 
divide DE in inverse proportion to P' and Q'. Inspection of Fig. 
59 will show that the triangles ACD and BCE are similar, hence 
EC:DC=BC:AC 
= P:Q 

from (i). It therefore follows that R' passes through the same point 
C. This point is called the centre of the parallel forces P and Q. 

E 



\ 


\ 


k 


\ 
\ 


V 


v 


y 


AE 

x" \ 

, Q X ^ 


\ ^ 


< \ 
\ 
\ 


.x \ 

x \ 


A 


N .'C 


V 


D 




(o) 



w 



FIG. 59. Centre of parallel forces. 



FIG. 60. Centre of gravity. 



If there are a number of parallel forces, it will be seen easily that 
their resultant always passes through the same point, whatever may 
be the inclination of the forces. A common example of this occurs 
in the case of the weight of a body. Each particle in the body 
possesses weight, hence gravitational effort on the body is really a 
large number of forces directed towards the earth's centre, and these 
will be parallel and vertical for any body of moderate dimensions. 
It is not possible to incline the directions of the forces in this case, 
but the same effect may be produced by inclining the body. The 
weights of all particles will still be vertical, but their directions will be 
altered in relation to a fixed line AB in the body (Fig. 60 (a and ^) ). 
Supposing the line CD of the resultant weight W to be marked on 
the body in Fig. 60 (a), and to be marked again as EF in Fig. 60 (), 
the intersection G of these lines of W would be the centre of the 



PARALLEL FORCES 



49 



weights of the composite particles. The name centre of gravity is 
given to this point 

Centre of gravity by calculation. The general method of calcu- 
lation will be understood by reference to Fig. 61. The body is 
supposed to be a thin sheet of material. Take two coordinate axes 




FIG. 61. Centre of gravity of a thin sheet. 



OX and OY. First let OX be horizontal; the weights of the 
particles being called ze/j, w 2 , ze> 3 , etc., and their coordinates 
)' (^3^3)' etc -> we nave 5 by taking moments about O, 
+ etc.) x = w l 



or, 



It is evident that 2w gives the total weight W of the sheet, hence 



Now turn the sheet round until OY is horizontal ; the lines of 
direction of the weights will be parallel to OX, and, by taking 
moments about O, we have 

+ etc) y = 



Draw a line parallel to OX, and at a distance y from it, and 
another parallel to OY at a distance x ; the intersection of these 
gives the centre of gravity G. 



D.M, 




50 MATERIALS AND STRUCTURES 

The position of the centre of gravity in certain simple cases may 
be seen by inspection. Thus for a slender straight rod or wire, it lies 
at the middle of the length. In a square or rectangular plate G 
lies at the intersection of the diagonals. 

A circular plate has G at its geometrical centre. The position of 
G in a triangular plate may be found by first imagining it to be cut 

into thin strips parallel to BC (Fig. 62). 
The centre of gravity of each strip will lie 
at its centre of length ; hence all these 
centres will lie in DA, where D is the 
centre of BC, and hence DA contains 
the centre of gravity of the plate. In the 
same way, by taking strips parallel to AB, 
the centre of gravity will lie in CE, where 
E is the centre of AB. Hence G lies at 
FIG. 62. Centre of gravity of a the intersection of DA and CE, and it 

triangle. . . 

is easy to show by geometry that DG is 

one-third of DA. Hence the rule that G lies one-third up the line 
joining the centre of one side to the opposite corner. 

Advantage is taken of a knowledge of the position of G in thin 
plates having simple outlines in applying equations (i) and (2) above. 
The following examples will illustrate the method. 

EXAMPLE i. Find the centre of gravity of the thin uniform plate shown 
in Fig. 63. 

Take axes OX and OY as shown and let the weight of the plate per 
square inch of surface be iv. For convenience of calculation the plate is 
divided into three rectangles as shown, the respective centres of gravity 
being G 1? G 2 and G 3 . Taking moments about OY, we have 

7/{(6x i) + (8x i) + (3>< i)}*=w(6x i x a)+w(8'x i 

- 26-5 
x= S 

I 7 
= 1^56 inches. 

Again, taking moments about OX, we have 



= 5-8 inches. 

EXAMPLE 2. A circular plate (Fig. 64) 12 inches diameter has a hole 
3 inches diameter. The distance between the centre A of the plate and 
the centre B of the hole is 2 inches. Find the centre of gravity. 



PARALLEL FORCES 



Take AB produced as OX, and take OY tangential to the circumference 
of the plate. It is evident that G lies in OX. Taking moments about 
OY, we may say that the moment of the plate as made is equal to that of 



YI 





r m 


--D - 




-- n 


f 






G, 


1- -r- 
:L 




X 


p-1'5 


6' 




3 


* 

'I'" 


G 


< 

! I 


oi 

? I 






G 3 


'^ , 













X 




FIG. 63. 



FIG. 64. 



the solid disc diminished by the moment of the material removed in 
cutting out the hole. Let w be the weight per square inch of surface, D 
the diameter of the plate and d that of the hole. Then 

D* 

4 ' 

7T<t 2 



Weight of solid disc 



Weight of piece cut out w . 

4 

w . *. c , /TrD 2 ird z \ 

Weight of plate as made = -ze/ ) 

\ 4 4 / 



Take moments about OY, and let OG = .r, 



_ 
= "D 2 -^ 2 

828 
= =6-13 inches. 

Other cases of symmetrical solids which may be worked out by 
application of the same principles are given below. 

Any uniform prismatic bar has its centre of gravity in its axis, at 
the middle of its length. 

A solid cone or pyramid has the centre of gravity one-quarter up 
the line joining the centre of tbe base to the apex. 



MATERIALS AND STRUCTURES 



A cone or pyramid open at the base and made of thin sheet metal 
has its centre of gravity one-third up the line joining the centre of 
the base to the apex. 

Graphical method for finding the centre of gravity. The follow- 
ing method of finding G by construction 
in the case of a thin sheet abed (Fig. 65) 
is sometimes of service. Join bd and find 
the centres of gravity ^ and c^ of the 
triangles abd and cbd; join <y 2 . Again, 
join ac, and find the centres of gravity c z 
and c of the triangles abc and adc\ join 
FIG. 65. Centre of gravity c% and c, cutting cfa in G, the centre of 

found graphically. ^^ Qf ^ ^^ 

States of equilibrium of a body. The equilibrium of a body will 
be stable, unstable or neutral, depending on whether it tends to return 





(a) 



(d) 



FIG. 66. Stable and unstable equilibrium. 



to its original position, to capsize, or to remain at rest when it is 
slightly disturbed from its original position. A body at rest under 
the action of gravity and supporting forces 
depends for its state of equilibrium on the 
situation of its centre of gravity. A cone 
gives an excellent example of all three 
states ; when resting on its base on a hori- 
zontal table the equilibrium is stable 
(Fig. 66 (a)), for on slightly disturbing it 
(Fig. 66 ()), R and W conspire to return 
it to its original position. If resting fR 

on its apex, the equilibrium is unstable 

/T-- ~/ V\ v i ^ j- . i /-r- FIG. 67. Neutral equilibrium. 

(Fig. 66 (c) ) ; a slight disturbance (r ig. 

66 (^)) shows that R and W conspire to overthrow it. If resting on 

its curved surface on a horizontal table (Fig. 67), the equilibrium is 




PARALLEL FORCES 53 



neutral, for, no matter what the position may be, R and W act in 
the same vertical line, and so balance. 

Reactions of the supports of a beam. In calculating the moment 
of the weight of a given body about a given axis, we may imagine 
that the whole weight is concentrated at the centre of gravity. This 
enables us to deal with problems on beams carrying distributed loads. 
The following example will make the method clear. 

EXAMPLE. A beam is supported at A and B (Fig. 68). The section 
of the beam is uniform and its weight is 200 Ib. per foot run. It carries 



* 



i , , Of 
!.~4'--Ji 10 



FIG. 68. Reactions of the supports of a beam. 

a load of 500 Ib. per foot run uniformly distributed over 9 feet of the 
length as shown. Find P and Q. 

The centre of gravity of the beam lies at G l at a distance of 7 feet 
from B. G 2 is the centre of gravity of the distributed load, and lies at 
9^ feet from B. 

Total weight of beam =W! = 200 x 14 = 2800 Ib. 

Total weight of load = W 2 = 500 x 9 =4500 Ib. 

Apply Wj at Gj and W 2 at G 2 , and take moments about B to find P : 



P = (2800x7) + (45 

10 
= 6235 Ib. 

Again, take moments about A to find Q : 



(2800 x 3)+ (4500 x^) 

10 
= 106 Ib. 



Checking the results, we have 



62354-1065 = 2800 + 4500, 
7300-7300. 

Parallel forces not in the same plane. In Fig. 69 are shown four 
bodies, one at each corner of the horizontal square ABCD. The 
weights of these bodies act vertically downwards, and hence are not 



54 



MATERIALS AND STRUCTURES 



all contained in the same vertical plane. Denoting the weights of the 
bodies by W A , W B , W c and W D , we may proceed to find the centre 
of gravity in the following manner. The resultant weight (W A + W B ) 
of the weights at A and B will act at G l , which divides AB in inverse 
proportion to W A and W B ; i.e. 

G 1 E:G 1 A = W A :W*. 

In the same way, the position of G 2 where the resultant (W c + W D ) 
of W c and W D acts may be found from : 

G 2 D:G 2 C = W C :W D . 
The resultant weight of all four bodies is equal to 



and will act at G which may be found from the following proportion 
G 2 G : G,G = (W A + W B ) : ( W c + W D ). 





FIG. 69. Parallel forces not in the same plane. 



FIG. 70. Pressure on table having 
three legs. 



Having thus determined the position of G, we may invert the 
problem and state the results in this way. Let ABCD be a square 
plate supported on legs at A, B, C and D. Let a weight having a 
magnitude (W A + W B + W c + W D ) be placed at the point G, the 
position of which has been calculated as above, then it will be evident 
that the pressure on the legs owing to this single load will have 
respectively the values given in the first problem, viz. W A , W B , W c 
and W D . Strictly speaking, this problem is indeterminate, depend- 
ing, as it does, on the exact equality of length of the legs, on their 
elastic properties and on the levelness of the floor on which they 
rest. A table having three legs gives a problem capable of exact 
solution independent of these conditions. 

Given a table resting horizontally on three legs at A, B and C, as 
shown in plan in Fig. 70. Let a weight W be placed at any point 
G, and let it be required to find the pressure on each leg due to W. 



PARALLEL FORCES 



55 



It will be noticed that if one of the legs, say A, be lifted slightly, the 
table will rotate in a vertical plane about the line BC. This indicates 
that the pressure on A may be calculated by taking moments about 
BC. Draw GM and AN perpendicular to BC, and let P A be the 
reaction of the leg A ; then, 
P A xAN = WxGM, 




In the same way, P B may be 
found by taking moments about 
AC, and P c by taking moments 
about AB. The results may be 
checked from 



EXPT. 6. Principle of moments. 
Fig. 7 1 shows a wooden disc which 
is free to rotate about its centre 
on a screw driven into a wall 
board. Attach cords to various points on the face of the disc and 
apply different forces by means of weights as shown. Let the disc 
come to rest under the action of these forces, and test the truth of 
the principle of moments by calculating the sum of the clockwise 
moments and also that of the anti-clockwise moments. 

EXPT. 7. Reactions of a beam. Fig. 72 shows an apparatus con- 
sisting of a wooden beam supported by means of two hanging spring 



FIG. 71. Apparatus for illustrating the 
principle of moments. 



n 





FIG. 72. Apparatus for determining the reactions of a beam. 

balances. Apply various loads and calculate the reactions of the 
supports. Make allowance for the weight of the beam and also 



MATERIALS AND STRUCTURES 



for any distributed loads by concentrating them at their respective 
centres of gravity. Repeat the experiment with altered loads and 
different positions of the points of support. Make 
a table showing in each case the calculated 
reactions and also those read from the spring 
balances. 

EXPT. 8. Centre of gravity of sheets. The 
centre of gravity of a thin sheet may be found by 
hanging it from a fixed support by means of a 
cord AB (Fig. 73) ; the cord extends downwards 
and has a small weight W, thus serving as a 
plumb-line. Mark the direction AC on the sheet 
and then repeat the operation by hanging the 
sheet from D, marking the new vertical DE. G 
will be the point of intersection of AC and DE. 
Carry out this experiment for the sheets of metal 
or millboard supplied. 

EXPT. 9. Centre of gravity of a solid body. The centre of gravity 
of a body such as a connecting rod (Fig. 74) may be found by 
balancing it on a knife edge, which may be arranged easily by use 
of V blocks and a square bar of steel. G will lie vertically over 




Ow 

FIG. 73. Experiment 
on the centre of gravity 
of a sheet. 



<r 



''/'////S///// 



FIG. 74. Experiment on the centre of gravity of a solid body. 

the knife edge when the rod is balanced. Carry out this experiment 
on the bodies supplied, in each case making a sketch of the body and 
recording on the sketch the dimensions necessary for indicating the 
position of G. 



EXERCISES ON CHAPTER III. 

1. A uniform horizontal rod AB is pivoted at its centre C, and carries 
a load of 12 Ib. at D and another of 20 Ib. at E. D and E are on 
opposite sides of C, CD and CE being 8 inches and 12 inches respectively. 
If balance has to be restored by means of a 14 Ib. weight, find where it 
must be placed. What will be the reaction of the pivot ? 

2. A rod AB carries loads of 3 Ib., 7 Ib. and 10 Ib. at distances of 2 
inches, 9 inches and 15 inches respectively from A. 'Find the point at 
which the rod will balance. Neglect the weight of the rod. 

3. Fig. 75 shows an arrangement of a right-angled bent lever ABC 
carrying a load of 40 Ib. AB and BC are 12 inches and 3 inches respec- 
tively and AB is horizontal. C is connected by a horizontal link CE to a 



EXERCISES ON CHAPTER III. 



57 



vertical lever DF, which is pivoted at D. DF and DE are 15 inches 
and 3 inches respectively. The arrangement is balanced by a cord FG 
passing over a pulley at G and carrying a load W. Find W, neglecting 
the weights of the various parts and also friction. 

4. A lever safety valve for a steam boiler has the following dimensions : 
Diameter of valve, 3 inches ; distance from fulcrum to valve centre, 4^ 
inches ; weight of valve and its attachment to the lever, 4^ Ib ; distance 
from fulcrum to centre of gravity of the lever, 14 inches ; weight of 
lever, 7 Ib. The weight on the end of the lever is 90 Ib. Find its 
distance from the fulcrum if the valve is to open with a steam pressure 
of 70 Ib. per square inch. 

5. A uniform beam 20 feet long weighs i| tons, and is supported at 
its ends A and B. A uniformly distributed load of ^ ton per foot run 
extends over 10 feet of the length measured from A, and a concentrated 
load of 3 tons rests at a point 4 feet from B. Find the reactions of the 
supports by calculation. 




40/6. 



FIG. 75- 



6. A beam 18 feet span carries loads of 2 tons, 4 tons and 8 tons at 
distances measured from one support of 3 feet, 8 feet and 12 feet respec- 
tively. Find graphically the reactions of the supports. Neglect the weight 
of the beam. 

7. A uniform beam weighs 2 tons and is 24 feet long. It is supported 
at a point A 6 feet from one end, and at another point B 4 feet from the 
other end. There is a concentrated load of i tons at each end and 
another of 3 tons at the middle of the beam. Find the reactions of the 
supports by calculation. 

8. Three weights of 4 Ib., 8 Ib. and 12 Ib. respectively are placed at 
the corners A, B and C of an equilateral triangle of 2 feet side. Find the 
centre of gravity. 

9. A letter L is cut out of thin cardboard. Height, 3 inches ; 
breadth, 2 inches ; width of material, inch. Find the centre of gravity. 

10. A solid pyramid has a square base of 3 inches edge and is 
5 inches high. It rests on its base on a board, one end of which may 
be raised. The edges of the base of the pyramid are parallel to the 
edges of the board, and slipping is prevented by means of a thin strip 



58 MATERIALS AND STRUCTURES 

nailed across the board. Find, by drawing or otherwise, the angle which 
the board makes with the horizontal when the pyramid just tips over. 

11. A flat equilateral triangular plate of 4 feet side is supported horizon- 
tally by three legs, one at each corner. A vertical force of 112 pounds is 
applied to the plate at a point which is distant 3 feet from one leg and 
1 8 inches from another. Determine the compressive force in each leg 
produced by this load. (B.E.) 

12. A scale-pan of a balance with unequal arms is weighted in such a 
way that the beam is horizontal when no masses are placed in the pans. 
A body when placed in the two pans successively is balanced by masses 
P and Q in the opposite pans. Prove that its mass is \/PQ. (L.U.) 

13. A horizontal platform is supported on three piers ABC forming a 
triangle in plan. AB = 6 feet ; AC = 8 feet ; BC = 8 feet. The centre of 
gravity of the platform and load carried is distant 5 feet from A and 
4 feet from B. Find the proportion of the load carried by each of the 
three piers. Show that, if there were four piers instead of three, the 
reactions could not be determined without further information (I.C.E.) 



CHAPTER IV. 

PROPERTIES OF COUPLES. SYSTEMS OF UNIPLANAR 

FORCES. 

Moment of a couple. Consider the couple formed by the equal 
forces Pj and P 2 (Fig. 76). Let d be the perpendicular distance, or 
arm, between the lines of the forces. 
It may be shown, by taking moments 
in succession about several points A, 
B, C, D, that the moment of the couple 
is the same about any point in its plane, 
and is given by Pd. 

Thus, taking moments about A, we 
have 

Moment of the couple 

= (Pj X o) - (P 2 X d) FlG ?6i _A couple has the same moment 

r> j / \ about any point in its plane. 

--tV*' .................. W 

the negative sign indicating an anti-clockwise moment. 
Taking moments about B, we have 

Moment of the couple = (P 2 x o) - (Pj x d) 




Taking moments about C gives 

Moment of the couple = - (P l x a) - P 2 (d- a) 

--V ....... .................. (3) 

Taking moments about D, we have 

Moment of the couple = (P 2 x V) - P l (d+ b] 

= -Pi* ........................... (4) 

As the forces are equal, the four results are identical, thus proving 
the proposition. 

Equilibrant of a couple. It has been seen (p. 43) that no single 
force can be the resultant of a couple, hence no single force can 



6o 



MATERIALS AND STRUCTURES 



equilibrate a couple. It will now be shown that another couple of 

equal opposite moment applied in the same plane, or in a parallel plane, 

will balance a given couple. 

In Fig. 77 are shown two couples, one having equal forces P l and 

P 2 , and the other couple having equal forces Q x and Q 2 . Produce 

the lines of these forces to intersect 
at A, B, C and D, and let a and b be 
the arms of the P and Q couples 
respectively. From A draw AM and 
AN perpendicular to P l and Q x re- 
spectively. Then AM = a, and AN = b. 
The triangles AMC and AND are 
similar, hence 

AC:AM = AD:AN, 
AC:AD=AM:AN 

~*W (i) 

Now if the couples have equal moments, we have 




FIG. 77. Two equal opposing couples 
are in equilibrium. 



or Q:P = a: (2) 

Hence, AC and AD may be taken to represent Q and P respec- 
tively to some scale of force. 

As ACBD is a parallelogram, it follows that the resultant of P, 
and Qj acting at B will be R = AB. 

Also the resultant R 2 of P 2 and Q 2 acting at A will be 

R 2 = BA. 




c 

FIG. 78. Equal opposing couples in parallel planes are in equilibrium. 

As R! and R 2 are equal, opposite, and in the same straight line, 
they balance; hence, the given couples are in equilibrium. We 



PROPERTIES OF COUPLES 61 

may therefore state that couples of equal opposite moment acting 
in the same plane are in equilibrium, and either couple may be 
said to be the equilibrant of the other couple. 

In Fig. 78 is shown a rectangular block having equal forces P x 
and P 2 applied to its vertical front edges AD and CB, and other 
equal forces P l and P 2 applied to the vertical back edges FG and 
HE. Let these forces be all equal, when the block will have a pair 
of equal opposite couples acting in parallel planes. That these 
couples balance may be seen by taking the resultant R x of the 
forces Pj, Pj, and also the resultant R 2 of the other equal pair 
P 2 , P 2 . These resultants are equal and opposite and act in the 
same straight line, and hence are in equilibrium. 

Eesultant of a couple. We have now seen that a couple can be 
balanced by the application in the same plane, or in a parallel plane, 
of a second couple having an equal opposite moment. Supposing 
the forces of the second couple to be reversed in sense, it is evident 



1 





FIG. 79. Single-handed tap wrench. FIG. 80. Double-handed tap wrench. 

that the effect of this couple on the body will be identical with that 
of the first couple. We may say now that either couple is the resultant 
of the other, i.e. the effect on the body as a whole will be the same, 
no matter which couple be applied to it. 

This proposition may be stated in a different way, viz. a couple 
may be moved from any given position to another position in the same 
plane or in a parallel plane, without thereby altering its effect on the body 
as a whole. 

Owing to the equality of the forces forming a couple, the applica- 
tion of a couple to any body will not tend to move it in any 
direction, but will merely tend to set up rotation. For example, 
in tapping a hole, the use of a single-handed tap wrench (Fig. 79) 
will tend to bend the tap and to spoil the thread ; a double-handed 
wrench enables a couple to be applied giving pure rotation to the 
tap (Fig. 80). It is evident that the same turning effort may be 



62 



MATERIALS AND STRUCTURES 



obtained by means of small forces and a large arm, or by larger 
forces and a smaller arm, a fact which we may state as follows : The 
forces of a couple may be altered in magnitude provided the arm be altered 
so as to make the moment the same as at first. 

The case of a ship having screw propellers affords an example of 
the balancing of couples in parallel planes. Referring to Fig. 81, 
couples are applied to the shaft at A by the engines and, neglecting 
the friction of the bearings, these couples are balanced by an equal 
opposite couple produced by the resistance of the water acting on 



*n A 




FIG. 8 1. Screw propeller shaft. 

the propeller at E. The planes of these couples are perpendicular 
to that of the paper and hence are parallel. The distance of A 
from E is immaterial so far as the equilibrium of the couples is 
concerned ; nor does the diameter of the propeller affect the problem 
of equilibrium. 

The law that every force must have an equal opposite force may 
now be extended by asserting that it is impossible for a couple to act 
alone ; there must always be an equal opposite couple acting in the same 
plane, or in a parallel plane. 

Substitution of a force and couple for a given force. In Fig. 82 
is shown a body having a force Pj applied at A. We will suppose 
that it would be more convenient to 
have the force applied at another point 
B. Apply equal opposite forces P 2 , 
P 2 , to B, each equal to P x and in a 
line parallel to P T ; these will be self- 
balancing and will therefore not affect 
the equilibrium of the body. Let d be 
the perpendicular distance between Pj 
and P 2 . Pj and the equal downward 

force P 9 at B form a COUple, the moment FIG. 82. Transference of a force to a 
. line parallel to the given line of action. 

of which is P 2 </; this couple may be 

moved to any convenient situation in the plane, leaving the upward 
force P 2 at B. A given force is therefore equivalent to a parallel equal 
force of like sense together with a couple having a moment obtained as above. 




PROPERTIES OF COUPLES 



Substitution of a force for a given force and a given couple. 

In Fig. 83 we have given a force P acting at A, together with a 

couple Q, Q, having an arm d. 

The moment of the couple is Q</. p ' 

Alter the forces of the couple so 

that each new force P', P' is equal 

to P, the new arm a being such that 




Apply the new couple so that 
one of its forces acts at A, in the 

Same line as P, and in the Opposite FIG. 83. Reduction of a given force and 

sense. These forces balance at A, :ouple to a single f 

leaving a single force P' acting at a perpendicular distance a from 

the given force P. 

EXAMPLE i. A single-handed tap wrench has a force of 30 Ib. applied 
at a distance of 15 inches from the axis of the tap (Fig. 84). The centre 
line of the wrench is at a height of 5 inches above the face of the work 
being tapped. Find the moment of the couple acting on the tap and also 
the moment of the force tending to bend the tap. 

Transferring P from A to B gives a force P acting at B, together with 
a couple having a moment given by 

Moment of couple = P x AB 

= 30 x 15=450 Ib.-inches. 

The force P acting at B tends to bend the tap about C. To calculate 
its moment we have 

Moment of P = PxBC 

= 30 x 5 = 150 Ib.-inches. 





FIG. 84. 



FIG. 85. 



EXAMPLE 2. A bent lever ACB (Fig. 85) is pivoted at C, and has 
forces P and Q applied at A and B respectively. Find the resultant 
turning moment on the lever and also the resultant force on the pivot. 

The solution may be obtained by drawing the lever to scale. Transfer 



6 4 



MATERIALS AND STRUCTURES 



P and Q to C as shown, giving forces P' = P and Q'=Q acting at C, 
together with a clockwise couple Q x CN and an anti-clockwise couple 
P x CM, CN and CM being perpendicular to Q and P respectively. The 
resultant turning moment may be calculated by taking the algebraic sum 
of the couples, thus 

Turning moment = (Q x CN) - (P x CM). 

This moment will be clockwise if the result is positive. 

To obtain the resultant force on the pivot, apply the parallelogram of 
forces as shown to find the resultant of P' and Q' acting at C. R gives 
the required force. 

Equilibrium of a system of uniplanar forces. Any system of forces 
acting in the same plane will be in equilibrium provided (a) there is 
no tendency to produce translational movement, 0) there is no tendency to 
rotate the body. These conditions may be tested either by mathe- 
matical equations or by graphical methods. To obtain the necessary 
equations we may proceed as follows. 

In Fig. 86, four forces P I} P 2 , P 3 , P 4 , are given acting in the 
plane of the paper at A, B, C and D respectively. Take any two 

Y 




FIG. 86. A system of uniplanar forces. 

rectangular axes OX and OY in the same plane and take components 
of each force parallel to these axes. Calling the angles made by 
the forces with OX <x 15 a 2 , a g and a 4 , the components will be 
(see p. 23) : 

Components parallel to OX : P l cos a lf P 2 cos a 2 , P 3 cos <x 3 , P 4 cos a 4 . 
Components parallel to OY : Pj sin a lf P 2 sin a 2 , P 3 sin a 3 , P 4 sin a 4 . 
These components may be substituted for the given forces. Now 



SYSTEMS OF UNIPLANAR FORCES 65 

transfer each component so that it acts at O instead of in its given 
position. This transference will necessitate the introduction of a 
couple for each component transferred. Let x l and y l be the 
coordinates of A, and describe similarly the coordinates of B, C 
and D. The couples required by the transference of the components 
of P! will be (P x cos c^)^ and (Pj sin a,)*, ; the other couples may be 
written in the same manner, giving 
Couples parallel to OX : 

(PjCOSa^jj, (P 2 COSa 2 )^ 2 , (P 3 COSa 3 )j 3 , (P 4 COSa 4 )j 4 . 

Couples parallel to OY : 

(Pjsinaj)^, (P 2 sina 2 )# 2 , (P 3 sina 3 )^ 3 , (P 4 sina 4 )# 4 . 

The couples parallel to OX may be reduced to a single resultant 
couple by adding their moments algebraically. Similarly, those 
parallel to OY may be reduced to a single couple, giving 

Resultant couple parallel to OX = 2(P cos a)y. 
Resultant couple parallel to OY = 2(Psina)#. 

Fig. 87 shows the reduction of the given system so far as we have 
proceeded, which now consists of a number of forces acting in OX 



p * 6Ut *+ Couple* Z(Pcos)y. 

P 3 sin 3 < - 



P, 5171 OC, 



*Couf>le~ 



FIG. 87. A system equivalent to that in Fig. 86. 

and OY, together with two couples. For equilibrium, there must be 
no tendency to produce movement in a direction parallel to OY, 
hence the algebraic sum of the forces in OY must be zero. This 
condition may be written : 

2P sin a = o ( i ) 

At the same time there must be no tendency to produce movement 
in a direction parallel to OX ; hence the algebraic sum of the forces 
in OX must be zero, a condition which may be written : 

2Pcosa = o (2) 

D.M. E 



MATERIALS AND STRUCTURES 



Further, there must be no tendency to produce rotation, a condition 
which may be secured provided (i) each of the couples is zero, in 
which case 2(Psina)a; + 2(Pcosa)j = o ; 

or, (ii) the couples may be of equal moment and of opposite sense of 
rotation, in which case their algebraic sum will again be zero. Hence 
the complete condition of no rotational tendency may be written : 

2(Psina)# + 2(Pcosa)^ = o (3) 

These equations (i), (2) and (3) being fulfilled simultaneously serve 
as tests for the equilibrium of any system of uniplanar forces. A 
little judgment must be exercised in the selection of the coordinate 
axes OX and OY in any particular problem so as to simplify the 
subsequent calculations. 

EXAMPLE. A roof truss, 20 feet span, 5 feet rise (Fig. 88), has a 
resultant wind pressure of 2000 Ib. acting at C, the centre of the right- 




f 


on 


: 3 


\ 


P 


x ' 






^ N ^^ 




FIG. b8. Reactions of the supports of a roof truss. 

hand rafter, in a direction perpendicular to that of the rafter. The truss 
is bolted down to the support at B, and rests on rollers at A, so that 
the reaction of the support at A is vertical. Find the reactions of the 
supports. 

In this case, BX and BY are the most convenient coordinate axes. 
First find H and V, the components of the load parallel to these axes, by 
drawing the triangle of forces abc. This triangle is similar to the triangle 
DBE, hence H : 2000 = 5 : 



V : 2000= 10 : V725, 



Let P and Q be the reactions of the supports, and let QH and Q v be 



THE LINK POLYGON 67 

the components of Q parallel to BX and BY respectively. Then, from 
the equations of equilibrium, we have 

2Psina=o; hence, P + Qv-V = o, 
or P + Q v = i79olb ................... (i) 

2Pcosa = o; hence, QH H=o, 
or Q H = 8 9 5 Ib ..................... (2) 

2(P sin a);tr+2(P cos a)y=o ; hence, (P x 20) - (V x 5) - (H x 2$)=o. 
It will be noted that the last equation is obtained by taking the algebraic 
sum of the moments of all the forces about B, Reducing it, we have 
2oP = (1790x5) + (895x2^), 



(3) 



Substitution of this value of P in (i) gives 
559'4+Qv=i79o, 

Qv=i23<>6 Ib ................................ (4) 

To find Q, we have Q = v/Q H 2 



= ^2311000 

= i2olb .................................. (5) 

To find the angle a which Q makes with the vertical, we have 

2 

Qv 

895 



1230-6 

= 0727 ; 
'. = 36 I' (6) 

Graphical solution by the link polygon. A convenient method of 
determining graphically the equilibrant of a system of uniplanar forces 
will now be explained. It is required to find the equilibrant of the 
given forces P lt P 2 and P 3 (Fig. 89 (a)). Take any point A on the line 
of P 1} and proceed to balance P 1 by the application of any pair of 
forces / t and / 2 intersecting at A. The triangle of forces abQ 
(Fig. 89 (^)), in which 

p i Pz'-p^ab'.bQ'.Qa, 

will determine the magnitudes of/ t and / 2 . Imagine p l and / 2 to 
be applied at A (Fig. 89 (a) ) through the medium of bars, or links, 
one of which, AB, is extended to a point B on the line of action of 
P 2 . To equilibrate this link, it must exert a pull/ 2 at B equal and 
opposite to the pull it gives to A. 

The forces f> and P 2 acting at B may be equilibrated by the 



68 



MATERIALS AND STRUCTURES 



application of a third force / 3 at B, / 3 being found in direction and 
magnitude from the triangle offerees Qbc (Fig. 89^)), in which 

A :P 2 : A = ^ : ^ : ^- 

Let / 3 be applied at B (Fig. 89(0)) by means of a link BC, 
intersecting P 3 at C and exerting at C a force equal and opposite to 
that which it exerts on B. 

The forces / 3 and P 3 acting at C are now equilibrated by means of 
a force / 4 applied at C, / 4 being found from the triangle of forces 
CW, in which / 8 : P 3 :/ 4 = O* : *rf : dO. 

Let the force / 4 be applied at C by means of a link, and let this link 
and that in which p l acts intersect at D. Each link will exert a force 
at D equal and opposite to that which it communicates to A and C 




FIG. 89. Graphical solution by the link polygon. 

respectively. The forces p l and/ 4 thus acting at D may be equilibrated 
by means of a third force E applied at D, E being found from the 
triangle of forces Qda, in which 



It will now be seen, by reference to Fig. 89 (a), that each of the given 
forces is balanced, that the closed link polygon A BCD is in equilibrium, 
and that the force E is also balanced. It therefore follows that the forces 
Pj, P 2 , P 3 and E are in equilibrium. 

Reference to Fig. 89 (b) will show that abed constitutes a closed polygon 
of forces for P lf P 2 , P 3 and E, and that the lines drawn from O to a, 
, c and */are parallel respectively to the links in Fig. Sg(a). As we 
had a liberty of choice of the directions of the first two links, viz. DA 
and AB, and as these directions, once chosen, settled the position of 



THE LINK POLYGON 



the point O in Fig. 89 (/;), we infer that the position of O is immaterial, 
the only effect of varying its position being to change somewhat the 
shape of the link polygon without altering the final value or position 
of E. It should also be noted that each link in Fig. 89 (a) is parallel 
to the line from O in Fig. 89 (b) which falls between the sides of the 
force polygon representing the two forces connected by the link in 
Fig. 89 (a). Thus, AB in Fig. 89 (a) is parallel to Ob in Fig. 89 (b\ the 
latter falling between ab and be which represent P l and P 2 respectively. 
In practice, Bow's notation is employed. Some examples are given 
to illustrate the method. 

EXAMPLE i. Given three forces of 3 tons, 4 tons and 2 tons respectively, 
find their equilibrant (Fig. 9o(fl)). 

The principles on which the solution is based are, as has been found 
above, (a) the force polygon must close, (b) the link polygon must close. 




E-3 25 tons 

(a) c < "i 

FIG. 90. An application of the link polygon. 

Naming the spaces A, B and C, and placing D provisionally near to the 
force of 2 tons, draw the force polygon ABC D (Fig. 90 (<)). The closing 
line DA gives the direction, sense and magnitude of the equilibrant. To 
find its proper position, take any pole O (Fig. 90 ()), and join O to the 
corners A, B, C, D of the force polygon. Choose any point a on the line 
of the 3 tons force. In space A draw a line ad, of indefinite length, 
parallel to OA in Fig. 90 (b). In space B draw a line ab parallel to OB ; 
and, in space C a line be parallel to OC. From c draw a line parallel to 
OD to intersect that drawn from a in the point d. Then E passes through 
d, and may now be shown completely in Fig. 90 (a). 

EXAMPLE 2. Four forces are given in Fig. 91 (a); find their resultant. 

The method employed consists in first finding the equilibrant and then 
reversing its sense. This example is of slightly greater :omplication, but 
the working does not differ from that illustrated above. In Fig. 91 (b) 
ABCDE is the force polygon, the closing side EA represents the equili- 
brant, hence AE represents the resultant. The position of the equilibrant 
is found by drawing the link polygon abcde, having its sides parallel to 



MATERIALS AND STRUCTURES 



the lines radiating from any pole O in Fig. 91 ( 
and de gives a point e on the line of action of R. 



The intersection of ae 




FIG. 91. The resultant determined by the link polygon. 

EXAMPLE 3. Given a beam carrying loads as shown (Fig. 92 (rt)); find 
the reactions of the supports. 

In this case, as all the forces are parallel, the force polygon becomes a 
straight line. The reactions AB and GA being unknown, begin in space 
B and draw the sides of the force polygon as BC, CD, DE, EF and FG. 
The corner A of the force polygon will fall on BG, and, its position having 
been determined, the segments GA and AB will give the magnitudes 
of the reactions. Choose any pole O and join it to the known corners 
of the force polygon, viz. B, C, D, E, F, G. Start constructing the polygon 
from a point a on the line of the left-hand reaction (Fig. 92(0:)) by 



\ c I I M 





(b) 



FIG. 92. Reactions of a beam by the link polygon method. 

drawing ab parallel to OB m the space lying between the reaction AB and 
the force BC. Then draw be, od, de, ef respectively parallel to OC, OD, 
OE, OF. Fromyj a point on the force FG, a line fg has to be drawn to 
intersect the reaction GA ; as these forces are in the same straight line, it 
is clear that fg is of zero length, and that the link polygon will conse- 
quently have a side short. Complete the link polygon by drawing^, and 
draw OA (Fig. 92^)) parallel to fa. The magnitudes of the reactions 
may now be scaled as AB and GA. 



EXPERIMENTS ON COUPLES 



EXPT. io, Equilibrium of two equal opposing couples. In Fig. 93 is 
shown a rod AB hung by a string attached at A and also to a fixed 
support at C. By means of cords, pulleys and weights, apply two 
equal, opposite and parallel forces P, P, and also another pair Q, Q. 
Adjust the values so that the following equation is satisfied : 

PxDE = Q-xFG. 

Note that the rod remains at rest under the action of these forces. 

Repeat the experiment, inclining the parallel forces P, P, at any 
angle to the horizontal, and inclining the parallel forces Q, Q, to a 
different angle, but arranging that the moment of the P, P, couple is 
equal to that of the Q, Q, couple. Note whether the rod is balanced 
under the action of these couples. 

Apply the P, P, couple only, and ascertain by actual trial whether 
it is possible to balance the rod in its vertical position as shown in 
the figure by application of any single force. 



FIG. 93. An experiment on couples. 




Plan of cord at B 

FIG. 94. Couples acting on a door. 



EXPT. ii. Couples acting on a door. Fig. 94 shows a board which 
may be taken as a model of a door hung on two hinges. The 
equal forces W and P form a couple, which is balanced by the equal 
opposing couple Q, Q. Weigh the board, measure a and b, and 
calculate Q from 



Apply the forces as shown and note whether the door is in 
equilibrium. 

EXPT. 12. Link polygon. Fig. 95 (a) shows a polygon ABCDEA 
made of light cord and having forces P, Q, S, T and V applied as 



MATERIALS AND STRUCTURES 



shown. Let the arrangement come to rest. Show by actual draw- 
ing (a) that the force polygon abcdea closes (Fig. 95 (<)), its sides being 




(b) 



FIG. 95. An experimental link polygon. 

drawn parallel and proportional to P, Q, S, T and V respectively ; 
(b] that lines drawn from 0, b, c, d and e parallel respectively to AB, 
BC, CD, DE and EA intersect in a common pole O. 

EXPT. 13. Hanging cord. A light cord has small rings at A, B, C 
and U and may be passed over pulleys E and F attached to a wall 
board (Fig. 96(0)). Weights W T , W 2 , W 3 and W 4 may be attached to 

a 

w, 

6 





FIG. 96. A hanging cord. 

the rings, and P and Q to the ends of the cord. Choose any values 
for W 1? W 2 , W 3 and W 4 and draw the force polygon for them as 
shown at abcde. Choose any suitable pole O, and join O to a, b, c, d 



HANGING CORDS AND CHAINS 



73 



and e. Oa and Oe will give the magnitudes of P and Q respectively. 
Fix the ring at A to the board by means of a bradawl or pin \ fix the 
pulley at E so that the direction of the cord AE is parallel to Oa ; 
fix the ring at B by means of a pin so that the direction of the cord 
AB is parallel to CM. Fix also the other rings C and D, and the 
pulley at F so that the directions of BC, CD and DF are parallel to 
<rO, dO and eO respectively. Apply the selected weights W 1? W 2 , W 3 
and W 4 , and also weights P and Q of magnitude given by Oa and 
Oe. Remove the bradawls and ascertain if the cord remains in 
equilibrium. 

EXPT. 14. Hanging chain. Fig. 97 (a) shows a short chain ACB in 
equilibrium under the action of forces V T , V 2 , H T and H 2 applied by 
means of cords, pulleys and weights. Find these forces by calcula- 
tion, as indicated below, first weighing the chain, and apply them as 
shown in the figure so as to test for the equilibrium of the chain. 




FIG. 97. Equilibrium of a hanging chain. 

Let D = the proposed dip or deflection of the chain in inches. 
S = the span AB in inches. 

It should be noted that D should not be too large when compared 
with S. Both may be measured conveniently by first stretching the 
chain between the two marked positions A and B on the wall board 
and then taking the required dimensions. It is assumed that A and 
B are on the same level. I 

Imagine the chain to be cut at its centre C, and consider the 
equilibrium of the right-hand half (Fig. 97 (b)). The weight of the 
whole chain being W lb., the weight of the half considered will be 

^W and will act at the centre of gravity G, which may be assumed 

g 

to be at - horizontally from B provided D is not too large. As a 

chain can only pull, the force H at C must be horizontal. Hence 
the portion BC is at rest under the action of two equal opposing 



74 



MATERIALS AND STRUCTURES 



couples, one formed by the equal forces V 2 and |W and the other 
by the equal forces H and H 2 . Hence 

V iw (j\ 

v 2 2 \ / 

H xD = JWx, 

4' 

WS 



and 



or 



00 



EXERCISES ON CHAPTER IV. 

1. A wooden gate weighs 100 lb., and has its centre of gravity situated 
21 inches from the vertical axis of the hinges. The hinges are 24 inches 
apart vertically, and the vertical reaction required to balance the gate is 
shared equally between them. Calculate the magnitude and direction of 
the reaction of each hinge and show both reactions in a diagram. 

2. A square plate of 2 feet edge has forces of 2, 3, 4 and 5 lb. applied 
as shown (Fig. 98). Find the force required in order to balance the 
plate. 

3. A plate having the shape of an equilateral triangle of 3 feet edge 
has forces of i, 2 and 3 lb. applied as shown (Fig. 99). Find the resultant 
force on the plate. 

,2 




4 3 

FIG. 98. FIG. 99. 

4. Suppose the plate in Question 3 to have equal forces of 2 lb. each 
applied along the edges in the same manner as before. What must be 
done in order to keep the plate in equilibrium ? 

5. A uniform beam 12 feet span and 18 inches deep weighs 900 lb. 

A load of 2 tons is applied to the top surface 
at 3 feet from the right-hand support at an 
angle of 45 to the horizontal (Fig. 100). 
Suppose the left-hand reaction to be vertical, 
and calculate the reactions of the supports. 

6. A beam AB rests against walls at A 
and B (Fig. 101). Vertical loads of 400 lb. 
and 600 lb. trisect the beam. Suppose the 

reaction at A to be horizontal, and calculate the reactions at A and B. 

Neglect the weight of the beam. 



FIG. 100. 



EXERCISES ON CHAPTER IV. 



75 



7. A triangular frame 15 feet span and 5 feet high (Fig. 102) carries 
loads of 400 Ib. bisecting AC, 600 Ib. at C and 800 Ib. bisecting BC at 
right angles. The reaction at B is vertical. Find the reactions of the 
supports by calculation. 




FIG. loi. 

8. Prove that two couples of equal opposing moment, acting in the 
same plane, balance. 

9. Show how a force acting at a given point may be moved to another 
point not in the original line of the force. Prove the method to be 
correct. 

10. Choose any three forces not meeting at a point and not parallel to 
one another. Show how we can find, graphically, their resultant or their 
equilibrant. (B.E.) 

11. Answer Question 10 in a manner suitable for calculation. 



600/4 




f 



FIG. 102. 



12. A number of forces act in a plane and do not meet in a point. 
Treating them graphically, what is the condition of equilibrium ? Prove 
your statement to be correct. (You are expected to choose more than 
three forces.) (B.E.) 

13. A uniform chain weighs 4 Ib., and is hung from two points on the 
same level. The span is 4 feet and the central dip is 6 inches. Calculate 
the pulls at the ends of the chain, and show the directions of the chain at 
the ends. 



76 MATERIALS AND STRUCTURES 

14. A beam AB of 24 feet span is supported at the ends, and carries 
vertical loads of 1-5, 2, 3 and 4-5 tons at distances of 3, 6, 12 and 18 feet 
from the support at A. Use the link polygon method and find the reactions 
of the supports. 

15. Answer Question 6 by construction. 

16. Answer Question 7 by construction. 

17. ABCD is a square of 2-inch side, BD being a diagonal. A force 
of 50 Ib. acts along BC from B towards C ; a force of 80 Ib. acts along CD 
from C towards D ; and a force of 6p Ib. acts along DB from D towards 
B. Replace these forces by two equivalent forces, one of which acts at A 
along the line AD. Find the magnitude of both these forces and the line 
and direction of the second. (I.C.E.) 

18. Prove that any system of coplanar forces may be replaced by a 
single force acting at any assigned point and a couple. Forces of I, 2, 3, 
4 Ib. weight act along the sides of a square taken in order. Find a point 
such that the forces may be replaced by a single force acting at that 
point. (L.U.) 



CHAPTER V. 

SIMPLE STRUCTURES. 

Some definitions. A structure is an arrangement of various parts 
constructed in such a manner that no relative motion (other than the 
small amounts due to the straining of the parts) takes place when the 
structure is loaded. The simple framed structures considered in this 
chapter consist of bars assumed to be connected by pin joints and 





(a) (b) (c) 

FIG. 103. Classes of structures : (a) deficient, () simply firm, (c) redundant. 

carrying loads applied at these joints. The bars under these conditions 
will be subjected to simple push or pull in the direction of their 
lengths, and our object will be to determine the magnitude of the 
force in each bar, and also whether the bar is under push or pull. 

Structures may be deficient, simply firm, or redundant. Deficient 
structures are really mechanisms, that is, the parts are capable of 
considerable relative motion. Fig. 103 (a) shows an example of a 
deficient structure, consisting of four bars connected by pin joints. 
The arrangement may be made simply firm by the introduction of a 
single diagonal bar (Fig 103 (^)), and will now be capable of preserving 
its shape under the load. The introduction .of a second diagonal bar 
(Fig. 1 03 (r) ) produces a redundant structure. In redundant structures, 
the length of any bar cannot be altered without either a correspond- 
ing alteration in the lengths of other bars of the structure, or the 
production of forces in the other b^rs. Good workmanship is 
essential in redundant structures to ensure the accurate fitting 
together of all parts, otheiwise some of the bars may require to be 



MATERIALS AND STRUCTURES 



(a) (b) 

FIG. 104. Effect of stiff joints. 



IS) 



forced into position. Unequal heating causes unequal expansion 
in redundant structures, and therefore introduces forces in the 
various parts. 

In simply firm structures, which form the subject of this chapter, 
the length of any part may be altered without thereby producing 

forces in the other parts. Con- 
sequently, the effects of unequal 
expansion are absent. A redundant 
structure may be converted into a 
simply firm structure by dropping 
out one or more of the redundant 
elements, or parts. Redundancy may 
be produced by stiff joints. For 

example, if the square in Fig. 103(0) i made with one stiff joint 
(Fig. 104(0)), the structure will now be simply firm. Two stiff 
joints (Fig. 104^)) will produce a redundant structure having one 
redundant element ; three and four stiff joints in this example give 
structures of two and three elements of redundancy respectively. 

Conditions of equilibrium. In solving problems concerning any 
structure, we may separate the forces ino two groups, external and 
internal. The external forces include all forces applied as loads, or 
reactions, to the structure. Obviously these forces, acting on the 
structure as a whole, must be in equilibrium independently of the shape 
of the structure, or of the form or arrangement of its parts. This con- 
sideration enables us to apply the principles of the foregoing chapters to 
such problems as the determination of the reactions of the supports. 

The internal forces include the pushes, or pulls, to which the various 
bars are subjected when the external forces are applied to the 
structure. Not only is the structure as a whole in equilibrium, but 
any bar, or any combination of selected bars in it, must be in 
equilibrium under the action of any external loads applied to the 
parts considered, together with the internal forces acting in the selected 
parts. Usually a joint is selected, when the principle just stated 
enables us to say that the forces acting at this joint, including external 
forces, if any, as well as the pushes or pulls of the bars meeting at 
the joint, are in equilibrium. Hence, the forces in these bars may 
be found by an application of the polygon of forces. 

It should be remembered in applying the polygon of forces that 
the solution depends on there being not more than two unknowns ; 
these may be either the magnitudes or the directions of two forces, or 
one magnitude and one direction. In cases where the forces do not 



SIMPLE STRUCTURES 



79 



all intersect at one point, there are three conditions of equilibrium to 
satisfy, and hence there may be three unknowns. 

The methods of obtaining the reactions have been explained fully in 
the preceding chapters ; hence in some of the following cases the con- 
structions, or calculations, for finding the reactions have been omitted. 

Simple roof truss. Fig. 105 (a) shows a simple roof truss consist- 
ing of five bars. There are three loads applied as shown, together 
1000/6 




(at 



FIG. 105. A roof truss for small spans. 

TABLE OF FORCES. 






Force in Ib. 




Force in Ib. 


Name of part. 




Name of part. 














Push. 


Pull. 




Push. 


Pull. 


Reaction AB 


1000 





AF 





1350 


Reaction EA 


IOOO 





GA 





1350 


CF 


1550 





FG 





575 


DG 


1550 












with two vertical reactions. To enable the forces to be named, 
letters are placed as shown for the application of Bow's notation. 
Thus, the left-hand reaction may be described as AB or BA, and the 
force in the vertical centre bar may be described as FG or GF, 
depending on the sense of rotation selected. 

As all the external forces are vertical, the polygon of forces for the 
equilibrium of the truss as a whole will be a straight line. In drawing 
it, we may proceed round the truss either clockwise or anti-clockwise ; 
but, once having settled on the direction, it should be preserved 
throughout the whole work of solution. Choosing a clockwise direc- 
tion, the straight line ABCDEA (Fig. 105^)) will be the polygon 
for the external forces. 



8o MATERIALS AND STRUCTURES 

Selecting the joint at the left-hand support, there are four forces, two 
of which are completely known, and other two of which the directions 
alone are known, viz. the forces CF and FA. Hence the polygon of 
forces can be drawn. In Fig. 105 ()), proceeding clockwise round 
the joint, AB and BC have been already drawn ; draw CF parallel to 
the rafter and AF parallel to the tie-bar ; these lines intersect in F and 
give the closed polygon of forces ABCFA. The force in the rafter may 
be scaled from CF and that in the tie-bar from FA. Taking these 
lines in order in relation to the joint under consideration, the sense of 
the force in the rafter in Fig. 105 (a) is CF in Fig. 105 (b), and hence 
is a push ; that in the tie-bar has a sense FA, and hence is a pull. 

Proceeding now to the top joint of the truss, we see that there are 
two unknowns, viz. the magnitudes of the forces in GF and DG, 
hence this joint may be solved by drawing the polygon of forces 
FCDGF(Fig. io5(J)). 

Taking now the joint at the right-hand support, and drawing the 
polygon of forces GDEAG, we find that the closing line AG has its 
position fixed already on the diagram. This fact provides a check 
on the accuracy of the whole of the preceding graphical work ; if on 
joining AG in Fig. 105 (<), it is found that this line is not parallel 
to the right-hand rafter, some error has occurred, and in order to 
eliminate it the work must be repeated. 

Rule for push or pull. The method of determining whether a bar 
is under push or pull may be simplified somewhat by developing the 
following rule from the principle explained above. 

Select any bar such as FG ; choose the joint at one end of it, say 
the lower ; cross the bar in the same sense of rotation in relation to 
this joint as was chosen in drawing the force diagrams in this case 
clockwise ; name the spaces in this order, viz. FG. FG in Fig. 105 (b) 
gives the sense of the force acting at the lower joint. As the force is 
upwards, the bar is pulling. 

It makes no difference in the application of the rule which end of 
the bar is selected. For example, choosing the top joint of the same 
bar and crossing it again clockwise as regards the upper end, the 
order is GF. GF in Fig. 105 (b) is downwards, hence the bar is 
pulling at the top joint. 

It is desirable to indicate on the drawing of the truss which bars 
are under push and which under pull. Probably the best way of 
doing this is to thicken the lines of the bars under push. If the whole 
line is thickened, the direction of the bar will be lost, hence, as shown 
in Fig. 105 (), a short piece at each end is left thin. 



SIMPLE STRUCTURES 



8 1 



A tabular statement of the forces in the bars should be made in 
the manner indicated on p. 79. 

Another form of roof truss. Fig. io6(a) shows a common type of 
roof truss carrying symmetrical loads. There will be no difficulty in 

1000 Lb 



500/6 



1000/6 




FIG. 106. Forces in a common type of roof truss ; weights only considered. 

TABLE OF FORCES. 



Name of part. 


Force in Ib. 


Name of part. 


Force in Ib. 


Push. 


Pull. 


Push. 


Pull. 


Reaction AB 


2000 





AH 





4225 


Reaction GA 


2000 





AL 





2480 


CH 


4675 





AN 





4225 


DK 


4250 





HK 


900 





EM 


4250 





KL 





1960 


FN 


4675 





LM 





1960 








MN 


900 






D.M. 



82 



MATERIALS AND STRUCTURES 



following the diagram of forces (Fig. 1 06 (/;)). The order in which 
the joints have been taken is indicated by the number placed against 
the joint. The sense of rotation employed is clockwise, and the 
closing check line is NA. 

The effect of wind pressure on the right-hand side of this truss is 
determined in Fig. 107 (a) and (6). It is assumed that the wind load 



500 to 



iqooto 




ft) 



FIG. 107. Wind acting on the right-hand side of the truss. 

TABLE OF FORCES. 



Name of part. 


Force in Ib. 


Name of part. 


Force in Ib. 


Push. 


Pull. 


Push. 


Pull. 


Reaction AB 


5 60 





AH 





1550 


Reaction GA,1 
inclined / 


1510 





AL 

AN 





1360 
3160 


BH 


1700 





HK 








BK 


1700 





KL 





260 


EM 


2770 





LM 





IQIO 


FN 


2770 





MN 


IOOO 






produces forces of 500 Ib. at the top and bottom ends of the rafter, 
and of TOGO Ib. at the middle, all three being perpendicular to the 



SIMPLE STRUCTURES 83 

rafter. As an example of the use of the link polygon, the reactions 
of the supports have been determined by this method. The left-hand 
reaction has been assumed to be vertical when that of the right-hand 
support will be inclined. Wind pressure only has been taken account 
of in the working. It will be noted that there are three unknown 
elements in the reactions, viz. the magnitude of the left-hand reaction 
and both the magnitude and direction of the right-hand reaction. In 
fact, all that is known of the latter reaction is that it acts through the 
point a. Now, in drawing the link polygon, one link must fall between 
this reaction and the force FG. As the line of the reaction is unknown, 
it will be impossible to draw this link unless the artifice is adopted 
of starting the drawing of the link polygon at the point a. The effect 
of this will be that the link in question will have zero length. 

First draw as much of the external force polygon as possible ; this 
is shown by BEFG in the force diagram. A will lie in the vertical 
through B as the reaction AB is vertical. Taking a convenient pole 
O and joining OB, OE, OF and OG, we start drawing the link poly- 
gon by making ab (Fig. 107 (<z)), which falls between FG and EF, 
parallel to OF. be falls between EF and BE, and is made parallel to 
OE. cd falls between BE and AB, and is made parallel to OB. 
The link parallel to OG is omitted, as it is of zero length, coinciding 
with a. Hence the closing line is da ; drawing OA parallel to ad to 
intersect the vertical through B in A gives the left-hand reaction as 
AB and the right-hand reaction as GA. 

The remainder of the diagram giving the internal forces is worked 
out in the usual manner, NA being the closing line. 

An application of the method of graphical moments. The effect 
of wind pressure on the left-hand side of this truss is determined in 
Fig. 1 08. The student will have noted, in applying the link polygon 
to the problem of finding the reactions, that the lines of the polygon 
have a tendency to obscure the drawing of the truss. In the case 
now before us, the method of graphical moments (p. 46) is employed 
and involves the drawing of very few lines on the truss. The resultant 
of the three wind loads has been taken as a single force of 2000 Ib. 
applied at c. Join ab, and with centre b and radius be describe an arc 
cutting ab in d. Make ae equal to 2000 Ib. to scale ; join be and 
draw df perpendicular to ab and cutting be inf. Draw fg parallel to 
ab y when ga will be the vertical component of the right-hand reaction. 
The horizontal component of this reaction will be equal and opposite 
to the horizontal component of the force of 2000 Ib. acting at c. 
Draw the triangle of forces dm, and make ha equal to me; the 



8 4 



MATERIALS AND STRUCTURES 



right-hand reaction will be the resultant ka of the components 
represented by ga and ha. 

The external force polygon (Fig. io8(^)) may now be drawn 
I 



500/6 




(b) 



FIG. 108. Wind acting on the left-hand side of the truss. 

TABLE OF FORCES. 



Name of part. 


Force in Ib. 


Name of part. 


Force in Ib. 


Push. 


Pull. 


Push. 


Pull. 


Reaction AB 


1200 





AH 





1870 


Reaction EA,\ 






AL 





320 


inclined / 


1020 




AN 





400 


CH 


2300 





HK 


IOOO 





DK 


2300 





KL 





1550 


EM 


1400 





LM 





100 


EN 


1400 





MN 









SIMPLE STRUCTURES 




FIG. 109. Combined dead and wind loads on the truss. 

TABLE OF FORCES. 



Name of part. 


Force in Ib. 


Name of part. 


Force in Ib. 


Push. 


Pull. 


Push. 


Pull. 


Reaction AB 


2560 





AH 





5830 


Reaction GA, } 
inclined / 


3350 





AL 

AN 





3900 
7450 


CH 


6400 





HK 


900 





DK 


5950 





KL 





2100 


EM 


7100 





LM 





3840 


FN 


7500 





MN 


IQOO 






.86 MATERIALS AND STRUCTURES 

for the three wind loads and the two reactions, and is shown by 
ABCDEA. The internal force diagram is completed as before. 
Notice that when the wind blows on the right-hand side, no force is 
induced thereby in HK, and when acting on the left-hand side there 
is no force in MN. This arises from the fact that there is no external 
load at the joint in the two cases respectively. Two forces acting in 
the same straight line, as is the case in the two parts of the rafter, 
balance, and it is impossible to apply a single inclined force at the 
point of action without disturbing the equilibrium. 

The total force in any bar of the frame due to the dead loads, 
i.e. the weights of the parts of the truss, and to the wind pressure 
jointly may now be determined by adding the results for the dead 
load (Fig. 1 06) and either those of Fig. 107 or of Fig. 108 depending 
on whether the wind is blowing on the right- or left-hand side. 

Combined dead load and wind pressure. As a further example of 
another method of obtaining the reactions, a diagram has been drawn 
in Fig. 109 for the combined dead loads and wind load on the right- 
hand side. The two forces acting at each joint of the right hand 
rafter have been combined by the parallelogram of forces, and the 
resultant used as a single force at each joint (Fig. 109 (a)). 

To find the reactions of the supports, we may take advantage of 
the principle that the external forces balance independently of the 
arrangement of the parts of the truss. Hence, any other convenient 
arrangement may be substituted for that given without disturbing the 
values of the reactions. The substituted frame chosen is sketched 
in Fig. 109 (b}. It will be seen that it is possible to determine all the 
forces in its parts without first determining the reactions. Thus, 
starting at the top joint i, where there are two unknowns only, we 
obtain DE/fc in the force diagram (Fig. 109^)). Proceeding to 
joint 2, we obtain EF/# ; at joint 3, CD/i/m is obtained, and at joint 
4 we obtain BOzA, thus determining the point A on the force 
polygon, and hence the reactions AB and GA. 

The internal force diagrams for the given arrangement of bars may 
now be proceeded with, the closing line being NA. It will be 
observed that the greater part of the lines drawn in the force diagram 
for the substituted frame are required for the actual frame, hence 
there has been but little wasted work. 

Another form of structure. In Fig. 110(0) is shown a structure 
intended to carry a load at its upper end. Since there is but one 
vertical load, the reactions of the foundation must reduce to one 
vertical upward force equal to the load applied. Hence, the polygon 



SIMPLE STRUCTURES 



for the external forces is completed by drawing AB downwards and 
BA upwards. It will be noted in this example that it is not necessary 
to determine the actual reactions of the foundations before finding 
the forces in the parts. A start can be made at the joint i, as there 
are only two unknowns there. The order of solution of the other 
joints is indicated by the numerals. In drawing the various polygons 

2 




FIG. no. A braced frame. 



TABLE OF FORCES. 



Name of part. 


Force in Ib. 


Name of part. 


Force in Ib. 


Push. 


Pull. 


Push. 


Pull. 


AC 


3540 





BC 





3400 


AE 


3960 





BD 





3350 


AG 


4050 





BF 





2850 


AK 


3440 





BH 





2440 


CD 


1420 





DE 





120 


EF 


1550 





FG 





450 


GH 


2220 





HK 









(Fig. 1 10 ()), anti-clockwise sense of rotation has been chosen. The 
student will observe that there is no force in HK, H and K coinciding 
in the force diagram. It is easy to see that this must be the case 
from consideration of the fact that the bars BH and AK are vertical, 
and therefore the vertical forces in them are capable of balancing the 
external load applied without any aid from the diagonal HK. In 
fact, the diagonal HK merely serves to steady the frame under the 



88 



MATERIALS AND STRUCTURES 



given loading. There would, of course, be a force in this bar if an 
inclined load were applied to the frame, or if there were a side effort 
caused by wind pressure. 

A larger roof truss. In Fig. 1 1 1 is shown a roof truss of a larger 
type and having a different arrangement of parts from those dealt 




FIG. in. A larger roof truss. 



with previously. This case presents no difficulties, and is included 
as an example which the student can work out for himself. 

The roof truss shown in Fig. 112(0} presents a difficulty which 
arises frequently. The external force polygon is drawn easily, but 





(b) 
iraoo 



FIG. ii2. A more difficult example of truss. 

in drawing the force polygon for the internal forces it will be found 
that it is impossible to proceed with the drawing after solving point i. 
All other points such as 2 and 3 have more than two unknowns, 
hence the solution cannot be obtained by application of the ordinary 
methods. We may proceed by either of two methods. 



EXERCISES ON CHAPTER V. 89 

(a) It will make no difference whatever in the forces in the remaining 
part of the truss if we imagine the left-hand portion (shown shaded in 
Fig. 112 (<$)) to be solid. Separating this portion as shown, we may 
calculate T, the force in the bar AP, by taking moments about point 6. 
Thus, taking the loads as shown, let the half-span be 15 feet and let 
the perpendicular from point 6 to the line of T be 7-5 feet, then 

(Tx7-5) + (4oox 5) + (4oox io) + (2oox I5)=i2oox 15. 
_ 1 8,000 - 2000 - 4000 - 3000 

7-5 
_9ooo 

~~r$ 

= 1200 lb. 

Having found T, the number of unknowns at the point 3 
(Fig. 112 (a) ), will now be found to be two only, hence this point 
may be solved. The solution for points 2, 5, 4, 6 may now be 
obtained in the usual manner. 

(b) It will make no difference whatever in the force in the bar AP 
if, instead of imagining the triangular portion above considered to be 
solid, we imagine it to have a different interior arrangement of bars. 
Thus, in Fig. 1 1 2 (c} is shown this portion with a new arrangement 
of bars substituted for that given. The force diagram for this 
substituted frame may be drawn as in Fig. 112 (d), and stopped 
directly the force in AP is found. The original arrangement of bars 
is now restored and the force diagram completed in the usual 
manner. The result for the force in AP is found graphically in 
Fig. 112 (d) to be 1200 lb. 

This method must be applied with caution. Care must be taken 
to ensure that the substituted arrangement of bars does nothing 
whatever to alter the force in the bar considered, viz. AP in 
Fig. 112(0). 

EXERCISES ON CHAPTER V. 

In each of these exercises the forces should be tabulated, distinguishing 
carefully push and pull members. 

' JOO/f 



400/6. 

E 



FIG. 113. 

1. Find the forces in all the bars of the roof truss shown in Fig. 113. 
The bars AF, FG, GH and HA are equal. 




MATERIALS AND STRUCTURES 



2. Find the forces in all the bars of the truss given in Fig. 1 14. The 
loads are in Ib. units. 

3. Find the forces in the roof truss shown in Fig. 106 ; apply the 
same loads, with the exception of that at the centre of the right-hand 
rafter, which in this case is 2000 Ib. Span 24 feet, rise 6 feet, rise of tie- 
bar i foot. Each rafter is bisected perpendicularly by the inclined strut. 



800 



800 



800 



600 



300 




4. Find the forces in all the members of the truss shown in Fig. 1 1 5. 
The loads are in Ib. units. 

5. Take again the roof truss given in Question i. Remove all the 
loads and apply wind loads of 400 Ib. at each end of the right-hand rafter, 
acting at right angles to the rafter. Find the forces in all the parts due 
to wind only. The left-hand reaction is vertical. 

6. Answer Question 5, supposing that the wind loads are applied to 
the left-hand rafter only. The left-hand reaction is vertical. 

7. From the results obtained in answering Questions i, 5 and 6, con- 
struct a table showing the maximum and minimum forces in each bar due 
to dead load and wind pressure combined. 

8. Find the forces in all the bars of the roof truss given in Fig. 116. 
The loads are in Ib. units. 

600 



600 



600 




9. A roof truss similar to Fig. 1 1 1 has a span of 30 feet ; the rise is 
8 feet and the height of the central horizontal part of the tie bar above 
the supports is 18 inches. If the truss carries a symmetrically distributed 
load of 4 tons, find the force in AO by calculation. 

10. In the roof truss given in Question 8, in addition to the stated 
loads, there are wind loads of 400, 800, 800 and 400 Ib. applied at the 
joints of the right-hand rafter and perpendicular to the rafter. The left- 
hand reaction is vertical. Find the reactions of the supports, using the 
link polygon. 



EXERCISES ON CHAPTER V. 



11. Answer Question 10 by application of the substituted frame 
method. 

12. Answer Question 10 by calculation. 

13. In Question 10 find the forces in all the parts due to combined dead 
load and wind pressure. 

14. A loaded Warren girder is shown in Fig. 117. Find the forces in 
all the members. The loads are in Ib. units. 



1000 



1000 



500 



500 




FIG. 117. 

15. A frame secured to a vertical wall has dimensions as shown in 
Fig. 118. The bars AD, AF, AH and AL are each 5 feet in length. 
Find the forces in all the parts produced bv the load of I ton. 



\t on 



2*--^ 20 




FIG. 118. 



16. Answer Question 15 if the load is moved horizontally so as to be 
vertically over the middle joint of the top member of the frame. 

17. Part of a pin-jointed frame, shown in Fig. 119, is loaded with a 
vertical dead load of 10,000 pounds and a normal wind pressure of 15,000 




MATERIALS AND STRUCTURES 



pounds, both being taken as uniformly distributed along AB. The sup- 
porting forces P, Q and R are shown by dotted lines. Find these forces 
and the forces in the bars which meet at C, indicating the struts 
and ties. (L.U.) 



2 tons 




18. A frame is loaded with 2 tons and supported as shown in Fig. 120. 
Find the reactions at A and D and the forces in the members, indicating 
which are struts and which are ties. (I.C.E.) 



CHAPTER VI. 

SIMPLE STRESSES AND STRAINS. 

Stress. If any section in a loaded body be taken, it will be found 
in general that the part of the body which lies on one side of the 
section is communicating forces across the section to the other part, 
and is itself experiencing equal opposite forces. The name stress is 
given to these mutual actions. The stress is described as tensile or 
pull if the effect is to pull the portions of the body apart, compressive 
or push if they are being pushed together, and shearing or tangential if 
the tendency is to cause one portion of the body to slide on the 
other portion. 

The stress is said to be distributed uniformly in cases where all 
small equal areas experience equal loads. Stress is measured by 
stating the force per unit area, the result being described as unitai 
stress, or stress intensity, or often simply as the stress. In the case of 
a uniform distribution of stress, the stress intensity will be found by 
dividing the total force by the area over which it is distributed. 
Should the stress vary from point to point, its intensity at any point 
may be stated by considering that the forces acting on a very small 
area embracing that point will show a very small variation and may 
be taken as uniformly distributed. Thus, if a be a very small portion 
of the area and/ the load on it, the stress intensity on a will be//. 

Units of stress employed in practice are pounds or tons per square 
inch or per square foot, or in the metric system, grams or kilograms 
per square centimetre. One atmosphere is sometimes used as a unit, 
being a stress of 14-7 Ib. per square inch ; it is useful to remember 
that a stress of one kilogram per square centimetre is roughly equal 
to one atmosphere.* 

* i kilogram per square centimetre = 2-205 Mb. P er -r square inch 

0-45 

= 14-19 Ib. per square inch. 



94 MATERIALS AND STRUCTURES 

EXAMPLE i. A bar of circular cross-section 2 inches in diameter is 
pulled with a force of 12 tons at each end. Find the tensile stress. 

Area of cross-section = = 3-1416 sq. inches. 

Tensile stress intensity = 

area 

12 



3-1416 
= 3-82 tons per square inch. 

EXAMPLE 2. Suppose the same bar to be in two portions connected 
by means of a knuckle joint having a pin i| inches in diameter (Fig. 121), 
and calculate the intensity of shearing stress on the pin. 



12 tons 




FIG. 121. 

It will be observed that the pin would have to shear at two sections 
for the joint to fracture by failure of the pin, hence : 

7ZY/ 2 

Area under shear stress = x 2 

4 



= 3-53 square inches. 



Shear stress intensity = -- 
area 

12 

~^S3 

= 3-39 tons per square inch. 

Stresses in shells. A shell is a vessel constructed of plates the 
thickness of which is small compared with the overall dimensions of 
the vessel, for example, a boiler of the cylindrical type. Such 
vessels have generally to withstand internal fluid pressure, and the 
plates are put under tensile stress thereby. Owing to the thinness 
of the plates, the stress on any section may be considered to be 
distributed uniformly. 

Taking a cylindrical shell (Fig. 122) in which there are no stays 
passing from end to end, 



STRESSES IN SHELLS 



95 



Let d= diameter of shell, inches, 

/ = fluid pressure, pounds per square inch, 
t = thickness of plate, inches, 
P = total pressure on each end of vessel, then 






> t *- 

* 
-< 
d 

i 


pi 

*- ^4- 




r 


a ^ 
J! 


w 
1 


1 





Section at AB ^Longitudinal Sect/oft 

FIG. 122. Stresses in a cylindrical shell. 

Owing to the forces P, P, any section such as AB will be under 
tensile stress. 

Sectional area at AB = circumference of shell x / 



Tensile stress intensity on AB = ; 

P*' 



= f- lb. per square inch. 

4/ 

The stress on a longitudinal section may be found in the following 
manner. Consider a ring cut from the shell by two cross-sections 
one inch apart (Fig. 123). It may be assumed 
that all other such rings will be under similar 
conditions, provided they are not taken too 
near to the ends of the shells where the staying 
action of the ends would interfere. The fluid 
pressure on the ring is shown by arrows in 
Fig. 123, everywhere directed perpendicular to 
the curved surface of the ring, i.e. radial. Com- 
ponents of these being taken, parallel and FIG. 123. A ring cut from 
perpendicular to a diameter AB, it will be seen 
that those parallel to AB equilibrate independently of the others. 
The upward and downward components perpendicular to AB will 
have resultants R T and R 2 respectively, which will have the effect 
of producing tensile stress on the sections at A and B. Clearly Rj 
and R 2 will be equal ; to obtain their magnitudes proceed thus : 




MATERIALS AND STRUCTURES 



There will be no difference experienced in the equilibrium of the 
ring if we imagine it to be filled up to the level of AB with cement 
(Fig. 124). The pressure on the surface of the cement will be 
perpendicular to AB, and the resultant force due to this will be 
Q =p x area of surface of AB 





FIG. 124. Resultant pressure on hah 
of the ring. 



FIG. 125. Stresses at A and B. 



R and Q now preserve the equilibrium of the ring, and must 
therefore be equal, hence R =pd. 

Imagine the material at A and B to be cut, and consider the equi- 
librium of the top half of the ring (Fig. 125). Forces T, T at A and 
B will be required, and are produced in the uncut shell by tensile 

stress at A and B. For equilibrium, we 

have R = 2 T, 

T = - = ^. 

2 2 

Also, 

Stress intensity at A or B x t x i = T ; 
.'. stress intensity on longitudinal section 

= ^- Ib. per square inch. 

Comparison of these results will show that 
the stress on a longitudinal section is double 
that on a circumferential section, a fact which 
explains why the longitudinal joints in boilers 
are made much stronger than the circum- 
ferential joints. 

A spherical shell may be worked out in a 
similar man-ner. Let the shell be filled up to the level of a horizontal 
diameter AB with cement (Fig. 126), then 




Sectional Plan 



FIG. 126. Stresses in a spherical 
shell. 



RIVETED JOINTS 9? 



The complete cross-section at AB is a ring of diameter d and 
thickness /, and is under tensile stress of intensity given by 

Tensile stress intensity x area of cross-section = R, 

-D 

.'. Tensile stress intensity = -r 



trdt 
= Ib. per square inch. 

As before, p = fluid pressure in Ib. per square inch, 
d= diameter of sphere in inches, 
/ = thickness of plate in inches. 

It will be noted that the stress intensity in a spherical shell is the 
same as that on the circumferential sections of a cylindrical shell of 
the same diameter and thickness, and subjected to the same fluid 
pressure. It will also be observed that a spherical shell is self-staying 
on account of the fact that its shape does not tend to alter when 
it is exposed to the internal fluid pressure. The same is true for 
the cylindrical portion of an ordinary boiler shell, but the flat end 
plates are liable to be bulged outwards unless supported or stayed 
in some effectual manner. 

Riveted joints. Plates may be connected permanently by means 
of riveted joints. In lap joints the edges of the plates overlap 
(Figs. 131 and 132) and are connected by one or two rows of 
rivets ; in butt joints the plates are brought together edge to edge 
(Figs. 133 and 134) and cover plates pass along the seam on both 
sides or on one side only. As the strength of the joint depends 
to a considerable extent on the workshop methods employed, it is 
necessary to make brief reference to these methods. 

Excepting in the case of very thin plates and small rivets, the rivets 
are heated before being inserted in the holes and are closed by the 
use of hand or pneumatic hammers, or by a hydraulic riveting machine. 
Owing to the great pressure exerted in the latter method, the rivets 
generally fill the hole better when finished and the plates are held 
together more firmly. In either method, the cooling of the rivet and 
consequent longitudinal contraction assist largely in binding the plates 
together, while at the same time the rivet is put under pull stress of 
an uncertain amount. 

P.M. G 



98 MATERIALS AND STRUCTURES 

Rivet holes may be punched or drilled. Punching injures the 
metal by overstraining the material round the hole, a defect which 
may be remedied by annealing, or by punching the hole about -^ inch 
smaller than the proper diameter, and then enlarging it to the size 
required with a reamer, thus getting rid of the 
overstrained material. The plates are punched 
separately, hence there is difficulty in ensuring 
that the holes shall come exactly opposite one 
r*?[ another when the plates are brought together; 
'-.4 drilling is effected with the plates together in 
'p.-.d-y position, and this method is to be preferred as 

FIG. i2 7 .-stress on a giving fair holes, as well as producing no injury 
to the plates. Punched holes may be brought 
fair by bolting the plates together before reamering. 

There is a lower limit to the diameter of hole which may be 
punched in a plate of given thickness, depending on the value of the 
stress under which the punch will crush. 

Let d= diameter of hole, inches (Fig. 127), 
/ = thickness of plate, inches, 

q = shearing stress of material of plate, in tons per square inch. 
p = crushing stress of material of punch, in tons per square inch. 
Area under shear stress = trd x /. 
Force P required to shear the material = qtr dt. 

Push stress on punch = P -5 



Equating this to / will give the limiting value of d, thus 



. 

P 

p for the material of the punch, tool steel, is about four times the 
value of q for mild steel, hence, the condition that the punch is on 
the point of crushing is ^_ t 

showing that the minimum diameter of hole which may be punched 
is equal to the thickness of the plate. If d is less than /, / must 
have a value greater than \q for punching to be possible. 



RIVETED JOINTS 



99 



Riveted joints should not be designed so as to load the rivets by 
tension, as the heads are not reliable under pull. The loading should 
be of the nature of pull or push along the direction of the plates, thus 
putting the rivets under shear stress. Lap joints (Fig. 128) and butt 
joints having a single cover plate (Fig. 129) are put under a bending 




FIG. 128. 



FIG. 129. 



action by reason of the forces being in parallel lines. Butt joints 
having double cover plates (Figs. 133 and 134) are free from this 
objection. In lap joints, the rivets will sustain equal shearing forces 
whether the plates be under pull or push ; in butt joints under push, 
the forces will be communicated from plate to plate along the edges 
in contact without putting the rivets under shear stress at all, provided 
the fitting is perfect. The cover plates and rivets in this case serve 
only to prevent the plates getting out of the same plane. For these 
reasons, both compression and tension members are best fitted with 
butt joints having double cover plates. 

Methods of failure of riveted joints. These may be described by 
reference to Fig. 1 30, showing a single riveted lap joint. 

(a) If the hole is situated too near 
the edge of the plate, the material may 
open out as at A during punching, or by 
reason of the bursting pressure exercised 
by the hot, soft rivet while being closed. 
To prevent this happening, the distance 
from the centre of the hole to the edge of 
the plate should not be less than 1-5 times 
the diameter of the rivet. 

(^) The material of the plate may crush 
at B owing to the rivet being too large 
in diameter. When the joint is loaded the 
rivet bears on one half of the cylindrical 
surface of the hole, producing a bearing 
stress which is calculated by dividing the 
load on the rivet by the " projected area " 
of the hole, the latter being calculated by taking the product of the 
diameter of the hole and the thickness of the plate. In girder work, 
the design of the riveted joints has to be based sometimes on the 




FIG. 130. Methods of failure of 
riveted joints. 



100 



MATERIALS AND STRUCTURES 



safe bearing stress; this stress ranges from 7 to 10 tons per square 
inch in practice. 

(<r) One of the plates may give way by tearing along the line CD. 
(d) The rivets may shear at EF. 

The most economical joint would be equally ready to fail by all 
four ways simultaneously. It is impossible to calculate (a) from first 
principles, but expressions giving the relations of the various quantities 
may be found by equalising the resistances of the joint to crushing, 
tearing and shearing. It is customary in this country to neglect the 
increase in strength owing to the frictional resistance to the plates 
sliding on one another. The precise conditions for any riveted joint 
cannot be stated definitely, hence empirical rules, or rules which are 
partly empirical, are often employed in practice. 

Lap joints. Lap joints may be single or double riveted ; it is rarely 
the case that there are more than two rows of rivets. The pitch is 
the distance from centre to centre of the rivets measured along the 
row. The strength of the joint may be considered by taking a strip 
equal in breadth to the pitch, as the conclusions arrived at for this 
piece may be assumed to be true for the entire joint. 

Let p = pitch of the rivets, inches ; 
d= diameter of the rivets, inches ; 
t= thickness of the plates, inches ; 
y~ = the ultimate tensile strength of 
the plates, tons per square inch ; 
f g = the ultimate shearing strength of 
the rivet, tons per square inch; 
fb = the bearing stress, tons per 
square inch of projected area, 
when the joint is on the point 
of failing by crushing. 




FIG. 131. Strength of a single-riveted 
lap joint. 



We have, for a single riveted lap joint (Fig. 131) : 
Least area of plate section under pull = (p- d}t y 

Resistance of joint to tearing =ft(p d)t tons ............. (i) 



Area of rivet section under shear 






Resistance of joint to shearing =/ - tons ................... ( 2 ) 

4 

Projected area = dt, 
Resistance of joint to crushing =fydt tons .................... (3) 



RIVETED JOINTS , v- I '*ta\ 



Equating (i), (2) and (3) gives : 



Taking 



d- 



/ft 



(4) 



- ......................... 

JS 

The diameter of the rivet may be found from this relation, and the 
pitch may then be calculated from 



P = ('l*S J i "-) + <*. v5) 

In double riveted lap joints there will be two rivet sections per 



; _iM^ 












1 


1 


/ 

o 

LJ ; 





FIG. 132. Strength of a double-riveted lap joint. 

pitch under shear (Fig. 132); there will also be two bearing areas 
per pitch. Hence 



= 1-27/4 



(6) 

(7) 



MATERIALS AND STRUCTURES 



Butt joints. The strength of butt joints may be calculated in a 
similar manner; it will be observed (Fig. 133) that, with two cover 






o I 

1 


t 

? 

* 





FIG. 133. Strength of a single-riveted butt joint. 

plates, the rivets are under double shear, i.e., each rivet would have to 
shear at two sections A and B for the joint to fail by shearing. Each 

rivet will thus have a shearing area of 2 . 

4 
For a single riveted butt joint (Fig. 133), 



.(8) 



7.' 



ft 




I 


o 





! O 

p 

O i 

* 





FIG. 134. Strength of a double-riveted butt joint. 

In the case of a double riveted butt joint (Fig. 134), we have 



.(10) 

.(II) 



RIVETED JOINTS 103 



Data from experiments. The ultimate tensile strength of iron 
plates may vary from 21 to 26 tons per square inch, and for steel 
plates may vary from 27 to 32 tons per square inch. Iron and steel 
rivets have an ultimate shearing strength of about 23 tons per square 
inch. Owing to the difficulty of stating precisely what the actual 
conditions are in a finished riveted joint, these stresses should be 
used with caution. Experiments on actual joints with iron plates 
and iron rivets show that the ratio f g /f t , is nearly i for drilled holes, 
and from 1-2 to 1-3 for punched holes which have been neither 
annealed nor reamered. For steel plates and steel rivets the values 
of the ratio appear to be about 0-75 for drilled holes and about 0-9 for 
punched holes neither annealed nor reamered. For either reamered 
or annealed punched holes the values are about the same as for 
drilled holes. Breakdown in experimental joints by crushing appears 
to take place for ratios of fb/fs of about 1-7 for rivets in single shear 
and about 2-35 for rivets in double shear. Provision against crush- 
ing is often made by employment of an empirical rule for the 
diameter of the rivet. A good practical rule is 

</=I-2>/7 tO I-4A//. 

When this rule is used, the diameter of the rivet is calculated 
first, and the pitch is then determined by equating the resistances 
to tearing and shearing. Afterwards, the bearing stress should be 
calculated in order to ascertain that its value is not excessive. 

In riveted joints designed under the Board of Trade rules, rivets 
under double shear are allowed if rivet sections per rivet only ; this 
is owing to the probability of the rivets not all bearing equally. This 
rule is often disregarded in other joints.* 

Efficiency of riveted joints. The efficiency of a riveted joint is 
the ratio of its actual strength to that of the solid plate. To calculate 
the efficiency, the ratios of the strength of the joint against tearing, 
shearing and crushing to the strength of the solid plate should be 
calculated separately, and the lowest value taken as the efficiency 
of the joint. It will be evident that all three ratios will be equal if 
the joint has been designed for equality of rupture by each of the 
three ways of failure, and the efficiency may be obtained then by 
consideration of the tearing resistance only. 

Resistance of joint to tearing = (P~ d)tft- 

Resistance of solid plate to tearing =ptf t . 

* For a full discussion of riveted joints, see Machine Design, Part I., by Prof. 
W. C. Unwin (Longmans, 1909). 



104 MATERIALS AND STRUCTURES 



P 

EXAMPLE i. A double riveted butt joint with double cover plates is 
used to connect steel plates of 0-5 inch thickness ; the holes are to be 
drilled. Find the diameter of the rivets from the empirical rule (p. 103), 
and also the pitch of the rivets, taking 



= | inch, nearly. 

</ (p. 102) 

= (3-142 x 075 x () 2 x 2) + 1 
= 4^ inches, nearly. 

EXAMPLE 2. Calculate the efficiency of the above joint. 
Efficiency =^-7 

^4-5-0-875 

4-5 

=0-805 
= 80-5 per cent. 

Or the efficiency may be calculated by considering the resistance to 
shearing. Thus : 

Area per pitch under shear stress = 4 - 

Strength against shearing = 7r</ 2 / s . 
Efficiency against shearing = 7rd' 2 f s -rfltf t 

nP fs 

= Pt ' ft. 
^3-142x49x0-75 

4-5x0-5x64 
= 0-802 
= 80-2 per cent. 

EXAMPLE 3. Calculate the bearing stress in the above joint when 
carrying a load which produces a stress of 4 tons per square inch in the 
solid plate. 

Area of solid plate per pitch pt 

= 4-5x0-5 
= 2-25 sq. inches. 
Load per pitch ^4x2-2 5 
= 9 tons. 



RIVETED JOINTS 



105 



This load is carried on the bearing surface of two rivets ; hence : 
Projected bearing surface per rivet = d r /. 

Bearing stress = -^ 

__9 

2x0-875 X0 '5 
= 10-3 tons per sq. inch. 

EXAMPLE 4. Two plates forming a tie-bar have to be connected end 
to end by a butt joint having double cover straps (Fig. 135). Each plate 

A, C. t, 




o o o o 
o 

6000 



6000 



o o o o 

o o o o o 

o 
o o o o 




B 1 D 1 F 1 



FIG. 135. Riveted joint for a tie-bar. 

is 10 inches wide and f inch thick ; the rivets are f inch in diameter. 
The stresses allowed are 6 tons per square inch pull, 4 tons per square 
inch shearing, and 10 tons per square inch bearing. Find the number 
of rivets required. 

Sectional area of each plate =iox = 7-5 square inches. 

Area abstracted by one rivet hole at the section AB = | x 1 = 0-56 sq. in. 

Net sectional area of plate at AB = 7-5 -0-56 

= 6-94 square inches. 
Total safe pull on the plate = 6-94x6 = 41 -64 tons. 



Sectional area of one rivet = 

4 



TT(t* 22 



7X4 



x r = 0-442 sq. in, 



Allowing if rivet sections for rivets under double shear, we have 
Shearing resistance of one rivet=o-442 x 1-75 X4 

= 3-09 tons. 

Projected area of one rivet = |x 1=0-56 square inch. 
Bearing resistance of one rivet = o-56x 10 = 5-6 tons. 
As the shearing resistance is lower than the bearing resistance, the 
shearing resistance must be taken in calculating the number of rivets 
required. Let N be the number of rivets on each side of the joint ; then 
Total safe pull on the plate = total shearing resistance of the rivets, 
41-64 = N X3-09, 
N = I4 rivets. 

To obtain a good arrangement of rivets, 15 rivets have been placed on 
each side of the joint in Fig. 135. 



io6 MATERIALS AND STRUCTURES 

At the section AB, the safe load which can be applied is that calculated 
above as 41-64 tons. At CD, the tearing strength of the plate is less 
than at AB, but to this must be added the resistance of the rivet on the 
left-hand side of CD, as this rivet would have to shear simultaneously 
with the plate tearing at CD for the joint to fail in this way. 

Sectional area of plate at CD = 7-5 -(2 xo-56) = 6-38 square inches. 
Resistance to tearing at CD = 6-38x6 = 38-28 tons. 
Adding the shearing resistance of one rivet to this, we have 
Safe load with reference to the section CD = 38-28 4-3-09 

= 41-37 tons. 

Considering the section EF, the three rivets on the left-hand side of 
EF would have to shear simultaneously with the plate tearing. 

Resistance to tearing at EF = { 7-5 - (3 x 0-56) }6 = 34-92 tons. 

Shearing resistance of three rivets = 3 x 3-09= 9-27 tons. 

Safe load with reference to the section EF = 44-19 tons. 

It is evident that the safe load with reference to any other section on 
the right-hand side of EF will have a greater value than that for the 
section EF. The minimum safe load is that calculated for the section 
CD, viz. 41-37 tons, which accordingly is the safe load which the joint 
will carry. 

Strain. Strain refers to the alterations of form or dimensions 
which occur when a body is loaded or subjected to stress. Thus a 
pulled or pushed bar is found to have become longer or shorter after 
the load is applied, and is said to have longitudinal strain. This kind 
of strain is measured by taking the ratio of the change in length to 
the original length. 

Let L = original length of bar, 

= alteration in length, both in the same units. 

Longitudinal strain = =-. 

Volumetric strain occurs when a body is subjected to uniform fluid 
pressure over the whole of its exposed surfaces. The volume will be 
changed somewhat under these conditions, and the volumetric strain 
is measured by taking the ratio of the change in volume to the 
original volume. 

Let V = original volume of body, 

v = change in volume, both in the same units. 

i) 
Volumetric strain = =-=. 

V 

Shearing strain occurs when a body is subjected to shear stress. 
Such a stress is distinguished from the other two just mentioned in 



TYPES OF STRAIN 



107 



that it produces a change in the shape of the body, while pull, push, 
and hydrostatic stress produce no such change. We may obtain an 
idea of what happens by holding one cover ot a thick book firmly 
on the table and applying a shearing force to the top cover (Fig. 136). 
The change in shape is evidenced by the square originally pencilled 




**.,. .p.,?.? 




FIG. 136. Shearing strain illustrated 
by a book. 



f 



FIG. 137. Measurement of 
shearing strain. 



on the end of the book becoming a rhombus. A solid body 
would behave in the same manner under similar conditions of load- 
ing, only, of course, in a minor degree (Fig. 137). The shearing 
strain is measured by stating the angle in radians through which the 
vertical edge has rotated on application of the shearing stress. 
Shearing strain = radians (Fig. 137). 

For metals 6 is always very small, and it is often sufficiently 
accurate to write, referring to Fig. 137: 

Shearing strain = 0, 

BE' 
~BC* 

Transverse strain. When a bar is pulled or pushed, not only is 
its length altered, but also its transverse dimensions. Thus a pulled 
bar becomes thinner, while a pushed bar becomes thicker. Such 
alterations are referred to as transverse strains and are measured in 
the same manner as longitudinal strains, viz. by taking the ratio of 
the alteration in transverse dimension to the original transverse 
dimension. 

H = a transverse dimension of the bar, 

h the change in H when the bar is loaded. 
h 



Let 



Transverse strain = 



H* 



For any given material, such as a metal, experiment shows that 






io8 MATERIALS AND STRUCTURES 

there is a definite ratio of longitudinal to transverse strain, ranging 
from 3 to 4 for common metals. 

Let a = longitudinal strain, 

b = transverse strain, 

a 
m-j. 

The value of m depends on the kind of material ; its reciprocal 

is called Poisson's ratio. Values of this ratio for common materials 
m 

are tabulated on p. 683. 

Elasticity. Elasticity is that property of matter by virtue of which 
a body endeavours to return to its original form and dimensions 
when strained, the recovery taking place when the disturbing forces 
are removed. Strain takes place while the loads are being applied 
to a body, hence mechanical work (see p. 325) is expended in pro- 
ducing strain, and is stored up, partly at any rate, in the body. 
The elasticity of any material is regarded as being perfect, provided 
the recovery of the original form and dimensions is perfect on 
removal of the loads, and provided also that the energy given out 
during recovery equals that expended while the body was being 
strained. 

The elasticity of a large number of materials is practically perfect 
provided they are not stressed beyond a certain limit, which depends 
on the kind of material and also on the nature of the stress applied. 
If loaded beyond this elastic limit of stress, the recovery of original 
form and dimensions is incomplete and the body is said to have 
acquired permanent set. 

Further, experiment shows that the strains are proportional to the 
stresses producing them provided that the elastic limit is not exceeded. 

This law was first discovered by Hooke, and bears his name. 
Most materials show slight divergencies from Hooke's law, but it is 
adhered to so closely in the case of common metals as to justify the 
assumption of its truth for nearly all practical purposes. 

Modulus of elasticity. Assuming Hooke's law to be true, and 
selecting any elastic material to which loads may be applied. 

Let / = the stress, 

s = the strain produced by /. 
Then p varies as s up to the elastic limit, hence the quantity will 

be constant for that material up to the elastic limit. The term 
modulus of elasticity is given lo this quantity. The value of the 



ELASTIC MODULI 109 



modulus of elasticity depends firstly on the nature of the material, 
and in the second place on the nature of the stress. For any given 
material there are three moduli of elasticity which should be under- 
stood. In each case the measurement is made by taking 

Modulus of elasticity = . 

The units of this expression will be governed by the unit of stress 
employed, as strain is simply a ratio. 

Young's modulus for a pushed or pulled bar is obtained by dividing 
the push or pull stress intensity on a cross section at 90 to the axis 
of the bar by the longitudinal strain. 

Let P = force of push or pull applied to the bar, 

A = area of the cross section, 
L = original length of the bar, 
= change of length of the bar, 
both the latter being in the same units. 
Then, writing E for Young's modulus, 

E = stress 
strain 

* ** 

A ' L Ae' 

The bulk modulus belongs to the case of a body subjected to hydro- 
static stress, which produces volumetric strain. 
Let / = the hydrostatic stress intensity, 

V the original volume of the body, 
v = the change in volume, 
both the latter being in the same units. 
Then, writing K for the bulk modulus, 



The rigidity modulus refers to the case of a body under shearing 
stress, and consequently changing its shape by shearing strain. 

Let q = the shearing stress intensity, 

6 = the shearing strain, in radians. 
Then, writing C for the rigidity modulus, 



The most convenient units to employ for the elastic moduli are 
tons or Ib. per square inch in the British system, and kilograms per 



no 



MATERIALS AND STRUCTURES 



square centimetre in the metric system. A table of values will be 
found on p. 683. 

Strains in a cylindrical boiler shell. It has been seen (p. 96) 
that the stresses in a cylindrical boiler shell on longitudinal seams 
and on circumferential seams are in the ratio of two to one. Suppose 
that in consequence of these stresses the circumference becomes 
greater by a small amount e. Let d be the original diameter of the 
shell, then the original length of the circumference will be ird, and 
the circumferential strain will be : 



Circumferential strain 



7T/ 



Also, new length of the circumference = ird+ e ; 
.'. new length of the diameter -. 

7T 

Hence, change in the diameter = - d 



d+--d 

7T 



to 



strain in the direction of a diameter 



(2) 



Comparison of (i) and (2) shows that the diametral and circum- 
ferential strains are equal. 



-6 




\P (d) (b) 

FIG. 138. Strains in a boiler shell. 

To obtain the circumferential strain, let / and \p be the stresses 
on the longitudinal and circumferential seams respectively. If / 
were to act alone (Fig. 138 (a)), the circumferential strain would be 
a (extension) and the transverse strain would be b (contraction). 
If \p were to act alone (Fig. 138^)), the longitudinal strain would 
be \a (extension) and the circumferential strain would be \b 



STRAINS IN SHELLS in 



(contraction). Hence, when both stresses act together, the strains 
produced will be : 

Circumferential strain = a - \b ...................... (3) 

Longitudinal strain = \a - b ....... ............... (4) 

Or, since m ^ ^' 



m 

And Circumferential strain = a --- 

2 m 



Longitudinal strain = -a - 

-, \ 

(6) 



2 m 

i 



Suppose m be taken equal to 4, then 

Circumferential strain = a ( i - ^) 

, =l (7) 

Longitudinal strain = a (^ - j) 

= > (8) 

Reference to Fig. 1 38 (a) shows that 

E-* 

a 

or = g (9) 

v Hence : Circumferential strain = \ ^. (10) 

O Ji 

Longitudinal strain = - ^ ( ll ) 

4 & 

ExAMPtE. A boiler shell 7 feet in diameter and 30 feet long is tested 
by hydraulic pressure (cold water) up to a stress of 6 tons per square inch 
on the longitudinal seams. Take = 13,500 tons per square inch and 
m = 4, and find how much water will escape when a test cock on the top 
of the boiler is opened. Neglect any bulging of the ends. 

[To answer this question, calculate the increase in volume of the shell 
while the pressure is being applied.] 

Circumferential strain = | x Tg f jo = 
The diametral strain is equal to this ; hence : 

Change in diameter =(7 x 12) x ^- 
=0-0327 inch. 
Final diameter of shell = 84-0327 inches. 



112 MATERIALS AND STRUCTURES 



Let JL) and d be the final and original diameters ; then 
Increase in cross-sectional area of the water = -(D 2 -</ 2 ) 



= - x 0-0327 x 2 x 84 nearly 

4 
= 4-32 square inches. 

.. increase in volume due to increase in sectional area = 4-32 x 360 

= 1555 cub. in, 
Again, Longitudinal strain = J x T sf(Rj = 2T IW 

/. change in length of the shell = 30 x 12 x ^7000 

= 0-04 inch. 

Sectional area of the water = - x 84 2 (nearly) 

4 
= 5542 sq. inches. 

.*. increase in volume due to increase in length =0-04x5 542 

= 222 cubic inches. 
Total increase in volume = 1550 + 222 

= 1777 cubic inches. 

The change in volume which occurs when charging cylinders for 
holding compressed gases is sometimes taken as a test of the 
soundness of the material of which the cylinder is constructed. 
The test is made by having the cylinder immersed in water contained 
in a closed vessel fitted with an external glass tube connected to the 
water space. In charging, the expansion of the cylinder will displace 
some of the water, which will therefore rise in the glass tube. An 
increase in volume of more than a prescribed limit, as indicated by the 
tube reading, affords evidence of defects in the material of the cylinder. 

Stresses in thick cylinders. In Fig. 139 (a) is shown a cylinder of 
considerable thickness under external and internal fluid pressures. 
Let push stresses be denoted positive, and let the external pressure 
be greater than the internal pressure. Consider a ring of unit length, 
having an inner radius r and outer radius (r + 8r) (Fig. 139 (<)). Let 
the radial stress on its inner surface be /, and let that on the outer 
surface be (p + Sp). The resultants of these stresses on the half ring 
(Fig. 139 (r)) will be 

P 1 =p x 2r (see p. 96), 



The resultant P of P x and P 2 is 

P = P 2 -P 1 

= (/ + 8p) x 2 (r + 8r) -p x zr. 



STRENGTH OF THICK CYLINDERS 



Let / be the tangential, or hoop stress on the ring ; the area over 
which this stress is distributed is 8r x i , and there are two horizontal 
sections, one at A and one at B (Fig. 139 (<:)) ; hence, 



or /. 8r=pr+r.8p+#.8r+8p.8r-pr 

by neglecting the product of the small quantities 8p and Sr. 

P* i P+$P 



(i) 




FIG. 139. Stresses in a thick cylinder. 

Another equation may be formed by consideration of the strains 
in the axial direction produced by p and f all over the cylinder. It 
may be assumed that cross sections of the cylinder remain plane 
when the fluid stress is applied, i.e. all fibres parallel to the axis of 
the cylinder lying between two cross sections change their lengths to 
the same extent. Hence the assumption that the axial strains are 
equal all over the cylinder. 

Axial strain produced by/ = -, 



As both/ and /are push stresses, both of these strains are exten 
sions, and the total axial strain will be 

+ = a constant, ' ;V ' ! 



or p +/= a constant. 

Taking 20, for the value of the constant, this gives 

p+f=2d 



(2) 



D.M. 



H 4 MATERIALS AND STRUCTURES 

From (2), f=2a-p. 

(i), (2a-p)Sr=r.8p+p.Sr, 

2a . r p . 8r=r. Sp+p . Sr, 

2a.8r=r.8p+2p.Sr. 
Multiply each side of this equation by r, giving 

2 ar . &r = r 2 . 8p + 2pr . Sr, 
or in the limit, when 8r becomes very small, 



2ar =**- + 2pr. 

dr 

The right-hand side is the differential coefficient of (pr*), i.e. 



Hence, d(pr & ) = 2ar . dr. 

Integrate, giving pr z = ar 2 + c. 

f = * + ? ................................. (3) 

and /= 20, -p 

= - ................................ (4) 

The solution of any particular problem may be obtained from (3) 
and (4) by first determining a and c from the given conditions. 
Take the ordinary case of a cylinder having an internal fluid pressure 
pi, the external pressure being regarded as zero (Fig. 140). We have 

p=pi when r=Rij /. A = + ^72 ................ (5) 



Hence, 



p = o when r=R ; ." 

P .2D 2 



n .2 



Substitution of these values in (4) gives 

/_ _%:__ io 

A R 2 -R, 2 ^Ro^Ri 2 '^ 
, R - 2 \ 



TEMPERATURE STRESSES 



This equation gives the hoop tension at any radius r ; the maximum 
hoop tension will occur where r has its smallest value, i.e. at the 
inner skin, where r=Ri. Hence, 

AT f R* 2 / . RO ! 

Maximum / = /trs =r-w i + 




It will be noticed from equation (8) that the maximum hoop 
tension is always greater than the internal fluid 
stress pi, independently of the thickness of the 
cylinder ; hence, it is impossible to design a 
solid cylinder to withstand a fluid pressure 
greater than a certain value for a given material. 
The difficulty may be overcome by shrinking 
one cylinder on the top of another, or by 
winding wire under strong tension over the 
outside of the cylinder. The effect is to put FlG> 14 * 

the inner parts under initial push hoop stress, and gives a distribution 
of stress more nearly uniform when the fluid pressure is applied. 

Stresses produced by change in temperature. If a metal bar be 
heated, its length will increase by an amount proportional to the 
increase in temperature, and to a coefficient, the value of which 
depends on the kind of material ; this is on the assumption that the 
bar is permitted to expand freely. 

Let L = the original length, in inches ; 
/ = the rise in temperature ; 

= the coefficient of expansion, i.e. the change in length 
per unit length produced by a rise in temperature 
of one degree. 
Then, change in length = L/e ; 

new length of the bar = L + Lfc 

(0 



Suppose that the bar is now cooled to its original temperature, and 
that forces are applied to its ends so as to prevent it from returning 
to its original length. Evidently these forces will have the same 
value as those which would be required to produce an elastic 
extension L/ in the bar had its original temperature been kept 
unaltered. 



n6 MATERIALS AND STRUCTURES 

Let P = the total force required in tons. 

A = the cross sectional area in square inches. 

p 
p = the stress produced by P in tons per square inch. 

E = Young's modulus in tons per square inch. 
Then Longitudinal strain = =fc. 



Ais > 



P = EA/tons, .............................. (2) 

p E/ tons per square inch ............. (3) 



EXAMPLE. If the bar be of steel for which =13,500 tons per square 
inch, and if the rise in temperature be 100 F., find the stress in the 
material under the conditions expressed above. Take 
= 0-000007, 

J>=Et 

= 13,500 x 100 x 0-000007 
= 9-45 tons per square inch. 

Suppose now that the bar be heated and at the same time held 
rigidly between abutments which prevent entirely any change in the 
length. These conditions may be imagined as follows : first allow the 
bar to expand freely on heating ; then apply forces to the ends and 
let these be sufficient to compress the bar back to its original length. 

Length of the bar before applying the forces = L(i 
Change in length produced by P = L/e. 

Elastic strain produced by P = 



i + U 

Now E= p .- 

stram 



E/e , x 

rz (4) 



The denominator will be nearly unity, as U is usually very small ; 
hence, (4) will have the same value nearly as (3). 

Effects produced by unequal heating. Fig. 141 illustrates three 
bars A, B and C attached to rigid cross pieces D and E ; E is fixed 



TEMPERATURE STRESSES 



117 



and D may rise or fall freely. B is centrally situated between A and 
C ; A and C have equal sectional areas and B may have a different 
sectional area. All three bars are of the same material. 

If all three bars be at the same tempera- 
ture at first, and if they be raised through 
the same range of temperature, all will 
attempt to expand equally in the direction 
of length, and no stress will be produced 
in any of them. Suppose, however, that B 
is raised to a certain temperature and that 
A and C are both raised to the same 
higher temperature, then B attempts to 
expand to a smaller extent than A and C. 
The cross pieces D and E will compel all 
three to come to the same length ; hence, 
B will be under pull and A and C will be FIG. 141. stresses due to unequal 
under push. This is indicated in the figure 

by the forces P and Q. As no force whatever is required from the 
outside in order to balance the arrangement under the altered 
conditions of temperature, it follows that 

1> = 2Q (i) 

Let the equal sectional areas of A and C be denoted by x and 
the sectional area of B by a 2 ; then 



V 


1 


{p 




| a 


"\ 


a f 


^ 


a- 


) 


A 




B 




C 


t< 


1 


{P 




1 



Stress in A = stress in C =/j = ; .'. Q =p l a l . 



Stress in B = = 



Hence, from (i), 



or 



This result indicates that if all three bars have the same sectional 
area, then the stress in B will be double that in A or C, irrespective 
of the actual values of the changes of temperature. 

To find the numerical values of/ x and/ 2 , proceed as follows : 

Let L = the original length of each bar. 

t } change in temperature of A and C. 
/., = change in temperature of B. 
= the coefficient of expansion. 
E = the common value of Young's modulus. 



I IS MATERIALS AND STRUCTURES 

First assume that all three bars expand freely ; then 
Extension of A = extension of C = L/^. 
Extension of B = L^. 
New length of A or C = L(i + ^) 

= ^L, 
where ^ = i + ^e. 

New length of B = L(i + t z c) 



where A 2 = i 

Let the bars now be compelled to come to the same final length 
L F by application of the forces P and Q. 

Shortening of A or C produced by Q = ^L - L,.. 
Extension of B produced by P = L F - ^ 2 L. 



Strain of A or C 

Strain 

Hence, for A or C, 
And for B, 



Lp 



or 




As the ratio of p l and p> 2 is known from (2), this result may be 
used for calculating L F , the final distance between the cross pieces; 
substitution in (3) and (4) will then give the values of/j and/ 2 . 

EXAMPLE. Take the following data for the arrangement shown in 
Fig. 141 and calculate the final distance between the cross pieces, also the 
stress in each mild steel bar. 

#!= i square inch. . L=ioo inches. 

a%=3 square inches. -30,000,000 Ib. per sq. inch. 

/i=iooF. = 0-000007. 



REINFORCED CONCRETE COLUMNS 



AlSO, 



From (5), 
whence 



;= i 4- ( ioo x 0-000007) : 
: = I + (50 x 0-000007) = 
iL= 1-0007 x ioo =100-07. 
A 2 L= 1-00035 x 100= 100-035. 
3-002 1 _ 1 00-07 - LF 
2-0007 



I -0007. 



LF- 100-035' 
L F = 100-04899. 

(Note, as the changes of length are calculated by taking the differences 
; n the lengths of the bars, it is necessary in examples of this kind to use a 
larger number of significant figures than that employed usually.) 

From (3), 30,000,000 : 
whence p l -. 

From (4), 30,000,000=^2 
whence 



100-07 



'* l 100-07 - 100-04899' 
= 6298 Ib. per square inch. 
100-035 



100-04899- 100-035' 
= 4195 Ib. per square inch. 
These stresses have the calculated ratio of 1-5. 

This problem may be varied by using bars of 
different materials and raising the temperatures 
of all to the same extent. The differences in 
the elastic moduli will produce a similar effect 
to that caused by unequal heating, and the 
calculation is effected in a similar manner, 
making use of the proper values of the co- 
efficients of expansion and of the elastic moduli. 

Reinforced concrete column. In Fig. 142 is 
shown a concrete column reinforced by steel 
bars arranged as shown in the plan. Appli- 
cation of an axial load to the column will cause 
both steel and concrete to shorten to the same 
extent ; as the lengths of both are equal, it 
follows that the strains are also equal. Using 
the suffixes c and s to denote the concrete and 
steel respectively, let s _ s 





FIG. 142. Reinforced con- 
crete column. 



Then 



the strains in the direction of the 

length of the column. 

f s and f c = the stresses in Ib. per square inch. 
Eg and E c = Young's moduli in Ib. per square inch. 
A 8 and A c = the sectional areas in square inches. 

T? _f* /T\ 



126 iMATEJEUALS AND STRUCTURES 

Dividing (i) by (2), we have 

E* = S* fc 

EC fc S 8 



The ratio of E s to E c varies somewhat ; the average value of 1 5 is 
usually taken. With this value, equation (3) shows that the stress in 
the steel will always be 15 times that in the concrete irrespective of the 
relation of the sectional areas of the concrete and steel. If 500 Ib. 
per square inch be taken as a safe stress for the concrete, then the 
stress in the steel will be 7500 Ib. per square inch. 

Suppose VV to be the load in Ib. applied to the column ; then 



..................... (4) 

a result which enables the safe load to be calculated if the sectional 
areas of the steel and concrete are given. 

The stresses produced in other composite bars under push or pull 
are calculated in a similar manner, making use of the proper values 
of Young's modulus. Such bars may take the form of a steel rod 
cased in some alloy such as gun-metal, or the arrangement may be 
as illustrated in Fig. 141, with A and C of one material and B of a 
different material. A central load applied to the top cross piece 1) 
will produce equal strains in all the bars, and the stresses will thus be 
proportional to the values of Young's modulus for the materials of 
the bars. 

Classification of stresses. Stresses may be either normal or tan- 
gential ; oblique stress is compounded of normal and tangential stresses. 
Stress is purely normal when its lines of direction are perpendicular 
to the surface over which it is distributed. Normal stresses may be 
either tensile or compressive. Stress is tangential or shearing when 



FIG. 143. Stress in a tie-bar. 



its lines of direction coincide with the surface over which it is 
distributed. Oblique stress may have its lines of direction inclined 
at any angle between o and 90 to the surface over which it is 
distributed. Normal tensile stress occurs in any section AB of a tie- 
bar subjected to axial pulls (Fig. 143), the section being perpendicular 



RELATIONS OF STRESSES 



121 



FIG. 144. Stress 
in a column. 



to the axis of the bar. The stress in this case will be uniformly 
distributed except for sections near the ends of the bar, and its 
intensity will be given by 

P 

p = : . 

f area of section AB 

Normal compressive stress will be found on any 
horizontal section AB of a vertical column (Fig. 144) 
carrying a weight W. If the line of W coincides with 
the axis of the column, the stress will be uniformly 
distributed and of intensity given by 

W 

* ~ area of section AB' 

Relation of oblique stress with normal and 
tangential components. Let ABCD (Fig. 145) represent the eleva- 
tion of a cube of unit edge, the top face being subjected to normal 
stress p n and also to tangential stress pi. On the supposition that 

these stresses are uniformly distributed, 
we may substitute resultant forces P N 
and P T , acting at the centre O of the 
top face, in a plane parallel to that 
of the paper, the values of P N and P , 
being p n and pt as the face is of unit 
area. The resultant of P N and P T 
will be 

R = VP N 2 + P T 2 , 

and will act at an angle 6 to the 
normal, the tangent of which is 




FIG. 145. Relation of stresses. tan " = ^ 

Now R may be taken to be the resultant of an infinite number of 
forces having the same direction as R, 
and uniformly distributed over the top 
face of the cube (Fig. 146), these forces 
constituting an oblique stress r, the value 
of which will be 




FIG. 146. 



R 



area of top face 



(0 



122 



MATERIALS AND STRUCTURES 



Other useful relations deduced easily from the figure are : 



pt = r . sin 0, 



(3) 
(4) 





Pn 

The angle is defined as the angle of obliquity of the stress. 

Some examples of oblique stress. A useful method of determining 
the stresses on any section of a loaded body consists in first imagining 
that the body has been actually cut at the given section. One portion 
only of the body is then taken, and the resultant forces are determined 
which must be applied to the section in order to produce equilibrium 
in this portion. The stresses and their distribution may then be found. 

Consider a column carrying a load P, the line of which coincides 
with the axis of the column (Fig. 147 (a)). Let the column be cut at 
a section AB and consider the upper portion (Fig. 147 (^)). For 
equilibrium, a resultant force P' = P must be applied in the same line 
as P. This will give rise to a stress which will be seen afterwards 
to be uniformly distributed over AB. Let the area of the section 

AB be S ; then p 

Stress intensity on AB=/=^ (i) 

o 



(a) 



, , 



tfflttt 

IP' 

(b) 



B 





FIG. 147. Normal and shear stresses in a column. 



Supposing the column to be cut along CD (Fig. 147 (a)), the angle 
between AB and CD being 0. Considering the equilibrium of the 
top portion (Fig. 147 (<:)), we see that a resultant force P' = P must be 
applied in the same line as P. P' will give rise to an oblique stress 
uniformly distributed over the section CD ; let ON be drawn normal 
to the section, when it will be evident that the angle between P' and 
ON, which is the angle of obliquity, is equal to 0. To find the 
stress intensity, we have 

P' 

Stress intensity =A= ? = 7^7^- 

J 2 area of section CD 



STRESSES IN COLUMNS AND TIES 



123 



Now 



cos 6 ; 



cos 6' 



area of section AB 
area of section CD 

.'. area of section CD = 

. = p , ^ S 
' cos 
P' 

= .COS0 

=/.cos (2) 

The intensity of the oblique stress on CD is therefore equal to the 
stress intensity on AB multiplied by the cosine of the angle between 
the two sections. 

It is of interest to determine the components of / normal and 
tangential to CD (Figs. 147 (c) and (d}}. From equations (2) and (3), 

p. 122, we have / n = A cos0, (3) 

/=/ sin0 (4) 

By substituting the value of / from equation (2) above, we obtain 

1 11 / * ^"^ ' \j/ 

/=/.sin0cos0 (6) 

Pn 
HO 



0-8 
O6 
0-4 
0-2 



30 



60 



90 
Q degrees 
FIG. 148. Variation ot normal stress in a column. 

It will be easily seen, from equation (5), that p n has its maximum 
value when is zero, the value being then p and the section AB 
in Fig. 147 (a). The value of p n diminishes as is increased, being 
zero when = 90. Equation (6) may be written as 

^ = i/.sin20, (7) 

an equation which shows that t>t has zero value when is zero, and 
that the value is again zero when is 90. The maximum value will 






I2 4 



MATERIALS AND STRUCTURES 



occur when 26 is 90, the value of the sine being then unity; will 
then be 45, and the value of/ will be 

Maximum value of p t = \p (8) 

The fact that the section at 45 to the axis of the column has 
maximum intensity of shearing stress explains the reason why some 

Pt 

0-5 
04 
0-3 
02 
O-l 




30 



60 



A C 




90 

6 degrees 

FIG. 149. Variation of shear stress in a column. 

materials, such as brick, stone or cement under compression, fracture 
along planes at 45 instead of simply crushing. Such materials are 

comparatively weak under shearing. 
The curves in Figs. 148 and 149 have 
been plotted from equations (5) and 
(6), taking the maximum value of p n 
as i ton per square inch, and illustrate 
the way in which p n and pi vary, 

I ft depending on the angle at which the 

" section is taken. 

__^ ; ___.. " r ^ 

P ^ p' The case of a rod under axial pulls 

I , - ft) may be worked out in a similar 

manner and the results will be iden- 
P P .s tical, with the substitution of normal 

^\+-^*e pull stress for the normal push stress 
~~oS~^ Jk ~*p' which occurs in the column. Fig. 150 
_Sr^~ ( c ) illustrates this case, and as it is 

lettered to correspond with the column 

D diagrams there will be no difficulty in 

rfa^Pn tracing the connection. 

-<-l "V'S Stresses which are not uniformly 

H K T*^^>^ 

I TJ; (d\ distributed. A varying stress may be 

realised by considering a horizontal 
surface ABC (Fig. 151), having a 
number of slender vertical heavy rods 
of varying heights standing on it. Some of these rods are shown 
in the figure. The effect on the surface ABC, which is supposed to 



Flu 150. Normal and shear stresses in 
a tie-bar. 



STRESS FIGURES 



125 




FIG. 151. Representation of a 
varying stress. 



be covered entirely by the rods, will be to produce stress of varying 

intensity. There is, however, no difficulty in seeing that the resultant 

force on ABC will be the total weight of 

the rods, and that the line of the resultant 

force must pass through the centre of gravity 

of the whole of the rods taken together. 

We may deduce from this that, if a stress 

figure be drawn for a given section by 

erecting ordinates at all points of the 

section, of length to scale to represent 

the intensity of normal stress at each point, 

the resultant force will pass through the 

centre of volume of the stress figure. The magnitude of the resultant 

force may be found thus : 

Let / = stress intensity at a given point, 
&z = a small area surrounding this point. 

Then Resultant force = ^p . Sa, (i) 

the summation being taken all over the section. 

Equation (i) may be interpreted as meaning the volume of the' 
stress figure, stress intensities being used for ordinates and square 
inches or other convenient units for units of area. 

EXAMPLE. A rectangular surface ABCD is subjected to normal 
stress, which varies uniformly from zero along AD to 4 tons per square 

inch along BC (Fig. 152). AB is 4" 
and BC is 3". Find the resultant force, 
and show where it acts. 

The stress figure will be drawn in this 
case by erecting ordinates BE and CF, 
each to scale, representing 4 tons per 
square inch. Join EF, AE and DF, 
thus giving a stress figure of wedge 
shape. To find the magnitude of the 
resultant force, calculate the volume of 
the wedge by multiplying the area of 




FIG. 152. A uniformly varying stress. 

the base by the ordinate of average height, viz. 2 tons per square inch. 
Resultant force =R=4X3X2 



Thus, 



The centre of volume of the wedge will lie vertically over a point O, 
found by the intersection of two lines GH and KL, G and H bisecting 
respectively AD and BC, and KB and LC being one-third of AB and CD 
respectively. R will then pass through O as shown. 



126 



MATERIALS AND STRUCTURES 



It will be clear that, in the case of a uniform normal stress, the 
centre of volume of the stress figure lies in the normal drawn from the 
centre of area of the section. It therefore follows that, if a resultant 
normal force acts through the centre of area of a given section, a 
stress which will be distributed uniformly over the section will result. 

In drawing stress figures, a useful convention is to draw the stress 
figure standing on one side of the section, for those parts of the section 
which are subjected to push stress, while pull stresses are represented 
by a stress figure standing on the other side of the section. 

Shearing stress. In Fig. i53(#) is shown a rectangular plate 
ABCD having shearing stress pt distributed over its top edge. Let 




(a) 



(b) ' 

FIG. 153- A plate under shearing stress. 



H P t 



the thickness of the plate from front to back be unity, then the total 

force along A B will be P=/,xAB (i) 

Substituting P as shown in Fig. 153^), the plate may be 
equilibrated horizontally by the application along CD of an equal 
opposite force P ; as P, P form a couple, equilibrium is completed by 
the application of equal opposite forces Q, Q along the edges AD and 
BC respectively, these forming a couple of moment equal and oppo- 
site to that of the first couple. For equilibrium we have 

Px AD = Qx AB (2) 

Let all these forces be produced from shearing stresses applied to 
the edges of the plate (Fig. 153 (r)), and let qt be the shearing stress 
which gives rise to Q, so that 

Q = #xAD (3) 

Substituting in (2), we have 

pt x AB x AD =q t x AD x AB, 
or pt = qt (4) 

For the general equilibrium of the plate it is therefore necessary that 
equal shearing stresses be applied to all four edges. 

Take any section EF of the plate as now stressed (Fig. 153^")), 
and consider the equilibrium of the portion ABFE (Fig. 154). From 
what has been said it will be seen by inspection of Fig. 154 that a 



SHEARING STRESSES 



127 



shearing stress.// must act along FE. Again, take another section 
GH (Fig. 153 (c)), and consider the equilibrium of the portion AGHI) 
(Fig. 155). Inspection shows that a shearing stress /< must act along 




FIG. 154. 

GH. We conclude that if any rectangular block be subjected to 
shearing stresses, such stresses must be equal on all four edges, and 
there will be an equal shearing stress on any section which is parallel 
to any edge of the block. 

Cube under shear stress. For simplicity, consider a cube of unit 
edge, the elevation of which is ABCD (Fig. 156). Let shearing 
stresses pt be applied as shown to those 
faces of the cube which are perpendicular 
to the paper. To find the stress on the 
diagonal section AC, cut the cube and 
consider the portion ABC (Fig. 157). 
The stresses along AB and CB produce 
forces /, p t9 acting at B; these will have 
a resultant r, acting at 45 to AB, and 
hence perpendicular to AC. The mag- 
nitude of r will be 

/=/. N/2. 

If r be produced it will evidently cut the diagonal AC at its middle 
point O, and may be balanced by an equal opposite force r applied 
at O as shown. Now r may be considered to be the resultant of a 
normal stress p n uniformly distributed over the diagonal section AC, 
the intensity of this stress being 




FIG. 156. Cube under shear 
stress. 



_ 

AB.x/ 

=Pt- 

This result shows that the diagonal AC is subjected to a normal 
pull stress of intensity equal to the given shearing stress. In the 
same way, by considering the portion ABD (Fig. 158), we may show 



128 



MATERIALS AND STRUCTURES 



that the diagonal BD is subjected to a normal push stress / w of inten- 
sity also equal to the given shearing stress. 

Supposing we have a rectangular plate ABCD (Fig. 159) having 
shearing stresses pt applied to its edges. Consider any square portion 





FIG. 157. The diagonal AC is under pure 
normal pull stress. 



FIG. 158. The diagonal BD is under pure 
normal push stress. 



abed having its edges parallel to the sides of the rectangle. We have 
already seen that these edges have equal shearing stresses p t acting 

on them. Hence the diagonal 
sections of the square have normal 
pull stress on ac and normal push 
stress on &d, the intensity of each of 
these being//;. We therefore infer 
that any section of the plate at 45 
to an edge will have normal stress of 
push or pull acting on it of intensity 
equal to the given shearing stress, 
intersecting at 90, and having purely 




'/J ' 

FIG. 159. Stresses in a rectangular plate. 

Two sections of a body 



normal stresses acting on them, are called principal axes of stress ; the 
stresses are called principal stresses. 

[For laboratory experiments on stress and strain, see Chapter XIII.] 



EXERCISES ON CHAPTER 



VI. 

Find the diameter if 



1. A round rod has to carry a pull of 15 tons, 
the safe stress is 6 tons per square inch. 

2. A short hollow cast-iron column is 6 inches in external and 4^ inches 
in internal diameter. Calculate the safe load if the stress allowed is 
7 tons per square inch. 

3. Plates 0-5 inch thick are to be connected by a double-riveted lap 
joint. Find the principal dimensions of the joint. Take d= \-2*Jt : /=6, 
/s = 5,/6= 10, in tons per square inch. Find the efficiency of the joint. 



EXERCISES ON CHAPTER VI. 129 

4. Answer Question 3 for a double-riveted butt joint with two cover- 
straps. The plates are | inch thick. Allow 1-75 rivet sections per rivet 
under shear. 

5. Two plates, each 16 inches by 0-5 inch thick, are to be connected 
by a butt joint having two cover-straps. The joint is to be under pull. 
Take stresses as given in Question 3, and find the required number of 
rivets |- inch in diameter. What would be the safe load for the joint ? 

6. A cylindrical boiler shell is 7-5 feet in diameter ; the working 
pressure is 150 Ib. per square inch. If the efficiency of the longitudinal 
riveted joint is 75 per cent., find the thickness of the plate for a safe stress 
of 5 tons per square inch. What will be the stress on a longitudinal 
section of the plate at some distance from the joint ? Find also the stress 
on a circumferential section of the plate. 

7. A spherical vessel, 6 feet in diameter, is subjected to an internal 
gaseous pressure of 120 Ib. per square inch. Find the thickness of plate 
required for a joint efficiency of 70 per cent, and a safe stress of 12,000 Ib. 
per square inch. 

8. A steel bar, 6 inches wide, 0-5 inch thick and 30 feet long, carries a 
pull of 1 8 tons. Find the extension in length and the contractions in 
width and thickness when the load is applied. Take E = 13,500 tons per 
square inch and m = y^. 

9. A vertical square plate of steel, 6 feet edge and 0-75 inch thick, has 
shearing forces of 200 tons acting along each edge. Suppose the lower 
edge to be horizontal and to be fixed rigidly, what will be the horizontal 
movement of the top edge when the load is applied ? Take C = 5500 tons 
per square inch. 

10. A cylinder for storing compressed oxygen under a pressure of 120 
atmospheres is 3 feet long and 5 inches diameter ; the thickness of the 
steel plate of which it is constructed is | inch. Find the alterations in 
diameter and length when the cylinder is being charged, and hence find 
the change in cubic capacity of the cylinder. Take = 13,000 tons per 
square inch and 7/2 = 4. 

11. A rod of brass 4 feet long and 0-5 inch diameter is cooled from 
1 50 F. to 60 F. Find what forces are required in order to prevent any 
change in the length. Take = 5700 tons per square inch and the 
coefficient of expansion = 0-00001. 

12. A steel boiler tube is 1 5 feet long, 3 inches internal diameter and is 
made of metal 0-3 inch thick. Supposing that half its natural expansion 
due to a range of temperature of 240 F. is prevented, what forces will the 
tube exert in the direction of its length ? What will be the stress in the 
tube ? Take E = 13,500 tons per square inch and = 0-000007. 

13. A tube of copper i 5 inch bore and 4 feet long, of metal o- 1 inch thick, 
has an internal steel rod 0-5 inch diameter, having swelled ends to which 
the tube is brazed. Suppose there to be no self-stressing at first, what will be 
the stresses in the copper and in the steel if both are raised in temperature 
to an extent of 100 F. ? Take E s = 13,500 and E c = 62oo tons per square 
inch ; coefficient of expansion of steel = 0-000007 and of copper = 0-0000096. 

14. A reinforced concrete column has a square section of 15 inches 
edge, and has four reinforcement bars of steel 1-5 inches diameter. Find 
the safe load if the stress in the concrete is 500 Ib. per square inch. How 
much of this load is carried by the steel ? Take the ratio of Young's 
modulus for the steel and for the concrete to be 15. 

D.M. I 



1 30 MATERIALS AND STRUCTURES 

15. A tie bar has a rectangular section 4 inches by 1-5 inches, and 
carries a pull of 30 tons. Find the normal and tangential stresses on 
sections making angles of o, 30, 45, 60 and 90 with the axis of the 
bar. Plot curves showing the relation of the stresses and angles. 

16. Draw the stress figure for a rectangular section 30 feet by i foot ; 
there is a normal push stress of 4 tons per square foot at one short edge, 
and the stress varies uniformly to a normal push stress of 0-5 ton per 
square foot at the opposite edge. What is the resultant force on the 
section ? Show where it acts. 

17. A ferro-concrete column is 14 inches square in cross section ; the 
main reinforcement consists of four longitudinal 2-inch diameter round 
steel rods, one rod being placed close to each angle of the cross section. 
The value of E (Young's modulus) for the steel is 29,000,000 Ib. per square 
inch and for the concrete 3,000,000 Ib. per square inch. If a gross com- 
pressive load of 60 tons is supported by this column, what is the gross load 
and the compressive stress per square inch in (a) the concrete, (b} the 
reinforcing bars ? (B.E.) 

18- A column which carries a load of 300,000 Ib. rests on a foundation 
whose area is 10 square feet ; find the normal and tangential components 
of the stress on a plane in the foundation, whose inclination to the 
horizontal is 15. Find also the inclination of the plane on which the 
tangential stress is a maximum, and calculate this maximum value. (L.U.) 

19. The London Building Act, 1909, allows stresses in steel of 5^ tons 
per square inch in shear and 1 1 tons per square inch of bearing area, but 
limits the shearing strength of a rivet in double shear to 1-75 times that of 
a like rivet in single shear. Prepare a table of rivet strengths, with these 
stresses, for i-inch rivets in single and double shear with plates of f inch, 
| inch, | inch, f inch and f inch in thickness. (I.C.E.) 

20. A cylinder, 8 inches external and 4 inches internal diameter, has an 
internal fluid pressure of 2000 Ib. per square inch. Find the maximum 
and minimum hoop tensions. 



CHAPTER VII. 



STRENGTH OF BEAMS. 

Some definitions. Beams are parts of a structure, usually supported 
horizontally, for the purpose of carrying loads applied transversely to 
their lengths. The term beam or joist is understood generally to refer 
to a structure of moderate size and constructed of one piece of 
material, such as the timber beams or joists used for supporting 
floors, or rolled steel beams also often used for floors. Beams of 
larger size and constructed of several parts secured together are 
called girders. 

Any beam will bend when loaded, owing to the strains which take 
place in the material. If straight initially, it will take the shape of 
some curve ; if curved initially, it will alter its curvature. The theory 
of the strength and stiffness of beams may be developed from the 
fundamental principles that (a) the beam as a whole is in equilibrium 
under the action of the external forces, which term embraces the 
applied loads and the reactions of the supports ; (b) any portion of 
the beam lying between two sections is in 
equilibrium under the action of any external 
forces applied to that portion, together with 
the stresses communicated across the sections 
from the other parts of the beam. 

Pure bending occurs when the following 
conditions are complied with, (a) There 
must be no resultant push or pull along the 
beam due to the action of the external forces ; 
this condition will be realised in the case 
of a horizontal beam carrying vertical loads 
and so supported that the reactions are 
vertical, (b) The external forces must be all 
applied in the plane in which the beam bends. 

T . -ii-i T FIG. 160. Unsymmetrical and 

111 Connection With the latter Condition, It symmetrical angle sections. 



(a) 




132 



MATERIALS AND STRUCTURES 



may be explained here that it does not follow necessarily that a 
beam carrying vertical loads will bend in a vertical plane. Side or 
horizontal bending as well as vertical bending will occur if the beam 
section be not symmetrical about a vertical line passing through the 





(e) 



FIG. 161. -Examples of symmetrical and unsymmetrical sections. 



w 















p 


(a) 


GL 



centre of area of the section. For example, the angle section shown 
in Fig. 1 60 (a) is not symmetrical about the vertical ab, and hence 
pure bending cannot occur with vertical loads. If the angle be 
situated as in Fig. i6o(), symmetry about ab is secured, and pure 
bending will occur, i.e. the beam when loaded vertically will bend in 

the vertical plane, of which ab 
is the trace. Figs. 161 (a\ (b) 
and (c) show other examples of 
symmetrical sections. An un- 
symmetrical bulb angle and Z 
bar are shown in Figs. 161 (d) 
and (e). Pure bending alone 
will be considered. 

Nature of the stresses in a 
beam. In practice, the problem 
which has to be solved first is 
generally that of finding the re- 
actions of the supports for given 
loading. In simple cases of pure 
bending, in which the beam rests 
on two supports, but is not fixed, 
the solution may be obtained by 
the methods given in Chaps. III. 
and IV. We now proceed to examine the stresses in the material of 
a loaded beam. The nature of these may be understood by considera- 
tion of the beam shown in Fig. 162 (a), which carries a single load W, 
and is supported at its ends. Supposing a number of saw cuts to be 
made in the lower portion of the beam (Fig. 162 (b} ), it is evident that 
these will tend to open out on the beam being loaded. Had the saw 




FIG. 162. Longitudinal tension and compression 
in a loaded beam. 



BENDING MOMENTS AND SHEARING FORCES 



133 



FIG. 163. Shearing tendency in a loaded 
beam. 



cuts been made in the upper portion (Fig. 162 (c)), it is clear that these 
would tend to close on loading the beam. We are therefore justified 
in concluding that longitudinal fibres situated in the lower portion 
of this beam are under pull, while 
those lying in the upper portion are 
under push. 

Again, it will be evident that if a 
vertical section AB be taken ( Fig. 1 63), 
there is a tendency for the left-hand 
portion to slide upwards and for the 
right-hand portion to slide downwards, 
indicating that th^re must be shear stresses acting on the section. 

Bending moment and shearing force. Let the beam shown in 
Fig. 164 (a) be cut at any section AB, and consider the problem of 
restoration to equilibrium of the left-hand portion (Fig. 164 (b) ). In 
general, the external forces will not be in equilibrium unaided, hence 
stresses will be required at the section AB. Whatever may be the 
magnitudes and directions of these stresses, they may be resolved into 
components along and perpendicular to AB, and their resultant 

forces X, Y and S substituted for the 
actual stresses. The problem may 
now be solved by application of the 
equations (p. 64), denoting horizontal 
and vertical forces by the suffixes x 
and y respectively : 

SP.-o, (i) 

2P y = o, (2) 

= o (3) 



K _K 







p B (a) 


Q 












p 


S' 



Y Since there are no forces other than 

X and Y acting along the beam, it 

FIG. 164. Bending moment and shearing follows from equation (l) that these are 
force at a beam section. 111 i r i 

equal, and hence they form a couple. 

Equation (2) shows that the algebraic sum of the forces parallel to 
AB must be zero, and hence S must be equal to the algebraic sum of 
the external forces applied to the portion of the beam under con- 
sideration. S is called the shearing force, and will produce shear 
stress distributed in some manner over the section AB. 

The meaning of equation (3) may be ascertained by taking 
moments about any axis in the section AB, the axis beipg perpendi- 
cular to the plane of bending and indicated by O in Fig. 164 (b). The 



134 MATERIALS AND STRUCTURES 

second term clearly refers to the resultant moment of the external 
forces applied to the portion of the beam considered (notice S has 
no moment about this axis) ; the first term refers to the moment of 
the couple produced by the equal forces X and Y. The equation 
shows that these moments must be equal. The resultant moment of 
the external forces is termed the bending moment, and the moment of 
the couple is termed the moment of resistance. 

Equation (3) may thus be read : 

Bending moment at AB = moment of resistance at AB. 

It will be evident, since the forces X, Y and S are communicated 
as stresses from the right-hand portion to the left-hand portion of the 
beam, and hence are mutual interactions, that their values would be 
unaltered had the calculation been performed by considering the 
right-hand portion of the beam instead of the left-hand portion. 
Hence the bending moment and shearing force at any section may 
be calculated from the loads and reactions applied to either portion of 
the beam. If the calculations be made for both portions the results 
should agree, thus affording a check on the accuracy of the work. 

Rules for bending moment and shearing force. The bending 
moment at any section of a beam means the tendency to rotate either 
portion of the beam about that section, and is calculated by taking 



FIG. 165. Positive and negative bending. FIG. 166. Positive and negative shear. 

the algebraic sum of the moments about the section of all the forces 
acting either on one or other portion of the beam. 

The shearing force at any section of a beam means the tendency 
of one portion of the beam to slide on the other portion, and is 
calculated by taking the algebraic sum of all the forces acting 
either on one or other portion of the beam. 

It is usual to call bending moments positive when the tendency is 
to cause the beam to become convex downwards, as in Fig. 165 (a). 
Fig. 165^) shows a case of negative bending moment. Shearing 
forces are denoted as positive if the tendency to slide is that shown 
in Fig. 1 66 (a), and negative if that in Fig. 166 (b). 



BENDING MOMENTS AND SHEARING FORCES 



135 



Bending-moment and shearing-force diagrams. Such diagrams 
are often required in the solution of beam and girder problems, and 
may be drawn by first calculating the values of the bending moments 
and shearing forces at a sufficient number of sections of the given 
beam. A horizontal datum line is chosen of length to scale to 
represent the length of the beam ; the calculated values are then set 

I ton per foot length 



rrr 


1 




i I 


TH^T 




b 2 4 

\\0tons 


6 


8 


10 12 

ftW 


14 \& 18 ^ 

I0ro7?5 


20 




FIG. 167. Bending moment and shearing force diagrams for a beam carrying a 
uniformly distributed load. 

off as ordinates, above or below the datum line according as they are 
positive or negative. The ends of the ordinates being joined by 
straight lines, or a curve depending on the circumstances, the result 
gives complete representations of the bending moments and shearing 
forces throughout the beam. 

EXAMPLE. A beam of 2o-feet span is supported at its ends and carries 
a uniformly distributed load of i ton per foot length (Fig. 1,67 ()). Draw 
bending-moment and shearing-force diagrams. 



136 



MATERIALS AND STRUCTURES 



To do this, first calculate the bending moments and shearing forces at 
sections 2 feet apart throughout the length of the beam. The reaction of 
each support will be 10 tons. Sample calculations are given below for the 
section 6 feet from the left-hand support, together with a complete table 
of the results from which the diagrams in Fig. 167^) and (d) have been 
plotted. As the loading is continuous, it is evident that both the bending 
moment and shearing force vary continuously ; hence neither diagram 
shows any break or sudden change in direction. 

For section 6 (Fig. 167 (b)\ 

Bending moment = (iox6)-(6x 3) 
= 60-18 

= 42 ton-feet, positive. 
Shearing force = 10 6 

= 4 tons, positive. 



Section. 


Bending moment, 
ton feet. 


Shearing force, 
tons. 


Section. 


Bending moment, 
ton -feet. 


Shearing force, 
tons. 








+ 10 


10 


+ 50 


O 


2 


+ 18 


+ 8 


12 


+ 48 


-2 


4 


+ 32 


+ 6 


H 


+ 42 


-4 


6 


+ 42 


+ 4 


16 


+ 32 


-6 


8 


+ 48 


+ 2 


18 


+ 18 


-8 


10 


+ 50 





20 





- 10 



Diagrams of bending moment and shearing force for four important 
cases are given in Fig. 168. These cases are of constant occurrence 
in practice, and should be worked out independently by the student. 

Shearing force at a concentrated load. Any difficulty which may 
occur in dealing with the shearing force at a concentrated load will 
disappear if it is remembered that there is never any case of 
a load being concentrated on a geometrical point, or line. This 
arises from the fact that such would produce an infinitely great stress, 
the area being zero. All loads are distributed really over a small 
portion of the length of the beam. In Fig. 169 (a), a load W is 
shown resting on a beam, and it may be convenient for some pur- 
poses, such as the calculation of the reactions, to speak of it as 
concentrated at its centre of gravity C ; actually it is distributed over 
a short length DE of the beam. The shearing force at any section 
lying between A and D will be positive and equal to P ; for any 
section between B and E the shearing force will be negative and 
equal to Q. For sections lying. between D and E, the shearing force 
will be + P at D, and will gradually diminish to zero, then will 



BENDING MOMENTS AND SHEARING FORCES 



change sign to negative, and will increase numerically to - Q at E. 
The section at which zero shearing force occirrs may be determined 




w 



+ 




-W 


s 


.i 




(a) 


Jw 




|*-.x-->| 


* 


^. . . , i w i" 



M 






W per 


unit 


length 












~l 











2 


/ 












wCiL-x) 




FIG. 168. Bending moment and shearing force diagrams for four important cases. 

from the consideration that the portion of W lying to the left of the 
section must be equal to P. Thus : 

PxAB = WxCB; 



Let F be the section of zero shear, then 

DF : P = DE : W ; 

P- DF W 
~DE' 



Hence 



or 



DE~AB 
DF:DE = CB": AB. 



.(3) 



133 



MATERIALS AND STRUCTURES 



We infer from this result, that the section of zero shear divides the 

load into segments which are 
inversely proportional to the 
segments into which the centre 
of gravity of the load divides 
the beam. The shearing-force 
diagram for this case is shown 
in Fig. 169 (I)}). 

In solving problems of this 
character, it is usually sufficient 
to state the shearing forces on 
each side of the load given as 

FIG. ^.-Shearing force at a load. Concentrated. 

EXAMPLE. Draw the bending-moment and shearing-force diagrams 
for the beam shown in Fig. 170 (<*). 

2 tons 



Shearing Forces 




5 ton s 
1*5 tons per foot 



* * 



10 



20i 
16- 
12 
61 
4 



P-ltttom 
Ton- ft. 



(a) 




Bending Moments 
Feet 



-16- 
-20 
-24 



Jons 



5- 



(b) 



-5 



S 'hearing Forces 




I0 



FIG. 170. Bending moment and shearing force diagrams for a loaded beam. 



BENDING MOMENTS AND SHEARING FORCES 



139 



Sections at 2-feet intervals have been chosen, and the calculations have 
been made in each case by considering the left-hand portion. Clockwise 
moments have been considered as positive and anti-clockwise as negative, 
thus giving the proper sign for the results of the bending-moment calcu- 
lations. Forces acting upwards have been taken as positive and down- 
ward forces as negative, giving the proper sign for the shearing-force 
results. The calculations are given in the table, and the diagrams have 
been plotted from the results as shown in Fig. ijo(b} and (c). Two 
results are given for the shearing force at the 6-feet and the 1 2-feet 
sections ; the first is that immediately to the left of the section, the 
second is that just to the right of the section. The shearing force at 
16 feet from P is that immediately to the left of the 5-ton load. 

BENDING MOMENTS AND SHEARING FORCES FOR A LOADED BEAM. 



Distance 
of section 
from P, 
feet. 


Bending moment, 
ton-feet. 


Shearing force, 
tons. 





O 


P-+7-33 


2 


(P x 2) - (3 x i)= + 11-67 


P-3= +4-33 


4 


(Px 4) -(6x2)= + 17-33 


P-6= + i-33 


6 


(Px 6) -(9x3)= + 16-99 


j P-9=-i.6; 


8 


(Px 8) -(12x4) -(2x2) =+6-67 


P- 12-2= -6-67 


10 


(P x 10) - (15 x 5) - (2 x 4) = - 9-67 


P-i5-2= -9-67 


12 


(Pxi2)-(i8x6)-(2x6)=-32 


j P-i8-2= - 12-67 
\P-i8-2 + Q= + ii-o 


14 


(P x 14) - (21 x 7) - (2 x 8) + (Q x 2)= - 13 


P-2I-2 + Q=+8-0 


16 


(Px i6)-(24x8)-(2x io) + (Qx4) = o 


P-24-2 + Q= +5-0 



Graphical methods of obtaining the bending-moment diagram. In 

Fig. 171 (a) is shown a beam carrying two loads W 1 and W 2 . The 
reactions of the supports P and Q have been determined by means 
of the force polygon shown in Fig. 171 (b), and the link polygon, 
Fig. 171 (c\ as has been explained on p. 70. It will happen usually 
that the closing link ab of the link polygon is not horizontal, and it is 
convenient for our present purpose that it should be so. To obtain 
this result, the pole O of the force polygon in Fig. 1 7 1 (b) has been 
moved vertically to O' in. the horizontal line through A. A new link 
polygon (Fig. i t ji(d)) is then drawn, having its sides parallel to the 
dotted lines radiating from O' in Fig. 1 7 1 (b) ; ab' will now be horizontal. 
The triangles a'ed' and O'AB are similar; hence 

^v = A:B 

a'e ~ O'A 
or d'exO'A = ARxa'e (i) 



140 



MATERIALS AND STRUCTURES 



Now AB represents the reaction P ; hence AB x de represents the 
moment of P about the section at W v i.e. represents the bending 
moment at W r Therefore the ordinate d'e of the link polygon, when 
multiplied by the horizontal polar distance O'A, gives the bending 
moment at Wj. 

In the same way, from the similar triangles c'fb' and DAO' we may 
show that c'fy. O'A represents the bending moment at W 2 . Therefore, 
the link polygon drfc'b' is the bending-moment diagram for the 
whole beam. 

To obtain the scale of the diagram, it will be noted that both d'e 
and AB in (i) above should be measured to the scale of force used 
in drawing the force diagram, Fig. 171 (b); also, both de and O'A 




-- -cc 

FIG. 171. Bending-moment diagram by the link polygon method. 

should be measured to the scale of length used in drawing the beam 
in Fig. 171 (a). Let these scales be/ tons per inch height of DB in 
Fig. 171 (b) and / feet per inch length in Fig. 171 (a). Then, if any 
ordinate y of Fig. 171(^0 be measured in inches, and if O'A be 
measured also in inches, the bending moment at the section of the 
beam vertically above y will be given by 

M x =y. O'A .//ton-feet (2) 

Another useful graphical method of obtaining the bending-moment 
diagram is illustrated in Figs. 172 (a) and (b). A base line OA is 
selected of length equal to that of the beam. Choosing a convenient 
scale of moments, AB is set off equal to PL, and is divided at E by 
setting off BE equal to W^. The remainder EA of BA will 
evidently be equal to W 2 o 2 , as is shown by the equation of moments 
about the right-hand support, viz. : 

(i) 



BENDING MOMENTS AND SHEARING FORCES 



141 



Join OB cutting the vertical through W l in C ; join CE cutting W 2 
produced in F; join FA. Then OCFA is the bending-moment 
diagram for the complete beam. 



w, 




,, , G A 

( & ) 

FIG. 172. Bending-moment diagram by the method of graphical moments. 

To prove this, take any ordinate y 1 . From similar triangles, we 
have A"R M * 



AB 



.x,. 



Now AB = P x L and OA = L ; hence 



that is, y 1 represents the moment of P about the section of the beam 
vertically over y l ; hence OCD is the bending-moment diagram for 
the portion of the beam lying between P and W 1 . In the same way, 
it may be shown that y z represents the moment of W 1 about the 
section vertically overj 3 ; y 2 represents the moment of P about the 
same section, and has the opposite sign to that of W 1 ; hence (y z -y s ) 
is the bending moment for this section. Similarly (y 4 y 6 y$) is the 
bending moment for the section vertically over jy 4 . 

In applying either of these graphical methods to the case of 
distributed loads, these loads may be cut up into portions of short 
length and the weight of each concentrated at its centre of gravity. 
The result will give a nearly equivalent system of concentrated loads. 



142 



MATERIALS AND STRUCTURES 



Bending of a beam. Suppose we have a beam consisting of a 
number of planks of equal lengths laid one on the other, and sup- 
ported at the ends. A load W, applied at the centre of the span, will 
cause all the planks to bend in a similar fashion, and, as their lengths 
will remain equal, the planks will overlap at the ends as shown 
(Fig. 173 (a) ). Strapping the planks firmly together will prevent this 




FIG. 173 (). Plank beam. 



FIG. 173 (). Strapped plank beam. 



occurring, and the beam will now bend as a whole, the ends of the 
planks remaining in one plane (Fig. 173 (b) ). The upper planks have 
become shorter and the lower planks longer ; hence, one intermediate 
plank will be unaltered in length. Assuming the middle plank to 
remain the same length as at first, it is clear that all planks above 
the middle must have become shorter, and all below the middle, 
longer than at first. It will also be evident that the change of 
length, and consequently the longitudinal strain, of any plank will 
depend on its distance above or below the middle, being greater as 
the distance is increased. 

For ordinary practical beams, it is assumed that no section is 
warped when loads are applied ; thus transverse sections which were 
plane in the unloaded beam remain plane when the loads are applied. 
While this assumption is justified on appeal to experiment, it must be 

noted that it is no longer true if the 
beam has been loaded excessively 
so that the elastic limit of the 
material has been exceeded. 

Some important definitions. In 
Fig. 174 is shown a portion of an 
unloaded beam. We have seen already that there will be one 
longitudinal section which will not suffer change of length when the 
beam is loaded; let NL represent this section, which is called the 
neutral lamina. Any plane transverse section, such as AB or CD, 
will intersect the neutral lamina in a straight line, which is shown by 
NA in the cross section; this line is called the neutral axis of the 
section. 

Longitudinal strains. In Fig. 175 (a) is shown a portion of a bent 
beam. Two adjacent and originally parallel sections AB and CD 



t 




FIG. 



B D 
174. Neutral axis of a beam section. 



STRAINS AND STRESSES IN BEAMS 143 

have been altered in position by the bending to A'B' and C'D'. ab 
is any longitudinal fibre parallel to the neutral lamina NL, and has 
been changed in length from ab to ab ', the change being one of 
shortening if ab lies on the concave side of NL and of extension if 
ab lies on the convex side. The actual change of length is made up 
of the two pieces aa and bb'. It is clear from the geometry of the 





B' B D 



FIG. 175. Longitudinal strains and stresses in a beam. 

figure that the combined length of these pieces will be proportional 
to the distance of ab from NL ; thus : 

(aa' + bb'} : (AA' + CC) = Ea : EA. 
The strain of ab will be given by 

aa + bb' 



Strain of ab = 



Also, Strain of AC = 



ab 
AA' + CC' 



AC 

Now all fibres lying between AB and CD were originally of equal 
lengths, viz. EF ; hence their strains are proportional simply to their 
changes in length, and hence to the distances of the fibres from NL. 
We may therefore write, taking y and m to be the distances respec- 
tively of ab and AC from NL : 

Strain of ab : strain of AC y : m, 

strain of any fibre 

or -T. r -, J c ^pp- = a constant. 

distance of fibre from NL 

Longitudinal stresses. Changes of length of any fibre must have 
been brought about by longitudinal stresses of push or pull, depend- 
ing upon whether shortening or extension has been produced. Thus 
ab' in Fig. 175 (a) must be under longitudinal push ; any fibre lying 
on the convex side of NL will be under longitudinal pull. Assuming 
the elastic limit not to be exceeded, these stresses will be proportional 
to the strains. Hence, from what has been said above regarding 
the strains, the longitudinal stress on any fibre will be proportional 
to its distance from NL. 






144 



MATERIALS AND STRUCTURES 



Let 
Then 



f= longitudinal stress on A'C' (Fig. 175 (^)), 

P 5> 



or 



m y 



a constant. 



The student will observe that fibres under longitudinal push stress 
not only shorten, but also expand laterally, while those under pull 
stress contract laterally. The ordinary theory of beams assumes that 
such lateral changes take place freely, the justification being that 
calculations based on the ordinary theory agree 
very closely with experimental results. The effect 
of the lateral changes on the section of a beam bent 
convex downwards will be understood by reference 
to Fig. 176, in which the lateral contractions of the 
lower fibres and the lateral expansions of those 
above the neutral lamina have the effect shown of 
causing the cross section apparently to be bent 
convex upwards, i.e. in the opposite sense to that of the length of the 
beam. The transverse curvature is called anticlastic, and may be 
observed very well if a rubber beam be experimented upon. The 
interference of anticlastic bending with the ordinary theory of beams 
will be most marked with'a very broad beam of little depth, a strip of 
clock spring, for example. 

Moment of resistance. Knowing the nature of the distribution of 
the stresses over the cross section, we may now proceed to find an 



FIG. 176. Anticlastic 
curvature. 




FIG. 177. Moment of resistance of a beam. 

expression for the moment of resistance. Referring to Fig. 177, 
showing a part side elevation and section of a loaded beam under 
pure bending, let a be the cross-sectional area of any fibre. 

Stress on a =p. 
Now p '.f=y ' m ; 



MOMENTS OF RESISTANCE 145 



Also, Force on a =pa = fa 



The force on any other fibre would be obtained in a similar 
manner, and, as these forces will be both push and pull when taken 
over the whole section, we may obtain the resultant force by 
summing algebraically. Thus : 

Resultant force on section ^ ay 

m J 



The factor 2ay simply means the moment of area of the whole 
section about NA, and, as in pure bending there is no resultant 
force along the length of the beam, we may equate equation (3) 

to zero. Now i- will not be zero ; hence 
m 

?ay = o ...................... (4) 

This latter result can only be true provided NA, the axis about 
which moments of area are to be taken, passes through the centre 
of area of the section and is perpendicular to the plane of bending. 
Hence, we have a simple rule for the position of the neutral axis of 
any section. The methods of finding the centres of gravity of thin 
sheets, discussed in Chapter III., may be applied. 

Again, taking moments about NA, and using equation (2) for the 
force on a, we have 

Moment of the force on a = ay x y 

m 



. 

A similar expression would give the moment of the force on any 
other fibre, and it will be noticed that all such moments will have the 
same sign independent of that of y, as the y has been squared in each 
case. The total moment may be obtained by summation, thus : 

Total moment of resistance = 20 y 2 .................. (6) 

wi 

In this result, ^ay 2 may be termed the second moment of area of the 
section, thus distinguishing it from the first moment, which would be 
?ay. The name moment of inertia is applied more commonly to Say 2 . 
arising on account of its similarity to the expression used in cal- 
culating the moment of inertia of a thin plate. 
D.M. K 



146 MATERIALS AND STRUCTURES 

The moments of inertia of many simple sections may be calculated 
easily by application of the methods of the integral calculus. Rolled 
sections are dealt with more easily by a graphical process, which will 
be explained later. Writing I NA for the moment of inertia of the 
section with reference to the neutral axis, and making use of what has 
been said on p. 134, we have 

Bending moment = moment of resistance, 



or 



This expression may be applied by first calculating the bending 
moment at the given section of the beam. It is useful to choose m 
as the greatest ordinate of the section, using NA as a datum line, 
when/ which is the stress on the fibre at a distance m from NA, will 
be the maximum value of the stress on the section. An example will 
render the method clear. 

EXAMPLE. A beam of i2-feet span carries a uniformly distributed 

load of 0-5 ton per foot run, together 

Jn t with a load of 2 tons at 3 feet from 
nS K ton per foot one end (Fig. 178). Given that the 

moment of inertia of the rectangular 
section is 180 in inch units, find the 
greatest stress on the section at the 
FIG. 178. middle of the span, which is 10 inches 

deep. 

To find the reactions, take moments about B (Fig. 178) : 
Total distributed load =6 tons. 



= = 4' tons. 



As a check, take moments about A : 
Qx i2 = (6 



42 

= 3-5 tons. 



= 8 ton s = total load. 

Now find the bending moment at C, thus : 

Mc = (Px6)-(2X 3 )-(3X 3 ) 
= 27-6-9 
= 12 ton-feet 
= 144 ton-inches. 

Or M c = (Qx6)-( 3 x 3 ) 

= 21-9 
= 12 ton-feet, as before. 



MODULI OF BEAM SECTIONS 147 

Again, taking m = 5 inches, we have 



i44= -180, 

/ . = 5 X U4 
J 1 80 
= 4 tons per square inch. 

In solving beam problems it is advisable to take all dimensions for 
bending moments and resisting moments in inches. 

Modulus of a beam section. The modulus of a beam section 
may be denned as the quantity by which the stress intensity at unit 
distance from the neutral axis must be multiplied in order to give 
the moment of resistance of the section. Taking the equation, 

Moment of resistance = I NA , 

let y be unity, and let/ x be the stress corresponding to this value of j>. 
Then Moment of resistance =/ x I NA 

=AZi> 

where Z l is a modulus of the section. 

Another modulus may be obtained by making use of the maximum 
stress form of the equation for the strength of a beam, viz. 

Moment of resistance = I NA 
m 



where Z is the modulus of the section, and is found from 

r 7 _ INA 
f_i - . 

m 

The latter is the more useful form of modulus in practice ; its 
value differs numerically from that of Z 1 . It will be noted that only 
sections which are symmetrical above and below the neutral axis 
will have equal values of m and / for tension and compression. 
Such sections have one value only for the modulus, all others having 
two values, one corresponding to the maximum tensile stress, the 
other to the maximum compression stress. 
Let ft = maximum tensile stress, 

mt = distance of/ from the neutral axis, 

f c = maximum compressive stress, 

m c = distance of f c from the neutral axis. 



148 



MATERIALS AND STRUCTURES 



Then, since the bending moment M at any section equals the 
moment of resistance at that section, we may write 



M - 



where Z = I NA /*0* is the tension modulus. 
Also, M = ^I NA 



where Z c = I NA /w c is the compression modulus. 
These results may be written 

M 



from which it may be inferred that the given safe stresses in tension 
and compression respectively must not be exceeded by the values 
obtained by dividing the bending moment at any section by the 
tension or compression modulus of the section. 

Graphical method of finding the neutral axis and moment of 
inertia of a section. Advantage is taken of the fact that the neutral 

axis passes through the 
centre of area of the sec- 
tion. To illustrate the 
method, reference is made 
to Fig. 179, in which is 
given an irregular figure, 
and it is required to draw 
a line through the centre 
of area parallel to OX, 
and also to find the 
1 moment of inertia of the 
~x figure about OX. 

Draw any convenient 



\ 
\ 


\ ' 
* i / 


/ > 

/ i 


\ 


\ i 

\ I / 




L 




/ (Y f 



e\ \m 



f 



\ I 



FIG. 179. Graphical method of finding the neutral axis and 
moment of inertia. 



JK axis OY perpendicular to 

OX, and take any narrow strip ab parallel to OX. Let the breadth 
of ab be 8y and let y be the distance of ab from OX. The area of 
the strip will be (ab . 8v) and its moment of area about OX will be 

Moment of area of strip = a. fy.y (i) 



POSITION OF THE NEUTRAL AXIS 



149 



Draw cd parallel to OX through the highest point on the figure ; 
draw ac and bd parallel to OY and join cQ and dO, cutting ab in e 
and /respectively. Then, from similar triangles, we nave 



or 



Substituting in (i) gives 






H 



Moment of area of strip = . ef. 8y. y 

= ef.Sy.H ................... (3) 

Now (ef. By) is the area of the strip ef; hence, if the whole section 
were cut into strips such as ab, and the construction repeated for 
each strip, the total moment of area would be given by the sum of 
the areas of the reduced strips such as ^multiplied by the constant 
factor H. In practice, a few breadths only are taken ; the reduced 
breadth for each is found by application of the above construction, 
and a fair curve is drawn through the ends. The area inclosed by 
this curve when multiplied by H will give the moment of area about 
OX of the given figure. Now the moment of area may also be 
found by taking the product of the area of the given figure and the 
distance of its centre of area from OX. 

Let A x = the area of the given figure, in square inches. 
A 2 = the area of the reduced figure, in square inches. 



Then 



y = the distance of the centre of area from OX, in inches. 
H = the height of the figure, in inches. 



-iiO TT / \ 

T-H (4) 



Fig. 1 80 shows the appli- 
cation of this method to a T 
section. The area A T of the 
section and the shaded area 
A 2 of the reduced figure were 
found by use of a planimeter. 
The neutral axis NA is drawn 
parallel to OX and at a dis- 
tance y from it. 




FIG. 180. Neutral axis of a T section. 



150 



MATERIALS AND STRUCTURES 



Referring again to Fig. 179, draw eg and fa parallel to OY, and join 
and hO, cutting ab in m and respectively. Then, from 
similar triangles : ^ H 

mn~ y 9 



or 



mn y 
.'. ef=mn. I 

Now, from the definition, 

IGX of strip #^ = area of strip xj/ 2 



H 



(from (2), p. 149) 



H . mn . . Sy . y 

y 



(from (5) above) 



(6) 

Again, (mn . Sy) is the area of the strip mn ; hence the total moment 
of inertia may be obtained by multiplying the sum of the areas of all 
, such strips by the constant 

factor H 2 . Choose a number 
of strips and repeat the con- 
struction on each, thus finding 
a number of points such as 
m and n. Draw a fair curve 
through them, when its area 
A 3 , multiplied by H 2 , will give 
the total moment of inertia. 

The moment of inertia of 
the same T section is worked 
out in Fig. 181. Greater 
accuracy is secured by using the neutral axis instead of OX in 
Fig. 1 80, thus producing two reduced figures, one for the original 
area above NA and another for that below NA. 
Let 

A 3 ' = shaded area of reduced figure above NA, in square inches. 
A 3 " = shaded area of reduced figure below NA, in square inches. 
Hj = height above NA, in inches. 
H 2 = height below NA, in inches. 
Then, 

Total moment of inertia about NA = A Z 'H\ + A 3 "H 2 2 (7) 




FIG. 181. Moment of inertia about NA of a T section. 



PROPERTIES OF SECTIONS 



PROPERTIES OF SECTIONS. 



Name of 
section. 



Rectangle 



Square 



Square 



Box 



I on side 



Cruciform 



Circle 



Hollow 
circle 



Section. 




IH...H 



4-i 

< I 




U---B ->| 



T-T 



uifi/T 

*- ~0 



e 











Area. 



BD 



BD-^ 



BD- 



2B/ + ^/ 



_/l 



7TR 2 



Distance of 
NA from 
bottom. 



I 
^' 



4B 



BD 3 

12 



BD 3 -^ 3 



12 



12 



(BD 2 -^ 2 ) 2 



7rR_ 4 
4 



D 2 

12 



12 



BD 3 - 



BD 3 - 



I2(2D/-/ 2 ) 



I 



R 2 
4 



152 MATERIALS AND STRUCTURES 

Radius of gyration. The radius of gyration of a section may be 
defined thus : Let k be such a quantity that the product of the area A 
of the section and k^ is equal to the moment of inertia of the section 
with reference to a given axis ; thus : 



Then k is called the radius of gyration of the section with reference 
to the stated axis. The square of its value may be found in any given 
case by first ascertaining the moment of inertia and then dividing by 
the area of the section. There are many cases where the use of k in 
preference to I is advantageous in the working of problems. 

Some commonly occurring sections and their properties are given 
in the Table, p. 151. No fillets or tapers have been taken into account 
in the tabulated results, which will therefore be of service in obtaining 
approximate solutions only in the case of ordinary practical sections. 

A rule, by use of which may be calculated the moment of inertia 
about an axis OX parallel to another axis CX passing through the 
centre of area, is expressed in the equation 



where A is the area of the section and d is the distance between the 
parallel axes. 

Proportional laws for the strength of beams. Suppose we have 
two beams of rectangular sections, both supported at the ends and 
carrying central loads, but of differing dimensions, the following 
equations will hold for the sections at the middle of the span : 

4 m 1 ] \d^ 12 
... 2 f^d* 
W > = 3 LT 



In the same way, 

Hence 

If the beams are made of the same material, the safe stress / x will 
be equal to/ 2 , and we may write 



Measuring the strengths of the beams by the central loads which they 
can carry safely, we may state this result as follows : The strengths of 




BEAMS OF I SECTION 



153 



beams of rectangular sections and of the same material are propor- 
tional to their breadths, to the squares of their depths, and are 
inversely proportional to their lengths. 

Proportional laws for beams of other sections may be obtained in 
a similar manner. Thus the strengths of solid circular sections 
are proportional to the cubes of the diameters, and are inversely 
proportional to the lengths. 

Approximate calculation for beams of I section. The following 
simple method is often used and has sufficient accuracy for many 
practical purposes. Fig. 182 
shows the section and part 
side elevation of a rolled 
beam of I section. The 
approximate moment of re- 
sistance is obtained by con- 
sidering the maximum stress 



K----6 



r 



(a.) 



(b) 



FIG. 182. Approximate moment of resistance for a 
beam of I section. 



intensity f due to bending 

to be distributed uniformly 

over the flanges only, the 

web being neglected excepting for its resistance to shearing. The 

width of each flange being b and the thickness /, the total stress P 

on each flange will be obtained by taking the product of/ and the 

flange area. 

Thus: P=/&/. 

Assuming each force P to act as though concentrated at the centre 
of area of the flange (Fig. 182 (ti)\ and that the distance between the 
centres is d, the moment of the couple formed by P, P, will give the 
moment of resistance. Thus : 

Moment of resistance = d 



This method may be used with fair results for rolled sections, and is 
used more extensively for built-up girders. To obtain the area of 
flange required at any section in such girders, the bending moment 
at the section is first calculated. Let this be M ; then 

M =fbtd 

=fd x area of flange ; 

.'. area of flange = -^ 

In order to secure the most economical results in built-up girders, 
f should be constant throughout the girder. This result may be 



154 MATERIALS AND STRUCTURES 

obtained by either of two methods : (a) d may be made proportional 
to M, in which case the area of the flange will be uniform throughout 
the length ; (ti) d may be constant and also the breadth b of the 
flanges; in this case the thickness of the flanges is increased by 
using two or more plates riveted together and extending along a 
portion of the length of the girder, more plates being used where the 
bending moment is greatest. 

It may be shown that, in beams of I section, the distribution of 
shear stress is practically uniform over the web \ hence, if S is the 
shearing force in tons and A w is the area of the web in square inches, 
then g 

Shearing stress = q = tons per square inch. 

AM; 

Beams of uniform strength. A beam is said to have uniform 
strength when the maximum stress intensity is the same for all 
cross sections. Considering the equation 

M = ^I, 

m 

/=Mf, 

m and I depend on the dimensions and shape of the section, and if 
these are constant throughout the beam, the only condition under 
which uniform maximum stress intensity f will occur is that M must 
be constant. Uniform bending moment may be produced in a por- 
tion of a beam by the application of couples. For example, if P 
and W be equal in the carriage axle shown on p. 186, then the 
bending moment throughout CD will be equal to the moment of 
the couple W x BD, and hence will be constant. 

More usually M is not constant, in which case uniform f may be 

obtained by varying the section in such a manner that M -- is 
constant. 

EXAMPLE i. In Fig. 183, let AB be a cantilever carrying a load W at 
B. Supposing that the cantilever has a rectangular section of uniform 
depth d) what must be the profile in the plan in order that uniform 
strength may be obtained ? 

d 
Here m = -, 

.}b& 

~I2 ' 

m 6 



BEAMS OF UNIFORM STRENGTH 



155 



Again, the bending moment at any section distant x from B is 



or 



For uniform strength, M-r- = a constant ; 

6 

.*. Wx . -7-75 = a constant. 
bd l 

-,=& constant ; 

/. b=xx a constant. 
The required profile in the plan will therefore be triangular (Fig. 183). 



U 




FIG. 183. 




FIG. 184. 



EXAMPLE 2. Supposing in Example i that the breadth had been 
uniform, and that it is required to find the profile in the elevation for 
uniform strength. As before, we have (Fig. 184) 

W;r. -7-75 = a constant ; 



:. -y2 = a constant, 

*/ 2 =.rxa constant, 

or d= *Jx x a constant. 

Hence the profile is parabolic (Fig. 184). 

EXAMPLE 3. Suppose in Example I that the load is uniformly dis- 
tributed and that the breadth is uniform. 

Find the profile in the elevation for uniform . w P er unit length 
strength (Fig. 185). 



hence iw^r 2 T-F, = a constant, 

bet 2 - 



or -Tg = a constant ; 

:.d=xx a constant. FIG. 185. 

The profile in the elevation is therefore triangular (Fig. 185). 




156 



MATERIALS AND STRUCTURES 



Other cases the student may work out easily for himself. It should 
be noted that, for practical reasons, the profile is often modified 

somewhat from that given by 
G (. AC ^ / 2 H calculation. 

Distribution of the shearing 
stress over a beam section. 
In Fig. 1 86, AB and CD are 
two cross sections of a loaded 
uniform beam, separated by a 
small distance &r. Let the 
shearing force at AB be S and 




FIG. 186. Stress figures due to bending. 



let 



and M 2 be the bend- 



ing moments at AB and CD 
respectively ; also, let M 2 be greater than M r Whatever may be the 
numerical value of the bending moment at AB, that at CD will be 
greater by an amount equal to the moment of S about any point on 
CD. Hence, M -M =S Sx d) 

This result will not be affected by any load which the beam may be 
carrying on AC, as the distance Sx is supposed to be taken of too 
small a value to permit either the magnitude of the load, or its arm 
in taking moments about any point on CD, to attain an appreciable 
value. The reader is here reminded again that all loads must be 
distributed over a definite area ; hence no concentrated load can be 
applied to AC. 

Owing to the bending moments Mj and M 2 , there will be push 
stresses yj and f% at A and C respectively. Let EF be a portion of 
the neutral layer and let m be the distance EA or FC ; then 

MT = I, (2) 

,* X / 



.(3) 



As I and m have the same value for both sections, and since M 2 is 
greater than M 15 / 2 will be greater than /j . The stress figures will be 
AEG and CFH for the portions of the sections AE and CF respec- 
tively. It is clear that there will be a resultant force acting on CF 
which will be greater than that acting on AE ; hence the net tendency 
will be to push the block AEFC towards the left. This block is 
shown separately in Fig. 187 in order that the question of restoring its 
equilibrium may be examined. 



DISTRIBUTION OF SHEAR STRESS 



157 



When the block forms a part of the beam it is clear that the only 
place where horizontal stresses may be applied in order to balance 
the resultants ~F l and F 2 is the horizontal section EF. Let Q be 
the total force produced by these stresses ; then, for equilibrium, 

Q = F 2 -F X (4) 






Q 



I 

i 

...*. 



FIG. 187. Equilibrium of the block AEFC. FIG. 188. Cross section of the block. 

To find the values of F 2 and Y l , let a be a small portion of the 
sectional area of AE (Fig. 188) situated at a distance y from the 
neutral axis and let p be the stress on a ; then 



m y 



or, 



Also, Force on a =pa = ^ 

- '^ ' 4 

.', total force on AE = 
m 



or 



where A is the area of the portion of the section lying above the 
neutral axis, and Y is the distance of its centre of area from the 
neutral axis. AY will be the moment of area about the neutral axis 
of that portion of the section lying above the neutral axis. In the 
same way : 

F = -4\Y (6) 



m 



Hence, 



m m 



(7) 



158 MATERIALS AND STRUCTURES 



Now, from (2) and (3), 

and t^ = 2. 

m I 

Substitution of these in (7) gives 



AV 
-^.(M.-M,) ................ (8) 

AV 
Hence, from (i), Q*=FJeS.to ................... (9) 

Let b be the breadth of the section at NA (Fig. 188) ; then the area 
of the horizontal section over which Q is distributed is (8x x b) ; hence, 
from (9), Q 

Shear stress on EF = F -^ T 
Sx .b 

SAY 



This expression gives the intensity of shearing stress along the 
neutral layer ; it also gives the shearing stress at points on the vertical 
sections AB and CD (Fig. 186) lying on the neutral axis. This may 
be understood by considering the thin rectangular block EFF'E' 
(Fig. 187); if there is a shear stress q on its lower face, there must 
be equal shear stresses on all its faces perpendicular to the 
paper (p. 126). 

UV (Fig. 189) is another horizontal section of the block AEFC. 
The shearing stress on this section arises from the fact that the stress 

figure CHXV for CV has a 
g reater volume than the stress 
figure AGWU for AU. The 
determination of the intensity 
of shear stress on this section, 




FIG. i8 9 .-Shear stresses at U and V. an ^ hence at the points U 

and V on the vertical sections, 

is proceeded with in the same manner as has been detailed above. 
AY in equation (5) will now mean the moment of area about the 
neutral axis of that portion of the section which lies above UV. 
b will be the breadth at U and V, and the final result will be 

SAY 



where I, as before, is the moment of inertia of the whole section. 



DISTRIBUTION OF SHEAR STRESS 



159 



The student will observe that if there is no variation in the bending 
moment, i.e. if M is constant, between two sections of a beam, there 
can be no shearing force and hence no shear stress on the sections. 

EXAMPLE. A beam has a rectangular section 4 inches broad and 
12 inches deep, and has a shearing force of 6000 Ib. (Fig. 190). Find the 



-*- 
6" 




i 




* N 




6" 




4- 






FIG. 190. Distribution of shear stress on a rectangular section. 

shearing stress at the neutral axis and at intervals of 2 inches from the 
BD 3 



neutral axis. 



1 = 



12 

4X 12 X I2X 12 
12 



= 576 inch units. 



At the neutral axis, 



= 24 square inches, 
= 3 inches, 
SAY 



4x576 
= 187-5 Ib. per square inch. 

At 2 inches from the neutral axis, 

A = 4X4=i6 square inches, 
Y=4 inches, 

_6ooox 16x4 
9 *~ 4X576 

= 166-7 Ib- per square inch. 

At 4 inches from the neutral axis, 

A = 2 X4 = 8 square inches, 
Y = 5 inches, 

_ 6000 x 8 x 5 
q *~ 4X576 
= 104-2 Ib. per square inch. 

At 6 inches from the neutral axis, 
A=o; 



i6o 



MATERIALS AND STRUCTURES 



These values have been used in constructing the diagram BCD 
(Fig. 190), the horizontal breadths of which show the shearing stress at 
any point of the section. The diagram is parabolic in outline. Fig. 191 
shows the diagram of shear stress distribution for an I section. The 




FIG. 191. Distribution of shear stress on an I section. 

quantities required for drawing it may be calculated by the same method. 
The result indicates the justification of ignoring the flanges and assuming 
that the web supplies the whole of the shearing resistance by means of a 
uniform shear stress (p. 154). 



EXERCISES ON CHAPTER VII. 

1. A beam 2o-feet span, supported at its ends, carries a load of 4 tons 
at the centre, another of 6 tons at 4 feet from one end, and a third load of 

2 tons at 6 feet from the other end. Calculate the bending moments and 
shearing forces at each load, and draw the diagrams of bending moment 
and shearing force. 

2. A beam AB, 16 feet long, rests on a support at A and on another 
support at C, which is four feet from B. The beam carries a uniformly 
distributed load of 0-5 ton per foot run, together with a load of 4 tons at 
6 feet from A and another of two tons at C. Calculate the bending 
moments and shearing forces at intervals of 2 feet, and draw diagrams 
of bending moment and shearing force. 

3. A beam AB, lo-feet span, supported at its ends, carries a distri- 
buted load which varies uniformly from 100 Ib. per inch run at A to 
200 Ib. per inch run at B. Find the bending moments and shearing forces 
at intervals of 2 feet, and draw diagrams of bending moment and shearing 
force. 

4. Making use of a graphical method, draw the bending-moment 
diagram for the beam given in Question i. State the scale clearly. 

5. Draw the bending-moment diagram for the beam given in 
Question 2, using a graphical method. Give the scale of your diagram. 

6. Find by calculation the neutral axis of a T section 4^ inches broad, 
5 inches deep, metal ^ inch thick. Neglect any fillets. 

7. A cast-iron beam has an I section, in which the top flange is 

3 inches broad, the bottom flange is 7 inches broad and the depth is 
10 inches over all. The metal has a uniform thickness of 0-75 inch. 
Neglect fillets and calculate the position of the neutral axis. 



EXERCISES ON CHAPTER VII. 161 

8. Draw the section given in Question 6 as it would be made in 
practice. Find the neutral axis and moment of inertia, using a graphical 
method. 

9. Answer Question 7 in the manner directed in Question 8, giving 
the neutral axis the moment of inertia. 

10. A timber beam of rectangular section, 3 inches broad by 9 inches 
deep by 12-feet span, carries a uniformly distributed load. Find the load 
if the stress due to bending is limited to 400 Ib. per square inch. 

11. A flat steel bar, section 2 inches by I inch, is 20 feet long, and is 
stored in a rack in which the two supports are each 4 feet from the end of 
the bar. Find the stress due to bending (a) at the middle of the length of 
the bar, (b) at the supports. Suppose the bar to be resting on its edge, what 
would be these stresses ? Take the weight of the material to be 0-28 Ib. 
per cubic inch. 

12. A beam of I section 10 inches deep, 6 inches wide, thickness of 
flanges | inch, thickness of web | inch, has a span of 1 5 feet and rests on 
the supports. If a load of 2 tons is carried at the centre, find the 
maximum stress due to bending (a) by an approximate method, (b) by 
first calculating the moment of inertia. Assuming the shearing force to 
be carried by the web and to be distributed uniformly, find the shear stress 
on the web. Neglect the weight of the beam. 

13. A pipe 24 inches internal diameter is constructed of mild steel plate 
| inch thick, and is full of water ; the ends are closed by blank flanges. 
If the pipe is supported at its ends, find the maximum span if the stress 
due to bending is not to exceed 5 tons per square inch. Take the weight 
of steel to be 0-28 Ib. per cubic inch and of water to be 62-5 Ib. per cubic 
foot. 

14. A timber beam of rectangular section, supported at its ends, carries 
a uniformly distributed load, and has been made to a certain drawing. 
Another timber beam has been made to the same drawing by simply 
altering the scale, so that span, breadth and depth are each multiplied by 
a constant factor n. Suppose both beams to be able to carry the same 
maximum stress due to bending, what will be the ratio of the uniformly 
distributed loads which may be applied ? 

15. A cast-iron bar of rectangular section is used as a beam of 3-feet 
span, supported at the ends, and carries a central load of 3000 Ib. The 
stress due to bending is not to exceed 1-5 tons per square inch. The bar 
is to have uniform strength, (a) Draw the profile in the elevation if the 
breadth is uniform and equal to 1-5 inches, (b) Suppose the depth to be 
uniform and equal to 3 inches, draw the profile in the plan. 

16. Take the data of Question 10, and find the maximum shearing 
stress in the beam. 

17. A beam of I section, 10 inches deep, 5 inches 
wide, metal | inch thick, has a maximum shear stress 
at a certain section of I ton per. square inch. Find the 
shear stress at places i, 2, 3, 4 and 4| inches from the 
neutral axis. Plot a shear-stress diagram. 

18. How is the section-modulus and radius of gyration 
of a section of a bar obtained, and how is this applied 
when ascertaining the strength of a beam ? Calculate 
the section-modulus and radius of gyration of the section 
given in Fig. 192 about the axis YY. (I.C.E.) 

D.M. L 




162 



MATERIALS AND STRUCTURES 



19. A girder AB, 25 feet long, carries three loads of 6, 1 1 and 7 tons 
respectively, placed at distances of 7, 16 and 21 feet from the end A. 
Find the reactions at either end and the bending moment at the centre. 

(I.C.E.) 

20. Fig. 193 represents a station roof, the centre pillars being 25 feet 
apart. The dead load can be taken as evenly distributed over the roof, 



I (on 



2 tons 




\ ton 



and of magnitude 15 Ib. per square foot of projected plan area. The 
wind pressure is to be taken as shown. Find the magnitude and direction 
of the resultant force on the roof, and give the bending moment at the base 
of the pillar. (L.U.) 



CHAPTER VIII. 
DEFLECTION OF BEAMS. 

Curve assumed by a loaded beam. Any beam when loaded will 
bend ; if the neutral lamina is straight, as seen in elevation in the 
unloaded beam, it will assume some curve when the loads are 
applied; any initial curvature of the neutral lamina will be altered 




FIG. 194. Curve of a beam supported at ends and loaded at middle. 

to a new curvature on applying the loads. A useful way of studying 
the curves of a loaded beam is to employ a thin steel knitting 
needle; this may be laid on a sheet of drawing paper and "loaded" 
by means of drawing pins pushed into the board. Figs. 194-196 
show some curves produced in this way. 

Examining Fig. 196, which represents the curve of a cantilever 
carrying a load at its free end, and taking two points P and P x lying 




95. Curve of a beam overhanging the supports. 



close together, two normals drawn from P and P l will intersect in 
O. It is evident that a short piece PPj of the curve could be drawn 
as a circular arc struck from O as centre with radius OP. If P and 
P x are taken very close together, O is called the centre of curvature for 
the curve at P, and OP = R is called the radius of curvature. It can 



i6 4 



MATERIALS AND STRUCTURES 



be seen readily in Fig. 196 that the radii of curvature for points near 
A are smaller than for others near B. In fact, as we shall see 




FIG. 196. Curve of a cantilever loaded at the free end. 

presently, the radius of curvature at any place is inversely proportional 

to the bending moment at that place. 

Curvature is a term used by mathematicians to express the rate of 

change of direction of a curve. Referring to Fig. 197, and taking 

points P and P : lying close together, O will be the centre of curvature 

and R = OP. Draw tangents PT and 
PjTj . The direction of the curve at P 
is along PT, and that at P x is along PjTj ; 
the change of direction between P and Pj 
will be the angle a in the figure. It will 
be evident that the angle PjOP is equal 
to , and stating its value in radians, 

_ PP i 

FIG. 197. Curvature. & -p 

The rate of change of direction may be expressed by dividing the 
change in direction by the distance PPj along the curve in which the 
change is effected ; hence 




Curvature = rate of change of direction = =5- 

"i 



RxPP x 
i 
R' 



CURVATURE OF BEAMS 



165 




FIG. 198. Slope and deflection of a cantilever. 



Curvature at a given point may therefore be stated as being the 
reciprocal of the radius of curvature. The units for curvature will 
be change of direction in radians per foot, or per inch, length of the 
curve according as R is in feet or inches. 

It will be understood that, for ordinary beams which are straight 
when unloaded, the radius of cur- 
vature at any place when the 
beam is loaded will be very large 
and that the curvature will be 
very small. 

Fig. 198 shows again the curve 
AB' of a loaded cantilever. 
Taking any point P on the curve and drawing a tangent PT, the 
angle i which PT makes with the original direction AB is called 
the slope at P ; i should be stated in radians. P is at a distance y 
below AB, andjy is called the deflection at P. For our purposes it 
is sufficient to be able to state R, i and y for any point. 

Curvature of a beam. Fig. 199 shows a portion of a loaded beam. 
Two cross sections occupying originally the positions AB and CD 

have been strained to A'B' and CD'. 
Assuming that they lie close together, the 
point of intersection O will be the centre 
of curvature for the portion EF of the 
neutral lamina. Bisecting EF in M and 
joining OM cutting AC in K, we have 
similar triangles OME and EAA'. Hence, 
using a similar method to that on p. 143, 

EM A A' 




or 



MO 

AK 
R 



EA ! 
AA' 

"3 

m 



8' B D D 

FIG. 199. Curvature of a beam. 



^ _ 

R~ AK' m 



.(i) 



Again, 



Also, 



Strain of AK 



AA' 
AK' 



strain of AK' 



strain <>t\\K 



1 66 MATERIALS AND STRUCTURES 

Substituting in (i), we have 

! = / - 
R E ' m 



Again, 



NA 



_ AB 
Substituting in (2) gives R = ET~ ............................... w 

We see therefore that the curvature at any place on the neutral 
lamina is proportional to the bending moment and inversely pro- 
portional to the moment of inertia of the section at that place. 

Mathematical expressions for the slope and curvature of a curve, 
such as that shown in Fig. 198, are : 



RlDl 

I \dxj j 



.(5) 



dy means the change in deflection as we pass along the beam by 
a small amount dx. For curves which are very flat, and hence for 
all beams, equation (5) simplifies by the denominator becoming 
unity; thus T & y 

R = ^ (6) 

dly _ MAB . , 

Hence, from (3) and (6), ^2~ ElNA ' W 

Slopes and deflections may be calculated from (7) by first evaluat- 
ing the bending moment ; integration of both sides will then give 
the slope ; further integration of both sides will give the deflection. 
The method is rather complicated, excepting in cases where the 
conditions of loading and supporting are simple. 

The following examples of an easier method are given as leading 
to a graphical solution which is simple in its application. It will be 
assumed that the bending is pure and that the beam is of uniform 
section unless the contrary be stated ; the latter assumption is made 
in order that I NA may be constant throughout the length of the beam. 



DEFLECTION OF CANTILEVERS 



167 



Cantilever having a load at the free end. In Fig. 200 (a) is shown 
a cantilever of length L and of uniform cross section, so that I NA is 
constant. We may consider for a moment that the whole of the 
material is perfectly rigid, excepting the portion lying between the two 
adjacent transverse parallel sections AB and CD. Supposing a load W 
to be applied to the free end (Fig. 200 (b)\ deflection of this end will 




FIG. 200. A rigid cantilever having a small elastic portion ABDC. 

take place by reason of the strains in the portion ABDC. AE and 
CF' will remain straight as at first, but C'F will now be inclined at 
an angle i to its original position. CD', the new position of CD, 
will be still perpendicular to C'F, so that the angle made by CD' 
with its original position CD will also be equal to i. Let the deflec- 
tion of F under these conditions be 8, and let NP be a portion of 
the neutral lamina. As both 8 and i will be exceedingly small in 
any practical beam, we may write 

. 8 .. 
2 = radians, 
oc 

or 8 = ix (i) 

The strain of AC, produced by a tensile stress / induced by the 
bending moment M^, will be CC divided' by AC ; hence we have 



(*) 



AC 

AC 
'CC' 

Again, from the general expression for the strength of a beam 
(p. 146) we have, noting that / is the stress intensity at a distance 
CP from the neutral plane, 

" = /~*n * 



1 68 MATERIALS AND STRUCTURES 



or 

<^r 

f=CP ^ t* 

' J T" \O/ 

Substitution in (2) gives 

W* AC 

p --r'cc 

CP 



2 = 



I WX.SX 

r i 

W^.S^c 



El 



W 
Hence, from (i), 8 = ~.x^.8x (5) 

Had any other portion been taken similar in properties to ABCD, 
we should have obtained a similar expression for the deflection due 
to its strains. Hence the total deflection A of F will be obtained by 
integrating (5) between the limits x = o and x = L. 

W 



WL" 



The slope / max at the end F may be obtained by integrating result 
(4), which gives the slope produced by the strains of the small portion 
ABDC of the cantilever. 

W 



WL 2 , N 

radians ............................ (7) 



Cantilever having a distributed load. The case of a uniform 
cantilever having a uniformly distributed load w per unit length 
may be worked out in a similar manner, the only difference being 
that wx * 

M* = - 2 - 

Inserting this value in place of Wx in (4) and (5) gives 

. wx 2 .Sx v 






DEFLECTION OF BEAMS 



169 



Integrating (9) to obtain A, we have 



A _ 

~ 



W 



fa 



8EI' ' 

The slope at the free end may be obtained from (8). 



.(10) 



wl* 
6EI' 



Beam having a load at the middle. The results now obtained 
enable the case of a uniform beam simply supported at its ends to 
be solved easily by considering the beam as a double cantilever 
held fixed at the middle of the span and deflected upwards by the 
reactions W at each end (Fig. 201). It is evident that the deflection 




FIG. 201. Deflection of a simply-supported beam carrying a central load. 

of C below AB will be equal to the elevation of A and B above the 
horizontal line through C when the beam is loaded. Hence, using 
the result obtained in (6) above, we have, by writing JW for W and 

^ LforL ' A ^W.(L) 3 

3EI 
WL 3 , x 

= 48ET < I2 > 

The slope at the ends may be obtained similarly from (7). 

* m ax= 2 *EI 

WL 2 



T6EI' 



Beam having a uniformly distributed load. This case (Fig. 202 (a) ) 
may be regarded also as a double cantilever fixed at the middle of 
the span. The loading will consist of a downward load \wL, on 



I 7 o 



MATERIALS AND STRUCTURES 



each half span together with a concentrated upward load of \wL at 
each free end. The solution may be derived from the results already 
obtained by use of the axiom that the resultant deflection and slope 
of a beam under a combined system of loads will be the algebraic 
sum of those produced by each load taken separately. 



W per unit Ungth, Q 



I 



(a) 




FIG. 202. Deflection of a simply-supported beam carrying a uniformly distributed load. 

In Fig. 202 (b] the effect of the distributed load may be examined. 
Let Aj be the downward deflection of the ends, when we have, by 
substitution in (10), 






, , 
~i28EI' 

Fig. 202 (c) shows the effect of the reactions considered alone in 
producing an upward deflection A 2 . From (6) we have 



3EI 



The resultant upward deflection A at the supports, and hence the 
downward deflection at the middle of the span under the proposed 



loading, will be 



A = A - A 



I28EI 



384 ' El ' 



.(16) 



DEFLECTION OF BEAMS 171 



Due to the distributed load there will be a downward slope at the 
supports the value of which, t\, may be obtained from (u). 




The upward reactions will produce an upward slope z' 2 , obtained 
from W- . _>L(JL)2 

2EI 

<"> 



Combining these results, we have for the slope z max at the supports : 

*max = *2 ~~ *1 

ze/L 3 



i6EI 4 8EI 



-24 EI 

Graphical solution. The method of obtaining the slope and 
deflection at the free end of a uniform cantilever, employed on p. 167, 
may be extended in such a manner as to enable the slope and 
deflection in more complicated cases to be found graphically. 

Referring to Fig. 203 (a), the slope i caused by a portion ACDB 
being elastic, while the remainder of the cantilever is supposed to 
be rigid, is equal to the angle CPC', and will be given by 



Now CC divided by AC is the strain of AC caused by the stress/; 
hence, 



AC 
or CC' = - 



E 

Substituting this value in (i) gives 



(a) 



1 7 2 



MATERIALS AND STRUCTURES 



Again, from the general expression for the strength of a beam, 

we have / 

M J T 
AB ~~ -"-NA) 



or 



/ M, 



m 



(4) 



Inserting this in (3) gives 



2 = 



EL 



(5) 



r 

i 

L. 



Let the diagram HKL (Fig. 203 (^)) be the bending-moment 

diagram for the cantilever. It 
will be clear that the product 
MAB &e is the area of the shaded 
strip of the diagram. Hence 
we may say that the change of 
slope i produced by the elastic - 
bending of the portion between 
AB and CD is given by the 
area of the strip of the bending- 
moment diagram lying under 
BD multiplied by the constant 





i 



Consider now the whole 



FIG. 203. Graphical method of deducing slope 
and deflection. 



cantilever to be elastic, then, 
the slope at E being zero, it 
follows that the slope at AB will 
be the sum of the areas of all 
strips between HL and QR 



multiplied by 



, or 



Slope at AB = z' AB = area HQRL x 



EL 



.(6) 



In cases where the bending-moment diagram has a simple outline, 
it may be possible to calculate the required area, otherwise it will be 
necessary to use a planimeter. If the area has been found in square 
inches, the result should be corrected by multiplying by the scale in 
inch-tons per inch used in setting out the ordinates such as QR, and 
by further multiplying by the scale in inches per inch used in setting 
out the abscissae such as HQ ; E should be taken in tons per square 
inch, and I NA in inch units. 



DEFLECTION OF BEAMS 173 



HK being divided at a convenient number of points, the slope at 
each point may be found by the above method and a diagram drawn 
showing the slope at all parts of the cantilever by means of plotting 
the results and drawing a fair curve through them (Fig. 203 (<:)). 

Again, referring to Fig. 203 (a), let y be the deflection at a point S, 
distant z from C, owing to the elasticity of ACDB. Then 

z* 

or y = iz. 

Now, i is given by the area of the shaded strip in the bending- 
moment diagram multiplied by - ; hence, 

y = tpj x area of shaded strip x z. 

That is to say, the increment y of the total deflection at S caused 
by the elasticity of ACDB is given by the moment about S of the 

shaded strip of the bending-moment diagram multiplied by =TJ 

To obtain the total deflection 3 at S, we must therefore evaluate the 
moment of area of HTVL about T (Fig. 20$ (ft)) and multiply the 

result by gj , paying regard to the scales in the manner already 

noted. Repeating the same operation in order to obtain the deflec- 
tion at several sections, data will be obtained from which the 
deflection curve (Fig. 203 (d)) may be drawn. 

Some applications of the graphical method. Taking again the 
case of a uniform cantilever carrying a load W at its free end (p. 167), 
and referring to Fig. 204, we have 



slope at P = i p = area CFGD x = 



The maximum slope will be at B, and may be obtained by writing 
x = o. 

WL 2 v 



174 



MATERIALS AND STRUCTURES 



To obtain the deflection at P, we have 

8 P = moment about F of area CFGD x =^ 

rLl 

= (moment of CFHD - moment of DGH)^ 



The maximum deflection will occur at B, and may be obtained by 
writing x = o. 

W 



The case of a uniform beam simply supported at both ends and 
carrying a central load may be worked out in a similar manner, and 
is left as an exercise for the student. 



A W per unit length 




FIG. 204. 



204. Graphical method applied 
cantilever loaded at the free end. 



FIG. 205. Graphical method applied to a 
cantilever uniformly loaded. 



A uniform cantilever carrying a uniformly distributed load w per unit 
length may be worked out easily so far as its maximum slope 
and deflection are concerned. The bending-moment diagram is 
parabolic (Fig. 205), and it may be noted that its area is one-third of 
the area of the circumscribing rectangle, i.e. one-third of CD multi- 
plied by CE, and that its centre of area G is distant horizontally 
three-quarters of CE from E. 



DEFLECTION OF BEAMS 



175 



Hence, 



L j_ 
2 ' 3 ' El 



~6E! <5) 

A B = moment about E of area CED x =|- 



L 3, j_ 
a TI H 






Also, 



^8EI < 6 > 

In the case of a uniform beam supported at both ends, and carrying 
a uniformly distributed load (Fig. 206), the bending-moment diagram 

k L . >J 

; Wper unit Length \ 







D (b) 

FIG. 206. Uniformly loaded beam. 

is also parabolic. The maximum bending-moment occurs at the 
middle of the span, and is given by 

ze/L 2 



FK = 



8 



The slopes at A and B will be equal, and may be found by apply- 
ing the rule to the area KFE, noting that the slope at the middle 
will be zero. 

*'A = 4 = area KFE x == 
rL.1 

= - area of circumscribing rectangle x ^= 



L i 
8 ' 2 "El 



1 7 6 



MATERIALS AND STRUCTURES 



Noting that the centre of area G of KFE is at a horizontal distance 
|FE from E, we have, reckoning the deflection of A or B upwards 
from the middle, 

A A = A B = moment of area KFE about E x =^= 



L 5L 
T'8l 



i 
El 



__ 
384 El 



.(8) 



A 


|W 


B 




c 














i 



Encastrd beams. We may now examine the case of a uniform 
beam which is fixed rigidly at both ends by being built into walls or by 
some other method (Fig. 207 (a)). In such cases it may be assumed 

that the sections at A and B, which 
are in the plane of the wall before 
loading, remain in the same plane 
after loading ; hence the slopes at 
A and B will be zero. In order 
that this may be the case it is 
necessary that the means used 
for fixing the ends should apply 
restraining bending moments at A 
and B. We may obtain a fair idea 
of the conditions by examining 
the beam shown in Fig. 207 (/;). 
Here the bending moments at A 
and B, applied by the loads WjWj 
on the overhanging ends, have the 
effect of keeping vertical the sections at A and B. Hence, in the 
beam shown in Fig. 207 (a), the walls must supply the bending 
moments at A and B, which in Fig. 207 (b) are given by the loads W r 
The curve of the bent beam will resemble Fig. 207 (c), and will be 
convex downwards between two points K and N, and convex up- 
wards between D and K and also between E and N. This comes 
about from the fact that the resultant curve is produced from two 
component curves, one (Fig. 208(0)) caused by the action of W 
tending to produce a curve wholly convex downwards, and the other 
(Fig. 208 (<)), caused by the action of the bending moments M A and M B 
(which are obviously equal and are transmitted uniformly throughout 
the length of the beam), tending to produce a curve which is wholly 
convex upwards. The resultant bending moment at any section may 




FIG. 207. Encastre beam loaded at 
middle. 



ENCASTRE BEAMS 



177 



be obtained by taking the algebraic sum of these moments for that 
section. 

Fig. 209 (a) gives the bending-moment diagram for a beam simply 
supported and carrying a central load W. Its ordinates give the 





(b) 



FIG. 208. Component curves of an 
encastre beam. 



FIG. 209. Component bending moment 
diagrams for an encastre beam. 



positive bending moments at any section of the beam under con- 
sideration due to W alone. Fig. 209^) shows the uniform negative 
bending moments due to the fixing of the ends. These diagrams 
may be combined as shown in Fig. 210 (a), when the shaded portions, 
which show the algebraic sum of the component diagrams, will give 
the resultant bending moments for the beam. 

The maximum bending moment due to W alone is represented by 

WL 

ch in Fig. 210(0), and is of value -- 

4 

To obtain the values of M A and M B , 
represented by ad and fa, we have the 
consideration that the slopes at D and E 
(see Fig. 207 (<:)) are zero. Hence the 
areas of the bending-moment diagram 
adf and fern (Fig. 210(0) must be equal, 
because the slope at the centre is given 
by their algebraic sum, and this must 
be zero for zero slope. For a similar 
reason the areas cmg and geb are equal ; FlG -,? I0 --Resuitant bending moment 

diagram for an encastre beam. 

hence it is easily seen from the figure 

that the triangular area acb must be equal to the rectangular area 

adeb. Thus, hm must be one-half of he, giving 




It will also be obvious from Fig. 210(0) that the points /and g, 

D.M. M 



1 78 



MATERIALS AND STRUCTURES 



Kwt it 
8 ' 2 4 



8 2 4 i2/ El 



where the resultant bending moments are of zero value, must lie at 
one-quarter span. 

The deflection upwards of E above F (Fig. 207 (<:)) may be obtained 
by taking the algebraic sum of the moments about e of the areas 
cmgaxid. geb (Fig. 210(6)), and dividing the result by El. This will 
give the central deflection A of the beam. 

A = (moment of area cmg - moment of area geb) ^ 

i L 5L\ /WL i L 
8 2 4 12 

i WL 3 , , 

"192 El ' 

An encastre* beam of uniform section carrying a uniformly distributed load 
(Fig. 2 1 1 (a) ) may be worked out in a similar manner. The parabolic 
curve afcgb (Fig. 211(6)) re- 
presents the bending-moment a w per unit Length B 

diagram for a beam simply sup- ^ 

ported at the ends and carrying L.| 
w per unit length. The maxi- 
mum bending moment will 
occur at the middle of the span, 

T 9 

and is represented by ch ^-. 

The rectangle adeb represents 
the uniform bending moment 
due to the fixing in the walls. 
The shaded area gives the resultant bending-moment diagram. 

Here, as in the last case, there is zero slope at A, C and B ; hence 
the areas adf, fcm, cmg and geb are equal ; consequently the parabolic 
area afcgb must be equal to the rectangular area adeb, giving 

2 





FIG. 211. An encastre beam uniformly loaded. 



(i) 

The bending moment at C will be given by 

-M A 

; Z/L 2 



24 



ENCASTRE BEAMS 179 



Thus, we see that the bending moment at the walls is double that 
at the middle of the span. 

To obtain the deflection at C, we must find the algebraic sum of 
the moments about h of the areas cmg and geb, or, since the result 
will be the same if the moment of the area mgbh be added to each, 
the calculation may be simplified by taking the algebraic sum of the 
moments about h of the areas chbg and hmeb. Hence, 

A c = (moment of area chbg - moment of area hmeb) 

El 



K2 
3 



L 3 



8 2 8/ 12 2 EI 

ze;L 4 \ i 
96 J El 



/-\ 
384 EI' 

The distance of /and g, the points of zero bending moment, from 
d and e respectively will be equal, and may be found by obtaining an 
expression for the bending moment at a distance x from the wall and 
then equating this to zero. Thus, 

MS = M A - bending moment at x for a beam 
simply supported 



12 



fwL zvx 2 \ 

I _ yy __ \ 

*Af 

V 2 2 / 



i 



12 2 2 

Equating this to zero gives 
x 2 Lx L 2 



or 




2 

= 0-2 1 iL or o-ySSL (4) 

Hence the points of zero bending moment lie at 0-21 iL from each 
wall. 



i8o MATERIALS AND STRUCTURES 

Points of contraflexure. The two last cases considered provide 
examples of beams in which the curvature is partly convex down- 
wards and elsewhere convex upwards. The centres of curvature for 
a portion of the length of the beam lie on the upper side, and for 
other portions lie on the lower side. Points on a beam where the 
curvature changes from convex upwards to convex downwards, i.e, 
where the centre of curvature changes from one side of the beam 
to the other, are called points of contraflexure. Curvature which is 
convex downwards may be called positive, and that which is convex 
upwards may be considered negative. The curvature changes sign 
at points of contraflexure, and hence must have zero value at such 
points. 

Considering the equation (p. 166), 

~ i M 

Curvature = ^ = ^, 
R H/l 

it is evident that, for the curvature to be zero, M must be zero. A 
point of contraflexure may hence be defined as a point of zero 
bending moment. Such points occur at quarter span for an encastre 

beam carrying a single load 
B at the middle of the span 

(p. 176) and at 0-2 uL from 
, , the walls in the case of a 

L J i w uniformly distributed load 

W per unit Length B (p. T 78). It should be noted 

777/f77777//// /////////////////// w ^ that encastre beams differ 

from beams which are simply 
supported at both ends in that 
one or both supports in the 

FIG. 212. Encastre beams. 

latter may suffer sinkage when 

the load is applied, or by reason of some alteration in the foundation 
conditions, without thereby affecting the distribution of bending 
moment along the beam. No such alteration in either of the walls 
fixing the ends of an encastre beam can occur without affecting the 
bending moments on the beam. For example, if the encastre beam in 
Fig. 212(0) should for any reason become loose in the holes in the wall, 
so that the fixing couples M A and M B disappear, the bending-moment 
diagram will change from that shown in Fig. 210(0) to that for a simply 
supported beam, and the maximum bending moment, and consequently 
the maximum stress due to bending, will be doubled. In the case 
of a uniform load (Fig. 2 1 2 (l>) ) such an alteration in the wall fixings 
would produce a change in the maximum bending moment from 





PROPPED CANTILEVERS 



181 



fr-A"* 7 *-^"* 

fn\ . 


A | 


C 


1 E 


B 


t 


p 









A C 


jw 


^ 


[1 i 


Pfl B 




^ D 


(b) E 


? 



FIG. 213. Beam cut at the points of 
contraflexure. 



_ to +~, that is, a numerical increase of 50 per cent. It 

12 

should also be noted that these alterations would be accompanied 
by very small alterations in the slope and deflection, the inference 
being that quite a small alteration in the shape or position of 
the fixing arrangements due to sinkage, or otherwise, will be 
sufficient to produce a large alteration in the bending moments and 
stresses. 

The difficulty may be overcome, if desired, by noting that at 
points of contraflexure there is zero bending moment, and that the 
beam may be cut at these points 
provided that means are provided 
there for taking up the shear. 
Fig. 2 13 (a) shows diagrammati- 
cally how this may be effected 
for an encastre beam carrying a 
central load W. The beam is 
cut at quarter span, and links CD 
and EF are used for suspend- 
ing the middle portion. These 
links will be under pulls of JW 
owing to the shearing force. Obviously no moderate changes in 
the supports can now affect the bending moments in the com- 
ponent parts of the beam. A practical method of designing the 
arrangement is shown in Fig. 213^), where the central portion is 
supported on a rocker at E and by a short column at CD. It will 
be observed that alterations of length, etc., due to expansion on 
heating, are taken up by this device without inducing stresses on 
the beam. The artifice of cutting a beam, or an arch, at places 
where it is desirable that there should be no possibility of any 
bending moment arising is often resorted to in practice. 

Propped cantilevers and beams. In Fig. 214(0) is shown a 
cantilever AB carrying a uniformly distributed load. The fixing in the 
wall at A is sufficient alone for the equilibrium of the cantilever, 
but an additional support or prop has been placed under B. 
The pressure on this prop depends on the elastic properties of the 
material of the cantilever and also on the level of the top of the 
prop. Assuming that the cantilever just touches the top of the prop 
before application of the load, the reaction of the prop may be 
calculated as follows. 

Supposing the prop to be removed (Fig. 214 (<)), the deflection Aj 



182 



MATERIALS AND STRUCTURES 



of the cantilever under the action of the distributed load would be 
given by w \j , 

1 = 8El (P ' I75 ^ W 

Now suppose that the distributed load is removed and that the 
prop is applied and pushed upwards until a deflection at B of the 

same magnitude as \ is obtained 
(Fig. 2 14 (<:)). The upward de- 
flection A 2 thus produced by the 
force P exerted by the prop will 




PT 3 
- (p. 174)- 



(d) 

FIG. 214. Propped cantilever. 



be 



If both P and the distributed 
load be applied simultaneously, 
p *.i_ 2 the levels will be the same at 
both A and B (Fig. 2 14 (</)), for 
Aj and A 2 are equal and opposite. 
Hence, 

PL 3 wL 4 WL 3 



where W is the total distributed load. 



3EI 

Hence, 



8EI SET* 



(3) 



The bending moment at any section C may be calculated now : 

<7/?l'V % 2 



(4) 



Points of contraflexure may be found by equating M c to zero 



(p. 180). Thus, 



wx 2 



- = o. 



Zero is one value of x satisfying this equation, hence B is a point 
of contraflexure. To obtain the other point, we have 



WX 

--- =o, 



or * = JL ............................... (5) 

The bending-moment diagram is shown in Fig. 215 (a). The bend- 
ing moment at the wall may be found by writing x = L in (4), giving 



.(6) 



BEAM HAVING THREE SUPPORTS 



183 



To obtain the bending moment at f L from B, we have, from (4), 



It will be understood that 
any vertical displacement of 
the prop, whether by reason 
of sinkage of the foundations 
or by changes in temperature, 
will alter the bending moment, 
and hence the stresses through- 
out the cantilever. The shear- 
ing-force diagram is given in 
Fig. 215 (b) ; the values of the 



(7) 




Bending Moments, 
fa) 



Shearing Forces 



w 



FIG. 215. Bending moment and shearing-force 
diagrams for a propped cantilever. 



shearing force are -f f W at the wall and - f W at the prop. 

The case of a beam resting on three supports at A, B and C is 
illustrated in Fig. 2 1 6 (a). The supports at A and B alone are 



W per unit length 




rWL 



Bending X *S 
Moments \\T-* 

Shearing 
Forces 



FIG. 216. A beam resting on three supports. 

sufficient for the equilibrium of the beam ; hence, in this case also, the 
reactions, bending moments, and stresses depend on the levels of the 
supports being preserved. 






1 84 MATERIALS AND STRUCTURES 

Suppose the supports to divide the beam into two equal spans and 
that the supports are all at the same level. If the support at C be 
removed (Fig. 2 1 6 (b) ), there will be a deflection A x at C given by 



Replace the support at C by pushing upwards until the level is 
restored (Fig. 2i6(r)). The upward deflection A 2 produced by P c 
will be given by p T 3 



Clearly A x and A 2 are equal. Hence, 
P C L 3 5 



48EI~38 4 El ' 
Pc-fwL 

= |W, .............................. ....(3) 

where W is the total load. 

It will be evident that P A and P B are equal. Hence, 

PA = PB = &W ........................... (4) 

The bending moment at D may be found from 






wx 






(5) 



Points of contraflexure occur where M D is zero ; to find these, we 
have * 



The value zero for x satisfies this equation ; hence, A is one point 
and, from symmetry, B is another point of contraflexure. To obtain 
others, wx 

T\^ L - = o, 

* = |L .................................. (6) 



Points of contraflexure therefore occur at f L from A and also at 
an equal distance from B. The complete bending-moment diagram 
is given in Fig. 2i6(</) and the shearing-force diagram appears in 
Fig. 2 1 6 (e). 

The beam here discussed is a simple illustration of continuous 
beams, t.e. beams continuous over several spans and resting on several 
supports. 



BEAMS OF UNIFORM CURVATURE 185 

Beams of uniform curvature. Considering again the equation 



it will be remembered that it has been assumed that the moment of 
inertia is uniform in all the cases considered. This is the case very 
often in practice ; for example, beams of comparatively short span 
generally consist of a rolled steel beam or of two or more similar 
beams placed side by side. When we consider larger beams, we 
find that the section in general is not uniform, but is varied so as to 
produce more nearly a beam of uniform strength (p. 154). The 
above equation may be modified so as to include a great number 
of such cases. Thus, 

M = ^I (p. 146); 

L= = _ L f 
* R El El' m 






Abeam of uniform strength is one having uniform maximum stress f. 
This may be secured by having constant depth and varying the breadth, 
in which case m will be constant. In equation (i) above, the right- 

hand side will contain nothing but constants, and therefore ^ will 

_K 

be constant. Such a beam will have constant radius of curvature, 
and hence will bend into the arc of a circle. In other cases of 
built-up plate girders having parallel flanges, the breadth is constant, 
and uniformity in f is secured by adjusting the thickness of the 
flanges, the number of flange plates becoming greater towards the 
middle of the span. Assuming that this variation of flange thickness 
does not alter the depth sensibly, we have a constant value of m, and 
the girder will have constant curvature. 

Constant curvature may also occur in a beam of uniform section. 
To obtain such a result, M must be constant in the equation 

JL_M 
R~ET 

This condition may be brought about by the loads and reactions 
being applied in the form of two equal opposing couples. A carriage 
axle is a common example (Fig. 217). Here AC = BD; equal 
loads W, W are applied at A and B, and the wheel reactions P, P at 



1 86 



MATERIALS AND STRUCTURES 



C and D will be each equal to W. The portion CD of the axle will 
therefore have uniform bending moment, given by 

M CD =WxAC, 
and hence will bend into the arc of a circle. The curvature in the 



w 

B 



TP p t 

FIG. 217. A carriage axle. 

overhanging portions AC and BD will vary, following the law for the 

cantilever worked out on p. 167. 

In Fig. 218 AB is a beam of length L bent into a circular curve 

ACB. Drawing the diameter EODC perpendicular to the chord AB, 

and remembering that the deflec- 
tion will be very small in practice 
we have, by application of the 
principle that the products of the 
segments of two intersecting chords 
in a circle are equal, 

EDxDC = ADxDB, 

or, very nearly, 

2 RxDC =(|L) 2 ; 

hence the deflection DC at the 
middle will be 

FIG. 218. Beam bent into a circular curve. \J\^> = TTFr- 







Substituting -^y for ^ in this result gives 



It will also be evident that the inclination of the tangents at A and 
B will be equal to the angle AOC. Expressing this in radians so as 
to obtain the slope at A and B, we have 

AC L 






AO 2R 
ML 

2fiT 



STRESS AND DEFLECTION IN BEAMS 187 



Relation of stress and deflection. In all the cases of deflection 
which have been considered, it will be noted that the expression for 
the maximum deflection has the form 

WL 3 



where c is a numerical coefficient, the value of which depends on the 
circumstances of the case. Hence we may write, 



Taking the general equation for the strength of a beam (p. 146), 



it will be noted that M is always proportional to WL, and that m is 
always proportional to d, the overall depth of the beam. Hence, 
from (2), , 



. 

a 
Substitution of this in (i) gives 



/L 2 
- 



Hence, in beams constructed of the same material, for which E 
will be constant, we may state that the maximum deflection will be 
directly proportional to the square of the length and inversely pro- 
portional to the depth when the beams are carrying loads which 
produce the same maximum value off. 

EXAMPLE. A steel bar of rectangular section is supported at its ends 
and carries a central load. The ratio of maximum deflection to span is 
not to exceed -%$$ ; the maximum stress is not to exceed 5 tons per square 
inch. Find the ratio of span to depth if E = 13,500 tons per square inch. 



1 88 MATERIALS AND STRUCTURES 

WL / T 2/1 
T~ = P I== ~^ ; 

8 (2) 



Hence, from (i), A = ^ . ^ 






jL 1 

YT7* 

6E A 



EXERCISES ON CHAPTER VIII. 

1. A bar of steel of square section, 2 inches edge, is used as a canti- 
lever, projecting 24 inches beyond the support, and has a load of 400 Ib. 
at its free end. Find the values of the radius of curvature for sections at 
3 inches intervals throughout the length. Plot R and the length of the 
cantilever. E = 13,500 tons per square inch. 

2. Find the slope and deflection at the free end of the cantilever given 
in Question i. 

3. A beam of I section, 8 inches deep, 4-5 inches wide, metal 0-5 inch 
thick, is simply supported on a span of 10 feet and carries a central load 
of 1-5 tons. Calculate the maximum deflection and also the slope at the 
ends. What will be the radius of curvature at the middle of the span ? 
Take =13,500 tons per square inch. 

4. Answer Question 3, supposing that the beam carries only a uniformly 
distributed load of 2 tons. 

5. An encastrd beam of I section has its ends fixed into walls 12 feet 
apart. The depth is 12 inches, and I is 50 in inch units. If the stress is 
limited to 5 tons per square inch, what central load would be safe ? 
Draw the diagrams of bending moment and shearing force. 

6. Answer Question 5, supposing that the load is to be distributed 
uniformly. 

7. Calculate the deflections at the middle of the span for the beams 
given in Questions 5 and 6. 

8. A cantilever projects 8 feet from a wall and carries a load of 
1-5 tons at 4 feet from the wall and another load of 075 ton at the free 
end. Draw the diagrams of bending moment, slope and -deflection, in 
each case giving the scale of the diagram. State the values of the slope 
and deflection at the free end. Take 1 = 350 in inch units and E= 13,500 
tons per square inch. 

9. Calculate the uniform bending moment which must be applied to a 
bar of steel of 0-25 inch in diameter in order to make it bend into the arc 



EXERCISES ON CHAPTER VIII. 189 

of a circle of 20 feet radius. = 30,000,000 Ib. per square inch. If the 
bar is 5 feet in length, what will be the deflection at its centre ? 

10. A girder is 40 feet span by 4 feet deep, and rests on its supports. 
The uniformly distributed load produces a maximum stress due to bending 
of 5 tons per square inch. Find the deflection at the middle of the span. 
=13,500 tons per square inch. 

11. Supposing the girder given in Question 10 to have uniform flange 
stress of 5 tons per square inch, what will be its radius of curvature ? 
Calculate the deflection at the centre. 

12. A cantilever of uniform section is built securely into a wall, and its 
outer end just touches a prop when there is no load. The cantilever is 
8 feet long, and carries a uniformly distributed load of 1000 Ib. per foot 
length. Find the reaction of the prop, and draw the diagrams of bending 
moment and shearing force ; give the calculations required for these. 

13. A beam 40 feet in length rests on three supports A, C and B at the 
same level ; the supports divide the beam into two equal spans. If there 
is a uniformly distributed load of 1-5 tons per foot length, find the reactions 
of the supports, and draw the diagrams of bending moments and shearing 
force, showing the necessary calculations. 

14. A piece of flat steel has to be bent round a drum 5 feet in diameter ; 
what is the maximum thickness which the strip can be made so that there 
shall be no permanent deformation when it is removed from the drum ? 
The steel has an elastic limit of 14 tons per square inch. E = 14,000 tons 
per square inch. (I.C.E.) 

15. Three rolled steel joists 6 inches deep are placed side by side 
spanning an opening of 10 feet ; the moment of inertia of the two outer 
joists is 20 and that of the inner one 44 inch-units. A central load of 
5 tons is so placed as to deflect each of the three joists equally ; state the 
amount of the load carried by each joist and the maximum unit stress (i.e. 
stress in tons per square inch) in the centre joist only. (I.C.E.) 

16. A beam is firmly built into a wall at one end, and rests freely at its 
other end on a vertical column whose centre line is distant 8 feet from the 
wall. The beam supports a wall, whose weight added to that of the beam 
itself is equivalent to a uniformly distributed load of 3200 Ib. per foot run 
of the beam. Find (a) the total load supported by the column ; (b} the 
bending moment and shear force at the section of the beam adjoining the 
wall ; (c) the position of the point of zero bending moment. Sketch 
complete bending moment and shear diagrams. (B.E.) 

17. A rectangular timber beam, supported at the ends, is of uniform 
section from end to end, and it carries a uniformly distributed load. If 
the working intensity of stress in the wood is not to exceed 2000 Ib. per 
square inch, and if the modulus of elasticity of the wood is 1,700,000 Ib. per 
square inch, determine the ratio of the depth of cross-section of beam to 
span of beam in order that the deflection may not exceed 5^ th part of the 
span. (B.E.) 

18. A horizontal beam, span 25 feet, is fixed at the ends. It carries a 
central load of 5 tons, and loads of 2 tons each at 5 feet from the ends. 
Determine the maximum bending moment, the bending moment at the 
centre of the span and the position of the points of contraflexure ; sketch 
also a diagram of shear force. (L.U.) 



I go MATERIALS AND STRUCTURES 

19. A floor, carrying a uniformly distributed load of 2 cwt. per square 
foot over a span of 20 feet, is proposed to be carried by either : (a) I joists, 
10 inches deep ; area, 12-35 square inches ; I (maximum), 212 inch-units ; 
pitch, 4 feet. Or, () I joists, 12 inches deep ; area, 15-9 square inches ; 
I (maximum), 375 inch-units ; pitch, 6 feet. Compare these two pro- 
positions by rinding the ratio of strengths, deflections and total weights of 
girders. Find the maximum skin stress in case (a). (L.U.) 

20. A uniform beam, 30 feet long, fixed at the ends, has a load of 
20 tons spread uniformly along it. It has also two loads of 3 tons, each 
hung from points which are 10 feet from the ends. What is the bending 
moment everywhere, and what is its greatest value ? (B.E.) 



CHAPTER IX. 



WORKING LOADS. BEAMS AND GIRDERS. 

Dead and live loads. The loads to which any structure is 
subjected may be divided into dead and live loads. The dead loads 
include the weights of all the permanent parts of the 
structure ; the live loads may consist of travelling weights 
and other forces, such as wind pressure, which may occur 
periodically. Dead loads produce stress of constant 
magnitude in the parts of the structure; the live loads 
produce fluctuating stresses; hence each part of the 
structure may be called upon to withstand stresses which 
fluctuate between maximum and minimum values. 

A load may be applied to a bar in three different 
ways : (a) in gradual application, the load on the bar 
is at first zero and the magnitude of the load is increased 
uniformly and slowly until the bar is carrying the whole 
load ; (b) sudden application may be realised by reference 
to Fig. 219, in which the load W is supported by short 
rods so that it is just touching the collar at the lower end of AB ; 
if the rods be knocked out, the load will suddenly rest on the 

collar; (c) impulsive application may be 
obtained by allowing W in Fig. 219 to 
drop from a height on to the collar. 

Eesilience. Fig. 220 illustrates the case 
of gradual application of pull to a bar. 
The bar extends by an amount propor- 
tional to the load, up to the elastic limit, 
and at any instance the resistance of the 

applied. When 




FIG. 219. 



Load 








; Ext ? 



* e 



FIG. 220,-DiagramJor a gradually oar [ s equa l to t h 

the load applied is P 19 the extension of 
the bar is e l and its resistance is equal to P r It will be evident 
from the figure, that the average value of the load is IP, and as 



192 MATERIALS AND STRUCTURES 

this force acts through a distance e, the work done will be given by 
the product of these quantities (p. 325). 

Work done in stretching the bar = ^P^ (i) 

As the resistance of the bar is at all times equal to the pull, it 
follows that the energy stored in the bar will be equal to |P<?. 
Let a ^e sectional area of the bar in square inches. 
P = the final pull in tons. 

p 
/= = the stress produced by P in tons per square inch. 

L = the original length of the bar in inches. 
e = the extension produced by P in inches. 
E = Young's modulus in tons per square inch. 

P L 
Then a" 7' 

* = = /- 

Also, P = fl/ 

Substituting these values in (i) gives : 

Energy stored in the bar = |/"./^ 

f 2 
= L^p inch-tons (2) 

This quantity is called the resilience of the bar. The resilience of 
the material is stated usually as the energy which can be stored in 
a cubic inch when stressed up to the elastic limit. This may be 
obtained from (2) by taking f to be the elastic limit stress and 
noting that aL is the volume of the bar in cubic inches. Hence 

f 2 
Resilience = j= inch-tons per cubic inch. 

Load suddenly applied. If the load be applied suddenly as in 
Fig. 219, and if the bar extends by an amount e, gravity is doing work 
on W throughout this extension. Hence, 

Work done on W = We inch-tons. 

This work may be represented by the rectangular diagram OKLM 
(Fig. 221), in which OK represents W and OM represents e. The 
resistance offered by the bar during the extension still follows the 
same law as before, i.e. at first the resistance is zero and it gradually 
increases, being proportional to the extension up to the elastic limit 



IMPULSIVE LOADS 



193 



, , 



This may be represented by the triangular diagram OMQ. At PN 

the resistance of the bar and the weight of the load are equal, but 

extension does not stop here, since more work has been done by 

gravity than can be stored in the rod. Extension will go on until 

the work done by gravity has been stored 

entirely in the rod, i.e. until the area 

OKLM is equal to the area OMQ. 

This will occur evidently when OM is 

equal to twice OP, or when MQ is 

twice OK. Now, had W been applied 

gradually, the stretch would have been o P M Ext? 

OP; hence the Sudden application Of FIG. 221. Diagram for a load applied 

W has produced a stretch double of 

this amount. Also PN would have been the final resistance of the 
bar had W been applied gradually; hence the sudden application 
has produced a resistance of twice this magnitude, and therefore 
also a stress equal to double of that which would have occurred 
with gradual application. 

The conditions are not attained easily in practice, but the effects 
of live loads in producing stress are often taken account of by 
estimating what the stress would be had the load been applied 
gradually and then taking double this stress as that which the part 
will be called upon to carry. 

Impulsive application of a load. In an impulsive application, 
let W be dropped from a height H inches (Fig. 219). Then 
Total work done by gravity = W(H + *) inch-tons. 

Extension will go on until the whole of this work is stored in the 
bar. From equation (2) (p. 192), we have 

ft 

(i \ j ip^ 

2E 

Hence, aL ~ 



Energy stored in the 






If e is small compared with H, as is the case generally, then 

2 EWH 



Working stresses. The stresses which may occur in any part of a 
structure are estimated by first calculating the stress produced by the 
dead load. Separate calculations are then made in order to determine 



D.M. 



194 MATERIALS AND STRUCTURES 

the stress produced by the live loads. The stress in the part under 
consideration will then fluctuate between known limits, and it remains 
to determine what ought to be the safe stress permitted in the part. 
The determination may be based on the known breaking strength of 
the material by taking the working stress as a fraction of the ultimate 
strength. The reciprocal of this fraction is called a factor of safety, 
and its value depends on the kind of material and the nature of the 
loads. Thus, for wrought iron and steel, the factor of safety may 
be 3 for a dead load, 5 for a stress which does not change from pull 
to push, 8 for a stress which alternates from a certain pull to an 
equal push and 12 for parts subjected to shock. Somewhat higher 
factors may be taken for cast iron and for timber, as these materials 
are less trustworthy. 

It may be noted here that a load which a piece of material may 
carry for an indefinite time, if applied steadily, will ultimately cause 
fracture if it is applied and removed many times. The effect is more 
marked if the load be alternated, i.e. applied first as a pull and then 
as a push, in the manner in which the piston rod of a steam engine is 
loaded. The experiments of Bauschinger, Wohler, Stanton and 
others, on the effects of repeatedly applied and alternating loads 
show that the strength to resist an indefinite number of repetitions 
depends on the range of stress rather than on the actual values of 
the maximum and minimum stresses. 

The following rule has been deduced by Unwin* from the results 
of Wohler's experiments, and applies to cases of varying stresses. 
Let f g the breaking strength of the material in tons per square 

inch under a load applied gradually, 
/j = the breaking strength of the same material in tons per 

square inch when subjected to a variable load which 

fluctuates from ,/j to / 2 and is repeated an indefinite 

number of times. Let this be of the same kind (push or 

pull) as/ s . 
/ 2 = the lower limit, in tons per square inch, to which the 

material is subjected, + if / 2 is of the same kind as 

/! and/*, - if/ 2 is of the opposite kind. 
r=/ 1 -/ 2 = the range of stress. 
Then Unwin's formula is 

(0 

* Machine Design, Part I. Prof. W. C. Unwin. (Longmans, 1909.) 



WORKING LOADS 



195 



n has the value 1-5 for wrought iron and mild steel. From 
equation (i), we have 



or 



w 



in which the negative sign has been disregarded. Equation (2) gives 
a dead-load stress yS which would produce, when applied steadily to 
the member, the same effect as the actual fluctuating stresses. 

If each side of (2) be multiplied by the sectional area of the bar, 
the stresses in the equation become total forces on the bar. Using 
capital letters to represent the total forces corresponding to the 
stresses f st rand/j, we have 



Equivalent stea dy load = F = ~ . ... (3) 

The Launhardt-Weyrauch formula also takes account of stress varia- 
tion. Let Fj and F 2 tons be the maximum and minimum forces 
to which the bar may be subjected, and \etf s tons per square inch 
be the breaking strength of the material under a gradually applied 
stress. Then 

Breaking stress = ^f s ( i + - ^ J tons per square inch ..... (4) 
Applying a factor of safety of 3 to this, we have 

Working stress = ifi\ T + ~ ^ ) tons per square inch ..... (5) 

EXAMPLE. A certain bar in a structure carries a pull of 80 tons due 
to the dead load ; the live load produces forces in the same bar varying 
from 20 tons pull to 40 tons push. Find the working stress and the cross- 
sectional area of the bar. The breaking strength of the material under a 
gradually applied pull is 30 tons per square inch. 

First Method. By doubling the live load pull and adding the result 
to the dead load pull, we have 

Equivalent dead load = 80 + (2 x 20) =120 tons. 
Taking 9 tons per square inch as the working stress, we have 
Sectional area of bar = 1 5 =13-3 sq. inches. 



IQ6 MATERIALS AND STRUCTURES 

Second Method. By Unwin's formula (3), 
R = 6o tons. 
F 1 = 8o + 2o=ioo tons 
n=i'$. 



o-i j ji j (1-5 x 60) W(?x 3600) + 4(100 -30)2 
Equivalent dead load = ^ 

= 128 tons. 

Again using 9 tons per square inch as the working stress, we have 
Sectional area of bar = 1 & = 14-2 sq. inches. 

Third Method. By the Launhardt-Weyrauch formula (5), we have 
fs = 30 tons per square inch. * ^ 

F 1 = 8o + 2o=ioo tons pull. 
F 2 =8o~4O = 4o tons pull. 
Working stress = f x 30(1 + /<&) 

= 8 tons per square inch. 

F, 100 
Cross-sectional area= g = ~o~ 

= I25 square inches. 

Wind pressure. If wind pressure be treated as a live load, then 
30 Ib. per square foot of vertical surface may be assumed to be the 
maximum. If treated as a dead load, then pressures up to 55 Ib. 
per square foot of vertical surface may be taken. Stanton's experi- 
ments at the National Physical Laboratory give, for small plates, 

p 0-0027 V 2 Ib. per square foot, 
or, for large plates, 

/^o-oo32V 2 Ib. per square foot, 

where V is the velocity of the wind in miles per hour. Hutton's 
formula may be used in calculating the normal pressure on inclined 
surfaces. 

Let p the pressure in Ib. per square foot on a surface perpen- 
dicular to the direction of the wind. 

/ n = the normal pressure in Ib. per square foot on a surface 
inclined at an angle to the direction of the wind. 

Then Hutton's formula gives 



where a is a coefficient depending on the value of 0. Values of a 



TRAVELLING LOADS 



197 



corresponding to different values of have been plotted in Fig. 222, 
and the value of a appropriate to any given surface may be taken 
from the curve. 



a. 
1-2 

10 
0-8 
0-6 
0-4 
O2 



10* 20' 30' 40* 50* 60* 70* 80* 90* 



FIG. 222. Values of a in Mutton's formula. 



1 
1 





TW 

C (D 




A i 


1 


1 


\ B 


f 


a --* 


(a) 


I- 


r 
>*x- 


/4sw 

W D 






Al 


1 




IB 


,F 


a * 


(b> 


la 



Travelling load. In Fig. 223(0;) AB is a beam simply supported 
at A and B and carrying a load W. The effects of W alone will be 
considered, any other, loads 
being disregarded. If W 
remains fixed in position, 
the reactions P and Q as 
well as the bending moment 
and shearing force at any 
section, such as D, have 
definite values. These values 
will alter if W travels along 
the beam, and it then be- 
comes necessary to determine 
what position W must occupy 
when a given section is sub- 
jected to the greatest bending 
moment it will be called upon 
to resist, as well as the value 
of this bending moment. 
The same questions must 
also be considered in relation to the shearing force at any given 
section. 

Let x = the distance of W from A. 

a = the distance of the given section D from A. 
L = the span of the beam, all in the same units. 





Max 

FIG. 223. Beam carrying a single rolling load. 



igS MATERIALS AND STRUCTURES 

Then, taking moments about B, we have 






Let W be on the right-hand side of D as shown in Fig. 2 23 ( 
Then the bending moment at D will be 



Hence, as x diminishes, i.e. as W travels towards the left and so 
approaches the section D, the bending moment at D increases. 

Now let W be on the left-hand side of D as shown in Fig. 223 (ti). 
Writing down the bending moment at D, we have 
M D -Pa-W(a-) 



(3) 



This result indicates that as x diminishes, i.e. as W, still travelling 
towards the left, recedes from the section D, the bending moment 
at D is becoming smaller. Therefore, the greatest bending moment 
which the section D will be called upon to resist will occur when W 
is immediately over the section. The value of this bending moment 
may be obtained by writing x = a in either (2) or (3) above, giving 



Maximum bending moment M n = i - -r 



If a be varied so as to obtain the maximum bending moments 
for other sections of the beam, and if the results of calculation 
from equation (4) be plotted, a parabolic curve will be obtained 
(Fig. 223 (c)), the ordinates of which will show the maximum bending 
moment for all sections of the beam. The centre section C is 

called upon to resist a bending moment -. It will, of course, be 

4 

understood that the values shown by the ordinates in Fig. 223 (c) are 
not attained simultaneously. The diagram must be interpreted as 
indicating that the bending moment at any section, say D, is zero 



TRAVELLING LOADS 199 



when W is off the beam, and increases gradually as W travels 
towards the section ; the maximum value M D is attained when W 
reaches the section. 

The shearing force at D, when W occupies any position on the 
beam lying on the right of D, will be positive and equal to P ; 
hence, from (i), 

(5) 



This result shows that the shearing force increases as x diminishes, 
i.e. as W approaches the section D from the right. 

Taking W in a position on the left of D, the shearing force will 
be negative, and will be given by 



We infer from this result that the negative shearing force at D 
diminishes as x becomes smaller, i.e. as W recedes from the section 
and approaches the left-hand support. 

The inferences from this discussion are that the shearing force 
at any section attains a maximum positive value when W lies close 
to the right-hand side of the section, that it becomes zero as W 
crosses the section, and attains a maximum negative value when 
W lies close to the left-hand side of the section. To obtain 
the values of these shearing forces, write x = a in equations (5) and 
(6), giving 

Maximum positive shearing force S D = 1 1 - =- J W .......... (7) 

Maximum negative shearing force S D = - y W ............... (8) 

L/ 

Varying a so as to obtain values of the shearing forces for other 
sections and plotting the values so found (Fig. 223 (d}\ we obtain two 
sloping straight lines. This diagram must be interpreted as follows : 
the shearing force at any section is zero when W is off the beam ; 
as W travels along the beam from right to left, the shearing force 
at any section D is positive and increases gradually, until the 
maximum value DE is attained when W is on the point of arriving 
at the section. As W crosses the section, the shearing force becomes 
negative and attains the maximum value DF when W reaches the 
other side of the section. 



206 



MATERIALS AND STRUCTURES 



W per unit Length 



Uniform travelling load. In Fig. 224 (a) is illustrated the case 
of a beam AB simply supported at A and B and subjected to a 
uniform load w per unit length. Taking the load of sufficient 

length to cover the whole span, 
it will be evident that the maxi- 
mum bending moment at any 
section will occur when the 
beam is loaded fully, i.e. when 
the whole span is covered by 
the load. The bending-moment 
diagram will therefore be para- 
bolic (Fig. 224^)) and of maxi- 




FIG. 224. Beam carrying a uniform travelling 
load ; maximum bending-moment diagram. 



mum height 



8 



In Fig. 224 (a) the nose E of the load is advancing towards a given 
section D. The shearing force at D is positive and is equal to P ; 
hence it increases as E approaches D. When E has crossed to the 
other side of D (Fig. 225 (a)), the shearing force is P diminished by 
the portion of the load lying between D and E. This shearing 
force will be less than that 
existing at D when the nose 
is vertically over D, for P 
will then have a certain 
value, and this value will 
be increased in Fig. 225 (a) 
by a fraction only of the 
portion of the load lying 
between D and E ; as the 
whole of the latter must be 
deducted from P in calcu- 
lating the shearing force in 
Fig. 225 (a), it follows that 
the positive shearing force 
at D in Fig. 225 (a) is 
diminishing. Hence the maximum positive shearing force at any 
section occurs when the whole of the part of the beam lying to 
the right of the section is covered by the load, the end of the load 
being vertically over the section. In the same manner it may 
be shown that the maximum negative shearing force at any section 
occurs when the part of the beam lying on the left of the section 
is covered by the load (Fig. 225 




FIG. 225. Beam carrying a uniform travelling load ; 
maximum shearing-force diagram. 



DEAD AND TRAVELLING LOADS 



201 



To obtain the values of these shearing forces, first let the load 
cover DB (Fig. 225 (a)) and take moments about B ; 






.'. maximum positive shearing force S D = P = -j- (L - a) 2 ....... (9) 

2 1 v 

Now let the load cover AD (Fig. 225 (0)) and take moments 



about A ; 



QL 






a 

wa- 

2 

wa 2 



Hence, Maximum negative shearing force S D = Q 

^ j_/ 

Variation of a in (9) and (10) so as to obtain values for other 
sections will evidently produce two parabolic curves when plotted 
(Fig. 2 25 (<:)). The interpretation of this diagram is similar to that 
of Fig. 223 (d)). The end ordinates are of magnitudes z#L. 

Combined dead and travelling loads. If, in addition to the 
travelling or live loads, the dead loads be considered, diagrams of 
bending moments and shearing 

forces may be drawn separately < f\ 

for the latter. Combined dia- 
grams of bending moments and 
shearing forces may then be con- 
structed by adding algebraically 
the corresponding ordinates of 
the diagrams. This has been 
done in Fig. 226 for a uniformly 
distributed dead load and a 
single rolling load. 

In Fig. 226(0), ACB is the 
bending-moment diagram for the 
dead load and ADB is that for 
the live load ; AEB is the com- 
bined diagram. In Fig. 226 (<r), 
FGKH is the shearing-force 
diagram for the dead load ; FGL 
and FGM are the shear diagrams 
for the live load; FGKRN is 

P 1G. 226. Diagrams for a beam carrying a dead 

the shear diagram lor the com- load and a ingi rolling load. 




202 



MATERIALS AND STRUCTURES 



bined loads, and shows the shearing force on any section lying close 
to the left of the live load ; GFHTQ is a similar diagram for sections 
lying close to the right of the live load. The construction consists in 
making FN = FH + FL and joining NK ; also make GQ = GK + GM 
and join QH. Zero shearing force occurs at T and at R. Sections 
lying between F and T are subjected to positive shear only, 
those lying between R and G have negative shearing force only ; 
sections lying between T and R have to resist both kinds of shearing 
force. 

Maximum bending moments for a non-uniform travelling load. 
In designing bridge girders, it is necessary sometimes to consider the 



IB. 




FIG. 227. Bending-moment diagrams for a system of rolling loads. 

effects of a non-uniform travelling load. Fig. 227 illustrates a con- 
venient method ; bending-moment diagrams for the girder when the 
load is occupying several given positions are obtained first ; from these 
diagrams the maximum bending moment at any section is determined. 

AB is the girder resting on supports at A and B ; sections at E, D 
and C divide the girder into four equal bays. Three loads Wj , W 2 
and W 3 at fixed distances apart have been chosen, but it will be 
understood that the method applies to any number of loads. 

First let W l be vertically over B, and let the distances of W 2 and 
W 3 from A be a and b respectively. 

Take moments about A and set these off along AN, which is 
drawn at right angles to AB. 

Moment of W l = WjL, represented by AF. 
W 2 = W , FH. 

W = WJ HN. 



TRAVELLING LOADS 203 

Let Q! be the reaction at B, then the sum of the above moments 
is equal to the moment of Q T about A ; hence the moment of Qj 
about A is represented by AN. 

Join BF and produce the line of W 2 to cut BF in G. Join GH 
and produce the line of W 3 to cut GH in K. Join KN and also 
BN. Draw the vertical ESTUV. 

From the laws of proportion applied to similar triangles, the fol- 
lowing statements may be made : 

Moment of Qj about E = EV. 

W x E = ES. 

W 2 E = ST. 

W 3 E = TU. 

Now, by taking moments about E, we have 

Bending moment at E = moment of Qj - moment of W l 

- moment of VV 2 - moment of W 8 
= EV - ES - ST - TU 

= uv. 

In the same way M D = WX and M C = YZ. Hence the bending- 
moment diagram for the given position of the loads is the shaded 
diagram in Fig. 227. 

To obtain the bending-moment diagram when W l is vertically over 
the section C, instead of moving the loads, leave them in their 
original position and shift the girder towards the right until C is 
vertically under Wj. The end B will then be at B 2 and A will 
coincide with the original position of E. The reaction Q 2 at B 2 will 
be obtained by taking moments about E, giving 

Q 2 x B 2 E = moment of W x + moment of W 2 + moment of W 3 



= EU. 

Join UB. 2 , when it will follow, by similar reasoning to that already 
employed, that the bending-moment diagram for Wj over C is 

B 2 BGKWUB 2 . 

Similarly, when W x is at D, the diagram of bending moments will 
be B 3 B.jBGKWB 3 ; also when Wj is at E, the bending-moment 
diagram will be B 4 B 8 B 2 BGYB 4 . 

Measure these diagrams so as to obtain from each the bending 
moments at E, D and C. If these are tabulated, there will be no 



204 



MATERIALS AND STRUCTURES 



difficulty in obtaining the maximum value of the bending moment at 
each section by inspection of the table. 



Position of 
load Wj. 


Bending moments at sections. 


C 


D 


E 


B 








C 








D 








E 








A 












It should be noted that the loads may run on to the bridge girder 
from either end, and that either W 1 or W 3 may lead. The effect 
of this on sections lying equidistant from the middle of the span, 
such as E and C, may be taken account of fully by choosing the 
maximum of the tabular values for C and E as being the bending 
moment to which both E and C may be subjected depending on 
which way the load runs on to the bridge. 

W per unit Length 




FIG. 228. Portion of a continuous beam. 

Continuous beams. Let ABC (Fig. 228) be a portion of a beam 
which is continuous over several supports ; three of the supports are 
situated at A, B and C respectively, the spans being / x and L 2 respec- 
tively. For simplicity, the load is taken as w per unit length, 
uniformly distributed throughout. 

There will be bending moments at each support owing to the 
beam being one continuous piece ; let these be M A , M B and M c 
respectively. Erect perpendiculars AD, BE and CF to represent 
these bending moments (Fig. 228 (It)) and join DE and EF. Draw 
also the parabolic bending-moment diagrams AGB and BHC for the 



CONTINUOUS BEAMS 205 



two segments AB and BC taken as cut at A, B and C, and simply 
resting on the supports. Then, as has been shown for an encastre 
beam (p. 176), the difference between the bending-moment diagrams, 
shown shaded, will be the bending-moment diagram for the portion 
ABC of the continuous beam. 

It is evident that the solution will depend on the determination of 
M A , M B and M c . To determine these, we have the principle that 
if all the supports are at the same level, then the deflections at A, B 
and C must be zero, whatever may be the changes in deflection 
occurring in the spans. Hence, taking moments of area about A, 
the moment of ADEB must equal that of AGB ; also taking moments 
of area about C, the moment of BEFC must equal that of BHC. 

Taking moments about A, and remembering that the parabolic 
area is two-thirds that of the circumscribing rectangle, we have 

2 Wl? , L , , 7 /l /,, r \l\ 2 7 

-. ^ L ./ 1 .- 1 = M A / ] ^ + (M B -M A )^--/ 1 . 

It will be noted that the right-hand side has been obtained by 
splitting ADEB into a rectangle of height AD and base AB, and 
a triangle of height (BE - AD) and base equal to AB. The equation 
is reduced as follows : 



^M^ + JMB/i ............................... (l) 

Taking moments about C in the same manner, we have 

2 Wlz 4 / 2 , ,4 2 

3' ^4.- = Mc4- + (M B -M c )---4, 



= JM C 4 + JM B 4 ............................... (2) 

Add (i) and (2) : 






(/i 3 + /a 8 ) = JM A A + iM B (I, + 4) + JM c / 2 , 

24 



or (/ 1 3 + 4 3 ) = M A / 1 + 2M B (/ 1 + 4) + M c 4 ................... (3) 

4 

Should the spans be carrying uniformly distributed loads of 
different values, let w l and w. 2 be the loads per unit length on AB 
and BC respectively. Then (i) and (2) will become 

7H 7 3 

24 =*M A / 1 + pt B / 1 , 



206 



MATERIALS AND STRUCTURES 



Adding these and reducing as before gives 



(4) 



Equations (3) and (4) are cases of Clapeyron's theorem of three 
moments. By use of these, an equation may be written down for 
any three successive supports of a continuous beam. If there are 
n supports, there will be (n - 2) equations ; other two equations may 
also be written from the data supplied for the ends of the beam. 
Thus, if the beam simply rests on the support at each end, the 
bending moments at these supports will be zero, and the other 
equations will be sufficient in order to obtain the complete solution. 



i ton per foot 




A 

Tons 
16 
12 
8 

4 



-4 
-8 
-12 
-16 

FIG. 229. A continuous beam having three spans. 

EXAMPLE i. A continuous beam rests on four supports on the same 
level and carries loads as shown in Fig. 229. Find the bending moments 
at the supports. 

Equation (4) applied to A, B and C gives 
x i 3 \ /i - X20 3N 





N. Shearing 
\. forces 


\ 


'^^ 


B X 

(c) \. 


C N 



CONTINUOUS BEAMS 207 

Also, M A = o; 

.'. 422 + 3000= 7oM B + 2oM c ............................ (i) 

Equation (4) applied to B, C and D gives 



Also, MD = O; 

/. 3000 + 375 = 2oM B + 6oM c ............................ (2) 

From (i) and (2), 2ioM B +6oM c = 10,266, 



M B = 6\3 ton-feet. 
From (2), 



6oMc = 2649, 

Mc = 44-i5 ton-feet. 

EXAMPLE 2. Find the reactions of the supports of the beam given in 
Example i. 

To find RA, write down an expression for the bending moment at B, 
obtained by calculating the moments about B of the forces acting on AB. 



56-25 - 1 5R A = 36-3, 

R A = i-33 tons. 

In the same way, R B may be found by writing down an expression for 
Me, taking moments about C of all the forces acting on ABC. 



306-25 + 200 - (1-33 x 35) - 2oR B =44-i5, 



R B = 20-77 tons. 
To find RC, take moments about D of all the forces acting on the beam. 

506-25 +45o-(45xi-33)-(3ox 20-77) -ioR c = o, 

ioR c = 273-25, 

Rc = 27-32 tons. 

Also, R A + RB + Re + RD = the total load on the beam, 

Ro = 52-5 -(1-33 + 20-77 + 27-32) 
= 3-08 tons. 

In order to check the accuracy, calculate RD by taking moments about 
C of the forces acting on CD. 

(1-5 x 10 x 1 o )-(R D x io) = M c =44-i5, 
ioR D = 30-85, 
R D = 3.08 tons. 



208 MATERIALS AND STRUCTURES 

EXAMPLE 3. Draw the diagrams of bending moment and shearing 
force. 

S on the right of A= + R A = + 1-33 tons. 
S on the left of B = R A - (0-5 x 15) 

= i-33-7-5 = -6vi7 tons. 
S on the right of B = R A + R B -(o-5 x 15) 

= i -33 + 2077 - 7- 5 = + U-6 tons. 
S on the left of C = R A + R B -(o-5 X35)-(i X2o) 
= 1.33 + 20-77-17.5-20 
= - 15-4 tons. 

S on the right of C = R A + R B + R C - 17-5-20 
= 1-33 + 2077 + 27.32-37-5 
= + 11-92 tons. 

S on the left of D = - R D = -3-08 tons. 

The shearing force varies uniformly between the supports ; the com- 
plete shearing-force diagram is given in Fig. 229 (c). 

The following quantities, together with the bending moments at the 
supports, are required for the bending-moment diagram. They are 
obtained by calculating the bending moments at the middle of each span, 
assuming that the beam is cut at B and C. 

,. 'Zfi/i 2 0-5XI5XI5 

Bending moment at the centre of A.B = ~ =. - Q J J 

o o 

= 14-06 ton-feet. 
Bending moment at the centre of BC=^^= 1 ' 5 * 2 X 2 

o o 

= 7j ton-feet. 
Bending moment at the centre of CD= | 3 = 5 

o o 

= 18-75 ton-feet. 

The bending-moment diagram is given in Fig. 229 (b\ and is drawn by 
making BE and CF equal to MB and Me respectively and joining AE, 
EF and FD. GH, KL and MN are then set up from the centres of AB, 
BC and CD, and are made equal to 14-06, 75 and 18-75 ton-feet respec- 
tively. The curves AGB, BKC and CMD are parabolic. The difference 
of these diagrams, shown shaded, is the bending-moment diagram for 
the beam. Points of contraflexure (p. 180) occur at O, P, Q and R, as 
the bending moments are zero there. 

Plate girders. Plate girders are used instead of rolled I sections 
when the dimensions of the girder become large. Such girders con- 
sist of top and bottom flange plates (Fig. 231) and a web plate 
secured to the flanges by riveted angles. The flange plates, as may 
be observed in Fig. 230, increase in number towards the middle of 



PLATE GIRDERS 209 



the span, where the bending moment is large. The web plate is 
generally of uniform thickness in girders of comparatively small span ; 
in very large girders, in which the web is built of several plates 
placed end to end, the plates near the supports may be made thicker 
than those at the middle, thus making allowance for the larger 
shearing forces near the supports. In calculating the dimensions of 



Flo. 230. Side elevation of a plate girder. 

the parts, it is customary to assume that the flanges supply the whole 
of the resistance to bending and that the web supplies the whole of 
the resistance to shearing. The web is liable to buckling, and 
requires to be stiffened at intervals. For this purpose vertical 
stiffeners are riveted to the web plate at intervals as shown in 
Fig. 230 ; these are of closer pitch near the supports, and may be 
constructed of angles as in Fig. 231, or may be of T section. 

The method of finding the principal dimensions may be understood 
by study of the following example : 

EXAMPLE. A plate girder of 30 feet span with parallel flanges has to 
carry a uniformly distributed dead load of 2 tons per foot length, including 
the weight of the girder. Find the principal dimensions. 

Taking the depth as ^j th of the span gives a depth of 2-5 feet. The 
breadths of the flanges may be B V h f the span, giving 10-5 inches 
for this dimension. 

The total load will be 60 tons. The maximum bending moment will be 

WL 60x30 

M max =--Q- = ^- = 225 ton-feet. 

o o 

Taking working stresses of 7 tons per square inch pull and 6 tons per 
square inch push, the sectional areas of the flanges at the centre of 
the span may be found. Let these be At and A c square inches for the 
bottom and top flanges respectively. The moment of resistance of the 
section to bending will be 7 At x the depth of the girder, or 6A C x the depth, 
according as the bottom or the top flange is considered. Equating these 
to the bending moment at the centre of the span gives 
?A t x 2^ = 225, 

At = --=12-85 square inches. 
17-5 2 

6A c x 2^=225, 

225 

Ac *== 15 square inches. 
^ 

D.M. O 



210 



MATERIALS AND STRUCTURES 




I0i"x|" 

FIG. 231. Section of a plate girder. 



It will be noted, from inspection of the section given in Fig. 231, that two 
rivet holes occur in each flange. In the case of the flange under push, it may 

be assumed that the rivets fill the holes 

|- rivets _ perfectly and that no compensation is 

necessary. In the case of the flange 
under pull the sectional area of the two 
rivet holes must be deducted from the 
total sectional area of the flange plates. 
The rivets in the present example may be 
taken as f" in diameter ; it is not customary 
to exceed this dimension to any extent on 
account of the difficulty of closing larger 
rivets by hand, as has sometimes to be 
done during erection. The sectional area 
of the horizontal limbs of the angles used 
for securing the flange plates to the web 
plate may be included in the flange area. Angles 3^" x 3^" x ^" are used 
in the present case. 

Taking the bottom flange first, in which the rivet hole allowance must 
be made, we have 

Net area of the horizontal limbs of two angles = 2(3^-})^ 

= 2-75 square inches. 
Using plates f" thick, 

Net area of one plate 10-5" x %" (io-$ - 1-5)! 

= 3-375 square inches. 

If three such plates are used, Net area = 3 x 3-375 

= 10-125 square inches. 

Adding this to the area provided by the angles, we have 

Sectional area supplied in bottom flange = 2- 7 5 + 10-125 

= 12-875 square inches. 

This is slightly in excess of the area actually required, viz. 12-85 square 
inches, and may thus be adopted with safety. 

Considering now the top flange, which is under push, and using 
the same dimensions of angles and also the same thickness of plates, 
we have 

Area of the horizontal limbs of two angles = 2 x 3^ x \ 

= 3-5 square inches. 

Area of one plate, 10-5" x " = 3-94 square inches. 
Area of three plates = 11-84 square inches. 
Total flange area = 3- 5 + 1 1 -84 

= 15-34 square inches. 

The area actually required is 15 square inches ; hence the assumed 
dimensions may be adopted. 



PLATE GIRDERS 



211 



The method of finding the lengths of the flange plates may be under- 
stood by reference to Fig. 232. The bending-moment diagram for the 
girder is drawn on a base AB, and is also redrawn inverted. The moment 
of resistance of the angle limbs is calculated, and also the moment of 
resistance of each plate separately, making allowance for rivet holes in 
the cases of those under pull. These are set off vertically from AB and 
horizontal lines ruled. The angles and the plates adjacent to the web 
must run the whole length of the girder. The other plates may stop at 



Ton- feet. 

250 



200 



150 



Upper flange 



Plate N?2 



100- 

50 

A 

50- 
100- 
150- 
200 



Plate NI 



Angle 



Angle 



Plate N? i 



Plate N2 



Plate N.3 



25QJ Lower flange 

FIG. 232. Construction for obtaining the flange-plate lengths in a plate girder. 

the points where their moment-of-resistance lines cut the bending- 
moment diagram, but are made a little longer in order that the riveting 
at the ends of the plates may be carried out properly. 

The thickness of the web plate may be found on the assumption that 
the shearing force is distributed uniformly over the section of the web. 
Assuming a shearing stress of 6 tons per square inch and taking a section 
close up to either support where the shearing force is a maximum and 
attains the value of 30 tons, the area required will be 

Sectional area of web = ^ = 5 square inches. 

For a plate 30 inches deep this would give a thickness of o 5 o = inch. 
To guard against the effects of rusting, no plate should be less than 
;4 inch thick ; further, buckling has to be considered ; hence the web may 
be taken as 2 inch thick 



212 



MATERIALS AND STRUCTURES 



The stiffeners should have a pitch not exceeding the depth of the girder, 
and the pitch may be halved near the supports. 

To find the pitch of the rivets connecting the flanges to the web, taking 
a section near the end of the girder, the shearing force is 30 tons, and as 
the girder is 2-5 feet deep this will be equivalent to an average shearing 
force of 304-2-5 = 12 tons per foot. Now the shearing force per foot of 
vertical section must be equal to the shearing force per foot of horizontal 
section (p. 126) ; hence the resistance which must be provided by the 
rivets will be 12 tons per horizontal foot. 

Taking rivets | inch in diameter, a shearing stress of 6 tons per square 
inch and a bearing stress of 10 tons per square inch, we have 

Bearing resistance of a f inch rivet in a f inch plate = | x |x 10 

= 2-81 tons. 

Shearing resistance, under double shear = if x x 6 



= 4-64 tons. 



Hence the bearing resistance must be taken. 



12 



Number of rivets per foot= ~- = 4 ; 

2 f o I 

.'. pitch = 3 inches. 

As the shearing force diminishes for sections taken nearer to the centre 
of the span, the pitch may be increased towards the centre. It is, how- 
ever, undesirable that the pitch should change too frequently. To find 
the section at which the pitch may be changed to 6 inches is equivalent 
to finding the section at which the shearing force is half the maximum, 
viz. 1 5 tons. This will occur evidently at quarter span ; hence the middle 
1 5 feet of the girder may have a rivet pitch of 6 inches. 

Parallel braced bridge girder. Fig. 233 shows in outline a bridge 
constructed of two Pratt girders A and B, one on each side of the 




FIG. 233. Bridge having two Pratt girders. 

bridge ; the roadway is supported by cross girders C which are 
attached to the main girders at the lower panel points a, b, d,f, etc., 
and transmit the road loads W 15 W 2 , W 3 , etc., to the girders at these 
points. The main girders each consist of two parallel booms, con- 
nected by inclined web bracings and vertical bars. The forces in 
the various parts of the girders are found generally by calculation 
in the following manner. 



BRIDGE GIRDERS 



213 



Forces in the top boom. Consider the bar ce (Fig. 233) ; if this bar 
were dropped out, the portion acd would rotate about d. Take 
moments about d of all the forces acting on 
acd which is shown separately in Fig. 234. 
T ce is the force in ce; W 3 has zero moment. 

(T ce x D) + (W l x ad) + (W 2 x bd) = P x ad, 
or T ce x D = (P x ad) - (W x x ad) - (W 2 x bd). 

The right-hand side of this expresses the 
bending moment at d\ writing this as 




the equation gives 



r~r* J - TJ -f-6 



(') 



In the same way, 



And 



There is no necessity for calculating the forces in the bars on the 
other side of k, as, with symmetrical loading, it is evident that the 
forces will repeat themselves. 

Forces in the bottom boom. It is evident that, as there are only 
horizontal and vertical forces at the joint b (Fig. 233), the force in 
ab will be equal to that in bd. If the bar bd be dropped out, then 
the portion abc will rotate about c. Taking moments about c of 
all the forces acting on abc (Fig. 235), we have 
(T 6d x D) + (W l xat) = Px ab, 



w 



b T 



bd 



W B 
FIG. 235. 



D 




Hence, 



dropped out (Fig. 233), adec will rotate about e (Fig. 236). 

(Tv x D) + ( Wj x ad) + (W 2 x bd) = P x */, 

T,// x D - (P x ^) - (Wj x ^) - (W 2 x ^) 



214 MATERIALS AND STRUCTURES 



J D"' ">' 

In the same way, T/y t = -^. 

Forces in the inclined braces. Considering the bar ae (Fig. 233), 
evidently the horizontal component of the force in it will 
be balanced by the force in abd. Let be the angle of 
inclination of the brace to the horizontal (Fig. 237). Then 

T a d T a c COS 0, 

or Tac = Tad sec (4) 

In the same way, the horizontal component of the FT ~j 




force in cd is balanced by the forces in bd and df 
-T w = 




cd 



(Fig. 238). Hence, 

bd df 



FIG. 238. 

or Ted = (Tc// - TM) sec ..................... (5) 

' 



In the same manner, T e /= (T// t - T,//) sec 



The force in gh requires special treatment. If the forces in gh 
and hm be both resolved vertically, the sum will be 
equal to W 5 (Fig. 239). Hence, 



W 
Tgh = *. cosec0 (7) 

FIG. 239. 2 

Forces in the vertical bars. The only force possible in be is the 
load W 2 applied at its lower end (Fig. 233). There can be no force 
in hk y as there is no load at its upper end. Consider 
the bar de\ the force in this bar is balanced by the ir 
vertical component of the force in the inclined brace 
ef which is connected to its top. Hence (Fig. 240), FIG. 240. 

In the same way, T/ r/ = T (J t t sin 0. 

The forces in the various members due to the dead load may be 




BRIDGE GIRDERS 



215 



found also by graphical methods, 
for the girder under discussion. 



P'ig. 241 shows the force diagram 



M 



ly/N 


X 


|\R 
J Q\ 


X 


X 


/<\X 


A \j 


L \ 


K 


1 H < 


- G i 


1 F i 


[ E i D J 


C 



v/ 


7j\ 


^ 

D 
E 

FM 
A 
BG 

H 

K 

1 


/ \ 


z s 


>1^ Yv 


^p \ 


XK 


S 2 


Q 


\ X 


\ X 


E 



FIG. 241. Graphical solution of a Pratt girder carrying a dead load. 

Live load forces. Suppose that a uniform live load, of length 
sufficient to cover the entire span, may run from either end on to 
the girder shown in Fig. 233. It is evident that maximum bending 
moment will occur at all sections when the span is covered wholly 
by the live load ; hence maximum forces will then occur in all the 
members of both top and bottom booms. If both live and dead 
loads are uniform, producing a ratio of live to dead load per foot 
length of girder equal to , then the bending moments at any section, 
produced by these loads, will also have the ratio , and the force in 
any boom member due to the live load will be n times the force in 
the same member due to the dead load. 

In finding the maximum live-load forces in the inclined bars of 
the web, it may be taken that the shearing force in any panel is 
balanced by the vertical component of the force in the inclined bar 
belonging to that panel. Maximum force in any inclined bar will 
therefore occur when maximum shearing force exists in the panel to 
which the bar belongs. The following simple practical rule gives 
results sufficiently accurate. Assume that maximum pull in the 
inclined bars cd, ef t gh (Fig. 233) occurs when each panel point 
situated on the right of the bar is carrying a load W, W being the 
live load per panel ; also that the maximum push in the inclined 
bars hm, fo, tiq, occurs when each panel point situated on the right 
of the bar is carrying a load equal to W. Under these conditions. 



216 



MATERIALS AND STRUCTURES 



the shearing force in the panel will be equal to the left-hand 
reaction P, P being calculated from the loads applied to the selected 
panel points. The force in the bar may then be found from the 
product P cosec 6. The maximum forces in the end bars ac and qr 
(Fig. 233) may be found by resolving vertically and horizontally the 
forces at a and r when the span is wholly covered by the live load. 

It will be noted that corresponding bars on each side of the 
middle of the span, such as cd and tiq^ undergo reversal from pull to 
push, owing to the condition that the live load may run on to the 
girder from either end of the bridge. In general it will be found, 
when the dead-load forces are combined with the live-load forces, 
that the inclined bars near the ends of the girder have forces fluctu- 
ating between maximum and minimum pulls, and that a few only 
near the middle of the span undergo actual reversal from pull to 
push. It is customary to design the inclined bars in such girders 
to withstand pull only, and to counterbrace those panels in which 
the inclined bars suffer reversal from pull to push as shown by the 
results of the calculations indicated above. Counterbracing is 
shown by dotted lines in the two centre panels of the girder shown 
in Fig. 233. It is assumed that the counterbraces fk and kl take 
as pulls the forces which would otherwise have to be carried as 
pushes by gh and km. 

Having found the maximum forces which may occur in the 
inclined bars due to the live load, the forces in the verticals may 
be found by considering the upper panel points c, e, g, etc. (Fig. 233). 
The force in any vertical bar will be equal to the vertical component 
of the force in the inclined bar which is connected to the same upper 
panel point. 

Bridge girder of varying depth. The principles underlying the 
solution of a bridge girder of varying depth may be understood by 




q 

FIG. 242. Bridge girder of varying depth. 



reference to Fig. 242. A single load W is alone considered, and 
P and Q are calculated first. 



BRIDGE GIRDERS 



217 



To find the force in the member be belonging to the bottom boom, 
it will be noticed that, if the bar be removed, Kbed will rotate about 
e. Taking moments about e, we have 

Force in be x be = P x Kb = M& ; 



. . force in vc=^. 
be 



(I) 

To find the force in the member ef of the top boom, the rotatiop 
point would be c if the bar were dropped out. Taking moments 
about c, first drawing ex perpendicular to ef, we have 
Force in efx cx= P x Ac= M c ; 

.'. force in ef= c (2) 

To find the force in ce, reference is made to Fig. 243, showing 
hbed together with all the forces acting on it. 1\ , T 2 and T are the 




FIG. 243. Construction for finding the force in a diagonal brace. 

forces in ef, be and ec respectively. T l and T 2 intersect, when pro- 
duced, at z, and hence have no moment about z. Draw zm perpen- 
dicular to the line of T and take moments about z. 
P x Az = T x zm, 




zm 



(3) 



To find the force in be, resolve horizon- 
tally and vertically all the forces acting at 

FIG. ^--Construction for finding the tO P J mt ' ( Fi g' 2 44)- For balance 
the force in a vertical member. o f the Vertical Components, W6 haVC 

T sin 7 + T! sin a = T 4 + T 3 sin ft 

T 3 sin/2 (4) 



218 



MATERIALS AND STRUCTURES 



Double Warren girder. As an example of another method of 
solution, consider the double Warren girder shown in Fig. 245 (a). 
The girder may be taken as made up of two component girders 
shown separately in Figs. 245 (b} and (c\ each carrying the loads which 
hang, in the complete girder, from panel points belonging to the 
component girder. Each component girder should be solved 
separately. The force in any member of the complete girder may 
.then be found by adding algebraically the forces in the corresponding 
bars of the component girders. 




FIG. 245. Double Warren girder and the component girders. 

Assume that the bracing is at 45, as is often the case in this 
type of girder ; also that the proportion of each load which is borne 
by each support takes the shortest route between the panel point and 
the support. Consider W x (Fig. 245 (b)) ; |W a is supported at & and 
-J-Wj is supported at n. The iWj arrives at g after traversing ha as pull 
and ag as push, thus producing forces -fWjv/2 pull in ah and W 1 
push in ag. The -J-Wj arrives at n after traversing /zras pull, cl as 
push, le as pull and en as push, and proauces forces equal to 
iW lN /2 in each of these bars. 

In the same manner, W 3 arrives at n by producing sW 3 \/2 pull in 
le and fW 3 \/2 push in en ; also |W 3 arrives at^by producing |W 3 \/2 
pull in le, |W 3 s/2 push in c/t, |-W 3 v/2 pull in ha and ?W 3 push in ag. 



REINFORCED CONCRETE BEAMS 



219 



The total forces in these bars in Fig. 245 (/;) may be found now by 
adding algebraically the results calculated for each. The forces in 
the boom members are best found by calculation from the bending 
moments in the manner described on p. 213. It will be noted that 
there are no forces in gh, e/Sindfti. 

The solution of the other component girder (Fig. 245 (<r)) is obtained 
in a similar manner. The force in any member such as be in 
Fig. 245 (a) will be found by adding algebraically the forces in ac 
(Fig. 245 (b)) and bd (Fig. 245 (c)). 

In girders of the double Warren type containing a large number 
of panels and uniformly loaded, the assumption may be made that 
the inclined bars in any panel share equally the shearing force in 
that panel. This assumption should not be made if the number of 
panels is small, as it then leads to absurd results. 

If vertical bars bh, ck, etc., be added to the girder shown in 
Fig. 245 (a), it may be assumed that each vertical bar transfers 
one half the load applied at the lower panel point to the upper 
panel point, and the solution may then be obtained in the same 
manner as before, with the vertical bars left out. 

Reinforced concrete beams. In Fig. 246 (a) is shown the section of a 
concrete beam having steel reinforcement bars near the bottom edge. 



k 6 



T" 
xd 

M 4 




T 
i 




/ 






(\-x)d 

* 




d 

\_ _-. 










1) 








/ 










(a) 




E !#- J 


B 


(c) 


B 



H 

d(Hx) 



(d) 



FIG. 246. Reinforced concrete beam. 



In making calculations regarding the strength of beams of this type, 
it may be assumed, as has been done for metal beams, that there is 
pure bending, that there is no resultant pull or push along the length 
of the beam and that cross sections which were plane in the unloaded 
beam remain plane when the beam is loaded. It follows from the 
last assumption, that the strains of longitudinal filaments will be pro- 
portional to the distance from the neutral layer (p. 1 43). Hence the 
strains all over any section AB (Fig. 246 (/;)) may be represented in the 
side elevation of the beam by a sloping straight line CE which passes 
through the neutral axis at O, giving the two strain diagrams AOC 



220 MATERIALS AND STRUCTURES 

and BOE, the horizontal breadths of which show the strain at any 
point. 

Let s c be the maximum strain in the concrete under compression, 
represented by AC (Fig. 2 46 (<)), and let s s , represented by DF, be 
the strain in the steel. Let d be the depth of the beam measured 
from the top to the centre of the reinforcement bars, and let xd be 
the distance of the neutral axis from the top. Then 

**- xd x / r \ 

s g (i-x)J~i-x 

Let c c be the push stress in the concrete corresponding to the 
strain s c and connected with it by 

E - Cc 
Ec ~V 

where E c is Young's modulus for the concrete. Also let t K be the 
pull stress in the steel corresponding to the strain s^ the connection 
being fi 

Es= v 

where E 6 is Young's modulus for the steel. Then 

E 6 . _ t^ f^. _ ^ ^_ 
E c ~ s s c c ~ c c ' s s 



The ratio of -^, denoted by m, is rather variable owing to the 

E c 

nature of concrete ; the average value of 15 is taken in practice ; 
hence the above result may be written 

*. *. s ........ ( ) 

c c (XT*) 

It is customary to allow safe stresses of 600 Ib. per square inch 
push in the concrete and 16,000 Ib. per square inch pull in the 
steel. Suppose that the section is so proportioned as to secure that 
these values occur simultaneously on a certain load being applied. 
Then, from (3), l6j000 x 

~~6oo~" F^ =I5 ' 
whence ^ = -^ = 0-36 ...................... (4) 

A section so designed is referred to generally as an economic 
section. 

In estimating the strength of the section to resist bending, it is 



REINFORCED CONCRETE BEAMS 221 

usual to disregard the stresses in that portion of the concrete lying 
below the neutral axis, and hence under pull stress. It follows that 
the stress diagram for the section will resemble that shown shaded in 
Fig. 246 (<r), in which push stress on the concrete is proportional to the 
distance from the neutral axis, c c being the maximum value, and the 
stress t g on the steel is assumed to be distributed uniformly over 
the steel. These stresses will give rise to equal resultant forces C 
and T on the concrete and steel respectively (Fig. 246 (</)), equal 
because there is no resultant force along the length of the beam. 

Let p be the ratio of the area of the steel to the rectangular area 
bd (Fig. 246 ()), and let A, 9 be the total sectional area of the steel bars 

in square inches. Then A s = pbd, (5) 

and T = t s p&d. (6) 

Also, Area of the concrete under push = bxd (Fig. 246 (a)). 
Average push stress in the concrete = |^ c . 

Total push in the concrete = C = \c ( bxd. (7) 

Also T = C; 



or t s p 

I Hi* (8) 

I , 1 ^^)< fr m ^)) (9> 

If the beam is of the economic section, then x is 036 from (4), and 
(9) becomes , = 0-00675 

= 0-675 percent (10) 

To obtain the moment of resistance to bending, we must calcu- 
late the moment of the couple formed by T and C. C acts at a 
distance \xd from the top : hence the distance between C and T is 

Hence, Moment of resistance = Cd( i - J#), ( 1 1 ) 

or =Td(i-x) (12) 

From (7) and (n), we have 

Moment of resistance = ^dbxd^ ( i - \x\ (13) 

or, from (6) and (12), 

Moment of resistance -f 8 pbd' 2 (\ - i.r) (14) 



222 



MATERIALS AND STRUCTURES 




FIG. 247.- Reinforced concrete T beam, NA below the 



Reinforced concrete T beams are much used. There are two 
cases, one in which the neutral axis falls below the slab (Fig. 247 (a)) 

and the other in which the 
neutral axis falls within 
the slab (Fig. 248(0)). As 
the concrete under pull is 
neglected, the stress diagram 
for the latter (Fig. 248^)) 
is identical with that for a 
beam of rectangular section 
(Fig. 246 (<:)); hence all the 
results already found apply 
to this case. In the former 
case (Fig. 247 ()), it is 
customary to disregard the shaded area, representing a small portion 
of the concrete under push. The stress diagram will then take the 
form shown in Fig. 247 (/;), 
and the equations become some- 
what altered. 

It should be noted that re- 
inforced concrete buildings are 
practically monolithic; columns, 
beams and floors are so con- 
structed as to form one piece. 
Hence all such beams must be 
regarded as fixed at the ends. It has been shown already (p. 177) 
that in such beams the bending moment reverses in sense near 
the walls ; hence the top sides of the beams near the walls will be 
under pull, and some of the reinforcement bars should be brought 
diagonally upwards and run near the top of the section over the 
supports. 

EXAMPLE i. A reinforced concrete beam is 9 inches wide, and is to 
have a moment of resistance of 200,000 Ib.-inches. The stresses of 600 Ib. 
per square inch on the concrete and 16,000 Ib. per square inch on the steel 
are to be attained simultaneously. Ratio of elastic moduli = 15. Find 
the depth of the beam and also the sectional area of steel required and 
the position of the neutral axis. 




FIG. 248. Reinforced concrete T beam, NA 
within the slab. 



From (10), 
From (4), 
From (13), 



0675 per cent. 



REINFORCED CONCRETE BEAMS 223 






2 X 2OO,CXX) 



0-36x9x600(1 -0-12) 

= 15-3 inches. 
Distance from the top to the neutral axis 

= ^=0-36 x 15-3 
= 5-51 inches. 
Sectional area of steel = A = />&j? 

= 0-00675 XQX 15-3 
=0-93 square inch. 

EXAMPLE 2. A reinforced concrete beam 9 inches wide by 18 inches 
deep has three steel reinforcement bars, each 075 inch in diameter. 
Find the position of the neutral axis and the moment of resistance. 
Neither of the stresses of 600 Ib. per square inch for the concrete and 
16,000 Ib. per square inch for the steel may be exceeded. Take the ratio 
of E to 0=15. 

Sectional area of steel = A, = 3 x gg = 3X22X9 

4 4x7x16 

= 1-33 square inches. 

From (3), ^^^ ................................. (0 

Also, T = C; 



Equating (i) and (2) above, we have 

I5(i-,r) = 8ur 
x 1-33' 



19-95 -19.95^ = 
8 ix* + 1 9'95-r -19-95=0; 
whence ;tr =0-387. 

Distance of the neutral axis from the top =^=0-387 x 18 

= 6-96 inches. 



or / 

Suppose t s be taken as 16,000. Then 

16,000 
c c = -g- -=673 Ib. per square inch, 



224 MATERIALS AND STRUCTURES 



a value inadmissible by the data. Take, therefore, c c as 600 it), per 
square inch, giving 

/ s = 6oox 23-8 

= 14,280 Ib. per square inch, 
and c c = 600 Ib. per square inch. 

From (14), 
Moment of resistance = t s pbd*( i - $x) 



= 14,280 x 1-33 XI8(I-^2?7) 

= 298,000 Ib.-inches. 

EXAMPLES ON CHAPTER IX. 

1. A steel bar is 20 feet long and has a sectional area of 4 square 
inches. Find the work done while a pull of 24 tons is applied gradually. 
Take = 13,500 tons per square inch. Find also the energy stored in a 
cubic inch of the bar. 

2. Suppose, in Question i, that the load is applied suddenly, and 
calculate the maximum stress produced. What will be the momentary 
extension of the bar ? 

3. It is found that a steady load of 400 Ib. resting at the middle of a 
beam produces a deflection at the centre of o-oi inch. What central 
deflection would be produced by a load of 100 Ib. dropped on to the 
middle of the beam from a height of 16 inches ? 

4. A certain steel bar in a girder carries a constant pull of 20 tons 
owing to the dead load. The live load produces in the same bar forces 
which range from 60 tons pull to 10 tons push. Find the working stress 
and the sectional area of the bar. Take ultimate tensile strength = 30 tons 
per square inch. 

5. A single load of 10 tons rolls along a girder of 30 feet span. Draw 
curves showing the maximum bending moments and shearing forces at 
every section. State the scales. 

6. Answer Question 5 for a uniformly distributed travelling load of 1-5 
tons per foot length which may cover the whole span. 

7. Supposing that the girder in Question 5 is uniform in section and 
weighs 8 tons. Draw the diagrams of M and S for the dead load. Then 
combine these diagrams with those already drawn for the single rolling 
load in order to show the effects of combined live and dead loads. 

8. A girder of 40 feet span is traversed by three concentrated loads of 
6 tons each at 7 feet centres, followed at an interval of 6 feet by a 
uniformly distributed load of 0-5 ton per foot. Find graphically the 
maximum bending moments at sections of the girder taken at 5 feet 
intervals. The load may run on to the girder from either end. 

9. A continuous beam of length 50 feet rests on four supports on the 
same level. The left-hand span is 20 feet and the others are 1 5 feet each. 
The left-hand span carries a uniform load of 2 tons per foot, the other 



EXERCISES ON CHAPTER IX. 225 

spans carry uniform loads of I ton per foot. Find the bending moments 
at the supports. 

10. In Question 9, find the reactions of the supports. 

11. In Question 9, draw diagrams of bending moments and shearing 
forces for the complete beam. State the scales. 

12. A plate girder 24 feet span, 2 feet deep, flanges 10 inches wide, 
carries a uniformly distributed load of 45 tons. The angle sections are 
3' 5 x 3'5 x *5 m inches. Take stresses as follows : pull, 7 tons per square 
inch ; push, 6 tons per square inch ; shearing, 6 tons per square inch ; 
bearing, 10 tons per square inch. Find the sectional area of each flange ; 
state the number and thickness of plates required for each flange at the 
middle of the span. What thickness of web plate would be suitable? 
If the rivets are 075 inch in diameter, what will be the pitch of those near 
the ends of the girder ? 

13. In Question 12, find the length of each plate in (a) the top flange, 
() the bottom flange. 

14. A Pratt girder (Fig. 241) 48 feet span has 6 equal bays of 8 feet 
each. The bracing bars make angles of 45 with the horizontal. There 
is a uniform dead load of i ton per foot length. Find the forces in the 
horizontal top and bottom bars of the two central bays ; also those in 
the two inclined bars nearest to one support. Find also the force in the 
vertical bar second from one support. 

15. In Question 14 a uniform live load of 1-25 tons per foot travels 
along the girder. Find the maximum forces it will produce in the same 
bars. The load is long enough to cover the whole girder. 

16. A model reinforced concrete beam 3-5 inches wide by 4-25 inches 
deep from the top to the centre of the reinforcement has to be made so 
that stresses of 600 and 16,000 Ib. per square inch will occur in the con- 
crete and in the steel respectively. Taking the ratio of the elastic moduli 
as 15, find the percentage of reinforcement required, the sectional area of 
the steel, the position of the neutral axis and the moment of resistance of 
the section. 

17. A reinforced concrete beam of rectangular section 12 inches wide 
by 1 8 inches deep has three steel reinforcement bars each 1-25 inches in 
diameter. Find the position of the neutral axis and the moment of 
resistance. Stresses of 600 and 16,000 Ib. per square inch respectively 
for the concrete and steel must not be exceeded. Take the ratio of the 
elastic moduli as 15. 

18. Experiments upon some wrought-iron bars showed that a per- 
manent set was taken when the bars were strained to a degree greater 
than that produced by a stress of 20,000 Ib. per square inch, but not when 
strained to a less degree. At that point the average strain was 0-0006 
foot per foot of length ; what was the resilience of this quality of iron in 
foot-pounds per square inch section per foot of length ? (I.C.E.) 

19. An iron bar 10 feet long having = 14,000 tons per square inch 
and a limit of elasticity = 14 tons per square inch is subjected to shocks of 
a total value of 224 foot-pounds. The bar is not to have any permanent 
set produced in it, this being guaranteed by the adoption of a factor of 
safety of 2. Find the required sectional area of the bar. (I.C.E.) 



226 MATERIALS AND STRUCTURES 

20. A vertical steel rod, 10 feet long, the cross section or which is 
i square inch, is fixed at its upper end and has a collar at its lower end. 
An annular weight of 300 Ib. is allowed to fall through a height of 
3 inches upon this collar. Determine the maximum intensity of stress 
produced in the steel rod if Young's modulus is 12,500 tons per square 
inch. (B.E.) 

21. Two bars, A and B, of circular section and the same material, are 
each 1 6 inches long. A is I inch in diameter for 4 inches of its length 
and 2 inches in diameter for the remainder ; B is i inch in diameter for 
12 inches of its length and 2 inches in diameter for the remainder. A 
receives an axial blow which produces a maximum stress in it of 10 tons 
per square inch. Calculate the maximum stress produced by the same 
blow on B. How much more energy can B absorb in this way than A 
without exceeding a given stress within the elastic limit of the material. ? 

(L.U.) 

22. A double Warren girder (Fig. 245) is 50 feet span and 10 feet deep 
and has five equal bays of 10 feet each. It is supported at the ends 
and carries a load of 12 tons at each of the four lower panel points 
(48 tons in all). Find the forces in the members. State the assumptions 
made. (L.U.) 



CHAPTER X. 

COLUMNS. ARCHES. 

Ties and struts. Those portions of a structure which are intended 
to be under pull are called ties ; parts under push are called struts, 
or columns. Columns are usually vertical pieces intended to carry 
weights. There is an essential difference which modifies greatly the 
method of calculating the strengths of ties and struts ; a loaded tie 
exhibits no tendency to bend if it is straight originally, and will tend 
to become straight if originally curved. A strut, if originally curved, 
will have its curvature increased 
by application of the load, and, P /- 9 r^\ P 



if straight at first, may very easily 

be under such conditions of load- FlG> 2 ^_ A straight tie> 

ing as will produce bending; want 

of uniformity in the elastic properties of the material may produce 

a similar effect. 

A straight tie AB is shown in Fig. 249, loaded with pulls P, P, 

applied in the axis of the bar. It is evident there is no tendency to 

bend the tie, and any cross section CD, at 90 to the axis of the bar, 

will have a uniformly distributed 
pull stress. A bent tie bar AB 
is shown in Fig. 250(0). The 
nature of the stresses on CD may 
be understood by considering the 
equilibrium of one half of the bar 
( c ) (Fig. 250 (^)). It will be observed 
that there is a bending couple of 

FIG. 25 o.-Bent ties and struts. clockwise moment ?d ; this is 

balanced by the moment of resist- 

ance at the section CD, the latter being represented by the forces 
Q, Q. It is apparent that the bending couple IV is endeavouring 
to straighten the bar. 




228 



MATERIALS AND STRUCTURES 



Had a strut of similar shape been chosen, the forces acting on one 
half of it would be as shown in Fig. 250 (c}. Here the couple Pd is 
anti-clockwise, and tends to produce further bending. 

In each of these cases there will be two kinds of stresses on the 
section CD : (a) a stress of uniform distribution due to the axial 
force P' ; (<) a stress due to the bending couple, varying from a 
maximum push stress at one edge to a maximum pull stress at the 
opposite edge. An initially straight strut which has been allowed to 
bend under the load will have a similar stress distribution. It may 
be taken that the effects of bending may be disregarded in axially 
loaded straight ties, but must be taken account of in all struts. 

Euler's formula for long columns. This formula may be deduced 
by considering the bending of a long flexible column of uniform 
cross section and carrying a load applied 
axially. If such a column is perfectly straight 
to begin with, and there are no inequalities 
in the elastic properties of the material, the 
application of an axial load will not tend to 
bend the column. On increasing the load, 
a certain critical load is reached, the mag- 
nitude of which depends on the method of 
fixing the ends of the column; under this 
load the material of the column becomes 
elastically unstable. This condition is evi- 
denced by the column refusing to spring back 
if slightly deflected from the vertical, while 
it does so readily for loads lower than the 
critical load. The slightest increase in the 
FIG. 251. Euler's theory of load beyond the critical value will cause a 
small deflection imparted to the column to 

increase without limit, and the column collapses. It will be evident 
from what has been said regarding the conditions to be realised, that 
it is not possible to obtain a column of such ideal material and 
construction as will show perfect agreement under test with Euler's 
result. But the formula is of service in enabling other more 
practical formulae to be devised. 

Considering a long column AB (Fig. 251(0)) of uniform cross 
section and length L. Let both ends be rounded, or pivoted, in such 
a manner that, if bending does occur, the column will assume a 
curve resembling a bow (Fig.25i (<)). The effect of a load P applied 
at A in producing stresses at any section D will be understood by 



(a) 



COLUMNS 229 



shifting P from A to D as shown by P'; the section at D will 
evidently be under an axial load P' = P producing uniform stress, 
together with a couple of moment Py. The couple gives a bending 
moment, and the effect of this alone is considered in the following ; 
the stress at D caused by the axial load P' is disregarded, as it is 
small compared with that produced by the bending moment in the 
case of a long column. The maximum bending moment will be 
found at the middle section C, at which the maximum value of j, 
viz. A (Fig. 251 ()), occurs, and will be given by 

Maximum bending moment = M C = PA (i) 

Taking the equation for the curvature of a beam (p. 166), we have 
for the curvature at D : i M 



p 

Since will be constant for a given load on a given column, 

we may write i 

It may be shown readily that a curve, plotted so that its ordinates 
y (Fig. 252) are the sines of the angles a represented by its 




JH TT radians 

FIG. 252. Curve of sines. 

abscissae, possesses the same property, viz. the curvature at any 
point is directly proportional to the ordinate j> = sina. It may 
thus be inferred that the curve of the bent column is a curve of sines 
to some scale. The scales of x and y may be stated by taking the 
origin at A (Fig. 251 (<)), when AB = L will represent TT radians, 
AE = -|L will represent \K radians; also CE = A will represent 

sin-= i, and y will represent the sine of an angle AF, which will 

have a value -- IT. Hence, 
!_/ 

A X 7T 

y : A = sin TT : sin -, 

~ : * (4) 



230 MATERIALS AND STRUCTURES 

It can be shown that, if a curve be given by an equation show- 
ing the relation of y and x, the curvature at any point may be 

d^v 

obtained by finding the second differential coefficient, viz. , pro- 

doc 

vided that the curvature is not too great. Application of this 
process to equation (4) will lead to a result which may be equated 
to that of equation (2) above. Thus, 



dy A TT TT 

~r = A T ' cos T~ 
dx L L 



2 

= -pJ- (from 4) ..................... (5) 

This gives the curvature at D (Fig. 251 (b}\ viz. . The negative 

RD 

sign may be disregarded, as its only significance has reference to the 
position of the centre of curvature. Equating (2) and (5), we have 

P 7T 2 

- y _ i> 

El y L 2> 
It is important to note that y cancels, giving 

P__7^ 

El L 2 ' 
P = ^ ..................................... (6) 

This is Euler's formula for a long column having both ends 
rounded. The meaning to be attached to the deflection y disappear- 
ing from the final result is that no deflection will occur until a 
certain load P given by (6) is applied. When the load attains 
this value, any small deflection will increase indefinitely with the 
consequent collapse of the column. 

A more general way of writing Euler's formula is 



................................... 

where / is a function of the length L of the column. The value of / 
depends on the method of fixing the ends, a point which we now 
proceed to examine. 



COLUMNS 



231 



Effect of fixing the ends of columns. In the case of the column 
discussed in finding Euler's formula, the ends were taken as rounded 
and the column bent as a whole. There was no bending moment at 
the ends, and these may be looked upon as points of contraflexure 
(p. 1 80). Fixing the ends will produce a stiffer and consequently 
stronger column. This may be taken account of in the formula by 
writing, instead of L, the length of the column, the distance / 




v-JL 



FIG. 253. Various methods of end-fixing in columns. 



between the points of contraflexure in the actual curve of the 
bent column. Some cases are noted below; reference is made to 
Fig- 253. 

CASE A. Both ends rounded. This is the case examined above ; 
/=L. 

CASE B. Both ends fixed and so controlled that the forces P, P 
remain in the same vertical line. Here /= ^L. 

CASE C. One end (the lower) fixed ; the other end guided so 
that the forces P, P remain in the same vertical, but the column is 
otherwise free at this end to take up any direction. In this case 
/=o-7L. 

CASE D. Both ends are fixed so that the directions at the ends 
of the curve of the column remain vertical, but one end is free to 
move horizontally relative to the other end, so that the forces P, P 
are not in the same vertical when the column bends. Only one 
point of contraflexure will occur in the column itself; the position of 
the second point may be seen by producing the curve of the column 
downwards (shown dotted in Fig. 253 D). In this case /=L. 



232 



MATERIALS AND STRUCTURES 



CASE E. One end fixed, the other end perfectly free. In this 
case, the free end is a point of contraflexure. The second point 
may be obtained by producing the curve of the column downwards. 
Here /= iL. 

Using Euler's formula, it will be noticed that, as / has to be 
squared, the effect of fixing both ends of the column as in Case B will 
be to give the column four times the strength of the same column 
having both ends rounded. 

Curve illustrating Euler's formula. Euler's formula may be 
modified by writing 



where A is the sectional area of the column and k is the least radius 
of gyration of the section, i.e. k is taken with reference to that axis 
containing the centre of the area of the section 
for which I has the minimum possible value. 
It is evident that the column will bend in a 
plane perpendicular to this axis. Two instances 
are given in Fig. 254 (a) and (b) ; in each of 
these OX is the axis perpendicular to the 
plane of bending, and k should be taken with 
respect to OX. Inserting this expression for I 
in equation (7), (p. 230), we have 




Let /= #L, where n is a coefficient depending 
on the method of fixing the ends. Then 

p? *EA/*y 

# \L/ 



FIG. 254. Plane of bending 

in columns. Or 



2 E 



The left-hand side of this expresses the collapsing load p per unit 
of sectional area. Hence, 2 i7 / 7, \ 2 

/-^F(f) (8) 

\Li/ 

This will be in tons per square inch, provided the following units 
are employed : 

E = Young's modulus, in tons per square inch. 
k = the least radius of gyration, in inch units. 
L = the length of the column, in inches. 



COLUMNS 



233 



In columns of a given material and having a stated method of 

op 

fixing the ends, the quantity ^ will be constant ; hence / may be 

calculated for different ratios of L to k, and a curve may be plotted 
from the results. Fig. 



r .,, Collapsing load 

gives such a curve for mild Ton / per ^ Ln ^ 

steel struts having both ends 250 
hinged. In this case = i 
and E has been taken as 200 
13,500 tons per square inch. 

It will be noticed from 150 
Fig. 255 that, for small ratios 
of L to /&, the collapsing stress 100 
obtained is absurd. Only 
when the ratio is large is 50 
a reasonable value obtained. 
This leads to the conclusion 
that agreement of Euler's for- 



50 100 150 



200 250 
Ratio 



with practical results Of FlG - 255. Euler's curve for mild steel struts, both 

ends hinged. 

tests should be looked for 

only in the case of struts which are very long as compared with the 

cross-sectional dimensions. 

Ewing's composite formula. Sir J. A. Ewing has suggested a 
composite formula made up of the crushing strength of a very short 
block together with the elastic instability load of a very long column, 
both composed of the same material as the actual column. 

Let f t = the crushing strength of a short block, in tons per square 

inch. 

P! = the crushing load, in tons, applied axially. 
A = the area over which P x is distributed, in square inches. 

Then Pi=/cA (i) 

Let P 2 = the elastic instability load of a long column, in tons; 

the column having the same sectional area A as the 

short block. 

Then 



7T 2 EI 



2 ~ / 2 

Combining these in accordance with Ewing's method gives the 
following formula for P, the collapsing load of the ordinary practical 
column when loaded axially : f . 

P- . /C ,, (3) 



7T 2 EI 



231 MATERIALS AND STRUCTURES 

This formula has the advantage of being continuous, and does not 
give an absurd result for a column of any practical length. If / is 
very short, the second term in the denominator becomes very small, 
and may be disregarded. The formula then reduces to the expression 
(i) for the crushing load of a short block. If / is very long, the 
formula reduces to Euler's formula by neglecting the unimportant 
terms. 

Rankine's formula for columns. This formula is the one in most 
frequent practical use. It is practically the same as Ewing's, although 
in a slightly altered form. Thus, 

P- 



f c A 



Now, I may be written as I = 

where A is the sectional area and k is the least radius of gyration. 
Hence, 



/ C A 
~ 



It is apparent that J/= will be constant for a given material, and 

TT^hi 

may be written c, the value of which is to be determined by experi- 
ments on the collapsing strength of columns. Hence, 




This is Rankine's form of the formula, and gives the total collapsing 
load on the column. The collapsing load per square unit of sectional 
area will be given by P 



This will be in tons per square inch, provided f c is in tons per 
square inch and / and k are in inch units. It is assumed that the 



COLUMNS 



235 



column is of uniform section and is loaded axially. The least value 
of k should be chosen as in the case of the Euler formula, and the 
value of / is the same as in the cases given on p. 231 for various methods 
of fixing the ends. Values of f e and c are given in the table below ; 
a suitable factor of safety should be applied in order to obtain the 
safe load : 

COEFFICIENTS IN RANKINE'S FORMULA. 



Material. 


f G for collapsing load. 


c* 


Lb. per sq. inch. 


Tons per sq. inch. 


Cast iron - - - 


80,000 


36 


Te'oo 


Hard steel - - - 


70,000 


31-2 


SOW 


Mild steel - 


48,000 


21-4 


7MH) 


Wrought iron 


36,000 


16 


9057) 


Timber (varies greatly) 


7,200 


3-2 


75(7 



It will be evident on inspection of the Rankine formula that 
allowance is made both for direct crushing and for bending. Owing 
to the radius of gyration entering into the formula, due regard 
has been paid to the distribution of the material in the section, 
i.e. the shape as well as the area of the section has been taken into 
account. 

It is useful to plot curves from equation (2) showing the collapsing 
load per square inch of sectional area for different ratios of L to /&, 
varying the material and the method of fixing the ends. Such a 
curve for mild steel, both ends hinged, is shown in Fig. 256, the 
corresponding Euler curve being given on the same diagram. 

Another useful set of curves is given in Fig. 257 ; here the safe loads 
per square inch of sectional area for mild-steel, wrought-iron and cast- 
iron columns have been plotted for different ratios of l\k. The 
factor of safety employed is 5. The curves indicate that mild steel 
may always carry a higher stress than wrought iron ; also that, at the 
ratio of // = 4o approximately, the safe stresses on mild steel and 
cast iron are equal; hence equal columns of cast iron and mild steel 
having this ratio of /// would carry equal safe loads. Wrought iron 
and cast iron have equal safe stresses at a ratio of Ijk of 65 approxi- 
mately. In designing a column to carry a given load, cast iron 
is the material calling for the smallest sectional area for ratios of L to 



* Note that / should be taken from the cases shown in Fig. 253. For values 
of 2 , see p. 151. 



236 



MATERIALS AND STRUCTURES 



k under 40, and mild steel demands the smallest sectional area for 
ratios of L to k above 40. Wrought iron would require a smaller 
sectional area than cast iron for ratios of L to k over 65. 



Breaking Load 
Tons per ay in. 



Sale load 
Tons per of In. 
6 



Ss 






' 
















\ 






\ 
















\ 




\ 
















\ 




\ 


















^ 




% 


















\ 


1 


















^ 


\ 


\ 


















S 


x^ 


\ 






















^ 






















""**- 






















2O 4O 6O 80 100 I2O 140 160 1 

J 


10 2C 

Rat ic 



40 




80 



200 
Ratio 



FlG. 256. Rankine and Euler curves for mild 
steel columns, both ends hinged. 



FIG. 257. Rankine curves for columns of 
different materials ; ends hinged. 



Gordon's formula. The formula bearing Gordon's name, and 
formerly in common use, is 




where f c and a are experimental coefficients and d is the least 
transverse dimension of the section. The formula is objectionable, 

from the fact that no allowance is 
made for the distribution of the 
material over the section. For ex- 
ample, referring to Fig. 258, in which 
are shown an I section and a box 
section of equal areas, and alsx) having 
equal over-all dimensions, Gordon's 
formula would have the same value of 
d for both sections, viz. B, and would 
give the same value of collapsing load 
for both. Rankine's formula would give fairer consideration to the 
box section, which is obviously the stiffer and stronger, from the 
fact that the radius of gyration of the box section with reference to 




..*. 



COLUMNS 



237 



OX is greater than that for the I section, and hence would give a 
greater load for the box section. 

Secondary flexure in columns. Professor Lilly has pointed out 
that columns constructed of thin plates are liable to fail by secondary 
flexure, i.e. the column may not fail by bending as a whole, but by 
the material buckling over a short length. Lilly has made many 
experiments in support of his views, and has proposed a formula in 
which account is taken of the ratio of the thickness of the plate to 
the radius of gyration. Further experimental 
work is required in order to settle the values 
of the experimental factors involved. 

Recent tests made at the University of 
Illinois on built-up columns indicate that the 
stress distribution may be very erratic ; 
especially in the neighbourhood of riveted 
joints. It is admitted that our knowledge of 
the strength of columns, struts and com- 
pression members generally is far from being 
complete. At present, most designers rely 
on the Rankine formula coupled with a 
liberal factor of safety. 

Effect of a non-axial load. In Fig. 259 (a) 
is shown a column the axis of which is AB, 
i.e. AB passes through the centres of area of 
all horizontal sections of the column. A 
load P is applied at C at a distance a from 
the axis. P may be moved from C to A 
provided a couple of moment Pa is applied. 
We have now an axial load P' = P together 
with a couple Pa which will give a uniform 
bending moment at all horizontal sections 
of the column. Let A be the area of the 
section, then P' will produce a uniformly 
distributed push stress ^ given by 







A 


C 






H 




























i 






(o.) 


















B 










..* 



FIG. 259. Column carrying 
a non-axial load. 



A=A- 



The bending moment Pa will give a stress distribution similiar to 
that of a beam under pure bending, which will vary from a pull stress 
ft at the edge DE (Fig. 259 ()), to a push stress /. at the edge FG. 
Let m t and m c be the distance of these edges respectively from OX. 



238 MATERIALS AND STRUCTURES 

Then, using the equation (p. 146) 

M = ^I, 

m 

we have Pa = ^l=^-l. 



Tin / x 

Whence p t = ....................................... (2) 

Pam c , . 

and C = ~~ ..................................... 



The stress due to the bending moment will vary uniformly between 
these values, being zero at OX. The stress figure for the direct 
stress combined with the bending stress may be drawn as shown in 
Fig. 259 (r), where HKML shows the uniformly distributed stress/! 
and MNRQL shows the varying stress due to the bending moment. 
The resultant stress figure is shaded, and shows that the maximum 
push stress occurs at the edge FG (Fig. 259 ()), and is given by 

Maximum push stress =p l +p c ................... (4) 

In the case shown there will be no stress at S (Fig. 259^)); the 
portion SK will be under push stress, and SH will be under pull 
stress, the maximum value of the latter occurring at the edge DE 
(Fig. 259 (I})) and given by 

Maximum pull stress =pt~Pi ................... (5) 

The presence of pull stress in a metal column is permissible, but 
is objectionable in a column of stone, brick, or other construction in 
which the jointing of the blocks of material is not considered to be 
trustworthy under pull. The extreme limit of stress distribution in 
such cases is taken usually to be zero stress at one edge and 
increasing gradually to a maximum push stress at the opposite edge. 
Taking a rectangular section (Fig. 260 (a)) of dimensions b and d, 
the values of m c and m t will be equal ; hence p c and p t will also be 
equal, and the stress figure (Fig. 260 (/;)) shows that the condition 
of no pull stress is 

A-A = ........................................... (6) 



Also ' 

and 



6Pa 



COLUMNS 



239 



Hence, 



6Pa P 
M 2 bd 



or 



-(9) 







T 










(a) o 



It therefore follows that P may be applied at a distance not 

exceeding \d from the centre of the section j^ ^ _ _^ 

in a direction parallel to d. Similarly P 
may be applied at a distance not exceeding 
\b in a direction parallel to b. We may 
thus state that P may be applied within the 
middle third of OX and OY (Fig. 260(0)) 
without giving rise to pull stress. 

In the same way it may be shown, for 
a column of solid circular cross section of 
radius r, that the load may be applied 
anywhere inside a circle of radius 0-25^, 
having its centre on the axis of the (fr) 

Column, Without the production Of pull FIG. 260. - Rectangular section 
s j. ress carrying a non-axial load. 

EXAMPLE i. A wrought-iron stanchion of square section 2 inches x 2 
inches is 8 feet high. Both ends are fixed. Find the safe axial load, 
using a factor of safety of 5. 



-JL 
K 



Here 



/=IL = 48 inches. 

I=A 2 = ; 
12 ' 

, s 2 4 i . , 

k* = = - -=- inch units. 

f c = 1 6 tons per square inch, 
i 



9oob' 



P = - 



l6x 2X2 



64 



+0768 



= 36-2 tons. 
Safe load =^^ =7-24 tons. 

EXAMPLE 2. A cast-iron column, of circular solid cross section 6 

inches diameter is bolted down firmly at its lower end and is perfectly 

free at the top. If the length is 15 feet, what axial load would cause 
rupture ? 



240 MATERIALS AND STRUCTURES 

Here /=2L = 36o inches. */ c = 36 tons per square inch. 



r 9 i 

= = - inch units. =-7 




K 

37x7 

= 27-5 tons. 

EXAMPLE 3. Taking a factor of safety of 5, find the diameter of a solid 
mild-steel strut 6 feet long to carry safely a load of 3 tons. Both ends 
are rounded. 

Let d= diameter of strut in inches. 

Then & = = ^ inch units. 

4 16 

/=L = 72 inches. y c = 2i-4 tons per square inch. 

A= V- C= 7^' 

Also, Collapsing load = P = 3X5 = i5 tons. 



15 = 




or 

i6-8i^ 4 - i$d' 2 - 165-9=0 ; 
whence d= 1^9 inches. 

Straight-line formula. Very fair approximation to the strength of 
a strut may be obtained by use of a straight-line formula, i.e. one for 
which the graph is a straight line, and the calculations required in 
designing a strut to fulfil given conditions become much simpler. 
The usual form of such formulae is 



ARCHES 



241 



where f c is the safe stress per square inch of sectional area of the 
column, /is the safe stress for a short block of the same material and 
c is a coefficient depending on the material, and ranging in value from 
0-005 for mild steel and wrought iron to 0-008 for cast iron and timber. 
Arches. In Fig. 261 (a) is shown a number of loads W 1} W 2 , etc., 
supported by an arrangement of links ABCDE, forming part of a 
link polygon. The construction necessary to determine the directions 
of the links is given in Fig. 261 (b) and has been explained on p. 67. 




|w. jw +w -I 

Aj-tyJ 




FIG. 261. Principle of the arch. 

The thrusts in the links are T 1? T 2 , etc., and may be scaled from the 
lines radiating from O. OF = P and NO = Q give the forces required 
to maintain the links in position. Instead of links we might have 
employed blocks (Fig. 261 (<:)), drawing the joints ab, cd, ef, etc., per- 
pendicular to the lines of P, T l and T 2 respectively. The arrange- 
ment now gives an arch such as might be constructed in masonry or 
brickwork. The original link polygon is called the line of resistance 
of the arch ; the forces acting at the joints of the blocks will have 
the same values P lt T 1? T 2 , etc., as in the link polygon. 

The best arrangement would be produced by having the line -of 
resistance passing through the centre of each joint and perpendicular 
to the joint Such would give a uniform distribution of stress over the 
joints, and there would be no tendency for any block to slide on its 
neighbours. Generally, it is not possible to secure these conditions, 
but it is usual to endeavour to satisfy the following conditions : 

(i) The line of resistance is arranged to come within the middle 
third of each joint ; this secures that there will be no tendency for 
the joints to open out either at the top or the bottom (p. 239). 
D.M. Q 



2 4 2 



MATERIALS AND STRUCTURES 




FIG. 262. Stress at an arch joint. 



(2) The stress on the joint produced by the forces P, T 1? T 2 , etc., 
is limited to a value which can be carried safely by the material. 

(3) The line of resistance should not be inclined to the normal to 
the joint at an angle greater than the limiting angle of resistance 
(see p. 363) ; this secures that there shall be no slip, independently 
of any binding effect owing to the mortar. 

Condition (2) above may be understood more clearly by reference 
to Fig. 262, in which are shown two of the blocks in equilibrium 

under the action of W 1} W 2 , P 
and T 2 . T 2 may be split into 
components Tj and S, normal 
and tangential respectively to 
the section ef. If T : acts at the 
centre of the joint, a uniformly 
distributed normal stress will be 
produced. Otherwise, as ex- 
plained for a column on p. 237, 
a varying normal stress will act 
on the section and may be represented by the stress figure efhg. 
The maximum stress p^ is limited to a safe value depending on the 
material of the blocks. 

Reference to Fig. 261 (b) will show that the horizontal component of 
any of the forces P I} Q, r l\, T 2 , etc., is given by OR = H. H is called 
the horizontal thrust of the arch, and is constant throughout a given 
arch carrying given vertical loads. 

It will be understood that the link polygon ABCDE (Fig. 261(0)) 
may have a greater or smaller vertical height depending on the position 
chosen for the pole O in Fig. 261 (b). The effect of this on the arch 
will be to give it a greater rise if O is nearer FN in Fig. 261 (b) ; H 
will be diminished thereby. Hence, an arch of given span and 
carrying given loads will have the horizontal thrust diminished by 
increasing the rise. 

Metal arches. From what has been said regarding the line of 
resistance falling within the middle third of the joints, it will be clear 
that the bending moment at any section of a masonry arch is limited 
to a small quantity only. The rule is unnecessary in the case of metal 
arches, as these are capable of withstanding large bending moments. 
Metal arches are of three principal types : (a) arches continuous 
from abutment to abutment, and firmly anchored to the abutments 
or springings; (b) arches continuous throughout their length, but 
hinged at the abutments by means of pin joints ; (c) arches having 



ARCHES 



243 




(a) 




pin joints at the abutments, and also a pin joint at the crown. These 
types are shown in outline in Fig. 263 (a), (b) and (c}. 

In the types (a) and (b) difficulties arise in the solution by reason 
of the inability of the arch to change its shape freely in order to 
accommodate changes in dimensions 
due to elastic strains of the metal, or to 
changes in temperature. In type (a), 
both the span and the directions of 
the tangents to the arch at the abut- 
ments are unaltered when the arch is 
under strain. In type (b) the direc- 
tions of the tangents at the abutments 
may alter, but the span remains con- 
stant. In type (c) the arch may rise 

freely at the crown to accommodate any strains of the metal ; hence 
this type is not liable to being self-stressed, nor can changes in 
temperature produce any stresses in the metal. Type (c) alone is 
considered here. 

Three-pin arch. In Fig. 264 (a) is shown an arch having pins A 
and B at the abutments, or springings, and one at the crown C. A 



w 




FIG. 264. Three-pin arch. 



single load W is being supported and all other weights are dis- 
regarded meanwhile. Let T A and T B be the abutment reactions. 
Acting on the arch are three external forces only, viz. W, T A and 
T B , and these are in equilibrium; hence their lines must meet at a 
point. Further, there will be two forces only acting on the portion 



244 MATERIALS AND STRUCTURES 

AC, viz T A and a reaction T c at C coming from the right-hand 
portion of the arch ; these forces are in equilibrium, and must there- 
fore act in the same straight line AC (Fig. 2 64 (/)). It follows that 
the line of T A in Fig. 264 (a) is AC, and production of AC to cut the 
line of W in D will give the point where T B must also intersect W ; 
therefore T B acts in the line BD. The equilibrium of the right-hand 
portion CB is indicated in Fig. 264 (c). T c (now reversed in sense), 
T B and W intersect at D and are in equilibrium. Both T A and T B 
may be found from the parallelogram of forces \)abc. 

It may be noted that W (Fig. 264(0)) might be supported by means 
of straight rods, or links, AD and BD jointed at A, D and B, and 
that these rods would be under thrust only. ADB is usually termed 
the linear arch. Again, if W were supported by a beam simply resting 
on supports at A and B, then ADB would be the bending-moment 
diagram for the beam to a scale in which the bending moment at 
E is represented by DE. 

The bending moment at any section of the arch may be found in 
the following manner. Let AB (Fig. 265) be a transverse section of 
an arch, let OX be the centre line of the arch, 
i.e. the line containing the centres of area of 
all transverse sections, and let OX intersect 
AB at C. Draw DC vertically to meet the 
T " T linear arch at D. The thrust T in the linear 
arch at D will act in the direction of the 
tangent DE to the linear arch at D, and may 
be transferred to C as shown by T' = T, 
provided a couple of moment T x CE be 
applied, CE being perpendicular to DE. 
The moment of this couple is the bending 

FIG. 265. Bending moment, . 

thrust and shear at a section moment at AB ; the normal thrust and 
shearing forces at AB may be obtained by re- 
solving T' into components respectively normal and tangential to AB. 
A convenient manner of expressing the bending moment may be 
obtained : Resolve T at D into horizontal and vertical components 
H and V by means of the triangle of forces DFE (Fig. 265). The 
triangles EFD and DEC are similar ; hence 

T_DE_DC 
H = FD = CE ; 
:. TxCE = HxDC. 

Now, since the linear arch is also the link polygon for the given 
loads, H is constant for any point in the arch (p. 242). Hence the 




ARCHES 



245 



intercepts DC (measured to the same scale as that used in drawing 
the arch), when multiplied by the constant horizontal thrust H, will 
give the bending moment at AB. It will be noted that DC 'is the 
vertical intercept between the arch centre line OX and the linear 
arch ; hence the area between these will represent the bending- 
moment diagram for the arch. 

The bending-moment diagram for the arch in Fig. 264 (a) is 
shaded. H may be found by first obtaining T A or T B and then 
taking the horizontal component. It will be noted that the diagram 
for AC falls below the arch centre line AFC ; the inference is that 
this portion of the arch is under negative bending. Reference to 
Fig. 264^) will render this point more clear; the forces T A and T c 
tend to increase the curvature of AC. The bending-moment diagram 
for CGB falls above the centre line ; CGB is under positive bending 
and will have its curvature diminished on application of the load. 

The following directions will be of service in dealing with more 
complicated loading ; reference is made to Fig. 266, in which ACB 
is the arch centre line. 




FIG. 266. Bending moments and reactions for a three-pin arch. 



Consider a simply supported beam having the same span as the arch 
and carrying the same loads. Draw the bending-moment diagram 
ADEB for this beam, using any convenient scale. The arch has 
zero bending moment at A, C and B ; hence the linear arch may be 
obtained by redrawing ADEB so that it passes through A, C and B. 
To do this, reduce all the ordinates of ADEB in the ratio of CF to 
EF, giving the linear arch AGCKB. The shaded area will be the 
bending-moment diagram for the arch. To obtain its scale, CF 
represents M F , the bending moment at F for the simply supported 
beam ; hence the scale of the shaded area is found by equating CF 
to this bending moment. 



246 



MATERIALS AND STRUCTURES 



The horizontal thrust H may be found from 
M F = HxCF, 



or 



CF' 



CF being measured to the scale used in drawing the arch. T A and 
T B may be found by compounding with H the reactions P and Q 
for the simply supported beam at A and B respectively. 

Suspension bridges. A simple type of suspension bridge is shown 
in Fig. 267, in which the roadway FG is supported by means of two 




FIG. 267. Suspension bridge. 

chains AB, one on each side of the bridge. The chains pass over 
rollers or sliding pieces on the tops of towers at A and B and are 
anchored securely at D and E. Suspending bars connected to the 
chain support the weight of the roadway. 

Assuming that the weight of the roadway is distributed uniformly 
and that the weight of the chain is small by comparison, also that 
the roadway is fairly flexible, the tensions at the points B and C 
may be found as shown in Fig. 268. The portion of the chain 




(a) 

FIG. 268. Tensions at B and C in a suspension bridge chain. 

hanging between B and C will support one quarter of the whole 
weight of the bridge, and this may be concentrated at its centre ot 
gravity. The horizontal pull H at C passes through the line of |\V at 
G ; T , the pull at B, must pass through the same point. The triangle 
of forces abc (Fig. 268 (<)) will then give the values of H and T . 

Let w (Fig. 269 (a)) be the load communicated by each suspender 
to the chain. T and H in this figure are the pulls at B and C 
respectively. To obtain the directions of the chain throughout, the 



SUSPENSION BRIDGES 



247 



pull in ab, together with H, supports the load carried by the four 
suspenders passing through fr, <r, d and e. The resultant of the four 
loads will intersect H at their centre G lt and the. pull in ab must 
pass through the same point, thus determining the direction of ab. 




FIG. 269. Shape of a suspension bridge chain. 

be will pass through G 2 , the point in which the resultant force in the 
suspenders c, d and e intersects the line of H. Similarly cd passes 
through G 3 and de completes the half-chain. If a curve were drawn 
to touch the lines ab, be, etc., its shape would be parabolic, owing to 
the geometrical property involved in the above construction. 

It will be evident that abcde is a link polygon capable of supporting 
the given loads. The pull in any link may be found from the force 
diagram (Fig. 269^)). 

The effect of a load passing along the bridge may be observed by 
inspection of Fig. 270. As both chain and roadway are flexible, the 
A 




FIG. 270. Effect of a load on a suspension bridge. 

chain alters in shape as shown. To avoid this undesirable effect, 
the roadway may be stiffened by the insertion of stiffening girders. 
The best type of such girders consists of two on each side of the road- 
way (Fig. 2 7 1 (a ) ), connected at the middle C by a hinge and also 
having hinges at the piers of the bridge, D and F. Girders of this 
type are free to rise or fall at the middle of the span and thus avoid 
any complications of stress which would result from any alteration in 
the length of the chain owing to changes of temperature or stretching. 
To understand the effect of a live load W on the chain in Fig. 271 (a\ 
it should be noted that the chain will alter its curve to a very small 
extent only, owing to the action of the stiffening girders; any 
alteration will be due to the elastic strains. Supposing the chain 



248 MATERIALS AND STRUCTURES 

to be parabolic initially, ancj to remain parabolic, it follows that the 
effect of W on the chain must be the same as would be produced by 
equal pulls in all the suspenders, this being the condition under 
which alone will the chain assume a parabolic curve. Hence, if 

W 

there be N suspending rods, the pull in each will be . The forces 

acting on the left-hand stiffening girder will be as shown in Fig. 271 (b); 
A B 




VV W W W W 

N IN IN IN IN 



J I -r-{ w w w w w 

TP *W (b) tN tN tN tN tN- 



fc 



(c) 

FIG. 271. Stiffening girders for a suspension bridge. 

those acting on the right-hand girder are indicated in Fig. 271 (c). It 
will be noted that a reaction P from the left-hand abutment together 
with another Q communicated through the pin at C from the right- 
hand girder are required for the equilibrium of the left-hand girder. 
The right-hand girder requires holding down against the pulls of the 
suspending rods ; hence the reactions Q and S act downwards. 
Knowing the loads, these reactions can be calculated, and the 
diagrams of bending moments and shearing forces for the girders 
may be drawn by application of methods already described. 

The length of parabolic chain required for a suspension bridge 
may be calculated approximately from the following formula : 

Let- L = the half length of the chain, in feet. 

S = the span, in feet. 
D = the dip, in feet. 

Then L = f + i 

EXERCISES ON CHAPTER X. 

1. Calculate the elastic instability load by Euler's formula for a bar of 
mild steel 10 feet long and 0-5 inch in diameter, fixed at both ends. 
Take = 13,500 tons per square inch. 

2. A mild-steel tube i-i inches in external diameter and i-o inch 
internal diameter and 8 feet long is used as a strut, having both ends 
hinged. What would be the collapsing load by Euler's formula? 
= 13,500 tons per square inch. 



EXERCISES ON CHAPTER X. 249 

3. A series of struts, having both ends rounded, have ratios of L to k 
of 40, 60, 80, etc., up to 200. Calculate the collapsing loads per square 
inch of sectional area, using Euler's formula, and plot these loads with the 
ratios of L to k. =13,000 tons per square inch. 

4. Answer Question 3 if both ends are fixed. 

5. Find the breaking load of the strut given in Question i by applica- 
tion of Rankine's formula. Take the coefficients from the table on p. 235. 

6. A solid mild-steel strut 2 inches diameter is 6 feet high. Use 
Rankine's formula and find the safe axial load if both ends are rounded. 
Factor of safety = 5. 

7. A wrought-iron tube is 4 inches in external diameter, and is made 
of metal 0-25 inch thick. It is used as a column 8 feet high, and has both 
ends fixed. Find the breaking load by use of Rankine's formula. 

8. A rolled I section of mild steel, flanges 5 inches wide, depth 
9 inches, metal 0-6 inch thick, is used as a strut 10 feet long, having one 
end fixed and the other end perfectly free. Find the safe axial load by 
Rankine's formula, taking a factor of safety of 6. 

9. A solid strut of mild steel is 1-5 inches in diameter and has both 
ends fixed. Find the length for which the breaking loads by Rankine 
and by Euler will be equal. Take E = 13,500 tons per square inch. 

10. The column given in Question 8 carries a load of one ton at the 
centre of area of one flange. Calculate the maximum and minimum 
stresses, and draw a stress diagram for a horizontal cross section of the 
column. 

11. Take the tube given in Question 7 and calculate at what distance 
from the axis a load may be applied without thereby producing tensile 
stress. 

12. A semicircular arch of 4 feet radius, hinged at the crown and 
springings, carries a uniform load of 500 Ib. per horizontal foot. Draw 
the bending-moment diagram. State from the diagram the maximum 
bending moment. 

13. The centre line of a three-pinned arch is a circular arc ; the 
horizontal distance from springing to springing is 150 feet and the rise is 
15 feet. There is a uniformly distributed load of 0-5 ton per horizontal 
foot together with concentrated loads of 10, 15 and 5 tons at horizontal 
distances from one springing of 20, 40 and 60 feet respectively. Draw 
the bending-moment diagram and state its scale ; find the horizontal thrust 
and the reactions at the springings. 

14. A suspension bridge is 100 feet span and the chains have a dip of 
12 feet. Suppose the uniform load on one chain to be 500 Ib. per 
horizontal foot, and find the maximum and minimum pulls in the chain. 

15. Find the length of chain required for the bridge in Question' 14. 
Suppose the chain were to stretch 0-25 inch, what will be the change in 
the dip ? 

16. A hollow cast-iron column, 12 inches in external diameter, 10 inches 
in internal diameter and 8 feet long, is subjected to a direct compressive 
load of 40 tons. A bracket bolted to the side of the column supports the 
end of a girder, which transmits to the bracket a load of 5 tons. The line 
of action of this load maybe assumed to be 12 inches from the axis of the 
column. Find the maximum and minimum stresses in a cross section of 
the column due to these loads. (B.E.) 



250 



MATERIALS AND STRUCTURES 



17. A horizontal link of rectangular section 4 inches deep and 2 inches 
thick is subjected to tension, the load being P tons. The line of action of 
the load is in the central plane of the thickness and 2-25 inches from the 
bottom face of the link, (a) Find the load P if the greatest tensile stress 
in the straight part of the link is 6 tons per square inch, (b) If the tensile 
stress on a cross section of the link varies uniformly from 6 tons per 
square inch at the top to 2 tons per square inch at the bottom, find P and 
the position of its line of action. (L.U.) 

x joist 1?. A. hollow cylindrical steel strut has to be 
-i J designed for the following conditions : Length 6 feet, 

16*6/015^ ax ial load 12 tons, ratio of internal to external 
diameter 0-8, factor of safety 10. Determine the 
necessary external diameter of the strut and the 
thickness of the metal if the ends of the strut are 
firmly built in. Use the Rankine formula, taking 
f2i tons per square inch, and a for rounded 



8 x6 joist 



19. Find the radii of gyration of a column con- 
sisting of three steel rolled joists, riveted together as shown in the sketch 
(Fig. 272), their properties being 





Area, 
sq. in. 


!* 

inch units. 


inch units. 


Thickness of web, 
inch. 


16 inch x 6 inch joist. 
8 inch x 6 inch joist. 


18-22 
IO-29 


726-0 

1 10-6 


27-0 
17-9 


0-55 
Not needed here. 



What would be the working load of such a column 24 feet long and 
with fixed ends, using the following straight-line formula : 



/ c =( 14560 -56- Jib., 



where fc is the working stress in Ibs. per square inch ; / is the length of 
the column in inches ; r is the least radius of gyration in inches. (I.C.E.) 



CHAPTER XL 

SHAFTS. SPRINGS. 

Twisting moment on a shaft. A shaft is a piece used for the 
transmission by rotation of motion and power. A moment tending 
to rotate the shaft is communicated at one place and is transmitted, 
by stresses in the material of the shaft, to the desired place. Con- 
sidering a shaft AB (Fig. 273 ()), having one end A fixed rigidly, and 





FIG. 273. Twisting moments on shafts. 

having an arm BC mounted at the other end. The effect of a force 
P applied at C may be examined by applying equal and opposite 
forces P' and P", each equal and parallel to P, at B so as to act 
through the axis of the shaft. These forces equilibrate and con- 
sequently do not interfere with P. The system now consists of a 
couple formed by the forces P and P", the sole tendency of which 
will be to rotate the shaft about its axis, together with a force P', 
the tendency of which will be to bend the shaft. The shaft as a 
whole would be equilibrated by the application of forces (not shown 
in the figure) at the rigid connection at A. 

A shaft is said to be under pure twist when there is no tendency to 
bend it, nor to produce push or pull in the direction of its axis. The 



2 5 2 



MATERIALS AND STRUCTURES 



shaft in Fig. 273 (a) would have been under pure twist had the couple 
formed by P and P" been applied alone. One method of securing 
this result is shown in Fig. 273^), in which a double arm CBD is 
used and two forces P, P, forming a couple of moment Pa, are applied 
at its ends. The moment of the couple is called the twisting moment, 
or torque, and is written T generally. Neglecting the weight of the 
shaft, the equilibrium of the whole as in Fig. 273^) requires the appli- 
cation at A of a couple having a moment equal and contrary to that 
of T. The condition to be fulfilled in order that a shaft may be 
under pure twist is that it must be equilibrated by two equal opposing 
couples acting in planes perpendicular to the axis of the shaft. 

Shearing stresses produced by torque. Consider the shaft to be 
cut at any cross section E in Fig. 273 (^), the section being perpendi- 
cular to the axis of the shaft. To equilibrate the outer portion of the 
shaft under the action of the applied couple P, P requires an equal 
contrary couple at the section E, and acting in the plane of the 
section. Such a couple can be brought about in the uncut shaft 
only by the existence of shearing stresses distributed in some 
manner over the section. The nature of the distribution may be 
understood by considering the straining of the shaft under the action 
of the couples. Experiment justifies the assumptions that, in a 
round shaft, sections such as that at E remain plane, i.e. unwarped, 
when the couples are applied, and that any 
radius of such a section changes its direction 
but remains straight ; it is assumed in this 
that the elastic limit is not exceeded. 

In Fig. 274, AB is a line drawn on the 
surface of the shaft parallel to the axis before 
straining. As A is fixed rigidly, it will remain 
unaltered in position, but the other end will 
rotate under the straining ; the result is that 
AB will change position to AB'. Any small 
rectangle such as CDFE drawn on the 
surface of the shaft will change its position 
and shape as shown at C'D'F'E'. The angle 
through which CD has rotated in order to 
assume the new position CD' is clearly equal to that through which 
AB has turned. This angle, BAB', equal to 0, is therefore the 
shear strain at all parts of the surface of the shaft. Had we been 
able to draw a rectangle inside the material at a radius OG, its 
circumferential movement and change of shape evidently would have 




B B 

FIG. 274. Torsional strains. 



SHAFTS 



253 



cross 



been proportional to its radius. Thus, on the outer end 
section, G would move to G' and B to B', and we have , 

GG':BB' = OG:OB. 

We may therefore state that, since the shear strain at any point on 
a cross section of the shaft is proportional to the radius, the shear 
stress at that point will also be proportional to the radius, provided 
the elastic limit is not exceeded. It will also be 
evident that the shear stress at any point on a cross 
section will have a direction perpendicular to the 
radius of the point. 

Moment of resistance to torsion. In Fig. 275 
is shown a cross section of a shaft under pure twist. 
Consider a small area a. 

Let R = radius of shaft, in inches ; 

r= radius of a, in inches ; 
pt intensity of stress at outer skin, in Ib. or 

tons per square inch ; 
pt = intensity of stress on a. 
Then /':/ = r:R, 




FIG. 275. 



and 



Force on a =ta = =. 



R _,=|. fl ,*. 

Taking the sum of such moments all over the section, we have 

pt _pt vrR 4 
R loz ~R' 2 

= Ib.- or ton-inches. 



Moment of force on a about 



Total moment = 




This expression is called the moment of resistance 
to torsion for a solid round shaft. The case of 
a hollow round shaft of external radius R x and 
internal radius R 9 (Fig. 276) would be worked out 
similarly, with the substitution of limits R 2 and Rj 
for o and R in the integration. Thus, 



^ , t> o 

Total moment = %-2 at* = ^ 

K R 2 K 






._ or ton-inches. 



254 



MATERIALS AND STRUCTURES 



B B' 



Any question regarding the safe strength of a round shaft under 
pure twist may be solved by equating the given torque to the proper 
expression for the moment of resistance to torsion. It will be clear 
that a hollow. shaft will have a greater strength than a solid one 
of the same weight. Apart from the practical 
consideration that the boring of an axial hole 
may lead to the detection of otherwise un- 
suspected flaws in the material, there are the 
considerations that the intensity of shear stress, 
as well as the arm for taking moments, are 
small near the axis in a solid shaft, so 
that material near the axis is being employed 
unprofitably. 

Torsional rigidity of a round shaft. It is 
often of importance to estimate the angle through 
which one end of a shaft will twist relatively 
to the other end briefly the angle of torsion. 




B B' 
FIG. 277. Angle of twist Referring to Fig. 277, one end of the shaft 

of a shaft. ' 

being fixed rigidly, the other end rotates through 
a small angle BOB', denoted by a, on application of a torque T, and 
the shear strain is given by the angle BAB', equal to radians. 
Taking the expression for the modulus of rigidity (p. 109), viz. 



Pt 



we 



i/ 
may substitute for p t and as follows : 



-(0 



2 
2T 



(*) 



Again, 



BB' 



, . _. 

= 6, in radians 



or 



Also, 



BB' 



or 



a, m radians ; 
' = BO.a; 



BO 



R 



-(3) 



SHAFTS 255 

where R and L are the radius and length of the shaft in inches. 
Substituting the values found in (2) and (3) in (i) gives 

2 T 



2TL 

radians 



The result for a hollow round shaft of external radius R x and 
internal R 2 may be found in a similar manner, using the expressions 



The final result is 

2 TL 



By substitution from (3) in (i) an expression may be obtained 
suitable for cases where the maximum shear stress is given. Thus, 
P AL 
-iR' 

or a = t , radians ............................... (6) 

CK 

This expression is applicable to both solid and hollow shafts by 
taking R as the external radius. 

The torsional rigidity, or stiffness, of a shaft may be measured by 
the reciprocal of a. 

Comparison of hollow with solid shafts. The relative strengths 
of two shafts may be estimated by comparing the torques, which may 
be applied without exceeding a given intensity of stress. Let two 
shafts, one hollow, the other solid, have the same external radius 
R. Let the internal radius of the hollow shaft be R, where n is a 
numerical coefficient. Let/ be the maximum intensity of shearing 
stress in each case. Then, for the solid shaft, 



(i) 

and for the hollow shaft, 

_/t7T(R*-*R*) 

2 R 

^) 



256 MATERIALS AND STRUCTURES 



Hence, =i-^ 4 .................................... (3) 

is 

For example, if the internal radius of the hollow shaft is one-third 
of the external radius, n J, and 



= 8 

81' 

Comparison may also be made of the strength of a hollow with 
a solid shaft having the same cross-sectional area, i,e, having the 
same weight per unit length. We have, for the solid shaft, 



(4) 



and for the hollow shaft, T A 

Putting R 2 = ^R 15 gives 

_ 



2 
Also, as the cross-sectional areas are equal, 



Hence > 



.-. R^R^i-w 2 ); ............................ (6) 

RL _i_ 

R 2 i-ri>' 



(7) 

V I - W 
TA T j_ I 

For example, if n = ^, =-' = 



V- 

1-18. 



SHAFTS 



257 



Thin tube under torsion. In the case of a thin tube under torsion 
it may be assumed that the shearing stress is distributed uniformly. 
Let p t = stress intensity, Ib. per sq. inch, 

R = mean radius of tube, inches, 
/= thickness of the tube walls, inch. 
Then Cross-sectional area = 27TR/*, 

Total shearing force = 

Moment of this force = 

Moment of resistance to torsion = 2 TrR 2 ^, Ib.-inches. 
Horse-power transmitted by shafting. Formulae connecting the 
horse-power (see p. 326) with the dimensions of the shaft and its 
speed are based on the average torque transmitted, and should there- 
fore be used with caution. The maximum torque, on which will 
depend the maximum intensity of shearing stress, may exceed the 
average torque considerably, leading to a result for the diameter of 
the shaft which may be much too small. 

Considering a solid round shaft under pure twist, let 
T = torque transmitted, Ib.-inches ; 
R = radius of shaft, inches ; 
pt = maximum shear stress, Ib. per sq. inch ; 
N = revolutions per min. 

T 
Then Work per revolution = . 2ir (p. 339) 



Now 

.'. work per re volution = 

12 

Work per minute = 




12 

. . H.P. = 



12 x 33,000 



40,081" 

EXAMPLE. Find the horse-power which may be transmitted by a 
shaft 2 inches in diameter at 180 revolutions per minute. The maximum 
shearing stress is 10,000 Ib. per square inch. 

_^R 3 N _ 10,000 x i x 1 80 
H ' P '~4o,o8i ~ 40,081 

=45 nearly. 
D.M. R 



258 



MATERIALS AND STRUCTURES 



.Equation (i) above may be altered so as to give the diameter of 
shaft required for a given power. Thus, 

40,081 x H.P. 





or 



x 40,081 x H.P. 



D = a coefficient x 



A common value of the coefficient is about 3-3 for steel shafts. 
Principal stresses for pure torque. In Fig. 278 is shown a shaft 




FIG. 278. Principal stresses for pure torque. 

under pure torque. A small square abcd^ having its edges ab and cd 
parallel to the axis of the shaft, has been sketched on the surface. 
Each edge of the square will be subjected to shearing stresses of 
magnitude pt\ hence the diagonals ac and bd have purely normal pull 
and push stresses respectively, the magnitude being also/* (p. 128). 
These diagonals are therefore principal axes of stress, and the 
stresses on them are the principal stresses for the case of the shaft 
being under pure torque. If the diagonals be produced round the 
shaft surface, it is evident that they will form helices having an 
inclination of 45 to the shaft axis. 

A shaft made of material weak under pull, and strong under both 
shear and push, would fracture along the helix of which ac forms a 
part. This fact may be illustrated by means of a stick of blackboard 
chalk ; on applying opposite couples by the fingers to the ends of 
the chalk, the fracture will be found to follow very closely a helix of 
45 inclination. Pure bending applied to the chalk will cause it to 



PRINCIPAL STRESSES 



259 



fracture across a section at 90 to the axis; it is therefore evident 
that bending and torque simultaneously applied will cause fracture 
to take place on some section intermediate between 45 and 90. 
A shaft made of cast iron would behave in a similar manner to the 
stick of chalk, as the stress properties are similar. 

Shafts made of ductile material, such as mild steel, behave in a 
different way. Fracture under pure torque takes place across a 
section at 90 to the axis, as the strength under pull and also under 
push is higher than that under shear. It may be shown that materials 
loaded in a complex manner have sections mutually perpendicular 
on which the stress is purely normal, i.e. the stresses are principal 
stresses. There is also a particular section which has a shearing 
stress greater than that on any other section. There is strong 
evidence for believing that brittle materials break down when the 
principal stress of tension reaches a certain value depending on 
the material ; many ductile materials break down, or yield, when 
the maximum shearing stress reaches a certain value. 

More general case of principal stresses. Let AB and BC be 
two sections of a body intersecting at 90 at B (Fig. 279 (a)). Let 



r.AC 



1 r. A C cos Q 




FIG. 279. Principal stresses and axes. 

AB and BC have normal stresses p l and / 2 respectively, and let each 
be subjected to equal shearing stresses pt. Let AC represent a third 
section of the body, cutting AB at an angle 0, and let the stress 
r on AC be purely normal. The wedge ABC will be in equilibrium 
under the action of these stresses, and it is required to determine 
from this condition the values of and of r. AC and r will then be 
a principal axis of stress and a principal stress respectively. For 
simplicity, let the thickness of the wedge from front to back be 
unity. 

Due to the given stresses / 15 / 2 and/*, the faces AB and BC will 
have resultant forces acting as shown in Fig. 279(^). The forces 



260 MATERIALS AND STRUCTURES 

pi . AB and / 2 . BC will act at the centres of AB and BC respec- 
tively. The forces p t . AB and p t . BC will act along AB and BC 
respectively. Due to r, a normal force r . AC will act at the centre 
of AC and will have an inclination to the vertical equal to 6. Hence 
its vertical and horizontal components will be r . AC . cos and 
r. AC . sin respectively. For equilibrium of the wedge, the sum 
of the vertical upward forces must be equal to the sum of the vertical 
downward forces ; also the sum of the horizontal forces acting towards 
the left must be equal to the sum of those acting towards the right. 
The algebraic expressions for these conditions are : 

r. AC. cos (9=^. AB+^.BC ................... (i) 

7-.AC.sin6>=/ 2 .BC+/,.AB ................... (2) 

To simplify (i), divide by AC, giving 

AB BC 



=p l . cos +pt . sin 0. 
Divide this by cos 0, giving 

*=/!+/. tan0 ...................... (3) 

Equation (2) may be simplified in a similar manner by dividing 
first by AC and then by sin &, giving 

>'=A+A- cot<9 ...................... (4) 

Equations (3) and (4) are simultaneous equations, from which the 
values of r and may be obtained by the ordinary rules of algebra ; 
thus, as the right-hand sicfes are equal, we have 
/!+/,. tan 0=/ 2 +/t. cot 6>, 
or /! -/ 2 =p t (cot 6 - tan 0) 

= 2p t . COt 20, 



or cot20= 2 ............................... (5) 

Again, writing equations (3) and (4) thus, 

r-^/, tan 0, ........................... (6) 

r-S z =f t cotO, ........................... (7) 

and taking products, we have 

(r-AX^-AHA 2 ......................... (8) 

The solution of this quadratic equation may be obtained in the 
usual manner, giving 



PRINCIPAL STRESSES 261 

The two roots of r in (9) indicate two principal stresses ; also 
equation (5) gives two values of 20 differing by 180 for which the 
cotangents are equal, and hence indicates two sections differing by 
90. The determination of which root of r acts on one section or 
the other may be obtained by inserting one of the calculated values 
of r in either (6) or (7) ; the resulting value of tan 6 or cot will 
indicate the particular section on which this value of r acts. 

In the above, both p and p* have been taken as pulls ; if either 
or both be pushes, the sign of p l and p. 2 or both should be reversed 
in (5) and (9). A positive value for r indicates pull and a negative 
value indicates push. If any of the given stresses p lt p% or p t be 
absent in the data, write zero where these missing values occur in 
the equations found above. 

For example, taking a cube having shearing stresses p t only (p. 127), 
Pi = o, A = o. 

Equation (5) gives cot 20 = = o ; 

.'. 20 = 90 or 270, 
= 45 or 135. 

/ T-g 

Equation (9) gives r = ^ 

= pt. 

The principal axes are therefore the diagonals of the cube, and 
the principal stresses are a push and a pull each equal to the given 
shear stress, thus agreeing with the results already obtained in a 
different manner. 

Stress on a section inclined to the principal axes. Having deter- 
mined the principal axes of stress and the principal stresses, the 
stresses on other sections may be found by the following construction. 
Reference is made to Fig. 280. 

Let OA and OB be the principal axes of stress (Fig. 280(0)), and 
let OA=/! and OB=/ 2 be the stresses acting on the sections OB 
and OA respectively. To find the stress acting on any other section 
OK, carry out the following construction. With centre O and radii 
OA and OB describe circles ; draw ON perpendicular to OK, cutting 
these circles in N and E respectively. Draw NC parallel to OB, 
and also ED parallel to OA and cutting NC in D. Join OD ; OD 
will represent the stress p acting on OK. 

To prove this, draw EF parallel to OB, and let the angle KOB, 
which is equal to the angle NOA, be called 0. Due to p there will 



262 



MATERIALS AND STRUCTURES 



be an oblique stress of magnitude p l cos 6 acting on OK (p. 121 ). 

OC 

Now cos is given by ^- in the diagram and ON is equal to p^ to 



scale ; hence OC represents the oblique stress. Again, due to / 2 

there will be an oblique stress of magnitude / 2 cos (90 - 6) =/ 2 sin 

p-p 
acting on OK. But sin 6 is given by ^, and OE is equal to / 2 to 

scale ; hence the latter oblique stress is given by EF, which is equal 
to CD. The resultant of these oblique stresses, represented by OC 
and CD respectively, will be OD, which accordingly gives the stress 
p on OK. The construction employed for finding D is a well known 



7" A 





- ' fa) 

FIG. 280. Ellipse of stress. 

method of finding points on the circumference of an ellipse having 
OA and OB for its semi axes. The ellipse is shown in Fig. 280 (a), 
and is called the ellipse of stress. 

Both principal stresses have been taken as pulls in the above con- 
struction. Had one been a push, as/j (Fig. 280 (//)), and the other a 
pull, then the construction is modified by producing ND to cut the 
remote circumference of the ellipse as shown. 

Maximum shearing stress. An important fact depends on the 
noting that the angle EDN (Figs. 280(0) and (&)) is 90, and that there- 
fore D lies always on the circumference of a circle having EN for its 
diameter. The radius of this circle will be |(ON - OE) = \(p\ -/ 2 ) 
for principal stresses of the same kind (Fig. 280 (a)); and will be 
J(ON + OE) = (/ 1 +/ 2 ) for unlike principal stresses (Fig. 2 So (/)). 
The stress / may be resolved into normal and shear stresses in each 
case (Fig. 281 (a) and (/;)), indicated by OG and OH respectively. It 
will be clear that the maximum possible value of the shear stress in 



MAXIMUM SHEARING STRESSES 



263 



both cases is represented by the radius of the circle having EN for 
diameter ; hence, for like stresses, 

Maximum shear stress = \(p^ -/ 2 )> ( T ) 

and for unlike stresses, 

Maximum shear stress = \(p\+ / 2 ) ( 2 ) 





FIG. 281. Normal and shearing stresses. 

These equations require further examination. Both (i) and (2) 
refer to sections taken perpendicular to the paper, and give the 
maximum shearing stresses for such sections. Fig. 282 (a) shows a 
bar under axial pull stress p l and transverse pull stress / 2 , both 
stresses in (a) being in the plane 
of the paper. The principal 
axes of stress are OX and OY ; 
the maximum shearing stress 
for sections perpendicular to the 
paper in (a) will be \ (p l -/ 2 )- 
Examine now Fig. 282 (), show- ' 
ing a side elevation of the bar ; 
/> 2 acts perpendicular to the 
plane of the paper, and it will 



uli 





* w 



F 'G. 282. Maximum shear stress in a bar under 
longitudinal and transverse pulls. 



be evident that the section AB, 

at 45 to the axis, has a shearing 

stress of magnitude J/ x acting 

on it (p. 124). Hence AB is 

the section of the bar which carries a shearing stress greater in 

magnitude than that on any other section. 

It will be noted therefore that with like principal stresses, e.g. the 
longitudinal and circumferential stresses in a boiler shell, the greater 
principal stress alone determines the value of the maximum shearing 



264 



MATERIALS AND STRUCTURES 



stress, and the latter has a value equal to one-half of the greater 
principal stress. In the case of unlike principal stresses the maximum 
shear stress must be calculated from J (p l +/ 2 )- 

The points above noted are of importance in dealing with crank 
shafts and other cases where the combinations of loading give rise to 
unlike principal stresses. The experimental work of Guest and 
others shows that elastic break-down occurs when the shearing stress 
attains a certain value in many ductile materials, as has been noted 
already, and the results above discussed enable us to determine the 
relation of the maximum shear stress to the loading. 

Shaft under combined bending and torsion. An example of this 
kind of loading will be found in any crank shaft. Considering a 
solid shaft : 

Let M = the maximum bending moment on the shaft, in Ib.-ifiches ; 
T = the maximum torque, in Ib.-inches ; 
R = the radius of the shaft, in inches. 

It is understood that M and T occur both at the same cross 
section. The stresses due to these may be found from : 



M 



4 
A7TR3 



or, 



or, pi = 



2T 



Reference to Fig. 283, in which a rectangle abed has been sketched 
on the shaft surface, shows that p z is 
absent. The principal stresses may be 
calculated from equation (9) (p. 260) : 




m 



(3) 



FIG. 283.- Shaft under combined 



This result indicates unlike principal 
stresses, as the quantity under the square 
root sign is greater than/j 2 . 

In the Rankine hypothesis, the maximum 
principal stress is the criterion of break-down, 
and this assumption may be applied to 
brittle materials. The Rankine equation 

Taking the 



torque and bending. may be OD tained as follows. 

larger principal stress, viz. 



SHAFTS 265 

and substituting from (i) and (2), we have 

' 



or 



The left-hand side of this result has the same form as the ex- 
pression for the moment of resistance of a shaft to torsion ; the only 
difference lies in the fact that r is a. push or pull stress, whereas, in 
the torque expression, a shear stress appears. It may be said that 
if a pure torque T e were applied to the shaft, of magnitude given 
by (4), a shear stress would be produced thereby equal in magnitude 
to the maximum principal stress. Hence, 

T, = MWM* + T 2 ............................ (5) 

The result is convenient for practical use, and is usually referred 
to as Rankine's formula. 

If the maximum shear stress be taken as determining the point of 
failure, the reduction is as follows : 

From equation (3) (p. 264), remembering that one value of r is 
push and the other pull : 




Also, Maximum shearing stress = q= - - - 



2 

(6) 
Inserting the values of/j and ft in terms of M and T, we have 







266 



MATERIALS AND STRUCTURES 



Let T e be a torque which, if applied alone, would produce a shear 
stress equal to q. Then 

(7) 




It will be noted that this expression gives an equivalent twisting 
moment of smaller value than that permitted by the Rankine 
equation (5). 

Springs. Springs are pieces intended to take a large amount of 
strain, and are used for minimising the effects of shocks, for storing 
energy, and for measuring forces. The load on any spring is kept well 
within the elastic limit; hence the change of 
length, or the distortion, of the spring will be 
proportional to the load applied. Springs vary 
in form, depending on the purpose for which they 
are intended ; a few common forms are discussed 
below. 

Helical springs. Helical springs are made by 
coiling a rod or wire of the material, generally 
steel, into a helix. If the spring is to be under 
pul^ the coils of the unstrained spring are made 
so as to lie close together; open-coiled helical 
springs are necessary in cases where the load is 
to be applied as a push, causing the spring to 
become shorter. Reference is made to Fig. 284, 
which shows a close-coiled helical spring under pull, and made of 
material having a round section. It may be assumed that the effect 
of the load is to put the material of the spring under pure torsion. 
Bending is also present, and must be taken into account in open- 
coiled springs, but is small enough to be disregarded in the close- 
coiled spring under consideration. 
Let P = load applied, Ib. ; 

R = the mean radius of the helix, inches ; 
r = the radius of the section, inches. 
Any cross section of the wire will be sub- 
jected to a torque given by 

T = PR Ib.-inches (i) 

Consider a short piece of the helix lying 
between two cross sections AB and CD 
(Fig. 285), and imagine AB to be fixed rigidly. 
Let F be the centre of the section CD, 
and take a horizontal radius which, when FIG. 285. 



FIG. 284. Helical spring 




F D 



SPRINGS 267 



produced, cuts the axis of the spring at O. The effect of the torque 
will be to cause CD to twist through an angle relative to AB, and 
FO will rotate into the position FO', the point O undergoing a 
deflection OO'. Let the mean length of the portion considered 
be /; then the angle of twist may be written from the equation 
for that of a shaft (p. 255). 

2T/ 2 PR/ 



A ' ' 

Again, a = 



Now OO' gives the extension of the spring along its axis owing to 
the straining of the small portion considered. The total extension 
will be the sum of the quantities such as OO' for the whole length of 
material in the helix, and can be obtained by writing the total length 
of wire instead of / in (20). In the case of a close-coiled spring the 
total length will be given with sufficient accuracy by multiplying the 
mean circumference of the helix by the number of complete turns N. 
Length of wire in helix = 2?rRN (3) 

2 PR 2 
Hence, Total extension of spring = ~ j. 2?rRN 

4 PR 3 N 



O 4 

8PD 3 N 



-(4) 
-(5) 



where D = mean diameter of helix, inches ; 

d= diameter of wire, inches; 
P = load applied, in Ib. ; 

C = the modulus of rigidity, Ib. per square inch ; 
N = number of complete coils. 

The result shows, as had been anticipated, that the extension is 
proportional to the load applied. 

An equation connecting the shearing stress with the extension may 
be obtained from (4). Thus, 

Total extension of spring = . 4 



26B MATERIALS AND STRUCTURES 



Now PR = T=(p. 253). 

Hence, Total extension = 



This result enables the maximum extension to be found for a 
given spring when a given safe shear stress pt Ib. per square inch must 
not be exceeded. 

Beginning with no load on the spring, the gradual application of 
a load P Ib., producing an extension e inches, will require the per- 
formance of a quantity of work given by (see p. 325) 
Work done = average force x e 

= JPxe. 
Inserting the value of e given in (6), we have 



. 

This work is stored in the extended spring, and represents the 
energy which can be given out when the spring is recovering its 
original length, on the assumption of perfect elastic qualities. 

The above formulae, being based on those for a shaft of round 
section, should be used only for helical springs made of round wire. 
A formula which may be used for the extension of a spring of square 
section, having sides equal to s inches, is 

Extension of spring = 44 4 ........... . ................ (8) 

V^O 

Helical spring under torsion. Helical springs are loaded occa- 
sionally under torsion in the manner indicated in Fig. 286, where a 
spring AB is subjected to equal opposing couples by means of forces 
applied to arms attached to the ends of the spring. It is evident that 
the material of the coil is subjected to bending and that the torque 
produced by the couples is balanced at any cross section of the wire 
by the moment of resistance of that section to bending. The. neutral 



SPRINGS 



269 



axis of any cross section will be parallel to the axis of the helix 
(Fig. 287). Further, the change of curvature of the helix produced 
by the application of the torque will follow the same law as that for 
a beam (p. 166). 



'ofjielix 






FIG. 286. Helical spring under torsion. 

Let T = Pa = torque applied, Ib.-inches ; 

R! = initial mean radius of helix, inches ; 
R 2 = final mean radius of helix, inches ; 
Nj = initial number of complete coils in helix ; 
N 2 = final number of complete coils in helix ; 
L = length of wire in helix, inches ; 
I NA = moment of inertia of section of wire, inch units. 

Then Initial curvature = =r- . 



Final curvature == ^-. 

K 2 

Suppose that the tendency is to increase the number of coils, 
then R 2 will be less than R r 

Change of curvature produced by T = ^- - ^-. 

K 2 KJ 

By use of the equation, 

Change of curvature = ^F (p. 166), 

Hi-LiMA 



we have 



I I 

RT> 
9 K l 



L NA 

T 
EI V 



-(0 



Again, assuming that the coils lie fairly close together, we have 
for the length of the helix, 

L=2irR 1 N 1 



Hence, 



2?rN 1 
L ' 

2irN 



270 MATERIALS AND STRUCTURES 

Substituting these values in (i) gives 
2?rN 27rN T 



_ 

TT ~TT~EI 



This expression gives the angle as a fraction of a revolution through 
which B will rotate relative to A when the torque is applied (Fig 286). 
To obtain the angle of twist in degrees we have 

TL 

Angle of twist = 360 



(3) 

NA 

It will be noted from this result that the angle of twist is pro- 
portional to the torque, a property which leads to the use of springs 
of this type in certain cases, for example, the hair spring controlling 
the escapement of a chronometer. The use of such a spring permits 
the balance wheel to alter its angle of swing somewhat without 
altering the time in which it vibrates. The same kind of spring is 
often used for controlling the movement of the drum in engine 
indicators, as its property produces a more even stretching of the 
string driving the drum, and in consequence a less erratic distortion 
of the diagram drawn on the paper surrounding the drum. 

The maximum torque which may be applied without exceeding a 
stated stress,^ may be found as in a beam (p. 146) from 

T-S 1 - <4> 

These results may be applied to helical springs under torsion and 
made of wire having circular, square or rectangular sections. 

Piston rings. Spring rings are often usfed for the purpose of the 
prevention of leakage past the piston in steam, gas and oil engines. 
A common way of making spring rings of moderate size is to turn a 
ring of uniform section, making the diameter somewhat larger than 
that of the cylinder. A piece is then cut out of the ring sufficient 
to allow the ring to be sprung into the cylinder, when the ends will 
come together. Cast iron is often used as the material. In this 
method of manufacture, the ring does not take a truly circular form 
when sprung to the diameter of the cylinder, and does not exert a 
uniform pressure all round the cylinder wall. To secure uniformity 
in the pressure and a truly circular shape, the thickness of the ring 
must be varied. The breadth will of course be uniform, as the ring 
fits accurately a groove turned in the piston. 



SPRINGS 



271 



Fig. 288 (a) shows a piston ring of varying thickness ; the split is 
situated at C, and the ring as drawn has been sprung into a cylinder 
so that the gap at C is closed. The ring is subjected to a uniform 
radial pressure as shown. 

Let </=the diameter of the cylinder, inches ; 

p = the pressure in Ib. per square inch of rubbing surface ; 
b = the breadth of the ring, inches ; 

/AB = the thickness of the ring at AB, diametrically opposite 
C, in inches. 



x-6 -H 




FIG. 288. Piston ring giving uniform bearing pressure. 

The half ring on the right-hand side between A and C is under 
similar conditions of loading to those of a boiler shell (p. 95). 
Hence, we may write for the resultant force on it : 

P r /lx' v 4 ttx 

This force will produce a bending moment on AB of amount 



The relation of the maximum stress f at AB and the thickness of 
the ring there will be given by 

M AB =-(seep. 152), 



, 



f 

f AB 



./ 



.(0 



272 MATERIALS AND STRUCTURES 

The relation between the thickness at any other section and that 
at AB may be found from the consideration that the ring is to be 
circular both before and after springing it into the cylinder. Hence 
the change of curvature all round it will be uniform. Now, 

M 

Change of curvature = ^y, 

and, since E is constant for a given material, it follows that for 
uniform change in curvature 

y = a constant ......................... (2) 

To obtain the bending moment at any section such as DE, con- 
sider the portion of the ring lying between C and DE (Fig. 288 ()). 
Join CD, and let the angle COD be a. The resultant pressure P 2 
acting on the arc CD may be found in the following way. A solid 
piece of the same breadth of the ring, viz. b, bounded by the chord 
and arc CD will be in equilibrium if subjected to hydrostatic stress/. 
The resultant pressure R on the chord produced by the hydrostatic 
stress is 



and this must be equal and opposite to P 2 . Hence, 



Again, M DE = P 2 x DF = *pb x DF 2 . 

Also, DF = DO sin Ja = |</sin Ja ; 



1 OL 

.'. M DE = 2p& sin 2 - 
4 2 



(3) 



Also, 1 = ................................... (4) 

12 

Hence, substituting the values of (3) and (4) in (2), we have 

- 

- = a constant, 



bt 



DE 



12 

or, since p, b and d are constant for a given ring under a given pressure, 

*t\ 

. 3 = a constant (5) 



SPRINGS , 273 



For the section AB, a is 180, and sin|a will be unity. Hence, 



2 sin 2 90" _ i 

Ts = 7a ~/3~~ 
*DE *AH 'AB 



(6) 



This result enables the thickness of any section to be calculated 
after first having determined the thickness at AB. 

Carriage spring. Carriage springs are constructed of a number of 
plates of gradually diminishing length, clamped together at the 



L 4, 

Cj* -a: J 1 




FIG. 289. Carriage spring. 

middle and loaded as shown in Fig. 289. Generally the strips 
have the same breadth and thickness. The material will be under 
bending. 

Let P = the load in lb., applied at each end ; 

L = the distance between the loads, inches ; 
N = the number of strips ; 
b the breadth of each strip, inches ; 
t= the thickness of each strip, inches. 

The maximum bending moment will occur at the middle section 
AB, and will be given by 

M AB = -|PLlb.-inches. 

This bending moment will be balanced by the total moment of 
resistance obtained by adding together the moments of resistance 
of all the strips. Assuming that each strip touches the strip immedi- 
ately above it throughout its whole length, both before and after 
loading, it follows that all the strips will experience equal changes in. 
P.M. s 



274 MATERIALS AND STRUCTURES 

curvature on the spring being loaded. Considering the curvature 
at AB, we have 

bending moment on strip 
Change of curvature of any strip = ., 

_ moment of resistance of strip 
El 

= a constant for all the strips. 

Hence, as E and I are both constant, it follows that all the strips 
have equal moments of resistance. 

Let f- maximum stress on any strip at the section AB, 
Ib. per square inch. 

f] /2 

Then, Moment of resistance of each strip = ~~, 

fbfi 

Total moment of resistance at AB = N * y - Ib.-inches. 

6 

Hence, M AB = N-, ........................... (i) 



, PL 
or ' = 



The profile of the spring in the elevation may be arranged so as to 
secure that this value of the maximum stress on any strip shall be 
constant throughout its length. Considering any section CD, let the 
number of strips be N CD . Then, from (i), 



* = 

If/ is constant, the only variables in this expression will be x and 
NCD- Hence, N CD c *, .................................... (4) 

or, the number of strips and hence the depth of the spring vary as 

the distance from the end. The 

""^^ ^*"^ profile in the elevation will there- 

^^^^^^ fore be triangular (Fig. 290). 

FIG. 29 o.-Ideal profile of a carriage spring. .As this is an awkward shape tp 



SPRINGS 



275 



produce, the ends of the strips are shaped usually as shown dotted 
in plan in Fig. 289, which produces practically the same result. 

The deflection of the spring may be calculated in the following 
manner : 

its moment of resistance 
Change of curvature of any strip = rT 



_ 

El 



As / is constant throughout the length of the strip, the change of 
curvature throughout will be uniform. Supposing the strips to be 
straight at first, each strip will bend into the arc of a circle when the 
spring is loaded. The conditions as regards any one strip might be 
attained by subjecting that strip separately to a uniform bending 

moment : - PL. Hence, 
N 2 




v' 

Now, for a beam bent into a circular arc, the deflection is given by 



Hence, 



4 

In the above, the frictional resistances of the strips rubbing on 
each other has been neglected. The effect of this will be to make 
the spring appear to be stiffer, as evidenced by a deflection smaller 
than that calculated, when the load is being increased. When the 
load is being removed, the deflection will be found to be somewhat 
larger than that calculated. Of course, work will be absorbed by 
these frictional resistances, with the effect that any vibrations 




276 MATERIALS AND STRUCTURES 



communicated to the spring by impulsive forces, or shock, will die 
out more rapidly than would be the case with a spring formed out 
of a single piece of material. 

EXERCISES ON CHAPTER XI. 

1. A mild-steel shaft is 6 inches diameter. If the safe shear stress 
allowed is 10,000 Ib. per square inch, what torque may be applied ? 

2. Find the diameter of a solid round shaft of mild steel to transmit a 
torque of 12,000 Ib.-inches with a safe shear stress of 9000 Ib. per square 
inch. 

3. A hollow shaft has an outside diameter of 18 inches and an inside 
diameter of 6 inches. Calculate the torque for a safe shear stress of 4-5 
tons per square inch. 

4. A solid shaft has the same weight and the same length as the shaft 
given in Question 3 and is made of similar material. Calculate the safe 
torque which may be applied. Give the value of the ratio Torque for 
the hollow shaft : torque for the solid shaft. 

5. What torque may be applied to a tube 3 inches in external diameter, 
of metal 0-125 mcn thick, if the stress is not to exceed 10,000 Ib. per square 
inch ? 

6. The shaft given in Question I is 60 feet in length. What will be 
the angle of twist when the maximum permissible torque is applied ? Take 
C = 13,000,000 Ib. per square inch. 

7. Find the angle of twist for the shaft given in Question 3 when the 
shear stress is 4-5 tons per square inch. The shaft is 100 feet in length. 
Take C = 55oo tons per square inch. 

8. What horse-power may be transmitted by a solid shaft 3 inches in 
diameter at 120 revolutions per minute ? The shear stress is 8000 Ib. per 
square inch. 

9. What diameter of steel shaft is required in order to transmit 
20 horse-power at 250 revolutions per minute? 

10. AB and BC are two sections of a body meeting at 90. Normal 
pull stresses of 5 and 4 tons per square inch act on AB and BC respec- 
tively. Shearing stresses of 3 tons per square inch act from A towards B 
and from C towards B. Find the principal stresses and the principal axes 
of stress. Draw a diagram showing the axes and stresses. 

11. Answer Question 10 if the normal stress of 5 tons per square inch 
on AB is a push. 

12. A mild-steel shaft 3 inches in diameter has a bending moment of 
4000 Ib.-inches together with a twisting moment of 6000 Ib.-inches. 
Calculate the following : (a) The equivalent torque according to Rankine ; 
(b} the equivalent torque on the maximum shear stress hypothesis ; (c) the 
maximum and minimum principal stresses ; (d] the maximum shearing 
stress. 

13. Supposing that a constant bending moment of 4000 Ib.-inches be 
applied to a shaft 3 inches in diameter, what torque may be applied if 
the maximum shear stress is limited to 10,000 Ib. per square inch? 



EXERCISES ON CHAPTER XI. 277 

14. A cylindrical boiler is 7 feet in diameter and is made of plates 0-5 
inch thick. The steam pressure is 100 Ib. per square inch, (a) Find the 
stresses on longitudinal and circumferential sections ; also the stresses on 
sections at 30, 45 and 60 degrees to the axis. (A) What is the maximum 
shear stress on the plate ? 

15. A helical spring is made of round steel wire 0-25 inch in diameter. 
The mean radius of the helix is 1-25 inches ; number of complete 
turns 120 ; the spring is close-coiled. Take C= 12,000,000 Ib. per square 
inch, and find the pull required to extend the spring one inch. 

16. A helical spring, material of circular section, has to extend i inch 
with a pull of 50 Ib. The mean radius of the helix is 2 inches, and the 
length of the helical part of the spring is one foot. Assume that the coils 
are close together, and find the diameter of the wire. C = 12,000,000 Ib. 
per square inch. 

17. Suppose that the spring given in Question 1 5 is put under torsion 
by couples applied at its ends. Find the torque required to twist the 
spring through one radian. = 30,000,000 Ib. per square inch. 

18. A helical spring is made of steel of square section, 0-3 inch edge, 
close-coiled. The mean radius of the helix is one inch, and there are 
20 complete turns. Take C = 12,000,000 Ib. per square inch, and find the 
pull required to extend the spring one inch. 

19. A piston ring for a cylinder 24 inches in diameter has to give a 
uniform pressure of 2 Ib. per square inch of rubbing surface. Find the 
maximum thickness of the ring if the stress is not to exceed 6000 Ib. per 
square inch. Find also the thickness at a section 90 from the split. 

20. A carriage spring of length 30 inches is made of steel plates 
2-5 inches wide by 0-25 inch thick. Find the number of plates 
required to carry a central load of 800 Ib. if the maximum stress is limited 
to 12 tons per square inch. Find the deflection under this load if 
E = 30,000,000 Ib. per square inch. 

21. A load is applied to the crank fixed to a wrought-iron shaft 6 inches 
diameter and 20 feet long, which twists the ends to the extent of 2 ; 
assuming the modulus of transverse elasticity (or coefficient of rigidity) to 
be 4000 tons per square inch, what is the extreme fibre-stress? (I.C.E.) 

22. A closely-coiled spiral spring has 24 coils ; the mean diameter of 
the coil is 4 inches and the diameter of the wire from which the spring is 
made is 0-5 inch. Determine the axial load which will elongate this 
spring 6 inches if the modulus of rigidity is 12,000,000 Ib. per square inch. 

(B.E.) 

23. A hollow steel shaft is to be used to transmit 1000 H.P. at 90 
revolutions per minute ; the internal diameter of the shaft is to be f of 
the external diameter. The maximum twisting moment exceeds the 
mean by 20 per cent. If the maximum intensity of shear stress is not to 
exceed 4-5 tons per square inch, find the external diameter of the shaft. 

24. At a certain point in a loaded body the principal stresses are a 
tension of 5 tons per square inch and a pressure of 3 tons per square inch, 
the latter acting in a horizontal direction. Another load is then applied 
to the body, giving rise to a second stress system, the principal com- 
ponents of which at the same point are a tension of 3 tons per square 



278 MATERIALS AND STRUCTURE 

inch and a pressure of 4 tons per square inch, the latter acting at an 
angle of 40 to the horizontal. Find the magnitudes and directions of 
the principal stresses of the resultant stress system. There is no stress 
at right angles to the plane of the paper. (B.E.) 



CHAPTER XII. 



EARTH PRESSURE. 

Earth pressure. Questions regarding the pressure of earth enter 
into the design of foundations and of retaining walls for holding 
back earth. It is not possible to obtain exact solutions owing to 
the variable properties of the material, 
and also to the fact that the pro- 
perties are altered very considerably 
by the presence or absence of water 
mixed with the earth. 

Referring to Fig. 291, if a mass of 
earth be cut to a vertical face OY, it 
will weather down by breaking away 
of the earth until a permanent surface 
OA is attained ultimately. Let < be 
the angle which OA makes with the 
horizontal, and consider a particle of 
earth resting on the slope at P. Its weight W may be resolved into 
two forces, one R perpendicular to the slope and another Q acting 
down the slope. Balance is obtained by the force of friction F acting 
up the slope, F being equal to Q. Defining the coefficient of friction 
/A as the ratio of F and R when sliding is just on the point of taking 
place (p. 353), i.e. F 




FIG. 291. Natural slope of earth. 



ab 



the triangle of forces Pal? gives 

Q=F 

Hence, ^ 

The coefficient of friction may range from 0-25 to i-o for earth 
sliding on earth, < ranging from 14 to 45 degrees. 



(i) 



MATERIALS AND STRUCTURES 



Rankine's theory of earth pressure. The effect of the weight W 
resting on the slope OA is to produce a stress on OA having an 
angle of obliquity equal to c/> when sliding is just possible. < may 
be called the natural angle of repose of the earth ; sliding will not 
occur if the angle of slope has any value less than <f>. 

In the Rankine theory, it is asumed that the shearing effects at 
any section in the earth follow the ordinary frictional laws, and that 
the obliquity of stress on any section of the earth cannot exceed the 
natural angle of repose of the earth. 

Referring to Fig. 292, AB is the horizontal earth surface and abed 
is a small rectangular block of earth having its top and bottom faces 

B 




FIG. 292. Conjugate stresses, earth FIG. 293. Conjugate stresses, earth 

surface level. surface sloping. 

horizontal. Let the area of the top face be one square foot, let y be 
the depth below the surface and let w be the weight of the earth in 
Ib. per cubic foot. The stress /j on the top face will be produced 
by the weight of the superincumbent column of earth, and will be 
given by p i = wy ib. per square foot (2) 

The stress / 2 acting on the vertical faces must be determined 
from the relation mentioned above, viz. </> must not be exceeded on 
any section of the block. 

In Fig. 293 the earth surface is sloping at an angle a to the 
horizontal, and ab and cd are at the same slope, be and ad being 
vertical. The stress p l will be given by 

p, = ? + = = wy cos a Ib. per square foot (*) 

area of top face ab 

It is evident that/!,/, acting on ab and cd respectively balance 
each other, neglecting the weight of the block ; hence / 2 and / 2 must 
balance independently, and must therefore act in the same straight 
line. It therefore follows that/ 2 must be parallel to ab. /, parallel to 
be and / 2 parallel to ab are called conjugate stresses. / 2 is determined 
by the same consideration as before, viz. $ must not be exceeded. 



EARTH PRESSURE 



281 



In Fig. 294, OA and OB represent principal stresses p^ and 
respectively, and the construction is shown for obtaining the stress 
on a section OK (p. 261). ON 
is perpendicular to OK, NP and 
MP are parallel respectively to 
the principal axes of stress OB 
and OA, and PO is the stress 
on OK. P lies always on the 
circumference of the circle de- 
scribed on MN as diameter. 
The angle of obliquity of/ as 
shown is PON; the maximum 
angle of obliquity will occur 
when OP is tangential to the 
circle NPM as shown by OT. 
The angle COT will correspond 
with the value of <$> in earthwork problems. 

CT 




*+* 
From Fig. 294, we have 



=AzA (4) 

A+A 

Pressure on retaining walls by Rankine's theory. The foregoing 

principles may be applied to give a simple graphical solution for the 

Y earth pressure on retaining 

walls. In Fig. 295 XY is 
the vertical earth face of a 
retaining wall, the earth sur- 
face being horizontal and level 
with the top of the wall. 
Produce the horizontal base 
of the wall and select a 
point O on it. Draw OA 
X p 2 E O " T> t D vertically and make it equal 

FIG. 295. Earth pressure on a wall, earth surface tO /j = Z#H lb. per Square 

foot. Draw OT, making the 

angle <f> with OA. Find, by trial, a circle having its centre C 
in OA, to pass through A and to touch OT. This circle will cut 
OA in B, and will correspond to the circle NPM in Fig. 294. 




282 



MATERIALS AND STRUCTURES 



Make OD equal to OB, and DO will represent the other principal 
stress p. 2 . The stress / 2 will be transmitted horizontally through the 
earth along OX, and an equal stress p. 2 will be produced on the 
wall at X. Make XE equal to / 2 , and join YE. The stress diagram 
for the face of the wall will be YXE. The average stress will be 
J/ 2 , and if one foot length of wall be taken, the total pressure P 




P will act at a point JH from the foot of the wall. 

Fig. 296 illustrates the procedure if the earth surface is surcharged, 
or inclined to the horizontal, at an angle a. Draw XO parallel to 

the earth surface. Draw 
OA vertically, and make 
OA equal to p l = wH cos a 
(p. 280). Draw OM per- 
pendicular to XO, and 
draw also OT, making the 
angle < with OM. Find, 
by trial, a circle having its 
centre C in OM, to pass 
through A and to touch 
OT. This circle cuts OM 
in M and B, and will 
correspond to the circle 

FIG. 296. -Earth pressure on a wall, earth surface sur- NPM in Fig. 294. Join 

MA and BA ; these will 

correspond with NP and PM in Fig. 294; hence the principal 
axes of stress will be parallel to MA and BA respectively, and the 
principal stresses will be represented by OM and OB respectively. 
Draw OD and OE parallel respectively to BA and AM ; make 
OD equal to OM and OE equal to OB. The ellipse of stress 
passes through D and E, and cuts XO produced in F. Hence FO 
is the value of / 2 . The quarter DFE alone of the ellipse need be 
drawn. 

Draw the stress diagram for the wall by making XG equal to / 2 
and joining GY. The average stress will be \p^ and for one foot 
length of wall we have p _ \p^ n 

P acts parallel to the earth surface, and is at a height JH above 
the foot of the wall. 

If the earth surface be not surcharged, a simple formula may be 
obtained for the stress on the wall at any depth : 



EARTH PRESSURE 



283 



Let p l = wh~\he earth pressure on a horizontal foot at a 

depth h feet. 
p^ = the pressure on the wall at the same depth. 

Then, from equation (4), p. 281, 



A+A 



i - sin </> 2/ 



And 



i - sin </> 

= -- r j O^. 

i + sm </> 

If the earth surface is surcharged at an angle to the horizontal 
equal to <, then p l = wh cos </>, and it may be shown that the other 
conjugate stress, / 2 , is equal to p^ and acts on the wall at an angle </> 
to the horizontal. 

If the angle of surcharge is a, the following equation may be used 
in order to find the value of : 



. 
wh cos a 



COS a - v/CGS 2 a - COSW 
- --- -^ \- 

COS a + >/COS 2 a - COS 2 <J 



Wedge theory of earth pressure. Let AB (Fig. 297) be the 
vertical face of a retaining wall, and let AC be the surface of the 
earth ; also let BC be a plane 
making the angle < with the 
horizontal. Considering the 
wedge of earth B AC, imagine 
that its particles are cemented 
together so as to form a solid 
body. Under this condition, 
the wedge would just rest 
without slipping on the in- 
clined plane BC if the wall 
were removed ; in other 
words, so far as the wedge 

-P. . . . . . FIG. 297. Wedge theory of earth pressure on a wall. 

BAG is concerned, there is 

no pressure on the wall. Again, considering an indefinitely thin 
wedge ABA', at rest between the plane BA' and the wall, as its 
weight is negligible, there will be no -pressure on the wall. Hence 
the pressure on the wall, being zero for the inclined planes BC 




28 4 



MATERIALS AND STRUCTURES 



and BA', will attain a maximum value for some plane such as BD 
lying between BC and BA'. If the wall were removed, the earth 
would break away at once along the section BD and the wedge ABD 
would fall, subsequently weathering would remove the wedge DBC. 
BD is called the plane of rupture ; the force acting on the wall may 
be obtained by considering the weight of ABD and the reaction 
of the earth lying under the section BD. 

The force P which the earth communicates to the wall may be 
assumed to be horizontal, thus ignoring any friction between the 

C 











G 


\ 


-1 


w 1 


r/ 


P 





/ 
O 


3 1 


/ 


v/ 

v 


K..A.. .- 


o 






FIG. 298. Equilibrium of the wedge ABD. 

vertical face of the wall and the earth ; also P may be assumed to act 
at ^H from the base of the wall (Fig. 298). W is the weight of the 
wedge ABD, and is calculated by taking account of one foot length 
of the wall. The reaction Q of the earth underneath BD acts at 
the angle < to the normal OE to the section BD. These three forces 
meet at O and are in equilibrium. If 6 is the angle DBC, the angle 
between the lines of W produced and Q will be equal to 0. abc is 
the triangle of forces for W, P, and Q, from which we have 

P 



or P = Wtan<9 (i) 

Draw DK and AL, each perpendicular to BC. Then, if w is the 
weight of the earth in Ib. per cubic foot, 

W = area ABD x w 

= (area BAG - area BDC) w 
= {1BC . AL - PC . DK) w 



EARTH PRESSURE 285 



Let DK be called x. Then 

W = o;BC(AL-*) (2) 

DK x 

Also, tan = == 



BK BC - KG ' 
and KC = DKcot(<-a) 

= X COt (< - a). 

*Y* 

Hence, tan Q = ^^ TT - \ ...................... (3) 

BC - x cot (< - a) 

Substitute the values of (2) and (3) in (i), giving 

p_i,. TJP (AL x)x / ^ 

- 



The whole of the quantities involved in this expression, with the 
exception of x, are constant for a given wall, the earth having a 
known value for < and a given slope at the surface. The maximum 
value of P may be found by differentiating the right-hand side and 
equating the result to zero. Thus, 

d f AL.x-x* } 
dx (VC-xcot(<j>-a)) 



_ (AL - 2x) {BC - x cot (< - a)} + (AL .x - x*) cot (ft - a) 

{BC-*COt(</>-a)}2 

This will be zero when the numerator is zero. Hence, 
AL . BC - AL . x cot (<#> - a) - 2 x . BC + 2X 2 cot (< - a) 

= - AL . x cot (< - a) + x* cot (< - a), 
AL.BC-2#.BC = - x 2 cot (< - a), 
AL . BC - *BC = x . BC - * 2 cot (<f> - a) 



By reference to Fig. 298, it will be noticed that this may be written 

or 2AABC - 2ABDC = 2ADKB, 

or ABAD = ADKB .(5) 

The condition for the maximum value of P is therefore that the area 
of the triangle BAD should be equal to the area of the triangle DKB. 
From (i), P = Wtan0 

= w . A BAD . tan 9 



(6) 



286 



MATERIALS AND STRUCTURES 



Graphical solutions by the wedge theory. The following geo- 
metrical constructions may be used for the determination of x : 

CASE i. Earth surface level with the top of the wall. Reference is 
made to Fig. 299. Draw BC making the angle < with the horizontal. 
Draw BE perpendicular to BC and cutting the earth surface produced 




FIG. 299. Graphical solution, wedge theory, earth surface level. 

in E. Make EF equal to EA. Then BF is equal to x. Draw FD 
parallel to BC and join BD; BD will be the plane of rupture. P will 
be found by measuring BF = x to the same scale as that used in draw- 
ing the wall and inserting the value in (6). Apply P horizontally at 
^H from the base. 

CASE 2. Earth surface surcharged at an angle a. Draw BC (Fig. 300) 
making the angle </> with the horizontal. Draw BE perpendicular to 




FIG. 300. Graphical solution, wedge theory, earth surcharged, 

BC and cutting the earth surface produced in E. On BE describe a 
semicircle, and draw AF perpendicular to BE. Make EG equal to 
EF ; then BG is equal to x. Draw GD parallel to BC and join DB ; 
DB will be the plane of rupture. Calculate the value of P and apply 
it as in Case i. 



EARTH PRESSURE 



287 



CASE 3. Earth surface surcharged at the angle </>. 
BE perpendicular to the earth surface and 
cutting it produced in E. Then BE is 
equal to x. P is calculated and applied 
as before. 

CASE 4. Earth surface surcharged at an 
angle a and friction between the earth and 
Draw BC (Fig. 302) 
the horizontal to cut 
in C. On BC as 
a semicircle. Make 



In Fig. 301, draw 



the wall considered. 
at the angle < to 
the earth surface 
diameter describe 




FIG. 301. Earth surface surcharged 
at <f>, wedge theory. 



AD equal to AB, and draw DE per- 
pendicular to BC. Make BF equal to BE, and draw FG parallel to 
AD. Make FK equal to FG. Join BG. Then the pressure P on one 
foot length of the wall is equal to the weight of the prism of earth 




FIG. 302. Wedge theory ; solution when friction of earth on wall is taken account of. 

having an area in square feet equal to the area FGK and a length of 
one foot. P will act at JH from the base of the wall, and will be 
inclined at an angle <\> to the horizontal. The plane of rupture is BG. 
It is assumed in the last case that the value of < is the same for 
earth sliding upon earth and for earth sliding upon masonry. 

Distribution of normal pressure on the base of the wall. Having 
found P by application of one of the above methods, the resultant 
pressure on the base of the wall may be found in the manner shown 
in Fig. 303. W is the weight of one foot length of the wall, acting 
vertically through its centre of gravity G. P and W intersect at O, 



288 



MATERIALS AND STRUCTURES 



and R is their resultant. For stability, R should pass within the 
middle third DE of the base of the wall (p. 259). 





FIG. 303. Resultant pressure on wall 
base. 



FIG. 304. Distribution of normal 
stress on the wall base. 



In Fig. 304, F is the point in which R intersects the base of the 
wall, and O is the middle of the base. R may be resolved into two 
forces, R v and R H ', the latter produces shearing stress on the base, 
having a somewhat indefinite distribution ; the former produces 
normal stress. To determine the latter, shift R v from F to O, and 
apply a compensating couple R v x FO = M. R v acting at O will 
produce a uniform normal stress p^ of value given by 

p, = f v .. , = = lb. per square foot. 

^ area of wall base BC 

M will produce a stress which will vary from a push / 2 at C to an 
equal pull / 2 at B. These may be found from 



M 



h 

m 



I, 



where m is |BC and I is the moment of inertia of i foot length 
of the wall base taken with reference to the axis passing through O 
and perpendicular to the plane of the paper. 

i x BC 3 



12 



Hence, 



Rv x FO = 



-%m^ 1 2 

6R V .FO 



A- BC 

A stress diagram CBED is drawn in Fig. 304, in which 
CD = A+A> and BE=A-A>- 



EARTH PRESSURE . 



289 



w > 


i 






Ifi 


^ * 


- 


"V 




F 


rf -r ^ 


D 


P,i 


/ 


^ ^ -'1 N 
,\ L i / It) 



Eankine's theory applied to foundations. In Fig. 305 is shown a 
wall the weight of which is supported by a vertical reaction coming 
from the earth on which it rests. Consider one foot length of the 
wall, and find its weight W lb. The vertical stress p l on the earth 
will be -yy 

1 area of base AB 
= -p5 lb. per square foot. 

A > 

The horizontal stress f> 2 acting on the vertical faces of a small 
rectangular block of earth immediately under the foot of the wall 
will be found from the con- 
sideration that the angle < 
must not be exceeded by 
the obliquity of the stress. 
Make OC to represent p l 
draw OT making the angle 
<f> with OC; find by trial a 
circle CTF having its centre 
E in OC. passing through C 
and touching OT ; make OG 
equal to OF; then GO is 
equal to / 2 . Part of the 
ellipse of stress has been 
drawn, although this is not 
required in the construction, 
tally through the earth, and will act on the vertical faces of a small 
rectangular block of earth at K. There will be a stress / 3 acting on 
the horizontal faces of this block and caused by the weight of the 
column of earth resting on the top face of the block. / 3 is found 
by a second application of the same construction. Make KH equal 
to/ 2 ; draw KL making the angle < with KH ; the circle HLM has 
its centre in KH, passes through H- and touches KL. Make KN 
equal to KM, when NK will be equal to/ 3 . 

Let D be the depth of the foot of the wall below the earth surface, 
and let w be the weight of the earth in lb. per cubic foot. Then 
/ 3 = wl) lb. per square foot ; 

/. Defect, 

w 

This result gives the minimum depth of the foundation, and 
represents the case of the earth surrounding the wall being just on 

D.M. T 




FIG. 305. Rankine's theory applied to foundations. 

The stress / 2 is transmitted horizon- 



290 MATERIALS AND STRUCTURES 

the point of heaving up. The actual depth of the foundation may 
be obtained by application of a factor of safety. 

D may be found by calculation from equation (4), p. 281. Thus, 




i - sin </> 2/ 2 

_ /i -sin</>\ 
P*P\ \^ + sm< y' 



Also, sin 4> 

-sin^ 



Again, 



sin MV x , . 

: 5 , from (2) ; 

+ sin </>/ 



z^ ze/ \i 
W /i -sin<^)\ 2 



>. AB\ 



i^^y ( 4) 

i + sin </>/ 



EXAMPLE. A wall carries a weight of 800 tons. The area of the foot 
of the wall is 200 square feet. Find the minimum depth of foundation if 
the weight of the earth is 1 20 Ib. per cubic foot and if </> is 30. 

b l = =4 tons per square foot. 
-sinew 



120 

= 8- feet. 



EXERCISES ON CHAPTER XII. 

1. Given principal stresses of 6 tons and 3 tons per square foot, both 
pushes, find the angle of greatest obliquity of stress. 

2. A retaining wall for earth, 12 feet high, has its earth face vertical. 
The surface of the earth is horizontal and is level with the top of the wall. 
Find the total force per foot length on the wall by Rankine's theory, taking 
the weight of the earth as 1 10 Ib. per cubic foot and < as 40. 

3. Answer Question 2 if the earth surface is surcharged at 20 to the 
horizontal. 



EXERCISES ON CHAPTER XII. 291 

4. Answer Question 2 by application of the wedge theory. 

5. Answer Question 3 by application of the wedge theory. 

6. Answer Question 3 by the wedge theory, taking account of the 
friction between the earth and the wall. It may be assumed that < 
has the same value for earth sliding on earth and for earth sliding on 
masonry. 

7. A masonry retaining wall for earth has its earth face vertical, and 
the earth is surcharged at an angle of 30 to the horizontal. The wall 
is 9 feet high, 2 feet broad at the top, and 5 feet broad at the base. The 
earth weighs 1 10 Ib. per cubic foot and < is 30. Find the total earth 
pressure on the wall by the wedge theory. 

8. In Question 7, the masonry weighs 120 Ib. per cubic foot. Find the 
resultant pressure on the horizontal base of the wall. Does it pass within 
the middle third of the base ? Find the maximum and minimum normal 
stresses on the base, and draw a diagram showing the distribution of 
normal stress. 

9. A wall and the load which it carries produce a stress of 3 tons per 
square foot on the earth underneath the wall. If the weight of earth is 
1 10 Ib. per cubic foot and if </> is 35, find the minimum depth of the 
foundation below the surface of the earth. 

10. A brick wall 25 feet high, of uniform thickness and weighing I2olb. 
per cubic foot, has to withstand a wind pressure of 56 Ib. per square foot. 
What must be the thickness of the wall in order to satisfy the condition 
that there shall be no tension in any joint of the brickwork ? (I.C.E.) 

11. Concrete exerts on earth at the bottom of a trench a downward 
pressure of 2 tons per square foot ; the earth weighs 130 Ib. per cubic foot 
and its angle of repose (in Rankine's theory) is 30 ; what is the least safe 
depth below the earth's natural surface of the bottom of the concrete ? 
Why are we unable to make much practical use of the theory of earth 
pressure? (B.E.) 

12. A concrete retaining wall is trapezoidal in cross section, 24 feet 
high ; thickness at top, 3 feet ; at base, 10 feet ; the back face, which is 
subjected to earth pressure, being vertical. The wall is not surcharged. 
If the concrete weighs 140 Ib. per cubic foot, the earth-filling behind the 
wall 125 Ib. per cubic foot, and if the angle of repose of the earth is 
22 degrees, investigate the stability of the wall. (B.E.) 

13. Give the assumptions upon which Rankine's theory of earth pres- 
sure is based. Show that the intensity of horizontal pressure on a 
retaining wall at a depth d feet below the horizontal earth surface is 

i sin <f> , 



where w is the weight of I cubic foot of earth and < is the angle of 
repose of the earth. A practical rule takes the pressure as equivalent 
to that given by a fluid weighing 20 Ib. per cubic foot. Find the angle of 
repose corresponding to this, assuming iv equals 100 Ib. per cubic'foot. 

(L.U.) 



CHAPTER XIII. 



TESTING OF MATERIALS. 

Wires under pull. A simple apparatus is illustrated in Fig. 306 
and will enable the elastic properties of wires under pull to be studied. 
Two wires, A and B, are hung from the same 
support, which should be fixed to the wall as high 
as possible in order that long wires may be used. 
One wire, B, is permanent and carries a fixed load 
W T , in order to keep it taut. The other wire, A, 
is that under test, and may be changed readily for 
another of different material. The test wire may 
be loaded with gradually increasing weights W. 
The extension is measured by means of a vernier 
D, clamped to the test wire and moving over a 
scale E, which is clamped to the permanent wire. 
The arrangement of two wires prevents any droop- 
ing of the support being measured as an extension 
of the wire. 




EXPT. 15. Elastic stretching of wires. See that 
the wires are free from kinks. Measure the 
FIG. 306. Apparatus for length L in inches from C to the vernier. Measure 

son wires. ^ diameter of the wjre gtate the materi al of 

the wire and also whatever is known of its treatment before it came 
into your hands. Apply gradually increasing loads to the wire A, 
and read the vernier after the application of each load. Stop the 
test when it becomes- evident that the extensions are increasing more 
rapidly than the loads. Tabulate the readings thus : 

TENSION TEST ON A WIRE. 



Load, Ib. 


Vernier reading. 


Extension, inches. 









TESTING OF MATERIALS 



293 






Plot the loads in column i as ordinates and the corresponding 
extensions in column 3 as abscissae (Fig. 307). It will be found that 
a straight line will pass through most 
of the points between O and a point 
A, after which the line turns towards 
the right. The point A indicates the *" 
break-down of Hooke's law. 

Let Wj = load in Ib. at A in Fig. 307. w , ! 

*/=the diameter of the wire in 

inches. 
Then, 

Stress at elastic break-down 
W 1 

~i^ 




OL 



^ Extension 



. per square inch. 



FIG. 307. Graph of a tensile test 
on a wire. 



Select a point P on the straight line OA (Fig. 307), and measure 
W 2 and e from the diagram. 

Let W 2 = load in Ib. at P, 

= extension in inches at P, 
L = length of test wire in inches. 

rp,, T7 . stress W L 

Then, Young's modulus = E = r- = , fe - 

strain \ird* e 

Several wires of different material should be 
tested in a similar manner. 

In Fig. 308 is shown in outline a simple form 
of machine for testing wires to breaking; the 
machine is fitted with an arrangement whereby an 
autographic diagram is produced, i.e. a diagram is 
drawn by the apparatus showing the loads and 
corresponding extensions. 

AB is the test wire, fixed at A and carrying a 
receptacle B at its lower end. The load is 
applied by means of lead shot, stored in another 
receptacle C, which is fitted with an orifice and 
a control shutter at its lower end ; D is a shoot 
for guiding the shot into B. C is hung from a 
helical spring E, which is extended when C is full 
and shortens uniformly as the weight is removed 
by the shot running out of C. A cord F is 
attached to E, passes round a guide pulley and 
FIG. 308. -Apparatus a i so two or three times round a drum G, and 

for testing wires to 

rupture. has a small weight H attached in order to keep 




294 



MATERIALS AND STRUCTURES 



it tight. A piece of paper is wrapped round G, and circumferential 
movements of this paper will be proportional to the load removed 
from C and applied to the test wire. A small guided frame carrying 
a pencil is attached to the test wire at P ; vertical movements of the 
pencil will indicate the extensions of the portion of test wire between 
A and P. In action, a curve is drawn on the paper which shows 
loads horizontally and extensions vertically. 

EXPT. 1 6. Tensile test to rupture. Arrange the apparatus and 
fit the test wire j see that all the arrangements are working properly. 

Draw the lines of zero extension and zero 
load by rotating the drum for the first 
and by moving the pencil frame vertically 
for the second. Measure the diameter of 
the test wire and the length from A to P. 
Allow the shot to run into B until the 
test wire breaks. To obtain the breaking 
load, weigh the receptacle B together with 
its contents. 

Let W = breaking load in lb., 

d= diameter of the wire in inches. 

W 

Breaking stress = -. - 2 lb. per square inch of original cross- 

i 7 sectional area. 



Load 



Extension 

FIG. 309. Autographic record of 
a test on copper wire. 



Then, 



In Fig. 309 is given a reproduction of a 
diagram after removal from a machine of this 
kind. The scale of loads may be found by 
placing different weights in C and observing the 
resulting movements of the paper on the drum. 
The diagram shown is for copper wire, and the 
point of elastic break-down may be stated 
roughly from it. 

Experiments should be made on several 
wires of different materials, such as copper, 
brass and iron. 

Wires under torsion. Apparatus by means 
of which may be measured the angle of twist 
produced in a wire by a given torque is illus- 
trated in Fig. 310. AB is a test wire, firmly 
fixed at A to a rigid clamp and carrying a heavy 
cylinder at B. The cylinder serves to keep the 
wire tight, and also provides means of apply- 
ing the torque. The torque must be applied 




FIG. 310. Apparatus for 



as a couple in order to avoid bending, and is torsion tests on wires. 



TESTING OF MATERIALS 



295 



produced by means of cords wound round B ; these cords pass over 
guide pulleys, and carry equal weights W x and W 2 at the ends. 
Pointers C and I) are clamped to the wire, and move as the wire 
twists over fixed graduated scales E and F. The angle of twist 
produced in the portion CD of the wire is thus indicated. 

EXPT. 17. Torsion test on wires. Arrange the apparatus as shown. 
State the material of the wire ; measure its diameter d and the length 
L between the pointers C and D, both in inches. Measure also the 
diameter D of the cylinder B, in inches. Apply gradually increasing 
loads, and read the scales E and F after each load is applied. Tabu- 
late the readings. 

EXPERIMENT ON TWISTING. 



Load, 
W! = W a> Ib. 


Torque, 
WiD, Ib. -inches. 


Angle of twist, 
degrees. 









Torque. 



Plot the torques in column 2 as ordinates and the corresponding 
angles of twist as abscissae. A typical diagram is given in Fig. 311, 
from which it will be observed that the graph is practically a 
straight line, indicating that the angle of 
twist is proportional to the torque. Select 
a point P on the straight line, and measure 
the torque T Ib.-inches and the angle a 
from the diagram. If the diagram is 
plotted in degrees, convert a to radians. 
The value of the modulus of rigidity of 
the material may be calculated. 
Let 

T = the torque, in Ib.-inches ; 

L = the length of the wire, in inches ; 

R = the radius of the wire, in inches ; 
a = angle of twist, in radians ; 




0}< a, *| Angle 



FIG. 311. Uraph of a torsion test 
on a wire. 



or 



C = the modulus of rigidity, in Ib. per square inch. 
Then, from equation (4), p. 255, we have 

_ 2TL 
~7rR4C' 
2 TL 



Several wires of brass, copper and steel should be tested. In each 
case, any information regarding the previous history of the wire 
should be noted. 



296 



MATERIALS AND STRUCTURES 



Helical springs under pull. The extensions of a helical spring 
under pull may be investigated by use of the apparatus illustrated in 
Fig. 312. A is the spring under test ; it is hung from a hook at the 
top of a stand. A graduated scale B is hung 
from the spring, and carries a hook on which 
loads W may be placed. The vertical move- 
ments of the scale indicate the extensions of 
the spring, and are read by means of a telescope 
atC. 

EXPT. 1 8. Extensions of helical springs. 
Make a helical spring by coiling round a round 
bar, or mandril, some wire for which you 
have found C previously, as directed on p. 295. 
Test this spring under gradually increasing 
loads, noting the extension produced by each 
load. Plot loads and extensions ; these should 
give a straight line if the extensions are pro- 
portional to the loads. Select a point on the 
plotted line, and read the load W Ib. and 
the corresponding extension e inches. 

Let 

D = the mean diameter of the helix, inches ; 
*/=the diameter of the wire, inches ; 
N = the number of complete turns in the 
helix. 




FIG. 312. Apparatus for 
testing helical springs. 



Then, from equation (5), p. 267, 
8WD 3 N 



Or 



C = 



8WI) 3 N 



Ib. per square inch. 



Calculate the value of C from this equation, and compare it with 
the value of C found by the direct method of applying torque. 

Other springs made of wire of circular section are supplied. Make 
similar experiments, and find the value of C for each spring. 

Springs of material having a square section are also supplied. If 
the side of the square is s in inches, find the numerical values of the 
coefficient c for each spring by inserting experimental values in the 
following equation : WD 3 N 



Maxwell's needle. A useful piece of apparatus for making vibra- 
tional experiments on wires is the Maxwell's needle shown in 
Fig. 313 (a). The wire AB is fixed firmly at A and is clamped at B 



TESTING OF MATERIALS 



297 



l el 




A 

N 




jB / C 


[DIE 


r | G | 




~~\ 


V V 




FIG. 313. Maxwells needle. 



to a brass tube C. Four inner tubes D, E, F and G of equal lengths 

can be pushed into C ; the total length of the four tubes is equal to 

the length of C. Two of the short tubes are empty, and the other two 

are closed at the ends and are 

loaded with lead shot. Experi- 

ments are made by first having the 

loaded tubes at I) and G and the 

empty ones at E and F. A few 

degrees of twist are given to the 

wire, and the needle is then allowed 

to oscillate horizontally. The time 

taken to execute, say 100 vibra- 

tions, is observed, and hence the 

time of one vibration is obtained. 

The tubes are then exchanged by 

placing the loaded pair at E and F 

and the empty pair at D and G, 

r J . 

and the experiment repeated in 

order to find the time of one vibration. The distribution of mass in 

the system has been altered without altering the actual quantity of 

matter, and the second time will be found to be shorter than the first. 

Let t l = the time in seconds to execute a vibration, the needle 

starting from the end of a swing and coming back 

again to the same position ; loaded tubes at D and G. 

/ 2 = the corresponding time in seconds when the loaded 

tubes are at E and F. 

m 1 = the mass in pounds of one loaded tube. 
m 2 = the mass in pounds of one empty tube. 
a = the half length of C in feet. 
L = the length of the test wire, in inches. 
d=\.\\Q diameter of the test wire, in inches. 
g=\he acceleration due to gravity = 32-2 feet per second 

per second. 
C = modulus of rigidity of material of wire, Ib. per sq. inch. 

Then c 

EXPT. 19. Determination of C by Maxwell's needle. Test several 
wires of different materials by this method, and calculate C for each. 
If wires of the same material have been tested for the values of C 
by other methods, compare the results. 



298 MATERIALS AND STRUCTURES 

Torsional oscillations of a helical spring. Maxwell's needle may 
be used for determining the value of Young's modulus for a wire of 
given material. The wire is first wound into a helical spring and 
arranged as shown in Fig. 3 1 3 (), where C is the Maxwell's needle. 
Take the same symbols as before, with the addition of the following : 

R = the mean radius of the helix, in inches. 

N = the number of complete turns in the helix. 

E = Young's modulus, in Ib. per square inch. 

Then E 

EXPT. 20. Determination of E by torsional oscillations of a spring. 
Twist the needle through a small horizontal angle, taking care not 
to raise or lower it while doing so. On being released, it will 
execute torsional oscillations. Ascertain the times as before for the 
loaded tubes in the outer position and also in the inner position. 
Measure the dimensions required, and calculate E from the above 
equation. No correction is required for the mass of the spring in 
this experiment. 

Longitudinal vibrations of a helical spring. Using the same 
apparatus, illustrated in Fig. 313^), the value of the modulus of 
rigidity may be found for the material of the spring. The spring, 
loaded with the needle, is pulled downwards a little and released ; 
it will then execute vibrations vertically. 

Let /=time in seconds to execute one vibration from the 

lowest position and back to the starting-point. 
M = the mass of the needle, or other load, hung on + one- 
third the mass of the spring, in pounds. 
N = the number of complete turns in the helix. 
R = the mean radius of the helix, in inches. 
</=the diameter of the wire, in inches. 
C = the modulus of rigidity, Ib. per square inch. 

Then C = 6 -* 

3 

EXPT. 21. Determination of C by longitudinal vibrations of a spring. 
Use a spring made of wire which has been tested already for the 
value of C by the direct method of torque (p. 295), and also by the 
method of torsional oscillations (p. 297). Find C for the material 
by application of the method described above, and compare the 
results by the three methods. 

The direct determination of Poisson's ratio, , and also of the 

m 



TESTING OF MATERIALS 



299 



bulk modulus K for a material presents considerable difficulty. 
These may be calculated easily from the known experimental values 
of E and C by use of the following relations : 



i E-2C 

Poisson's ratio = - = ^ . 

EC 



m 



Bulk modulus = K = 



3(3C-E) 

Take the results for E and C which you have obtained for wires of 
the same material, and calculate and K for each material. 

M 

EXAMPLE. A series of tests on steel wires gave average values as 
follows : = 13,500 and = 5500 tons per square inch. Find the values 
of Poisson's ratio and of the bulk modulus. 



= E-2C 

2C 

_ 13,500- i i,ooo_ 



K = 



11,000 

EC 



i 

41 



3(3C-E) 

n,t;oox 5 coo 

=-T r = 825o tons per sq. inch. 

3(3x5500-13,500) i- 

Elastic bending of beams. The apparatus shown in Fig. 314 is 
capable of giving very accurate experimental results on the elastic 




FIG. 314. Apparatus for elastic bending of beams. 

bending of beams. The test beam A rests on steel knife-edges 
supported by blocks B, B. The blocks may be bolted at any distance 
apart on a lathe bed C. The load W is applied by means of a 
shackle D having a steel knife-edge which rests on the beam. The 
piece E, carried by the shackle, is pierced by a hole which is covered 



300 MATERIALS AND STRUCTURES 

by a piece of transparent celluloid having a fine line ruled on it. 
This line is observed through a micrometer microscope F, and will 
travel over the eyepiece scale as the beam deflects. The value of a 
scale division of the eyepiece scale may be ascertained by use of a 
scale engraved on the vertical pillar of the microscope ; a rack and 
pinion movement permits of vertical movement of the microscope up 
or down the pillar. 

For testing beams fixed at the ends, the knife-edges at B, B are 
removed ; these are merely dropped into V grooves on the top of the 
blocks. The test beam now rests on the top of the blocks (Fig. 315), 
and is held down firmly at each end by a strong cast-iron cap and 
four studs. 

The angle of slope at any position of the beam may be measured 
by means of the arrangement shown in Fig. 316. A is a small three- 




Pic. 315. Test beam fixed at ends. FIG. 316. Apparatus for measuring 

the slope of a beam. 

legged stool carrying a mirror and rests on the test beam. B is a 
reading telescope having a hair line in the eyepiece, and is used to 
observe the reading of a scale C reflected to the telescope by the 
aid of the mirror at A. 

The apparatus may be used for a large number of experiments ; 
the following indicates some of the more simple. 

EXPT. 22. Take a bar of mild steel of rectangular section about 
2 inches x i inch and about 3-5 feet in length. Arrange it as a beam 
simply supported on a span of 3 feet and loaded at the centre of the 
span. Apply gradually increasing loads, and measure the deflection 
at the centre of the span after the application of each load. Verify 
these readings by removing the loads, one at a time, and observing 
the deflections after the removal of each load. Tabulate these 
readings, and plot loads and deflections. If the resulting diagram is 
a straight line, then the deflections of the beam are proportional to 
the load. Select a point on the plotted line, and note the load W Ib. 
and corresponding deflection A inches ; also let L be the span in 
inches. Calculate the value of Young's modulus for the material, 
using the equation given on p. 169, viz.: 



. 
48EI 



TESTING OF MATERIALS 



301 



For the given section, breadth B and depth D, both in inches, 

T _ BD3 - 
1 - j 



or 



WL 3 
= 



E = 



WL 3 
ABI) 



Ib. per square inch. 



EXPT. 23. Use the same piece of material (a) as a cantilever, (ft) 
as a beam fixed at both ends. In each case measure the deflections 
for loads gradually increased and gradually diminished. Plot the 
results and determine Young's modulus, making use of the following 
equation for case (a) : 

WI 3 



WL 3 
For case (b) use A = 7^I ^ P ' I?8 ^ 

Compare the values of E obtained by the three methods employed. 

EXPT. 24. Arrange a test bar as a cantilever (Fig. 317). Let the 
load be applied at B, and arrange the three-legged mirror stool at C, 




FIG. 317. Slope of a cantilever. 

a scale divided decimally in inches at D, and a reading telescope at E. 
On loading the cantilever a certain angle of slope will occur at B ; 
as there is no load, and consequently no bending moment between 
B and C, whatever slope exists at B will occur uniformly between B 
and C. Hence the slope measured at C will be the slope at the 
point of application of B. The slope at B may be calculated from 

WL, 2 
* B = =y radians (p. 168). 

If the piece of material used in the previous bending experiments 
is employed in this experiment, E is known, and hence i B may be 
calculated for any given load. To verify the calculation, observe the 
scale readings for gradually increasing and gradually diminishing 
loads ; plot the results, and select from the diagram the value of / B 
corresponding to the value of W used in the calculation. 



3 02 



MATERIALS AND STRUCTURES 



In reducing the scale readings to radians, it must be noted that if 
the mirror at C tilts through an angle /, the ray of light CD will 
travel through an angle of magnitude 22. Let a be the change of 
scale reading due to an increment of load, and let b be the distance 
from the mirror to the scale, both in inches. The angle turned 
through by the ray CD will be 

DCD' = -. radian ; 
o 



.. 

t = r radian. 
2b 

Ten-ton testing machine. In Fig. 318 is shown in outline the 
principal parts of a testing machine constructed by Messrs. Joshua 




FIG. 318. Ten-ton Buckton testing machine. 

Buckton to the design of Mr. J. H. Wicksteed. As illustrated, the 
machine is arranged for applying pull. The test piece, AB, is held 
by grips in two crossheads C and D ; D is guided by the main column 
E of the machine, and may be drawn downwards by means of a screw 
F and a wheel G ; the latter serves as a nut for F, and is prevented 
from moving vertically. The rotation of G is effected by gearing and 
belt drive from some source of power; open and crossed belts permit 
of either direction of rotation being given to G. The belts are under 
the control of the operator by means of striking gear. The upper 
crosshead C is hung from a knife-edge H fixed in the beam K. The 



TESTING OF MATERIALS 



303 



beam is supported by a knife-edge L resting on the top of the 
column E. Its movement in a vertical plane is limited by spring- 
buffer stops M and N. A counter-poise P can be moved along the 
beam by means of a screw and hand-wheel under the control of the 
operator until the pull transmitted through the test piece to the beam 
is equilibrated. The magnitude of the pull is shown by the position 
of the counterpoise in relation to a scale of pounds which is attached 
to the beam. 

For applying push to the test piece, the machine is modified as 
shown in Fig. 319. The specimen AB is placed between crossheads 





,01 




FIG. 321. Shearing 
device. 





FIG. 319. Arrangement FIG. 320. Arrangement for bending 
for applying push. tests. 



FIG. 322. Punching 
device. 



Q and R, the former being connected to the screw F and the latter 
being hung from the beam. 

In carrying out bending tests, arrangements are made as shown in 
Fig. 320. The test beam AB rests on supports T, V, which in turn 
are carried by a beam S secured to the crosshead R. The weighing 
beam on the machine thus supports the beam under test. A central 
load is applied by means of a ram attached to the crosshead Q and 
drawn downwards by means of the screw F. 

Simple shearing tests are carried out by means of the appliance 
illustrated in Fig. 321. The piece X may slide inside W; the test 
piece AB is pushed into cylindrical steel dies carried by W ; X has 
another steel die which bears on the central portion of the test piece. 



304 



MATERIALS AND STRUCTURES 



The machine is arranged for pull as in Fig. 318; W is attached to C 
and X to D. On operating the machine, the test piece is put under 
double shear under much the same conditions as a rivet in a 
double-strapped butt-joint (p. 102). 

Fig. 322 shows an appliance which may be used for 

n punching tests. The upper block a can move vertically 

relative to the lower block b, and is guided by pins c 
and d. a carries a punch Y and b has a die Z. 
AB is the test piece. The machine is arranged for 
compression as shown in Fig. 319, and the punching 
appliance is placed between the crossheads Q and R. 
. The same machine may be used for torsion tests, 

but it will be found more convenient to have a 
separate torsion machine. One such is described on 

FIG. 3 2 3 .-Flat 6 

test piece. P' 6 1 u - 

A flat bar tension test piece is shown in Fig. 323. 
The enlarged ends ensure that fracture shall not take place in 
the grips. Fig. 324 shows the pair of steel wedge grips used for 
holding each end of this test piece. The grips have serrated 
faces for gripping securely the specimen. Round test pieces may 
be gripped in a similar manner, but a better plan is to have each 



B , , C 







FIG. 324. Wedge grips. 



FIG. 325. Spherical seated 
screwed grip. 



FIG. 326. Grip for 

brittle materials. 



end of the specimen A screwed into a holder B (Fig. 325) ; the 
holder has a nut C resting in a spherical seat formed in D, and 
permits of better alignment of the specimen in the machine than is 
possible with wedge grips. Both patterns of grip are used for ductile 
materials. 

For holding hard non-ductile materials like cast iron, the holder 
shown in Fig. 326 is employed. The specimen A is round, and has 
each end enlarged as shown at B. A split nut C screwed into the 
holder D supports A. 



TESTING OF MATERIALS 



305 




The arrangement shown in Fig. 327 will be found to give satis- 
factory working in compression tests. The ends of the specimen AB 
are screwed into holders C and D. Hard 
steel balls are placed at E and F in conical 
depressions, and enable the load to be applied 
very nearly axially. 

Columns made of cycle tubes provide a 
large range of useful tests. The arrangement 
when both ends are rounded is shown in 
Fig. 328. Conical hard steel plugs C and 
D are inserted in the ends of the tube AB 
and bear on hard steel seats E and F. It 
will be found useful to carry out a series of 
tests on specimens having a range of ratios 
of L to k. The breaking loads for these 
should be plotted in the manner described 
on p. 235. Great care should be taken in 
order to secure initial straightness, and the 
load should be applied as smoothly as .possible 
in order to avoid shocks which would pre- 
cipitate rupture. 

Autographic recorder. The autographic 



B 



FIG. 327. Test piece arranged 
for compression. 



/////tVf////i 

ft? 



recorder fitted to the machine in the laboratory at West Ham was 
designed by Professor Barr of Glasgow University, and is shown in 
I outline in Fig. 329. AB is the test piece under pull, 

and has two clamps D and E attached to it at a 
measured distance apart. A cord F is attached to D, 
passes over a pulley at E and thence to a drum C. 
Any extension of the test piece between D and E 
will be shown by rotation of the drum. The drum C 
has a paper wrapped round it on which the diagram 
of loads and extensions is drawn by a pencil G. 
Horizontal distances on this paper will represent 
extensions of the portion DE of the test piece. 

The pencil G is given vertical movements propor- 
tional to the load on the specimen by means of 
the following mechanism. The counterpoise of the 
machine is driven along the beam by means of the 
operating wheel H and gearing connected to the 
spindle K. The same spindle is connected also by gear wheels L to 
a screwed spindle M, on which is threaded a guided frame N carrying 
D.M. u 



^A 



FIG. 328. Tubular 
test column. 



3 6 



MATERIALS AND STRUCTURES 



the pencil G. Vertical movements of the pencil will therefore be 
reduced copies of horizontal movements of the counterpoise, and 
thus will represent to scale the load on the specimen. 

In testing ductile materials there are generally two points where 
the piece stretches so rapidly that the beam of the machine is certain 
to drop on to the lower buffer-stop; these are the yield point and 
the part of the test where local contraction is occurring preparatory to 
fracture. Should the beam drop on the buffer-stop, a portion of the 




H 



FIG. 329. Autographic recorder. 

diagram will be lost, as the load on the specimen is no longer re- 
presented by the position of the counterpoise on the beam. To 
obtain the complete diagram, a special spring Q is suspended from 
the end column of the machine (Fig. 329). Adjustable lock nuts 
are provided at R, and a bracket S is fixed to the end of the machine 
beam P. 

As the beam descends, S will come into contact with R and 
the spring Q will extend, thus removing some of the load from 
the test piece. The movement of the beam while extending Q is 



TESTING OF MATERIALS 307 

utilised for lowering the pencil G by an amount proportional to the 
load removed from the test specimen. The screwed spindle M is 
capable of vertical movement, and is held up in normal circum- 
stances by means of a lever T and balance weight W, the collar U 
thus being pressed against the fixed bracket V. A rod X is con- 
nected to the lever T, and has its end hooked to engage a pin Y 
fixed to a lever Z which is secured to the machine beam P. As the 
beam descends, S comes into contact with R and Y arrives at the 
hooked end of X simultaneously. Further movement of the beam 
will extend Q and lower M, and so will lower the pencil by an 
amount proportional to the load taken off the specimen by the spring. 

It will be evident that the apparatus can be used for the produc- 
tion of an autographic record of any of the tests made in the machine; 
the cord which rotates the drum is connected in each case to the 
part the movements of which are to be recorded as horizontal dis- 
tances on the paper. 

Extensometers. In tension tests which do not exceed the elastic 
limit, it is necessary to attach some form of extensometer to the 
specimen for the purposes of detecting and measuring the very small 
extensions which occur. The instrument devised by Sir J. A. Ewing 
is probably the most useful in general practice, and is shown in outline 
in Fig. 330. The test piece AB has clamped to it two blocks or levers 
C and D, by means of pairs of pointed pinching screws at E and F. 
C and D are connected by a bar G which is pivoted to D at H and 
is pulled against C at its upper end by means of a spring M ; the 
end of G has a ball K formed on it which beds in a conical recess in 
the end of the micrometer screw L. At the other end of C is sus- 
pended a rod N having a ball at its upper end ; this ball is pulled 
upwards into a conical recess by means of a light spring O. The lower 
end of N is guided by pins on D and carries a fine hair line at P. 
This hair line is observed through a micrometer microscope Q. 

Suppose that the test piece extends under pull and that the rod G 
remains unaltered in length. The hair line P will be displaced up- 
wards relative to the microscope, and so will appear to travel over 
the eyepiece scale. Each scale division represents approximately one 
five-thousandth of an inch, and it is easy with a little experience to 
subdivide each division into ten parts, thus enabling readings to be 
taken to the nearest fifty-thousandth of an inch. 

The precise value of a scale division of the microscope is ascertained 
as follows : After focussing the instrument and reading the scale, the 
micrometer L is given one complete turn. As its pitch is 0-02 inch, 



308 



MATERIALS AND STRUCTURES 



the effect is to change the length of G by this amount. The arms of 
the lever C on either side of the specimen are equal ; hence P will be 
moved relative to the microscope by 0-02 inch. The microscope 
scale is read again, and the difference between this and the original 
reading corresponds to a movement of P of 0-02 inch. Now 
in use, G remains unaltered in length, and the movement of P is 
produced by the extension of the specimen ; the effect of the levers 
is to produce a movement at P equal to double the extension of the 
specimen. Accordingly o-oi inch extension of the specimen will 
produce a movement of 0-02 inch at P. Hence the scale divisions 



D 





B 
FIG. 330. Ewing's extensometer. 

movement in the microscope found as directed above correspond to 
o-o i inch extension. Once focussed and calibrated, the instrument 
requires no further adjustment during the test unless the extension is 
sufficiently large to run the risk of moving the hair line beyond the 
limit of the microscope scale. In this event, it may be brought to a 
working position again by giving the micrometer L one turn, when 
the test will proceed as before. The loads are applied best in equal 
increments, and the reading of the microscope taken after application 
of each increment. The following record of a tensile test may assist 
in indicating the methods of noting the observations and of reducing 
the results : 



TESTING OF MATERIALS 



309 



TENSILE TEST ON A MILD-STEEL SPECIMEN. 

Laboratory No. A, M.S., 14.10.10. 

Form of test piece ; round, with swelled ends ; ends rough turned 
to 0-75 inch diameter; body turned and polished. A length of 
10 inches of the body was marked off at i inch intervals by light 
centre punch dots. 

Diameter of specimen, 0-445 inch- 
Area of cross section, 0-1556 inch. 

Elastic test with Ewing's extensometer. 

CALIBRATION OF INSTRUMENT. 





Microscope scale. 




Original. 


Final. 


Difference. 


i turn 


30 


77-2 


47-2 


i turn 


50 


97-2 


47-2 


i turn 


40 


87-2 


47-2 



47-2 microscope scale divisions are equivalent to an extension of 
o-oi inch. 

.'. i microscope scale division = jyVtf = 0-0002 1 1 8 inch. 
The load was applied in increments of 100 Ib. A number of these 
readings are omitted in the following table in order to economise 
space. None of the omitted readings depart from the plotted curve 
(Fig- 330- 

LOG OF TEST. 



Load, Ib. 


Microscope scale. 


Load, Ib. 


Microscope scale. 


100 


30-0 


4700 


65-3 


500 


33-o 


4800 


66-1 


IIOO 


37-9 


4900 


67-1 


1600 


41-9 


5000 


68-0 


2000 


44-8 


5100 


68-8 


2500 


48-3 


52OO 


69-3 


3000 


52-1 


5300 


70-2 


3500 


56-0 


5400 


71-1 


4000 


59-8 


5500 


72-1 


4500 


63-8 


5600 


73-o 


4600 


64-6 


5700 


74;5 








creeping to 


* 






83-5 



3io 



MATERIALS AND STRUCTURES 



The loads and scale readings are shown plotted in Fig. 331. It 
will be observed that the line ceases to be straight at A, which point 
accordingly indicates the break-down of Hooke's law. 

Load at elastic break-down, 4700 Ib. 
Stress at elastic break-down = 



0-1556 x 2240 
= 13-48 tons per sq. inch. 
Load at which creeping started = 5700 Ib. 

5700 



0-1556 x 2240 
= 16-35 tons P er sc l- i 



Stress 



The creeping of the hair line in the instrument marks the com- 
mencement of a stage in which the beam of the testing machine 
would exhibit a tendency to descend slowly towards the lower buffer- 



Lb 

6000 

5000 
4000 
3000 

2000 
1000 













/B~ 










/ 










/ 










/ 


/ 








/ 


/ 










/ 













30 40 50 60 70 80 90 
Microscope scale 

FIG. 331. Elastic limit tensile test ; mild steel. 

stop while a constant load is maintained on the specimen. When 
this stage is developed fully, the material is said to be in a plastic 
state, and the point is called the yield point. The autographic record 
(Fig. 332) shows the yield point clearly. 

To determine Young's modulus : 

From the diagram (Fig. 331), a load of 4000 Ib. produced an ex- 
tension corresponding to 29 microscope scale divisions. 

Extension of specimen = 29 x 0-0002 1 18 
= 0-00614 inch. 



TESTING OF MATERIALS 



Gauge points of extensometer are 8 inches apart. 

. 0-00614 
Strain = = 0-000768. 

o 

Stress = -^^-=2 5, 7 oo Ib. per sq. inch. 



Young's modulus = E = 



= 33,500,000 Ib. per sq. inch. 



000700 

= 14,900 tons per sq. inch. 

Test to maximum load. The extensometer being removed, the 
autographic recorder was connected to the specimen at 10 inch 
gauge points, and the load was increased from zero until the test 
piece began to form a neck preparatory to breaking. The resulting 
diagram is shown in Fig. 332, and gives the yield load as 6600 Ib. 
From this we calculate 

Yield stress = - z = 18-93 tons P er S Q- mcn - 
0-1556 x 2240 

The maximum load which the specimen could carry was 9600 Ib. 



Hence, 



Breaking stress 



9600 



0-1556 x 2240 



27-5 tons per sq. inch. 



The autographic record (Fig. 332) shows an interesting point 
regarding the effects of overstrain (i.e. straining beyond the yield 
point) on the elastic properties of 
the material. After the specimen -L,b. 
had been stretched 1-2 inches on 



4000 



2000 



10000- 

a length of 10 inches, the load 
was removed. Reapplication of 8000- 
the load ca.used the diagram to 
rise from zero along a practically 6000- 
straight line until the former 
curve was reached again at a 
load of about 9500 Ib. Yielding 
along the curve then continued 
as before. The overstraining had 
hardened the material and raised 
the yield load from 6600 Ib. to 
about 9500 Ib., i.e. only slightly 
below the ultimate load. 

The test piece was removed from the machine and the lengths of 
the intervals between the centre punch dots were measured ; also the 
diameters at each dot. From these the curves in Fig. 333 were 
plotted. It will be noted that both extensions and diameters vary 
considerably, and illustrate the necessity for stating the distance 
between the gauge points as well as the percentage extension of a 



IQ 



2'0 



3 : 
Inches 



FIG. 332. Autographic record ; mild steel 
under tension. 



312 



MATERIALS AND STRUCTURES 



test piece. A good method is to measure the total extension on a 
length of 10 inches, also the extension on the 2 inches interval 




0-38 




039- 
040- 

041 
Inch,. 

FIG. 333. Dimensions of a mild steel specimen after a tension test. 

which includes the fracture ; the difference between these will be 
the general extension on the remaining 8 inches of the specimen. 

These extensions, expressed as per- 
centages, give useful information re- 
garding the ductility of the material. 
Another measure of the ductility may 
be obtained by measuring the cross- 
sectional area of the fracture. The 
loss in area may be found from this 
measurement, and may be expressed 
as a percentage of the original sectional 
area. 

Fig. 334, copied from the autographic 
record of a specimen of Delta metal 
under pull, is given as illustrating 
totally different characteristics from the 
mild-steel diagram shown in Fig. 332. 
In particular, the absence of any yield 
point will be noticed. 

Bending tests. The records given 
in Figs. 335 and 336 illustrate an 
instructive test made on a mild-steel bar having a span of 36 
inches, breadth 2-01 inches and depth 2-015 inches. The bar was 




05 l-O 1-5 Inches 



FIG. 334. Autographic record ; Delta 
metal under tension. 



TESTING OF MATERIALS 



313 



arranged as shown in Fig. 320 and bent by application of a central 
load until the deflection was 2-6 inches. The record (Fig. 335) 
shows that yielding was reached at about 7200 Ib. The test was 
arrested at about 2-1 inch deflection, and the load brought to zero 
and then reapplied ; the diagram shows that the new yield load is 
about 9000 Ib. 

Then the load was removed entirely and the bar turned over; 
central loading was applied again so as to straighten the bar. The 
diagram given in Fig. 336 shows the result. There is practically no 
part of this test where the elastic law is followed, a fact which will 
be understood readily when it is realised that the bar came out of 
the former test badly overstrained both on its compression side and 




10 20 s-o 

Inchef 

FIG. 335. Mild steel bar under bending ; 
first test. 



1-0 



2-0 



3-0 
inches 

FIG. 336. Mild steel bar under bending ; 
second test. 



on its tensile side, and, in the effort to recover some of the deflection 
imposed on it, the material became self-stressed throughout. The 
second test began therefore with the material in a complicated state 
of stress. This test was also arrested at about 2-2 inch deflection. 
On reapplication of the load, a yield load of about 10,000 Ib. will be 
observed in the diagram. Had the bar been annealed after straighten- 
ing, it is probable that a diagram somewhat resembling Fig. 335 for 
the first test would be obtained. The tendency of the annealing is 
to remove self-stressing from the material. 

Fig. 337 has been copied from the autographic record obtained in 
testing a cast-iron bar under bending. The bar was rectangular in 
section, 2 inches wide and i^V inches deep, span 20 inches. Rupture 
occurred with a central load of 3200 Ib., the maximum deflection 
recorded being 0-2 inch. It will be noted that the load and 



314 



MATERIALS AND STRUCTURES 



deflection remain approximately proportional up to fracture. The 
contrast of the ductile and brittle materials is rendered clear by 
inspection of Figs. 335 and 337. The mild-steel bar could not be 
broken by bending ; the cast-iron bar could take a very small 
deflection only. 

The usual test for timber is by bending under similar conditions 
to those noted above. The specimens should be of as large size 
as is possible, then the effect of any local flaws such as shakes and 
knots will not be emphasised, as would be the case with a smaller 
specimen containing the same flaws. In Fig. 338 are given copies of 




2000 



o 0-2 04- Inch 

FIG. 337. Cast-iron test bar under bending. 



W Ib 

12000 



8000 



4000- 



Yellow deal 



Teak 




Oak 



10 20 Inches 

Deflection 

FIG. 338. Bending tests on timber. 



records of bending tests on yellow deal, teak and oak. The yellow 
deal specimen was arranged with the annual rings nearly horizontal, 
and failed by shearing horizontally along the fibres and round the 
annual rings. The dimensions and results are given in the following 
table: 

BENDING TESTS ON THREE TIMBER SPECIMENS. 



Material. 


Span X breadth X depth, 
inch units. 


Central breaking- 
load, Ib. 


Deflection at 
centre, inch. 


Yellow deal - 
Teak - 
Oak - 


24x2-9 X3-4 
60x3-14 x 4-26 

24x1-35x2-25 


12,600 
6,720 
3,500 


0-8 
0-92 



In reducing the results of tests on cast-iron and timber specimens, 
it is usual to state the value of the coefficient of rupture. This 
coefficient represents the value which the maximum stress at rupture 



TESTING OF MATERIALS 



315 



due to bending would have if Hooke's elastic law were followed 
throughout. For a beam of rectangular section supported at the 
ends and having the load applied at the middle of the span, the 
calculation will be as follows : 

Let W = maximum load, in Ibs. 

L = the span, in inches. 
b = the breadth, in inches. 
^=the depth, in inches. 



Then 



WL 
4 



m 

L 

\d 



12 



and 



Coefficient of rupture =/= - -r-^. 



Shearing tests. Autographic records of two shearing tests carried 
out in the apparatus de- 
scribed on p. 303 are given 
in Figs. 339 and 340. The 
former is for a mild-steel 
specimen and the latter is 
for a specimen of gun- 
metal. It should be noted 
that pure shear is not ob- 
tained with this apparatus, 
the specimen being under 
bending as well as shearing. 
In Older to minimise the 
bending effect, the speci- 
mens should be turned to 
fit the bored holes in the 
dies. The results of the 
tests are given below. 




4000 



FIG. 339. Mild steel under 
shearing. 



0'5 Inch 

FIG. 340. Gun-metal 
under shearing. 



SHEARING TESTS. 



Material. 


Diameter, 
inch. 


Cross-sectional area, 
square inch. 


Shearing load, 
Ib. 


Shearing strength, 
tons per sq. inch. 


Actual. 


Under shear. 


Mild steel - 
Gun-metal 


0-496 
0-500 


0-194 
0-196 


0-388 
0-392 


l6,4OO 
10,190 


1 8-9 
1 1-6 



MATERIALS AND STRUCTURES 



Punching tests. In punching a hole in a piece of material, 
the action of the punch is first to increase the pressure on the 
material until the plastic stage is reached ; 
in this stage, some of the metal flows from 
under the punch into the surrounding plate, 
the plate immediately under the punch be- 
coming thinner. This effect continues until, 
partly by the increasing force on the punch 
and partly by the diminishing thickness of 
the plate, the rupturing shear stress on the 
material is attained and a wad is pushed out. 
The following results of a punching test may 
be of interest; the autographic record is 
given in Fig. 341. 

PUNCHING TEST ON A WROUGHT-!RON 
PLATE. 

O'-l 0'2 /rich 




4000- 



FIG. 341. Punching test on 
wrought iron. 



Thickness of the plate = 0-265 mcn - 
Diameter of the wad punched out = 0-38 
inch. 
Area under shear stress = ^dt=^ xo-38xo-265 

= 0-317 square inch. 

Maximum load on the punch = 15,750 Ib. 

r 5>75 



Maximum shearing stress 



22-2 tons per sq. inch. 



0-317 x 2240 

Thickness of the plate round the hole after punching = 0-268 inch. 
Thickness of the wad = 0-2 5 7 inch. 
Loss of thickness of material in the wad = 0-008 inch. 
Gain of thickness of material round the hole = 0-003 inch. 
Total work done in punching the hole, represented by the area of 
the autographic record, is about 810 inch-lb. 

Avery torsion machine. An outline diagram of this machine is 
given in Fig. 342, where AB is the test piece. The end B is con- 
nected to a worm wheel C, which may be rotated by means of a 
worm D and hand wheel E. The wheel C has 90 teeth ; hence each 
quarter turn of the hand wheel twists the specimen through one 
degree. The torque is measured by means of a system of levers 
FG, MK and NP, connected to the end A of the specimen. NP 
carries a counterpoise Q, which may be run along the lever by means 
of a screw and hand wheel, and shows the torque by its position 
relative to a scale attached to the lever. The scale reads from zero 
to 1000 Ib.-inches: to obtain higher torque the counterpoise is run 



TESTING OF MATERIALS 317 

back to zero, and a load R is suspended from the end of the lever 
and is sufficient to give a torque of 1000 Ib.-inches. The test then 
proceeds to 2000 Ib.-inches by traversing the counterpoise. This 
process repeated enables 5000 Ib.-inches torque to be obtained, the 
lever MK resting on the knife-edge M during this stage. To increase 
the torque further, the knife-edge M is lowered and L is raised 

P 




F 

FIG. 342. Arrangement of Avery torsion machine. 

simultaneously by means of a lever ; the effect of this is to double 
the value of the scale divisions on the lever NP. The effects of the 
loads at R also are doubled. To reset the torque at 5000 Ib.-inches, 
hang two weights at R (equivalent now to 4000 Ib.-inches) and set 
the counterpoise at 500 (equivalent to 1000 Ib.-inches). The test 
then proceeds as before, the capacity of the machine being now 
10,000 Ib.-inches. 

In testing a piece to destruction, readings are taken of the torques 
and of the corresponding angles of twist by counting the number of 
teeth passed by the worm-wheel ; each tooth represents 4 degrees of 
twist. Plotting these readings will give a curve such as is illustrated 
in Fig. 343. The principal results of this test are given below. 

TORSION TEST ON A MILD-STEEL SPECIMEN. 
Laboratory No. 6, M.S., 22.3.10. 
Original diameter, 0-756 inch. 
Original length, 5-625 inches. 
Diameter after fracture, 0-754 inch. 



MATERIALS AND STRUCTURES 



Length after fracture, 5-750 inches. 

Yield torque, estimated by the dropping of the beam, 2000 

Ib.-inches. 

Breaking torque, 6400 Ib.-inches. 

Angle of twist at yield, 3-5 degrees on 5-75 inches length. 
Angle of twist at breaking, 1896 degrees on 5-75 inches length. 
Mean torque, from diagram, 5650 Ib.-inches. 
Total work done in fracturing specimen, 187,000 inch-lb. 
Work done per cubic inch of material, 74,200 inch-lb. 



Lb- inches 

7000 



5000 



4000 
3000 
8000 



1000 



400 



800 



1200 



1600 2000 
Degrees 

FIG. 343. Graph of a torsion test on mild steel. 

The form of the test specimen is indicated at AB in Fig. 344 ; the 
same diagram also illustrates an appliance whereby angles of twist 

G 




FIG. 3<v. Apparatus for measuring angles of twist within the elastic limit. 

within the elastic limit may be measured. C and D are two pieces 
of wrought-iron steam tube turned and bored at L to an easy fit. 



TESTING OF MATERIALS 



319 













X 










y 


^ 








/ 


r 








/ 










/ 










/ 













Three steel pinching screws nip the specimen at E and other three 
screws engage it at F. The angle of twist is measured on the length 
of the specimen between E and F. C carries a micrometer micro- 
scope G, balanced by means of a weight K, and D carries a rod 
having a small piece of transparent celluloid at H. A radial line is 
scratched on the celluloid, and is sighted through the microscope. 

Lb.-inches 

2000 

1600 

1200 
800 

400 



Angle of twist 

FIG. 345. Graph of an elastic torsion test, mild steel. 

The circumferential movement of the line is given by the scale 
readings of the microscope ; these reduced to inches, and divided by 
the radius of the mark sighted on the scratched line, will give angles of 
twist in radians. These numbers may be plotted as shown in Fig. 345, 
which illustrates the results obtained in testing the following specimen. 

ELASTIC TORSION TEST ON A MILD-STEEL SPECIMEN. 

Laboratory No. 2, M.S., 17.2.10. 

Diameter, 0-753 inch. 

Gauge points, 7-5 inches. 

Value of one scale division of the microscope, 0-000377 radian. 

Hooke's law broke down at 1750 Ib.-inches of torque (point A in 

Fig- 345)- 

Angle of twist at break-down of Hooke's law, 0-0358 radian. 
Maximum stress at break-down of Hooke's law, 9-3 tons per 

square inch. 
From the diagram, torque T=i65o Ib.-inches, when the angle of 

twist a is 0-0336 radian. 
Hence, 

TPT 

Modulus of rigidity = C = 4 = 5200 tons per square inch. 

" Creeping " of the specimen was noticed first distinctly when the 
torque was 2000 Ib.-inches, i.e. at this load the machine beam would 
begin to show an inclination gradually to droop under a steady load. 
The point is marked B in Fig. 345, and, as will be seen, occurs a 
considerable interval beyond the point A of elastic break-down. 



320 MATERIALS AND STRUCTURES 

The total work done up to the elastic limit will be found by taking 
the product of the mean torque and the angle of twist, and is 30-3 
inch-lb. 

To obtain the resilience, i.e. the work done per cubic inch of 
material, divide the total work by the volume of the specimen 
between the gauge points. The result is 9-3 inch-lb. 

Cement testing. Portland cement is made by mixing chalk and 
fine clay in certain proportions, burning the mixture at a clinkering 
temperature, and finely grinding the resulting product. Cement of 
this kind is much used for making concrete for constructional pur- 
poses. Concrete consists of an aggregate of clean broken stones, 
etc., to which sufficient clean sand is added to fill completely the 
voids between the stones. A quantity of Portland cement is inti- 
mately mixed with these, sufficient to coat the surface of every stone 
and every particle of sand with cement. Water is added, and the 
whole is mixed thoroughly in order to produce a plastic mass, which is 
rammed into moulds prepared to give the required structural shape. 

The qualities which Portland cement should possess have been 
laid down by the Engineering Standards Committee, and the tests 
should be carried out in accordance with the terms of their specifica- 
tion, 1 a copy of which should be in the hands of the experimenter. 

The fineness to which the cement has been ground is of great 
importance, and is tested by means of sieves, one having 5776 holes 
per square inch and another having 32,400 holes per square inch. 
These sieves are made in a special way of wire having standard 
diameters in terms of the specification. The residue left on the first 
sieve should not exceed 3 per cent., and on the latter 18 per cent. 

The specific gravity of the cement is taken now in place of 
weighing the cement in bulk. This may be ascertained by use of a 
specific gravity bottle having a graduated stem and containing a 
measured quantity of turpentine (cement will not set in turpentine). 
A measured weight W of cement is introduced into the bottle, 
and its volume V may be observed from the rise in level of the 
turpentine in the stem of the bottle. Then 



where w is the weight of a cubic unit of water and p is the specific 
gravity of the cement. The specific gravity should be not less than 
3-15 for cement freshly burned and ground. 3-10 is permitted at a 
period not less than four weeks after grinding. 

1 British Standard Specification for Portland Cement, Crosby, Lock wood & Son. 
Revised 1910. 



TESTING OF MATERIALS 321 

The strength of cement is determined usually by means of tensile 
tests, although cement in practice is generally under compression. 
Tensile tests may be carried out in a comparatively small machine, 
while compression tests require a machine capable of exerting great 
pressure. When compression tests are made, the test briquettes are 
generally cubical ; briquettes for tensile tests are prepared in moulds 
having a shape in accordance with that laid down in the standard 
specification. Considerable experience is required in order satisfac- 
torily to gauge or mix the cement intended for test briquettes. The 
quantity of water to be used depends on the kind of cement, and 
greatly influences the strength of the cement. The student can 
test this easily by preparing several briquettes having water per- 
centages of from 1 8 to 25, and testing these for tensile strength. 

The moulds should rest on an iron plate while being filled; no 
severe mechanical ramming should be necessary if the correct per- 
centage of water has been used. During the first 24 hours after 
filling, a damp cloth should be placed over the moulds. The 
briquettes are removed from the moulds then and placed in clean 
water until the strength test is carried out. The temperature 
throughout should be near 60 Fah. The tensile strength of neat 
cement briquettes (i.e. briquettes made of cement alone, without sand 
or other material) at 7 days from gauging should not be less than 
400 Ib. per square inch. Briquettes consisting of one part by weight 
of cement to three parts by weight of Leighton Buzzard sand, prepared 
in accordance with the terms of the standard specification, should 
have a tensile strength of 150 Ib. per square inch at 7 days after 
gauging. 

The setting time is tested by means of a standard needle having a 
flat point one millimetre square and having a total weight of 300 
grams. The cement 'is taken as set when the application of the 
needle fails to make an impression. 

The soundness of the cement is tested by the Le Chatelier method. 
A cylindrical mould having an axial split and furnished with two 
long pointers is filled with cement, as directed in the standard 
specification. This is kept in water for 24 hours, and then the 
distance between the ends of the pointers is measured. The mould 
and. cement are then boiled for 6 hours and allowed to cool. The 
distance is measured again, and the increase should not exceed a 
stated amount. 

Cubical cement and concrete blocks, bricks and stones are tested 
under compression. It is best to prepare the top and bottom 
P.M. X 



322 



MATERIALS AND STRUCTURES 



surfaces by smoothly coating them with plaster of Paris in order to 
give level parallel surfaces for the testing machine plates to bear 
upon. Generally, the fracture is by shearing on planes roughly at 
45 to the horizontal. Broken cement compression briquettes 
generally resemble two square based pyramids standing apex on apex. 



EXERCISES ON CHAPTER XIII. 

1. The following is the experimental record of a test on a specimen of 
cast iron. The object of the experiment was to determine the compres- 
sion value of E for the material : Ewing's extensometer was employed. 
The specimen was turned and polished. 

Diameter of specimen, 0-474 inch ; gauge points, 8 inches ; calibration 
of extensometer, i scale division = j^ inch. 



Load, Ib. 


200 


300 


400 


500 


600 


700 


800 


900 


IOOO 


IIOO 


Scale readings, \ 
load increasing / 


50-0 


48-2 


467 


45-o 


43-5 


41-9 


40-1 


38-6 


37-2 


36-1 


Scale readings, \ 
load decreasing / 


50-0 


48-2 


47-0 


45-6 


44-o 


42-0 


40-1 


38-5 


37-2 


36-1 



Find the value of E. 

2. Tensile tests were carried out on a turned and polished specimen of 
gun-metal. The following observations were made : Diameter of speci- 
men, 0-534 inch ; gauge points, 8 inches ; calibration of Ewing's exten- 
someter, i scale di vision = 4 gigft inch. The extensometer readings are 
given below : 



Load, Ib. 


o 


100 


200 


300 


400 


500 


600 


Scale readings 


40-0 


42-0 


43-4 


44-8 


46-2 


48-0 


49-1 


Load, Ib. 


700 


800 


900 


IOOO 


IIOO 


1 200 


1300 


Scale readings 


50-8 


52-0 


53-5 


55-o 


56-3 


58-1 


60- 1 


Load, Ib. 


1400 


1500 


1600 


1700 


. 175 


1800 





Scale readings 


61-8 


63-3 


66-0 


68-3 


69-3 


71-0 


44-5 



Creeping was first observed at 1600 Ib. load. 

Find the value of E. What is the stress when Hooke's law breaks 
down for this material ? How much permanent set was given ? 

3. The gun-metal specimen given in Question 2 was tested to breaking 
under tension after five weeks rest. The following observations were 
made : 

Breaking load, 5480 Ib. ; load at which the beam of the machine 
dropped, 3600 Ib. ; stretch on a length of 8 inches, 0-85 inch ; stretch on 



EXERCISES ON CHAPTER XIII. 



323 



a length of 2 inches, including the fracture, 0-3 inch ; diameter at fracture, 
0-479 inch. Reduce these observations, following the procedure indicated 
on p. 311. 

4. A mild-steel bar of square section 2 inches x 2 inches was arranged 
as a beam of 60 inches span, simply supported ; the load was applied at 
the middle of the span, and the deflections at the load were measured by 
means of a micrometer microscope, the calibration of which gave one 
eyepiece-scale division =0-065 mm - The following observations were 
taken : 



Load, Ib. 


500 


IOOO 


1500 


2000 


2500 


3000 


350 


Eyepiece-scale \ 
divisions J 


o 


29 


51 


72-8 


95 


"7-5 


139-7 


Load, Ib. 


3700 


3800 


3900 


4000 


4100 


4200 




Eyepiece-scale ^ 
divisions J 


149 


153-5 


158.5 


163-5 


169-2 


175-5 





Find the value of E for the material, also the maximum stress in the 
bar when Hooke's law broke down. 

5. A mild-steel bar 1-478 inches broad x 0-091 inch deep was arranged 
as a cantilever, the load being 16-3 inches from the support. Deflection 
and slope at the load were measured by means of the apparatus illustrated 
in Figs. 314 and 317. The calibration of the micrometer microscope used 
for observing the deflections gave one eyepiece-scale division = 0-6 mm. 
In the slope observations a scale of millimetres was used ; distance from 
the mirror to the scale = 566 mm. The following observations were 
taken : 



Load, Ib. 


o 


2 


4 


6 


8 


10 


12 


14 


Deflection scale 


7-93 


7-63 


7-3 


7-0 


6-7 


6-4 


6-1 


5-8 


Slope scale 


103-2 


IO2-5 


101-9 


IOI-I 


100-4 


99-8 


99-0 


98-2 



The beam theory gives for the ratio of deflection to slope of a cantilever 
carrying a load at its free end : 

A_WL 3 2EI _2 
z~3EI X WL 2 3 
Compare the experimental ratio of A : z with that calculated. 

6. A cast-iron test bar was tested under bending ; span 36 inches, 
simply supported ; breadth 1-02 inches ; depth 2-04 inches. The obser- 
vations gave the central breaking load = 41 20 Ib. and the maximum 
central deflection =0-5 inch. Find the coefficient of rupture. 

7. The following particulars relate to tests on a model reinforced 
concrete column. Height of column 24 inches ; section square, of 
3 inches edge ; main reinforcement, four mild-steel bars, each 0-31 inch 
diameter, arranged at the corners of a square of i| inches edge ; secondary 



324 



MATERIALS AND STRUCTURES 



reinforcement, thirteen horizontal lacings of iron wire, 0-067 inch diameter, 
about 2 inches pitch. Concrete mixture, cement 7 lb., granite chips 14 lb., 
water 3 lb. The column was made in a wooden mould, removed five days 
after making and tested fourteen days after making ; it was kept damp 
throughout this time. 

Observations taken : Hooke's law broke down sensibly at 9000 lb. load ; 
8000 lb. load shortened the column to the extent of 0-0154 inch ; the 
column ruptured when the load reached 20,670 lb. 

Taking 7/2=15, find the stress in the steel and in the concrete when 
Hooke's law broke down. Assuming the elastic laws to hold up to 
rupture, find these stresses at rupture. What is the value of E for the 
complete column ? 

8. The following observations were made during a torsion test on a 
mild-steel specimen : Diameter of specimen, 0-714 inch ; gauge points of 
strain indicator, 7-81 inches ; calibration of indicator, one scale division 
= 0-04 degree twist. 



Torque, lb. -inches 


o 


200 


400 


600 


800 


IOOO 


Scale divisions 


576 


569 


5 62 


554 


546 


538 


Torque, lb. -inches 


1 100 


I2OO 


I3OO 


1400 


1450 


1500 


Scale divisions 


534 


530 


526 


520 


517 


514 



Find the value of C ; also the stress at break-down of Hooke's law and 
the resilience in inch-lb. per cubic inch of material. 

9. A test was made in order to determine C for a copper wire by the 
torsional oscillation method, using Maxwell's needle. Employing the 
symbols explained on p. 297, the following observations were taken : 



mi , pounds. 


;2 > pounds. 


a, feet. 


d, inch. 


L, inches. 


t\ , sec. 


*2> sec. 


0-457 


0-053 


o-5 


0-048 


25 


4-3 


2-73 



Find the value of C for this material. 

10. Using the same Maxwell's needle, particulars of which are given in 
Question 9, the following observations were made during a test for the 
determination of E for steel wire by the torsional oscillations of a helical 
spring. Diameter of wire, 0-081 inch ; mean radius of helix, 0-4945 inch ; 
number of complete turns in helix, 133; ^ = 4-21 seconds; t^ = 2-666 
seconds. Find the value of E. 

11. The spring given in Question 10 was tested by the longitudinal 
vibration method in order to determine C. Mass of load hung from 
spring, 1-575 pounds ; mass of spring, 0-6048 pound ; time of one complete 
vibration, 0-631 second. Find the value of C. 

12. Use the results obtained in Questions 10 and 11, and calculate the 
value of the bulk modulus K for the material of the spring ; find also 
the value of Poisson's ratio. 



PART II. 
MACHINES AND HYDRAULICS. 



CHAPTER XIV. 
WORK, ENERGY, POWER, SIMPLE MACHINES. 

Work. Work is said to be done by a force when it acts through a 
distance. If a body A (Fig. 346) is at rest under the action of two 
equal forces P and R, no work is being done by either force ; if the 
body is moving with constant speed towards the right, work is being 
done by P against the resistance R. Work is measured by the pro- 
duct of the magnitude of the force and the distance through which it 
acts, the latter being measured along, or parallel to, the line of the 
force. In the case of a car travelling along a level road (Fig. 347) no 



W 





FIG. 346. 



FIG. 347. 



FIG. 348. 



work is done by the weight W, nor by the reactions of the ground, as 
none of these forces advance through any distance in the directions of 
their lines of action. Work is done by the weight of the car in 
descending an incline (Fig. 348). If the total height of descent is H, 
then the work done by W will be WH. Or, the solution may be 
obtained by resolving W into components P and Q respectively, 
parallel and at right angles to the incline. Q does no work while the 
car is descending ; P does work to the amount P x AB. 

The unit of work in general use in this country is the foot-pound, 
and is performed when a force of one pound weight acts through a 
distance of one foot. The foot-ton, inch-pound, and inch-ton are 



326 MACHINES AND HYDRAULICS 

used occasionally. Metric units of work are the gram-centimetre, 
the kilogram-centimetre and the kilogram-metre. 

Energy. Energy means capability of doing work, and is measured 
by stating the units of work capable of being performed. There are 
many different forms of energy, such as potential energy, said to be 
possessed by a raised body in virtue of the fact that its weight may 
perform work while the body is descending ; kinetic energy, which a 
body possesses when in motion and gives up while coming to rest ; 
elastic energy, possessed by a body under strain and given out while 
coming back to its original form or dimensions ; heat energy, which a 
body may give up in cooling to a lower temperature ; chemical energy, 
which may be present in a substance owing to its constituents being 
capable of combining in such a way as to liberate energy in the form 
of heat; electrical energy, possessed by a body by virtue of its 
electric potential being higher than that of surrounding bodies. 

Conservation of energy. The experience of all observers shows 
that the following general law is true : Energy cannot be created nor 
destroyed, but can be converted from one form into another form. This law 
is known as the conservation of energy. If no waste of energy were to 
occur during the conversion, a given quantity of energy in one form 
could be converted into an equal quantity in a different form. Exact 
equality never is obtained in practice ; there is always waste, some- 
times to a very large extent. For example, in converting the energy 
available in coal into mechanical work by means of a steam boiler 
and engine, it is common to find wasted 90 per cent, of the energy 
available, only 10 per cent, appearing in the desired form. 

In measuring heat energy, the British thermal unit may be used, 
one such unit being the quantity of heat required to raise the tem- 
perature of one pound of water through one degree Fahrenheit. The 
pound-calorie unit is likely to be used more extensively- in future, 
and is the quantity of heat required to raise the temperature of one 
pound of water through one degree Centigrade. The experiments of 
Dr. Joule and others show that an expenditure of 778 foot-pounds 
of energy will produce one British thermal unit ; 1400 foot-pounds is 
the energy equivalent to a pound-calorie unit. Mechanical energy 
may be converted into heat without very large waste occurring (for 
example, in mechanically stirring water), but the reverse operation is 
always accompanied with great waste. 

Power. Power means rate of doing work. The British unit of 
power is the horse-power, and is developed when work is being done 
at the rate of 33,000 foot-pounds per minute. The horse-power in 



MACHINES 



327 



any given case may be calculated by dividing by 33,000 the work 
done per minute in foot-pounds. 

The electrical power unit is the watt, and is the rate of working 
when an electric current of one ampere flows from one point of 
a conductor to another, the potential difference between the points 
being one volt. The product of amperes and volts gives watts. 
746 watts are equivalent to the mechanical horse-power. When the 
amperes and volts are stated, we have 

amperes x volts 



Mechanical horse-power 



746 



The Board of Trade unit of electrical energy is- one kilowatt main- 
tained for one hour. One horse-power maintained for one hour 
would produce 33,000 x 60 or 1,980,000 foot-pounds. The kilowatt- 
hour would therefore produce energy given by 
Energy = 1,980,000 x -Vrrr 

= 2,654,000 foot-pounds. 

Machines. A machine is an arrangement designed for the purpose 
of taking in energy in some definite form, modifying it, and delivering 
it in a form more suitable for the purpose 
in view. Machines for raising weights are 
arranged conveniently in most mechanical 
laboratories, and experiments on such are 
very instructive. Fig. 349 shows, in outline, 
a small crab which may be taken as a type 
of such machines. A load W Ib. is suspended 
from a cord wrapped round a drum, and is 
raised by the action of another load P Ib. 
attached to a cord coiled round an operating 
wheel. The wheel and drum are connected 
by means of toothed wheels, so that P descends as W ascends. 

The velocity ratio of such a machine is defined as the ratio of the 
distance moved by P while W ascends a measured distance. Let H 
and h be these distances respectively in inches (Fig. 349) ; they may 
be measured direct in the machine. Then 

TT 

Velocity ratio = V = -- r ( i ) 

Let P be so adjusted that it will descend with steady speed on 
being started by hand, thus raising a load W. The mechanical 
advantage of the machine is defined by 

Mechanical advantage = -= (2) 




FIG. 349. Outline diagram of 
an experimental crab. 



328 MACHINES AND HYDRAULICS 

By the principle of the conservation of energy, if no waste of 
energy occurs in the machine, the work done by P would be equal 
to the work done on the load. Suppose, in these circumstances, 
that the same working force P is employed, a larger load W 1 Ib. could 
be raised than would be the case in the actual machine. W 1 may be 
calculated as follows : 

Work done by P = work done on W 1 , 
PH^Wj/z, 

W l = P - = P V 

= P x the velocity ratio ............. (3) 

The effect of frictional and other sources of waste in the actual 
machine has been to diminish the load from W 1 to W. Hence, 
Effect of friction = F = Wj - W 

-FV-W ............................ (4) 

Efficiency of machines. The energy supplied to the machine is 
PH inch-lb. ^Fig. 349), the energy actually given out by the machine 
is Wh inch-lb. The efficiency of the machine is denned by 

. energy given out 

Efficiency = p =- 

energy supplied 



__ 

PH~P V 
_ mechanical advantage 
velocity ratio 

The efficiency thus stated will be always less than unity. Efficiency 
is often given as a percentage, obtained by multiplying the result 
given in (5) by 100. 100 per cent, efficiency could be obtained only 
under the condition of no energy being wasted in the machine, a 
condition impossible to attain in practice. 

From equation (3), we have 

w,,pH, 



A result which shows that the mechanical advantage of an ideal 
machine having no waste of energy is equal to the velocity ratio. 

For machines of the type described above, the following equation 
may be stated : 
Energy supplied = energy given out + energy wasted in the machine. 



MACHINES 



329 



Occasionally machines have to be considered in which there are 
internal springs or other devices for storing energy. In such cases 
the equation becomes : 
Energy supplied = energy given out + energy stored in the machine 

+ energy wasted in the machine. 

A machine is said to be running light when no energy is being 
given out. If no energy is being stored in a machine running light, 
then the energy supplied must be sufficient to make good the energy 
wasted in overcoming the resistances in the machine. 

Reversal of machines. A machine in which the frictional re- 
sistances are small may reverse if P is removed. To investigate this 
point, consider the machine when W i being raised (Fig. 349) : 
Energy supplied = PH, 
Energy given out = W/, 
Energy wasted =PH-W> (7) 

Now let P be removed and let the conditions be such that W is 
just able to reverse the machine. Let W descend through a height h. 
1 hen Energy supplied and wasted in the machine = Vfh (8) 

Assuming that this waste has the same 
value as when W is being raised, we 
have, from (7) and (8), 




FIG. 350. A small lifting crab. 



W* i 

Or, Efficiency = p^y = - 

Hence, when W is being raised, the 
efficiency will be 50 per cent, for reversal 
to be possible if P is removed. Any 
value of the efficiency exceeding 50 per 
cent, would be accompanied by the 
same effect. 

The following record of tests on a lifting crab will serve as a 
model for carrying out experiments on laboratory machines. 

EXPERIMENT ON A SMALL LIFTING CRAB. 
Date of test, loth February, 1911. 

The machine used was constructed by students in the workshops 
of the West Ham Technical Institute. Its general arrangement in 
" single-gear " is shown in Fig. 350. A weight W is suspended from 



330 



MACHINES AND HYDRAULICS 



a cord wrapped round a drum A. Motion is communicated to A by 
means of toothed wheels B and C ; these are of gun-metal with 
machine-cut teeth. Energy is supplied by a descending weight P, 
which is attached to a cord wrapped round a wheel D. 

The object of the experiments was the determination of the 
mechanical advantage and efficiency for various loads. 

By direct measurement of the distances moved by P and W, the 
velocity ratio was found to be = 8-78. 

The weight of the hook from which W was suspended is 1-75 Ib. 
The weight of the scale pan in which were placed the weights 
making up P is 0-665 Ib. 

The machine having been first oiled, the weights W and P were 
adjusted so as to secure descent of P with steady speed. The 
results obtained are given below. 

RECORD OF EXPERIMENTS AND RESULTS. 



.w't 

including 
weight of hook. 


(2) 

. Plb -.> 

including 
weight of 
scale pan. 


Load Wj if no 
frictional 
resistances, 
W x =PVlb. 


(4) 

Effect of 
friction, 
F=(W a -W)lb. 


(5) 

Mechanical 
advantage, 
W 
P' 


(6) 
Efficiency, 
per cent., 

^XXOO. 


8-75 


-785 


157 


6-95 


4-9 


55-8 


1575 


2-665 


23-4 


7-65 


5-9 


67-2 


22-75 


3-565 


31-3 


8-55 


6-38 


72-6 


29.75 


4-405 


387 


8-95 


6-74 


76-6 


3675 


5-335 


46-8 


10-05 


6-89 


78-5 


4375 


6-215 


54-6 


10-85 


7-04 


80-0 


575 


7-115 


62-5 


11-75 


7-14 


81-2 


5775 


8-065 


70-8 


13-05 


7-16 


81.6 


64-75 


8.915 


78-4 


13-65 


7-26 


82-7 


7175 


9.815 


86-2 


14-45 


7-30 


83-2 


78-75 


10-705 


94-1 


15-35 


7-36 


83-7 


8575 


11.59 


101-8 


16-05 


7-40 


84-3 


92-75 


12-515 


I IO 


17-25 


7-41 


84-4 


9975 


13-405 


118 


18-25 


7-43 


84-6 


106-75 


14-285 


125-4 


18-65 


7-47 


85-0 


H375 


15-205 


133-8 


20-05 


7-48 


85-2 


120-75 


16-065 


141 


20-25 


7-5i 


85-5 


127-75 


16-965 


149 


21-25 


7-53 


85-7 



Curves are plotted in Fig. 351 showing the relation of P and W 
and also that of F and W. It will be noted that these give straight 
lines. Curves of mechanical advantage and of efficiency in relation 
to W are shown in Fig. 352. It will be noted that both increase 
rapidly when the values of W are small and tend to become constant 
when the value of W is about 120 Ib. The efficiency tends to 
attain a constant value of 86 per cent. 



MACHINES 



331 



PA 
LB 


noF 




20 
18 
16 

14 
12 
10 
8 
6 
4 

2 


























.y 


X 






















^x 


? 




















. 


x^ 






x 
















/ 






^ 


x 














X 


^ 






X 
















X 






/*. 


i^ 














x 


J?^ 






^ 


^ 












/ 


X 








? 






















x 






















X 


* 






















X 




























10 20 30 40 50 60 70 80 90 100 HO 120 130 LB 



FIG. 351. Graphs of F and W, and P and W, for a small crab. 



EFFICIENCY 
PER CENT 
100 



MECHANICAL 
ADVANTAGE 





10 20 30 40 50 60 70 80 90 100 MO 120 
FIG. 352. Graphs of efficiency and mechanical advantage for a small crab. 

As both of the curves showing the relation of P and of F with W 
are straight lines, it follows that the following equations will represent 

these relations : p = a \y + b, ( i ) 

F = <W + 4 (2) 

where a, b, c and d are constants to be determined. 



332 



MACHINES AND HYDRAULICS 



Select two points on the PW graph, and read the corresponding 
values of P and W. 

P= 3-5 Ib. when W= 22-7 Ib. 
P=i6-olb. when W=i2o-olb. 
Hence, from (i), 3.5 = 22-7^ + ^, 
16= i2oa + l>. 

Solving these simultaneous equations, we obtain 
= 0-128, 
^ = 0-64; 

Similarly, 

When F= 8 Ib., W= 20 Ib. 
WhenF=i81b., W=ioolb. 
Hence, from (2), 8= 2oc + d, 
1 8 = i ooc + d. 

The solution of these gives 
;= 0-125, 



FIG. 353. Pulley 
blocks. 



Hence, F = o-i25W + 5'5 (4) 

Suppose the machine to be running light, then 

W = o, and the corresponding values of P and F obtained from 

(3) and (4) are p = . 64 ft., 

F= 5 . 5 lb. 

The interpretation is that a force of 0-64 Ib. is required to work 
the machine when running light, and that, if there 
were no frictional waste, a load of 5-5 Ib. could be 
raised by this force. 

Hoisting tackle. The fact that the mechanical 
advantage of a machine, neglecting friction, is equal 
to the velocity ratio, enables the latter to be cal- 
culated easily in cases of hoisting tackle. A few 
such appliances, which may be found in most 
laboratories, are here given. 

In the pulley-block arrangement shown in Fig. 353, 
let n be the number of ropes leading from the 
lower to the upper block. Neglecting friction, 

i th 
each of these ropes will support - of W, and this 

will also be the value of P. Hence, 

W_W_ TW 

V -p- "TIT" ^* FIG. 354- Weston's 

* differential blocks. 




MACHINES 



333 



A set of Western's differential blocks is shown in outline in Fig. 354 ; 
the upper block has two pulleys of different diameters, and a chain, 
shown dotted, is used. The links of the chain passing round these 
pulleys engage with recesses which prevent slipping. Neglecting 
friction, each of the chains A and B will support JW. Taking 
moments about the centre C of the upper pulleys, and calling the 
radii R and r respectively, we have 

!\V x CD = (P x CF) + (|W x CE), 



Instead of R and r, the number of links which can be fitted round 



B 




W 

FRONT ELEVATION END ELEVATION 

FIG. 355. Wheel and differential axle. 




FIG. 356. Helical blocks. 



the circumferences of the pulleys may be used : evidently these will 
be numbers proportional to R and r. 

The wheel and differential axle (Fig. 355) is a similar contrivance, 
but has a separate pulley A for receiving the hoisting rope. Taking 
moments as before, we have 



y.- W = 

p 



A set of helical blocks is shown in outline in Fig. 356. A is 



334 



MACHINES AND HYDRAULICS 



operated by hand by means of a hanging endless chain and rotates 
a worm B, which in turn advances the worm wheel C one tooth for 
each revolution of A. If there be n c teeth on C, then A will rotate 
n c times for one revolution of C, and P will advance a distance 
C L A , which is equal to n c times the length of the number of links 
of the hanging chain which will pass once round A. The chain 
sustaining the load W is fixed at E to the upper block, passes round 
F, and then is led round D, which has recesses fitting the links in 
order to prevent slipping. Let L D be the length of the number 
of links which will pass once round D. Then in one revolution of 
D, W will be raised through a height equal to4L D . Hence, 



EXPTS. 33 to 37. Experiments on the hoisting appliances de- 
scribed above should be carried out and the results reduced by 
methods similar to those explained for a small crab on p. 329. 

Diagram of work. Since work is measured by the product of 
force and distance, it follows that the area of a diagram in which 
ordinates represent force and abscissae represent distances will 

p 



Oi. 



-D -J 




FIG. 357. Diagram of work 
done by a uniform force. 



o o 

FIG. 358. Diagram of work 
done by a varying force. 



represent the work done. A uniform force P pounds acting through 
a distance D feet does work which may be represented by the area 
of a rectangle (Fig. 357). To obtain the scale of the diagram : 

Let i inch height represent p Ib. ; 

i inch length represent d feet. 

Then one square inch of area will represent pd foot-pounds of 
work. If the area of the rectangle is A square inches, then 
Work done =pdK foot-lb. 

In the case of a varying force, the work diagram is drawn by 
setting off ordinates to represent the magnitude of the force at 
different intervals of the distance acted through (Fig. 358). A fair 
curve drawn through the tops of the ordinates will enable the force 
to be measured at any stage of the distance. The work done is the 
product of the average force and the distance, and as the average 



DIAGRAMS OF WORK 



335 



force is given, to scale, by the average height of the diagram, and 
the distance, to scale, b^ the length of the diagram, we have, as 
before, the work done represented by the area of the diagram. 
Using the same symbols as before, one square inch of area represents 
pd foot-pounds of work, and the total work done will be given by 

Work done =d A foot-lb. 



The area A may be measured by means of a planimeter, or by use 
of any convenient mensuration rule (p. 6). 

The case of hoisting at steady speed a load from a deep pit is of 
interest (Fig. 359). Let W l Ib. be the weight of the cage and load, 
and let W 2 Ib. be the total weight of the vertical rope when the cage 
is at the bottom, a depth of H feet. At first the pull P Ib. required 
at the top of the rope will be (W l + W 2 ) Ib. P will diminish gradually 




FIG. 359. Diagram of work 
done in hoisting a load. 




FIG. 360. Work done in raising 
a body. 



as the cage ascends, and will become equal to \V l when the cage 
is at the top. The work diagram for hoisting the cage and load alone 
is a rectangle ABCD, BC and AB representing Wj and H respec- 
tively ; the diagram for hoisting the rope alone is DCE, in which 
W 2 is represented by CE. From the diagrams, we have 

Total work done = ( W l + 1 W 2 ) H foot-lb. 

Work done in elevating a body. It will be shown now that the 
work done in raising vertically a given body may be calculated by 
concentrating the total weight at the centre of gravity. Referring to 
Fig. 360, let ze^, z# 2 , etc., be the weights of the small particles of 
which the body is composed, and let h^ // 2 , etc., be their initial 



336 



MACHINES AND HYDRAULICS 



heights above ground level, and let //, /* 2 ', etc., be their final 
heights. Then 

Work done on w l = w l (h-{ - h-^ 

and Work done on w^ w^(h^-h^^ etc. 

Hence, 
Total work done = (w^ + w^ + etc.) - (wji^ + w 2 /i 2 + etc.) 




FIG. 361. Thomson indicator. 

Let G and G' be the initial and final positions of the centre of 
gravity of the body, situated respectively at heights H and H', and 
let W be the total weight of the body. Then 

WH' = 2ze/#(P- 49) 
and 



INDICATED HORSE-POWER 337 



Hence, Total work done = WH' - WH 

= W(H'-H). 

Therefore the total work done in raising a body may be calculated 
by taking the product of the weight of the body and the vertical 
height through which the centre of gravity has been raised. This 
method is equivalent to concentrating the total weight at the centre 
of gravity. 

Indicated work and horse-power. An indicator is an instrument 
used in obtaining a diagram of work done in the cylinder of an 
engine. The essential parts of an indicator are shown in Fig. 361. 
A small cylinder A is fitted with a piston B, which is controlled by a 
helical spring C. Connection is made at D to the engine cylinder ; 
E is a stop cock. The piston B is connected by means of a piston 
rod to a lever system having a pencil fixed at P ; the function of the 
lever system is to guide P in a straight vertical line, and to give it an 
enlarged copy of the motion of the piston B. As the spring follows 
Hooke's law, it follows that the movement of P will represent a 
definite pressure in pounds per square inch for each inch of vertical 
travel. The pencil moves over a piece of paper wrapped round a 
drum F. The drum is rotated in one direction by means of a cord 
G, and is brought back again by means of an internal spring. The 
cord G is actuated by some reciprocat- 
ing part of the engine which gives it 
a reduced copy of the motion of the 
engine piston. Hence a diagram will be 
drawn on the paper showing pressures 

in the engine cylinder by its ordinates, * L 

and distances travelled by the engine FIG. 362. -Work done during the 
piston by its abscissae (Fig. 362). The 

curve abc is for the forward travel of the piston, actuated by the 
steam or other pressure, and the curve cde is for the backward travel, 
and shows the exhaust. 

The work done during the stroke may be found by first obtaining 
the average height of the diagram inclosed by the curves in inches 
and multiplying this by the scale of pressure ; the result gives the 
average pressure on the piston in pounds per square inch. 

Let A = the effective area of the piston, in square inches. 

L = the length of the stroke, in feet. 
p m = the average pressure, in Ib. per square inch, 

Then Work done per stroke =/ m AL foot-lb. 

P.M. Y 




338 



MACHINES AND HYDRAULICS 



In the case of a double-acting steam engine, the diagram of work 
for the other side of the piston will resemble Fig. 363. The effective 

T^\2 

area of one side of the piston (Fig. 364) will be - , and of the other 

IT 4 

side - (D 2 - d^\ Let p' m be the average pressure for the latter side 
4 

of the piston. Then 

Work done per revolution = ( / m +/ m - (I) 2 - </ 2 )) L ft.-lb. 

144 J 

The work done per minute will be obtained by multiplying by N, 
the revolutions per minute, and the indicated horse-power, written I.H.P., 
by dividing the result by 33,000. 

Rough calculations are made often by neglecting the piston rod; 



T-J2 

thus A will be assumed as -- for each side of the piston. A mean 

4 
pressure / is taken as \(pm,+p'ni) and used for both sides of the 




FIG. 363. Work done during the 
return stroke. 



Id 



r 



FIG. 364. Double-acting steam 
engine cylinder. 



piston. The calculation for indicated horse-power will be given 
approximately by 2 MLN 

I.H.P. = -^ , 
33000 

where N, as before, is the revolutions per minute. 

In the case of a gas or oil engine, in which one side only of the 
piston is used, the other side being open to the atmosphere, the 
indicator diagram (Fig. 365) is used in the same manner to obtain 
the mean pressure. The work done during the cycle will be given 

b y Work done =/AL. 

Let N E = number of explosions per minute. 

Then ,. H .P.=-^k 

33000 

The indicated horse-power may be taken as a measure of the 
energy given to the engine piston during a stated time. A fraction 
only of this can be given out by the engine, the difference represent- 



BRAKE HORSE-POWER 



339 



ing horse-power expended in driving the engine itself and overcoming 
the fractional resistances of its mechanism. 




FIG. 365. Indicator diagram from a gas engine cylinder. 

Brake horse-power. Provided the engine is not too large, the 
horse-power which the engine is capable of giving out in doing 
useful work may be measured by means of a brake. The result is 
called the brake horse-power, written B.H.P. It is evident that the 
efficiency of the engine mechanism will be given by 

a. . power given out 

Mechanical efficiency = rr - 

power supplied 

B.H.P. 



I.H.P. 

This may be expressed as a percentage by multiplying by 100. 
The horse-power expended in overcoming the frictional resistances 
of the mechanism will be 

H.P. wasted in the engine = I.H.P. -B.H.P. 

Work done by a couple. Let equal forces P l and P 2 lb., forming 

a couple (Fig. 366), act on a body free to rotate about an axis at O 

and let the body make one revolution. As P ] does not advance 

through any distance, it does no work. P 2 

advances through a distance 2ird feet, where d 

is the arm of the couple in feet. Hence, 

Work done by the couple per revolution 

= Po x 2ird 



27T 




FIG. 366. Work done 
by a couple. 



= moment of couple x angle of rotation 

in radians. 

The units of this result will be foot-lb. if the moment of the couple 
is stated in Ib.-feet units. It is evident that any axis of rotation 



340 MACHINES AND HYDRAULICS 

perpendicular to the plane of the couple may be chosen without 
altering the numerical result, because a couple has the same moment 
about any point in its plane (p. 59). Since the work done will be 
proportional to the angle of rotation, we have 

Work done = (moment of couple in Ib.-feet x a) foot-lb., 
where a is the total angle turned through in radians. 
Let N = revolutions per minute, 

T = moment of couple, in Ib.-feet. 

Then Angle of rotation = 2?rN radians per minute ; 

Work done per minute = T x 2?rN foot.-lb. 

Advantage is taken of this result in estimating the brake horse- 
power of an engine. 

Brakes. In the more usual form of brakes, frictional resistance is 
applied to the flywheel of the engine by means of a band. Rotation 
of the band is prevented by means of pulls applied by dead weights, 
or by spring balances. From the observed values of the pulls, the 
moment of the applied couple may be calculated. This, together 
with the revolutions per minute, enables the work done per minute 
and the horse-power to be calculated. 

As the work done against the frictional resistance of the band is 
transformed into heat, and thus will cause the temperature of the 
wheel to rise, it is often necessary to adopt some means of cooling 
the wheel. 

Rope brakes. A simple form of brake is shown in Fig. 367, and 
consists of two ropes passed round the wheel and prevented from 
slipping off sideways by means of wooden brake blocks, four of which 
are shown. A dead weight W is applied to one end of the ropes 
and a spring balance applies a force P to the other end. The net 
resistance to rotation will be (W - P), and this constitutes one force 
of the couple. The other equal force is Q, and arises from a pressure 
applied to the wheel shaft by its bearings. The forces W and P are 
applied at a radius R, measured to the centre of the rope. Hence, 
the moment of the couple applied is (W - P)R. 

Let W = dead load, in Ib. 

P = spring balance pull, in Ib. 

R = radius to the rope centre in feet. 

N = revolutions per min. 



ROPE BRAKES 



341 



Then Work done per revolution = (W - P) R . 2ir foot-lb. 
min. = ( W - P) 27rRN foot-lb. 

(W-P)27rRN 

Brake horse-power = . 

33000 

In using a brake of this pattern, it is advisable to have W attached 
by a loose rope to an eyebolt fixed to the floor. This device will 
prevent any accident should the brake jam or seize. 



Section of 
wheel rim 





FIG. 367. Simple rope brake. 



FIG. 368. Rope brake for small powers. 



In cases when the power is small, it may be better to pass the 
ropes round half the circumference of the wheel only (Fig. 368), 
using a spring balance at each end. The brake horse-power may be 
calculated from 

(P 1 -P 2 ) 2 7rRN 
B.H.P. = V * . 

33000 

This plan has an advantage in the fact that 
both spring balances are assisting to sustain the 
weight of the wheel, and thus partially relieve 
the shaft bearings of pressure. Hence there 
will be lower frictional resistances in the engine 
and a slightly improved mechanical efficiency. 
A leather strap may be substituted for ropes 
in this kind of brake. 

Cooling of the wheel may be effected by 
having its rim of channel section (Fig. 369) and 
running cold water in through a pipe A having a regulating valve. 
Centrifugal action maintains the water in the rim recess, provided 




FIG. 369. Arrangement for 
cooling the brake wheel. 



34* 



MACHINES AND HYDRAULICS 



the speed of rotation be sufficient. The heated water is removed 
gradually by means of a scoop pipe B having a sharpened edge, 
.and thus a continuous water circulation is maintained. 

Band brake. An excellent form of brake has been designed by 
Professor Mellanby of the Royal Technical College, Glasgow. An 
application of this brake to the flywheel of a steam engine of about 1 5 
horse-power in the author's laboratory is shown in Fig. 370. A number 
of wooden blocks A are arranged round the circumference of the wheel, 



B 



View of the 
toggle joint 




FIG. 370. Mellanby type of band brake. 

and are held in position by hoop iron bands B, B. The brake bands 
are in halves, connected at C by means of long adjusting bolts fitted 
with lock nuts, and at D by means of toggle joints, by use of which 
the tension of the bands may be adjusted. A dead load W is hung 
from a pin E, which is attached to the brake hoops by four rods. A 
spring balance applies a pull P through a similar arrangement on 
the other side of the brake. There is a short column G fixed to the 
floor and slotted at its top end so as to restrict the movements of the 
pin F. Details of the toggle joint are shown separately. Two 
blocks Kj and K 2 are connected by four links H and pins to the 



HIGH-SPEED BRAKE 



343 



brake bands B. The blocks Kj and K 2 may be made to approach one 
another and thus shorten the brake bands by means of the long bolt 
and the hand wheel M ; a feather key in K x prevents rotation of the 
bolt. Helical springs N, N assist the adjustment, 

In use, P and W are adjusted very easily so as to be equal. Hence 
a pure couple is applied to the wheel, and the shaft bearings are 
relieved of carrying any of the dead load W. The toggle-joint 
adjustment is very good, and enables the frictional resistance of this 
particular brake to be adjusted within very fine limits. In the ori- 
ginal large form, a dash-pot is introduced at G to subdue oscillations 
of the brake. This has not been found necessary in the smaller 
brake used by the author. 

It will be noted that both P and W offer resistance to rotation. 

Let d= horizontal distance between P and W in feet. 



Then 



B.H.P. = 



33000 33000 

provided P and W are adjusted so as to be equal. If they are not 
exactly equal, then their mean, J(W + P), should be taken, giving 



B.H.P. = 



33000 



33000 




FIG. 371. Brake for high speeds of rotation. 



High-speed brake. In Fig. 371 is shown a brake for high speeds 
of rotation. The brake wheel consists of a flanged wheel mounted 



344 MACHINES AND HYDRAULICS 

on the second motion shaft of a De Laval steam turbine, and runs 
at 3750 revolutions per minute. The brake blocks A, A are made of 
Wood, and are pressed to the wheel by means of two bolts B, B fitted 
with wing nuts and helical springs C, C, the latter rendering it easy 
to adjust and maintain the desired pressure, A steel band 1) is fixed 
to the brake blocks, and serves to keep the parts together when 
the brake is removed and also for the application of the loads W x 
and W 2 . The whole contrivance is balanced when Wj and W 2 are 
removed (leaving the suspending rods F, F in position); hence the 
effective force is (W l - W 2 ) Ib. at a radius R feet. 

(W,- W 2 )2jrRN 
B.H.P. =- 3 . 

33000 

Other methods of estimating effective horse-power. Hydraulic 
brakes have been used for fairly high powers. The principle of 
such brakes is to fit a badly designed centrifugal pump to the 
engine shaft. The pump wheel and casing are so constructed as 
to set up violent eddies in the water, with the result that there is 
considerable resistance opposed to rotation of the wheel. The pump 
casing is capable of rotating with the wheel, but is prevented from so 
doing by an attached lever and dead weight. The moment of this 
weight gives the couple required for the estimation of the energy 
absorbed. Brakes of this type were introduced by Professor Osborne 
Reynolds, and in his hands served not only for determining the horse- 
power of the engine, but also for the determination of the mechanical 
equivalent of heat. The latter experiment was carried out by 
observing the quantity and rise of temperature of the water passed 
through the brake in a given time. 

The brake or effective horse-power of very large engines cannot be 
determined experimentally by use of a brake. If electrical generators 
are being driven, a close estimation may be made from the electrical 
energy delivered from the generator, making allowance for electrical 
and mechanical waste in the machine. 

In electrical installations driven by steam turbines, the electrical 
horse-power alone can be measured, as no indicator diagrams can be 
obtained from turbines. 

Shaft horse-power. Where turbines are adopted on board ships 
for driving the propellers, the shaft horse-power is measured, and corre- 
sponds to the brake horse -power. The method consists in measuring 
the angle of twist in a test length of the propeller shafting by means 
of a torsion-meter. The test length is calibrated carefully before 
being placed on board, and should be re-calibrated at intervals, so 



SHAFT HORSE-POWER 



345 



that a curve is available showing the moment of the couple required 
to produce a given angle of twist. The turning moment on the shaft 
is obtained from the angle of twist indicated by the torsion-meter. 
Let T = turning moment, in Ib.-feet. 

N = revolutions per min. 

TX 27TN 

Then Shaft horse-power = 

33000 

Shaft calibration In Fig. 372 is shown the method employed 
at the Thames Iron Works engine department for calibrating the 




M 



P 

Pf 




FIG. 372. Arrangement for calibrating a shaft for shaft horse-power. 

test length of shaft ; the view is a plan. The shaft AB has 
flanged ends solid with the shaft, and is bolted at A to a 
very rigid bracket C ; a bearing at D supports the other end. A 
beam EF is bolted to the end B of the shaft, and couples may 
be applied by means of the upward pull of a 5 -ton Denison 
weigher at E, and the equal downward force applied by placing 
weights in a skip hung from F. GH, KL and MN are balanced 
arms fixed to the shaft, and have verniers and scales at the ends 
H, L and N which serve to measure the angle of twist inde- 
pendently of the torsion-meter. The arm GH is bolted to the 
flange at A, and indicates any yielding of the bracket C or of the fix- 
ing of the shaft to the bracket. The difference of the readings at L 



34 6 



MACHINES AND HYDRAULICS 



and N will give the angle of twist of the shaft between the arms KL 
and MN, and will not be affected by any yielding of the bracket or 
other fixings. 

The torsion-meter is fixed to the shaft at OP, and is of the 
Hopkinson-Thring type ; the lamp and scale are situated at Q. In 
Fig. 373 is shown the arrangement of the torsion-meter. C is a 
sleeve made in halves and clamped to the shaft AB, which it grips at 
its left-hand end. D is a collar, also made in halves, and clamped to 
the shaft. The angle through which C twists relative to D is 
measured by means of a small mirror at E. The mirror may rotate 
slightly about a radial axis on the collar D, and is controlled by a 

short rod attached to the sleeve 
C at F. A ray of light from the 
lamp H is reflected and changed 
in direction horizontally by the 
mirror. Two mirrors are used 
at E, placed back to back, and 
the ray is reflected to the scale 
when E arrives at the top and 
also when it is at the bottom ; 




I 

1 
1 
1 

1 
1 

1 
J 

1 

t 


\ 

\ 

\ 

( \ 


i 


r 



FIG. 373. Hopkinson-Thring torsion-meter. 



when at the top, the ray is re- 
flected to the left part of the 
scale, and is reflected to the 
right part when E arrives at the 
bottom. Owing to the rapid 
rotation of the shaft, these inter- 
mittent rays produce practically a continuous light on the scale. A 
separate fixed mirror (not shown in the illustration) is attached to the 
sleeve and serves as a zero .pointer on the scale. The scale and lamp 
are carried on trunnions to facilitate the preliminary adjustment 
required in order to secure that both zero mirror and movable mirror 
give the same scale reading when there is no torque on the shaft. 

The following records were obtained by Mr. C. H. Cheltnam 
during a calibration test with the apparatus described above : 

CALIBRATION OF A PROPELLER SHAFT. 

External diameters of the shaft between the vernier arms : 
12-25 inches for a length of 6 inches ; 
11-375 inches for a length of 24-75 inches; 
11-25 inches for a length of 134 inches. 

Diameter of the hole in the shaft, 6-75 inches. 



SHAFT CALIBRATION 



347 



Distance between the clamping planes of the vernier arms, 164-75 
inches. 

Radius of the vernier arms, 114-6 inches. 

(Since one radian = 5 7-3 degrees, a movement of 2 inches at the 
vernier represents one degree twist on a length of shaft of 164-75 
inches.) 

External diameter of the shaft at the torsion meter, 11-25 inches. 

Diameter of the shaft hole at the torsion meter, 6-75 inches. 

Distance between the clamping planes of the meter, 33-625 inches. 

One division on the torsion-meter scale corresponds to an angle of 
twist of -- degrees. 

LOG OF TEST. 



No. 


Torque, 
Ib.-feet. 


Vernier readings, inches. 


Angle of twist, degrees, 
by verniers. 


Torsion- 
meter 
readings. 


No. i. 


No. 2. 


I 


16,800 


0-075 


0-300 


0-II25 


15-2 


2 


33,600 


0-150 


0-610 


0-2300 


30-2 


3 


50,400 


0-225 


0-915 


0-3450 


45-5 


4 


67,200 


0-300 


1-225 


0-4625 


61-0 


S 


84,000 


0-380 


1-535 


0-5775 


77-0 


6 


IOO,8OO 


0-460 


1-850 


0-6950 


92-5 


7 


117,600 


0-545 


2-160 


0-8075 


1 08-0 


8 


134,400 


0-620 


2-480 


0-9300 


124-0 



The torques and angles of twist obtained from the vernier readings 
are plotted in Fig. 374; in Fig. 375 the torsion-meter readings and 
torques have been plotted ; both give straight lines. 



Lb-feet Lb.-feet 






















f 




























^- 




















2 


























/ 




















( 




























/ 




















/ 






100000 




















/ 






















/ 


























/ 




















f 










8Q000 
6QOOO 
















/ 












60POO 
40QO.O 












/ 






















/ 






















/ 






















/ 


r 




















/ 
























/ 






















7 
























^ 






















/ 
























/ 






















/ 
























/ 

















































/ 
























~/_ 






















y 


























0-2 04 06 08 I'O 2 


4O 


G 


10 


00 120 



Deynu Scott readings 

Fiu 374. Graph of the vernier readings. FIG. 373. Graph of the torsion-meter readings. 



348 MACHINES AND HYDRAULICS 

To check the meter readings, the modulus of rigidity is calculated 
(a) from the vernier readings, (b) from the meter readings. 

(a) The torque at No. 6 is 100800 x 12 = 1,209,600 Ib.-inches, and 
gives an angle of twist of 0-6950 degree = 0-012 13 radian. Using 
equation (5) (p. 255) for the angle of twist of a hollow shaft, viz. 

2 TL 

a = /T. d T) 4\r- radians, (i) 



and modifying it to suit the case of a shaft of three different external 
radii R a , R&, R c , and corresponding lengths L a , L&, L c , the same 
torque being applied throughout, we have 

2TL C 



or 



4 



L c 1 

e 4 - R 2 4 J 



Ra 4 - R 2 4 R& 4 - R 2 4 
= 2X 12096001" 6 24-75 , T 34 "I 

~ TT x 0-01213 [>i25 4 - 3-375 4 s-687 4 - 3-375' 5-625^ - 3'375 4 J 

= 11,774,000 lb. per square inch. 

(b) The torque at No. 6 is 1,209,600 Ib.-inches, and gives a scale 
reading on the meter of 92-5. Hence, 

a = 638 * 92 ' 5 * 5^3 = ' 00253 radian> 



From (i), C 

2 x 1209600 x 33-625 
~7r(5-625 4 3'375 4 ) x 0-00253 
= 11,742,000 lb. per square inch. 

The agreement of these values of C is close enough testimony to 
the accuracy of the meter. To obtain the shaft horse-power constant, 
we have ^ XT 

ou r u T X 27rN 

Shaft horse-power = , 

33000 

where T is the torque in Ib.-feet and N is the revolutions per minute. 
At No. 6, the torque is 100,800 Ib.-feet and the meter reading is 
92-5 scale divisions. Hence, 

100800 
Torque for one scale division = - 

92-5 

= 1090 Ib.-feet. 



TRANSMISSION DYNAMOMETERS 



349 



Let the meter reading be n scale divisions. Then 
T= logon Ib.-feet. 

109072 x 27rN 
33000" 

i 

#N 

4-82 



Hence, 



Shaft horse-power = 



EXAMPLE. At the steam trial of the vessel to which this shaft was 
fitted, the mean meter reading was 98 scale divisions at 300 revolutions 
per minute. Find the shaft horse-power. 



Shaft horse-power = - N 
4-02 




Transmission dynamometers are sometimes used for estimating 
the horse-power required to drive a given machine. The principle of 
the Froude or Thorneycroft dynamometer is shown 
in Fig. 376. A is a pulley on the line shaft ; B is 
a pulley on a shaft connected to the machine to be 
driven. A drives B by means of a belt passing 
round pulleys C and D which are mounted on a 
frame pivoted at F. When power is being trans- 
mitted, the pulls Tj, Tj of the belt are greater than 
T 2 , T 2 ; hence a force P applied to the frame at G 
is necessary in order to preserve equilibrium. Taking 
moments about F, we have 

P x GF = (2T X x FC) - (2T 2 x FD). 

The arms FC and FD are usually equal. Hence, 
PxGF = 2 FC(T 1 -T 2 ), 




or 



FIG. 376. Froude or 
Thorneycroft dyna- 
mometer. 



Let 



Then 



R = the radius of pulley B in feet. 
N = revs, per min. of pulley B. 

(T 1 -T 2 ) 2 7rRN 

H.P. = * * K . 

33000 



The more usual method now adopted is to drive the machine 
direct by means of an electro-motor and measure the electrical 
horse-power consumed. 



350 



MACHINES AND HYDRAULICS 



EXERCISES ON CHAPTER XIV. 

1. Calculate what useful work is done in pumping 1000 gallons of 
water to a height of 60 feet. If this work is done in 25 minutes, what 
horse-power is being developed? Suppose that the efficiency of the 
pumping arrangements is 55 per cent., and find what horse-power must be 
supplied. 

2. A load of 4000 Ib. is raised at steady speed from the bottom of a, 
shaft 360 feet deep by means of a rope weighing 10 Ib. per yard. Calcu- 
late the total work done, and draw a diagram of work. 

3. A loaded truck has a total weight of 15 tons. The frictional 
resistances amount to 12 Ib. per ton. Calculate the work done in hauling 
it a distance of half a mile (a) on a level track, (b) up an incline of 
i in 80. 

4. Find the price in pence per 1000 foot-lb. of energy purchased in 
the following cases : 

(a) Coal, of heating value 15,000 British thermal units per pound, at 
16 shillings per ton. 

(b) Petroleum, of heating value 19,500 British thermal units per pound, 
at lod. per gallon weighing 8-2 Ib. 

(c) Gas, of heating value 520 British thermal units per cubic foot, at 
2-25 shillings per 1000 cubic feet. 

(d} Electricity, at i-$d. per Board of Trade unit. 

5. In a machine used for hoisting a load the velocity ratio is 45, and 
it is found that a load of 180 Ib. can be raised steadily by application of a 
force of 12 Ib. Find the mechanical advantage, effect of friction and the 
efficiency. Would there be any danger of reversal if the force of 12 Ib. 
were removed ? 

6. A load of 1200 Ib. is raised by means of a rope provided with an 
arrangement for indicating the pull at any instant. The following obser- 
vations were made : 



Height above ground, 1 
feet J 


o 


10 


20 


35 


50 


65 


80 


Pull in rope, Ib. 


2000 


1950 


1880 


1800 


1750 


1650 


1500 



Find approximately the work done on the load. 

7. The cylinder of a steam engine is 30 inches in diameter, and the 
stroke of the piston is 4 feet. The piston rod is 5 inches in diameter. 
Suppose the mean pressure for both sides of the piston to be 65 Ib. per 
square inch, what will be the horse-power at 75 revolutions per minute? 

8. A rope brake is fitted to a flywheel 3 feet in diameter to the rope 
centre and running at 220 revolutions per minute. It is desired to absorb 
7 brake horse-power. What should be the difference in the pulls at the 
two ends of the rope ? 

9. A shaft 6 inches in diameter runs at 180 revolutions per minute 
and transmits 900 horse-power. Assume the torque to be uniform, and 
calculate its value. 



EXERCISES ON CHAPTER XIV. 



351 



10. In Question 9 a torsion-meter is fitted to a shaft at points 6 feet 
apart. Taking C to be 5500 tons per square inch, what angle of twist, in 
degrees, will be indicated by the instrument ? 

11. In calibrating a propeller shaft by use of the apparatus illustrated 
in Fig. 372, the following observations were made : External and internal 
diameters of the hollow shaft, 7 inches and 4-017 inches respectively ; 
distance between the clamping planes of the vernier arms, 51 inches ; 
distance between the clamping planes of the torsion-meter, 25-5 inches ; 
radius of the vernier arms, 105 inches. 24,000 divisions on the meter scale 
correspond to an angle of twist of i radian. 



Test No. 


Torque, 
Ib. -inches. 


Difference 
in vernier 
readings, 
inches. 


Angle of twist 
on a length of 
51 inches, 
by verniers, 
radian. 


Torsion-meter 
readings. 


Angle of twist 
on a length of 
25.5 inches, 
by meter, 
radian. 


I 





0-0000 









2 


80,640 


0-I725 




19-85 




3 


108,864 


0-2300 




26-75 




4 


l8l,440 


0-3875 




44-60 




5 


254,016 


0-5425 




62-65 




6 


338,688 


0-7200 




82-75 




7 


43M24 


0-9250 




105.50 





Fill in the blank columns. Plot (a) torque and angle of twist by 
verniers, (b] torque and angle of twist by meter. Find and compare the 
torques required to produce o-ooi radian twist on a length of 51 inches, 
(c] by verniers, (d) by meter ; (e) find the modulus of rigidity from the 
vernier readings ; (/) find the shaft horse-power constant from the meter 
readings. On steam trials the mean torsion-meter reading was 98-5 and 
the revolutions per minute 665 ; (g) find the shaft horse-power. 

12. Estimate in ton-inches the maximum torsion of a shaft driven by 
an engine of 500 I.H.P. at a speed of 200 revolutions per minute, allowing 
an efficiency of 85 per cent, and a ratio of maximum to mean turning 
effort of 1-25. (I.C.E.) 

13. A destroyer has a solid circular propeller shaft, 9^ inches in 
diameter, which makes 400 revolutions per minute. A torsion-meter, 
fixed to the shaft, shows that the angle of twist over a length of 20 inches 
is 0-15. If the modulus of rigidity is 5000 tons per square inch, find the 
horse-power transmitted through this shaft. (B.E.) 

14. Explain how the work done by a varying force can be measured by 
means of an indicator diagram. The pressure on a piston P working in 
a cylinder AB of length 3 feet is proportional to its distance from A. If 
the pressure on the piston at B is 150 Ib. weight, draw a diagram showing 
the pressure in any position, and find the work done as the piston moves 
from B to A. (L.U.) 

15. Describe a differential pulley block. The diameters of the two 
grooves are 12 and 11-5 inches, what is the velocity ratio? Experiments 
are made on this pulley block when a load W is lifted by an effort E. 
When W was 600 Ib. E was 26 Ib., and when W was 300 Ib. E was 18 Ib. : 
what is E probably when W is 800 Ib.? What is the efficiency when W 
is8oolb.? (B.E.) 



352 MACHINES AND HYDRAULICS 

16. An electrometer lifts 80 tons of grain 100 feet high ; the electric 
energy costs 40 pence at the rate of 2 pence per unit. How much electric 
energy is used ? What is the efficiency of the lifting arrangements ? 

(B.E.) 



P ^1 
>tA 



.F 



CHAPTER XV. 

FRICTION 

Definitions. When two bodies are pressed together it will be 
found that there is a resistance offered to the sliding of one upon the 
other. This resistance is called the force of friction. The force which 
friction offers always acts contrary to the direction of motion of the 
body, or, if the body is at rest, the force tends to prevent motion. 

Let two bodies A and B (Fig. 377 (a)) be pressed together so that 
the mutual pressure perpendicular to the surfaces in contact is R. Let 
B be fixed, and let a force P, 
parallel to the surfaces in >[,R 

contact, be applied. If P is A 

not large enough to produce //)//////! ;// . 

sliding, or, if sliding with B jR ^/AI! R 

steady speed takes place, B '*' 

will apply to A a frictional 

force F equal and opposite to P (Fig. 37 7 ()). The force F may 
have any value lower than a certain maximum, which depends on the 
magnitude of R and on the nature and condition of the surfaces in 
contact. If P is less than the maximum value of F, sliding will 
not occur ; sliding will be on the point of occurring when P is 
equal to the maximum possible value of F. It is found that the 
frictional resistance offered after steady sliding conditions have been 
attained is less than that offered when the body is on the point 
of sliding. 

Let Y S = frictional resistance in Ib. when the body is on the 

point of sliding. 
Ffc = frictional resistance in Ib. when steady sliding has 

been attained. 
R = perpendicular pressure in Ib. between the surfaces 

in contact. 
D.M. z 



354 MACHINES AND HYDRAULICS 



Then 



are called respectively the static and kinetic coefficients of friction. 

Friction of dry surfaces. For dry clean surfaces, experiments 
show that the following laws are complied with approximately : 

The force of friction is practically proportional to the perpendicular 
pressure between the surfaces in contact, and is independent of the extent 
of such surfaces and of the speed of rubbing, if moderate. Another way 
of expressing the same laws is to say that for two given bodies, the 
kinetic coefficient of friction is practically constant for moderate pressures 
and speeds. It is very difficult to secure any consistent experimental 
results on the static coefficient of friction ; it is roughly constant for 
two given bodies. 

The value of the coefficient of friction in any given case depends 
on the nature of the materials, especially on the hardness and ability 
to take on a smooth regular surface, and on the state of the rubbing 
surfaces as regards cleanliness. Rubbing surfaces are made usually 
of fair shape and are well fitted to one another. If clean and dry, a 
film of air may be present between the surfaces and prevent actual 
contact. Pressure and working may squeeze this film out, and the 
bodies will then adhere strongly together, or seize. Seizing takes 
place more rapidly with bodies of the same than with those of 
different materials. 

Considerable increase in the speed of rubbing and also heating of 
the bodies tend to lower the value of the coefficient of friction. For 
this reason, the frictional force produced by the application of the 
brakes to the wheels of a locomotive running at high speed is higher 
during the first few seconds than is ultimately the case after the 
temperature has risen owing to the conversion of mechanical work 
into heat. The coefficient rises again when the speed becomes very 
slow, and may become sufficiently high to cause the wheels to skid 
just before stopping. The coefficient of friction for light pressures 
on large areas is a little greater than for heavy pressures on small 
areas. 

The value of the coefficient of friction to be expected in any given 
case cannot be predicted with accuracy on account of the erratic 
nature of the conditions. The following table gives average values 
only; experimental results will often show considerable variance 
with the tabular values. 



LAWS OF FRICTION 355 

COEFFICIENTS OF FRICTION. 

AVERAGE VALUES. 

Metal on metal, dry, 0-2 ; oiled continuously, 0-05. 

Metal on wood, dry, 0-6; greasy, 0-2. 

Wood on wood, dry, 0-2 to 0-5 ; greasy, o-i. 

Hemp ropes on metal, dry, 0-25; greasy, 0-15. 

Leather belts on iron pulleys, 0-3 to 0-5. 

Leather on wood, 0-3 to 0-5. 

Stone on stone, 07. 

Wood on stone, 0-6. 

Metal on stone, 0-5. 

Fluid friction. For liquids such as water and oils flowing in a 
pipe, the following laws are followed approximately : 

The Motional resistance is independent of the pressure to which the 
liquid is subjected, and is proportional to the extent of the surface wetted 
by the liquid. 

The resistance is very small at slow speeds ; below a certain critical speed 
the motion of the liquid is steady and the resistance is proportional to the 
speed; at speeds above this, the liquid breaks up into eddies, and the 
resistance is proportional approximately to the square of the speed. 

The critical speed depends on the nature of the liquid and on its tem- 
perature. Rise of temperature of the liquid diminishes the resistance. 
The resistance is independent of the material of which the pipe or channel 
is made, but the wetted surface should be smooth ; rough surfaces increase 
the resistance. 

Friction in machine bearings. The frictional laws for lubricated 
machine bearings are intermediate between those for liquids and for 
dry surfaces. The ideal bearing would have a film of oil of uniform 
thickness, and would run at constant temperature. There would be 
no metallic contact anywhere, and the resistance would be that of 
metal rubbing on oil. In such a bearing, the laws of liquid friction 
would be followed, and the resistance would be independent of the 
load and proportional to the speed of rubbing. In ordinary bearings 
the resistances experienced depend on the success which is achieved 
in getting the oil into the bearing and in preserving the oil film ; 
the working load is kept sufficiently low to avoid the danger of the 
film being squeezed out and seizing occurring. 

Friction of journals. The value of the coefficient of friction to 

be expected in any given case depends largely on the method of 

lubrication. In Beauchamp Tower's* experiments, one method 

of lubrication adopted was to have an oil bath under the journal 

* Proc. Inst. Mechanical Engineers, 1883 and 1884. 



356 



MACHINES AND HYDRAULICS 



(Fig. 378). Remarkably steady conditions were obtained, and it was 
found that the coefficient of friction could be expressed by 

c-Jv ( v 

^ = -7-' <0 

where c is a coefficient the value of which depends on the kind of 
lubricant used, v is the speed of rubbing in feet per second, / is the 
pressure per square inch of projected area of the journal. 





Let 



FIG. 378. Oil bath lubrication. FIG. 379. Projected area of a journal. 

P = the total load on the bearing, in Ib. 
d= diameter of bearing, in inches. 
L = length of bearing, in inches (Fig. 379). 
Projected area of bearing = 



Then 

p=z Ib. per sq. inch ...... (2) 

d Li 

The following table gives some of Tower's results : 

JOURNAL FRICTION, OIL-BATH LUBRICATION. 



Lubricant. 


A 

Ib. per sq. inch. 


feet per sec. 


f 


Cy 

mean value for range 
of loads and speeds given. 


Olive oil - 


/520 

lioo 


2-6l 

7-85 


0-0008) 
0-0089 / 


0-29 


Lard oil - 


{520 


2-61 


0-0009) 


0-28 5 




lioo 


7-85 


0-009 J 





Mineral grease 


[625 


2-61 

7-85 


0-00 1 ) 

0-008 3 J 


0-425 


Sperm oil 


(310 

lioo 


2-61 

7-85 


O-OOIl) 

0-0064 / 


O-2O5 


Rape oil - 


(415 

U53 


2-61 

7-85 


o-ooc>9\ 
0-004 / 


O-2I5 


Mineral oil 


/3io 
lioo 


2-61 
6-99 


0-0014) 
0-0073 f. 


O27 



FRICTION OF JOURNALS 357 

It will be noted in (i) above that the coefficient of friction is 
inversely proportional to /, and hence is independent of the total 
pressure P on the bearing with oil-bath lubrication. It also follows 
that the frictional resistance of the bearing will be constant for all 
working loads, and will vary as the square root of the speed. Thus, 
referring to Fig. 380 : 

Let F = the frictional resistance of the bearing, Ib. 

P = the load on the bearing, Ib. 

Then, from (i) and (2) above 

F F fjv 



............ ................................. (3) 

an expression which is independent of P. 

With less perfect systems of lubrication, there is a tendency for the 
oil film to be broken partially, and higher coefficients of friction are 
obtained. In some cases the coefficient may 
reach values from 0-03 to 0-08. 

Heating of journals. Work is done against 
the frictional resistance and is converted into 
heat. Referring to Fig. 380, in which D is the 
diameter of the journal in feet, we have 




Work done in one revolution = /xPTrD foot-lb. 

T T\XT r A. iu FIG. 380. Friction of a 

per minute = /xP?rDN foot-lb., J journal. 

where N is the number of revolutions per minute. 

Heat produced = ^ British thermal units per minute. 

This heat is dissipated by conduction and radiation, but the 
temperature of the bearing will rise during the early period of 
running. At higher temperatures the oil possesses lower viscosity, 
i.e. it flows more easily and offers less resistance to rubbing; hence 
less work will be done, and consequently less heat will be produced 
as the temperature rises. The tendency is therefore to attain a steady 
temperature, in which condition the heat developed will be exactly 
balanced by the heat carried away by conduction and radiation. It 
must be noted, however, that the lower viscosity possessed by the 
oil at the higher temperatures increases its liability to be squeezed 
out ; hence, if steady conditions are to be attained, the load must not 
be too great and the oil must be of suitable quality. 100 Fahrenheit 
may be regarded as a safe limit of temperature under full working 



358 



MACHINES AND HYDRAULICS 



load. Occasionally the bearings are made hollow, and a water 
circulation is provided in order to keep the temperature low. Bearing 
pressures up to 3000 Ib. per square inch are used, depending on the 
nature of the materials, method of lubrication, means of cooling, 
speed of rubbing, and on the consideration of whether the load 
always pushes on one side of the bearing or is alternately push and 
pull. Forced lubrication is often used, the oil being supplied under 
pressure to the bearings by means of a pump.* 

Friction of a flat pivot. The case of a flat pivot or foot-step 
bearing (Fig. 381) may be worked on the assumption that the 
coefficient of friction /* is constant for all parts of the rubbing surface; 
the resultant frictional force F will then be found from 



If it is assumed that the distribution of bearing pressure is uniform, 

we have p 

Load per unit 





FIG. 381. Flat pivot bearing. FIG. 382. 

Consider a narrow ring (Fig. 382) having a radius r and a breadth 6>. 
Area of the ring = 2irr . &>. 



Load on the ring = -^ 27rr - * 
Frictional force on the ring = -=sf? . r. o>. 



R 2 



2? 



Moment of this force = -^- ' 

To obtain the total moment, this expression should be integrated 
over the whole rubbing surface ; thus : 

2 p fR 2 p R3 

Total frictional moment ^^oT 1 r ^-^ r ~~w^ ' 

Jo o 

=SR.A*P (0 

= Fx|R (i') 

* An excellent discussion of the theory of lubrication and design of machine bear- 
ings will be found in Machine Design, Part I. , by Prof. W. C. Unwin. Longmans, 
1909. 



FRICTION OF PIVOTS 359 

It is seen thus that the resultant frictional resistance F may be 
taken to act at a radius equal to two-thirds the radius of the bearing. 

The frictional moment of a flat pivot may also be solved on the 
assumption that the wear is uniform and is proportional to the 
product pv, where / is the bearing pressure in Ib. per square inch 
and v is the speed of rubbing at any given part of the rubbing 
surfaces. Thus, p v = a constant. 

Also the velocity v at any point varies as the radius r. Hence, 
pr = & constant = a say ; 



Considering the narrow ring (Fig. 382) : 

Load on the ring =/. 2irr. Sr= 2ira . &r. 

Friction on the ring = 27r//^ .r. 

Integrating over the whole rubbing surface, we have 

f R 
Total frictional resistance = F = 2^^a . I dr 

Jo 

= 27T/AflR (2) 

Again, 

Moment of the friction on the ring = 2ir^ . r. 8r. 
The total moment will be obtained by integration of this expression 
over the whole rubbing surface ; thus : 

f R R 2 

Total moment of friction = 27r/xa .1 r.dr 2ir^a . 

Jo 2 

= TT/Ztf R 2 

= FxlR(from( 2 )) (3) 

= i/*PR (30 

It is probable that the actual value of the moment of friction will 
fall between the limits expressed in (i') and (3). 

In the case of a collar (Fig. 383), no great error will be made by 
assuming that F acts at the mean radius i(Rj + R 2 ). Hence, 

Moment of F = -|/*P(R 1 + R 2 ) Ib. inches (4) 

Tower's experiments on collar friction show that ft is independent 
both of speed and pressure unless the pressure I p 

is very small. The average value of /* found \ ^. 

was 0-036. The bearing pressure should not J [ L 

exceed 50 Ib. per square inch. i |*" R i 

Tower also experimented with a flat pivot _^ a 
bearing 3 inches in diameter. If Tower's 
results obtained lor the moments of friction 
be reduced from equation ( i ), thus : 



i 

> i 

= moment of F x 



> 



FIG. 383. Collar bearings. 



36o 



MACHINES AND HYDRAULICS 



values of /A are found which vary from about 0-015 at 5 revolutions 
per minute and 40 Ib. per square inch, bearing pressure to about 
0-006 at 350 revolutions per minute and 100 Ib. per square inch 
bearing pressure. 

Schiele pivot. In the Schiele pivot (Fig. 384), the bearing is 

curved so as to secure uni- 
form axial wear all over the 
surface. There is thus less 
likelihood of the oil film 
being squeezed out. 

Supposing the bearing to 
wear so that the point F 
descends ultimately to E, 
then EF is the axial wear, 
and is constant for any point 
on the bearing. EG is normal 
to the surface at E and FG 
is parallel to the tangent EH 
at E. The normal wear is 
EG, and may be assumed to 
be proportional to the speed 
of rubbing. The intensity of 
normal pressure at E is p ; it 
is assumed that p is constant for all points on the surface. The 
velocity of rubbing v feet per second at any point evidently will be 
proportional to the radius r. Hence, 

EG oc v oc r ; 




where k is a constant. The triangles EFG and HEK are similar. 
Hence, EF_EH_EH. 

EG~EK~ r ; 

EGxEH Jr.EH 

. . xL r = = 

r r 

= /.EH =a constant (i) 

Hence EH is constant ; a curve such as AB having this property 
is called a tractrix. 

Considering a narrow ring having a radius r and a horizontal 
breadth 8r (Fig. 384), we have 

Horizontal projected area of the ring = 2irr. Sr. 
Actual area of the ring = 2irr.&r. 



sin<9 



Also, 



sin 



EH 
EK 



EH . 
r 



ROLLING FRICTION 361 

Actual area of the ring = 21?. r. EH. 

Normal pressure on the ring=/x 27r.8r.EH (2) 

Friction on the ring = 27r.pn.8r. EH. 

Moment of this friction = 2-n-p^. EH . r. Br. 

Total moment of friction = 27r/^EH I r.dr 

JR S 
A .,-jR^-R, 2 

= 27T^/xEH 1 



-R 2 2). ...(3) 
Again, (2) gives 

Normal pressure on the ring = 2?r/. EH ,8r. 
The vertical component of this = 2irp. EH. Br. sin 



= 2Trp.r.r. ............... (4) 

The sum of the vertical components for all the rings composing 
the curved surface of the bearing will be equal to W. Hence, 

fRi 
r.dr 



-R 2 2) (5) 

Substitution of this in (3) gives 

Total moment of friction = /*W. EH (6) 

The Schiele pivot is not much used in practice on account of the 
difficulty of manufacture. 

Rolling friction. In rolling friction, such as that of a wheel or 
roller travelling on a flat surface, the frictional resistances are roughly 
proportional to the load and inversely proportional to the radius of 
the wheel or roller. The resistance also depends on the hardness of 
the materials, and is comparatively small for very hard surfaces. In 
ball bearings, both balls and ball races are made of hardened steel ; 
the races are best made concave, to a radius about 0-66 the diameter 
of the balls. This plan both reduces the resistance and enables a 
heavier load to be carried. In such bearings, the value of /* is 
practically constant through wide ranges of speeds and loads; 0-0015 
is an average value. 

Fig. 385 shows a heavy pattern of ball bearing made by The 
Hoffmann Co. and applied to a shackle B for holding one end of a 



362 



MACHINES AND HYDRAULICS 



test piece A undergoing both tension and torsion. The test piece is 
screwed to the shackle, the end of which is furnished with a nut C, 

which rests on the top ball race D. 
The bottom ball race E has its lower 
face made spherical to fit the corre- 
sponding spherical bottom of the cup 
F. This arrangement permits the test 
piece to accommodate itself to any 
want of alignment. A cage G made 
of two thin plates, secured together by 
means of four distance pieces, holds 
the balls in position and prevents them 
from coming into contact with one 
another ; the cage also prevents any of 
the balls being lost when the bearing 
is taken to pieces. A similar bearing 
is applied to the other end of the test 
piece. The moment of friction is very 
small ; with a tensile load of four tons 
and a test piece i inch in diameter, 
it is possible to rotate the whole by 
simply gripping the test piece with one 
hand. 

Resultant reaction between two 




Plan of Cage 

F,a.38s.-Ho ffm ann thrust ball bearing. 



g 



block A resting on a horizontal table BC. The weight W of the 
block will be constant, and will act in a line perpendicular to BC 





(b) 



FIG. 386. Reactions at the surfaces in contact. 



Let a horizontal force P 1 be applied to the block ; P l and W have a 
resultant R x . For equilibrium, the table must exert a resultant force 



ANGLES OF FRICTION 363 

on the block equal and opposite to Rj and in the same straight line ; 
let this force be E 15 cutting BC in I). E x may be resolved into two 
forces, one Q perpendicular to BC and the other F x along BC. Let 
</>! be the angle which E x makes with GD. Then 

F HG 



Now, when P x is zero, <j and hence tan <^ will also be zero, and 
Q will act in the same line as W. < x will increase as P l increases, 
and will reach a maximum value when the block is on the point of 
slipping. It is evident that Q will always be equal to W. Let <J> be 
the value of the angle when the block slips, and let F be the corre- 
sponding value of the frictional force. Then 

-p 
Coefficient of friction = /A = = tan <. 

There will be two values of tan </> corresponding to the static and 
kinetic coefficients of friction respectively. When the block is on the 
point of sliding, < is called the friction angle or the limiting angle of 
resistance ; when steady sliding is occurring, < is lower in value, and 
may be termed the angle of sliding friction. 

It is evident from Fig. 386 (a) that P 1 and Fj are always equal 
(assuming no sliding, or sliding with steady speed), so also are 
W and Q. These forces form couples having equal opposing 
moments, and so balance the block. The force 
Q acting at D will give rise to normal stress of 
a distribution as shown by the stress figure in 
Fig. 386 (b). The action is partially to relieve 
the pressure near the right-hand edge of the 
block and to increase it near the left-hand edge. 
With a sufficiently large value of /x, and by 
applying P at a large enough height above the 
table, the block can be made to overturn instead f'T 
of sliding. The condition of overturning may ' / 
also be stated by reference to Fig. 387. Here FIG. 3 8 7 . -Condition that 

. , a block may overturn. 

the resultant R of P and W may fall outside the 
base AB before sliding begins. Hence E, which must act on AB, 
cannot get into the same line as R, and the block will overturn. 
For overturning to be impossible, R must fall within AB. 

EXAMPLE. A wall of rectangular section 2 feet thick is subjected to a 
uniform normal pressure on one side of 50 Ib. per square foot (Fig. 388). 
Taking the weight of material as 150 Ib. per cubic foot and /A=O7, find 




MACHINES AND HYDRAULICS 



whether sliding at the base is possible. For what height of wall would 
overturning just occur ? 

Consider a portion of wall one foot in length, and let H feet be the 
height. Then 

W = 2H x i5o = 3OoH Ib. per foot run, 
F 



*- 

i 
i 



w 



= 2ioH Ib. per foot run. 

This result represents the maximum possible 
value of F. 
Again, P = 5oH Ib. per foot run. 

Hence, as P will always be much less than the 
maximum frictional resistance possible, the wall 

FIG. 3 88.-Stability of a w j u nQt ^^ 

When overturning is just possible, the resultant 

of P and W will act through O, and the moments of P and W about C 
will be equal. Taking moments about O, we have 



Moment of P 



5oH x = 



Ib.-feet. 



Moment of W=3ooH x i = 3ooH Ib.-feet. 
Equating these moments gives 



H = r2 feet. 

Friction on inclined planes. In Fig. 389 (a) is shown a block of 
weight W Ib. sliding steadily up a plane of inclination a to the 





FIG. 389. Friction on an incline ; P horizontal. 

horizontal, under the action of a horizontal force P Ib. Draw AN 
perpendicular to the plane ; then the angle between W and AN is 
equal to a. Draw AC making with AN an angle < equal to the 
angle of sliding friction ; the resultant reaction R of the plane will 
act in the line CA. The relation of P to W is deduced from the 
triangle of forces ABC. 



FRICTION ON INCLINED PLANES 



or 



P _ tan a + tan < 
W ~ i - tan a tan <k 



P = 



(P. 8), 



i - /A tan a/ 

The case of the block sliding down is shown in Fig. 389 (). 
R acts at an angle <f> to AN, but on the other side of it. 

P"D/~' 
-DV^ / , x 



...(I) 

Here 



P _ tan <j> - tan a 
W ~ i+ tan a tan </>' 
p_ w />-tanaY 



+ /Uana/ 

It will be noted in the last case, that if <f> is less than a, the block 
will slide down without the necessity for the application of a force P. 
Rest is just possible, unaided, if a is equal to the limiting angle 
of resistance. 

When P is applied parallel to the incline, the forces are as shown 
in Fig. 390 (a) and (<). For sliding up (Fig. 390 (a)), we have 
P EC = sin BAG 
W~AB~sin ACB 

sin (a -f ft) _ sin a cos <ft 4- cos a sin (ft 
~ sin (90 - <f>) ~ cos <j> 

= sin a + cos a tan < ; 




FIG. 390. Friction on an incline ; P parallel to the incline. 

For sliding down (Fig. 390 ()), we have 
P _ EC _ sin BAG 
W~AB~sinACB 

sin (< - a) sin cos a - cos < sin a 



sin (90 - <f>) 
= tan (f) cos a - sin a ; 
P = W (/* cos a - sin a). 



cos 



(4) 



366 



MACHINES AND HYDRAULICS 



Friction of a screw. The results for a block on an inclined plane 
and acted on by a horizontal force may be applied to a square threaded 
screw. Such a screw may be regarded as an inclined plane wrapped 
round a cylinder. In Fig. 391 are shown two successive positions 
A and B of a block of weight W Ib. being pushed up such an inclined 




FIG. 391. Inclined plane wrapped 
round a cylinder. 




1 

FIG. 392. Friction of a square threaded 
screw. 



plane by means of a horizontal force P Ib. In the actual mechanism, 
the load is applied over a considerable portion of the surface of the 
incline (Fig. 392), and P may be assumed to act at the mean radius 
R inches of the screw. Let p inches be the pitch of the screw, and 
let one turn of the thread be developed as shown in Fig. 393. 

^_ p 

~27rR 
Using equation (i), p. 365, we have, for raising W, 

P-wf tanct +/*\ 

\i -/A tan a/ 

'T / / 

P ___/7ru+/^ 




= w 27rR 



I - 



FIG. 393. Development of one turn of a 
screw thread. 



27TR, 



For lowering the load, P in Fig. 393 will be reversed in direction 
usually, and equation (2), p. 365, should be used : 



i + /x tan 



\ 

a/ 



FRICTION OF SCREWS 367 

If p = tan a, or if 27rR/z =p t P will be zero, and the load will be 
on the point of running down unaided. Running down with 
continually increasing speed may occur if tan a is less than /*, and 
may be prevented by application of a force P given by (2) above 
and applied in the same sense as for raising the load. If rotation 
of the screw is produced by means of a force Q Ib. applied to a 
spanner at a radius L inches, the nut being fixed, we have 

= PR, 

p (3) 



The above solution is applicable to the case of a screw-jack 
(Fig. 411), the friction of the screw alone enters into the problem. 

The efficiency of such an arrangement may be calculated by con, 
sidering the screw to make one revolution in raising the load. Then 

^ r/ . . Work done on W 

Efficiency = 

Also, 

.'. Efficiency = 

Substituting for W/P from equation (i) above, we have 

,- . / 2?rR pli<\ p / x 

Efficiency = ( ~- ) -*-=; (4) 

V/+27TR/V 27TR 

If n be the ratio of the mean circumference 2?rR to the pitch /, 
so that 2?rR == np) equation (4) may be written : 

Efficiency = 




p + npp. n 

l 



EXAMPLE. In a certain square threaded screw, n=io and ^=0125. 
Find the efficiency while raising a load. 

10-0-125 



=0-44 

= 44 per cent. 

In tightening a nut on a bolt (Fig. 394), not only has the moment 
of the friction of the screw to be considered, but also the friction 
between the nut and the part against which it is being rotated. 



368 



MACHINES AND HYDRAULICS 



Let R x be the mean radius of the bearing surface in inches and W Ib. 
be the pull on the bolt. Then 



Moment of F = /xWRj Ib.-inches 



(6) 





FIG. 394. Friction of a nut. 



FIG. 395. Friction of a V thread. 



Let P Ib. be the frictional resistance of the screw, found from 
equation (i), p. 366, and let R be the mean radius of the threads. 
Then QL = /xWR 1 + PR (7) 

In V threaded screws (Fig. 395), the pressure between the bearing 
surfaces of the nut and the bolt threads is increased. If W is the 
load, it should be resolved into a force S perpendicular to the thread 
section and another horizontal force. Then 
W 



s- w 

, o 7) 

COS/5 

where /? is half the angle of the V. All the results found for square 
threaded screws may be used for V threaded screws by writing 
fi sec /3 instead of /x. in the equations. 

Friction circle for a journal. It is useful to consider the friction 
of a journal A resting on a loosely fitting bearing B (Fig. 396(0)). 

I I 




FIG. 396. Friction of a loose bearing. 

If there is no rotation, the load W on the journal will be balanced 
by an equal opposite reaction Q applied by the bearing. 



Let a 



FRICTION CIRCLE 



369 



couple of moment T be applied to the journal, of sufficient magni- 
tude to produce steady rotation in the direction shown (Fig. 396 (b) ). 
The journal will roll up the bearing until the place of contact is , 
at which steady slipping will occur. The condition which fixes the 
position of b is that the vertical force Q acting at b must make an 
angle <f> with the normal ab, <f> being the angle of sliding friction. Q 
and W being still equal, form a couple of moment W x fa, and this 
couple balances T, the couple applied. Hence, 



be , be . 

= tan <P = , very nearly ; 
ac ab J J ' 



Also, 



(I) 

This will be in Ib.-feet if r is in feet and W is in Ib. 
It will be noted that Q is tangential to a small circle of radius f, 
drawn with centre a. This circle is called the friction circle, and its 
radius is equal to be. Hence, 

Radius of friction circle =/= r tan < very nearly 

= nr feet, (2) 

where ft = tan <, is the coefficient of friction, 

r= radius of journal in feet. 

The same result is true for a closely fitting bearing (Fig. 397). 
Here R Ib. is the resultant reaction of the bearing, the components 
of which are Q, the resultant vertical reaction and F the resultant 
frictional force. R acts at an angle </> to Q for the direction of 
rotation as shown, or on the other side of Q for i 

the contrary direction of rotation In either 
case, R is tangential to the friction circle, and 
gives a moment R/ Ib.-feet opposing rotation. 
To obtain the work done, we have 

Frictional couple = R/ Ib.-feet, 
Work done in one revolution = R/x 2?r foot-lb. 
Let N = revolutions per minute. 
Then Work done per minute = 27rNR/foot-lb., 

, 27TNR/ 

H.P. wasted = - 

33000 

The value of R is given actually by FIG. 397 .-Friction circle 

a bearing. 

but as the coefficient of friction and hence the frictional resistance is 
very small for well lubricated journals, no great error is made by 
taking R equal to Q, the load on the journal. 




D.M. 



2 A 



MACHINES AND HYDRAULICS 



EXAMPLE i. In a machine for raising a load W the load is suspended 
from a rope wound round a drum A, 8 inches in diameter, to the rope 
centre (Fig. 398). The axle on which the drum is fixed has journals 
1-5 inches in diameter, and is rotated by a toothed 
wheel B, 18 inches in diameter, to which a force 
P is applied. Find the mechanical advantage 
and efficiency of the machine, taking the co- 
efficient of friction of the bearings to be o-i and 
W to be a load of 500 Ib. 

Neglecting friction, and taking moments about 
the centre of the drum, we have 
W x 4 = P x 9, 

W 9 

-p = 7 = 2 ' 2 5 (0 




FIG. 398. Friction of a simple 
machine. 



(2) 



Taking account of friction, and assuming that R is equal sensibly to 
(P-f-W), we have, by taking moments about the centre of the drum, 



Also, 



inch. 



.'. 2000 + (P+ 500) 



Hence, 



p ^ 2037- 5x40 

357 
= 228-3 Ib. - 

W 500 

Mechanical advantage =-5- = -^-Q 
r 220-3 



(3) 



(4) 



Let the drum make one revolution. Then 

Work done by P = Px;r. 18 inch-lb. 
Work done on W = W x TT . 8 inch-lb. 

, . 87rW 8x500 

Efficiency-^- i 



=0-97 

= 97 per cent ............................ (5) 

EXAMPLE 2. The mechanism shown in Fig. 399 consists of a crank OA 
fixed to a shaft having OZ for its axis of rotation. The crank is driven in 
the direction of rotation shown, by means of a slotted bar B ; a block C 
may slide in the slot, and has a hole to receive the crank pin. The force 
P pushes during the stroke from right to left, and pulls during the return 
stroke. Show by drawing how the turning moment on the crank, as 
modified by friction, may be obtained. Give the construction for each 
quadrant, assuming /x tan <> is the same for both block and pin. 



FRICTION IN A SLOTTED-BAR MECHANISM 



371 



In answering this question it is essential to remember that the force 
which the block gives to the crank pin must be tangential to the friction 
circle, and must act so as to oppose the motion of rotation of the pin. 




PLAN 
FIG. 399. Crank and slotted-bar mechanism. 

Further, the force which the slotted bar gives to the block must act at an 
angle <f> to the normal, and must be applied so as to oppose the sliding 
motion of the block. For ordinary values of the coefficient of friction 
these forces, shown by R in Fig. 400, will be very nearly equal to P. 



M _.- 






FIG. 400. Friction of the block and crank pin in Fig. 399. 



The constructions required are shown in Fig. 400 (a) to (d}. In the first 
and fourth quadrants (a) and (d} the block is sliding upwards, and in the 



372 MACHINES AND HYDRAULICS 

second and third quadrants (b) and (c) it is sliding downwards. In each 
case the turning moment is RxOM, OM being drawn perpendicular to 
R from O, the centre of the crank shaft. 

Friction of the crank pin and the crosshead pin. Figs. 401 (a) 
and (b) show the application of the friction circle method to the 
determination of the line of thrust along a connecting rod, when 
account is taken of the friction at the crank pin and the crosshead 
pin. The diameters of the friction circles at A and B are calculated 
and the circles drawn. The line of thrust Q will be a common 
tangent to these circles for any given crank position. Draw OM 
perpendicular to the line of Q; the turning moment on the crank will 
be Q x OM. No difficulty will be experienced in choosing the line 
of Q if it is remembered that the frictional moments at A and B both 
tend to reduce the turning moment on the crank; hence the common 
tangent which gives the line of Q must be so drawn as to make OM 



S'3- 




FIG. 401. Friction of the crank pin and crosshead pin in a crank and connecting rod 
mechanism. 

a minimum. Thus, in Fig. 401 (a), Q touches the top of the circle at 
A and the bottom of that at B; in Fig. 401 (b\ Q touches the bottom 
of both circles. The change in the line of Q from the top to the 
bottom of the circle at A takes place when the crank makes 90 with 
the centre line OA ; in this position, the connecting rod makes its 
maximum angle with the centre line OA and has no angular motion 
for an instant, i.e. at this point the crosshead pin is not rubbing in its 
bearing. The solution for other positions is given at Q' in Figs. 401 
(a) and (b). 

In Fig. 401 (b), it will be noted that, as B' approaches the inner 
dead point, the line of Q' will pass through O. In this position there 
is no turning moment oh the crank. Further, the crank must rotate 
through a small angle beyond the dead point before the line of Q will 
pass above O. There will, therefore, be a small crank angle near 
each dead point in which there will be no turning moment tending to 
rotate the crank in the direction of rotation of the crank shaft. These 
angles may be determined approximately as follows: In Fig. 402, O is 
the crank shaft centre and A is the crosshead pin at the end of the 



FRICTION IN A CRANK AND CONNECTING ROD 373 

stroke. Draw the friction circle at A ; draw the lines of Q and Q' 
touching the circle at A and passing through O. Draw the friction 




FIG. 402. Angle of zero turning moment due to friction at the crank and crosshead pins ; 
inner dead point. 

circles at B and B', touching the lines of Q and Q' ; then BOB' is the 
angle within which there is zero turning moment near the inner dead 




B 



FIG. 403. Angle of zero turning moment due to friction at the crank and crosshead pins ; 
outer dead point. 

point. The construction for the outer dead point is given in Fig. 
403, and will be followed readily. 

Friction of the crank-shaft bearings. The loads producing 
frictional resistances at the crank-shaft bearings include the weight 
of the shaft and its attachments, belt 
pulls or other forces due to the driving 
of machinery and a reaction owing 
to the thrust of the connecting rod. 
Considering the latter alone, and 
referring to Fig. 404, Q is the thrust 
of the connecting rod, making allow- 
ance for the friction of the crosshead 
pin and the crank pin as illustrated 
in Fig. 401 (a). A force Q', equal, 
opposite and parallel to Q is applied by the crank-shaft bearing to 
the shaft, Q and Q' together forming a couple which causes the shaft 




FIG. 404. Friction of the crank-shaft 
bearings. 



374 



MACHINES AND HYDRAULICS 



to rotate. During rotation, Q' will be tangential to the friction circle 
for the shaft and is so shown in Fig. 404. Draw OM perpendicular 
to Q and cutting the shaft friction circle at O'. The effective turning 
moment will be Q x O'M, and has been diminished by an amount 
Q x OO' by reason of the friction produced by Q in the shaft bearings. 
Near the dead points, the effect of the friction at the crosshead pin, 
crank pin and crank-shaft bearings is that there will be a small angle 
embracing each dead point within which no force, however great, 
along the piston rod will cause the shaft to rotate if at rest. These 




FIG. 405. Dead angle at inner dead point. 

angles are called dead angles and are shown at BOB' in Figs. 405 and 
406. The construction is similar to that in Figs. 401 (a) and (b\ 
with the addition of the friction circle at O for the crank-shaft 
bearings. The lines of the forces Q and Q' are tangential to the 
friction circles at A and O, and the friction circles at B and B' are 
drawn to touch the lines of the forces, produced if necessary. 




FIG. 406. Dead angle at outer dead point. 



The student will note that the dead angles so found take account 
only of the friction at the crank-shaft bearings produced by the 
thrust of the connecting rod. If at rest, the crank shaft will not 
commence rotation until the turning moment Q x O'M (Fig. 404) is 
large enough to overcome the resisting moment due to the total 
.friction at the crank-shaft bearings together with the resistances 
offered by any machinery to be driven. 

Experiments on friction. Experiments have been described in 
Chapter XIV., in which the general effect of friction in the complete 
machine was one of the factors to be determined. The following 
additional experiments may be performed usefully. 



EXPERIMENTS ON FRICTION 



375 



EXPT. 38. Friction of a slider. AB is a wooden board or flat piece 
of metal having its top surface brought horizontal by means of a spirit 
level (Fig. 407). A slider C, of wood or metal, may be drawn along 
AB by means of a horizontal force P applied by using a cord, pulley 
and scale pan. The upper 
surface of AB and the under 
surface of C should be clean 
and dry. Weigh the slider C 
and also the scale pan. R is 
the perpendicular reaction of 
the surfaces in contact, and 
is equal to the weight of the 




P 



1 


1 F ^- 


A 


IR a* 



FIG. 407. Friction of a slider. 



slider together with the load 

placed on it. Add loads to 

the scale pan, tapping AB gently after each load is applied, until the 

slider is drawn steadily along AB. P will be nearly equal to the 

weight of the scale pan together with the loads placed in it, and 

the kinetic friction F will have the same value. Calculate the kinetic 

coefficient of friction from 

F 



The experiment should be repeated with several different loads on 
the slider, and F and R should be tabulated for each. Plot F and R ; 
if this gives a straight line, find the average value of ^ from the graph. 

Repeat the experiment, using different materials for the board and 
for the slider. It is useful to have a set of sliders, all of the same 
material, but having the under sides cut away so as to give different 
areas of contact. 

EXPT. 39. Determination of the angle of sliding friction. In Fig. 408 

AB is a board which may be 
set at different angles to the 
horizontal. A block C is 
placed on it, and the angle 
is varied until the block will 
slide steadily down after being 
assisted to start. Measure 
the angle BAD which AB 
makes with the horizontal ; 
this will give the value of the 
,,,,,,,,,,, ,,',,,,,.,,,,,,,,,,,,,,, f ^n an gl e of sliding friction. Cal- 

FIG. 408. Apparatus for determining the angle of CU ^ ate A* from * 

sliding friction. ^ = fan BAD. 

Repeat the experiment using different materials. 

EXPT. 40. Rolling friction. In Fig. 409 is shown apparatus similar 
to that of Fig. 407, but having a small carriage mounted on wheels 




376 



MACHINES AND HYDRAULICS 



having bearings constructed to reduce friction as much as possible. 
The board should be levelled carefully, and the tractive effort P 
required to draw the carriage steadily along should be found for 
different loads on the carriage. It is useful to have three or four 
different roads for the carriage to run on; these may be of plate glass, 

SH 



FIG. 409. Apparatus for rolling friction. 

metal, wood and rubber. The effect of the varying degrees of hard- 
ness should be contrasted by comparing the results for the different 
roads, and this may be done easily by plotting tractive effort and load 
for each road on the same sheet of squared paper. 

EXPT. 41. Effect of speed of rubbing. In Fig. 410, A is a wheel 
which may be rotated at different speeds by some source of power. 
B is a block which is pressed on the rim of the wheel by means of a 
shackle C and a load D. The block B is restrained from rotation 





FIG. 410. Apparatus for investigating the 
effect of speed of rubbing. 



FIG. 411. Experimental screw-jack. 



by a cord and another load at E. The perpendicular pressure 
between the block and the wheel will be the weight of the block, 
together with those of the shackle, scale pan and load. The force 
of friction will be nearly equal to the combined weights of the scale 
pan and load at E. Hence /* may be determined for different speeds 
of rubbing. It will be observed that the friction is greater on 
starting with both wheel and block cold, and diminishes after a few 
seconds as the rubbing parts become warm. The experiment should 
be repeated with blocks of different materials. 



TESTING OF LUBRICANTS 



377 



EXPT. 42. Friction of a screw. The screw-jack shown in Fig. 411 
may be experimented on in the same manner as that explained in 
Chapter XIV. for other types of lifting machines. 

Testing of lubricants. The mechanical testing of lubricants is 
performed usually by feeding the lubricant into a test bearing, which 
may be loaded and run at varying speeds. Provision is made for 
measuring the torque required to rotate the shaft and also for 
measuring the temperature of the oil. There are many different 
forms of machine. One which has given useful information in the 
hands of Messrs. W. W. F. Pullen and W. T. Finlay at the South- 
western Polytechnic* is shown in Fig. 412. A shaft AB is loaded 




FIG. 412. Pullen's machine for testing lubricants. 

with two equal flywheels C and D ; the central enlarged portion of 
the shaft runs in a bearing and is lubricated by means of a loose 
ring G, which hangs freely on the shaft and dips into an oil bath ; 
the ring revolves slowly as the shaft rotates. The oil leaving the 
bearing is spun off by collars F, F fixed to the shaft and having several 
sharp edges to prevent the oil travelling axially along the shaft ; the 
oil is thus returned to the oil bath and is used again. K is a gauge 
tube indicating the quantity of oil in the bath. The temperature of 
the oil is controlled by a U tube H, through which water may be 
circulated. A gas flame in the space J under the bath can be used 
to raise the temperature of the oil. The temperature is measured 
by a thermometer suspended in the oil. The machine is direct 
driven by an electromotor arranged as shown in Fig. 413. The 
motor A has its bearings supported by rollers B, C and D, and is 

* Proc. Inst. Meek. Eng., 1909. 




37 8 MACHINES AND HYDRAULICS 

therefore free to rock about its axis. A balance weight is fitted at E, 
and a counterpoise F serves to measure the torque. The shaft 
runs in the direction of the arrow, and the rotor of the machine 

applies a torque of opposite 
sense to the stator; this torque 
is balanced by the counter- 
poise, and is equal to the 
torque required to drive the 
oil-testing machine. This 
type of machine is very useful 
for testing oils under steady 
load and under different con- 
ditions as regards speed and 

FIG. 413. Arrangement of electromotor for diiving 

Pullen's machine. temperature. 



EXERCISES ON CHAPTER XV. 

1. A shaft journal is 4 inches in diameter and has a load of 4000 Ib. 
If the coefficient of friction is 0-06, find the torque resisting the motion. 
Calculate also the energy absorbed in foot-lb. per minute in overcoming 
friction ; to what heat in B.T.U. is this energy equivalent ? The shaft 
revolves 150 times per minute. 

2. A vertical shaft is supported on a flat pivot bearing 2 inches in 
diameter and carries a load of 150 Ib. The shaft revolves 300 times 
per minute. Take ^=0-03, and calculate the moment of the frictional 
resistance, (a) assuming that the distribution of bearing pressure is uniform, 
(b] assuming that the wear is uniform. In each case calculate the horse- 
power absorbed by the pivot. 

3. The thrust of a propeller shaft is taken by 6 collars, 12 inches 
diameter, the rubbing surface inner diameter being 8 inches. The shaft 
runs at 120 revolutions per minute. Take /x = o-o5, the bearing pressure 
60 Ib. per square inch of rubbing surface, and find the horse-power 
absorbed by the bearing. 

4. A block weighing W Ib. is dragged along a level table by a force 
P Ib. acting at an angle 6 to the horizontal. The coefficient of friction 
may be taken of constant value 0-25. Obtain the values in terms of W, 
(a) of P, (b) of the work done in dragging the block a distance of one foot. 
Give the results when 6 is o, 15, 30, 45, 60, and 75 degrees. Plot graphs 
showing the relation of P and 0, and also the relation of the work done 
by P and 6. 

5. A block weighing W Ib. is pushed up an incline, making an angle 
with the horizontal. The coefficient of friction has a constant value of 
0-25. Find in terms of W (a) the values of the force P Ib., parallel to the 
incline, (&) the work done in raising the block through a vertical height of 
one foot. Give the results for 6 equal to o, 15, 30, 45, 60, 75 and 90 
degrees. Plot graphs for each case, (a) and (^ 



EXERCISES ON CHAPTER XV. 379 



6. Answer Question 5 if P is horizontal. What is the value of 6 when 
P becomes infinitely great ? 

7. In a screw-jack the pitch of the square threaded screw is 0-5 inch 
and the mean diameter is 2 inches. The force exerted on the bar used in 
turning the screw is applied at a radius of 21 inches. Find this force if 
a load of 3 tons is being raised. Take ^=0-2. What is the efficiency of 
this machine ? 

8. With the screw-jack given in Question 7, find the force required at 
the end of the bar in order to lower the load of 3 tons. 

9. Show that the horizontal force required to move a weight W up a 
plane whose slope is i is W % , where fi is the coefficient of friction. 

A right- and left-hand square-threaded screw (pitch 0-25 inch, mean 
diameter of thread I inch) is used as a strainer. Find the couple required 
to tighten against a pull of looo Ib. ^=0-15. (I.C.E.) 

10. In a i -inch Whit worth bolt and nut take the dimensions as 
follows : pitch, 0-125 inch ; angle of the V thread, 60 degrees ; mean 
diameter of the thread, 0-8 inch ; mean radius of the bearing surface of 
the nut, 0-9 inch. Take the coefficient of friction to be 0-2 for both the 
screw and the nut. Find the force required at the end of a spanner 
15 inches long in order to obtain a pull of 1000 Ib. on the bolt. 

11. A horizontal lever, instead of having a knife edge as a fulcrum, is 
pivoted on a pin 2 inches in diameter. The arms of the lever are 8 inches 
and 5 feet respectively. The coefficient of friction for the pin is 0-2. 
What load at the end of the short arm can be raised by a vertical pull of 
loo Ib. at the end of the long arm ? (B.E.) 

12. The arms of a bent lever ACS are at right angles to one another ; 
AC is 12 inches long and is horizontal ; BC is 27 inches long, and B is 
vertically above C. The lever may turn on a fixed shaft 3 inches in 
diameter at C. A load of 2000 Ib. is hung from A. Find what horizontal 
force is required at B (a) if A is ascending, (b) if A is descending. Take 
the coefficient of friction for the shaft to be o-i. 

13. In the mechanism shown in Fig. 399, the crank OA is 6 inches 
long and has anti-clockwise rotation ; the crank pin at A is 2 inches in 
diameter and the width of the slot in the bar is 2-75 inches. Take the 
force P as constant and equal to 1000 Ib. ; find the turning moment on 
the crank in each of the four positions when the crank makes 45 degrees 
with the line of P, (a) neglecting friction, (b) taking account of friction. 
The coefficient of friction for all rubbing surfaces may be taken as o-i. 

14. In the crank and connecting-rod mechanism of an ordinary steam- 
engine, the crank and connecting-rod are 7 inches and 30 inches long 
respectively. The diameter of the crank pin is 3-5 inches and that of 
the crosshead pin is 3 inches. When the crank has travelled 45 degrees 
from the inner dead point the total force urging the crosshead is 
3000 Ib. Find the turning moment on the crank for this position, taking 
/z for the crank pin and crosshead pin to be 0-06. Find both angles in 
which there is zero turning moment on the crank. 

15. Determine an expression for the work absorbed per minute in 
overcoming the friction of a collar bearing. State the assumptions made 



380 MACHINES AND HYDRAULICS 

in deriving the formula. The thrust in a shaft is taken by 8 collars 
26 inches external diameter, the diameter of the shaft between the collars 
being 17 inches. The thrust pressure is 60 lb.- per square inch, the 
coefficient of friction is 0-04, and the speed of the shaft is 90 revolutions 
per minute. Find the horse-power absorbed by the friction of the thrust 
bearing. (L.U.) 



CHAPTER XVI. 

VELOCITY. ACCELERATION. 

Velocity. The Velocity of a body may be defined as the rate at 
which the body is changing its position. The four elements which 
enter into a body's velocity are : (a) the distance travelled, (b, the 
time taken to travel this distance, (c) the direction in which the body 
is moving, (d) the sense along the line of direction ; the sense may be 
described as positive or negative. A body having uniform velocity will 
travel equal distances in equal intervals of time, and the velocity 
may be calculated by dividing the distance by the time. In the 
case of varying velocity, the result of this calculation will be the 
average velocity of the body. 

The units of time employed are the mean solar second, minute, or 
hour. The unit of distance may be the foot, mile, centimetre, metre 
or kilometre. Common units of velocity are the foot per second, the 
mile per hour, the centimetre per second, and the kilometre per hour. 

Let s = distance travelled in feet, 

t = time taken, in seconds, 
v = the velocity. 

Then v = - feet per second. 

This will be the velocity at any instant if the rate of travelling is 
uniform, and will give the average velocity if the rate is varying. 

Distance-time diagrams. In Fig. 414, the distances travelled by 
a given body have been set off as ordinates on a time base. Thus 
i A is the distance travelled during the first second of the motion, 
26 is the distance travelled in the first two seconds, and so on. 6F 
is the total distance travelled in 6 seconds. Drawing AG, BH, CK, 
etc., horizontally, it is evident that BG is the distance travelled 
between the end of the first second and the beginning of the third, 
CH is the distance travelled during the third second, DK, EL and 



MACHINES AND HYDRAULICS 



FM are the distances travelled during the fourth, fifth and sixth 
seconds respectively. If all these distances were equal, the velocity 
would be uniform and the line OF would be straight (Fig. 415). A 




FIG. 414. Distance-time diagram. 



4 5 ^ 

Seconds 

FIG. 415. Distance-time diagram, 
velocity uniform. 



straight-line distance-time diagram therefore represents the case of 
uniform velocity. 

Referring again to Fig. 414, the average velocity during the six 
seconds would be obtained by dividing 6F in feet by 6 seconds. 
The average velocity during any second such as the fourth may be 
calculated by dividing DK in feet by i second. 

In Fig. 416, let AB = 5 > 1 and CD = J 2 be the distances in feet 
travelled during the times t^ and / 2 seconds respectively. Drawing 

AE parallel to OD, the distance travelled 

during the interval / 2 ~^i will be 

CE = s 2 - s l . Hence, 

Average velocity during the interval BD 



Feet 




Sa 



Seconds 



The actual velocity at any instant of 
the interval may differ somewhat from 
this. If the interval be made very small 

we may write the difference in the distances by the symbol 8s and 

the difference in the time by 8t. 

Average velocity during a small interval = ^-. 

Now let S/, represented by BD or AE in Fig. 416, be reduced 
indefinitely until finally it gives us the conception of an " instant." If 
dt is its value when so reduced, and if ds is the distance travelled, 
then, at the instant considered, 

ds v 



VELOCITY 



383 



The velocity of a body at any instant may be described as the 
distance which would be travelled during the next second had the 
velocity possessed at the instant considered remained uniform. 

The mathematical calculation involved in (i) above is performed 
by use of the rules of the differential calculus (p. 9). 

EXAMPLE. Suppose the equation connecting s and / for the motion of 
a given body to be _ ! , 2 

where a is a constant. Find the velocity at any instant. 

ds d 



= at. 

If the time up to the required instant be inseited in this result, the 
velocity at that instant will be obtained. 

In dealing with a moving point in a machine the space-time 
diagram may be drawn by setting out the mechanism in a number 
of positions differing by equal intervals of time, and then measuring 
the distances travelled by the point in question. The average 
velocity over each of the intervals may be obtained very closely 
from the diagram. 

EXAMPLE i. A rigid bar AB, 3-1 feet in length, moves so that one 
end A is always in OX (Fig. 417), and the other end B is always in OY, 
which is perpendicular to OX. A is 
at first 3 feet from O and travels to 
O in 6 seconds with uniform velocity. 
Draw the space-time diagram for B. 

Divide AO into six equal intervals 
as shown. A will traverse each 
interval in one second. 

Velocity of A (uniform) 

= 1=0-5 foot per sec. 

Find the positions of B for each 
position of A ; these are numbered 
i', 2', 3', etc. to correspond with the 
numbering of the positions of A. 
Measure Bi', 62', 63', etc.; these X 
will be the distances travelled by B 
in I, 2 and 3 seconds respectively. 




FIG. 417. A rigid bar AB ; A moves in OX ; 
B moves in OY. 



Choose suitable scales and draw the space-time diagram (Fig. 418), by 
setting off the distances travelled by B up to the stated times. The 
numbering i', 2', 3', etc. in this diagram agrees with that in Fig. 417. 



MACHINES AND HYDRAULICS 



EXAMPLE 2. Find from Fig. 418 the average velocity of B for each 
interval of time and draw a velocity-time diagram. 

The average velocity during the third second may be obtained by 
dividing H3' in feet by 0-5 second. It is preferable to measure 33' and 

Feet 



20 






^c__ 


K 


! 






9*>^ 






I 


i-5- 




./T 


H 




I 
j 


1-0- 


/ 








i 

i 




/ 


\ 






1 


0-5 


/ 


\ 






I 




/ 


\ \ 






i 



O I 2 3 4 

FIG. 418. Space-time diagram for the point B in Fig. 417. 



5 6 
Seconds 



22' and take the difference as the value of H3'. The average velocity so 
calculated may be taken to be the actual velocity at the middle of the 
interval, and is set off as BC in Fig. 419, which is the velocity-time 
diagram. It is best to set out the quantities in a table thus : 



Interval No. 


Ordinate. 


Distance 
in feet. 


Difference in 
distance, feet. 


Average velocity 
= difference-:- 1 sec. 
Feet per sec. 





















I -06 


I -06 


I 


II* 


I -06 












0-53 


o-53 


2 


22' 


i*59 












o-35 


0-35 


3 


33' 


1-94 












O-22 


0-22 


4 


44' 


2-16 












Oil 


O-II 


5 


55' 


2-27 












0-03 


0-03 


6 


66' 


2-30 







The last column is plotted at the middle of the intervals in Fig. 419 ; a 
fair curve through the plotted points gives the required velocity-time 
diagram. 

A useful property of the velocity-time diagram is that its area 
represents the distance travelled. The distance is equal to the 
average velocity multiplied by the time, and the average velocity 
evidently will be given to scale by the average height of the diagram. 



ACCELERATION 



385 




Seconds 



FIG. 419. Velocity- time diagram for the point B 
in Fig. 417. 



while its base represents the time to scale. The area of the diagram 
is its average height multiplied by its base and therefore represents 
the distance travelled. To 
obtain the scale: Feet per sec. 

Let 
i inch of height represent v 

feet per second, 
i inch of length represent / 

seconds. 
Then 

i square inch of area repre- 
sents vt feet. 

Hence, the area of the velo- 
city-time diagram, in square 
inches, multiplied by vt will give the distance travelled. 

Acceleration. Acceleration means rate of change of velocity ; it is 
measured by dividing the change in velocity by the time in which 
the change is effected. The change in velocity may be either positive 
or negative, depending on whether the velocity is increasing or 

diminishing, and the accelera- 
tion will have the same sign. 
If the change in velocity is 
stated in feet per second, and 
if the time in which the change 
takes place is stated in seconds, 
__ then the units of the accelera- 
tion will be feet per second per 
second. 

Acceleration may be studied 
from the velocity-time diagram. 
Fig. 420 shows such a diagram 
in which a velocity z/ x occurs at the end of a time / x and a velocity 
v 2 at the end of / 2 ; these velocities are represented by AB and 
CD respectively. The change in velocity an increase in this 
case is DE. 

Change in velocity = v* - v l . 

Time in which this change is effected = / 2 - 
.'. Acceleration during the interval AC 

This expression will be strictly correct if the gain in velocity is 
D.M. 2 B 



Velocity 




FIG. 420. Deduction of acceleration from a 
velocity-time diagram. 




386 



MACHINES AND HYDRAULICS 



acquired uniformly throughout the interval, in which case BD would 
be straight. If BD is curved, then the value given by (i) will be the 
average acceleration over the interval. The acceleration at any 
instant may be calculated by diminishing / 2 - ^ indefinitely, when 



dv 

Acceleration = a = . 
at 



<*) 



Feet per sec 
2 -ix 



In the interval FH (Fig. 420), the change in velocity is a decrease, 
shown by GL. If the acceleration at A is positive, that at F will be 

negative. At M, where the tan- 
gent to the curve is horizontal, 
there is no change in the 
velocity over an indefinitely 
small interval of time, and 
hence there is no acceleration. 
An acceleration-time diagram 
may be deduced from the 
velocity-time diagram by the 
method already applied in 
Example 2, p. 384, for obtain- 
ing a velocity-time diagram from a space-time diagram. The average 
acceleration over any interval is set out as an ordinate at the middle 
of the interval. 

EXAMPLE. Taking the data of Example i, p. 383, and the velocity- 
time diagram (Fig. 421, redrawn from Fig. 419) from Example 2, p. 384. 
draw an acceleration-time diagram. 

The tabular form of calculation may be adopted as follows : 




Seconds 

FIG. 421. Velocity-time diagram for the point B 
in Fig. 417. 



Interval No. 


Ordinate. 


Velocity, 
feet per sec. 


Change in vel. , 
feet per sec. 


Average acceleration 
= change in vel.-:-i sec., 
feet per sec. per sec. 





oo' 


I-8 7 












-1-20 


- 1-20 


I 


II' 


0-67 












-0-25 


-0-25 


2 


22' 


0-42 












-0-15 


-0-15 


3 


33' 


0-27 












-0-10 


-0-10 


4 


44' 


0-17 












-0-09 


-0-09 


5 


55' 


0-08 












-0-08 


-0-08 


6 


6 


o 







ACCELERATION 



387 



The last column is plotted at the middle of the intervals as shown in 
Fig. 422, and a fair curve is drawn through the plotted points, thus 
obtaining the acceleration-time diagram. 



6 Seconds 




-1-0- 



Feet per sec. per sec. 

FIG. 422. Acceleration-time diagram for the point B in Fig. 417. 

If the distance travelled is given by an equation connecting s and 
/, the acceleration may be found by two successive differentiations. 
Thus : Let - 



where c is a constant. Then 



ds 



dv 



The indices in simply indicate that s has been differentiated 

twice with respect to / (p. 12). 

Equations for uniform acceleration. Reference is made to Figs. 
423 and 424, the former showing the velocity-time diagram for a 



Vetoed 



Velocity 




k ------- 



\Time 




FIG. 423. Velocity-time diagram, 
starting from rest. 



FIG. 424. Velocity-time diagram, 
starting with velocity z>j. 



body starting from rest ; Fig. 424 shows the diagram if the body starts 
with a given velocity v l ; in both cases the acceleration is uniform. 



388 MACHINES AND HYDRAULICS 

Starting from rest (Fig. 423) : 

Let v = the velocity in feet per second, 

/ = the time in seconds taken to acquire , 

s = the distance travelled, in feet, 

a = the acceleration, feet per sec. per sec. 

By definition, a = -> 

or, v = at ........................................... ( i ) 

s = the average velocity x t ; 
'. s = Jvt ........................................... (2) 

Or, s = the area of the diagram 

= / x ^v t x \at (from ( i ) ) ; 
'. - = JaA ....................................... (3) 

From(i), / = - 

v 2 v 2 
Inserting this in (3), s = \a-^ = ; 

.'. v 2 = 2as ........................................... (4) 

Starting with a velocity i\ (Fig. 424) : 

Let ^ and v. 2 = the initial and final velocities respectively 

in feet per second, 

/=the time in seconds in which v\ increases 
to 2 , 

s = the distance travelled, in feet, 
a = the acceleration, in feet per sec. per sec. 
Then v 2 -v l = &t ........................................... (5) 

s = the average velocity x / 



Or, s = the area of the diagram 

= rectangle OCD A + triangle CDB 
= !/+(/x JDB). 

Also, BD = /; 

/. B = v 1 t+Jat2 .................................. (7) 

From (5), ; = - 



TRIANGLE OF VELOCITIES AND ACCELERATIONS 389 




Inserting this in (6), s = 



(8) 

The case of a body falling under the action of gravity is one of 
nearly uniform acceleration. The acceleration would be quite 
constant, but for the resistance offered by the atmosphere, and for 
the fact that a body weighs less when at a height above the surface 
of the earth. The symbol g is used to denote the acceleration of a 
body falling freely, that is, neglecting atmospheric resistances. The 
value of g varies to a small extent, being about 32-088 feet per second 
per second at the equator and about 32-252 at the poles. The value 
32-2 may be taken for all parts of the British Isles. The equations 
found above may be modified to suit a body falling freely, by writing 
g instead of a, and the height h feet instead of s. 

Composition and resolution of velocities and accelerations. A 
given velocity is a vector quantity and may be represented in the 




B a 

FIG. 425. Triangle and parallelogram of velocities. 

same manner as a force by a straight line having an arrow point. 
Hence problems involving the resolution or composition of velocities 
may be solved in the same way as for forces by the application of the 
triangle or polygon of velocities. 

Let a point A have component velocities V x and V 2 in the plane 
of the paper (Fig. 425). The resultant velocity may be found from 
the triangle abc in which ab represents V ]5 be represents V 2 and ac 
gives the resultant velocity V which should be shown applied at A. 
A parallelogram of velocities, ABDC, may be used by making AB = V l 
and AC = V 2 ; the diagonal AD gives the resultant velocity. 

Rectangular components of a given velocity V (Fig. 426) along two 
axes OX and OY may be calculated from 

V^ = V cos a, 
V y -Vsina. 



390 



MACHINES AND HYDRAULICS 



EXAMPLE. A body slides down an incline of 30 (Fig. 427) with a 
velocity of 10 feet per second. Find the horizontal and vertical com- 
ponents of its velocity. 

V/i = V cos 30 = 10 x -$ = 8-66 feet per sec. 
V v = Vsin 30= 10x^ = 5 feet per sec. 

It will be understood that, as acceleration has magnitude, direction 
and sense, this quantity can be represented also by a straight line 





FIG. 426. Rectangular components 
of a velocity. 



FIG. 427. 



having an arrow point. Problems involving the composition and 
resolution of accelerations may be solved by use of the same con- 
structions as for velocities. 

EXAMPLE. A body slides down an inclined plane with an acceleration 
a feet per second per second (Fig. 428). If the plane makes an angle a 
to the horizontal, find the component accelerations () normal to the plane, 
(b} vertical. 

Make OA to represent a to scale and draw the 
parallelogram of accelerations OBAC, OB being 
normal to the plane and OC being vertical. The 
angle OBA will be equal to a. Hence, 




and 



OB 

OA = C ta ' 

Normal acceleration = a n = a cot a. 
OC 



FIG. 428. 



OA 



= cosec a, 



Also, 

and Vertical acceleration =a v = acosec a. 
The relation of a n and av is given by 

= ^rr^ = cos a ; 
.'. a n =avCQsa. 

Angular velocity. When a body is rotating about a fixed axis, 
the radius of any point in the body turns through a definite angle 
in unit time. The term angular velocity is given to the rate of 



ANGULAR VELOCITY 



39i 




describing angles, and may be measured in revolutions per minute 
or per second, or, more conveniently for the purposes of calculation, 
in radians per second ; the symbol to is taken usually to denote the 
latter. 

Since there are 2-rr radians in a complete revolution, the connection 
between <o and N, the revolutions per minute, will be ' 

N TrN 

to = 27r = radians per second. 
60 30 

Let a line OA (Fig. 429) have uniform speed of rotation in the 
plane of the paper about O as centre. The point A will have a 
uniform linear velocity v feet per second in 
the circumference of a circle ; let r be the 
radius of the circle in feet. It is evident 
that the length of the arc described by A 
in one second will be v feet, and the angle 

subtended by this arc will be - radians. OA 

turns through this angle in one second, hence 
its angular velocity is 

v FIG. 429. Relation of angular 

= - radians per second. and linear velocities. 

It will be noticed that the linear velocities of other points in the 
line OA will be proportional to their radii, hence such velocities will 
be unequal. The same numerical result will be obtained for the 
angular velocity by dividing the linear velocity of any point by its 
radius. It is obvious that, under given conditions of speed of 
rotation, all radii of a body turn through equal angles per second, 
hence only one numerical result is possible for the angular velocity. 

Equations of angular motion. In uniform angular velocity equal 
angles are described in equal intervals of time. The total angle a 
described by a rotating line in a time / seconds will be, if the angular 
velocity is uniform, a = wtradians . 

If the angular velocity varies, the body is said to have angular 
acceleration. Angular acceleration is measured in radians per second 
per second and is written 6. Suppose a line to start from rest with 
a uniform angular acceleration 0, its angular velocity at the end of t 
seconds will be w = 0t? mdians per second (j) 

The average angular velocity will be |o>, hence the total angle 
described will be a= i_ wt ^ 



MACHINES AND HYDRAULICS 



Substituting for <o from (i) gives 

=4 2 ............................... (3) 

a 

Again, from ( i ), /=^, .'. / 2 = ^- 

Substituting this value in (3), we obtain 



.' 2 = 20a ................................... (4) 

It will be observed that the above results are similar to those given 
on p. 388 for rectilinear motion with the substitution of w for v, and 
# for a. Making these substitutions, we may obtain the corresponding 
equations for angular motion when the body has an initial angular 
velocity <o r W2 - Wl = 0t .................................. (5) 



*-? = ** ................................... (8) 

The relation between the linear acceleration of a point in a revolving 
line and the angular acceleration of the line will be given by 

6 = - radians per sec. per sec., ..................... (9) 

where a = linear acceleration of A (Fig. 429) in feet per sec. per sec., 

r = radius of A in feet. 

Denning the angular velocity of a rotating line as its rate of 
describing angles, and its angular acceleration as the rate of change 
of angular velocity, suppose a line to describe a small angle So, in 
an interval of time S/. The average angular velocity during the 
interval will be * 

w " = r 

If Sa be taken indefinitely small and written da, the time dt in 
which it is traversed will be our conception of an instant, and the 
angular velocity at this instant will be 

da, . . 



If the angular velocity alters by a small amount Sw during an 
interval of time S/, then 

Average angular acceleration = 8 a = -- 



ANGULAR VELOCITY 393 

If these be reduced indefinitely, the result will give the angular 
acceleration at the instant considered, viz., 



The results (10) and (n) are suitable for application of the rules 
of the differential calculus (p. 9). 

EXAMPLE i. An engine starts from rest and acquires a speed of 
300 revolutions per minute in 40 seconds from the start. What has been 
its angular acceleration ? 

300 

(0=4 -277=107? 

60 
= 31-41 radians per sec. 

= w = 3J^41 
/ 40 

=0-785 radian per sec. per sec. 

EXAMPLE 2. The driving wheel of a locomotive is 6 feet in diameter. 
Assuming no slipping between the wheel and the rail, what is the angular 
velocity of the wheel when the engine is running at 60 miles per hour. 



Velocity of engine = - = 88 feet per sec. 



As the distance travelled in one second is 88 feet, we may find the 
revolutions per second of the wheel by imagining 88 feet of rail to be 
wrapped round the circumference of the wheel. 

OQ 

Number of turns = -, ; 
ird 

88x7 , 

.'. Revolutions per sec. = --- -^A-by. 
22x6 

(0 = 4-67x277 
= 29-33 radians per sec. 

Or the following method may be used. Referring to Fig. 430, if there 
is no slipping, the point A on the rim of the wheel is in contact with the 
rail for an instant and is therefore at rest. Hence the whole wheel is 
rotating about A for an instant. The angular velocity will therefore be 

ivenb >' velocity of O 

= --JL 

oo 

= 29-33 radians per sec. 

EXAMPLE 3. Using the data of Example 2 and referring to Fig. 431, 
find the velocities of the points on the rim of the wheel marked B, C and 
D, supposing no slipping. 



394 



MACHINES AND HYDRAULICS 



In answering this, it will be assumed that the whole wheel is rotating 
about A for an instant, and that the velocity of any point is proportional 




FIG. 430. Angular velocity 
of a rolling wheel. 




FIG. 431. Velocities of points in 
a rolling wheel. 



to the radius of that point frgm A as centre and has a direction perpen- 
dicular to that radius. 

The angular velocities of AB, AC and AD are equal and are given by 
the angular velocity of OA in Fig. 430, viz. 

00 = 29-33 radians per sec. 
Also, 'Z/^wR ; 

* 2> B = 29-33 xAB_ 

= 29-33 X3V/2 

= 124-4 feet per sec. 
?/ c = 29-33 x AC 

= 29-33x6 

= 176 feet per sec. 
^D = 29-33 x AD 

= 29-33x3^2 

= 124-4 feet per sec. 

Angular velocity and acceleration diagrams. Diagrams showing 
the angle traversed, the angular velocity, and the angular acceleration, 
all three on bases representing time may be drawn by the same methods 
as have been explained on pp. 381-387 for linear velocities, etc. 
The angle traversed is treated in the same manner as the distance 
travelled and an angle-time diagram is drawn. The angular-velocity 
diagram is then deduced from the angle-time diagram and an angular- 
velocity-time diagram is drawn. The angular-acceleration-time 
diagram may then be deduced from the angular-velocity-time diagram. 

Velocity changed in direction. Hitherto the acceleration due to 
changes in the magnitude of a body's velocity alone have been con- 
sidered. There may be also changes effected in the direction of the 
velocity, and such will give rise to accelerations. 



MOTION IN A CIRCLE 395 

Let a point move along a straight line AB (Fig. 432 (a)) with a 
velocity TJ I ; on reaching the point B, let the point move along BC 
with a velocity v 2 . To determine the change in velocity which has 
taken place at B, the following method may be used. Stop the 
point on reaching B by applying a velocity equal and opposite to v^ ; 
this is represented by DB in the figure. The point now being at 
rest can be dispatched along any line with any velocity ; to comply 
with the given conditions, give it a velocity v<> in the line BC, repre- 
sented by EB in the figure. The total change in velocity has com- 
ponents represented by DB and EB ; hence the parallelogram BDFE 
gives FB = v c as the resultant or total change in velocity. 




FIG. 432. Velocity changed in direction. 

A convenient construction is shown in Fig. 432 (&). Take any point 
O, and draw OA and OC to represent completely v l and v.-, respec- 
tively. Then the change in velocity will be AC = v c . The sense of 
the change in velocity may be found from the rule that it is directed 
from the end A of the initial velocity toivards the end C of the final 
velocity (Fig. 432 (/;)). 

For reasons that will be apparent later, it is not possible to make a 
body take a sudden change in velocity ; the transition from AB to 
BC in Fig. 432 (a) will take place along some curve, such as GHK. 
This makes no difference in the construction for finding the total 
change in velocity. Suppose that the body takes / seconds to pass 
from G to K along the curve, then this gives the time taken to effect 
the total change in velocity v c - Hence, 

7) 

Resultant acceleration = -y, 

and has the same direction and sense as v c . 

Motion in a circle. A small body moving in the circumference of 
a circle with uniform velocity v is continually changing the direction 
of its velocity. At any point of the circumference the direction of 
the velocity will be along the tangent ; at P 1 (Fig. 433 (a)) the velocity 
will be #j = v, and at P 2 the velocity will be z> 2 = v. To obtain the 
change in velocity between P 1 and P 2 , draw the triangle OAB 



396 



MACHINES AND HYDRAULICS 



(Fig. 433 (<)). AB = v c will be the change in velocity, and is shown in 
Fig. 433 (a) passing through the point C where v and u 2 intersect. It is 

y. 




(b) 



FIG. 433. Motion in a circular path. 

evident that if v c be produced it will pass through O l , the centre of 
the circle, and this will be the case no matter what may be the 
positions chosen for P l and P 2 . The acceleration due to v c will also 

pass through O r The follow 
ing method may be used to 
obtain the acceleration. 

Referring to Fig. 434, in 
p which a point P is moving in 
the circumference of a circle of 
radius R with uniform velocity 
v. At Pj the velocity v l is 
along the tangent, and its hori- 
zontal and vertical components 
will be #! sin a x and v^ cos a t 
respectively, where a, is the 
angle OP l makes with the hori- 
zontal diameter AB. Similarly 




FIG. 434. Acceleration of a point moving in the 



circumference of a circle. 

at P 2 , the components will be 

V 2 sin a 2 and v. 2 cos a 2 respectively. As v l = v. 2 = v, we have 
Change in horizontal velocity = v sin a 2 - v sin a l 

= V (sin a., - sin a T ) 



OP 2 OP l 



=|.P,K. 



(0 



MOTION IN A CIRCLE 397 

Again, the time, /, taken to pass from Pj to P 2 will be the time in 
which this change in velocity has been effected, and may be cal- 
culated from p p tf = V f t 

Hence, Horizontal acceleration of P = - 

v 2 P 2 K 
= R' pV &) 

If ctj and a 2 are very nearly equal, the angle P!OP 2 will be very 
small and the arc P X P 2 will be a straight line practically. The angle 
PjP 2 K will be equal to a r Hence, 

P 2 K = 

.*. horizontal acceleration of P = =? cos ctj (4) 

This acceleration will be directed always towards the vertical 
diameter NS, as the sign of the acceleration will be the same as that 
of cos a. 

Let P be at A. Then a = o, cosa= i, and the acceleration will be 

""' (5) 

and will be directed along AO. As any reference diameter might 
have been taken instead of AB, it follows that for any position of P, 
the acceleration towards the centre of the circle will be given by (5). 
The result will be in feet per second per second if 
v = the velocity in feet per second, 
R = the radius of the circle in feet. 
The acceleration may be stated in terms of the angular velocity by 

writing v 

to = , or v = wR. 
R 



From (5), tf = L=:o,2R ............................ (6) 

EXAMPLE i. A motor car is travelling at 20 miles per hour round a 
curve of 600 feet radius. What is the acceleration towards the centre of 

thecircle? 



z/ 2 88 x 88 , 

# = --= -. = 1-434 feet per sec. per sec. 
R 3x3x600 -^ 



MACHINES AND HYDRAULICS 



EXAMPLE 2. What is the acceleration, towards the centre, of a point 
on the rim of a wheel 4 feet diameter and running at 300 revolutions per 
minute ? <o = ^ x 27r = IOTT radians per second, 

a = w' 2 R = loo x Y x Y x 2 
= 1975 feet per sec. per sec. 

Simple harmonic motion. In Fig. 435, the point P travels in the 
circumference of the circle ANBS with uniform velocity v. Drawing 
PM perpendicular to the diameter AB, it will be noticed that M, the 
projection of P on AB, will vibrate in AB as P rotates. The velocity 
and acceleration of M at any instant will be the horizontal com- 
ponents of the velocity and acceleration of P, viz. 

a =wRsina, (i) 



a = =. cos a = a> 2 R cos a, (2} 

K 

where R is the radius of the circle. 

The vibratory motion of M is called simple harmonic motion. One 
of its properties is that the acceleration is directed always towards 

the middle point O of the vibration. Again, since cos a = -y^-, and 
is therefore proportional to OM, the acceleration is proportional to 
N VV 




FIG. 435. The motion of M is 
simple harmonic. 



FIG. 436. Acceleration diagram for M, 
on a space base. 



OM, the distance of M at any instant from the middle of the 
vibration. When M is at A, the acceleration is proportional to OA 
and is positive, i.e. directed towards the left ; when M is at B, the 
acceleration has the same value, but is directed towards the right and 
js negative. M has no acceleration when at O. An acceleration 
diagram may be drawn by erecting ordinates A A' and BB', each 
equal to R, on the diameter AB and joining A'B' (Fig. 436). Any 



SIMPLE HARMONIC MOTION 



399 



ordinate MC will then give the acceleration for the position M. The 
scale of this diagram is obtained from the consideration that when P 
is at A (Fig. 435), cos a = i and a = a> 2 R ; hence the scale is such 
that AA' = o>' 2 R. As the diagram has been drawn on a distance, not 
a time, base, it may be called a distance-acceleration diagram. 

The velocity of M at any instant is proportional to sin a. Now 

PM 
sina = - (Fig. 435), and is therefore proportional to PM ; hence 

the velocity is proportional to PM. When M is at A the velocity 
is zero, and has also zero value at B. Maximum velocity is attained 
at O, when V = z/. The circle in Fig. 435 is a velocity diagram on a 
distance base AB, as any ordinate PM will give the velocity of M at 
the instant considered, the scale being such that ON represents v. 
V is positive, i.e. towards the left, if PM is above AB, and negative 
if PM is below AB. 

Velocity-time and acceleration-time diagrams may be drawn by 
noting that, as the velocity of P in Fig. 435 is uniform, equal angles 
will be described by OP in equal times. Divide the circle into 
twelve equal angles of 30 each, and calculate V = v sin a, and also 

a = R COS a * r ea k P os ^^ on f p - Set off a base of angles from 
o to 360 (Fig. 437), and erect ordinates having the calculated values 






FIG. 437. Velocity-time and acceleration-time diagrams for simple harmonic motion. 

V and a. The base represents angles or time to different scales ; the 
scale of time is such that the total. length of the base line represents 
the time of one revolution of OP in Fig. 435. 

Fig- 399 (p- 371) shows a well-known mechanism, used in pumps, 
which realises simple harmonic motion. The slotted bar has a 
sliding block, in which is bored a hole to receive the crank pin. 
The vertical components of the velocity and acceleration of the crank 



4 oo MACHINES AND HYDRAULICS 

pin .are thus eliminated, and the horizontal components alone are 
communicated to the piston rods. 

The time of a complete vibration in simple harmonic motion from 
A to B and back again (Fig. 435) may be estimated from the fact 
that it will be equal to that of a complete revolution of P. 
Let T = the time of one vibration in seconds. 

v = the velocity of P, in feet per second. 
R = the radius of the circle = the amplitude of the 

vibration, in feet. 
Then z;T = 27rR, 

T = 2:r ...................................... (3) 



27T / v 

, .............................. (4) 

o> 



where w is the angular velocity of OP in radians per second. 

EXAMPLE. A point is describing simple harmonic vibrations in a line 
4 feet long. Its velocity at the instant of passing through the centre of 
the line is 12 feet per second. What is the time of a complete vibration ? 



V 

where R is 2 feet and v is 12 feet per second. Hence, 

2X22X2 

7 x 12 
= 1-05 seconds. 

Change in angular velocity. A given angular velocity may be 
represented by means of a vector in the following manner. In 
Fig. 438 (a) is shown a wheel rotating about an axis OA with an 
angular velocity w. A person situated on the right-hand side sees 
the wheel rotating in the clockwise direction, and may represent 
the angular velocity by drawing a line OA perpendicular to the 
plane of rotation of the wheel and on the same side of this plane 
as the person is situated. OA is made, to scale, of length to 
represent o>. A person situated on the left-hand side will see the 
wheel rotating in the anti-clockwise direction, and may represent 
the angular velocity by means of a perpendicular to the plane of 
rotation drawn on the opposite side of this plane. Both observers 
will thus agree in erecting the perpendicular on the same side of the 
plane of rotation. The perpendicular represents the magnitude and 
direction of rotation of an angular velocity in a plane perpendicular to 



CHANGE IN ANGULAR VELOCITY 



401 



the line, and will thus obey the same laws as a vector. Two or more 
component angular velocities represented in this way may be dealt 
with and their resultant found by means of a triangle or polygon of 
velocities. 

In Fig. 438 (a) the wheel is revolving in a vertical plane ; at 
the same time its axis is revolving in a horizontal plane as indicated 
by the arrows at the ends of the axis. A plan of the wheel is given 
in Fig. 438 (b) ; OA represents the angular velocity of the wheel at 
one instant, and OA' represents its angular velocity after a short 
interval of time during which the wheel has turned horizontally into 
the position indicated by dotted lines. Since OA and OA' represent 
the initial and final angular velocities respectively, it follows, by the 
same reasoning as for linear velocity (p. 395), that the change in 





FIG. 438. Change in angular 
velocity. 



FIG. 439. Change in angular velocity 
by method of linear velocities. 



angular velocity is represented by AA'. The actual change in angular 
velocity takes place in a plane perpendicular to AA', i.e. a vertical 
plane in the given case, and is anti-clockwise to an observer situated 
at B. 

It may be of assistance to the student to consider the problem from 
the point of view of linear velocities. In Fig. 439 (a) is given a plan 
of the wheel. OA represents v lt the initial velocity of a point on the 
top of the wheel ; OA' represents the final velocity v. 2 of the point on 
the top of the wheel; A A' represents v c , the change in velocity of 
this point. In the same way BB' represents v c , the change in linear 
velocity of a point at the bottom of the wheel. Fig. 439 (b) shows 
these changes in linear velocity in their proper positions, and indicates 
that a change in angular velocity is taking place in a vertical plane 
containing the axis of the wheel. 

D.M. 2 c 



4 02 



MACHINES AND HYDRAULICS 




In Fig. 440 OA represents o>, the angular velocity of the wheel. It 
will be evident that the successive additions of small changes in 
angular velocity such as that represented by AA' will cause A to 
describe a complete circle. The total change 
in angular velocity during one rotation of the 
wheel axis in the horizontal plane will be the 
circumference of the circle, and will be given by 

Change in angular velocity = 27r<o. 
If this result be divided by the time taken by 
A in describing the complete circle, i.e. the time 
in which the wheel axis makes one complete 
rotation in the horizontal plane, the result will 
give the angular acceleration. It is evident that the angular accelera- 
tion will take place in the same plane as that in which the change in 
angular velocity occurs, viz. a vertical plane containing the wheel axis. 
Relative velocity. The relative velocity of two bodies may be defined as 
the velocity which an observer situated on one of them would perceive in the 
other. An observer in one of two trains, moving side by side with 
equal velocities of the same sense, would perceive no velocity in the 
other and would therefore say that the relative velocity is zero. If 
the train carrying the observer has a velocity of 30 miles per hour, 
and the other, one of 35 miles per hour, he will see the other train 
moving past him at a rate of 5 miles per hour, which velocity he 



FIG. 440. Plan of the wheel 
shown in Fig. 438. 




(a) V B (b> 

FIG. 441. Velocity of B relative to A. 

would call the relative velocity of the trains. Had the trains been 
moving in opposite directions, the relative velocity would be 65 miles 
per hour. A stream of water moving at 8 feet per second reaching a 
water wheel, the buckets of which are moving away from the stream 
at 6 feet per second, will enter the buckets with a relative velocity of 
2 feet per second. 

If two bodies A and B have velocities as shown at V A and V B 
(Fig. 441 (a)), their relative velocity may be obtained in the following 



EXERCISES ON CHAPTER XVI. 



43 




manner. Stop A by giving both A and B a velocity V A equal and 
opposite to that originally possessed by A ; this artifice will not alter 
their relative velocities. B has now component velocities V B and 
V A , the resultant of which is V R . As A is at rest, the velocity of B 
relative to A will be V R . 

In Fig. 442, B has been brought to rest by giving both A and B a 
velocity V B equal and opposite to that originally possessed by B. 
The resultant velocity of A will now 
be V R , and, as B is at rest, this will be 
the velocity of A relative to B. 

It will be clear that V R in Fig. 441 (a) 
is equal and opposite to V R in Fig. 442, 
showing that the velocity of B relative 
to A is equal and opposite to the 
velocity of A relative to B. A 

FIG. 442. Velocity of A relative to B. 

A convenient construction is given 

in Fig. 441 (b). From any point O draw OA and OB to represent 
respectively V A and V B , both being placed so that the senses are 
away from O. Then AB represents the relative velocity of sense 
from A towards B if the velocity of B relative to A is required, 
and of opposite sense if the velocity of A relative to B is required. 



EXERCISES ON CHAPTER XVI. 

1. In a crank and connecting-rod mechanism, the crank is i foot and 
the connecting rod is 4 feet in length. The line of stroke of the cross- 
head pin passes through the axis of the crank shaft. Find, by drawing, 
the distances of the crosshead from the beginning of the stroke for crank 
intervals of 30 throughout the revolution. Plot a distance-time diagram." 

2. Use the data obtained in the solution of Question i, and calculate 
the mean velocity of the crosshead for each interval. The crank rotates 
uniformly at 180 revolutions per minute. Draw a velocity-time diagram. 

3. Using the results of Question 2, calculate the mean acceleration 
for each interval. Plot an acceleration-time diagram. 

4. Answer Questions i, 2 and 3 for the case in which the line of 
stroke of the crosshead passes the axis of the crank shaft at a distance of 
6 inches. 

5. The distance between two stations is 1-6 miles. A locomotive, 
starting from one station, gives the train an acceleration of 25 miles per 
hour in 0-5 minute until the speed reaches 30 miles per hour. This speed 
is maintained until brakes are applied and the train is brought to rest at 
the second station under a negative acceleration of 3 feet per second 
per second. Find the time taken to perform the journey. 



404 MACHINES AND HYDRAULICS 

6. The distance travelled by a body is given in feet by the equation 
,y=o-02/ 2 + 3, / being the time in seconds from the start. Find the 
distance travelled, the velocity and the acceleration at the end of 
4 seconds, starting from rest. 

7. A body, falling freely under the action of gravity, passes two 
points 30 feet apart vertically in 0-2 second. From what height above 
the higher point did it start to fall ? 

8. A body is thrown upwards from the foot of a cliff 40 feet high and 
reaches a height of 12 feet above the cliff. It finally alights on the cliff 
top. Find the total time of the flight and the initial velocity. 

9. A body slides down a plane inclined at 10 degrees to the horizontal 
under the action of gravity. What is the acceleration in the direction of 
the motion, neglecting frictional effects ? Suppose the body to start from 
rest, what will be the velocity after it has travelled 1 2 feet ? 

10. A boat is steered across a river 100 yards wide in such a way that, 
if there were no current, its line of motion would be at 90 degrees to the 
banks. Actually it reaches a point 40 yards further down stream, and 
takes 3 minutes to cross. What is the speed of the current ? 

11. A pistol fires a bullet with a velocity of 1000 feet per second. 
Suppose it to be fired by a person in a train travelling at 60 miles per 
hour, (a) forward in the line of the motion of the train, (fr) backward 
along the same line, (c) in a line parallel to the partitions of the com- 
partments, and calculate in each case the resultant velocity of the bullet. 

12. A wheel slows from 120 to no revolutions per minute. What has 
been the change in angular velocity in radians per second ? If the 
change took place in 2 minutes, find the angular acceleration. 

13. A wheel starts from rest and acquires a speed of 150 revolutions 
per minute in 30 seconds. Find the angular acceleration and the revolu- 
tions made by the wheel while getting up speed. 

14. Starting from rest, a wheel 2 feet in diameter rolls without 
slipping through a distance of 40 yards in 8 seconds. Find the angular 
acceleration and the angular velocity at the end of the given time. Plot 
an angular velocity-time diagram. 

15. Water travels along a horizontal pipe with a uniform speed of 
4 feet per second. The pipe changes direction to the extent of 30 degrees. 
Find the change in the velocity of the water. 

16. A wheel 12 inches in diameter revolves 18,000 times per minute. 
Find the central acceleration of a point on the rim. 

17. Calculate the central acceleration of a train running at 50 miles 
per hour round a curve having a radius of 0-75 mile. 

18. A point describes simple harmonic vibrations in a line 2 feet long. 
The time of one complete vibration is 0-3 second. Find the maximum 
velocity. 

19. A wheel revolves in a vertical plane 300 times per minute. The 
plane keeps vertical, but rotates through an angle of 90. Find the change 
in angular velocity, and show it in a diagram. If the change took place 
in 2-5 seconds, find the angular acceleration. 



EXERCISES ON CHAPTER XVI. 405 

20. A carriage wheel is 4 feet in diameter and is travelling at 6 miles 
per hour. What is the velocity of a point at the top of the wheel relative 
to (cz) a person seated in the carriage, (b} a person standing on the ground. 
Answer the same regarding a point at the bottom of the wheel. 

21. A railway line A crosses another B by means of a bridge, the 
angle of intersection, as seen in the plan, being 30 degrees. A train 
on A is approaching the point of intersection with a velocity of 40 
miles per hour and another train on B is receding from the intersection, 
on the same side of it, with a velocity of 20 miles per hour. Find the 
relative velocity of the trains. 

22. A particle moves with simple harmonic motion ; show that its 
time of complete oscillation is independent of the amplitude of its motion. 
The amplitude of the motion is 5 feet and the complete time of oscillation 
is 4 seconds ; find the time occupied by the particle in passing between 
points which are distant 4 feet and 2 feet from the centre of force and are 
on the same side of it. (L.U.) 

23. At midnight a vessel A was 40 miles due N. of a vessel B ; A 
steamed 20 miles per hour on a S.W. course and B 12 miles per hour 
due W. They can exchange signals when 10 miles apart. When can 
they begin to signal, and how long can they continue ? (I.C.E.) 



CHAPTER XVII 
INERTIA. 

Inertia. Inertia is that property of matter by virtue of which a 
body tends to preserve its state of rest or of uniform velocity in a 
straight line, and offers resistance to any change being made in the 
velocity possessed by it at any instant, whether the change be one of 
magnitude or of direction of velocity. Hence, in order to effect any 
such change, it will be necessary to employ force to overcome the 
inertia of the body. There will be no resultant force acting on any 
body which is travelling with uniform velocity in a straight line ; in 
such a case the external forces, if any, applied to the body are in 
equilibrium. The existence of acceleration in a body implies the 
presence of a resultant external force, and this force must be applied 
in the line of, and must have the same sense as the proposed 
acceleration. 

The estimation of the magnitude of the force required to produce 
a given acceleration may be obtained from an experimental law. All 
bodies at the same place fall freely with the same acceleration g. 
Now their weights are proportional to their masses, and as these 
weights are the resultant forces producing acceleration, it follows that 
the force required to produce a given acceleration is proportional to 
the mass of the body. It may also be shown by experiment that the 
force required to produce acceleration in a body of given mass is 
proportional to the acceleration. Hence, we have the law that the 
force required is proportional jointly to the body's mass and accelera- 
tion, and consequently will be measured by the product of the 
mass and the acceleration. 

From the case of a body falling freely we know that a force of i Ib. 
weight acting on a mass of i pound gives an acceleration of g feet 
per second per second. It follows that the algebraic statement of 
the above law will be 

P = Ib. weight, (i) 



INERTIA 



407 



where m = the mass of the body in pounds, 

a = its acceleration in feet per sec. per sec. 

The result of the calculation by use of equation (i) will vary to a 
small extent depending on the value of g at the particular place. An 
absolute unit of force may be employed which does not vary, and is 
denned as the force required to give unit acceleration to a body 
having unit mass. The British absolute unit of force is the poundal, 
and is the force that would give an acceleration of one foot per 
second per second to a body free to move and having a mass of one 
pound. The metric absolute unit is the dyne ; a force of one dyne 
acting on a body free to move and of mass one gram would produce 
an acceleration of one centimetre per second per second. Using 
these units, equation (i) becomes 

F = ma, in absolute units, (2) 

the result being in poundals, or dynes, respectively if 

m = the mass of the body in pounds, or grams, 

a = its acceleration in feet, or centimetres per sec. per sec. 

The weight of a body expressed in absolute units will be given by 

W = mg (3) 

Also, a force stated in poundals or dynes may be converted into 
Ib. weight or grams weight by dividing by the proper value of g, 
which may be taken as 32-2 feet per second per second in the 
British system, or as 981 centimetres per second per second in the 
metric system, for all parts of the British Isles. 

In using the above equations, it must be understood clearly that 
each side of the equation represents a force ; the left-hand side 
represents the resultant force applied to the body from the outside ; 
the right-hand side represents the force due to the collective resist- 
ance of all the particles of the body to any change being made in 
the velocity. The whole equation expresses the equality of these 
forces. The student would do well to recall again the fact that a 
force cannot act alone ; there must always be an equal opposite force, 
and if the latter is not wholly supplied by some resistance given by 
an outside agency such as friction, etc., it must be supplied in part 
by the inertia of the body. Equality of the forces is an invariable 
law. 

EXAMPLE i. A train has a mass of 200 tons. If frictional resistances 
amount to 12 Ib. weight per ton, what steady pull must the locomotive 
exert in order to increase the speed on a level road from 20 to 40 miles 
per hour, the change to take place in i^ minutes ? 



MACHINES AND HYDRAULICS 



Let 



Then 
Also, 



T = pull required, in Ib. weight units. 
F = total frictional resistance, Ib. weight. 
P = resultant force producing acceleration, Ib. 
weight. 



r>_T T? __"** 

JL 1 JT " 

F = 200 x 1 2 = 2400 Ib. weight. 



(I) 



, . . , . . ZU X kJOU OO ,. 

Initial velocity = -7 ^ = feet per sec. 
7 60x60 3 

Final velocity = feet per sec. 

/i;6 88\ 
Acceleration = <2 = l-rQO 

\ 3 3 / 

= foot per sec. per sec. 
270 

Substituting these values in (i) gives 



200 x 2240 x 88 



32-2 x 270 



7-2400 = 



= 6935 Ib. weight. 

EXAMPLE 2. The mass of a train is 250 tons and frictional resistances 
amount to 1 1 Ib. weight per ton. The speed 
on reaching the top of an incline of i in 80 
is 30 miles per hour, and the train runs 
down with steam shut off. If the incline is 
is 0-5 mile long, what will be the speed at 
the bottom ? 

Referring to Fig. 443, the weight of the 
train, W, may be resolved into two forces T 
and R respectively, parallel and perpendicular to the incline. Let a be 
the angle made by the incline with the horizontal. Then 
T = W sin a = W tan a, very nearly, 
= 250 x 2240 x^j 
= 7000 Ib. weight. 

Also, Friction = F = 25ox 11=2750 Ib. weight. 

P = T-F 

= 7000-2750 = 4250 Ib. weight. 
ma . 




Also, 



Again, Initial velocity = v 



a = 1^ = 4250x32-2 
m 250x2240 
= 0-244 foot per sec. per sec. 
30 x 5280 
60x60 



44 feet per sec. 



KINETIC ENERGY 409 



And 2/ 2 a - v^ = 2as (p. 389) ; 

.-. Z/2 2 - (44X44)= 2X0-244X^-2, 



feet per sec. 
38-8 miles per hour. 



Kinetic energy. In Fig. 444 is shown a body of mass m pounds 
able to move freely. Let the body be at rest at A and let a force 
P Ib. -weight be applied, in consequence of which the body moves 
with continually increasing velocity to B, a horizontal distance of 
s feet. Work wall be done by P 
against the resistance due to the 
inertia of the body. p 

Work done by P = Pjfoot-lb. '*- ..... ~ 5 ...... *' 

As there has been no external FlGl ^-Kinetic energy of a body. 
resistances of any kind, it follows that the whole of the work done 
by P will be stored in the body at B in the form of kinetic energy. 
Let v feet per second be the velocity at B, and let a be the accelera- 
tion in feet per second per second. 

Then P = Ib. weight. 

,2 

Also, s = feet (p. 388). 

Hence, Work done by P = -- or, 

g 2a 

cy 

Kinetic energy of body = foot-lb ............................... (i) 

2g 

Note that the velocity is squared in this result, hence its sign, 
positive or negative, is immaterial. The interpretation of this is that 
kinetic energy is not a directed, or vector, quantity, and a body 
moving in any direction will have kinetic energy which may be 
calculated by use of the expression found above. The kinetic energy 
may be expressed in absolute units by omitting g. 

9 

Kinetic energy = - foot-poundals ................... (2) 

EXAMPLE i. A railway truck of mass 20 tons moving at 6 feet per 
second comes into collision with buffer stops and is brought to rest in a 
distance of 9 inches. What has been the average resistance of the 

buffers? m tf 2 ox6x6 

Kinetic energy = - = ^ - = 1 1 1 8 foot-tons. 

2g 04-4 



4 io MACHINES AND HYDRAULICS 

Let P = the average resistance in tons weight. 
Then, Work done against P = P x ^ foot-tons. 

Hence, Px^=u-i8, 

ii-i8x 12 

9 
= 14-9 tons weight. 

Average forces calculated in this manner are described sometimes as 
space-average forces. 

EXAMPLE 2. A vessel of mass 10,000 tons and having a speed of 
30 feet per second is slowed to 10 feet per second in travelling a distance 
of 3000 feet. Calculate the average resistance to the motion. 
Here we have 

. , . . mi>? mv>? 

Change in kinetic energy = -^ ----- r, 

"v> **<S 



10,000 



= 124,100 foot-tons. 

Let P = the average resistance in tons weight. 
Then, Work done against P = P x 3000 foot-tons. 
Hence, 3000 P = 1 24, 100, 

P = 4jj37 tons weight. 

Momentum. The momentum of a body in motion is measured by 
the product of its mass and velocity. The units will be stated by 
giving the units of mass and velocity employed ; thus, if the pound 
and the foot-second units are employed, then 

Momentum = mv pound-foot-seconds. 

Suppose a body of mass m pounds, free to move, to be acted on 
by a force P Ib. weight during a time / seconds, and that the body is 
at rest at first. An acceleration a feet per second per second will be 

produced, such that p w<2 . 

= Ib. weight ........................ (i) 

Since P acts for a time / seconds, the velocity of the body at the end 
of this time will be 

v = at feet per second (p. 388) ; 

i) 
.'. a - feet per second per second. 



MOMENTUM 



411 



And from (i), by substitution, 

mv 
P Ib. weight (2) 

Now mv is the momentum possessed by the body at the end of 
the time t seconds, consequently - will be the momentum it acquires 

each second, i.e. the rate of change of momentum. Hence, the 
force in Ib. weight generating momentum will be numerically equal to the 
rate of change of momentum in pound-foot-second units divided by g. 

Or, we may write F = poundals, (3) 

t 

showing that the force in absolute units is equal to the rate of change 
of momentum. 

Suppose equal forces P, P, to act during the same interval of time 
on two bodies A and B, free to move and initially at rest. Let the 
masses be m A and m E respectively, and let V A and V B be the velocities 
acquired at the end of the time /. 

From (2) above, P = A A , for the body A; 



5 , for the body B ; 
<s* 

*A^A = ^B_B 
g* g* 



It may be stated therefore that equal forces, acting during the same time, 
generate equal momenta irrespective of the masses of the bodies. 

Impulsive forces. Supposing a body in motion to possess a 
momentum mv, which is abstracted by the body encountering a 
uniform resistance P. If this is accomplished in / seconds, then 

P = 
gt 

It will be noticed that if / becomes very small, P will become very 
large, and is then said to be impulsive. If P be not uniform, its 
average value may be found from the above equation. In the case 
of impulsive action, P is called the average force of the blow. 

Change of momentum. Momentum depends on the velocity of 
a body, and, since velocity has direction, momentum will also be a 
directed quality and so can be represented by a vector. Momentum 
differs in this respect from kinetic energy which depends on v 2 - 



412 MACHINES AND HYDRAULICS 

Change of momentum must be estimated always by taking the 
change in the body's velocity, paying attention to both magnitude 
and direction. Having found the magnitude and direction of the 
change in momentum, the force required may be calculated and will 
act in the same line of direction. 

EXAMPLE i. A locomotive picks up a supply of water from a long 
trough laid between the rails (Fig. 445) while travelling at 40 miles per 

hour. Suppose the speed to remain un- 
altered, what additional resistance is offered 
if 5 tons of water be picked up in 50 
seconds ? 

The water in the trough has no momen- 
tum ; after it is picked up it has the same 
velocity as the train, hence 

p_mv_ 5x40x5280 

FIG. 445. Locomotive picking up gt $2-2 X 50 X 60 X 60 

water. . 

= 0-182 ton weight. 

EXAMPLE 2. A gun discharges 350 bullets per minute, each of mass 
0-025 pound, with a velocity of 2000 feet per second. Neglecting the mass 
cf the powder gases, find the backward force on the gun. 




Mass of bullets ejected per second = -|^ **5 =0-146 pound. 
Momentum generated per second 
Force required to eject the bullets 



Momentum generated per second =0-146 x 2000 pound-foot-sec. 

0-146 x 2000 

32-2 
= 9*07 Ib. weight. 

It is evident that the backward force acting on the gun will be equal to 
the force required to eject the bullets, viz. 9-07 Ib. weight. 

EXAMPLE 3. A hammer head of mass 2 pounds and having a velocity 
of 24 feet per second is brought to rest in 0-005 second. Find the average 
force of the blow. m7 j 2 x 24 

~ gt "32-2x0-005 
= 298 Ib. weight. 

Average forces calculated in this manner are described sometimes as 
time-average forces. 

Centre of mass. It will be understood that every particle in a 
body offers resistance, due to its inertia, to any attempted change 
in its velocity. In Fig. 446 is shown a body travelling in a straight 
line towards the left and having an acceleration a. There being 
no rotation of the body, every particle will experience the same 



ROTATIONAL INERTIA 413 

acceleration a. Calling the masses of the particles m v ;#.,, etc., 
the resultant resistance will be 

R = m^a + rn^a + m^a -f etc. 

This force will act through the centre C of the parallel forces 
m^ m. 2 a, etc. (p. 48), a centre which is called the centre of mass of 
the body. It may be assumed for all practical purposes that the 
centre of mass and the centre of gravity of a body coincide. 

Let C be the centre cf mass of a body (Fig. 447), and let R 
acting through C be the resultant inertia resistance. The resultant 
external force F producing acceleration must clearly act in the same 
straight line as R if there is to be no rotation of the body. Hence, 
we have the principle that if the external forces acting on a body free to 





FIG. 446 Centre of mass of a body. FIG. 447. 

move are to produce no rotation, their resultant must pass through the 
centre of mass of the body. The truth of this may be tested easily by 
laying a pencil on the table and flicking it with the finger nail. An 
impulse applied near the end will cause the pencil to fly off rotating 
as it goes ; an impulse applied through the centre will produce no 
rotation. 

Eotational inertia. In Fig. 448 is shown a body which is capable 
of turning freely about an axis OZ perpendicular to the plane of the 
paper. In order to produce rotation, without 
tendency to displace or translate the body, a 
couple must be applied. Let the two forces 
P, P, form a couple, one of the forces being 
applied at the axis, and let the forces rotate 
with the body so that a constant moment 
is exerted. The forces being in Ib. weight 
and the arm D being in feet, the moment F 'G. 4 48.-Rotationai inertia, 
will be T = PD Ib.-feet (i) 

As the body is free to rotate, the only resistance which will be 
opposed to the couple must be due to the inertia of the body 
causing it to endeavour to rotate with uniform angular velocity. For 
inertia resistance to be possible there must be angular acceleration, 




414 MACHINES AND HYDRAULICS 

consequently each particle of the body will have a linear acceleration 
in the direction of its path of motion. 

Considering one such particle m 1 pound at radius r^ feet ; its 
linear acceleration will be a^ feet per sec. per sec., and the resistance 
which the particle will offer is 

/^^Ib. weight. 
(*> 

Now, flj = Or-L , 

where 0is the angular acceleration in radians per sec. per sec. (p. 392). 



. 

To obtain the moment of this resistance, multiply p l by r^ giving 

m+Qrf 

Moment of resistance of particle =/ 1 ^ 1 = 

A 
i\ 

= --Vi 2 ................ (2) 

Now had any other particle been chosen, a similar expression for 
its moment of resistance would result. Hence 
Total moment of resistance due to inertia of body 

= ; r 2W, ................................. (3) 

o 

the summation being taken throughout the body. The quantity 
2mr 2 may be called the second moment of mass, or more commonly, 
the moment of inertia of the body, written I. Using a suffix OZ to 
indicate the axis about which moments must be taken, (3) becomes 

n 

Total moment of resistance = - I oz ................. , . . (4) 

o 

Clearly this moment must be equal to the moment of the couple 
applied to the body. Hence equating (i) and (4) we have 

T = PD = ^ ............................... (5) 

s 

If the couple is measured in absolute units, say L poundal-feet, 
(5) becomes L = I OZ ....... , ............................. (6) 

The analogy between this equation and the corresponding one for 
rectilinear motion may be noted ; viz. 

F = ma ..................................... (7) 

In (7) a force appears on the left hand side, and in (6) the moment 
of a force ; in (7) the product of mass and linear acceleration are on 



MOMENTS OF INERTIA 



4*5 



the right hand side, and in (6) the product of second moment of 
mass or moment of inertia and angular acceleration. 

The following common cases of moments of inertia may be noted 

MOMENTS OF INERTIA. 

The results are all in pound-(foot) 2 units if the mass M is taken in 
pounds and the linear dimensions in feet. 

I. A slender uniform rod. 

(a) Axis OX parallel to rod and at a distance D from it (Fig. 449). 

i ox =MD 2 . 

(b} Axis OX perpendicular to rod through one end (Fig. 450). 

ML 2 



(c) Axis OX perpendicular to rod through its centre of gravity 
(Fig. 451). ML2 





12 






* -L --H j 


IT- T" 


Y 


U.---B -i 


t 


L 






p 


u T 


~X 


H 


V 


V 




| 



O x X 

FIG. 449. FIG. 450. FIG. 451. 



FIG. 452. 



II. A thin uniform rectangular plate. 

(a) Axis OX coinciding with a long edge (Fig. 452). 

! = MH 2 
3 

(^) Axis OY coinciding with a short edge (Fig. 452). 

MB 2 



(c) Axis GX through centre of gravity and parallel to long 
edge (Fig. 453). 






(d) Axis GY through centre of gravity and parallel to short 
edge (Fig. 453)- MB , 






4*6 



MACHINES AND HYDRAULICS 



(e) Axis OZ through one corner and perpendicular to plate 

(Fig- 454). i = M(H^ B 2 ) 

3 

(f) Axis GZ through the centre of gravity and perpendicular to 
plate (Fig. 454). M(H 2 + B 2 ) 

fcr '-^- 

III. A thick uniform plate. 

(a) Axis OY coinciding with one edge (Fig. 455). 

= M(B 2 J 1 T 2 ) 
3 

(b) Axis GZ parallel to OY and passing through the centre of 

gravity (Fig. 455). 

M(B 2 + T 2 ) 



L 



12 





T 


^"i ! 


^ 


S* -B > 




i 




G 


f 

M 

I 







.'**' 


i 


.--' 


^^i 



FIG. 454. 






FIG. 



IV. A thin circular plate. 

(a) Axis OX forming any diameter of the plate (Fig. 456). 

I _ MR2 

4 

(b) Axis TV forming any tangent to the plate (Fig, 456). 

_5MR 2 

1 TV 

(f) Axis OZ passing through the centre and perpendicular to 
the plate (Fig. 456). 

MR 2 



Axis TZ touching the circumference and perpendicular to 
the plate (Fig. 456). 

3 MR2 

2 



MOMENTS OF INERTIA 



417 



V. A thin circular plate having a concentric hole. 

(a) Axis OX forming any diameter of the plate (Fig. 457). 

M(R 1 2 + R 2 2 ) 

~T 

(I)) Axis OZ passing through the centre and perpendicular to 
the plate (Fig. 457). 

Ioz = 




FIG. 457. 



VI. A solid cylinder. 

Axis OX coinciding with axis of cylinder. 

MR 2 
lox- 

VII. A hollow concentric cylinder. 

Axis OX coinciding with axis of cylinder. 

M(R 1 2 + R 2 2 ) 
Iox= 2 

VIII. A solid sphere. 

Axis OX forming any diameter. 

2 MR 2 
IQX- 

The following rules are useful in calculating moments of inertia. 

(a) Given I ox and I y for a thin uniform plate, to find I OZt OZ 
being perpendicular to the plane containing OX and OY : 

IQZ = Iox + IOY- 

(b) Given I GX for a thin uniform plate, GX being an axis passing 
through the centre of gravity, to find I x> OX being parallel to GX 
at a distance D : I ox = J GX + MD 2 . 

(c) Routh's rule : If a body is symmetrical about three axes 
which are mutually perpendicular, the moment of inertia about one 
axis is equal to the mass of the body multiplied by the sum of the 
squares of the other two semi-axes and divided by 3, 4, or 5 according 
as the body is rectangular, elliptical or ellipsoidal. 



D.M. 



2 D 



4 i8 



MACHINES AND HYDRAULICS 



EXAMPLE i. A rectangular plate, as shown in Fig. 455, is symmetrical 
about GZ and other two axes passing through G, and parallel to B and T 
respectively. 

IGZ 



12 



EXAMPLE 2. A solid cylinder (special case of an elliptical body) is 
symmetrical about the axis of the cylinder OX and about other two axes 
forming diameters at 90 and passing through the centre of gravity 
of the cylinder. Hence : 



Iox = 



M(R 2 +R 2 ) 



EXAMPLE 3. A solid sphere is symmetrical about any three diameters 
which are mutually perpendicular. Hence, about one diameter, OX : 

M(R 2 + R 2 ) 2 MR 2 
Iox= 

The radius of gyration of a body is defined as a quantity k such that, 
if its square be multiplied by the mass of the 
body, the result gives the moment of inertia of 
the body about a given axis. Taking the case 
of a solid cylinder as an example, the moment 
of inertia about OZ, the axis of the cylinder, is 

MR 2 




Let 



Then 



l 



= , or 



***, 
/a 



FIG. 458. An experimental 
flywheel. 



which gives the value of the radius of 
for this particular axis. 

EXAMPLE i. A flywheel has a moment of inertia of 8000 in pound and 
foot units, and is brought from rest to a speed of 180 revolutions per 
minute in 25 seconds. What average couple must have acted ? 

Final angular velocity = w = -j X27r = 6ir radians per sec. 
Angular acceleration =0 = - = radians per sec. per sec. 



T== = *m^ =l8? i b ._f eet 
g 25x32-2 

EXAMPLE 2. In a laboratory experiment, a flywheel of mass 100 pounds 
and radius of gyration 1-25 feet (Fig. 458) is mounted so that it may be 
rotated by a falling weight attached to a cord wrapped round the wheel 
axle. Neglect friction and find what will be the accelerations if a body 



KINETIC ENERGY OF ROTATION 419 

of 10 pound weight is attached to the cord and if the radius of the axle is 
2 inches. 

Let M = mass hung on, in pounds. 
Mg-\ts weight, in absolute units. 
T-=pull in cord, in absolute units. 
r= radius of axle, in feet. 
I = moment of inertia of wheel 

= ioox 1-25 x 1-25 = 156-2 pound and foot units. 
= the linear acceleration of M, in feet per sec. per sec. 
0=the angular acceleration of the wheel, in radians per sec. 

per sec. 
Then, considering M, we have : 

Resultant force producing acceleration = Mg-Tg-=Ma (i) 

Considering the wheel, we have : 

Couple producing acceleration = Tgr = 10 (2) 

Also, e=- r . (3) 

These three equations will enable the solutions to be obtained. Thus : 
From (2) and (3), Tgr\- ; 

.'. 1g-\% (4) 

Substitute this value in (i), giving 
T a 



10x32-2 



Mg- 
"", . I ~~ 



io + (i56-2 x6x6) 

=0-0572 feet per sec. per sec. 
From (3), = =0-0572x6 

=0-343 radian per sec. per sec. 

Kinetic energy of rotation. In Fig. 459 is shown a body rotating 
with uniform angular velocity w about an axis OZ 
perpendicular to the plane of the paper. Con- 
sidering one of its particles m ly the linear velocity 
of which is v l , we have 

Kinetic energy of particle = 1 1 . 
Now, v i = <*>r lt 

FIG. 459. Kinetic 
1)-^ = (D f] . energy of rotation. 




420 MACHINES AND HYDRAULICS 

to 2 
Hence, Kinetic energy of particle = m^ (i) 

A similar expression would result for any other particle, hence 

o> 2 
Total kinetic energy of body = 2mr 2 , or, 

Kinetic energy = I oz (2) 

In using this equation with o> in radians per second, g should be in 
feet per second per second and I in pound-(foot) 2 units to obtain the 
result in foot-lb. The corresponding equation for foot-poundals 

would be W 2 

Kinetic energy = - I oz (3) 

EXAMPLE i. A flywheel has a mass of 5000 pound and a radius of 
gyration of 4 feet. Find its kinetic energy at 1 50 revolutions per minute. 
o> = 1 ^> . 27r = 5?r radians per sec. per sec. 
I = M/ 2 = 5000 x 1 6 = 80,000 pound-(foot) 2 . 

o> 2 T 25 x?r 2 x 80,000 

Kinetic energy = I= z 

"2.g 64-4 

;oo foot-lb. 



EXAMPLE 2. The above flywheel slows from 150 to 148 revolutions 
per minute. Find the energy which has been abstracted. 

Change in kinetic energy = ^ -- ^j = (o^ 2 - o> 2 2 ). 

Also, &! = 57T, 

o> 2 = W.27r = 4-93377; 

/. Energy abstracted = (<DI - a> 2 )(a> 1 + o> 2 ) 



= 8160 foot-lb. 

Energy of a rolling wheel. The total kinetic energy of a wheel 
rolling along a road will be made up of kinetic energy of rotation 
together with kinetic energy of translation. 

Let o> = the angular velocity, radians per sec. 

v the linear velocity in feet per sec. of the carriage to which 
the wheel is attached (this will also be the velocity 
of the centre of the wheel). 
M = the mass of the wheel in pounds. 
k its radius of gyration, in feet, with reference to the axle. 



ENERGY OF ROLLING WHEELS 421 

Then, Kinetic energy of rotation = - = foot-lb. 



Kinetic energy of translation = -- foot-lb. 



<o 2 ! 

Total kinetic energy = -- 1 -- .............. (i) 

2.T 2g 

Again, if there be no slipping between the wheel and the road, we 

have (p. 393) v ................................. (j) 

-K 

where R is the radius of the wheel in feet. 

Substituting in ( i ), we obtain, for perfect rolling : 

_ . . . . v*M& Mv* 

Total kinetic energy = =^ + 

- 2 2g 



Energy of a wheel rolling down an inclined plane. The 

principle of the conservation of energy may be applied to the case 
of a body rolling down an inclined plane (Fig. 460). In rolling from 
A to B, the body descends through a vertical height H feet ; hence, 
if M is the mass in pounds, 

Work done by gravity = M^H foot-poundals .............. (i) 

Assuming that none of this is wasted, 
the total kinetic energy at B will be 
equal to the same quantity. The energy 
at B is made up of translational kinetic 
energy owing to the linear velocity v feet 
per second of the mass centre and of 
rotational kinetic energy owing to the FIG. 460. Energy of a wheel rolling 

. , . . , down an incline. 

angular velocity w radians per second. 

Hence, Total energy at B = ( - + I oz j foot-poundals, ....... (2) 

I z being the moment of inertia in pound and feet units about the 
axis of rotation passing through the mass centre of the body. 
Equating (i) and (2), we have 




Writing M> 2 for I z> this will give 

Mz/ 2 eo 2 
------- + 

or ^H = J(^ + o,2#) ......................... (3) 



4 22 



MACHINES AND HYDRAULICS 



If there is no slipping between the wheel and the plane, we have 



v 
= 



where R is the radius of the body in feet. 
Hence, 



or 




(4) 



Motion of a wheel rolling down an incline. The following 
way of regarding the same problem should be studied. Fig. 461 



Couple FR 




Q = My. co set 

Mg.cosa 

FIG. 461. Forces acting on a wheel rolling down an incline. 

shows a body rolling down a plane inclined at an angle a to the 
horizontal. The weight M^ may be resolved into two forces respec- 
tively parallel to and at right angles to the plane ; these will be 
M^sina and M^cosa. The normal reaction Q of the plane will 
be equal to M^cosa. If there is no friction, all these forces act 
through the mass centre O, and there will be no rotation, i.e. the 
body will slip down the plane without any rolling. Suppose that 
a maximum frictional force F may act between the plane and the 
body and that /A is the coefficient of friction, then 

F = fJ<Mg cos a. 

To investigate the effect of F, transfer it to the mass centre O as 
shown and apply an anti-clockwise couple of magnitude FR. Then 

P = Resultant force at O acting down the plane 
= Mg sin a - F 
= M^sin a - /xM^cos a. 



WHEEL ROLLING DOWN AN INCLINE 423 

Let a be the linear acceleration of O down the plane. Then 



or M^-(sin a - /x cos a) = M#, 

.'. a=g(sma -/x cos a) ................... (5) 

Also, owing to the couple FR, an angular acceleration 6 will be 
produced, to be obtained from 



M/fc 2 



If the rolling is perfect, i.e. no slip, we have 

9- a 

Hence, from (5) and (6), 

g / . v wRcosa 

^- (sin a - /z cos a) =g' I ' 

uR 2 cos a 
Sin a - [JL COS a = ! -; , 



Sina = /xcosa(-^ 

tan a . . 

~A2 "*" ^ 

This expresses the minimum value of the coefficient of friction 
consistent with perfect rolling. Ass'uming that the rolling is perfect, 
the value of the linear acceleration a may be calculated as follows : 

From (6), p cos a = = ^ 

a & /0 \ 



Substituting this value in (5) gives 



..'sin a 
. . a 



424 MACHINES AND HYDRAULICS 

Suppose that the body starts from rest at A (Fig. 462) and rolls 
^ to B. The linear velocity v of the mass 

' A centre at B may be calculated thus, 




FIG. 462. 



?. 388). 
Also, y- = sin a ; or, L 
H 



sin a 
Inserting the value of a from (9), we have 

2^ sin a H 2^-H 

1r = W~ = Z5 J 

* sin a k z 

I + R2 I+ R2 



Comparison of (4) with (10) will show that the same result has 
been obtained by both methods. 

EXAMPLE. In a laboratory experiment, a small steel ball was allowed 
to roll down a plane of length 6 feet and inclination i 40'. The average 
time taken (six experiments) was 4-25 seconds. Compare the experimental 
and calculated accelerations of the ball. 

To obtain the experimental acceleration, we have 

s = \a^ 
where s is the length of the incline and t is the time of descent. Hence, 

_2_ 2x6 
* 1 ~72~4.25X4-25 

= 0-664 feet per sec. per sec. 
To calculate the acceleration, take equation (9), p. 423. 



For a ball, 



Hence, 

= f x 32-2 x 0-02908 

=0-669 feet per sec. per sec. 

The experimental and calculated accelerations differ by about three- 
quarters of one per cent.; the agreement is good. 




CENTRIFUGAL FORCE 425 

Centrifugal force. It has been shown (p. 397) that, when a small 
body moves in the circumference of a circle of radius R feet with 
uniform velocity v feet per second (Fig. 463), there is a constant 
acceleration towards the centre of the circle given by 

a = ^- feet per sec. per sec. 

To produce this acceleration requires the application of a uniform 
force P, also continually directed towards the centre of the circle, 
and given by ma 



or. 



g 

ft. weight (i) 



This force overcomes the inertia of the body, which would other- 
wise pursue a straight line path, and may be called the central force. 





FIG. 463. Central and centri- FIG. 464. Resultant 

fugal forces. centrifugal force. 

It is resisted by an equal and opposite force Q (Fig. 463), produced 
by the inertia of the body. Q is called the centrifugal force. 

Expressed in terms of the angular velocity w radians per second, 



--. , 

2 

p = . mR lb. weight ................... (2) 

A large body rotating about an axis may be considered as being 
made of a large number of small bodies ; for each of these, rf/g will 
be the same, hence the total central force will be 

2 

P = ~ ( W 1 R 1 + W 2 R 2 + W 3 R 3 + etC ')' 
o 

The quantity inside the brackets would have the same numerical 
value if the whole mass were concentrated at the centre of mass. 
Let M = mass of whole body, in pounds, 

Y = radius of the centre of mass, in feet (Fig. 464). 



4 26 



MACHINES AND HYDRAULICS 



-(3) 



Then P = Q = MY lb. weight 

g 

= 2 MY poundals (3') 

It follows from this result that if a body rotates about an axis 
passing through its centre of mass (in which case Y = o), there will be 
no resultant pull on the axis due to centrifugal action. There may 
be a disturbance set up if the body is not symmetrical about an axis 
at right angles to the axis of rotation, and passing through the centre 
of mass. For example, in Fig. 465 is shown a rod rotating about an 







FIG. 465. An unsymmetrical load produces rocking couples. 

axis GX, G being the centre of mass. The rod is not symmetrical 
about GY, hence, considering the halves separately, there will be 
centrifugal forces as shown by Q, Q, forming a couple tending to 
bring the rod into the axis GY. If this tendency is to be balanced, 
forces S, S, forming an equal opposite couple must be applied by the 
bearings. These forces will, of course, rotate with the rod and 
produce what is called a rocking couple. In Fig. 466 is shown a 





FIG. 466. A balanced 
symmetrical body. 



FIG. 467. Balancing a piece of 
work in a lathe. 



body symmetrical about GY and consequently having neither rocking 
couple nor resultant centrifugal force. 

In Fig. 467 is shown the face plate of a turning lathe with a piece 
of work B attached to the face plate by means of an angle plate C. 



EFFECTS OF CENTRIFUGAL FORCE 



427 



-ThL"^-- R --- 



To the other side of the face plate is attached a balance weight 
which is adjusted until there is no tendency to rotate the spindle 
of the lathe from any position of rest, i.e. the centre of gravity 
of the whole falls on the axis of rotation. This is called static 
balancing and will serve very well for low speeds. It will be 
observed, however, that the bodies attached to the face plate are not 
symmetrical. G l and G 2 , the centres 
of gravity of the work and of the balance 
weight, are not in the same vertical 
line, hence the centrifugal forces P x 
and P 2 , being equal, form a rocking 
couple which will set up troublesome 
vibrations if the speed be increased. 
The effect may be reduced by having 
the balance weight further from the 
face of the plate. 

In Fig 468 is shown a motor car travelling in a curved path. To 
prevent side slipping, the road is banked up to such an extent that 
the resultant Q of the centrifugal force and the weight falls perpen- 
dicularly to the road surface. 

Let M = mass of car, in pounds. 

v = its speed, in feet per sec. 
R = radius of curve, in feet. 




FIG. 468. Section of a banked 
motor track. 



Then, Weight of car = M^ poundals. 
Centrifugal force = -=r- poundals. 

Also, ABG is the triangle of forces. Hence, 
Centrifugal force _ M^ 2 _ v 2 _ AB 



Now, 



Weight of car 
AB 
BG 



EG 



= tan a, 



and a is also the angle which the section of the road surface makes 
with the horizontal ; hence, 



Railway tracks are also banked up in a similar manner ; the super- 
elevation of the outer rail prevents the flanges of the outer wheels 
grinding against the rail in rounding a curve. 



428 MACHINES AND HYDRAULICS 



EXERCISES ON CHAPTER XVII. 

1. A body of mass 200 pounds has an acceleration of 150 feet per 
second per second at a given instant. Calculate the resistance due to 
the inertia of the body. 

2. A resultant force of 1220 dynes acts on a body of mass 1-25 grams. 
Calculate the acceleration in cm. per sec. per sec. 

3. A train has a mass of 250 tons, and starts with an acceleration of 
i-i feet per second per second. Frictional resistances amount to n Ib. 
weight per ton. Find the pull which the locomotive must exert. 

4. A body slides down a plane inclined at 20 degrees to the horizontal. 
The coefficient of friction is o-i ; find the acceleration and the time taken 
to travel the first 20 feet. 

5. A load of 10 pounds is attached to a cord which exerts a steady 
upward pull less than 10 Ib. weight. Starting from rest, the load is found 
to descend 6 feet vertically in 4 seconds. Find the pull in the cord. 

6. A shot has a mass of 20 pounds and a speed of 1500 feet per 
second. Find its kinetic energy in foot-tons. Supposing an obstacle to 
be encountered and that the shot is brought to rest in a distance of 
12 feet, what is the average resistance? 

7. Calculate the momentum of the shot given in Question 6. Suppose 
that the shot had been brought to rest in 0-02 second, and calculate the 
average resistance. 

8. A man stands on a small truck mounted on wheels which are 
practically frictionless. If the man jumps off at the rear end, what will 
happen to the truck ? Take the masses of the man and the truck to be 
150 and 200 pounds respectively, and assume that the man is travelling 
at 8 feet per second immediately he has left the truck. 

9. A jet of water delivers 50 pounds of water per second with a 
velocity of 35 feet per second. The jet strikes a plate which is fixed with 
its plane at 90 degrees to the jet. Find the pressure on the plate. 

n 10. Suppose in Question 9 that the plate had been curved in such 
v a manner that the jet slides on to it and has the direction of its velocity 
on leaving the plate inclined at 90 degrees to its original direction. Find 
the change in velocity, and hence find the pressure on the plate. 

11. A wheel has a moment of inertia of 10,000 in pound and feet units, 
and is brought from rest to 200 revolutions per minute in 25 seconds. 
Calculate what steady couple must have acted on it. 

12. An iron plate 4 feet high, 2 feet wide and 2 inches thick is hinged 
at a vertical edge. Calculate its moment of inertia about the axis of the 
hinges. Take the density of iron to be 480 pounds per cubic foot. 

13. Find the moment of inertia about the axis of rotation of a hollow 
shaft 20 inches external and 8 inches internal diameter by 60 feet long. 
Take the density as given iri Question 12. 

14. A solid ball of cast iron is 12 inches in diameter ; density of metal 
450 pounds per cubic foot. Find the moment of inertia about an axis 
which touches the surface of the ball. 



EXERCISES ON CHAPTER XVII. 429 

15. Referring to Question 5 : The upper part of the cord is wrapped 
round a drum 6 inches diameter measured to the cord centre, and a 
flywheel is attached to the same shaft as the drum. Find the moment 
of inertia of the flywheel. 

16. A solid disc of cast iron is 4 feet in diameter and 6 inches thick, 
and rotates about an axis at 90 degrees to its plane and passing through 
its centre. For the density, see Question 14. Speed 150 revolutions per 
minute. Find the radius of gyration and the kinetic energy of the disc. 

17. If the disc given in Question 16 slows to 140 revolutions per 
minute, how much energy will be given up ? 

18. Suppose that the disc given in Question 16 were to roll without 
slip down an incline of I in 10, what would be the linear acceleration of 
its centre ? 

19. A blade of a small steam turbine has a mass of 0-05 pound and 
revolves in the circumference of a circle 8 inches in diameter 24,000 times 
per minute. Find the centrifugal force. 

20. An oval track for motor cycles has a minimum radius of 80 yards, 
and has to be banked to suit a maximum speed of 65 miles per hour. 
Find the slope of the cross section at the places where the minimum 
radii occur. 

21. A tramcar weighs 12 tons complete. Each of the axles with its 
wheels, etc., weighs 0-5 ton, and has a radius of gyration of I foot. The 
diameter of the wheel tread is 3 feet, and the car is travelling at 12 miles 
per hour. Find (a) the energy of translation of the car ; (b) the energy 
of rotation of the two axles ; (c) the total kinetic energy of the vehicle. 

(B.E.) 

22. Prove the formula for the acceleration of a point moving with 
uniform speed in a circle. Find in direction and magnitude the force 
required to compel a body weighing 10 Ib. to move in a curved path, the 
radius of curvature at the point considered being 20 feet, the velocity of 
the body 40 feet per second, and the acceleration in its path 48 feet per 
second per second. (I.C.E.) 

23. A motor car, whose resistance to motion on the level is supposed 
to be the same at all speeds, has been running steadily on the level at 
20 miles per hour; it now gets into a rise of i in 12. What is the 
maximum length of this rising road which may be traversed by the car 
without changing gear ? (B.E.) 

24. A train weighing 300 tons, travelling at 60 miles per hour down a 
slope of i in no, with steam shut off, has the brakes applied and stops in 
450 yards. Find the space-average of the retarding force in tons exerted 
by the brakes ; if the time that elapses between the putting on of the 
brakes and the moment of stopping is 36 seconds, find the time-average 
of the retarding force in tons. (I.C.E.) 



CHAPTER XVIII. 

INERTIA CONTINUED. 

Angular momentum. The angular momentum or moment of momentum 
of a particle may be defined by reference to Fig. 469. A particle of 
mass m pounds revolving in a circle of r feet in 
n radius has a linear velocity of v feet per second 
' r/ " \ at any instant in the direction of the tangent. 

/ / \ Hence its linear momentum at any instant will be 

i -t****2 Linear momentum of particle = mv, 

t O i , 

; ' and v = ur\ 

\ / .'. Linear momentum of particle = umr. (i) 

v ^ -'' The moment of this about OZ (Fig. 469) may 

FIG. 469 -Angular mo- be obtained by multiplying by r> the result being 
called the moment of momentum, or angular 
momentum of the particle. 

Angular momentum of particle = umr 2 (2) 

A body having many particles would have a similar expression for 

each. Hence, v 

Angular momentum of a body = vtLmr 1 

I. (3) 

Consider now a body free to rotate about a fixed axis, and, 
starting from rest, acted on by a constant couple T Ib.-feet. The 
constant angular acceleration being 0, we have, as in equation (5) 
P- 4H, T ^I OZ 

Let T act during a time / seconds, then the angular velocity w at 

the end of this time will be 

(j) 
w = 0t, or, 9 = -. 

Hence, T = " * Ib.-feet, (4) 

gt 

or L^^'poundals-feet (5) 



r.YKOSTATIC ACTION 



431 



Now, (ol (}7 is the angular momentum acquired in the time /seconds, 
hence <ol oz // will be the gain of angular momentum per second. 
We may therefore state that the couple in Ib.-feet acting on a body free 
to rotate about a fixed axis is numerically equal to the rate of change of 
angular momentum divided by g ; or, omitting g, the couple will be in 
poundal-feet. This statement should be compared with those for 
linear momentum given on p. 411. 

It will be evident that the applied couple must be acting in the 
plane of rotation of the body ; should this not be the case, then 
rectangular components of the couple should be taken, and. that 
component which is in the plane of rotation used in applying 
equation (4). 

Gyrostatic action. In Fig. 470 is shown a cycle wheel suspended 
by means of a long cord attached at C to one end of the bearing pin. 




n 



i-Q w. 



FIG. 470. A cycle wheel showing 
gyrostatic action. 



FIG 471. Angular velocities of 
the wheel shown in Fig. 470. 



If the wheel be at rest, it cannot maintain the position shown without 
assistance, but, if set revolving, it will be found to be capable of 
maintaining its plane of revolution vertical. It will be noticed, 
however, that the wheel spindle slowly revolves in azimuth, i.e. in 
a horizontal plane ; the vertical plane of revolution of the wheel will, 
of course, be always perpendicular to the wheel axis. The effect is 
owing to the action of the couple formed by the equal forces T, the 
pull of the cord, and Mg, the weight of the wheel. This couple acts 
in a vertical plane containing the wheel axis, and will produce 
changes in the angular velocity of the wheel ; these changes must 
occur in the plane containing the couple. 

In Fig. 471 is shown a plan of the wheel ; as it is revolving clock- 
wise when viewed from the right hand side, Oa may be drawn to 



432 



MACHINES AND HYDRAULICS 



represent w, the angular velocity of the wheel in its present position. 
The couple acting is clockwise when viewed from the front side, 
hence ab will represent the change of angular velocity occurring in 
a brief interval of time (p. 401). Hence the angular velocity of the 
wheel at the end of the interval will be represented by Ob. The 
vertical plane of revolution of the wheel will turn from the position 
OA to OA' during the interval, and the wheel spindle which 
occupied the position Oa at first will revolve clockwise when viewed 
from above. 

Let L = the couple applied, poundal-feet. 

I = the moment of inertia of the wheel about its axis, pound 

and foot units. 
Wj = the angular velocity of the wheel about its axis, radians 

per second. 
co 2 = the angular velocity of the wheel axis in the horizontal 

plane, radians per second. 

/=the time in which the axis makes a complete revolution in 
the horizontal plane, seconds. 

Then, in one horizontal revolution of the axis, change in angular 
velocity of the wheel = 2Tro> r (p. 402.) 

Also, L- 




*I. (p. 414.) 



Also, 



= 27T 



In Figs. 470 and 471, 
L = M^xCO; 



R 



or 



(*) 



where CO is the horizontal distance 
between the centre of gravity of the 
wheel and the suspending cord. 
; ; Gyrostatic action in motor cars. 

FIG. 472. Gyrostatic action in an ordinary In Fig. 472 IS shown a motor Car 

travelling round a curve ; when 

looked at from the front, the engine flywheel has a clockwise angular 
velocity. Let OA represent the angular velocity of the flywheel 
when the car is in the position shown, and let OB represent the 



GYROSTATIC ACTION 



433 



angular velocity after a short interval of time ; then the change in 
angular velocity will be represented by AB, and indicates that a 
clockwise couple must be acting on the car as seen in the side 
elevation. This couple can come only from the reactions of the 
ground, hence the front wheels are exerting a greater pressure and 
the back wheels a smaller pressure than when the car is running in 
a straight line. If the car is turning towards the right instead of 
towards the left, there will be an increase in pressure on the back 
wheels and a, diminution in pressure on the front wheels; the 
student should make a diagram of this case for himself. 

Let P = the change in pressure on each axle of the car, in poundals. 
Wj = the angular velocity of the engine, in radians per second. 
I = the moment of inertia of the revolving parts of the engine, 

in pound and foot units. 
V = the velocity of the car, in feet per second. 
R = the radius of the curve, in feet. 
w 2 = V/R = the angular velocity in azimuth of the engine shaft, 

radians per second. 

D the distance, centre to centre of the wheel axles, in feet 
Then, \ Couple acting = 



P = ~Fyp- poundals 

-Jg height 

In Fig. 473 is shown a car having a wheel at O rotating in a 



<f QJ 







FIG. 473. Gyrostatic action of a revolving wheel in a car. 

vertical plane parallel to the planes of revolution of the back wheels 
of the car. In the side elevation the wheel rotates clockwise ; the car 
D.M. 2 K 



434 



MACHINES AND HYDRAULICS 



is shown in plan turning towards the right. Oa will represent the 
initial angular velocity of the wheel, Qb represents the angular velocity 
after a brief interval, and ab represents the change in angular 
velocity during this interval. Viewed from the front of the car, 
the change in angular velocity is anti-clockwise, hence an anti-clock- 
wise couple P, P, must act on the wheel in a vertical plane con- 
taining the wheel axis. This will give rise to an equal opposite 
couple Q, Q, acting on the car, and will cause the wheels at AA to 
exert greater pressure on the ground and those at BB to tend 
to lift. It will be noted that in this case both centrifugal force and 
gyrostatic action conspire to upset the car. If the wheel at O be 
made to rotate anti clockwise, the gyrostatic couple will have the 
opposite sense to that in Fig. 473, and will tend to equilibrate the 
effects of centrifugal force on the car. The student should sketch 
the diagram for this case, and also for the case of the car turning 
towards the left. 

Further points regarding gyrostatic action. A wheel revolving 
in a given plane may be shifted to any parallel plane without any 




FIG. 474. An experimental gyrostat. 



gyrostatic action being evidenced. This is in consequence of such a 
change in position being unaccompanied by any change in the 
angular velocity of the wheel ; hence there is no change in angular 



GYROSTATIC ACTION 



435 



momentum, and therefore no couple is required. A couple is 
required in every case where the new plane of revolution is inclined 
to the initial plane. 

In Fig. 474 is shown a common form of gyrostat by use of which 
useful information regarding the behaviour of gyrostats may be 
obtained. The revolving wheel A rotates on a spindle BC, the 
bearings of which at B and C are formed in a ring which has freedom 
to rotate about an axis DE. The ring has bearings at D and E in 
another semicircular ring DFE, which has freedom to rotate about 
the vertical axis FG. The spindle FG is dropped into a vertical 
hole in a heavy stand. The effect of a weight W hung from C will 
be to cause the axis BC to rotate in a horizontal plane, accompanied, 
of course, by the whole frame ; the direction of this rotation, as 
viewed from above, will be either clockwise or anti-clockwise, depend- 
ing on the direction of rotation of the wheel. 

It will be noted that the original vertical plane of rotation gradu- 
ally becomes inclined to the vertical as the motion goes on. This is 
owing to the action of a horizontal couple acting 
on the frame, and produced by the frictional 
resistance offered by the stand to the rotation 
of the spindle FG. In Fig. 475 an elevation of 
the wheel is shown, rotating in the vertical plane 
OA. Oa represents the angular- velocity of the 
wheel ; ab represents the change in angular velocity 
in a given interval of time in the horizontal plane 
containing the wheel axis, and produced by the frictional couple. 
Qb is the altered angular velocity of the wheel at the end of the 
interval. The wheel will now be rotating in the plane OA', per- 
pendicular to Ob and inclined at an angle AOA' to the vertical. 

While the wheel is rotating, if the free motion of the semi-circular 
ring DFE be impeded by application of a finger, it will be noted that 
the wheel turns over. This effect is precisely the same as the effect of 
the frictional couple, only it is more marked, as the horizontal couple 
produced by the finger is larger than that produced by the friction. 
If DFE be held forcibly from rotating, the wheel will assume 
a horizontal plane of rotation instantly. In fact, the wheel is only 
capable of exerting a couple which will equilibrate the couple applied 
by means of W, provided its motion in azimuth is allowed to take 
place freely. 

Schlick's anti-rolling gyrostat. Fig. 476 illustrates in outline 
the method used by Schlick for reducing the rolling of a ship 




436 MACHINES AND HYDRAULICS 

among waves. The view is a cross section of the ship ; A is a 
heavy wheel revolving in a horizontal plane about the axis BC 
The frame in which the wheel rotates can rock about a hori- 
zontal axis DE, which works in bearings secured to the ship's 
frames. DE is perpendicular to the direc- 
tion of length of the ship. When the ship 
B rolls, the axis DE is forced out of the hori- 

zontal, and the axis BC will be inclined 
by either B or C coming out of the paper. 
A couple, applied by vertical forces acting 
at D and E, is required to give the wheel 
FIG. 47 6. Schiick's anti-rolling frame this motion, and an equal opposite 

gyrostat. . n . 

couple acts on the ship, tending to give 

to it a motion opposite to the rolling motion. In consequence of 
this reaction on the ship, the rolling effect is made much smaller. 
Freedom of motion about DE must be provided, otherwise the wheel 
is incapable, as has been shown above, of offering any resistance to 
rolling. 

There are many other applications of the principle of the gyrostat, 
such as in the gyro-compass used on board ships, in the Brennan 
monorail cars, and in steering torpedoes. 

Simple harmonic vibrations. It has been shown (p. 398) that a 
body, in describing simple harmonic vibrations, possesses at any 
instant an acceleration directed towards the centre of the vibration, 
and proportional to the distance of the body from the centre of the 
vibration. A force will be necessary in order to produce this acceler- 
ation, and the force will evidently follow the same law as the 
acceleration, i.e. it will be constantly directed 

towards the centre of the vibration, and K-- R ---- +{ 

will be proportional to the distance of the _ i 9 < F * 
body from the centre. B O m A 

( In Fig._ 477 a body of mass m pounds Fir " 477 \^^ harmonic 
vibrates with simple harmonic motion in the 

line AB. Let v feet per second be the velocity when the body is 
passing through the centre O, then the accelerations at A and B will be 

2,2 

a i=f ^ feet per second per second, 

R being the length of OA in feet (p. 397). 

Let Fj be the force in poundals required at A and B, then 



= -=r- poundals ............................ ( i ) 



SIMPLE HARMONIC VIBRATIONS 437 



Supposing the body to be situated at C, its acceleration a may be 
found from ^_OC_OC 

<^~OA~ll ' 

OC 

.. a = ai . 

Also, the force F required to produce the acceleration may be 
found from jr QC OC 

F! = OA = ^: ; 

. F _ F OC_^ OC 
*' R ~ R ' R 

mv^ ~~ 
= ~RZ~ poundals (2) 

Suppose OC to be one foot, and that /x represents the value of the 
force required when the body is at this distance from O, then 

JiT P un dals (3) 

The time of one complete vibration from A to B and back to A is 
given by (p. 400) R 

T = 27T . 
V 

From (3), -*-=; 

. R Im 



Substitution of this value gives 

T = 2irA/ seconds (5) 

EXAMPLE. A body of mass 2 pounds executes simple harmonic 
vibrations. When at a distance of 3 inches from the centre of the 
vibration, a force of 0-4 Ib. weight is acting on it. Find the time of 
vibration. 

The force required at a distance of one foot from the centre will be 
four times that required at 3 inches. Hence, 

fji = 0-4 x 4 = i -6 Ib. weight 
= i -6 g poundals. 



2X22 

Hence > 



1-238 seconds. 



438 MACHINES AND HYDRAULICS 

Simple harmonic torsional oscillations. A body will execute 
simple harmonic torsional oscillations if it is under the influence of a 
couple which varies as the angle described by the body from the 
mean position, the couple having a sense of rotation always tending 
to restore the body to the mean position. Thus, a body secured to 
the lower end of a vertical wire, the upper end of which is fixed 
rigidly, will hang, when at rest, in a position which may be described 
as the mean position. As has been explained on pp. 255 and 295, if the 
wire be twisted by rotation of the body, it will exert a couple which will 
be proportional to the angle of twist ; this couple will be constantly 
endeavouring to restore the body to the mean position, hence the 
body will describe simple harmonic torsional oscillations. The time 
of vibration may be deduced by analogy from equation (5), p. 437, 
showing the time of simple harmonic rectilinear vibrations ; the 
moment of inertia of the body about the wire axis must be substituted 
for m, and the couple acting at unit angle (one radian) from the mean 
position must be substituted for /x. Thus 

Let M = the mass of the body, in pounds. 

k = its radius of gyration about the axis of vibration, in 
feet. 

I = M/fc 2 = the moment of inertia about the same axis, in pound 
and foot units. 

A = the couple acting at one radian displacement from the 
mean position, in poundal-feet. 

T = the time elapsing between successive passages of the 
body through the same position. 



Then, T = 2^ 

V 



Mk 2 
2-n-A / seconds. 

A 



EXAMPLE. A flywheel having a mass of 1000 pounds and a radius of 
gyration of 2 feet, is fixed to the end of a shaft 4 feet long and 3 inches in 
diameter. It has been found from a separate calculation that the shaft 
has an angle of twist of 0-0005 radian when a torque of 1000 Ib.-inches is 
applied. Find the time of a free torsional oscillation. Take ^-=32-2. 

The term " free " indicates that the frictional effects of the bearings and 
of the atmosphere are to be disregarded. 



= looo x 2 x 2 = 4000 pound and foot units. 



SIMPLE PENDULUM 439 



The angle of twist is proportional to the torque, and if this were true up 
to one radian, we have 

Torque at one radian _ i . 
looo "0-0005 ' 



.'. Torque at one radian = 



0-0005 
= 2,000,000 Ib.-inches ; 

= 2,000,000 Xg 

12 

= 5)3^7,000 poundal-feet. 



Hence, 



= 44 J~ 4ocx 
7 ^5>3 6 7,< 



Suppose n to be the number of torsional oscillations per minute ; then 

60 
= o-i7is 



If this shaft were driven by means of an engine connected to a 
crank fixed to the shaft at the end remote from the flywheel, and if 
the shaft were to have a speed of 350 revolutions per minute, the 
engine would be delivering impulses to the shaft which would keep 
time with the free oscillations of the shaft. In 
these circumstances, the angle of oscillation would 
rapidly increase in magnitude. As the stress in 
the shaft is proportional to the angle of twist, a 
very large stress would be produced and the shaft 
would be in danger of breaking. A somewhat 
higher or lower speed of revolution is necessary 
in order to avoid these effects ; in no case should 
the impulses given to the shaft synchronise with 
the free torsional oscillations. 

V tTli 

The simple pendulum. A simple pendulum may 
be realised by suspending a small heavy body at FIG. 47 8.-A simple 
the end of a very light thread and allowing it to 
vibrate through small angles under the action of gravity. In Fig. 478 
the body at B is under the action of its weight mg and the pull T 
of the thread. The resultant of these forces is F, a force which 




440 



MACHINES AND HYDRAULICS 



is urging the body towards the vertical. The triangle of forces 
will be ABD, and we have 

JF = BD 

mg~ A& 



T7 BD 

F = m? --T-. =r- 



or 



Now, if the angle BAD is kept very small, AC and AD will be 
very nearly equal. Let L be the length of the thread in feet ; then 

F = w <r =-^-BD. ...(i) 

Hence we may say that F is proportional to BD. For very small 
angles of swing BD and BC coincide practically, therefore the body 
will execute simple harmonic vibrations under the action of a force F 
which varies as the distance from the vertical through A. To obtain 
the value of /*, the force at unit distance, make BD equal to one foot 
in(i); then 

poundals. 



Now, 




= (p. 437) 



EXAMPLE. 



Find the time of vibration of a simple pendulum of length 
4 feet at a place where g is 32 feet per 
second per second. 

2X22 

7 
= 2-222 seconds. 



The compound pendulum. Any 

body vibrating about an axis under the 
action of gravity and having dimensions 
which do not comply with those re- 
quired for a simple pendulum may be 
called a compound pendulum. In 
Fig. 479 (a) is shown a compound 
pendulum consisting of a body vibrating 




FIG. 479. A compound pendulum and 
an equivalent simple pendulum. 



COMPOUND PENDULUM 441 

about A. G is the centre of mass of the body, and the line AG 
makes an angle a with the vertical passing through A in the position 
under consideration. In Fig. 479 (b} is shown a simple pendulum 
CD vibrating about C ; at the instant considered CD makes the 
same angle a with the vertical passing through C. Both pendulums 
will execute small vibrations in the same time provided that their 
angular accelerations in the given positions are equal. 
Considering the compound pendulum, 
Let M = its mass, in pounds. 

Y = the distance AG, in feet. 

I A = M^A = its moment of inertia about A, in pound and 

foot units. 

1 = the angular acceleration in radians per sec. per sec. 
in the given position. 



r~, n couple applied M^ x GB 

Inen "\ = ~ ^r = - 2 

IA M/ A 

_"x GAsina . . 

~~# '^ 

A 

Considering now the simple pendulum, 
Let m = its mass, in pounds. 
L = its length, in feet. 
I c = wL 2 = its moment of inertia about C, in pound and 

foot units. 
2 = its angular acceleration in radians per sec. per sec. in 

the given position. 
Then couple applied 



mgx DE _g x DC sin a 
L 2 



L 
To comply with the required conditions, we have 



. N 







.(3) 



442 



MACHINES AND HYDRAULICS 



The length L of the corresponding simple pendulum may be 
calculated from this result, and hence the time of vibration of both 
pendulums may be found. If AG be produced to Z (Fig. 479 (#)), 
making AZ equal to L, the point so found is called the centre of 
oscillation. The centre of oscillation may be denned as the point at 
which the whole mass of a compound pendulum may be concen- 
trated without thereby altering the time of vibration. 

Centre of percussion. If a body is capable of rotating freely 
about a fixed axis, it will be found that, in general, a blow delivered 
to the body will produce an impulse on the axis. 
There is, however, one point in the body at which 
a blow will produce no impulse on the axis ; this 
point is called the centre of percussion. 

In Fig. 480, C is the axis about which the body 
may turn freely and G is the centre of mass. Let 
an impulse F be delivered to the body at a point Z. 
The effects of F may be examined by transferring 
F to the centre of mass, applying at the same time 
a clockwise couple of moment F x GZ. The force 
F acting at G will produce pure translation, and 
if the mass of the body is M pounds, every point in it will have an 
acceleration a l feet per second per second, found from 




FIG. 480. Centre of 
percussion. 



or 



M 



In particular, C will have this acceleration a^ towards the left. 
Further, the couple F x GZ will produce a clockwise angular 
acceleration 6. found from T? v r-y 

ya JT X \J/_J 

~~IG~~* 

where I G is the moment of inertia of the body about an axis passing 



through G and parallel to the axis at C. 
moment of inertia, we have 

FxGZ 

: ~ 



Writing M/ G for this 



As a consequence of this angular acceleration, C will have a linear 
acceleration a 2 feet per second per second towards the right, to be 
found from a ., = x CG 

F x GZ x CG 



EQUIVALENT DYNAMICAL SYSTEMS 443 

If there is to be no impulse on the axis at C, there must be 
equality of a t and 2 . Hence, 

F = FxGZxCG 

GZxCG 

or i= - z ; 

k G 

.'. c = GZxCG (3) 

Also, I c = I c + M . CG 2 , (p. 417.) 

2 o 

.*. G = c-CG 2 (4) 

Also, GZ = CZ-CG (5) 

Substituting these values in (3) gives : 

/ C -CG 2 = (CZ-CG)CG, 



Comparison of this result with that found for the position of the 
centre of oscillation (p. 441, equation (3)), indicates that the centre 
of percussion of a body coincides with the centre of oscillation. 

Reduction of a given body to an equivalent dynamical system. 
It is often convenient to substitute for a given body two separate 
bodies connected by means of 
an imaginary rigid rod, and 
arranged in such a way that the 
substituted bodies behave under 
the action of any force or forces 
in exactly the same manner as 

the given body. In Fig. 481 (a) (ft-)- . / 

is shown a body of mass M, /L\ 

and having its centre of mass at FlG> 48l ._ Equivalent dynamical system . 
G ; Fig. 48 1 (I)) shows an equi- 
valent system, consisting of two bodies at A' and B', having masses 
m l and m 2 respectively, and having their centre of mass at G'. A 
and B in Fig. 481 (a) correspond with A' and B'. The conditions 
of equivalence may be stated as follows : 

(i) The mass M must be equal to the sum of m^ and m 2 , and the 
points G and G' must divide AB and A'B' respectively in the same 
proportion. A force applied at G or at G' will then produce pure 




444 MACHINES AND HYDRAULICS 

translation with equal accelerations in the given body or in the 
substituted system. Hence, 

m 1 + m 2 = 'M t (i) 

m l x AG = #z 2 x BG, 
or m^a = mj) (2) 

(2) The moment of inertia of the given body about any axis pass- 
ing through its centre of mass must be the same as the moment of 
inertia of the substituted system about a similarly situated axis. 
This condition ensures that the given body and the substituted 
system shall possess equal angular accelerations when acted on by 
equal couples. Hence, 

m l d 2 + m i >P = Ukl (3) 

These equations may be reduced as follows : 

b a 

From (2), m^-m^ m 2 = ~l m \- 

Substituting these in (i) gives : 

m 1 + -rm l = M ; 

M M 



Also, - 



-J-s 



M Ma 


a 



Inserting these values in (3), we have 



or 



.' ab = & (6) 

The required equivalent system may thus be obtained by first 
selecting a. b may then be calculated from (6), having first deter- 
mined the value of ^ G . m^ and m% may now be calculated from (4) 
and (5). 



EQUIVALENT DYNAMICAL SYSTEMS 



445 



EXAMPLE. A connecting rod (Fig. 482) 4 feet long has its centre of 
mass G at 2-8 feet from the small end. The mass of the rod is 200 



u 3-5 





m, 

<P 



rb*?- ------- a ------ * 

FIG. 482. Equivalent dynamical system for a connecting rod. 

pounds, and an equivalent system is required in which one of the two 
masses is to be situated at the small end. k* G is 2 in foot units. Find the 
system. 

Here a is 2-8 feet ; hence, from (6), 



/ 
From (4), 



From (5), 



=0714 foot 



200 X 0-7 14 



200 X 0-7 14 
3-514 



= 40-6 pounds. 



Ma 



200X2-8 



3-5I4 
= 159-4 pounds. 

If it is desired to give the body shown in Fig. 481 (a) any assigned 
motion, the forces required may be obtained as follows : Find the 
linear accelerations at A and B, both in direction and magnitude ; 
let these be a l and a 2 respectively. In Fig. 481 (<), showing the 
equivalent system, m l and m 2 will have accelerations <Zj and a 2 
respectively, and forces will be required acting in the lines of these 
accelerations, and given by 



both in absolute units. The same forces applied at A and B respec- 
tively in Fig. 481 (a) will give the proposed motion to the body. 



44 6 



MACHINES AND HYDRAULICS 



EXPT. 43. The law F = ma may be verified roughly by means of 
the apparatus illustrated in Fig. 483. A and B are two similar scale 
pans connected to a fine cord C which passes over two aluminium 
pulleys D and E. A cord Cj , of the same kind as C, is attached to 
the bottom of each pan, and compensates for the 
extra weight of cord on the B side of the pulleys. 
A fall of about 10 feet should be arranged for 
the scale pans. 

Place equal masses in the scale pans, and 
find by trial what additional mass placed in A 
will cause it to descend with uniform velocity 
when given a start. Any additional mass placed 
in A will now give A an acceleration downwards 
and B an equal acceleration upwards. Let A 
have a total fall of H feet, and make several 
experiments without changing the masses, 
noting the time in seconds for each descent by 
means of a stop-watch. Take the average time 
/ seconds and calculate the acceleration a, from 





FIG. 483. Apparatus for TT 
verifying the law F = ma. a l = ~^ feet per S6C. per S6C ( I ) 

The acceleration should also be calculated as follows : 
Let M s = mass of each scale pan, pounds. 

M w = the equal masses added, pounds. 
M e = the additional mass in A required to secure uniform 

velocity, pounds. 

M a = mass added to A for the purpose of producing 
acceleration, pounds. 

The total mass to which acceleration has been given, neglecting 
the cord and pulleys, is 



The force F which has produced the acceleration is the weight 
of M a , i.e. M a g in poundals. Hence, 



or 



M a g= {2(M S + M lo ) + M + M a }a 2 , 

feet er sec ' 



sec ........ 



This value should agree fairly well with that experimentally found 
and given by 1 in equation (i). 

Repeat the experiment two or three times with different masses. 



MOMENT OF INERTIA OF A FLYWHEEL 447 

EXPT. 44. To find the moment of inertia of a small flywheel by the 
method of a falling load. The apparatus used consists of a small 
flywheel (Fig. 458, p. 418) having a drum on its shaft and capable of 
being rotated by means of a cord wrapped round the drum, and 
having a scale pan containing a load attached to its end. The cord 
is attached to the drum in such a manner that it drops off when the 
scale pan reaches the floor. 

Allow the scale pan to descend slowly through a measured height, 
and note the number of revolutions made by the wheel during this 
operation. Wind up the scale pan to the same height, place a load 
in it, then allow the wheel to start unaided, at the same moment 
starting a stop-watch. Stop the watch at the instant the scale pan 
reaches the floor, and note the time of descent. Allow the wheel to 
go on revolving until friction brings it to rest, and note the total 
number of revolutions which it makes from start to stop. 
Let m l = the mass of the scale pan, in pounds. 

m% = the mass placed in the scale pan, in pounds. 
M = m l + m 2 = the total falling mass, in pounds. 
H = the height of fall of the scale pan, in feet. 

^=the time of fall, in seconds. 

N x = number of revolutions made by the wheel during the fall. 
N 2 = the total number of revolutions from start to stop. 
The total work done by gravity will be M^H foot-poundals, and, 
up to the instant that the scale pan is on the point of touching the 
floor, this work has been expended as follows: (a) in giving kinetic 
energy to the falling mass M; (b) in overcoming frictional resistances; 
(f) in giving kinetic energy to the wheel. If v be the velocity of M 
when the scale pan arrives at the floor, the average velocity of 
descent will be \v feet per second. Hence, 



2H , 

. . v = feet per second. 

. Mv* M 4 H 2 

. . Kinetic energy acquired by M = - = ~- 

2MH 2 f 

= -g - foot-poundals. 

The difference between M c ^H and the kinetic energy acquired by 
the falling mass M represents the energy reaching the drum, and is 
expended in overcoming friction and in giving kinetic energy to 
the wheel. 2 MH 2 

Energy reaching the drum = M^H - 2 

MHf -- 2~ ) foQt;-poundals . 



448 MACHINES AND HYDRAULICS 

Ultimately, the whole of this energy is dissipated in overcoming 
frictional resistances throughout the entire motion of the wheel, i.e. 
in N 2 revolutions. Assuming that the frictional waste per revolution 
is constant, we have 

Energy wasted per revolution = MHuf- -rj- J -f- N 2 , 

Energy wasted while M is falling = MH(^- ^- j^ 1 foot-poundals. 

Y t /JN 2 

Let E = the kinetic energy possessed by the wheel at the instant 
the scale pan reaches the floor. 

rp, ,-, AT TT/ 2H\ ,,/ 2HXN, 

Then E = MH- -- - MHg- .J 



-- 1 foot-poundals 
N 



The angular velocity of the wheel at the instant the scale pan 
reaches the floor may be calculated as follows : 
Revolutions described in / seconds = N\ 

= average revs, per sec. x t\ 

.'. Average revolutions per sec. = - . 

2N, 

And, maximum revolutions per sec. = - ; 

2N 

.". Maximum angular velocity of wheel = w= 1. 2^ 

4 7r N 1 ,. 
= ^ 1 radians per sec. 

Now, 

o 

Maximum kinetic energy of the wheel = I = E foot-lb. ; 



MH 



_ 2 - 

N/ foot units. 



The experiment should be repeated several times with different 
masses m 2 and with different heights of fall H ; the values of I 
should be calculated for each experiment and the mean value taken. 



CENTRE OF OSCILLATION 



449 



EXPT. 45. To find the centre of oscillation, or the centre of percussion, 
of a given body. A connecting rod has been selected as a useful 
example (Fig. 484). The rod AB is suspended from a knife edge 
consisting of a square bar of tool steel 
CD, passing through the hole in the G 
small end and resting on V blocks at 
E and F. The rod can vibrate now 
in the same plane as that in which it 
will vibrate when built into the engine. 
GH is a simple pendulum consisting 
of a small heavy bob and a light cord. 
Cause both rod and simple pendulum 
to execute small vibrations, starting 
both together at the end of a swing. 
Adjust the length of GH until both 
vibrate in the same time. Measure 
GH and mark a point on the con- 
necting rod at this length from its axis 
of vibration. This will give the centre 
of oscillation or percussion when the 
rod is vibrating about the upper edge 
of the tool steel bar. 

EXPT. 46. Take a uniform bar of metal about 3 feet long and of 
section about i inch by f inch. Referring to p. 443, it will be seen 
that the centre of percussion Z for this bar will be at a distance from 
C given by 




B B 

FIG. 484. Centre of oscillation by 
experiment. 



Let L be the length of the bar. Then 



Mark clearly the position of Z on the bar ; allow the bar to hang 
vertically, using a finger and thumb at C. Use another short bar 
and strike the bar sharply at different points. The absence of any 
jar on the fingers when the bar is struck at Z will be observed 
readily, and gives confirmation of the calculated position of Z. 

EXPT. 47. To find the radius. of gyration of a given body about an axis 
passing through its centre of mass. In Fig. 485 is shown a flywheel 
arranged in the same manner as the connecting rod in Fig. 484. 
Find the length of the corresponding simple pendulum as directed 
previously, being careful to cause the flywheel to vibrate in the same 
plane as that in which it will rotate subsequently. Measure BK, the 
distance from the axis of vibration to the centre of mass of the wheel. 
Weigh the wheel in order to estimate its mass. 
D.M. 2 F 



450 



MACHINES AND HYDRAULICS 



Taking equation (3), p. 441, and applying it to the present case, 
we have ,2 



where 



Hence, 
Now, 



L = GH, in feet; 

Y = BK, in feet; 

B = the radius of gyration about B, in feet. 

4-LY. 

I K = IB -MY 2 , 



or 



/& K = s/Y(L- Y) feet (i) 

G 





<> 



FIG. 485. Radius of gyration of a flywheel by experiment. 

The moment of inertia of the wheel about its axis of rotation 

willbe I K = M/ 

= MY(L - Y) pound and foot units. . . .(2) 

This experiment therefore provides a means of finding the data 
required for estimating the kinetic energy and the rotational inertia 
of a given flywheel. 

EXPT. 48. To find the velocity acquired by a wheel in rolling down an 
incline. In Fig. 486 is shown a long incline AB consisting of two 
angle bars with a gap between them. The angle bars are pivoted to 
a bracket at A, and a prop at F enables the angle of inclination to 
be altered. 

The wheel D has a spindle projecting on each side of the wheel, 
and has a collar E on each side secured by a nut to the spindle. 
The collars are coned slightly for the purpose of keeping the wheel 
centrally in the gap as it rolls down and to prevent the wheel from 
rubbing on the angles. A fixed stop is fitted at C. The object of 
the arrangement is to increase the time taken in rolling down the 
incline. 



EXERCISES ON CHAPTER XVIII. 451 

First determine the square of the radius of gyration of the wheel 
and its attachments about the axis of the spindle, by the method 
explained in the last experiment. Let this be & in foot units. 




FIG. 486. Apparatus for investigating the motion of a wheel rolling down an incline. 

Set the incline to a suitable angle by means of the prop. Measure 
the difference in level between the centre of the wheel spindle when 
in the starting position and when in the stopping position ; let this be 
H feet. Measure also the distance travelled, parallel to the incline, 
by the wheel centre ; let this be L feet. Let the wheel start unaided, 
and note the time taken in rolling down ; let this be / seconds. Let 
the linear velocity of the wheel centre at the instant of arriving at 
the bottom be v feet per second. Then 

L = the average velocity x / 



2L 

.'. v = feet per second ......................... (i) 

Taking equation (4), p. 422, and writing r instead of R, where r 
is the mean radius of the collars E in feet, 



v= / -75 feet per second. * ...... . ........ (2) 




(i) and (2) are expressions for the velocity found by entirely 
independent methods, and the results obtained from them should 
agree. Give the results for v by both methods ; repeat the experi- 
ment, using different angles of inclination and collars having a 
different diameter. 



EXERCISES ON CHAPTER XVIII. 

1. A wheel has a moment of inertia of 24,000 in pound and foot units, 
and runs at 90 revolutions per minute. Find its moment of momentum. 
Suppose that the speed changes to 88 revolutions per minute in 0-5 second, 
what couple must have acted ? 

2. A wheel has a moment of inertia of 20 in pound and foot units, 
and has a speed of 90 revolutions per minute ; the plane of revolution is 
vertical. The wheel is mounted so that its axis is capable of turning in a 
horizontal plane (i.e. in azimuth). The axis is found to have an angular 



452 MACHINES AND HYDRAULICS 

velocity of - radian per second in azimuth. Calculate the couple 
10 

acting. Show the couple and the directions of both angular velocities 
clearly in a diagram. 

3. A cycle wheel has a mass of 5 pounds and its radius of gyration is 
one foot. It is suspended as shown in Fig. 470, the distance between 
the suspending cord and the mass centre being 2-5 inches. The wheel is 
spun and revolves with its plane vertical 120 times per minute. Find 
the angular velocity in azimuth. 

4. A body having a mass of 12 pounds vibrates in a straight line 18 
inches long with simple harmonic motion. The time of one complete 
vibration is 0-25 second. Find what force must act on it at the end of 
each stroke and the velocity at the middle of the stroke. 

5. A small wheel having a moment of inertia of 0-4 in pound and foot 
units has its plane horizontal and is attached firmly at its centre to a 
vertical steel wire, the top end of which is fixed to a rigid bracket. The 
wheel can execute torsional oscillations under the control of the wire. 
The wire is 0-06 inch in diameter and 36 inches long, and its modulus of 
rigidity is known to be 11,000,000 Ib. per square inch. Find the time 
of one complete oscillation. 

6. A thin disc 24 inches in diameter can execute small vibrations under 
the influence of gravity about a horizontal axis at 90 degrees to the plane 
of the disc and bisecting a radius. Find the length of the equivalent 
simple pendulum and the time of one complete vibration. 

7. A thin uniform steel rod 3 feet long hangs freely from its top end. 
Find the centre of percussion. 

8. A uniform bar of mild steel, section 2 inches by I inch, 4 feet long, 
has masses of 4 and 2 pounds attached at distances of I foot and 3-5 feet 
respectively from one end. Take the density of the bar to be 0-28 pound 
per cubic inch. Find the mass centre and the moment of inertia about 
an axis at 90 degrees to the flat face of the bar and passing through the 
mass centre. 

9. Take the system given in Question 8 and reduce it to an equivalent 
dynamic system having a mass situated at the end of the bar adjacent 
to the given 2 pound mass. 

10. Take the equivalent dynamic system found in answer to Question 
9. A force of 100 Ib. weight is applied at 90 degrees to the bar (a) at the 
mass centre, (^) at 3 inches from the mass centre. Find, in each case, 
the translational acceleration of the mass centre and the angular acceler- 
ation, if any, of the bar. 

11. Explain what is meant by moment of momentum. Calculate the 
moment of momentum of a body weighing 300 Ib. rotating at 1250 revolu- 
tions per minute, the radius of gyration of the body about the axis of 
rotation being 1-7 foot. What property is measured by rate of change of 
moment of momentum ? (I.C.E.) 

12. A body of 40 pounds hangs from a spiral spring, which it elongates 
2-5 inches. The body is then pulled down a short distance and let go. 
Determine the number of complete oscillations the body will make per 
minute, assuming that the weight of the spring is 20 Ib. (B.E.) 



EXERCISES ON CHAPTER XVIII. 453 

13. A body weighing 161 Ib. has a simple harmonic motion, the total 
length of one swing being 2 feet ; the periodic time is i second. Make a 

1 diagram showing its velocity and another showing its acceleration at 
every point of its path. What force is giving to the body this motion ? 
What is its greatest value ? (B.E.) 

14. A heavy circular disc is supported on a shaft 3 inches in diameter, 
carried on roller bearings ; a cord is wrapped round the shaft. It is 
found by experiment that a weight of 6 Ib. suspended from this cord is 
just sufficient to overcome the friction of the roller bearings and maintain 
a uniform speed of rotation of the disc. When a weight of 30 Ib. is 
suspended from the cord, it is found that this weight descends vertically 
14 feet in 2 seconds of time. Determine the moment of inertia of the 
disc in pound-foot 2 units. Neglect the inertia of shaft and cord, and 
assume that the speed of rotation of the disc increases at a uniform rate 
in the second experiment. (B.E.) 

15. Obtain the magnitude and position of the single force which when 
applied perpendicularly to the axis of a uniform bar (48 inches long, 
weighing 200 Ib.) will give it a translational acceleration of 40 feet per 
second per second and a rotational acceleration of 10 radians per second 
per second. (I.C.E.) 

16. In a hoisting gear a load of 300 Ib. is attached to a rope wound 
round a drum, the diameter to the centre of the rope being 4 feet. A 
brake drum is attached to the rope drum and fitted with a band brake. 
The combined weight of the two drums is 720 Ib., and the radius of 
gyration of the two together is 20 inches. The weight starts from rest 
and attains a speed of 10 feet per second. The brake is then applied and 
the speed is maintained constant until the load reaches 20 feet from the 
bottom, when the brake is tightened so as to give uniform retardation 
until the load comes to rest. The total descent is 100 feet, find the time 
taken for the descent and the tension in the rope during slowing. (I.C.E.) 

17. Show that the natural period of vertical oscillation of a load 
supported by a spring is the same as the period of a simple pendulum 
whose length is equal to the static deflection of the spring due to the load. 
When a carriage underframe and body are mounted on the springs, 
these are observed to deflect i^ inch. Calculate the time of a vertical 
oscillation. (I.C.E.) 

18. Show that a body having plane motion may be represented by two 
masses supposed concentrated at points. A rocking lever (mass 600 pounds) 
has a radius of gyration about its centre of gravity of 18 inches, and the 
centre of gravity is distant 6 inches from the axis round which the lever 
rocks. Find the magnitude of the equivalent masses if one is supposed 
to be concentrated at the axis, and find also the distance of the other 
mass from the axis. Find the torque required to give the lever an 
acceleration of 10 radians per second per second. (L.U.) 

19. The revolving parts of a motor car engine rotate clockwise when 
looked at from the front of the car, and have a moment of inertia of 400 
in pound and foot units. The car is being steered in a circular path of 
400 feet radius at 12 miles per hour, and the engine runs at Sop revolutions 
per minute, (a) What are the effects on the steering and driving axles due 
to gyroscopic action ? The distance between these axles is 8 feet, (b} 
Suppose the car to be turned and driven in the reverse direction over the 
same curve at the same speed, what will be the effects on the axles ? (L.U.) 



454 MACHINES AND HYDRAULICS 

20. A hollow circular cylinder, of mass M, can rotate freely about an 
external generator, which is horizontal. Its cross-section consists of 
concentric circles of radii 3 and 5 feet. Show that its moment of inertia 
about the fixed generator is 42 M units, and find the least angular velocity 
with which the cylinder must be started when it is in equilibrium, so that 
it may just make a complete revolution. (L.U.) 



CHAPTER XIX. 



x B 




LINK MECHANISMS. 

Link mechanisms. Links are used for transmitting motion from 
one point to another in a mechanism. In any complete mechanism 
containing links, usually each 
part is constrained so as to 
move always over the same path / 
in the same definite manner; f 
the whole may then be defined \ 
as a kinematic chain. The slider- 
crank-chain is a well known 
example of complete restraint 
(Fig. 487) here the crank CB revolves about an axis C and forms 
one link in the chain ; the connecting rod AB is connected to the 
crank at B, hence this end of the rod revolves about the centre C ; 
its other end A is constrained by the sliding block D and slotted 
frame E so as to move always in a straight line. 

Fig. 488 (a) shows a case of incomplete restraint ; there are two 
cranks AB and CD, capable of rotation about A and C respectively, 



FIG. 487. Slider-crank-chain. 





FIG. 488. Examples of incompletely and completely restrained mechanisms. 

and connected by two links BE and DE, jointed at E. It is impossible 
to make any calculations in a case such as this. Complete restraint 
may be secured by having a block at the joint E and guiding it to 
move in a definite line (Fig. 488 ()) ; the addition of another crank 
GF and a connecting rod FE will secure definite motion for every 



456 



MACHINES AND HYDRAULICS 



part of the mechanism. In cases of complete restraint, problems 
regarding the path, velocity and acceleration of any point may be 
solved, and calculations made regarding the effects of inertia in pro- 
ducing stresses in the parts and in modifying the forces given to the 
mechanism by outside agencies. 

The path of any point in a mechanism is found best by drawing 
the mechanism in several different positions and marking in each 
the position of the point under consideration ; a fair curve may be 
drawn through these points and will give the desired path. The path 




FIG. 489. Path of a point in a connecting-rod, 
i' 

of a point D in the connecting rod of a slider-crank-chain is shown in 
Fig. 489 as an illustration of the method. A simple method of 
obtaining velocity and acceleration diagrams has been given in 
Chapter XVI. ; some special methods will now be examined. 

Velocity of any point in a rotating body. In Fig. 490 is shown a 
body rotating about an axis at C which is perpendicular to the plane 

of the paper. The direction of the 
velocity of any point, such as A or B, 
will be perpendicular to the radius. 
To calculate the velocity of B, if the 
\ velocity of A is given, let the body 
\ make one revolution ; then 
| Distance travelled by A = 2ir . CA. 
i Distance travelled by B = 2?r . CB. 
i As these distances are travelled in 
' the same time, we have 




V 2= 27T.CB 

~ 



27T 



FIG. 490. Velocities of points in a 
rotating body. 






This result shows that the velocities of different points in a body having 
motion of rotation only are proportional to the radii. 



INSTANTANEOUS CENTRE 



457 



Possible velocities in a link. Let AB be a rigid rod or link 
(Fig. 491), and let A have a velocity V A at a given instant. V A will 
have components V A cos a and V A sin a along and perpendicular to the 
rod respectively. Let B have a velocity V B at the same instant, the 
components of V B in the same directions will be V B cos/2 and 
V B sin/3. As the rod is rigid, i.e. cannot bend or alter its length, it 
follows that V B cos ft and V A cos a must be equal, otherwise the rod 
is becoming shorter or longer. The other component of the velocity 
of B may be of any magnitude and of either sense along BY. The 
result may be expressed by saying that the velocity of B relative to A, or 
of A relative to B must be perpendicular to the line AB. 




I 9CT-i 
-, 




FlG. 491. Possible velocities of the 
ends of a link. 



FIG. 492. Instantaneous centre 
of a link. 



Instantaneous centre. The relations of the velocities of the ends 
of the rod AB may be examined by the following method, which is 
suitable for graphical solutions. Reference is made to Fig. 492. 
Here the velocity of A is along V A , but for an instant it might be 
imagined that A is rotating about any centre in AI which is perpen- 
dicular to V A ; this will not alter the direction of the velocity, which 
will still be along V A . In the same way, we may imagine that B is 
rotating about any centre in BI for an instant, Bl being perpendicular 
to V B . Hence I, the point of intersection of AI and BI may be 
looked upon as a centre about which both A and B are rotating for 
an instant, and is called the instantaneous centre. We have, therefore, 



= 
V B BI* 

The application of this method to a crank and connecting- rod is 
shown in Fig. 493 (a). Given the velocity of B, equal to V B , to find 
the velocity of A draw AI perpendicular to V A , i.e. to AC, and also 



458 MACHINES AND HYDRAULICS 

produce CB, which is perpendicular to V B , to cut AI in I. Then 

y* IA 

V B IB' 

A more convenient construction is to produce, if necessary, the line 
of the connecting rod AB to cut CN in Z. The triangles IAB and 
CBZ are similar. Hence, 

CZ_IA_XA 
CB~IB~V B ' 

CZ=^.V A . 

V B 

If the crank is rotating with uniform angular velocity, V B will be 
constant and CZ may be taken to represent the velocity of A to a 
scale in which V B is represented by CB, the length of the crank. It 
is evident that V A is zero when the crank pin is at either L or R ; 





360Y 270* /ISO* 




FIG. 493. Velocity diagram for the point A, deduced by the instantaneous centre method. 

also, when the crank is at 90* to LR, Z coincides with N, and CZ 
will be equal to the crank, therefore V A and V B will be equal. 
Fig. 493 (b) shows a velocity-time curve for A, drawn by setting off 
the values of CZ on a base of equal crank angles. 

Four-bar chain. Fig. 494 shows an example of a double crank and 
connecting rod. Two cranks, one AB, revolving about A, and another 
CD, revolving about C, are connected by a link BD ; the frame 
forms the fourth bar of the chain. For the position shown, I is the 
instantaneous centre, obtained by producing BA and CD. As before 

VD_ID 

V B IB* 

If V B is given, V D may be found from this construction, and the 
angular velocity of CD may be calculated from 

Angular velocity of CD = 7- 



In Fig. 495 is shown another pair of cranks AB revolving about A, 



FOUR-BAR CHAIN 



459 



and CD revolving about C ; BD is the connecting link, and I is its 
instantaneous centre. As before 

VB_IB . , 

V D "IP' 

o>j = the angular velocity of AB, 
o> 2 = the angular velocity of CD. 
V B = o> 1 .AB, 
V D = <o 2 .CD. 
(Q 1 .AB_V B _IB 
w 2 .CD~V D ~ID' 

IB CD , x 

ID 'AS < (2) 



Let 



Then 



Hence, 




a D 




r 

FIG. 494. A four-bar chain. 



V..V 

FIG. 495. Angular velocities in a 
four -bar chain. 



Produce DB and CA to meet in Z, and \nark the angles a, ft, 6 
and < as shown ; then, in the triangle ZAB, 



and, in the triangle ZCD, 

ZC sin ( 1 80- a) _ sina 

ZD " sin ~~ sin ' 
AB ZC sin a sin^> 



Hence, 



ZD' ZA~sin6>' sin^' 
. ZC _ sin a sin <ft ZD 
" ZA~sin' sin^' AB ' 



Now, in the triangle IBD, sin a/sin ft = IB/ID; and in the triangle 
ZCD, sin <ft/sin 6 = CD/ZD. Hence, 

ZC_IB CD ZD 
ZA~ID' ZD' AB 



. 

ID AB 



460 



MACHINES AND HYDRAULICS 



Therefore, from (2) and (3), 

Wl ZC 'f 

* 2 = ZA <4> 

The result shows that the angular velocities of AB and CD are inversely 
proportional to the segments in which CA is divided toy DB produced. 

Wheel and racks. As a further example of the use of the instan- 
taneous centre, Fig. 496 (a) shows a wheel between two racks. If the 
wheel is moving towards the left with a velocity V c , and if the rack 
AD is fixed, then A will be the instantaneous centre of the wheel. 
Hence, V R BA 



_ 

V C ~CA 



_ 

~ 



showing that the velocity of the top rack is twice that of -the centre 
of the wheel. 

If the racks are moving as shown in Fig. 496 (b\ then I may be 
found from the given values of V A and V B ; thus 

ZA = IA 

V B IB' 

Having found the position of I, the velocity V c of the centre of 
the wheel may be calculated from 




W (b) 

FIG. 496. Wheel and racks. 



FIG. 497. Scott-Russell parallel motion. 



Parallel motions. By the term parallel motion is meant an 
arrangement for constraining a point to move in a straight line. In 
the Scott-Russell parallel motion (Fig. 497), .a link AP has one end A 
guided so as to move in the straight line AB. Another link BC is 
pivoted at B, and is connected by a pin to the centre C of AP. 
AC = CP= : BC, hence P, B and A will always lie on a semicircle 
which has AP for diameter. The angle ABP will always be 90, and 
hence P will move in a straight ^vertical line passing through B. 



PARALLEL MOTIONS 



461 



It will be noted that the instantaneous centre for AP is at the 
intersection of BC produced and AI drawn perpendicular to AB. 
Further, from the geometry of the figure, P will lie always on a 
horizontal line drawn from I, and will, therefore, be moving vertically 
in any position of the mechanism. This confirms the result already 
noted. 

In practice it is often convenient to guide A as shown in Fig. 498. 
The short arc in which A now moves interferes with the straight line 
motion of P to a small extent only. 
This modification of the Scott- 
Russell parallel motion is used 
sometimes in indicators for guiding 
the pencil in a straight line. The 
arrangement permits of P having 
a magnified copy of the motion of 
the piston G. The instantaneous 
centre I for AP is the point of 
intersection of DA and BC when 
produced, and IP, drawn hori- 
zontally through I, gives the position 
of P on the link AP. Joining AB, 
it will be seen that the triangles 

ABC and ICP are nearly similar. Also AC and 1C are nearly equal 
for all practicable 'positions of the mechanism. Hence, 

IC = BC 
CP AC' 
AC = BC 
CP AC' 
AC 2 




FIG. 498. Parallel motion for an 
indicator. 



or 



CP = 



BC ' 

a result which enables CP to be calculated when AC and BC are 
given. 

In the Watt parallel motion (Fig. 499), two equal links AB and 
DC are pivoted at A and D respectively, and connected by a 
third link BC. It is evident that movement of the mechanism 
will cause B and C to deviate to the left and right respectively ; 
hence P, the centre of BC, will move in a straight vertical line 
for a considerable distance. If the movement of the mechanism 
continues, P will describe a curve resembling a rough figure eight 
(the lemniscate). 



4 62 



MACHINES AND HYDRAULICS 



In Fig. 500 is shown a Watt parallel motion in which AB and DC 
are not equal. I will be the point of intersection of AB and DC, 




FIG. 499. Watt parallel motion ; equal arms. 



FIG. 500. Watt parallel motion 
unequal arms. 




and P may be found by drawing IP horizontally from I. From 
the geometry of the figure, it may be shown that 
BP:PC = CD:AB. 

In Fig. 501 is shown the arrangement of Watt's parallel motion 
used in beam engines. AB and DC are 
equal, and P, the centre of BC, moves in a 
straight vertical line. DC is extended to E, 
CE being equal to DC, and bars EF and 
FB are added so as to form a parallelogram 
CEFB. EF will then be double of CP, and 
F, P and D will lie in a straight line always. 
FD will be double of PD, consequently, if 
P is moving in a straight vertical line, so also 

will F. In the engine, F and P serve to guide the ends of the low 
pressure and high pressure piston rods respectively. 

Inertia effects in a mechanism. In investigating problems re- 
garding the forces or turning moments which may be delivered by a 
machine, it is often necessary to consider the effects of the inertia 
of the parts of the machine. The following case of a slotted bar 
mechanism (Fig. 502) giving simple harmonic motion to a piston A 
should be studied. Frictional effects have been considered already 
partly (p. 371), and are disregarded here. 

Fig. 502 (a) is a diagram showing the effective pressure on the 
piston throughout the stroke ; any ordinate such as p^ gives the 
difference in pressure on the two sides of the piston at the moment 
considered. Hence, the net force P urging the piston towards the 

left is 7TY/2 

P=/!--lb. weight, 



INERTIA EFFECTS IN MECHANISMS 



463 



where d is the diameter of the cylinder in inches and p l is the 
pressure in pounds per square inch. 

But for the inertia of the piston, piston rod, and slotted bar, the 
whole of this force would be transmitted to the crank pin. These 
parts will all have equal accelerations in this mechanism. 

Let M = mass of reciprocating parts, pounds. 

a = their acceleration, feet per sec. per sec. 

Then the force required to overcome inertia will be 

^ Ma ., . . 
F = Ib. weight. 

<5 

The force Q actually reaching the crank pin in the position 
considered will be given by 

Q = P-F 

Ma 



The acceleration a may be found for any position by the method 




(c) 



360* 270 180' 90 

FIG. 502. Diagrams for a slotted bar mechanism, taking account of inertia. 

explained on p. 398. Fig. 502 (b) is a diagram in which the 
ordinates/ 2 , etc., have been calculated from 



. 

S 4 

These ordinates will then represent the forces required to over- 
come inertia per square inch of piston area. The scales used in 
Fig. 502 (b} are the same as for Fig. 502 (a), hence a combined 
diagram (Fig. 502 (c)) may be drawn by simply adding the ordinates 



464 



MACHINES AND HYDRAULICS 



algebraically, the result showing ^, the force per square inch of 
piston area which is transmitted to the crank pin. 
The turning moment on the crank pin will be 

T = QxOM, 

where OM is perpendicular to the line of Q. A polar turning-moment 
diagram may be drawn by producing the crank OB and making BC 
equal to T to a convenient scale. This being done for a number 
of crank angles, a fair curve drawn through the ends will give the 
required diagram. Or a turning-moment diagram may be drawn as 
in Fig. 502 (d) by using a base of equal crank angles and setting off 
the values of T at the chosen angles. 

Locomotive side rod. In Fig. 503, A and C are the centres of 
two driving wheels of a locomotive ; the equal cranks AB and CD 




I, 'a 

FIG. 503. Motion of a locomotive side rod. 



are connected by the side rod BD. The velocities V B and V D for 
the given position may be found by taking I x and I 2 as the instan- 
taneous centres of the wheels, assuming that there is no slipping 
between the wheels and the rails. V A and V c will be equal to the 
velocity of the locomotive. Hence, 

V R LB IjB 

t T A A 



or 



Also, 



IjD 
I 2 C 



The velocities of B and D being equal in all respects, it follows 
that the velocity of any point in the side rod will be equal to that 
of B or D ; thus V G is equal to V B or V D . 

Assuming that the speed of the locomotive is constant, and that 
the consequent angular velocity of each wheel is w radians per 



LOCOMOTIVE SIDE ROD 



465 



second, the accelerations of B and D will be unaltered if we 
imagine that the wheels rotate with an angular velocity w, and that 
their centres remain fixed in position. B and D will therefore have 
accelerations directed towards A and C respectively, and of amount 
w 2 R feet per second per second, R being the crank radius in feet. 
These accelerations are equal in all respects ; hence the acceleration 
of any point in the rod will have an equal value and will have the 
same direction. 

Since the side rod is always moving parallel to the rail and has no 
angular motion, the resultant force required to give it its motion 
must act through its centre of mass, and must be in the same line as 
the acceleration of the mass centre. If the rod is uniform, the 
centre of mass G will bisect BD ; let M be the mass of the rod in 
pounds, then the resultant force R x required to overcome the inertia 
of the rod will be 



Mw 2 R 
- 



, . 
Ib. weight. 



Obviously Rj is the resultant of two equal and parallel forces, one 
acting at each crank pin. 

The force R I} reversed in sense, gives the effect of the inertia 
resistance of the rod on the wheel bearings at A and C. It is 
evident that there will be a lifting 
effort when the side rod is in its 
highest position (Fig. 504^)), and 
an additional pressure on the rails 
when the rod is in its lowest position 
(Fig. 504 (a)). Rj acts towards the 
right (Fig. 504^)), or towards the 
left (Fig. 504 (d} ), when the cranks 
are horizontal. 

Fig. 504 also indicates the effect 
of Rj in producing a transverse load 
on the rod. For a uniform rod, R x 
is the resultant of an inertia load 
which has a uniform distribution per 
unit length of the rod, and in this 
respect resembles the weight W of 

the rod. As will be Seen by inspec 

tion of Figs. 504 (a) and (^), R T and 
W conspire when the rod is in the lowest position, and are opposed 
when the rod is in its highest position. The maximum bending 
D.M. 2 G 




*>.- 



* * * % 

D t * i I A 

!%>*' tw (d; > r ? 

FIG. 504. Inertia effects in a locomotive 
side rod. 



466 MACHINES AND HYDRAULICS 

effect on the rod will, therefore, occur in the position shown in 
Fig. 504 (a). If W is the weight of the rod in lb., the total uniformly 
distributed load producing bending moment will be 

Total distributed load = R x + W 

M<o 2 R ... 

= + W lb. weight. 

& 

The calculation of the maximum bending moment and stresses 
produced by this may be performed by the methods explained in 
Chapter VII. 

Crank and connecting rod. The inertia of the moving parts in 
the crank and connecting rod mechanism produces effects similar to 
those in the slotted bar mechanism (p. 462), but the problem is 
somewhat more complicated owing to the oblique action of the 
connecting rod. In the slotted bar mechanism, the piston has 
simple harmonic vibrations, and hence has equal accelerations when 
at equal distances from the centre of the stroke ; the connecting rod 
causes the accelerations to be unequal to an extent which is more 
marked if the connecting rod is short. A very long rod produces 
nearly equal accelerations, a rod of infinite length would give simple 
harmonic motion ; hence the name infinite connecting rod mechanism 
sometimes given to the slotted bar arrangement. Further, the piston, 
piston rod and crosshead have straight-line motion, and hence are 
dealt with easily, while the connecting rod has one end moving in 
a straight line and the other end in a circle. For simplicity, it is 
customary to treat the rod in two parts, a fraction, say one-half, of its 
mass being assumed to be concentrated at the centre of the crank pin 
and rotating with it, while the remainder of the mass is assumed to 
move in a straight line with the crosshead. The mass of the recipro- 
cating parts will then include the piston, piston rod, crosshead and 
the assigned part of the connecting rod, and this mass will require 
forces in order to overcome its inertia. 

The acceleration diagram may be drawn by the method described 
for another mechanism on p. 386. The work may be made more 
accurate by first drawing a velocity-time diagram for the piston by 
the instantaneous centre method (p. 458) ; then the average accelera- 
tions over equal intervals may be calculated, and the results set off at 
the centres of the intervals. .Or Klein's construction may be used as 
follows to obtain an acceleration diagram direct on a base repre- 
senting the stroke. 

Klein's construction. In Fig. 505 is shown a crank CB ot radius 



KLEIN'S CONSTRUCTION 



467 



R feet and a connecting rod AB in a given position. On AB as 
diameter describe a circle ; produce AB, if necessary, to cut NS 
in Z; describe another circle with centre B and radius BZ, cutting 




FIG. 505. Klein's construction for the acceleration of the piston. 

the first circle in D and E. Join DE, cutting AB in F and AC 
in K, producing DE if necessary. Then, as will be proved later, 
KC represents the acceleration of the piston to a scale in which the 

v l 
central acceleration of B, viz. ^ is represented by BC. It is assumed 

usually that B is moving with uniform velocity v feet per second. 




FIG. 506. Klein's construction, crank in second quadrant. 

The construction should be repeated for crank angles differing by 
30 ; KC should be measured for each position, and the results set 
off as at AL on a base GH, which represents the stroke of the 
piston. The acceleration diagram is obtained by drawing a fair 
curve through the ordinals, and is shown at GMLPH. Fig. 506 
shows the construction when the crank is in the second quadrant, 



4 68 



MACHINES AND HYDRAULICS 



and in Fig. 507 is given the construction when the crank is on the 
dead points B and B'. 



D 


' C 


),' 




r W~ I "*-x~ 7 

' N ' ,' N / 

\:/ 

^ ( 




\ 
\ x 


K' 1C BJ! 

/ 1 V y' * 

/ II x 


K A x A 



. -. .,. 
" e \ T E \ 

FIG. 507. Klein's construction, crank at dead points. 

Accelerations at ends of the stroke. The accelerations of the 
piston when the crank is at the dead points may be found also by the 
following method. In Fig. 508 the crank is shown at a very small 
angle from the dead point ; imagine that the connecting rod is so 
guided that it is moving parallel to the line of the stroke, i.e. BA' 
is parallel to AC. Every part of the connecting rod will have the 

9 

same acceleration as B, viz. -=r towards the left. Now, the connect- 

K. 

ing rod is actually moving in such a manner that one end, B, has a 
velocity v at right angles to the rod ; to bring A' into the centre line 

AC, give A' a velocity v as shown. Owing to this, A will have 
v i 

an acceleration =- towards the left ; hence total acceleration of A 
LJ 

will be 2,2 v i tf/ R \ 

* = R+ L = RA I + L/ Per SeC * P6r SeC ' 5 (*' 

when v = the velocity of the crank pin, feet per sec. ; 

R = the radius of the crank, in feet ; 

L = the length -of the connecting rod, in feet. 




FIG. 508. Acceleration of the piston at the 
inner dead point. 



FIG. 509. Acceleration of the piston 
at the outer dead point. 



At the outer dead centre (Fig. 509) a similar method may be used, 

v i v i 

but now ^ is towards the right and -=- is towards the left. Hence 
K. Li 



TURNING MOMENT 469 



the resultant acceleration of A will be towards the right, and will be 
given by 



^ tf & ( R\ . 
* = g- L = RVL/ feet P ersec P er 



sec 



These results, (i) and (2), are of service in making preliminary 
calculations of the accelerations of the piston when at the ends of 
the stroke. 

The effective force Q acting on the crosshead in the line of the 
stroke may be estimated now. 

Let d=i\\Q diameter of the cylinder, in inches. 

p l = the effective pressure on the piston at a given position, 

Ib. per square inch. 
M = the mass of the reciprocating parts, including the 

assigned part of the connecting rod, pounds. 
flj = their acceleration in the given position. 

Then Q= A ^-Ma i b . wei ght .................................... (3) 

4 <5 . 

Turning moment. The turning moment produced by Q may be 
calculated as follows, reference being made to Fig. 510 and friction 




FIG. 510. Turning moment on the crank. 

being neglected. I is the instantaneous centre for the given position, 
from which it appears that rotation of the rod round I is produced 
by Q and resisted by the crank pin with a force S. Hence, 
QxIA = SxIB, 

|i ;>;;;;:; s=Q ra (4) 

Produce AB to cut CN in Z; then the triangles ABI and BZC 
are similar. Hence, IA CZ CZ 
IB = BC = 1T 
Substitution in (4) gives CZ . 



470 



MACHINES AND HYDRAULICS 



S is the reaction of the crank pin, and, if reversed, will be the 
crank effort given by the connecting rod to the crank pin. Hence, 

Turning moment = T = S x R 
~CZ . 



= Q.CZ (6) 

This result will be in Ib.-feet if Q is in Ib.-weight units and CZ is 
measured in feet to the same scale as that used in drawing the 
mechanism. 

With the alterations and additions noted above, the method of 
obtaining a turning moment diagram used on p. 463 may be employed. 




360* 180* 

FIG. 511. Diagrams for a slider-crank-chain, taking account of inertia. 

The various diagrams required are shown in Fig. 511, and will be 
followed readily. 

General effects of inertia. The student will observe that the 
general effect of the inertia of the moving parts is to produce a 
more uniform turning moment on the crank. During the early 
part of the stroke, the gaseous pressure on the piston is high, but is 
absorbed partly in accelerating the moving parts, hence the turning 
moment is smaller ; later in the stroke, the gaseous pressure is low, 
but the moving parts are losing velocity now, and their inertia assists 
the gaseous pressure in making the turning moment larger. 

Greater uniformity in the turning moment may be obtained by 
having two or more cylinders with pistons operating on separate 
cranks. If there are two cylinders, the cranks are placed generally 



MOTION OF THE CONNECTING ROD 



471 



at 90 to each other ; in the case of three cylinders, the cranks are 
generally at 120; with a larger number of cranks, the precise crank 
angles cannot be stated, as other considerations are involved. Usually 
an attempt is made in such cases to 
produce a self-balanced machine, 
i,e. one in which the inertia effects 
balance one another without pro- 
ducing disturbances in the frame or 
foundation. 

Turning moment diagrams are 
given in Fig. 512 for two cylinders 
similar to the case illustrated in 
Fig. 511. The cranks are at 90, 
and the turning moment diagrams 
for each crank separately are shown 
by ABCDA and EFGHE ; these 
are displaced relatively to each other 
by 90. Summing the corresponding ordinates, the combined 
turning moment diagram is HKBLFMDNH. Greater uniformity 
has been obtained, and there is no point where the turning point 
is zero. 

Further points regarding the motion of the connecting rod. It 
has been explained that, for positions near the dead points, the 
motion of the connecting rod may be assumed to be compounded of 




FIG. 512. Turning moment diagram fot 
two cranks at 90. 




FIG. 513. Analysis of the motion of a connecting rod. 

a motion of translation together with another motion of rotation 
round the crank pin (p. 468). The same assumptions may be made 
when the rod is in any other position (Fig. 513). The first of these 



472 MACHINES AND HYDRAULICS 

motions would cause the rod always to move parallel to the centre 
line AC, and it would occupy the position A'B when the crank is at 
CB ; the latter motion produces the effect of rotating the rod into its 
proper position AB. Owing to the first of these component motions, 
all points in the rod will possess the same velocity and acceleration as the 
crank pin B ; in this respect, the motion is precisely the same as that 
of the side rod of a locomotive (p. 464). The acceleration thus pro- 

V 2 

duced at A will be _?, and may be represented by the length of the 
R 

crank BC. Hence, 



or V B = BCxR = BC 2 ...................... (i) 

The point A will possess other accelerations owing to the com- 
ponent motion of rotation of the rod about B ; in consequence of 
this angular motion, A will have a variable velocity v in a direction 
at right angles to the rod. The value of v will depend on the posi- 
tion of the crank, and hence will be undergoing change in most 
positions of the mechanism. Owing to this, there will be an acceler- 
ation of A in the line of z>, i.e. at right angles to the connecting rod. 
Further, A will possess the ordinary central acceleration towards B, of 

v 2 - 
magnitude given by =-. Hence in all A possesses three component 

.L/ 

accelerations, and the resultant of these must have a direction coinciding 
with that of AC. 

To find an expression for #, reference is made to Fig. 513, show- 
ing I, the instantaneous centre of the rod. The angular velocity of 

V v 

the rod will be , and will be given also by T -. Hence, 

J..D LJ 

*_VB 
L IB' 



Also, 

v 1 i V 2 

Central acceleration of A towards B = y- = ^ ' f^^ 2 

\-i !_/ IJj 

-SBT Vl ...................... (3) 

Referring to Fig. 514, showing Klein's construction together with 



MOTION OF THE CONNECTING ROD 



473 



the position of I, join BD and DA. The triangles BFD and BDA 
are similar. Hence, -gp -g^ 

BD = AB' 
BD 2 



or 



BF = 



AB' 



Also, by the construction, BD is equal to BZ, and AB is equal to 
L. Hence, 



\ 




FIG. 514. Proof of Klein's construction. 



Again, the triangles BCZ and IAB are similar. Hence, 
BZ AB , L 
BC~ IB~IB ; 
LxBC 



BZ = 



From (4), 



IB 

L 2 .BC 2 i BC 2 x L 
"L = 



IB S 



IB 2 



(5) 



Comparison of (3) and (5) will show that FB represents the 
central acceleration of A towards B to the same scale in which BC 

V 2 

represents 5. 

The resultant acceleration of A along AC may be found now by 
means of a polygon of accelerations. In Fig. 514, FB is the central 

V 2 

acceleration of A towards B ; BC is the component -=5 the com- 
ix. 

ponent acceleration at 90 to the rod owing to variation in v is 



474 MACHINES AND HYDRAULICS 

represented by KF, and the closing line KC gives the resultant accelera- 
tion of A along AC. It will be noted that this result proves the truth 
of Klein's construction. 

Since KF gives the linear acceleration of A in the direction at 90 
to that of the rod, it follows that the angular acceleration of the rod 
will be given by 

Angular acceleration of the connecting rod = -= (6) 

LJ 

Acceleration image of the connecting rod. In Fig. 5 1 5 KF and 
FB have been copied from Fig. 514. The resultant of these 




FK;. 515. Acceleration image of a connecting rod. 

accelerations will be KB, the closing line of the triangle of accelera- 
tions FBK. The acceleration of A along AC may be taken to be the 
resultant of the accelerations KB and BC, and is represented by the 
closing line of the triangle of accelerations KBC. Consider any 
other point in the connecting rod, such as G. Its velocity at 90 
to the rod, and hence its accelerations, owing to the rod rotating 
about B, will be simply proportional to BG, that is, 

acceleration of G : acceleration of A = BG : BA. 

Draw.GH parallel to AC, and cutting KB in H ; then 
Acceleration of G : acceleration of A = BH : BK. 

Now KB represents the resultant of the two component accelera- 
tions of A which are respectively along and at 90 to AB hence H.B 
will represent the resultant of the similar components of G. The 
component acceleration of G, owing to B rotating about C, remains 

V 2 
of unaltered value -^, and is represented by BC. Hence the re- 

Jx 

resultant acceleration of G will be the closing line HC of the triangle 
of accelerations HBC. The resultant acceleration of any other point in 
the rod may be found in a similar manner by drawing a line from the point 
parallel to AC to cut KB, and joining the point so found on KB to C. On 
account of this property of KB it is called usually the acceleration 
image of the connecting rod. 



INERTIA OF THE CONNECTING ROD 475 

Eesultant force required to give acceleration to the connecting 
rod. In Fig. 516, let G be the mass centre of the connecting rod. 




FIG. 516. Resultant force required to overcome the inertia of the connecting rod. 

The acceleration of G is HC, and might be produced by a force R 
acting at G in a line parallel to HC. The magnitude of R, if the 
rod has a mass M pounds, will be 

M . HC , V .. V. / \ 

R= - Ib. weight (i) 

> 

This force would not produce any angular acceleration in the 
connecting rod on account of its line of action passing through the 
mass centre of the rod. In order to obtain the actual motion of 
the rod, which includes angular acceleration in most positions, R will 
require to be shifted from G, thereby giving a couple which will 
produce the required angular acceleration. A convenient way is to 
use an equivalent dynamic system by substituting two masses, m l 
and m 2 pounds (Fig. 517), for the actual mass of the rod. One of 



' /T5N G m 

c 'B \T T 9/ 

" * ^ i 



FIG. 517. A dynamical system equivalent to the connecting rod. 

these, m l , may be situated at the centre of the crosshead pin A, at a 
distance a from G ; the other mass, m. 2 , will be at a distance b on 
the other side of G. For this arrangement to be equivalent to the 
actual rod, the following conditions must be complied with (p. 444) : 

m l + m 2 = M. (2) 

m^a = m.J) (3) 

^2 + ^2 = M> G . (4) 



MACHINES AND HYDRAULICS 



k c is the radius of gyration of the connecting rod about an axis 
passing through G and parallel to the crank shaft. The solution of 
these equations gives a& = A*G 9 ................................... (5) 

* = f .................................... (6) 

From this result b may be calculated when a and k G are known ; 
the masses m L and m 2 may be determined then from equations (2) 
and (3). 

Reference may be made now to Fig. 516, which shows the crank 
and connecting rod, the latter being represented by the equivalent 
masses m l and m 2 . To accelerate m 1 requires a force R T acting in 
the line of the acceleration of A, viz. AC. To accelerate m 2 requires 
a force R 2 acting in the line of the acceleration of D ; this line may be 
found by drawing DE parallel to AC and cutting BK in E ; the 
acceleration of D will be represented then by EC. R 2 will be 
parallel to EC, and cuts the line of Rj produced in F. Hence the 
resultant of R t and R 2 , which will be the resultant force R required 
to accelerate the rod, must pass through F. The line of R will be 
parallel to the acceleration of the mass centre G, viz. HC, and the 
magnitude of R will be given by equation (i) (p. 475). The couple 
giving angular acceleration to the connecting rod will be R x GM, 
GM being the perpendicular from G to the line of R. 

Reactions on the engine frame produced by the inertia of the 
connecting rod. In Fig. 518, R is the resultant force required to 

D 




FIG. 518. Components of R at the crank and crosshead pins. 

overcome the inertia of the connecting rod in the given position. 
R is actually the resultant of two forces, one of which, P, is applied 
by the guide bar to the pin at A ; the other force, Q, is applied to 
the rod at B by the crank pin. If the friction of the slipper be 
neglected, P will act at right angles to AC, and its line will intersect 



INERTIA OF THE CONNECTING ROD 



477 



the line of R at D ; hence Q also acts through D. . The magnitudes 
of P and Q may be determined by drawing the parallelogram of 
forces DGEF, in which DE is made equal to R, and P and Q will 
be represented by DG and DF respectively. The reaction on the 
guide bar at A will be obtained by reversing the sense of P (Fig. 519) ; 
in the same diagram, the reaction on the crank pin is shown by 
reversing the sense of Q. 

The force Q in Fig. 519 is equivalent to an equal and parallel 
force Q, of the same sense, acting at C together with a couple ot 
moment Q x CH, CH being perpendicular to the line of Q. Q 




FIG. 519. Reactions of the engine frame due to the inertia of the connecting rod. 

acting at C produces a pressure on the main bearing and hence on 
the engine frame ; the couple Q x CH modifies the turning moment 
on the crank. 

Bending moment on the connecting rod produced by its inertia. 
Assuming that the inertia effects on the connecting rod will be 




FIG. 520. Transverse inertia load on the connecting rod. 

greatest when the crank and connecting rod are at 90 to each other, 
the bending-moment diagram may be drawn as follows : In Fig. 520 



47 8 



MACHINES AND HYDRAULICS 



ABC is 90, and- BD is the acceleration image of the rod. The 
acceleration of B will be towards C and will be o> 2 R feet per second 
per second, where w is the angular velocity of the crank in radians 
per second and R is the radius of the crank in feet. The acceleration 
of A will be represented by DC, and its component perpendicular 
to AB will be found by drawing DE perpendicular to CB. CE 
will be the required component. Neglecting the very small acceler- 
ation represented by CE, the acceleration of any point on the 
connecting rod may be found by making BF perpendicular to AB 
and equal to o> 2 R and by joining FA. The acceleration normal to 
AB of any point H in the rod will be represented by HK, perpen- 
dicular to AB. 

Let m be the mass of the rod in pounds per inch length at H ; 
then the transverse inertia load on the rod at H will be 

Inertia load per inch length = - Ib. weight, 

cb 

HK being measured to the acceleration scale in feet per second per 
second. A similar calculation should be made for a number of 




FIG. 521. Bending moment diagram produced by the inertia load on the connecting rod. 

points on the rod ; the calculations are somewhat simpler in the case 
of a rod of uniform cross section, and in other caset may sometimes 
be simplified by taking m to be the average mass per inch length. 
A load curve is constructed then by drawing AB (Fig. 521 (a)) to 
represent the length of the rod and setting off the calculated loads 
per inch at the various points chosen, as at PQ, Considering the 
portion BP, the average load per inch will be |(BN + PQ) and the 
load on BP will be Wl = l(BN-r-PQ)BP. 



SLIDE-VALVE GEAR 



479 



Wj will act through tfie centre of area of BNQP. Carrying out 
the same process for the other portions of the rod, we obtain the 
equivalent system of concentrated loads W 1? W 2 , W 3 , etc. The 
bending-moment diagram may be drawn now by the link polygon 
method (p. 140) ,the construction being shown in Fig. 521 (fi) and (c). 

Simple slide-valve gear. In ordinary reciprocating engines, the 
valve employed to distribute the steam to each end of the cylinder 
consists of an inverted rectangular box V (Fig. 522), which slides on 





VlL 

^vC^&x<^^css>s 



Bi 




FIG. 522. Simple slide-valve for a steam engine. 

a flat face formed on the cylinder. Two steam passages, or ports Sj 
and S 2 , lead to each end of the cylinder, and another E leads to the 
atmosphere or to the condenser. Movement of the valve to the 
right will permit steam to flow through S x into the left-hand side of 
the cylinder, and at the same time permits the steam in the right-hand 
side to flow out through S 2 and E. Movement of the valve to the 
left will admit steam to the right-hand side through S 2 , and will 
permit exhaust from the left-hand side through S x and E. 

CD is the centre line of the valve and AB is the centre line of the 
cylinder ports ; the valve is in its mid-position when these lines are 
coincident vertically. In this position, the valve generally laps over 
the edge of the ports ; / is called the outside lap, and gives an earlier 
cut-off than if there were no outside lap ; e l is called positive inside 
lap ; if the inside lap is made as shown at e% it is called negative ; 
the inside lap determines the point at which the exhaust steam is 
stopped from flowing out of the cylinder. Cut-off of the steam supply 
is effected when some fraction of the piston stroke has been com- 
pleted; the remainder of the stroke is then completed under the 
expansive action of the steam. Closing of the exhaust is effected 
before the end of the return stroke in ordsr to entrap some of the 



480 



MACHINES AND HYDRAULICS 



exhaust steam in the cylinder ; this is compressed by the returning 
piston, and acts as a cushion in bringing it to rest. To understand 
the complete distribution of the steam, it is necessary that the 
displacement of the valve from its mid-position should be known 
for any given crank position, and hence for any piston position. 

The slide valve is driven generally by means of an eccentric, con- 
sisting of a disc A secured to the crank shaft B and revolving with it 
(Fig. 523). The hole in the disc is bored at a small distance from 




FIG. 523. Eccentric driving a slide valve. 

the centre of the disc. A strap C surrounds the disc and is connected 
by an eccentric rod DE to the valve rod EF. The eccentric is 
equivalent to a crank having a radius equal to the distance from the 
centre of the shaft to the centre of the disc. This radius is generally 
very small compared with the length of the connecting rod ; hence it 
may be assumed that the motion of the valve is simply harmonic. 

Let the circle ABCD (Fig. 524) represent the path described by 
the centre of the eccentric, which is rotating in the direction of the 

arrow. AC may be taken to represent 
the travel of the valve, which will be 
in its mid-position O when the eccentric 
is at OB or at OD. Assuming simple 
harmonic motion, when the eccentric 
is in any position, such as OE, if EM 
and EN are perpendicular to AC and 
to BD respectively, OM or NE will be 
the displacement of the valve. If the 
angle separating the eccentric and 
the crank be EOK, then OK will be 
the corresponding crank position 

Since the valve displacement towards the right must be equal to / at 
admission, if OL, equal to /, be measured and LE X drawn perpen- 
dicular to AC, OEj will be the position of the eccentric at admission. 
By setting off the angle EjOKj equal to EOK, the position of the 
crank at admission OK X may be found. Producing EjL to E 2 will 




VALVE DIAGRAMS 481 




determine the eccentric position OE 2 at cut-off. Release and 
cushioning are controlled by the inner edge of the valve. Make 
OQ equal to e, measuring it to the left of O if positive and to the 
right if negative, and draw E 3 QE 4 perpendicular to AC ; OE 3 and 
OE 4 will be the eccentric positions at release and cushioning 
respectively. In each case, the corresponding crank position may 
be obtained by setting back angles equal to EOK. 

Valve diagrams are based on the idea of obtaining direct the 
displacement of the valve by merely drawing the crank in any given 
position. In the Reuleaux valve diagram (Fig. 525), OSj and OS 2 
are the crank positions at admission and 
cut-off; it is evident that the angle S-f>S 2 
in this figure is equal to EjOEg in Fig. 524. 
Hence DjDg, parallel to S^, will cor- 
respond to BD in Fig. 524, and the valve 
displacement for any crank position OK 
will be KN, which is drawn perpendicular 
to DjDg, and hence corresponds to EN 
in Fig. 524. Draw 8^2 and E^ parallel FlG . 525 .^~~ x ^ diagt3im . 
to DjDo and at distances from it equal 

to / and e respectively. Then, of the total displacement KN, SN is 
equal to the outside lap, and hence KS will be the amount by which 
the valve edge has opened the port to steam. Similarly, at OK', 
K'N' will be the displacement on the left of the mid position and EK' 
will be the amount by which the inner edge of the valve has 
uncovered the port to exhaust. At admission and cut-off, the 
crank will be at OS l and OS 2 respectively, as has been used in the 
construction ; in these positions the port opening is zero. At release 
and cushioning the crank will be at OE 2 and OE l respectively, because 
in these positions the exhaust opening will be zero. 

There are many other types of valve diagrams ; any standard text 
book may be consulted for the principles on which they are based.* 

Component eccentrics for a valve gear. In Fig. 526 (a), OC is 
the crank on the dead centre and OE is the corresponding eccentric 
position ; the angle EOV is called the angle of advance and is denoted 
by a. Suppose that, instead of using OE, component eccentrics OV 
and OH are employed, these being found by drawing EV and EH 
respectively parallel and at right angles to CO. It is to be under- 
stood that the actual motion of the valve is to be obtained by adding 

* A complete discussion is given in Valves and Valve Gear Mechanisms, 
by Prof. W. E. Dalby ; (Arnold). 

D.M. 2 H 



482 



MACHINES AND HYDRAULICS 



together the displacement produced by the component eccentric OH 

and that produced by OV. OV and OH are called the 90 and the 

1 80 components respectively, and are given by 

OV = OE cos a. 

OH = OEsina. 

In Fig. 526(^)5 the crank has moved through an angle 6 from the 
dead point. We have 

Displacement produced by OE = OM = OE sin (a + 0) 

= OE sin a cos 6 + OE cos a sin 6. 

Displacement produced by OV = ON = OV sin = OE cos a sin 6. 
Displacement produced by OH = ON' = OH cos 6 = OE sin a cos 6. 
Sum of displacements produced by OH and OV = OE sin a cos 

+ OE cos a sin 0. 

Hence the resultant displacement produced by the component 
eccentrics is equal to that produced by the actual eccentric. 
Let OH and OV be written a and b respectively ; then 

Displacement of the valve = a cos + ^sin B (i) 




FIG. 526. Component eccentrics. , , m . 

a and b are both constant in a simple valve motion having a single 
eccentric. In many cases of more complicated valve gears the 
motion of the valve may be represented approximately by an 
equation of form similar to (i), showing that the valve may be 
imagined to derive its motion from a single eccentric acting direct on 
the valve. 

Hackworth valve gear. An example is shown in Fig. 527 of a 
Hackworth valve gear: BC is the crank and AB is the connecting 
rod. The eccentric CE is at 180 to the crank and is connected 
to a rod EF, the end F of which may slide on a guide GH. 
GH is pivoted at K, and the angle /3 which it makes with CK may 



HACKWORTH VALVE GEAR 



483 



be adjusted by hand. Alterations in the cut-off and reversal of the 
direction of motion of the engine are effected by altering /?. The 
valve rod LN is connected to EF at L. 




FIG. 527. Hackworth valve gear. 

The displacements of the valve will be very nearly equal to the 
displacement of L vertically. 

Let r= the radius of the eccentric. 



</=LF. Then 
Displacement of E vertically from CK = r cos 9. 

If F were to move in the straight line CK, the displacement of L 
vertically owing to the above displacement of E would be 

Vertical displacement of L = - r cos 6 ......................... ( i ) 

Again, Displacement of E horizontally from AC = r sin 0. 

In Fig. 528 CP represents this displacement, and F is supposed to 
be connected direct to P. KQ will be very nearly equal to CP. 



4 8 4 



MACHINES AND HYDRAULICS 



Hence, 



r sin 6 tan /?. 






Hence, Vertical displacement of L = - . r sin 6 tan ,8 



. ( 2 ) 




FIG. 528. Approximations in the motions in the Hackworth valve gear. 

To obtain the total displacement of L, take the sum of (i) and (2), 
noting that the result of (2) may be either positive or negative, de- 
pending on the setting of the guide bar HG. Thus 

Displacement of L = ( - Acos 9 ft- rte,n/3}sin& (3) 

The result shows that the radii of the component eccentrics are 

Radius of the 1 80 component eccentric = a = ~r. (4) 

Radius of the 90 component eccentric = b = r tan /?. ... (5) 

The first of these, a, is evidently constant. The other, /;, depends 
on the value of /?. To obtain the maximum value of b take the 

maximum values of /3, positive and 
negative, and hence of tan/:?, and 
obtain the numerical value of (4) 
and (5). In Fig. 529 CY is made 
equal to the 180 component eccentric 
a, and CX and CX' are respectively 
equal to the maximum positive and 
negative values of the 90 component 
eccentric b. The centre of the re- 
sultant eccentric will lie on E'YE, 
drawn parallel to XX', for all settings of the guide bar GH. Its 
limiting positions will be CE and CE' respectively. The resultant 
eccentric for any other setting of the guide bar may be obtained by 
calculating CX" from (5) ; the resultant eccentric will then be CE". 

The motion of the valve may be obtained thus for any setting of 
the guide bar GH by connecting it direct to the resultant eccentric 
found in this manner. E'YE in Fig. 529 is called the characteristic 
line of the gear. 




FIG. 529. Characteristic line for the 
Hackworth valve gear. 



OSCILLATING ENGINE MECHANISM 



485 



Oscillating engine mechanism. This mechanism is illustrated in 
Fig. 530 (a) ; a crank BC revolves about B, and the cylinder is 
capable of oscillating about an axis or trunnion at A. There is no 
connecting rod ; the piston rod is connected direct to the crank pin 
as shown. In Fig. 530 (ft) is shown a slider crank chain, in which the 





FIG. 530. Oscillating engine mechanism. 

crank BC revolves about C and A moves in the straight line AC. 
Comparison of (a) and (ft) will show that (a) has been produced 
from (ft) by an inversion of the mechanism. In (ft), AB is capable of 
swinging about AC, while in (a) AB is fixed and AC is capable of 
swinging about AB. 

In Fig. 530 (<r), let the lengths of AB and BC be denoted by L and 
R respectively in feet, and let the velocity of the crank pin be uni- 
formly equal to v feet per second. The- angular velocity of the 
crank pin will be given by 



w 1 = : radians per second. 



To obtain the angular velocity of the cylinder, resolve v into 
velocities respectively along and perpendicular to AC by means of 
the parallelogram CEFB. CD = V will be the component per- 
pendicular to AC. It is evident that the angular velocities of the 



4 86 



MACHINES AND HYDRAULICS 



cylinder and of the piston rod AC about A will be equal. Hence the 
angular velocity of the cylinder is 

V 

to 2 = ^ radians per second (2) 

ALx 

Draw BG perpendicular to AC and GH parallel to BC ; the 
triangles BCA and HGA will be similar. Hence, 

CA^BA^ L 
CG~BH~BH ; 



Substitution in (2) gives 



BH 
V.BH 



W2 ~L.CG* 

Again, the triangles BGC and FDC are similar. Hence, 
CG^CD^V 
CB CF v ' 

V V 

.-. CG= --R = . 

V (Wj 

Substitute this value in (3), giving 



-(3) 



V.BH wj 



BH 



L 



V 



(4) 



This result shows that the angular velocity of the cylinder is repre- 
sented by BH to a scale in which the angular velocity of the crank 
pin is represented by the constant length L. 
The cylinder has zero angular velocity in the 
two positions in which the crank and piston rod 
are at right angles. For positions of the crank 
below these, the cylinder is swinging towards 
the right, and its angular velocity may be called 
positive ; for crank positions above the zero 
positions, the angular velocity will be of op- 
posite sense, and may be described as negative. 
A polar diagram of angular velocity (Fig. 531) 
may be drawn by carrying out the construction 
for crank angles differing by 30. BH (Fig. 
530 (<:)) is measured for each position and set 
FIG. 531. Angular velocity off from C along the radius BC, towards B for 
%. f r " sdllating negative and away from B for positive values. 




QUICK-RETURN MOTIONS 



487 



The points D and E may be found by describing a circle on AB as 
diameter ; the angles BEA and BDA will then be each 90. At G 
and F the angular velocities of the cylinder will be v/AG and vjAF 
respectively. 

Examining Fig. 531, it will be noted that the crank, rotating 
uniformly, takes a longer time to traverse the arc DOE than it does 
to traverse the arc EFD. Hence the average angular velocity of the 
cylinder towards the left will be less than that towards the right. In- 
spection of the angular- velocity diagram will illustrate the same point. 

Quick-return motions. Advantage is taken of these facts in the 
quick-return motion fitted sometimes to shaping machines (Fig. 532). 




FIG. 532. Quick-return motion for a shaping machine. 

The tool G cuts the work H on the stroke towards the right only, 
and it is advantageous that this cutting stroke should be executed 
at a slower speed than that of the idle return stroke. The tool is 
fixed to a sliding ram F, guided so as to move horizontally. F is 
connected by a short link ED to the top of a slotted bar AD, which 
may oscillate about A. AD is driven by means of a uniformly- 
rotating crank BC, the crank pin of which engages a block at C 
which may slide in the slot of AD. The tool will be at the ends of 
its travel when BC is in the positions BK and BL, both of which are 
at 90 to AD (Fig. 533). 

Neglecting the effect of the obliquity of DE (Fig. 532), the travel 
of the tool may be found thus 

Half travel of tool = DO (Fig. 533). 



MACHINES AND HYDRAULICS 



A1 DO AD 

A1S ' BK=AB> 

. BK.AD AD 

~AB~ = AB' R ' 

AD 

and Travel of tool = d = 2 -^ . R, 

AJo 

where R is the radius of the crank BC. 




FIG. 533. Quick-return motion ; mechanism at the ends of the travel. 

The average speed on the cutting and return strokes may be 
calculated by first finding the times taken by the crank to traverse 
the arcs LMK and KNL respectively. 

Let T = time of i revolution of crank, in seconds. 

/ c = time to traverse arc LMK, in seconds. 

/ r = time to traverse arc KNL, in seconds. 
Then t c : t r : T = arc LMK : arc KNL : 2:rR. 

arc LMK 



27TR 






_arc KNL 

~~ 



Also, Average cutting speed = - . 

*c 

d 

Average return speed = -. 
IT 



The maximum cutting and return speeds may be obtained easily 
from Figs. 534 (d) and (ft) respectively. 



QUICK-RETURN MOTIONS 



489 



V c AD 



AD 



And 




A 

(a) (b) 

FIG. 534. Quick-return motion ; maximum cutting and return speeds. 

The Whitworth quick-return motion (Fig. 535) is produced by 
another inversion of the slider-crank-chain. A slotted link CD 
revolves on an axis at C, and is connected to the ram of the shaping 
machine by the rod DK. Motion is given to CD by means of a 
crank AB revolving round an axis at B ; its crank pin A has a block 




FIG. 536. Outline of Whitworth quick- 
return motion. 



it 

FIG. 535.^ Whitworth quick-return motion. 

bearing which slides in the slot of CD. Inspection of the outline 
diagram (Fig. 536) will show that BC is the crank in the slider-crank- 



490 



MACHINES AND HYDRAULICS 



chain, and now is fixed ; AB was formerly the connecting rod, and 
ACD was the line of stroke. 

The travel of the tool will be twice CD, and the tool will be at the 
ends of the stroke when CD is passing through its horizontal positions 
CE and CF. The arc through which BA turns during the cutting 
stroke is FHE, and the arc during the return stroke is EGF. The 
average speeds may be calculated in the same manner as for the 
quick-return motion discussed above. The maximum speed during 
the cutting stroke will occur when A is at H, and that during the 
return stroke will occur when A is at G. 

Let v = velocity of A, assumed uniform. 

V c = maximum cutting velocity. 

V R = maximum return velocity. 

TU V c CD _, CD , x 

Then = ^^, or, V c = ->v (i) 



CH' 



Also, 



or, 



CD 



V. 



v CG' 
Comparison of (i) and (2) shows that 

V C :V R = CG:CH. 

Cams. Cams are employed when the reciprocating motion to be 
given to the end of a rod or lever is of an irregular character. In 
Fig. 537 (a) the rod OA reciprocates vertically in the line OA, and is 

;A 





FIG. 537. Two examples of cams. 

driven by a disc B fixed to a revolving shaft C. The rim of B is 
shaped so as to give the required motion to the rod. In Fig. 537 (/>) 
is shown another example, in which the direction of motion of OA 
does not pass through the axis of the shaft C. 



EXERCISES ON CHAPTER XIX. 491 

The outline of the cam in Fig. 537 (a) is drawn by setting off equal 
angles oCi, iC2, 2C3, etc., and marking off the distances along OA, 
through which the rod is to travel while the cam is describing these 
equal angles. The points of division on OA are then brought to the 
corresponding radii by means of arcs described with C as centre. 
The outline in Fig. 537 (ft) is obtained by first drawing a circle with 
centre C and touching the line AO produced. Tangents are then 
drawn to this circle at equal angular intervals, and the distances 
along OA to be described during each of these intervals are brought 
to the corresponding tangent by means of arcs drawn with centre C. 



EXERCISES ON CHAPTER XIX. 

1. A link AB, 2 feet long, is horizontal at a given instant and is 
moving in a vertical plane. The velocity of the left-hand end A is lofeet 
per second upwards to the right at 45 degrees to AB. The velocity of B 
relative to A is 4 feet per second downwards. Find the actual velocity 
of B. 

2. In Question i, find the instantaneous centre of the rod and the 
velocity of the centre of the link. 

3. In a crank and connecting-rod mechanism, the crank CB is i foot 
long and the connecting rod AB is 4 feet long. The velocity of the 
crank pin B is uniform and equal to 10 feet per second. Divide the 
crank circle into intervals of 30 degrees, and find the velocity of the cross- 
head A for each of the crank positions so determined. Draw diagrams 
of velocity for a complete revolution (a) on a time base, (b) on a space 
base, of length to represent twice the stroke of the crosshead. 

4. In a double crank and connecting rod (four-bar chain) the cranks 
are AB, 1-5 feet long, and CD, 2-5 feet long; the connecting link BC 
is 2 feet long ; the bar AD is fixed and is 2-25 feet long. B has a linear 
velocity of 2 feet per second. For the position in which AB makes 
60 degrees with AD, find (a) the angular velocity of AB, (b) the velocity 
of C, (c) the angular velocity of DC. Use the instantaneous centre 
method, and check the result for (c) by the ratio given on p. 460. 

5. A four-bar chain has cranks AB and DC 2 and 175 feet long 
respectively ; the connecting link BC is 1-5 feet long and the bar AD is- 
fixed horizontally and is 1-5 feet long. The cranks are crossed. Draw 
the mechanism when AB makes 45 degrees with AD, and find a point E 
in the link BC (produced if necessary) which is moving in a vertical 
direction in this position of the mechanism. 

6. In the parallel motion shown in Fig. 498, AC is i inch and BC 
is inch long. The line joining AB is horizontal and ijj inch long.- AD 
is 1-5 inch long and is vertical when AC is horizontal. Find the length 
of CF, and confirm the result by drawing for the positions when AC 
makes 15 degrees with AB. 

7. In the mechanism shown in Fig. 502, the crank is 3 inches long 
and has a uniform speed of 60 revolutions per minute. The mass of the 



492 MACHINES AND HYDRAULICS 

reciprocating parts is 80 pounds. Find the forces in Ib. weight required to 
overcome inertia for crank positions differing by 30 degrees throughout the 
revolution. Plot these forces on a base line representing twice the stroke. 

8. In Question 7 the steam cylinder driving the mechanism is 

5 inches in diameter ; the net pressure of the steam urging the piston 
is 50 Ib. per square inch up to half stroke, and is 35, 24 and 20 Ib. per 
square inch at 0-6, 0-8 and the end of the stroke respectively. Draw the 
pressure diagram, and find the turning moment on the crank for each 
position given in Question 7, making allowance for inertia. Draw the 
turning-moment diagram. 

9. Two parallel shafts, 30 inches axis to axis, have each a crank 

6 inches long connected by a uniform steel rod 30 inches long, 2 inches 
deep and i inch wide. The shafts are driven at equal speeds. Take the 
density of steel to be 0-28 pound per cubic inch, and find the maximum 
upward and downward forces due to the inertia and weight of the rod 
when the speed is 150 revolutions per minute. 

10. In Question 9, the stress due to bending of the rod is limited to 
10,000 Ib. per square inch. What is the maximum permissible speed of 
the shafts in revolutions per minute ? 

11. Take the data of Question 3 and find the acceleration of the 
crosshead in the line of the stroke when the crank has travelled 
60 degrees from the inner dead point. Do this by use of Klein's con- 
struction. Calculate also the accelerations when the crosshead is at 
each end of the stroke. The mass of the reciprocating parts is 
500 pounds ; calculate the forces required to overcome their inertia in 
these positions. 

12. In Question 11, the total effective steam pressure on the piston 
when the crank is 60 degrees from the inner dead point is 9000 Ib. 
What will be the turning moment on the crank, (a) neglecting inertia, 
(&) taking account of inertia ? 

13. With the data of Question 3, find the acceleration image of the 
connecting rod when the crank is at 90 degrees to the connecting rod, 
and hence find the acceleration of the centre of the rod. 

14. In Question 13, take the connecting rod to be of uniform section 
and of mass 3 pounds per inch length. Find the resultant force which 
must act on the rod in order to overcome its inertia. 

15. Draw the bending-moment diagram for the connecting rod as given 
in Question 14, and state the value of the maximum bending moment. 

16. In an oscillating-engine mechanism, the crank is 2 feet long and 
makes 50 revolutions per minute. The distance between the centre of 
the crank shaft and the cylinder trunnion is 5 feet. Find the angular 
velocity of the cylinder when the crank is passing each dead point. 
Answer the same when the crank is at 45 degrees from the outer dead 
point. 

17. In Question 16, if the crank rotates clockwise, find the time of 
swing of the cylinder (a) from left to right, (b) from right to left. Cal- 
culate the angle through which the cylinder oscillates. 

18. The axis of a vertical rod passes through the axis of a horizontal 
shaft when produced downwards. The rod has a roller I inch diameter 



EXERCISES ON CHAPTER XIX. 493 

at its lower end and is driven vertically by a cam fixed to the shaft. The 
minimum radius of the cam is 2 inches, and it gives simple harmonic 
motion to the rod during the upward travel of 2 inches, followed by a 
period of rest while the shaft rotates through 45 degrees. The downward 
travel is simple harmonic. Draw the outline of the cam. 

19. Answer Question 18, supposing that the axis of the rod passes 
0-5 inch from the shaft axis when produced. 

20. Describe the character of the straining actions to which the 
coupling-rod of a locomotive engine is subject, and sketch an appropriate 
form of transverse section. In a locomotive having driving wheels of 
6 feet 6 inches diameter, the coupling-rod is 8 feet long between its 
centres, and is attached to cranks of I foot radius. Suppose the locomo- 
tive to travel at 60 miles an hour, and the weight W of the coupling-rod 
to be uniformly distributed along the 8 feet of its length, estimate the 
maximum bending moment to which it will be subjected. (I.C.E.) 

21. Describe, without proof, a construction for determining the 
acceleration of the slider in the slider-crank mechanism. Apply the 
construction to find the acceleration of the piston of an ordinary direct- 
acting engine when the crank is 30 from the inner dead centre. Length 
of crank, 8 inches. Length of connecting rod, 36 inches. Speed of 
crank shaft, 200 revolutions per minute. State the answer in feet per 
second per second. (L.U.) 

22. In a slider-crank chain AB is the connecting rod, 30 inches long, 
BC the crank and AC the horizontal line of stroke. In AB produced 
beyond B a point P is taken, BP being 18 inches. If the locus of P 
is an approximately vertical straight line, while AB travels through angles 
from o to 30 with the line of stroke, find a suitable length for BC. A 
load of 2000 pounds at P acts at right angles to the line of stroke ; find 
the pressure on the crosshead required to equilibrate, and find also the 
thrusts on the guides and crank when BAG = 30. (L.U.) 

23. In a four-bar chain ABCD, AB is the driving, CD the driven 
crank, and BC the coupler, DA being fixed. BC, produced if necessary, 
cuts AD in P. Show that the ratio of the angular velocity of CD to that 
of AB is PA/PD. Draw the velocity diagram for this chain when AB, 
BC, CD and DA are i, 6, 3 and 7 feet respectively, the angle BAD being 
90 and AB and CD being on the same side of AD. If the velocity of B 
is i foot per second, find the velocity of C, and check by using the ratio 
given above. (L.U.) 

24. A simple slide valve driven by an eccentric has a travel of 5 inches. 
The cut-off is at of the stroke of the piston, and the release takes place 
at | of the stroke. The lead is J inch. Assuming that the valve and 
piston have simple harmonic motions, find the outside and inside laps 
of the valve and the position of the piston when compression begins. 

(L.U.) 

25. A connecting rod is 5 feet long and 5 inches in diameter, assumed 
uniform throughout its length. Stroke of piston, 2 feet 6 inches. 
Revolutions per minute, 180. Weight of material, 480 Ib. per cubic foot. 
When the crank angle is 60 measured from the inner dead centre, draw 
the load and bending-moment curves on the connecting rod due to its 
inertia, and state the value of the maximum bending moment. (L.U.) 



CHAPTER XX. 

FLYWHEELS. GOVERNORS. 

Fluctuations in angular velocity. It frequently is the case that 
it is important to secure uniformity of angular velocity in some shaft 
belonging to a machine. This condition is usually very desirable in 
engines supplying motive power. In such cases there may be two 
kinds of disturbance owing to lack of equality in the rates of supply 
of energy to the engine and of abstraction of energy for driving 
purposes. In any machine we have the following equation (p. 328) 
for the balance of energies during a stated time : 

Energy supplied = energy abstracted + energy wasted in the machine. 

Suppose the energy supplied exceeds that required in order to 
satisfy the above equation, then the machine must be increasing 
its speed, as the excess energy must be disposed of, and the only 
method available is by the storage of additional kinetic energy in the 
parts of the machine. The converse will be the case if the energy 
supplied falls below that required in order to satisfy the equation. 

For simplicity, suppose the moment of resistance to rotation of 
the engine shaft, supplied by the machinery to be driven, to be 
uniform. There will be a demand then for a constant amount of 
energy during each revolution of the shaft. But the rate of supply 
of energy to the shaft is never uniform, depending as it does on the 
turning-moment diagram (p. 470), which may show great lack of 
uniformity. The result would be evidenced in rapid alterations in 
the angular velocity of the shaft, a jerky action which it is the 
function of the flywheel to remedy. This the flywheel does by storing 
the excess energy in the kinetic form, and its large moment of inertia 
enables it to do so with comparatively small changes in its speed. 
It is evident that, if the energy supplied during the revolution is 
exactly sufficient to satisfy the equation, the angular velocity of the 
flywheel at the end of the revolution will be equal to that at the 



FLYWHEELS 495 






beginning, i.e. there is no net gain or loss of speed despite inter- 
mediate small fluctuations. 

A second kind of variation in the angular velocity may occur 
during a period extending over several revolutions of the shaft. 
This would be owing to the supply of working fluid to the engine 
being too la-rge or not enough, and would be evidenced by a steady 
rise or fall in the speed of rotation. The flywheel alone is incom- 
petent to deal with this matter, which must be remedied by a 
contrivance called a governor. The governor is driven by the 
engine, and is constructed so that the relative positions of its parts 
alter with change of speed. These movements may be made to 
operate valves which control the supply of working fluid, and thus 
the shaft is kept rotating at a speed which may vary only slightly 
above and below the mean speed. Absolute steadiness of speed 
cannot be attained, for change of speed must occur before the 
governor will move into another position, and so operate the control 
valve. 

Kinetic energy of a flywheel. Some calculations regarding the 
capacity for energy of a given wheel and of its change of angular 
velocity in giving up a stated amount of energy will be found on 
p. 420. 

Let I = the moment of inertia of the wheel, pound and foot units. 
<o = its angular velocity, in radians per second. 

o> 2 
Then Kinetic energy of wheel = I foot-lb ................ (i) 

Let the wheel change its speed from Nj to N 2 revolutions per 
minute. Then 

o 9 

Change in kinetic energy = L I 2_ j 



Al N l 

AlSO, (0, = ^ . 27T = - 

60 30 



TT 

. Hence, 
30 



30 

I /7T 2 N 

Change in kinetic energy = ( -- l 
2g\ 900 

-^- 



= 0-00548 (Nj 2 - N 2 2 ) foot-lb. . . .( 2) 

<b 

Again, the kinetic energy at any speed N revolutions per minute 
varies as N 2 ; hence, if M be the kinetic energy at one revolution per 



496 MACHINES AND HYDRAULICS 

minute, the kinetic energy at N revolutions per minute will be 
MN 2 . Hence, the change in kinetic energy in passing from Nj to N,, 
revolutions per minute may be written : 

Change in kinetic energy = M(Nj 2 - N 2 ' J ) ............. (3) 

The fluctuation in speed of the wheel may be defined as (N\ - N 2 ), 
and the coefficient of fluctuation of speed is taken as the ratio which the 
fluctuation in speed bears to the mean speed. It is sufficiently 
accurate to write 

Mean speed = J(Nj + N 2 ). 

N N 
Hence, Coefficient of fluctuation of speed = -j-r^f - 



In practice, values of the coefficient of fluctuation of speed are 
found varying from 0-05 to 0-008 depending on the type of 
machinery. 

Dimensions of an engine flywheel. In estimating the dimensions 
of a flywheel for an engine, sufficient information must be given or 
assumed to enable the fluctuation of energy during a complete cycle 
to be ascertained. The process consists in reducing the driving 
effort on the piston to a tangential force acting on the crank pin, 
making proper allowance for the inertia of the reciprocating parts of 
the engine. A crank-effort diagram showing the values of this force 
throughout a cycle is drawn. Another diagram is drawn on the same 
base, showing the driving resistances to be overcome reduced to 
another tangential force acting at the crank pin. Comparison of 
these diagrams will enable the fluctuation of energy to be obtained. 

The turning-moment diagram may be used in order to calculate R, 
the tangential force acting on the crank pin. Thus, 

T = RxBC, 



where BC is the length of the crank in feet and T is the turning 
moment in Ib.-feet. 

Values of R for a steam engine having a single cylinder are set off 
in Fig. 538 on a base having a length equal to the circumference 
of the crank-pin circle. The resulting crank-effort diagram OCBDA 
shows the work done on the crank pin per revolution, neglecting 
friction. 

The whole of the work done on the crank pin is utilised in 
overcoming (a) frictional resistances in the engine, (b) the external 
resistances which are opposed by the machinery being driven. 



FLYWHEELS 497 

Assuming both of these to be uniform when reduced to a tangential 
force at the crank pin, it is evident that both will be taken account 
of by constructing a rectangular diagram OEFA of height equal to 
the average height of the crank-effort diagram. This rectangle will 
express graphically the equality of energy supplied and energy 
abstracted during the revolution. The driving effort and the resist- 
ance to be overcome are equal at G, H, K and L. 



F Q 




A L K B H GO 

FIG. 538. Fluctuation in energy in a steam engine. 

Consider the portions of the diagram showing the energies while 
the crank pin travels through the arc represented by GH. Work is 
done on the crank to an amount represented by the area GMCNH, 
and the work abstracted is represented by the area GMNH. Hence, 
surplus work, represented by the area MCN, has been done, with 
the result that the flywheel will have its angular velocity increased 
while the crank is passing from G to H. 

In the same way, while the crank is passing from H to K, energy 
represented by the sum of the areas HNB and BPK has been given 
to the crank and energy represented by the rectangle HNPK has 
been abstracted. Insufficient energy, represented by the area NBP, 
has been supplied during this interval and the flywheel will be 
decreasing its angular velocity. Hence, maximum speed will occur 
when the crank is at H and minimum speed when the crank is at K. 
Assuming that the speeds of the flywheel at G and K are equal, it 
follows that the excess energy represented by MCN will be equal to 
the deficient energy NBP, and it may be said that the energy repre- 
sented by the area MCN has been given to the flywheel and taken 
away again while the crank is passing from G to K. The fluctuation 
of energy is therefore given by either of the equal areas MCN or 
NBP. In the same way, the area PDQ is equal to the sum of the 
areas QAF and EMO, and represents the fluctuation of energy 
during the remainder of the revolution. 

The coefficient of fluctuation of energy may be defined as the ratio of 
the maximum fluctuation in energy to the total work done in one 
cycle, the cycle occupying one revolution in a steam engine and 
two revolutions in an internal combustion engine working on the 
four-stroke cycle. 

P.M. 2 1 



498 MACHINES AND HYDRAULICS 

Let E = the area MCN, expressed in foot-lb., representing 

the maximum fluctuation. 
a>! and CD., = the angular velocities of the flywheel at H and K 

respectively, in radians per sec. 

I = the moment of inertia of the wheel, in pound and 
foot units. 



Then 

*g *g 

i(<V-<^) = E ...................... ........... (i) 

Let a> 2 = TztOp 

where n is a fraction expressing the minimum permissible ratio of to 2 
to w,. Then T 

- / o o o\ T-^ 

-W-a-VHE. 



or I- ( 2 \ 

A o / .-)\ ..................... \*t 

u*(i-n 2 ) 

The moment of inertia which the flywheel must possess may be 
calculated from this equation. 

Centrifugal tension in flywheels. In Fig. 539 is shown the rim 
of a revolving flywheel, the other parts of the wheel being disregarded 
in what follows. The centrifugal forces produce radial loads on the 





FIG. 539. Rim of a revolving flywheel. FIG. 540. 

rim of a kind similar to those produced by internal pressure on a 
cylindrical shell (p. 94). 

Let v = the velocity of the rim, in feet per second. 
r = the mean radius of the rim, in feet. 
m = the mass of the rim, in pounds per foot circumference. 

Then Centrifugal force per foot circumference = Ib. 



GOVERNORS 499 



The resultant centrifugal force for half the wheel, corresponding to 
x d) in the cylindrical shell, will be 

mv' 1 2ml) 1 n 

R= X2r = - Ib. 

& g 

Let a = sectional area of the rim, in sq. inches. 

f/ = tensile stress on , Ib. per sq. inch. 
Then, assuming q to be distributed uniformly, 
R = 2qa (Fig. 540), 



or --- = 2qa, 



mv' ,, . , 

= Ib. per sq. inch. 



Let p = the density of the material, in pounds per cubic foot. 

Then m = px i x --- pounds per foot circumference, 
144 



pav 1 
= 



and 





or - Ib. per sq. inch .................................. (i) 

J 44A r 

This result shows that the stress due to centrifugal force is inde- 
pendent of the sectional area of the rim and of the radius of the 
wheel. Equation (i) may be written 



(2) 



We may deduce from this result that, for a given material having 
a density />, there is a maximum speed of rim corresponding to a safe 
stress q for the material in question, and that this speed is independent 
of the dimensions of the wheel. 

Governors. In Fig 541 is shown a simple type of governor such 
as was used by James Watt for controlling the speed of steam engines. 
Two heavy revolving masses A l and A 2 are suspended by links AjB 
and A 2 B to the upper end of a shaft BF ; the joints at B permit of 
A l and A 2 moving outwards or inwards in circular arcs. Another 
pair of links A l C l and A 9 C 2 connect A l and A 2 to a sleeve D, which 
will move upwards if A 1 and A 2 move, to a larger radius. The sleeve 
is connected by means of a bent lever, pivoted at G, and a rod H 
to a throttle valve K, which is situated in the pipe supplying steam 



500 



MACHINES AND HYDRAULICS 



to the engine and controls the supply of steam. The shaft BF is 
driven by the engine by means of bevel wheels at F, and hence the 

masses A 1 and A 2 will revolve 
about the axis of BF. The 
action of the centrifugal force, 
the weight, and the pull of the 
links on each revolving mass, 
will cause the mass to take up 
a definite radius depending on 
the engine speed. The working 
positions of the revolving masses 
are settled by the considera- 
tions that when they are at the 
extreme outer or inner working 
radius, the throttle valve should 
be closed completely or opened 
fully respectively. Each of these 
radii will correspond to a definite 
speed of rotation, and the engine 
controlled by the governor will 
be capable of a range of speed 
between these limits. In order 
that the range of speed should not be too great, the difference 
between the extreme working radii of the revolving masses should 
not be too large. 

In Fig. 542, three forms of simple governor are shown in outline ; 
these differ merely in the position of the point of suspension of the 




FIG. 541. Simple unloaded governor. 




(c) 



FIG. 542. Forms of simple governors. 



upper links. In (a\ the joint B is on the axis of rotation ; in (/;), the 
joints B x and B 2 are outside the axis, and are situated at the ends of 
a short cross piece ^E 2 which is fixed to the shaft; the same 
arrangement is used in (*), but the links are open in (b) and are 



GOVERNORS 501 



crossed in (c). The following argument applies equally to each of 
these cases : 

Let w = the angular velocity of the governor shaft, in radians 

per sec. 

m = t\\Q mass of either A l or A 2 , in pounds. 
r=the radius in feet of the revolving masses, corresponding 

to (o. 

h = the height in feet of the cone of revolution described by 
the links and shown by YB in (a) and by YO in (b) 
and (c). 

T = the pull in each link. 

Considering one revolving mass, it will be in equilibrium under 
the action of three forces, viz. its weight mg poundals, the centrifugal 
force u 2 mr poundals, and the pull T. It is evident that AjYB in (a) 
and AjYO in (ff) and (c) will be the triangle of forces for these 
forces in equilibrium. Hence, 

F r taPmr r 

= T> or > * = 7 

mg h mg h 



and h = .................................... (i') 

(0- 

This result neglects the effects of the mass of the link and also 
friction, and shows that the height is independent of the weight of 
the revolving masses. 

Such governors can be used for low speeds only. For example, 
if (o = 47r radians per second, corresponding to 120 revolutions per 
minute, h would be 0-2 foot nearly, a height which is not practicable. 
Running at low speeds, comparatively small forces will be available 
for operating the throttle valve unless the revolving masses are made 
heavy. Accordingly, simple governors usually have revolving masses 
of large dimensions, and are run at speeds rarely exceeding 60 
revolutions per minute. 

Loaded governor. The speed may be increased and the revolving 
masses kept small by the addition of a load on the sleeve (Fig. 
543 (a)). If M be the mass of this load, half its weight M^ will be 
carried by each pin Cj and C 2 . Cj is in equilibrium under the 
action of the three forces JMjf, the pull P in C 1 A 1 and a horizontal 
force Q supplied by the sleeve (Fig. 543 (&)). The pull P is trans- 
mitted to Aj by the link, and applies a force P to the revolving 



5 02 



MACHINES AND HYDRAULICS 



mass which may be resolved into a vertical force ^M^ and a 
horizontal force Q. If all four links be equal, the triangle of forces 
AjDjCj will give 1M> O = h r 

where h and rare equal respectively to BY and A T Y in Fig. 543 (a). 
Hence, Q 




(ft) (b) 

FIG. 543. Loaded Porter governor. 

The revolving mass A l is now subjected to three forces, viz. T, the 
resultant (F- Q) of the centrifugal force F and Q, and the resultant 
(mg+ Mg) of the weights (Fig. 543 (a)). AjBY will be the triangle 
of forces. Hence, 

F-Q_ * 
h! 



or 




m 






() 



mm 



m 



(2") 



Comparison of these results with equations (i) and (i') for the 
simple governor will show that, for the same value of w, h in the loaded 
governor will be greater than that for the simple governor in the 

r /M + m\ 
ratio of - - ). 
m / 






EFFECT OF THE GOVERNOR ARMS 503 

Equation (2") for the loaded governor shows that h will be made 
larger or smaller by an increase or diminution of M without alteration 
in the values of m and to. The arrangement therefore admits of 
adjustment of the working radii of the revolving masses by means of 
varying the load. 

Frictional effects in the governor mechanism may be taken into account 
by the artifice of eliminating the frictional forces and applying instead 
a force to the sleeve, which will have the same effect. This force 
must be applied always of sense opposite to that of the direction of 
motion of the sleeve. The effect might be produced by imagining 
a mass M F pounds to be added to the load M if the sleeve is rising, 
and to be abstracted if the sleeve is falling. Equation (2) then 
becomes : ^M F ^ g_ /MM F \g > 

m )h \ m ) 1i 






the positive sign being used if the sleeve is rising or attempting to 
rise, and the negative sign if the sleeve is falling or attempting to fall. 
Two extreme values of h may thus be calculated from (3), indi- 
cating that, owing to friction, the governor may remain at any height 
intermediate between these extreme values while running at a given 
steady speed w. The effect on the engine is to permit some variation 
in speed to occur before the governor will begin to respond by 
altering its height. 

Effect of the governor arms. In Fig. 544, AB is a rod hinged at 
A and rotating about the vertical axis AK. Centrifugal force and 
gravity will compel the rod to assume an angle a to the vertical, the 
value of a depending on the speed of rotation. Steady conditions 
will be obtained when the total moment of gravity about A is equal 
to the total moment of the centrifugal force. Consider a small 
portion of the mass of the rod at P, and let the rod be uniform. 

L et m = the mass of the small portion, in pounds. 
r = the radius of the small portion, in feet. 
y = its distance from A, in feet. 
M = the total mass of the rod, in pounds. 
Y = the distance of the centre of mass G from A in feet. 
L = the length of the rod, in feet. 
R = the radius of B, in feet. 
H = the vertical height of A over B, in feet. 



MACHINES AND HYDRAULICS 



Then, taking moments about A of the forces acting on the small 
P9 rtion, we have mg x Np = m ^ r x AN? 

(i) 



or 



mgxysma ma 2 }' sin a xycos , 



g , my = (o cos a . my. 
The total moments for the whole rod 




will be obtained by 
integrating both sides of this equation, 
giving 



g 



my = w 2 cos a 



i: 



= (o' 2 cosal 



where I A is the moment of inertia of 
the rod with respect to A, viz. 4ML 2 
(p. 415). Hence, 



.(2) 



( R jj 

FIG. 544. A revolving uniform load. 



This result determines the required 
relation of a and w for a given uniform 
rod. 
In the actual governor the arm is constrained by the action of the 

revolving mass to rotate at an angle to the vertical, differing from 

that given for a free arm in (2) above. In this 

case, the moment of the weight of the arm may be 

calculated still as above by imagining that the whole ; 

arm is concentrated at the centre of gravity. The 

moment of the centrifugal force may be calculated - 

by first imagining that the whole arm is concentrated 

at the centre of mass and calculating the centrifugal 

force f produced thereby. Then find the position 

of f (Fig. 545) in order that its moment may agree with the integrated 

result of the right-hand side of (i). Thus, 




R f L 

Moment off=fx = Mw 2 x = w 2 sin a cos a I my 2 

= w 2 sin a cos a -..-- 
3 

,,R H ML 2 , _,. x 

= 0,2 _ . __ . ___ ; ( S ee Fig. 544) 

2 MR RHM 

. . or x = or - 5 

2 3 

2TJ 

x Trri 



(3) 



EFFECT OF THE GOVERNOR ARMS 



505 



Effect of the arms in a loaded governor. This result may be 
applied to a loaded Porter governor having equal arms (Fig. 546 (#)). 
mg is the weight of each arm, and in the case of AB is equivalent to 
a force \mg at A and an equal force at B (Fig. 546 ()). /for each 
arm is equivalent to a force f / acting at B, together with a force y 
acting at the vertical spindle AC. The mass of the revolving mass 
being M, its weight will be M,^ and the centrifugal force will be 





FIG. 546. Forces in a loaded Porter governor, including effects due to the arms. 

Mw 2 R. The mass of the load is L, and half its weight, viz. 
will be borne by the right-hand arms as shown. The forces at A are 
balanced direct by the reactions of the -pin securing AB to the 
spindle at A. The force J/ at C is balanced by the reaction on the 
sleeve produced by the spindle. Draw CD perpendicular to AC, 
and produce AB to cut CD at D (Fig. 546 ()). 

Take moments about D of the remaining forces, remembering 

that /=ww 2 . 



(MVR 

0)2 (M 



) R 



2 R, 



or 



} H = 

/M 

i V 



(4) 



Stability of a governor. Considering one revolving mass of a 
governor, the centrifugal force is given by 

(i) 



5<>6 



MACHINES AND HYDRAULICS 



Suppose a> to be increased by a small amount Sw, and that, in con- 
sequence, r increases by 8r and F by 8F. For the new position 

we have 



m(^r-\- 2wr. &w -f- w 2 . 8rJ, (2) 

by neglecting the square of 6w, and also the term involving the pro- 
duct of the small quantities Sw and 8r. Subtraction of (2) and (i) 
gives 

8 = m(2(ur. Sa> + w 2 . 8r) (3) 

Dividing (3) by (i), we have 

8F 2<D) 



= 2 H , 
10 r 



or 



Sto 8 



(4) 



If 8(D is an increase in angular velocity, the left-hand side is posi- 
tive ; hence -pr must be greater than , i.e. the rate of increase of 

the centrifugal force must be greater than the rate of increase of the 
B radius. The result expresses 

the condition of stability in a 
governor, i.e. the moving to a 
definite new radius and remain- 
ing there when the revolving 
masses suffer a change in speed. 
An interesting example of a 
governor which exhibits neutral 
equilibrium is produced by 




FIG. 547. Parabolic governor. 



arranging that the revolving masses move about . B in parabolic 
instead of circular arcs (Fig. 547). Here the pull of the link on 
A is supplied by a normal pressure T given by the guide. Hence 
we have, as before, 



or, 



mg 



K 

Now h BY is the subnormal to the dotted parabola, and it is 
known from geometry that the subnormal to a given parabola is 
constant ; hence h is constant, and therefore w must also be constant. 



HARTNELL GOVERNOR 



507 



A governor of this type is isochronous, i.e. it will run at one speed 
only if friction be absent, and any change from this speed will send 
the revolving masses immediately to one or other extreme end of the 
range. 

The question of sensitiveness of a governor is allied closely to its 
stability. The change of radius for a given fractional change in 
speed is large in a sensitive governor, but if too large, as in the para- 
bolic governor, the stability may disappear. 

Hartnell governor. In the spring loaded Hartnell governor (Fig. 
548), the revolving masses A l and A 2 are supported by bent 
levers, which are pivoted at B x and B 2 on 
pins supported by a bracket (not shown in 
the illustration) which is fixed to and driven 
by the shaft. A spring E bearing on a sleeve 
1) presses downwards the ends CjC 2 of the 
bent levers. The revolving masses travel a 
small distance only from the verticals passing 
through Bj and B 2 ; hence the effect of their 
weights in exercising control may be neglected. 
Supposing, for simplicity, that A l E l and BjCj 
are equal, then, by taking moments about Bj, we see that the 
centrifugal force F acting on A x will be equal to one-half of the 
total force 2Q exerted by the spring. Provided that the adjustment 
of the spring is correct, this governor will possess great sensitiveness, 
but easily may be made unstable. 

Suppose that the revolving masses are displaced from r^ to a 
slightly greater radius r^ without changing the angular velocity u>, the 
centrifugal force will be increased to F + SF, and Q will be increased 
to Q + SQ, owing to the additional compression of the spring. 
Assuming that these forces are equal, we have 



or 




B, 



FIG. 548. Hartnell governor. 



o,v 2 = Q + 

Also, initially, mvrr^ Q ; 

.'. m>' 2 (r. 2 -r l ) = SQ. 



Hence, 



Q ' 
<V_SQ 



The condition, therefore, that the governor may remain in the 
new position with the speed unaltered is that the rate of increase of 



508 MACHINES AND HYDRAULICS 

the force in the spring is equal to the rate of increase of the radius 
of the revolving masses. This condition may be secured by adjusting 
the spring; but as the stability of the governor then would be 
neutral, the practical adjustment is made so as to disagree with the 
condition above expressed. 

Effort of a governor. Suppose that the speed of a governor is 
increased from <0j to w. 2 , and that the sleeve is held so as to prevent 
outward movement of the revolving masses. There will be additional 
centrifugal force, and consequently an effort will be exerted on the 
sleeve which may be utilised in overcoming the resistance offered by 
the control valve mechanism. It is evident that the effort will 
diminish if outward movement of the revolving masses be permitted, 
and will attain zero value when they reach the position corresponding 
with the new speed of rotation. The effort of the governor may be 
defined as the average effort exerted on the sleeve during a given 
change of speed executed in the manner described above, and may 
be taken as 0-5 of the maximum effort. 

Taking a simple governor (p. 501) for which 



if P is the maximum effort in lb., it may be imagined that P is 
produced by the weight of a load M, arranged as in the loaded 
governor (p. 502). Hence, 

w 2 2 = (- -U (2) 

Hence, from (i) and (2), = 
o> 2 2 - Wl 2 _ 






M H^?V (3) 



Let P = weight of M, in lb. 

w = weight of m, in lb. 

Then 



_ /w.S-wA /a> 2 2 \ 

P = ( -* l - ) w = ( -\ - i I w. 

\ w, / VWj 2 / 



Let Wo = wwj , 

so that n expresses the fractional change in speed. Then 



P = ( -i)w 

\ <v 

= (2 -i) W ; 

.". effort of a simple governor = 4- (n 2 - i)ze/lb .................... (4) 



BALANCING 



59 



In the case of a loaded governor (p. 502), P may be taken as being 
equivalent to the weight of an additional mass M l applied to M. 
Similar reasoning to that employed for the simple governor will give 






<5) 



where w and W are the weights of one revolving mass and of the 
load respectively in lb. 

Balancing. The complete treatment of the principles of balancing 
the moving parts of an engine or other machine is beyond the scope 
of this book. * Reference will be made to some 
of the easier principles. 

Two rotating bodies may be made to balance 
each other if both have their centres of mass 
in the same plane which is perpendicular to the 
axis of rotation, and if the centre of mass of 
the combined bodies is in the axis of rotation 
(p. 426). 

Thus, in Fig. 549, m l and m., will balance, 
provided the forces F, F are equal and are in 
the same straight line. This will be the case 
if m- [ r 1 = m z r. 2 and if the line joining G 1 and G. 2 
passes through the axis at right angles. 

Three revolving bodies may balance, provided the resultant centri- 
fugal force of two of them, F l and F 3 in Fig. 550, is equal and 
opposite to the centrifugal force, F 2 , of the other. It is thus evident 




FIG. 549. Balance of two 
revolving bodies. 







FIG. 550. Balance of three revolving bodies. 

that all three centres of mass must be contained by the same plane 

*For a complete discussion on this subject, the student is referred to The 
Balancing of Engines, by Prof. W. E. Dalby ; (Arnold). 



5io MACHINES AND HYDRAULICS 

which also contains the axis of rotation, and that the bodies must be 
disposed as shown in Fig. 550. 

Taking dimensions as indicated in Fig. 550, 

F 1 + F 3 = F 2 ................................. (i) 

and F 1 1 = F 3 # 3 ............................... (2) 

These equations indicate the conditions of equilibrium to be 
fulfilled, and may be reduced thus : 

From ( i ), to 2 w 1 r 1 + w 2 w 3 r 3 = to 2 /0 2 ?2 , 

or m^ + m 3 r 3 = m^r* ............................ (3) 

This result shows that the centre of mass of the combined bodies 
falls on the axis of rotation. 

From (2), iD 2 m l r l a l = <t) 2 m s r 3 a 3J 

or 0^0! = vyz 3 ............................. (4) 

This equation secures that there shall be no rocking couple set up. 

In this case, there are two equations, (3) and (4), and eight quantities 
involved ; hence six of these must be given or assumed. 

The balancing of four or more revolving masses is capable of 
many solutions, and graphical or semi-graphical methods are best. 

Locomotive balancing. As an illustration of the method by 
means of which the balancing of four revolving masses may be 
carried out, the following example of a locomotive should be studied. 

Equal masses m^ and m are given (Fig. 551), rotating at equal 
radii r, r, and symmetrically disposed in relation to the wheels A 
and B in the planes of which balance weights are to be placed. 
The term " balance weights " is used to denote bodies which must be 
attached to the mechanism for the purpose of obtaining balance. 
a is the distance of m^ from A and of m z from B ; $ is the distance 
of m l from B and of m. 2 from A. 

Balance the centrifugal force of m l separately by attaching balance 
weights to A and B at the same radius r ; if these balance weights 
be represented by A l and Bj respectively, their masses will be 
given by A^K^m, ........ ' ......................... (,) 

and Ajfl^B/ ............................... (2) 



i i , 

From which, A l = ^ 



-T) \, , x 

and B i=ri> ............................... (4) 

In the same way, balance F 2 by attaching balance weights to A 
and B at the same radius r. Let these be A 2 and B 2 respectively. 
Then it is evident from symmetry that A l and B 2 are equal ; also A 2 



LOCOMOTIVE BALANCING 



is equal to B x . These balance weights are shown in the elevations 
of the wheels in Fig. 551 ; the views are taken in the directions of 
the arrows c and d shown in the plan. Find the centres of gravity 
of AjA 2 and also of B T B 2 by joining their centres and dividing the 
distances in G A and G so that 



and 




Elev? of B 



FIG. 551. Balancing in an inside cylinder locomotive. 

Now, since A l + B x = m l and Bj = A 2 , it follows that A l + A 2 = m l , 
and for a similar reason B 1 + B 2 = w 2 . Hence, if instead of A lt A 2 , 
Bp B 2 , a mass equal to m l be placed at G A , and if another equal to 
m. 2 be placed at G B , the four masses will be in balance. Or the 
ordinary practical solution may be obtained by applying balance 
weights having their centres of mass in OG A and OG B produced. 
Let M A and M B be their masses respectively. Then balance will be 
secured if M A x K A O = m l x G A O 

and M B x K B O = m. 2 x G B O. 

M A and M B will be equal, provided their radii K A O and K B O are 

equal. 

Graphical solution of balancing problems. This solution depends 
on the principles that the centrifugal forces must not produce (a) a 
resultant force ; () a resultant couple. Reference is made to 
Fig- 55 2 - 



512 



MACHINES AND HYDRAULICS 



Since the angular velocities of all the bodies are equal, the centri- 
fugal forces Fj, F 2 , etc., may be represented by the products m^r^ 
w 2 r. 2 , etc. Each force, such as F lt lies in a plane which also con- 
tains the axis of rotation, and may be moved along this plane until 
it comes into a reference plane OZ which is perpendicular to the 
axis of rotation. To leave matters unaltered, with F l acting in OZ, 
a couple must be applied in the plane containing F x and the axis of 
rotation; the moment of this couple will be L 1 = F 1 a 1 , where a x is 
the distance of the plane of rotation of F 1 from the reference plane. 
z 






- i*2~~ 




J 





a 3 








FIG. 552. Graphical solution of balancing four revolving bodies. 

The couple can be represented by an axis, or vector similar to that 
used in representing angular velocities (p. 400), and this may be 
drawn as Lj from O in the reference plane. Treating similarly the 
other forces, we have four forces F 19 F 2 , F 3 and F 4 acting in the 
reference plane at O, together with four couples represented by 
the axes L lf L 2 , L 3 and L 4 also in the reference plane. 

The first condition of equilibrium will be satisfied if the polygon 
of forces ABCD closes. The second condition of equilibrium will 
be satisfied if the polygon of axes of couples EFGH closes. 

It will be clear that, in order to satisfy these conditions, some 



BALANCING OF ROTATING MASSES 



513 



attention must be paid to the data. The polygon of forces will be 
impossible or insoluble if there be less or more than two unknown 
quantities. These may be the magnitudes of two of the forces, or 
the direction of two forces, or one magnitude and one direction. 

Similarly, the polygon of couples requires two unknowns, viz. the 
magnitudes of two couples, or the directions of two axes, or one 
magnitude and one direction. 

Apparatus for testing balance. The above solution may be 
applied to any number of revolving masses exceeding three in 
number. A convenient apparatus for testing its truth is illustrated 
in Fig. 553. A wooden frame is 
slung from a support by three 
chains and carries a shaft having 
four discs. Various weights may 
be attached to the discs, which 
may be placed at any angle rela- 
tive to one another and may be 
fixed at any place on the length 
of the shaft. The shaft is driven 
by means of a small electro-motor 
also carried by the frame. If the 
revolving masses are in balance, 
no vibration of the frame will 
occur when the machine is 
running. A problem worked out 
on paper therefore can be tested 

easily. Another interesting point FIG. 553. Apparatus for experiments on the 
n j i_ . i balancing of revolving bodies. 

illustrated by this apparatus may 

be noticed as the speed rises ; if there be want of balance, violent 
vibrations will occur at a certain speed, viz. that speed at which 
the natural period of oscillation of the whole apparatus is equal to 
the speed of rotation of the shaft. 

EXPT. 49. The following data will serve to illustrate one of the 
many problems which may arise. Let there be four revolving 




masses, m l 



m 3 , m all known and of values selected from the 



weights supplied with the apparatus. Let the radii be equal. Then 
m lt m z , m s and m may be taken to represent F 15 F 2 , F 3 and F 4 
respectively. Assume the directions of F l and F 2 , and find, by the 
polygon of forces, the remaining two directions. This will also 
settle the directions of all the axes of couples, and, as there must be 
still two unknowns, assume values for a^ and a 2 in Fig. 552, and 
find the remaining axes by use of the polygon of couples. 
VM. 2 K 



514 



MACHINES AND HYDRAULICS 



The values taken to represent the couples may be m-^a^ m^a^ etc., 
as the radii are equal. The polygon gives the values of m B a 3 and 
#/ 4 4 , and a B and # 4 may be found by dividing these by m 3 and m 4 
respectively. Having worked out the solution on paper, arrange the 
apparatus in accordance with your solution, and then test by actually 
running it. 



Balance of reciprocating masses. 



is shown a set of four balanced rotating masses, 
fi 



Referring to Fig. 554, in which 
Wo, wio and m A . 




[t]m 



FIG. 554. Components of the centrifugal forces in a set of four balanced revolving bodies. 

the balance will not be affected if we imagine F 1? F 2 , F 3 and F 4 to 
be resolved horizontally and vertically. It will be evident now that 

the horizontal components must 
balance independently, and so 
also must the vertical com- 
ponents. 

Let the masses be removed 
and arranged so that they may 
be driven in vertical lines by 
means of cranks taking the places 
of the discs, and connected by 
rods of length sufficient to give 
the masses practically simple 
harmonic motion (Fig. 555). It 
is evident that we have got rid 

FIG. 555. The bodies in Fig. 554 arranged as simply of the horizontal COm- 
reciprocatmg masses. r J 

ponents of the forces F lt F 2 , etc., 

and retained the vertical components. Hence, if the masses were 
in balance in their original positions on the discs, the forces 
due to their inertia will also be in balance when the same masses 
vibrate in the manner illustrated in Fig. 555 with simple harmonic 



MOTION OF THE RECIPROCATING PARTS 515 

motion. This leads to the rule that primary balance (i.e. balance 
neglecting the oblique action of the connecting rods) may be secured 
by imagining the reciprocating masses to be attached to discs, and 
treating the problem as one in revolving masses. 

Approximate equations for the velocity and acceleration of the 
reciprocating parts. In Fig. 556, let the crank and connecting rod 
be R and L feet long respectively, and let the crank make an angle 



,-r w 

i-'H5 


D ~t~~ 


\ i 


/ 
/ 
/ 



I FIG. 556. Motion of the reciprocating masses. 

a with the centre line, the angle which the connecting rod makes 
with the same line being /3. BD is perpendicular to AC and x is 
the distance between A and C. 

Let the angular velocity to of CB be uniform. Then 



= R COS a + L COS /3 ( I ) 

Also, BD = R sin a = L sin ft 

a R 

. . sin p = :=r- sm a ; 



and cos ft = \/i - sin 2 /? 

/ R2 

= (i - j-2sin 

Substituting this value in (i) gives 

/ R2 

x = R cos a -f L i - -o sin 2 a 



/ R2 
( i - y-o 
V L 



On expanding the factor in brackets by the binomial theorem, 
two terms only need be taken, as, for ordinary ratios of R to L, 
the remaining terms are negligible. Hence, 

x = R cos a -f L 

\ 

R 2 
-Rcosa + L-- r sin (3) 



516 MACHINES AND HYDRAULICS 

To obtain the velocity of A in the direction of AC, differentiate x 
with respect to the time, giving 

dx d& R 2 . da 



.. da. 

Also, ^ = w 

and 2 sin a cos a = sin 2 a. 

dx . . wR 2 . 

Hence, V A = = - wR sin a -- - sin 2a ................... (4) 

d t 2 .L/ 

From this equation, the velocity of A may be calculated for any 
crank position. To obtain the acceleration of A in the line of AC, 
differentiate (4) with respect to the time, giving 

da co 



9T> w 2 R 2 

= - cofK COS a -- - COS 2<x ...... , ............ (5) 

The acceleration of A may be calculated from this equation for 
any crank angle. Let M be the mass of the reciprocating parts. 
Then the force required in order to overcome their inertia when the 
crank is at an angle a is 

MV 2 R 2 
P= -Mw 2 Rcosa- cos 20, ................... (6) 

I j 

Suppose that the mass M is concentrated at the crank-pin centre B 
I / (Fig- 55?)- Then the central force 

I VMU^R required will be Mo> 2 R, and the 

_ _ , fid/ component of this force parallel to 

" Mw*RcoscL t* 16 centre nne AC will be Mw 2 R cos a. 

Evidently this is equal to the first 
term of (6). The factor cos 2 a in 
tne second term, having reference to 
/ an angle 2 a, which will be double of 

/ the crank angle in all crank positions, 

*"- -*' may be interpreted by reference to 

FIG. 557. Equivalent imaginary primary an imaginary crank rotating at twice 

the angular velocity of the real crank, 
i.e. o> for the imaginary crank would be equal to 2w. Hence, 



4 

Therefore the second term in (6) becomes 

'-j COS2a = ^ COS 2a ..(7) 



PRIMARY AND SECONDARY BALANCING 517 

Let a mass equal to M be concentrated at a crank radius r 
(Fig. 558), set at an angle 2 a and rotating 
with angular velocity w . Then 

Central force = Mo> V. - _>__^ ,.2oi \ 



Component of this force parallel to the a; r ^ _^s^_ A_ .\_ 
line of stroke = M<o V cos 2 a. ^ y 

If this be made equal to (7), we have \ / 



Mw 2R 2 

- j . 

4L secondary mass. 



w 

Mto VcOS2<X = - - cos 20tj FlG 55 8. Equivalent imaginary 



or r = .................................................. (8) 

4L 

Hence the second term in equation (6) would be produced by a 
mass M equal to that of the reciprocating parts, concentrated at 

R 2 

a crank radius , its crank rotating at an angular velocity double 
4-L 

that of the engine crank and making an angle with CA in Fig. 556 
double of that made by the engine crank. The complete equivalent 
system is shown in Fig. 559, where CB is the real crank and CD is 
the imaginary crank. The balancing of the effects of M at B is 



Mti/Rcosa 







FIG. 559. Effects of the reciprocating masses produced by an imaginary revolving system. 

called primary balancing, and balancing the effects of M at D is called 
secondary balancing. The disturbances produced in the direction of 
the line of the stroke, if no attempt at balancing is made, may be 
calculated easily from the first term of equation (6) for primary 
disturbances and from the second term of the same equation for 
secondary disturbances. It will be understood, of course, that, if 
the disturbances on the engine frame are being calculated, the senses 
of the forces shown in Fig. 559 must be reversed. 

EXAMPLE. A horizontal engine, stroke 2 feet, mass of reciprocating 
parts 300 pounds, has a speed of 240 revolutions per minute. Find the 



MACHINES AND HYDRAULICS 



primary and secondary disturbances on the frame when the crank is at o, 
45> 9j 1 3S arj d 1 80 from the inner dead point. The connecting rod is 
4 feet long. 

Primary disturbance : P l = cos a Ib. weight. 



T- 



. 27r = 87r radians per sec. 



300 X 6ATT 2 X I 

= cos a 



5880 cos a Ib. weight. 



a 





45 


90 


135 


1 80 






j 




I 




cos a - 


+ I 


+ >75 


o 




~ l 


PU Ib. weight 


+ 5880 


+4160 


o 


-4160 


-5880 



P! is denoted positive when the disturbance on the frame is in the sense 
from B towards A (Fig. 556), and negative when of the opposite sense. 
Secondary disturbances : 

_ Mft> 2 R 2 
P 8? __c6s2a 

300 x 6471-- x i 
= J cos 20, 

32-2x4 

= 1470 cos 2a Ib. weight. 

The same convention regarding signs being adopted, the disturbances 
will have values as given below : 



a 


o 


45 


90 


135 


1 80 


2tt 


o 


90 


1 80 


270 


360 


COS 2tt - 


+ 1 





- I 


o 


+ 1 


P 2 , Ib. weight 


+ 1470 





-1470 


o +1470 



The combined primary and secondary disturbances will be obtained by 
taking the algebraic sum of the corresponding values of P, and P 2 : 



a 





45 90 


135 


1 80 


(P^Psj), Ib. weight 


+ 7350 


+ 4160 


-1470 


-4160 


-4410 



WHIRLING OF SHAFTS 



510 



Whirling of shafts. In Fig. 560 is shown a vertical shaft AB 
running in swivel bearings at A and B ; these bearings do not in any 
way restrain the directions of the shaft axis 
at A and B ; hence, bending of the shaft 
will correspond to the case of a ' ^am simply 
supported at the ends. A heavy wheel is 
mounted on the shaft midway between the 
bearings, and it is assumed that its centre 
of mass C does not fall quite in the shaft 
axis. The effects of centrifugal force may 
be examined as follows : 

Let M = the mass of the wheel, in pounds. 
R = the distance in feet of the centre 

of mass of the wheel from the 

shaft axis. 

A = the deflection produced by cen- 
trifugal force, in feet. 
w = the angular velocity, in radians 

per sec. 

L = the length of the shaft, in inches. 
I = the moment of inertia, or second 

moment of area, of the shaft 

section about a diameter, inch 

units. 
E = Young's modulus, Ib. per sq. inch. 




FlG " s6 - W S ng of a loaded 



Centrifugal force = P 



Also, 



PL 3 



Mo> 2 (R + A) 

S 
inches (p. i 



Ib. weight 



PL 3 

-- feet, 
57 6EI 



and 



L 3 



Equating (i) and (2), we have 



L 3 



or 



Mw 2 L 3 R = 

. 



- a> 2 ML 3 A 









It is evident that a critical speed will occur when the denominator 
of this fraction becomes zero ; the deflection will become very large 



520 MACHINES AND HYDRAULICS 

then, and the shaft is said to whirl. To obtain this speed, we have 



If the shaft is of steel, this equation will reduce to the following 
form by using the usual values of the coefficients : 

o>= 746000^^ ...................... (5) 

If the wheel in Fig. 560 be removed, the plain shaft will whirl, but 
at a much higher speed of revolution. The effect is owing to some- 
what similar conditions to those which produce elastic instability in a 
long strut (p. 228), viz. want of perfect straightness and of perfect 
uniformity in elastic properties. Any slight deflection will be in- 
creased indefinitely when the whirling speed is attained. 



-*- K 




fb) 

FIG. 561. Whirling of a uniform shaft having swivel bearings. 

Fig. 561 (a) shows a uniform shaft in swivel bearings at A and B 
and deflected to the curve ACB by whirling. 

Let m = the mass in pounds per inch length. 
L = the length of the shaft, in inches. 
y = the radius in inches at P, distant x inches from O. 
A = the maximum radius OC, in inches, 
w = the angular velocity, in radians per second. 
F'= the centrifugal force at any point, in Ib. weight per inch 

length. 

M = the bending moment at any section, in Ib. -inches. 
I = the moment of inertia of the shaft section, in inch units. 
E = Young's modulus, in Ib. weight per square inch. 
g= acceleration due to gravitation, inches per second per 
second. 



WHIRLING OF SHAFTS 521 



Then, at P, F = Ib. weight per inch length; ...(i) 

<b 

.'. Focj. 

This result indicates that the curve in Fig. 561 (a) not only repre- 
sents the deflection, but also the load per unit length to another 
scale. Hence, we may write 

F-y, 

where c is a numerical coefficient rectifying the scale. 

Now, if the coordinates y and x refer to a given deflection curve, 
the second differential coefficients, when plotted, will represent a 
curve of bending moments, and the fourth differential coefficients 
will represent a curve of loads which would produce the given deflec- 
tion curve. Hence, in the present case, 



From this expression, the shape of the deflection curve has to be 
obtained, and may be inferred to be a curve of cosines. Thus, take 
the equation ,jy = cos and obtain the fourth differential coefficient : 

y = cos ; -j- = - sin 6 -= = _ C os 6 -~ = sin ; -^- = cos 0. 
' dx dx* dx* dx^ 

Therefore, in the curve representing the equation y = cos 0, 

d*y 

-~ = cos 9 =y. 

dx* 

In Fig. 561 (a\ there is zero deflection at A and B and maximum 
deflection at C. Hence the corresponding cosine curve (Fig. 561 (/>)) 

7T 7T 

will have the origin at O, OE and CD will represent +- and - 

respectively (for which the cosines are zero), and OG will represent 
cos 0=1. HK, corresponding to y in Fig. 561 (), will represent 
cos 0, where is the angle represented by OH, corresponding to x in 
Fig. 561 (a). From the diagrams, we have 

0- x 7r _ 7r r 
-L'2-L*-. 



y cos TT 

Also, TT = - - = cos x ; 

A coso L ' 



Obtaining the fourth differential coefficient of this, 

d^V . 7T 4 7T 



(3) 

, , 

(4) 



522 MACHINES AND HYDRAULICS 

d^y M 
N W ' = > 



AISO, ^ =F ; 

.-. g=... (V) 

Hence, from (i) and (5), 

* = ~^; ... (6) 

dx^ EI^- 
Equating (4) and (6), 

2 4 

= A , cos^-^. ...(7) 



This equation is true for any corresponding values of y and x, 
Take the value x o, when 



7T 
COS -,T = COS O = I 



and = A. 



. 7T 4 

Hence, = A ...................................... ( 8 ) 



The deflection cancels from both sides of this equation, indicating 
that a critical speed w has been attained, and giving the result 






< 



This result expresses the whirling speed. If the bearings restrain 
axially the directions of the shaft at A and B (Fig. 561 (#)), then it 
may be shown that the whirling speed is given by 

' 2 



EXERCISES ON CHAPTER XX. 

1. Find the M of a flywheel which, when running at 200 revolutions 
per minute, will increase its speed by i per cent, while storing 5000 
foot-lb. of energy. 

2. A solid disc of cast iron, density 450 pounds per cubic foot, is 
8 inches in diameter by 2 inches thick and runs at 2500 revolutions per 
minute. What percentage increase in speed will occur if it is called upon 
to store an additional 200 foot-lb. of energy ? 



EXERCISES ON CHAPTER XX. 523 

3. A cast-iron flywheel is 30 feet in mean diameter. The safe tensile 
stress is 2000 Ib. per square inch. Find the maximum permissible speed 
of revolution of the wheel. Take the density as 450 pounds per cubic foot. 

4. A mild-steel hoop is 18 inches in mean diameter and the elastic 
limit of the material is 18 tons per square inch. At what speed of revolu- 
tion would permanent damage begin to occur ? Take the density as 480 
pounds per cubic foot. 

5. M/Tiat is the limit to the velocity of the rim of an ordinary flywheel? 
Does it depend on the diameter? Prove your statements (B.E.) 

6. In the manufacture of a large drum for a steam turbine a hollow, 
red-hot steel billet is, at a high gpeed, rolled between internal and 
external rollers, which effect a gradual increase of diameter and diminu- 
tion of thickness. Show that the intensity of the tangential stress of 
the material of the drum remains constant during the rolling operation, 
assuming a constant speed of the rollers. Determine the limiting speed 
of the rollers to keep the tensile stress within i ton per square inch. 
(Weight of steel, 485 Ib. per cubic foot.) (I.C.E.) 

7. The indicated horse-power of a steam engine is 100 ; the mean 
crank shaft speed is 200 revolutions per minute. The energy to be 
taken up by the flywheel of the engine between its minimum and maxi- 
mum speeds is 10 per cent, of the work done in the cylinders per revolu- 
tion of the crank shaft. If the radius of gyration of the flywheel is 2 feet 
6 inches, determine its weight in order that the total fluctuation of speed 
may not exceed 2 per cent, of the mean speed. (L.U.) 

8. Show from first principles that two flywheels of the same dimensions 
but of materials of different densities will have equal kinetic energies 
when run at the speeds which give equal hoop stresses. Calculate the 
kinetic energy stored per pound of rim in a cast-iron flywheel, when the 
hoop stress is 800 pounds per square inch. Cast iron weighs 450 pounds 
per cubic foot. (L.U.) 

9. In a simple Watt governor, the height of the cone of revolution is 
4 inches. What is the speed in revolutions per minute ? 

10. A Porter governor has revolving masses of 2 pounds each. The 
arms are all equal and 8 inches long. If the height of the cone of revolution 
is to be 5 inches at 180 revolutions per minute, find the dead load 
required. 

11. In Question 10, the throttle valve is full open when the height is 
5-5 inches and closed entirely when the height is 4-5 inches. Find the 
limits of speed of revolution controlled by the governor. State the 
total variation in speed as a percentage of the mean speed of 180 revolu- 
tions per minute. 

12. A uniform rod 8 inches long, mass 4 pounds, is hinged at its upper 
end to a vertical axis of revolution. Find the speed at which the arm 
will describe a- cone of semi -vertical angle 45 degrees. Supposing this 
speed to be doubled without alteration in the position of the rod, what 
controlling couple must be applied to the rod ? 

13. The mass of each of the balls of a spring-loaded governor 
arranged as in Fig. 548 is 5 pounds. When the radius of the balls is 
6 inches the governor makes 250 revolutions per minute. Find the total 



524 



MACHINES AND HYDRAULICS 



compressive force in the spring, and, neglecting friction, find the stiffness, 
i.e. the force per inch compression, of the spring that the governor may 
be isochronous. Show that the effect of friction would be to make the 
governor stable. (L.U.) 

14. A Porter governor has equal links 10 inches long, each ball weighs 
5 pounds and the load is 25 pounds. When the ball radius is 6 inches 
the valve is full open, and when the radius is 7-5 inches the valve is 
closed. Find the maximum speed and the range of speed. .If the 
maximum speed is to be increased 20 per cent, by an addition to the 
load, find what addition is required. (L.U.) 

15. Three bodies of 2, 3 and 5 pounds mass respectively revolve at 
equal radii round a horizontal axis. The axial distance between the 
outer pair of bodies is 18 inches. Arrange the bodies so that they shall be 
in balance. 

16. The four weights iv lt o/ 2 , -ze/ 3 , w 4 (Fig. 562) rotate in one plane 
about an axis, their magnitudes and the radii at which they act being 

given in the table : 




Weight. 


Magnitude in Ib. 


Radius in feet. 


W l 


10 


0-5 


7/ 2 


8 


I-O 


7/ 3 


6 


1-25 


W 4 


12 


0-75 



FIG. 562. 

Find graphically the equivalent single mass in magnitude and direction, 
acting at a radius of i foot ; and calculate the total displacing force on 
the shaft when the revolutions are 200 per minute. (I.C.E.) 

17. A shaft runs in bearings A, B, 15 feet apart, and carries three 
pulleys C, D and E, which weigh 360, 400 and 200 pounds respectively, 
and are placed at 4, 9 and 12 feet from A. Their centres of gravity are 
distant from the shaft centre line by amounts : C ^ inch, D inch and 
E I inch. Arrange the angular positions of the pulleys on the shaft so 
that there should be no dynamic force on B, and find for that arrange- 
ment the dynamic force on A when the shaft runs at 100 revolutions per 
minute. (L.U.) 

18. Find the positions and magnitudes of the balance weights required 
to balance all the revolving and of the reciprocating masses in a simple 
inside cylinder locomotive specified as follows : masses per cylinder at 12 
inch radius, revolving 720 pounds, reciprocating 630 pounds ; centre to 
centre of cylinders, 26 inches ; planes of balance weights, 58 inches 
apart ; radius of balance weights, 32 inches. (L.U.) 

19. The reciprocating masses for the first, second and third cylinders 
of a four-cylinder engine are 4, 6 and 8 tons, and the centre lines of these 
cylinders are 13, 9 and 4 feet respectively from that of the fourth cylinder. 
Find the fourth reciprocating mass, and the angles between the various 
cranks, in order that these may be balanced. (B.E.) 

20. Show that the disturbing effect of a reciprocating mass connected 
to a crank by the equivalent of an infinite connecting rod is the same as 



EXERCISES ON CHAPTER XX. 525 

that produced in the line of stroke by an equal mass placed at the crank 
pin. An engine has three cylinders A, B and C whose axes are parallel. 
The axis of B is at a distance a from the axis of A and a distance c from 
the axis of C. The mass of the reciprocating parts of B is M. Assuming 
that all the pistons have harmonic motion and the same length of stroke, 
show how the cranks on the crank shaft must be placed, and find the 
masses of the reciprocating parts of A and C in order that all the 
reciprocating parts may be completely balanced. (L.U.) 

21. A four-cylinder vertical engine, cranks at right angles, has its 
cranks equally spaced between the bearings, the pitch being 

4>. Taken from the left, the order is A, B, C, D. The 
revolving mass for each cylinder is M t and the reciprocating 

mass M 2 , and the speed is w radians per second. The 

crank radius is r and the connecting-rod length /. Examine D 

the primary and secondary balance, forces and couples 

when (a) the cranks are as shown in Fig. 563, (b} the cranks C 

are at 45 degrees to the line of stroke. (L.U.) FIG. 563. 

22. A vertical steel shaft i inch in diameter runs in swivel bearings 36 
inches centre to centre. A wheel of mass 20 pounds is mounted at the 
centre of the shaft, and its centre of mass is at a small distance from the 
shaft axis. At what speed of revolution will whirling occur? Take 
E = 30,000,000 Ib. per square inch. 

23. A steel shaft 2 inches in diameter runs in swivel bearings 9 feet 
centre to centre. At what speed will whirling occur ? Take = 30,000,000 
Ib. per square inch and the density 0-28 pound per cubic inch.. 

24. Answer Question 24 if the bearings constrain the directions of the 
shaft at its ends. 

25. In Question 23, the speed of the shaft is 600 revolutions per 
minute. Find the limiting distance centre to centre of the bearings. 



CHAPTER XXI. 



TRANSMISSION OF MOTION BY BELTS, ROPES, CHAINS 
AND TOOTHED WHEELS. 

Driving by belt. Motion may be transmitted from one shaft to 
another by means of a belt running on the rims of pulleys which are 
fixed to the shafts. The driving effort is transmitted from the belt 
to the pulley by the agency of the frictional resistance to slipping of 
the belt on the pulley. A will drive B in the same direction of 
rotation if the belt is open (Fig. 564), and in the opposite direction 
if the belt is crossed (Fig. 565). In the latter case, each portion of 





FIG. 564. Open belt. 



FIG. 565. Crossed belt. 



the belt is given a half turn in order that the same side of the 
material may bear against the rims of both A and B. In these 
diagrams the shafts are parallel, and both pulleys are arranged so that 
their planes of revolution coincide ; if this condition be not attended 

to, the belt will not remain on the 
pulleys. It is customary also to 
round slightly the rims of the pulleys 
(Fig. 566), with a view to enable 
the belt to ride on the centre of the 
rim ; the action will be understood 
by reference to Fig. 567, which 
shows the exaggerated case of two 
frusta of cones placed base to base. 
The belt, in bedding down on the conical surface, bends as shown ; 
consequently the points a and a will be higher up the cone than b 





FIG. 566. 



FIG. 567. 



DRIVING BY BELTS 



527 



and //, which came into contact a little before a and a. Hence 
the belt will climb to the highest part and remain there. 

It will be evident that the part of the belt which is advancing 
towards the pulley must be moving in the same plane as that in 
which the pulley is rotating. The part receding from the pulley may 
do so in a plane which does not coincide with the plane of rotation. 
Advantage is taken of these conditions in the case of two shafts 
having directions at right angles (Fig. 568). A is so arranged on 
the lower shaft that the part C of the belt leaving it is moving in the 
plane in which B rotates ; similarly B is so arranged that the portion 
D of the belt leaves B in the same plane as that in which A is 
rotating. The belt will ride safely on both pulleys, provided that 
the directions of rotation are not reversed at any time. Reversal of 





FIG. 568. Two shafts at 90 connected 
by a belt. 



FIG. 569. Use of jockey pulleys. 



direction must be preceded by a rearrangement of the pulleys. The 
distance between the shafts should not be small enough to render 
excessive the angle at which the belt leaves the pulleys. 

In Fig. 569 is shown an arrangement in which A drives B by 
means of a belt which is guided into the proper planes by jockey 
pulleys running freely at C and D. 

Velocity ratio of belt pulleys. A certain amount of slipping is 
always present in belt driving ; in the best cases there may be i to 2 
per cent, of the motion of the driven pulley lost in slipping. The 
belt usually comes off the pulleys if the slip exceeds 10 per cent. 
Neglecting slipping, it will be evident that the speed of the belt will 
be equal to the speeds of the rims of both pulleys. Referring to 
Fig. 57> 



528 



MACHINES AND HYDRAULICS 



Let D A = the diameter of A, in feet. 

D B = the diameter of B, in feet. 
V = the velocity of the belt, in feet per minute. 

N A = revolutions per minute of A. 

N B = revolutions per minute of B. 
Then, Distance travelled by rim of each pulley = V feet per minute. 



-^rv 

N V 
B ~7rD B 

and - = -. 




FIG. 570. Velocity ratio of belt 
pulleys. 



Hence the speeds of revolution are inversely proportional to the 
diameters of the pulleys. 

Strictly speaking, the diameters should be measured to the mean 
thickness of the belt, i.e. the thickness of the belt should be added 
to D A and D B . The presence of slip usually renders this correction 
an unnecessary refinement. 

In Fig. 571, A is an engine pulley driving a line shaft pulley B; 
a countershaft has a pulley D driven from a pulley C on the line 




FIG. 571. A belt pulley arrangement. 



shaft ; a machine pulley F is driven from the countershaft puiiey E, 
Assuming that there is no slip, 

NB = D_A. N = D_A N 
N A D B ' '!>* 

N=Pc. - N = DC 

N c P ' D V D c ' 



Also, 



DRIVING BY BELTS 







DC D A 


But 


N = N * 








D= IVIV A> 


Again, 


N F D E 

N E ~D F ; 


N F = ^ E .N E . 
Jjp 



And N E = N D ; 

D A x D r x D, 



.N, 



D B x D D x Dp 

Now A, C and E are drivers and B, D and F are driven pulleys ; 
hence we have the rule : To obtain the speed of revolution of the last 
wheel, multiply the speed of the first wheel by the product of the diameters 
of all the drivers and divide by the product of the diameters of all the 
driven pulleys. 

Supposing that each pair of pulleys connected by a belt experiences 
a percentage slip /, i.e. the driven pulley loses by slip p revolutions 
in every 100; then 

N B =- A 



A 
D B \ 100 



Since N B = N C and N D = N E , these reduce to 



100 



and N^Acn - 

D B xI) D xD F A \ ioo 

Friction of a belt on a pulley. The greatest possible difference 
which can exist between the pulls on the tight and slack sides of a 
belt will depend on the maximum frictional resistance to slipping of 
the belt on the pulley. In Fig. 572 (a) is shown a pulley having a 
belt embracing it over an arc of contact AB. Let T l and T 2 be the 
pulls at the ends when the belt is on the point of slipping, and let T a 
be the larger pull. Let the angle subtended by AB at the centre of 
the pulley be radians, and consider a small arc CD subtending a 
small angle 8a radian. The portion CD of the belt will be in 
equilibrium under the action of forces T and T 4- <$T, these being the 
pulls at D and C respectively (Fig. 572 (/;)), together with a normal 
reaction p from the pulley rim and also the frictional resistance to 
slipping. 

D.M. 2 L 



530 MACHINES AND HYDRAULICS 

Resolve T into components along and at right angles to/; these 
will be T sin JSa and T cos JSa respectively. In the same manner, 
T + ST will have components (T + ST)sinJSa and (T + ST)cosSa 




(a) (b) 

FIG. 572. Friction of a belt on a pulley. 

respectively along the same lines. The sum of the components 
along / must be equal to /, hence 

/ = T sin JSa + (T + 8T) sin J8a 

= (2T + 8T) sin J8a. 
Neglecting the products of small quantities, this reduces to 



Again, the difference between the sine of a very small angle and 
its radian measure is infinitesimal. Hence, 

p = 2T . |8a 
= T.8a ........................................... (i) 

Let the coefficient of friction be /*. Then 

Frictional resistance of arc CD = /*/ 

= /xT.Sa ............. (2) 

This frictional resistance must be equal to the difference in the 
components of T and T + 8T taken at right angles to /, hence 

/xT . Ba = (T + 6T) cos JSa - T cos JSa 

= 8TcosiSa. 

The angle J8a being very small, its cosine may be taken as unity 
and the equation reduces to 



(3) 



FRICTION OF BELTS 



In the limit, writing da. and dTT, and integrating both sides, we 
have ,T^ T ,* 

-T/M d ^ 

JTo L Jo 



or 



(3') 



This equation may be written 



-(4) 



_^ _ fHO- pf p 
r~r\ C 1 Vx * 




T 

~ = a constant. 
AD 



FIG. 573. Tensions in the belt at 
different parts of the arc of contact. 



where e is the base of the hyperbolic logarithms (p. n). 

The physical meaning of this equation may be understood by 
dividing the total arc of contact into a number of equal arcs AB, BC, 
CD, etc. (Fig. 573). Let each arc sub- 
tend an angle a at the centre, and let the 
tensions in the belt at B, C, D, etc., be 
denoted by T B , T c , T D , etc. Equation 
(4) above applies to each arc. Hence, 

T 

1 1 = MO. 

T*- nr ~"* T 1 
B AC A D 

As the right-hand side is constant in 
each of these expressions, the ratios of 
the tensions will be constant, i.e. 

T 1= T B 

T B T c 

Hence, if the value of the constant for a given angle is known, the 
ratio of the tensions for any angle when slipping is about to occur 
may be calculated easily. 

EXAMPLE i. A rope is coiled round a fixed drum over an arc of 
contact of 90. It is found that slipping occurs when the ratio of the 
pulls is f . Find the ratio of the pulls for an arc of contact of 270. 

To = T9o = Ilso = 3. 
TISO T 2 7o 4 

T> !i 

Ti80 T 27 o 

T 0= 27 
T 270 64' 

EXAMPLE 2. A leather belt laps 180 round a cast-iron pulley. 
Taking ft =0-5, calculate the pull on the slack side when slipping is about 
to occur, if the pull on the tight side is 300 Ib. 



-no 



OO 



'=2 X 3 X 3, 
444 



or, 



532 MACHINES AND HYDRAULICS 



log* 1 = ^ = o-57r= 1-5708 ; 



Here 6 1 80 = TT radians. Hence, 

1 



or T 2 =_ = 2. 4 . 

4-01 

Horse-power transmitted by a belt. It will be observed that the 
diameter of the pulley does not enter into the expression for the 
ratios of the pulls of a belt or rope. For example, in the last result, 
the pulls would be 300 Ib. and 62-4 Ib. when the belt is embracing 
a pulley 3 feet in diameter or 6 feet in diameter, provided the arc of 
contact is 180 in each case. Some other cause must be looked for 
to explain the known fact that a belt which constantly slips on a 
certain drive may be remedied by substituting pulleys of larger 
diameter on both shafts, keeping the ratio of the diameters as at first 
so as not to alter the speeds of the shafts. The explanation lies 
in the fact that the belt is now running at a higher 
speed, and will therefore do the same work per 
minute, or will transmit the same horse-power, with 
a smaller difference in pulls. Thus, 
Let Tj = pull on tight side, Ib. 
T 2 = pull on slack side, Ib. 
V = velocity of belt in feet per minute. 
Considering the driven pulley (Fig. 574), T l is 
urging it to turn and T 2 is tending to prevent rotation ; hence the 
net driving force is (T l - T 2 ). 

Work done per minute = (T l -T 2 )V foot-lb. 

(T T )V 
Horse-power transmitted = * - - ...................... (i) 

33,000 

Now let V be increased to V 2 feet per minute by the substitution 
of larger pulleys running at the same revolutions per minute. The 
horse-power transmitted being the same as at first, we have 
(T 1 -T 2 )V_(T/-T 2 QV 2 

33,000 33> 

where T/ and T 2 ' denoted the altered pulls in the belt. This gives 
(T^T^V-Ciy-T^V, ............. . ........ (2) 

As V 2 is greater than V, it follows that (T/ - T 2 ') must be less 
than (Tj-Tg). Hence there is now less frictional resistance to 
slipping called for, and consequently the risk of slipping is reduced. 




DRIVING BY ROPES 



533 



Equation (i) above for the horse-power may be written in terms of 
the maximum pull Tj in the belt. Thus, 



' T - 
* 2 - 

Substituting in (i) gives 

Horse-power transmitted = 



(3) 



33,000 



From this equation the dimensions of a belt suitable for trans- 
mitting a given horse-power may be obtained. The strength of a 
belt is stated in pounds per inch of width generally. 

Let b = width of belt in inches. 

p = safe pull per inch width of belt. 

Then T l ----pb 



and 



Horse- 



(5) 



The width b may be calculated from this result when the other 
quantities involved are given. 

Driving by rope. Ropes of cotton, hemp, manila or steel wire 
may be used for transmitting motion. In such cases the rims of the 
pulleys are grooved to receive the ropes. The section of a pulley 





FIG. 575. Section of the rim of a rope 
pulley. 



FIG. 576. Pressures on the 
groove of a rope pulley. 



rim suitable for ropes of cotton or similar material is given in 
Fig. 575. The ropes bear on the sides of the wedge-shaped grooves, 
thus increasing the frictional resistance to slipping. In Fig. 576, 
Let a = half the angle of the wedge. 

p = the normal force on a small arc of the rim, in Ib. 
ju. = the coefficient of friction. 



534 



MACHINES AND HYDRAULICS 



Then / will be equal to the sum of the vertical components of 
the normal pressures q, q on the sides of the groove. Hence, 
p=2q sin a, 
q = \p cosec a .................................. ( i ) 

Now the frictional resistance to sliding on the small arc considered is 
f= 2^q = 2\*\p cosec a 
=p , fji cosec a ................................. (2) 

Had the case been that of a flat belt on an ordinary pulley, the 
frictional resistance would be /x/. Hence, the results already 
obtained for flat belts may be used for ropes which bear on the sides 
of the groove by writing ft cosec a instead of /x. Thus, from equation 
(4), p. 531, and equation (4), p. 533, we have 



In the case of wire ropes, the rope should not bear on the sides of 
the groove, as it would suffer injury thereby. Fig. 577 shows a 
suitable form of rim, in which the rope beds on the bottom of the 
groove ; it is found advantageous to line the bottom of the groove 





FIG. 577. Section of a wire 
rope pulley. 



FIG. 578. Section of an idle 
pulley for a wire rope. 



with leather, with a view of increasing the frictional resistance. 
Where the ropes are very long, idle bearer pulleys may be used at 
intervals to support the ropes. These run loose on their bearings, 
and may have rims of a section shown in Fig. 578. 

The equations for a flat belt apply without alteration to the case of 
a wire rope bedding on the bottom of the groove. 

Centrifugal tension in belts and ropes. The portion of a belt or 
rope which laps on the pulley is subject to centrifugal forces when 
the belt is running (Fig. 579). 




BELT STRIKING GEARS 535 

Let m = the mass of the belt per foot run, in pounds. 

v = the velocity of the belt, in feet per sec. 

r = radius, in feet, to the centre of the belt. 
Then the centrifugal force per foot length of arc will be given by 

/= Ib. weight. 

gr 

These radial forces will have a 
resultant R directed towards the 
left in Fig. 579, and will be balanced 
by tensions T, T in the belt, which 
are in addition to those required 
for driving purposes. The case is v 

... .... FIG. 579. Centrifugal tension in a belt. 

analogous to a boiler shell sub- 
jected to internal pressure, and may be solved by the method given 
on pp. 95 and 96. R=/ X 2 r 

mV* 21HV 1 ., . , 

. 2r = - Ib. weight 
gr g 

Also, 2T = R ; 

.*. T = - Ib. weight. 

For leather belts, m may be taken as 

m = o4A pounds per foot run, 

where A is the cross-sectional area of the belt in square inches. 

The general effects of centrifugal force are to increase the pulls in 
the belt, and also to reduce partially the radial pressures on the rim 
of the pulley. As the latter are relied on for the production of the 
frictional driving effort, it follows that excessive slipping will occur at 
speeds which are too high, and the power transmitted will be reduced 
thereby. 

Belt striking gears. The intermittent motion required for driving 
many classes of machines may be obtained by means of two pulleys 
on the countershaft driving the machine. In Fig. 580 a pulley A on 
the main or line shaft drives a countershaft having two pulleys, one 
Bj running loosely on the countershaft and the other B 2 fixed to the 
shaft. The belt may be moved from one pulley to the other by 
means of forks C, C, which loosely embrace the belt. The forks are 
operated by a sliding bar D and a handle E, carried to a suitable 
position for the operator. The pulley A is made specially wide, so 



536 



MACHINES AND HYDRAULICS 



as to permit the belt to ride on either Bj or B 2 ; in the former case, 
the countershaft and machine will be at rest. 




FIG. 580. Belt striking gear. 

Another arrangement is shown in Fig. 581. The countershaft B 
has two loose pulleys Lj_ and L 2 , and also a pulley F fixed to the 
shaft. There are two belts, one D open and one E crossed ; these 
are operated by the belt-striking forks and bar shown at C. No 
motion will be transmitted to the countershaft if both belts are on 
the loose pulleys, and motion in either one or the other direction will 



n 


ELEVATION 







i 




m > 


1 


D 

PLAN 


cU| 


h 

u 



U A 





FIG. 581. Arrangement for reversing 
a machine. 



FIG. 582. Stepped cones. 



occur, depending on which belt is made to ride on F. The arrange- 
ment forms a convenient reversing gear. 

Variation in the velocity of rotation of the driven shaft may be 
accomplished by means of stepped cones or speed pulleys (Fig. 582). 
These consist of a number of pulleys of different diameters mounted 
on the shafts so as to oppose the smallest and the greatest. The belt 
may ride on any corresponding pair. 



DRIVING BY CHAINS 



537 



The length of belt required enters into the question of stepped 
cones, as the belt has to fit any corresponding pair without alteration 
being made in its length. For a crossed belt it may be shown that 
the sum of the diameters of any corresponding pulleys should be 
constant for the whole set. With an open belt there is a small 
divergence from this rule, which becomes negligible if the distance 
between the shafts is large compared with the pulley diameters ; such 
is usually the case. 

Transmission of motion by chains. In cases where the driving 
effort is too large to be transmitted by a belt or rope, or where 
slipping is inadmissible, chains may be used in combination with 
toothed or sprocket wheels. A few patterns of suitable chains are 
given in Fig. 583. (a) is a block chain in which a number of small 



(a) 



(b) 




FIG. 583. Types of driving chains. 



FIG. 584. Sprocket wheel for chain driving, 
showing the effect produced by the chain 
stretching. 



blocks are connected by pairs of links and riveted pins. Chains of 
this pattern are used for conveyors, as the carriers are attached 
readily to the blocks. () is a similar pattern, but made entirely of 
links, (c) is a better form, and works more easily. The inner links 
are connected by a tube riveted over at its ends, and a roller runs 
on the tube ; the outer links are connected by a pin passing through 
the tube and riveted over at its ends. 

A sprocket wheel is shown in Fig. 584. The centres of the chain 
pins lie at the corners of a polygon having sides equal to the pitch p 
of the chain. The driving force P may act at radii which will vary 
from Rj to R 2 , and thus cause variations in the turning moment and 



533 MACHINES AND HYDRAULICS 

in the velocity ratio. These variations will be small, provided the 
number of teeth on the sprocket wheel be sufficiently large. 

The form of the teeth may be constructed by first drawing semi- 
circles of radius r equal to that of the chain pin, or roller. Using 
radii slightly smaller than (p-r) and centres nearly coinciding with 
the adjacent pin centres, the sides of the teeth may be drawn. These 
will be such as to enable the chain to leave the wheel at the top 
without the pin or roller touching the face of the tooth. 

In general there is practically no pull on the slack side of a chain ; 
hence, the work done per minute is given by the product of P and 
the velocity of the chain in feet per minute. The chain is liable to 
stretching of the links and to wear at the pins, both of which tend to 
increase the pitch. The effect of this will be ultimately that the top 
pin, or roller alone, as is shown in Fig. 584, will be bearing against 
its tooth, and this tooth accordingly will carry the whole load. The 




FIG. 585. Renold's silent chain. 

effect is manifest in the chain grinding on the teeth, thus introducing 
additional frictional resistance and also wearing the teeth. These 
effects may be obviated somewhat by using a roller chain, and by 
making the spaces between the teeth wider than the diameter of the 
chain pin or roller. Increase in the pitch is provided for perfectly 
in the Renold's silent chain. The links are of the form shown in 
Fig. 585 ; any increase in the pitch, caused by wear or stretching, 
has simply the effect of causing the links to ride on the teeth at a 
larger radius from the centre of the wheel. Speeds of 1250 feet per 
minute and horse-powers up to 500 have been attained with these 
chains. 

Friction gearing. In cases where the shafts are close enough 
together, motion may be communicated from one to the other by 



FRICTION GEARING 



539 



means of friction gearing. In Fig. 586 two parallel shafts have wheels 
A and B fixed on them ; A is pressed against B by application of 
forces P, P, and the frictional resistance between the rims enables a 
driving effort F to be communicated from A to B. B may be made 
of cast iron and A of compressed millboard or leather ; the coefficient 
of friction is thus increased somewhat. There will always be a certain 
amount of slipping, but such gear is advantageous where heavy parts 
connected to B have to be brought from rest to a high speed. The 
slipping which occurs enables the desired speed to be attained without 
giving impulses or shocks to the mechanism. Further, owing to the 
small movement required to bring A out of gear with B, the driving 
effort can be got rid of quickly. As in belt pulleys, the angular 
velocities are inversely proportional to the diameters of the wheels. 





U 

FIG. 586. Friction wheels for parallel shafts. 



X-R,--:. 

FIG. 587. Bevel friction wheels. 



If the shafts are not parallel but have their axes intersecting, the 
friction wheels must form part of conical surfaces in order that 
perfect rolling may be possible at all parts of contact (Fig. 587). 
The vertex of each cone coincides with O, the point in which the 
axes of the shafts intersect. Let R x and R 2 be the largest radii of A 
and B respectively. These radii, GF and HF, come into contact at 
F ; hence the revolutions per minute of the wheels, neglecting slipping, 
will be given by N ^ RF R ^ 

N; = GF = RV 

Further, for any other point of contact C, the geometry of the 
figure shows that R p -^^ -^ 

GF = EC = ~Ri' 

Hence the relative angular velocities communicated at C will be 
the same at C as at F, showing that, if there be no slip at F, there 



540 



MACHINES AND HYDRAULICS 



will be no slip anywhere, i.e. the rolling will be perfect. Wheels of 

this kind are called bevel wheels. 

Driving by toothed wheels. Motion lost by reason of slipping may 

be eliminated entirely by the addition of teeth to the rims of friction 

wheels. Fig. 588 shows two toothed wheels in gear; the original 

friction wheels are shown dotted, 
and come into contact at a point 
on the line joining the centres 
of the wheels. This point is 
called the pitch point, and the 
circles are called pitch circles. 
The length of the arc on the 
pitch circle between the centres 
of an adjacent pair of teeth is 
called the circular pitch of the 
teeth. It is evident that the 

pitch must be the same for both wheels. For practical purposes, 

the diametral pitch is used often, and is the result of dividing the 

diameter of the wheel by the number of teeth. 




FIG. 588. Toothed wheels in gear. 



Let 



Then 



D = the diameter of the wheel. 
n = the number of teeth. 
p = the circular pitch. 
pd = the diametral pitch. 



or 




Also, 



Unless otherwise specified, the term "pitch " 
will be taken to mean the circular pitch. 

Referring to Fig. 589, other definitions are 
as follows : The portion EFGC of the tooth 
which lies outside the pitch circle is called the addendum; 
dotted circle FGL is the addendum circle ; the working sides of the 
tooth at EF and CG are called faces. The portion EH KG which 
lies within the pitch circle is called the root of the tooth ; the dotted 
circle HKM is the root circle; the working sides EH and CK are 
called flanks. EC is the thickness of the tooth and CD is the width 
of the space between the teeth. 



FIG. 589. Proportions of 
wheel teeth. 



the 



DRIVING BY TOOTHED WHEELS 541 

Ordinary proportions of teeth may be stated. Reference is made 
to Fig. 589, and p is the circular pitch. 

Thickness of tooth = 0-48^. 
Space between teeth = o-52/. 
Total length of tooth = (a + ) = 0-7^. 
Length of addendum = a = o-$p. 
Length of root = b = o-^p. 
Width of tooth = 2p to 3 \p. 

These proportions allow of a clearance equal to 0-04^ between 
the thickness of the tooth and the space into which it enters on the 
other wheel; also a clearance of o-ip between the point of the tooth 
and the bottom of the space. With accurate machine-cut teeth, these 
clearances are often made smaller. 

Power transmitted by toothed wheels. Let P Ib. be the driving 
effort applied to a toothed wheel tangential to the pitch circle, and 
let R feet be the radius of the pitch circle. In one revolution, work 
will be done equal to 2?rRP foot-lb. If the wheel makes N revolu- 
tions per minute, we have 

Work done per minute = 2?rRPN. 

27TRPN 



Horse-power transmitted 



33000 ' 



p _ 33000 x horse-power 

27TRN 

If the horse-power be given, P may be calculated, and hence the 
dimensions of the tooth may be estimated in order that sufficient 
strength may be secured. It is best to use the rules of proportional 
strength. Suppose it is known that a certain wheel made of a 
given material has transmitted a force Pj successfully, and that 
the width, length and thickness of its teeth are ^, /j and t } respec- 
tively. The connection of these dimensions with those of the teeth 
of another wheel of the same material which has to transmit a 
force P 2 will be given by (p. 152) 



It has been assumed here that P a and P 2 are applied at the extreme 
point of the tooth, as in practice might be the case by accident. 
Also that the whole of the driving effort may act possibly on one 
tooth. 



542 



MACHINES AND HYDRAULICS 



Angular velocity ratio of toothed wheels. It is evident from 
Fig- 590 that two toothed wheels in gear revolve in opposite 
directions ; also that the speeds of the circumferences of the pitch 



circles will be equal. Hence, 




NA_DB 
N B D A ' 
Also, 

Number of teeth on A = n A = '** - ; 

f 

Number of teeth on B = n E = - \ 
D R ;Z R 



FIG. 590. Angular velocity ratio of 
toothed wheels. 



and = "2. 

N B n A 



Hence, the revolutions per minute are inversely proportional to the 
numbers of teeth. It will be obvious, from what has been said on 
p. 539 regarding friction bevel wheels, that the same rule applies also 
to such wheels. 

If the wheels A and B are required to revolve in the same direction, 
an idle wheel C may be interposed (Fig. 591). Since the velocities 




FIG. 591. Use of an idle wheel. 



FIG. 592. Two idle wheels. 



of all three pitch circle circumferences must be equal, it follows that 
there will be no change in the angular-velocity ratio of A and B. 
Hence, N 



Any number of idle wheels (Fig. 592) may be inserted without 
affecting the angular-velocity ratio of A and B. 

Trains of wheels. Fig. 593 shows a train of toothed wheels. In 
this case we have : 



N R 



Also, 



p ' N C 

N E = N D ; 



^C N_B 

r J N A 

N C =N B . 



TRAINS OF WHEELS 



543 



Hence, 



or 



N E N c N A 

NF 
N A 



n v 



B 




PLAN 

FIG. 593. Train of wheels. 



If F, D and B be called drivers, and E, C and A followers, the 

above result gives us the rule that the angular-velocity ratio of the 

first and last wheels in the train is equal 

to the product of the numbers of teeth on 

the followers divided by the product of 

the numbers of teeth on the drivers. 

Fig. 594 shows the gearing wheels 

used in the Wolseley motor cars for 

enabling the car to run at different 

speeds. The shaft AB is driven by 

the engine, and has a wheel C fixed 

to it and gearing always with a wheel 

G on the secondary shaft EF. When 

the clutch between the engine and 

AB is "in," the secondary shaft EF 

will be revolving. H, K and L are wheels of different sizes mounted 

on, and revolving with, EF. The shaft RS is connected at S to the 

road wheel axle by means of gearing 
not shown in Fig. 594 ; this shaft 
runs freely in the hollow shaft AB, 
and is made square between R and 
S. M, N, P and Q are wheels which 
may slide on the square shaft RS, 
and are under the control of the 
driver by means of an arrangement 
of interlocking bars (not shown in 
the figure). The wheel C is hollow, 
and is furnished with internal teeth 
at D. M may be slid into C, and, 
when so situated, AB will drive RS 
direct, the secondary shaft EF then 
running idle. Other speeds may be 
obtained by withdrawing M from C 
and gearing N with H, or P with 
K, or Q with L. The lever system 

FIG. S94.-Gear wheels for a motor car. for sliding the wheels is SO devised 




544 



MACHINES AND HYDRAULICS 




FIG. 595. Bevel wheels. 



as to prevent two pairs of wheels being in gear simultaneously. 

Reversal of the car is obtained by sliding idle wheels (not shown in 

the figure) on another secondary shaft. 

Bevel wheels. If the directions of the shaft axes intersect, it has 

been shown (p. 539) that cones may be used for driving; hence 

conical pitch surfaces are em- 
ployed for toothed wheels on 
intersecting shafts. In Fig. 
595, the axes of the shafts 
intersect at O, and OAB and 
OBC are the conical pitch 
surfaces. The dimensions 
are settled from the relation 

NOD = BC 

NOE AB' 

To obtain the shape of 
the teeth, ADB and EEC 
are other conical surfaces 

obtained by drawing BD and BE perpendicular to OB. These 
conical surfaces are developed by describing arcs BF and BG, using 
D and E respectively as centres. The teeth may then be drawn on 
these arcs as pitch circles by ordinary methods. The teeth are 
tapered along the conical surfaces AOB and BOC, 'and finally vanish 
at O; hence portions only of the conical surfaces are used, shown 
in the figure at BKHC and BKLA. 

Mitre wheels are bevel wheels of equal size on shafts meeting at 
90, and are used in cases where the shafts are to have equal speeds 
of rotation. 

In Fig. 596 is shown an example of the use of mitre wheels. A 
is a continuously revolving shaft having a mitre wheel B fixed to it, 
and driving other two mitre wheels C and D which run loose on the 
shaft EF. Each of the mitre wheels C and D has projecting claws on 
its inner face, which may engage with the claws of a clutch G. G 
may slide on the shaft, and has a long feather key which compels it 
to rotate with the shaft ; a pivoted lever H enables the clutch to be 
operated. In the position shown no motion will be communicated 
to the shaft EF ; motion of either sense of rotation may be obtained 
by causing G to engage with either C or D ; the arrows indicate the 
directions of rotation. The arrangement thus provides for inter- 
mittent motion and for reversal. 

Fie. 597 illustrates a common type of differential gearing used for the 



DIFFERENTIAL GEARING 



545 



driving axle of a motor car. A toothed wheel A, shown in section, 
runs loose on the axle EF, and has two bevel wheels B, B mounted on 
radial spindles. EF is the axle to which the road wheels are attached, 




FIG. 596. Arrangement for intermittent motion and for reversal. 

and is made in two pieces. A bevel wheel C is fixed to the portion 
E, and another bevel wheel D is fixed to F ; C and D gear with the 
bevel wheels B, B. The wheel A is driven by the engine, and, if 
both road wheels are rotating at the same speed, the wheels B, B 




M 




FIG. 597. Differential gear for a motor car. FIG. 598. Milne's- Daimler differential gear. 

will not rotate on their spindles. In rounding a curve, the inner 
road wheel must rotate at a lower speed than the outer wheel, and 
this difference in speed is permitted by the bevel wheels B, B 
D.M. 2 M 



546 



MACHINES AND HYDRAULICS 



rotating on their spindles. It will be evident that, if C were held 
fixed, D would rotate at twice its former speed. 

Fig. 598 shows the application of the same arrangement in the 
Milne's-Daimler differential gear.* AB is a shaft driven by the 
engine and carries mitre wheels D, D, running loose on cross 
spindles C, C. These wheels gear into mitre wheels E and K at 
the inner ends of sleeves which run loose on AB, thus permitting 
differential motion to the sleeves. F and L are bevel wheels at the 
outer ends of the sleeves, and gear with wheels G and M fixed 
respectively to the halves H and N of the road-wheel axle. 

Epicyclic trains of wheels. In trains of this kind there is usually 
one fixed wheel A (Fig. 599) i.e. A does not rotate together with 

one or more wheels mounted on an 

,153. jcn. P -go, arm D which may rotate about the 

centre of A. The solution of such 
trains may be obtained by the 
following method. Imagine the 
whole set of wheels to be locked 
and that the bracket carrying A 
is free to rotate. Give the whole 
arrangement one rotation in the 
clockwise direction, then, keeping 
the arm fixed in position, apply 
a correction by giving A one revolution in the anti-clockwise 
direction. Calling clockwise rotation positive, the process may be 
tabulated thus : 




PLAN 
FIG. 599. An epicyclic train of wheels. 



Part 


A 


B 


C 


D 


Wheels all locked 


+ i 


+ i 


+ i 


+ i 


Correction - 


- i 


+^ 


-j* 





Final result - 





l+ ^ 


I-| 


t' 



The result shows that, if A and B have the same number of teeth, 
B will rotate twice clockwise for one clockwise rotation of the arm. 
If A and C have the same number of teeth, C will not rotate on its 
spindle ; a radial arrow sketched on the upper side of C will point 
always in the same direction as the arm D is rotated. 



* Proc. Inst. Mech. Eng., 1907. 



EPICYCLIC REDUCING GEARS 



547 



Epicyclic reducing gears. In Fig. 600, showing an arrangement 
for reducing the speed of rotation, the wheel D is fixed and has 
internal teeth ; E is an arm fixed 
to the shaft F, and carries two 
wheels B and C fixed together 
so as to revolve as one wheel. 
C gears with the internal teeth 
of D, and B is driven by a 
wheel A. It will be noted that, 
if D drives C with the arm E 
fixed, both wheels will have the 
same sense of rotation. The 



solution is as follows : 




FIG. 600. Speed reduction gear. 



Part - 


A 


B 


C 


D 


E 


F 


Wheels all locked 


+ i 


+ i 


+ i 


+ i 


+ i 


+ i 


Correction - - 


+ ^B .^D 


_^D 


_D 


_ r 










n\ nc 


nc 


nc 








Final result - 


( BD\ 


l-g 


i - 





+ i 


+ i 


\ iip^nc' 



In Fig. 60 1, showing another type of speed reduction gear, the 
shaft AB is driven by a wheel at A, and has an arm C fixed to it 




FIG. 601. Another type of speed reduction gear. 

carrying a loose bevel wheel D. D gears with two bevel wheels E 
and G runnfng loose on the shaft AB. E may be a fixed wheel, or 
may be rotated in the same or in the opposite sense to that of AB. 
It is evident that D does not rotate on C during the locked operation, 
and that E and G will rotate in opposite directions during the 



MACHINES AND HYDRAULICS 



correction operation with AB and C fixed, D being an idle wheel 
during this operation. Supposing E to be a fixed wheel, the solution 
will be as follows : 



Part - 


AB 


C 


E 


G 


Wheels all locked 


+ i 


+ i 


+ i 


+ 1 < 


Correction - 








- 1 


+ 


Final result - 


+' 


- 





+(1+3 



Suppose now that E is not a fixed wheel, but is rotated N E times 
during + i revolution of AB. The solution will be : 



Part - 


AB 


C 


E 


G 


Wheels all locked 


+ i 


+ i 


+ i 


+ i 


Correction - 








-IN E 


+ T N E 


Final result - 


- 


+' 


N E 


.+<.**.> 



In the result for G, the - sign is to be taken if A and E are 
driven in the same direction, and the + sign if they are driven in 
opposite directions. 

The Humpage gear is shown diagram matically in Fig. 602. A is 
the driving shaft and has a bevel wheel B fixed to it Two bevel 

wheels C and D, made in one 
piece so as to rotate together, 
run on an arm E fixed to a sleeve 
F, which runs loose on the shaft 
A. C gears with a fixed bevel 
wheel G, and D gears with a 
bevel wheel H, which is secured 
to the driven shaft K. For the 
sake of obtaining balance and of 
producing practically a driving 

FIG. 602. Humpage gear. , , ""'__* iv''"' ' i_ i 

couple, the arm E and the wheels 

C and D are duplicated. The solution may be obtained by giving 
the whole gear + i revolution with the wheels locked ; apply a 
correction by keeping the sleeve F and the arm E fixed and giving 




SHAPE OF TEETH 



549 



- i revolution to G. During this correction, C will act as an idle 
wheel between G and B ; also G will drive H in the anti-clock- 
wise sense through the wheels C and D ; the ratio of the revolutions 
of G and H during this operation will be 



Tabulating the operations, we have 



Part - 


A 


F 


G 


K 


Wheels all locked 
Correction - 


-fi 
, *c 


+ i 




+ i 


+ 1 
n G x# D 


+ n B 




n c x H 


Final result - 


' + ? 

B 


+ 1 





j fn G xn D \ 


W XtfH/ 



Hence, 



NA 
N K 



in G x n D \ 
\n c x nj 



Shape of teeth. The shape of the teeth must be such as to fulfil 
the condition of a uniform ratio of angular velocities in the wheels 
which gear together. If this condition be neglected, the teeth will 
work together badly, producing excessive wear and rattling owing to 
back lash. 

Referring to Fig. 603, let P be the point of contact of two teeth, 
one on the wheel which has its centre at A and the other on the 
wheel which revolves about B. At P a point on wheel A is moving 




FIG. 603. Condition for securing a constant angular velocity ratio in toothed wheels. 



550 MACHINES AND HYDRAULICS 

at right angles to AP, and a point on wheel B is moving at right 
angles to BP. Let V A and V R be these velocities, represented by DP 
and CP respectively. Let PO be the direction of a common normal 
to the tooth surfaces at P ; it is clear that, if contact is to be main- 
tained, and if there is to be no interpenetration of the teeth on A and 
B, the components of z/ A and V B along PO must be equal. Resolving 
V B along and at right angles to PO by means of the triangle CEP, the 
equal normal components of z> A and z> B will be represented by 
EP = z/ N . Produce the line of # N , and draw AM and BN perpen- 
dicular to # N . Then, if W A and <O B are the angular velocities of the 
wheels A and B respectively, 

W A = J/N BN = BN 
o> B AM z> N AM' 

Again, from the similar triangles AMO and BNO, 
BN = BO = (o A 
AM~AO~<o B > 

If O be selected as the pitch point, the ratio BO/AO will be 
constant, as O is then a fixed point. Hence, 

W A BO R B 

= -r-pr = =r = a constant. 

O>B AO R A 

Thus, the condition to be fulfilled in order to maintain a constant 
angular-velocity ratio is that the common normal at any point of contact 
of two teeth must pass through the pitch point. Theoretically, for a 
given design of tooth on one wheel, the teeth on the other wheel may 
be shaped so as to enable the common normal to comply with this 
condition. In practice, however, cycloidal teeth and involute teeth 
alone are used, and, in modern machine-cut wheels, the teeth are 
generally of the involute form. 

Cycloidal teeth. The cycloid is a curve traced by a point P on 




FIG. 604. A cycloid. 



the circumference of a circle which may roll along a straight line 
(Fig. 604). In any given position, the point of contact I is the 



CYCLOIDAL TEETH 



instantaneous centre for the rolling wheel ; hence the direction of the 
cycloidal curve at P is perpendicular to IP ; therefore the normal at 
P passes through the point of contact I. 

If the rolling circle having a centre Q (Fig. 605) rolls on the 
circumference of another circle having A for centre, an epicycloid OD 
will be traced. If the rolling circle rolls on the inside of the circum- 
ference' of the circle, a hypocycloid OE will be traced (Fig. 605). In 




FIG. 605. Epicycloid and hypocycloid. 

the epicycloid, if N\ is the point of contact of the circles, and Pj is the 
corresponding position of the tracing point, it will be clear that 
the direction of the epicycloidal curve at P l is at right angles to NjPj, 
as Nj will be the instantaneous centre of the rolling circle in the 
given position. Hence N^ is the normal to the curve at P r For 
similar reasons, N 2 is the instantaneous centre of the rolling circle 




FIG. 606. Mechanical construction of an epicycloid. 

when the tracing point is at P 2 on the hypocycloidal curve, and N 2 P 2 
is the normal to the hypocycloid at P 2 . 

In Fig. 606 is shown a useful way of producing an epicycloid. The 
wheel A and the rolling circle revolve about fixed centres at A and C, 



552 



MACHINES AND HYDRAULICS 



and drive one another in the same manner as friction wheels but with- 
out slip, A piece of paper D is fixed to the wheel A and revolves 

with it, and a pencil P on the 
rolling circle bears on the paper. 
The result is the epicycloid P P. 
It is evident that the normal at 
P passes through the pitch point 
O. In Fig. 607 is shown a 
similar method of drawing a 
hypocycloid by means of another 
wheel having its centre at B, 
and the same rolling circle 
revolving about a fixed centre 
C. A piece of paper attached 
to B will have drawn on it a 
hypocycloid P 'P. If there has 
been no slip, in each of these 
figures the arcs OP and OP, 
on the wheels and on the 
rolling circles respectively, will 
be equal. Let the arcs OP in each figure be equal, and imagine 
that the two figures are superposed, so that the wheels A and B 
come into contact at the pitch point O (Fig. 608). The arcs OP 
on the rolling circles in Figs. 606 and 607 will also be equal, and 
the points P will coincide in Fig. 608. OP will now be simul- 




FIG. 607. Mechanical construction of a 
hypocycloid. 




FIG. 608. The constructions of Figs. 606 and 607 superposed. 

taneously the normal to the epicycloid and also to the hypocycloid, 
and these curves will be in contact at P. Therefore the curves comply 
with the condition that the common normal must pass through the 



CYCLOIDAL TEETH 553 






pitch point, and thus may be used for the faces of the teeth on A, and 
for the flanks of the teeth on B. The flanks of the teeth on A and 
the faces of those on B may be produced in the same manner. It is 
evidently essential that the same rolling circle must be used both for 
the faces of A and for the flanks of B ; the rolling circle used for the 
flanks of A and for the faces of B may be of the same or of another 
diameter. It should be noted that the hypocycloid becomes a 
straight line, forming a diameter of the wheel if the rolling circle has 
a diameter equal to the wheel radius ; hence the flanks of the teeth 
would be radial lines. Any larger diameter of rolling circle would 
produce teeth thin and weak at the roots. In designing a set of 
wheels, the rolling circle should not have a diameter larger than the 
radius of the smallest wheel of the set. 

Path of contact. From Figs. 606, 607 and 608, it will be evident 
that P and P' on the cycloidal curves were initially in contact at O, 
and that the point of contact has travelled along the arc OP of the 
rolling circle. Contact will cease when the circumference of the 
tolling circle passes outside the addendum circle. In Fig. 609, EFG 



p s v, ty 
-UL 1 

M I" To IF T B 



FIG. 609. Path of contact in cycloidal teeth. 

and LMN are parts of the addendum circles of the wheels A and B 
respectively. These intersect the rolling circles at P and Q respec- 
tively ; hence the complete path of contact is POQ, and is formed of 
two circular arcs. 

In Fig. 610 two teeth are just starting contact at P. The point C 
will be in contact when it reaches O, and the arc CO on the pitch 
circle is called the arc of approach. In the same figure, two teeth are 
just finishing contact at Q ; E was in contact when passing through 
O, and OE is called the arc of recess. PO and OQ are called the 
paths of approach and of recess respectively. Let the arc OF be equal 



554 



MACHINES AND HYDRAULICS 



to the arc OE. Then COF is the length of arc which passes the 
pitch point while a tooth on A remains in contact with one on B, and 




FIG. 610. Arcs and paths of approach and recess. 

may be called the axe of contact. If the condition is to be fulfilled 

that two pairs of teeth are to be in contact always, the arc of contact 

should be twice the pitch. 

Involute teeth. Fig. 611 shows an involute P P 4 to a circular 

curve PQOfii ', the curve may be drawn by wrapping a string round 

the circular curve and having a tracing 
pencil attached at its end P . On the 
string being unwrapped, the pencil will 
trace out the involute P P 4 . It is evident 
that the string, in any position such as 
O 2 P 2 , is perpendicular to the direction of 
motion of the pencil ; O 2 is therefore the 
instantaneous centre of the string O 2 P 2 
and O 2 P 2 is normal to the involute at P 2 . 
In Fig. 612 is shown a mechanical method of drawing an involute 

to the circle having A for its centre. Let a crossed belt be passed 




FIG. 611. Involute to a circle. 




FIG. 612. Mechanical construction of an involute to the circle having centre at A. 

round two wheels revolving about A and B respectively, and let a piece 
of paper be fastened to wheel A and revolve with it. A tracing 



INVOLUTE TEETH 555 



pencil secured to the belt at P will draw an involute on the paper. 
It is evident that CP, the normal to the involute at P, passes always 
through a point O on AB, and the two parts of the belt intersect in 
the same point. An involute to the wheel B may be drawn in a 
similar manner (Fig. 613) by securing the paper to wheel B. The 




FIG. 613. Mechanical construction ot an involute to the circle having centre at B. 

normal at P' passes through the same point O. If the diagrams 
(Figs. 612 and 613) be superposed so that P and P' coincide, it is 
evident that the two involute curves fulfil the condition that the 
common normal passes through a fixed point O, which accordingly 
may be taken for the pitch point of a pair of wheels having teeth 
shaped to the involute curves. 

Let v be the velocity of the belt. Then, in Fig. 612, 

W A v DB DB 

- = -r-p, . = -r-^ = a constant. 
W B AC v AC 

Also, from the similar triangles AOC and BOD, 

DB = BO 
AC~AO ; 



Hence the radii AC and DB of the generating circles should be 
inversely proportional to the angular velocities of the wheels. 

It is clear that part of the straight line CD (Fig. 612) will be 
the path of contact. Practical considerations rule that this line 
should make about 15 with the common tangent to the pitch circles 
at O (Fig. 614). The intersections P and Q of CD with the 
addendum circles of the wheels will determine the length PQ of 
the path of contact. 

Using the same pair of generating circles connected by a belt as in 



556 



MACHINES AND HYDRAULICS 



Fig. 612, the same involute curves will be produced irrespective of the 
distance AB separating the wheel centres. As the resulting teeth will 
be of the same shape as at first, it follows that the distance apart of a 



To A 




to 



Fio. 615. An involute rack. 



FIG. 614. Path of contact in involute teeth. 

pair of involute toothed wheels may be varied to a small extent 
without interfering with their correct working. This may be advan- 
tageous for taking up back lash. 

If one of the two wheels in gear becomes of infinitely large radius, 
A the case of a rack is obtained 
(Fig. 615). The pitch line CD is 
straight, and the involute is a straight 
line perpendicular to the line of 
contact OA. Hence the sides of 
the teeth in involute racks are straight lines. 

Helical and screw gearing. Greater smoothness of running may 
be obtained by using wheels possessing several sets of teeth (Fig. 616), 
each set stepped back a little from the adjacent set on one side. If 
the steps are made indefinitely narrow (Fig. 617), we obtain a helical 
wheel. Single helical wheels would produce axial thrusts on the 
shafts, and this objection is obviated, as is indicated in Fig. 617, by 
employing double helical teeth sloping in opposite ways. Such 
wheels, with machine-cut teeth, run with remarkable smoothness, 



HELICAL GEARING 



557 



and are equally suitable for low and high speeds of running and for 
heavy loads. When the speed is high, it is best to run the wheels in 
an oil bath. 

A pair of screw wheels is shown diagrammatically in Fig. 618. The 
cylindrical pitch surfaces of two wheels A and B touch at O. CD 







FIG. 616. Stepped teeth. 




FIG. 617. Double helical teeth. 



and EF are the axes of A and B respectively. Imagine a sheet of 
paper having a straight line GOH drawn on it to be placed between 
the cylinders. If the paper is wrapped round A, GOH will map 




FIG. 618. Pair of screw wheels. 

out a helix, and, if wrapped round B, a corresponding helix will be 
described by GOH. These helices define the shape of screw teeth ; 
the other teeth may be produced by having a number of lines parallel 
to GOH and drawn on the sheet of paper. In Fig. 619 GOH and 
ed show two of these lines. The perpendicular distance Qb separating 
these lines is called the divided normal pitch, and is evidently the 



558 



MACHINES AND HYDRAULICS 



same for both wheels A and B. Oa, measured along the circum- 
ference of A, and Or, measured along the circumference of B, are 
called divided circumferential pitches ; it will be clear that these pitches 
must divide evenly into the circumferences of A and B respectively.* 





FIG. 619. Pitches in a screw wheel. 



FIG. 620. Pawl and ratchet wheel. 



Ratchet wheels. In Fig. 620, a wheel A is to have intermittent 
motion to be derived from an arm B which vibrates about the axis 
of A. A pawl C is pivoted to B, and will 
engage the teeth of A when B is moving 
anti-clockwise ; the pawl slips over the teeth 
of A when B is moving clockwise. Clockwise 
rotation of A may be prevented by a pawl D 
pivoted to some fixed part of the machine. 
It will be noted that lost motion to the extent 
of one tooth may occur between A and B ; 
this may be reduced by means of a second 
pawl E pivoted to B. The possible lost motion 
will now be half the former amount. In cycle 
free wheels, several pawls are often fitted so as 
to reduce lost motion to a minimum. 

Couplings for shafts. The Oldham coupling 
is illustrated in Fig. 621. A flanged coupling 
A is fixed to a shaft B, and has a groove cut 
in its face. Another similar coupling C is fixed 
to the shaft D. The axes of the shafts B 
and D are parallel. A plate E is interposed 
between the faces of these couplings, and has 
a projection on each side which is a sliding 
FIG. 621.- Oldham coupling, fit in the grooves ; the projections are at 90 to 

* For a complete discussion on toothed wheels, see Machine Design, Part /., 
by Prof. W. C. Unwin. Longmans, 1909. 




HOOKE'S COUPLING 



559 



each other. One shaft can thus drive the other, and as the grooves 
will always make 90 with each other, the shafts will have equal 
angular velocities in all positions. 

Hooke's coupling is illustrated in Fig. 622, and is used for con- 
necting shafts in which the axes OA and OB intersect, but are not 
necessarily in the same straight 
line. The end of each shaft is 
formed with a jaw, and the con- 
nection is made by means of a 
cross C, which is free to swivel on 
the set-screws D. The arrange- 
ment is shown in outline in 




FIG. 622. Hooke's coupling. 



Fig. 623, in which the ends of the arms OC and OC 1? attached to the 
shaft A (Fig. 623 ()), rotate in the circle YC'XQ' (Fig. 623(0)); 
the ends of the arms OD and OD 15 attached to the shaft B (Fig. 
623^)), also rotate in a circular path, but this path projects as an 
ellipse YjXYg (Fig. 623 (a)) owing to the inclination of the shaft 
axes OA and OB. 

Suppose OC rotates from OY to OC' through an angle 6 (Fig. 
623 (a)), and that the shafts OA and OB are in the same straight line. 
Then OD would rotate through an equal angle from OX and D 
would be situated at D", OD" being at 90 to OC'. If OB makes 
an angle a with OA (Fig. 623 (/;)), then D will occupy a position on 





FIG. 623. Diagram of a Hooke's coupling. 



the ellipse, obtained by rotating the cross about C'C/ (Fig. 623 (a)) ; 
this operation will cause D" to move at 90 to C'OC/, and gives the 
position of D as D' on the ellipse, i.e. OC' and OD' are still at 90. 
The angle XOD' is not the true magnitude of the angle through 



560 MACHINES AND HYDRAULICS 

which OD' has rotated from OX ; the true angle may be obtained by 
drawing D'E parallel to OY, cutting the circle at E ; XOE = < will 
then be the angle from OX through which OD rotates while OC 
rotates through an angle from OY. Produce ED' to cut OX in M, 
and let a be the angle between the directions of OD and OC X as 
seen in Fig. 623 (ft). Then 

D'M = ME cos a. 

D'M ME 
Also, =OM = OM COS 

= tan <j> cos a, 

tan , x 

or tan</> = - ..................................... (i) 

cos a 

This gives the relation of <f> and 6. The relation of the angular 
velocities of A and B may be obtained by differentiating both sides 
of (i) with respect to time. Thus, 

, d$ cosasec 2 dd sec 2 dO 

cpr*(h - = _ __ = _ _ 
9 dt COS 2 a dt COS a dt 

d$> dd . 

Now, O>B = ^p and u> A = , 

sec 2 






sec 2 



<D A SCC 2 </> COS a 

Now, sec 2 < = i + tan 2 < 

tan 2 



WB 
W A 



COS a 



COS a 



COS a 



cos 2 6/(i - sin 2 a) 
COS a 

i - sin 2 a cos 2 ^ 



This ratio will have maxima values when cos 2 is a maximum ; 
this will occur when cos 6 is + i or i, i.e. when 6 is o or 180. 
The minimum value of the ratio will occur when cos 2 has its 
minimum value ; this occurs when cos is o, i.e. when 6 is 90 or 



EXERCISES ON CHAPTER XXI. 561 

270. Equality in the angular velocities occurs when the numerator 
and denominator in (2) are equal, giving 

i - sin 2 cos' 2 = cos a, 
sin' 2 a cos' J # = i - cos a, 

2f) _ 




cos<9= . = ......................... (3) 



The student will find it a useful exercise to plot values of the ratio 
of W B to W A for values of from o to 360. 



EXERCISES ON CHAPTER XXI. 

1. An engine runs at 200 revolutions per minute and drives a line 
shaft by means of a belt. The engine pulley is 24 inches diameter and 
the line-shaft pulley is 20 inches diameter. A dynamo is driven from a 
pulley 36 inches diameter on the line shaft by a belt running on a pulley 
8 inches diameter on the dynamo shaft. Find the speeds of the line shaft 
and of the dynamo (a) if there is no slip, (b) if there is 5 per cent, slip at 
each belt. 

2. A line shaft runs at 150 revolutions per minute. A machine has 
to be driven at 1800 revolutions per minute by belts from the line shaft ; 
the pulley on the machine is 6 inches in diameter. In this particular case 
it is not desirable to use pulleys exceeding 36 inches or less than 6 inches 
in diameter, and it may be assumed that there will be 4 per cent, slip at 
each belt. Sketch a suitable arrangement giving the diameters of the 
pulleys and the speeds of any counter shafts employed. 

3. A belt laps 180 degrees round a pulley rim. The larger pull 
applied is 400 Ib. and the coefficient of friction is 0-5. Find the smaller 
pull, T 2 , when slipping just occurs. Find also the pull in the belt at 
intervals of 30 degrees round the half-circumference of the pulley, and 
plot these on a base representing angles. 

4. A rope is wound three times round a rough post, and one end of 
the rope is pulled with a force of 20 Ib. If the coefficient of friction 
between the rope and the post is 0-35, what pull at the other end of the 
rope would cause it to slip round the post? (B.E.) 

5. Find, from the following data, what width of leather belt is needed 
to transmit 25 horse-power to a certain machine : (a) Diameter of belt 
pulley, 30 inches, (b) The belt is in contact with ^ of the circumference 
of the pulley, (c) Revolutions of pulley per minute, 150. (d) Coefficient 
of friction between belt and pulley 0-22. (e) Safe maximum tension per 
inch width of belt, 80 Ib. (B.E.) 

6. A factory engine develops 400 horse-power, which is transmitted 
to the line shafting in the various mill floors by 20 hemp ropes. Find, from 
the following data, the maximum tension in any one of the ropes, if they 

P,M, 2 N 



562 MACHINES AND HYDRAULICS 

all transmit an equal share of the total power : (a) Diameter of grooved 
flywheel on which ropes work, 20 feet. (&} Angle of groove, 60 degrees. 
(c) Angle of contact of ropes with flywheel rim, 240 degrees, (d) 
Coefficient of friction, 0-18. (e) Revolutions of flywheel per minute, 80. 

(B.E.) 

7. A compressor is driven by a gas engine of 18 indicated horse- 
power, running at 240 revolutions per minute, by means of a belt 0-5 inch 
thick from the engine pulley, which is i foot in diameter. The com- 
pressor is double-acting, mean pressure 50 Ib. per square inch, cylinder 
diameter 8 inches, stroke 14 inches. If the mechanical efficiency of the 
engine is 82 per cent., of the compressor 86 per cent., and if the slip of 
the belt is 5 per cent., find the maximum speed at which the compressor 
can be run and the minimum diameter of the pulley fitted to it. (L.U.) 

8. A rope drives a grooved pulley, the speed of the rope being 5000 
feet per minute. Find the horse-power transmitted by the rope from the 
following data : ^=0-25 ; angle of groove 45 ; angle of lap 200 ; weight 
of rope per foot run 0-28 pound ; maximum permissible tension in the 
rope 200 pounds. (You are expected to make allowance for centrifugal 
effects on the rope.) (L.U.) 

9. A machine demands 6 horse-power, and is driven by means of a 
spur wheel 18 inches in diameter and running at 150 revolutions per 
minute. Find the tangential driving effort on the teeth of the spur wheel. 

10. In a turning lathe, the slide-rest holding the tool is driven by a 
leading screw having 3 threads per inch. It is desired to cut a screw of 
1 8 threads per inch. Give suitable numbers of teeth for a wheel train 
connecting the lathe mandrel to the leading screw. 

11. A watch is wound up at the same time each night and the main 
spring spindle receives 3-5 turns during the winding. What is the velocity 
ratio of the train of wheels connecting the hour hand with the main spring 
spindle ? What is the velocity ratio of the train connecting the minute 
hand with the hour hand? Give suitable numbers of teeth. for the latter 
train, no wheels to have more than 36 nor less than 8 teeth. 

12. The driving wheels of a motor car are 3-5 feet in diameter, and 
the engine runs at a constant speed of 900 revolutions per minute. Find 
the velocity ratios of wheel trains suitable for car speeds of 20, 12 and 5 
miles per hour respectively. 

13. Sketch and discuss the use of a differential gear (a) as a suitable 
means of connecting the driving wheels on a motor car, (b} as a speed- 
reducing gear : show how to calculate the speed ratio. (L.U.) 

14. In the epicyclic train shown in Fig. 600, the wheels have teeth 
as follows: D, 48; B, 10 ; C, 12 ; A, 30. If F makes one clockwise 
revolution, find the revolutions of A. 

15. In the gear shown in Fig. 601, the numbers of teeth are : D, 40 ; 
E, 20 ; G, 40. If E is fixed, find the revolutions of G for one clockwise 
revolution of A. Answer the same if E is driven at the rate of 3 anti- 
clockwise revolutions for one clockwise revolution of A. 

16. In the Humpage gear illustrated in Fig. 602, the wheels have teeth 
as follows : B, 25 ; C, 30 ; G, 45. Calculate the numbers of teeth on D 
and H, so that the ratio of the rotational speed of A to that of K is 56 : =;. 

(L.U.) 



EXERCISES ON CHAPTER XXI. 563 

17. State and prove the geometrical condition which must be satisfied 
in order that a pair of spur wheels may gear together with a constant 
angular- velocity ratio. (I.C.E.) 

18. The centres of two spur wheels in gear with one another are 
12 inches apart. One wheel has 40 teeth, and the other has 20 teeth. 
Neglecting friction, the line of pressure between the teeth in gear makes 
a constant angle of 75 with the line of centres. The teeth are designed 
so that the path of contact of a pair of teeth in gear is 2 inches long, and 
is bisected by the line of centres. Draw full-size a side elevation of two 
teeth in gear. (L.U.) 

19. The axes of two shafts intersect at an angle of 1 50. The shafts 
are connected by a Hooke's coupling. On a straight base 8 inches long, 
representing 360, draw a curve whose ordinates represent the angular 
velocity of the driven shaft for one revolution, the angular velocity of the 
driving shaft being constant and represented by an ordinate 2 inches long. 

(L.U.) 



CHAPTER XXII. 

HYDRAULIC PRESSURE. HYDRAULIC MACHINES. 

Some properties of fluids. A fluid may be defined as a substance 
which cannot offer permanent resistance to forces which tend to change its 
shape. Fluids are either liquid or gaseous ; gases possess the 
property of indefinite expansion. Liquids alter their bulk but 
slightly under pressure, and such small changes may be disregarded 
usually. Gases exist either as vapours or as so-called perfect gases ; 
the perfect gas was supposed to exist as a gas under all conditions of 
pressure and temperature ; but it is now well known that all gases 
can be liquefied by great pressure and cold. A vapour may be 
defined as a gas near its liquefying point, and a perfect gas as the 
same substance far removed from its liquefying point. 

Liquids are said to be mobile when they change their shape very 
easily ; chloroform is an example showing great mobility, a property 
which renders it useful for delicate spirit levels. Viscous liquids are 
those which change their shape with difficulty ; examples of such are 
cylinder oil and treacle. 

Change of shape of a body always occurs as a consequence of the 
application of shearing stresses. A rectangular block under the 
action of equal push stresses on all its faces will have its volume 
diminished, but will remain rectangular; shearing stresses applied 
to the block would alter its shape (p. 107). Hence, if shearing 
stresses be applied to a fluid, change of shape of the fluid will go on 
continuously during the application of the stresses, i.e. the fluid will 
be in motion. Conversely, if the fluid is at rest, there cannot be any 
shearing stresses acting on it ; the stresses must be normal at all parts. 
Frictional forces always occur as tangential or shearing forces, and 
hence must be absent from any fluid at rest. 

The principal liquid in use in hydraulics is water, and it. will be 
understood that water is being referred to in the following sections 
unless some other liquid is specified, 



STRESS ON IMMERSED SURFACES 



565 



X 




7 




X 


1 


i - - 


1 




! 




; i 




H 




:i 




1 






fifH 



FIG. 624. Stress on a horizontal immersed 
surface. 



Stress on horizontal immersed surfaces. In Fig. 624 is shown a 
tank containing a liquid at rest. Consider the equilibrium of a 
vertical column of the liquid stand- 
ing on one square foot of the tank 
bottom. The forces acting will be 
(a) the weight of the column, (b) 
an upward reaction from the tank 
bottom, (c) normal forces which the 
surrounding liquid applies to the 
vertical sides of the column. The 
normal forces, being horizontal, can- 
not contribute to the support of the 
weight of the column, which is a 
vertical force ; hence (a) and (ft) 
equilibrate each other. 

Let H = the height of the column, in feet. 

w = the weight of a cubic foot of the liquid, in Ib. 
A = the total area of the tank bottom, in square feet. 
Then Weight of the column = H x i x i x w 

= ?e/H Ib., 

and this will be the reaction of one square foot of the tank bottom. 
Hence, 

Stress on the tank bottom = zvH Ib. per square foot. 
The tank bottom being horizontal, the stress on any other square 
foot will be 7t>H ; hence the total pressure on the bottom may be 
calculated by multiplying the area of the bottom by ze/H. Thus, 

Total pressure on the tank bottom = ze/H A Ib. 

It will be noted that this result is quite independent of the shape 
of the tank, provided the bottom is horizontal. All tanks having 
horizontal bottoms of equal areas and charged with the same liquid 
to equal depths will have equal total pressures on their bottoms 
irrespective of the actual weights of liquid in the tanks. The student 
should guard against the error of supposing that the weight of liquid 
in the tank gives the pressure on the bottom. 

Stress on inclined immersed surfaces. Let abc (Fig. 625) be the 
end elevation of a triangular prism immersed in a liquid and having 
its axis horizontal. The triangular ends are taken perpendicular to 
the axis and are vertical. Considering the equilibrium of the prism, 
the fluid pressures on the ends will evidently equilibrate each other. 



566 MACHINES AND HYDRAULICS 

If the sides of the prism be taken to be very small, then the weight of 
the prism may be disregarded, and the fluid stresses/, q and r, acting on 
ab, be and ca respectively, may be assumed to be distributed uniformly 




FIG. 625. Stress on inclined immersed surfaces. 

over the faces of the prism. Let the length of the prism be unity, 
when the resultant forces on the three faces will be given by 

P =p . ab. 



For equilibrium of the prism, these forces must balance. It will 
be noted that P, Q and R meet at the centre of the circle passing 
through a, b and c, and hence comply with the condition that the 
three forces must pass through the same point. ABC (Fig. 625) is 
the triangle of forces, in which AB, BC and CA represent P, Q and 
R respectively. As these sides are drawn perpendicular to ab, be 
and ca respectively, the triangles ABC and abc are similar. Hence, 

P:Q:R=/-.a:?. &:*-. <ra ..................... (i) 

= AB : BC : CA ....................... (2) 

= ab : be : ca ........................ (3) 

From (i) and (3), 

p . ab : q . be : r . ca = ab : be : ca ; 
.'. p = q = r. 

We may say therefore that the fluid stresses on the faces of the 
prism are equal. Considering the limiting case of the end elevation 
of the prism being reduced practically to a point by reason of the 
sides being taken indefinitely small, the law may be stated thus : 
The stress at a point in a fluid is the same on all planes passing through 
that point, or fluids transmit stresses equally in all directions. We have 
already seen that the stress on a horizontal plane is ze>H Ib. per 



TOTAL FLUID PRESSURE 



567 



square foot, and it follows that the stress at a point H feet deep on 
any plane will be given by the same expression. 

It will be noted that the stress at any point is proportional to the 
depth, and hence varies uniformly from zero value at the free surface 
of the liquid, i.e. the surface exposed 
to the atmosphere. Stress diagrams 
may be employed with advantage ; 
such a diagram is given in Fig. 626 
for the water stresses on each side of 
a lock gate having differing depths of 
water on the two sides. The stress at 
B will be/ 1 = ?^H 1 and that at E will 
be/ 2 = wH 2 , and these are represented 
by CB and FE respectively. The 
complete stress diagrams are ABC and 
DEF, and their breadths will give the stress at any depth. The 
term stress at a point in a fluid may be defined as the pressure which 
would be exerted on unit area embracing that point if the stresses were 
distributed uniformly. 

Total pressure on an immersed surface. In Fig. 627 (a) and (ft) 
are shown front and end elevations of immersed surfaces, the former 
being vertical and the latter inclined. The method of finding the 




F D ; E B o. 

FIG. 626. Stresses on a lock gate. 





FIG. 627. Total pressure on immersed surfaces. 

total pressure applies equally to both surfaces. Consider a small 
area a at a depth y. The stress on a will be 

p = ivy 
and Force on a = wya 

The total force on the surface may be found by integrating this 
expression over the whole area. Thus, 

Total force w^ay. 

If the total area is A square feet and the depth of the centre of 
area is Y feet, then 2aj' = AY (pp. 49 and 145). Hence, 
Total force =>w AY Ib. 



568 MACHINES AND HYDRAULICS 

The expression wY is the stress at a depth Y, and may be defined 
as the average stress on the immersed surface. Hence the rule : 
To find the total pressure on an immersed surface multiply the average 
stress (which will be found at the centre of area) by the total area. 

It will be noted that the force acting on the small area a, given 
above as wya, will have the same value in the case of the whole 
surface being curved. The rule for the total pressure is therefore not 
confined to flat surfaces, but may be applied to any immersed surface. 

Distinction should be made between the terms total pressure and 
resultant pressure. The latter term refers to the resultant of all the 
fluid stresses acting on a surface, and is obtained by resolving 
these stresses along chosen axes and then reducing by the methods 
explained in Chapter IV. Usually the operation is simple; for 
example, the resultant pressure on any vessel containing a liquid is 
evidently equal to the weight of the contained liquid. A method of 
dealing with the resultant pressure on floating or immersed bodies 
will be explained below. 

EXAMPLE i. A cylindrical tank, diameter 7 feet, contains water to a 
depth of 4 feet. The bottom is horizontal. Calculate the total pressure 
and the resultant pressure on the wetted surface. Take 62-5 Ib. per 
cubic foot for the weight of water. 

Total pressure on the bottom wA i y i 

ir<P 
= 62-5 x X4 

= 62-5x( 2 7 *x Y)X4 
= 9625 Ib. 
Total pressure on the curved surface = wA 2 Y 2 

= 62-5 X(?JY/X4)X2 

= 62-5 x( 2 T - x;x4)x2 
= ii,ooo Ib. 

Total pressure on the wetted surface = 962 5 + 1 1,000 

= 20,625 Ib. 

The stresses on the curved surface will equilibrate each other ; hence 
the resultant pressure is simply the total pressure on the bottom, or 
Resultant pressure = 9625 Ib. 

EXAMPLE 2. A spherical vessel 3^- feet in diameter is sunk in sea 
water, its centre being at a depth of 40 feet. Calculate the total pressure 
on its surface. Sea water weighs 64 Ib. per cubic foot. 
Total pressure = wA Y 

= 64 x 47rr 2 x 40 

= 64 x 4 x 2 7 2 x J x | x 40 

= 98,560 Ib. 



RESULTANT FLUID PRESSURE 569 

Resultant pressure on a floating or immersed body. When a 
body is floating at rest in a fluid which is also at rest, it is subjected 
to two resultant forces its weight and the resultant fluid pressure 
on its surfaces. The weight is a downward vertical force acting 
through G, the centre of gravity of the body (Fig. 628 (a)). The 
resultant fluid pressure balances W, and therefore must be an 
upward vertical force R = W, and must act in the same straight line 
with W. R is due to the buoyant effect of the fluid, and is called 
the buoyancy. 

Imagine for a moment that the surrounding fluid becomes solid, 
and so can preserve its shape, and let the body be removed, leaving 
a cavity which it fits exactly (Fig. 628 (<)). Let this cavity be filled 
with the fluid, and let the surrounding fluid return again to its original 
condition. The pressures on the fluid now filling the cavity will be 
identical with those which acted on the body, and the effect will be 



V 5 



W () 

FIG. 628. Resultant pressure on a floating body. 

the same the weight of the fluid will be supported. Hence the 
weights of the fluid filling the cavity and of the body must be equal, 
as each is equal to R, the resultant pressure of the surrounding fluid. 
Further, R must act through the centre of gravity of the fluid filling 
the cavity ; this centre is called the centre of buoyancy, and from 
what has been said it will be clear that the centre of buoyancy B 
(Fig. 628 (/>)) and G (Fig. 628 (a)) must be in the same vertical line. 
We may state, therefore, that when a body is floating at rest in still fluid, 
the weight of the body is equal to the weight of the fluid displaced, and that 
the centres of gravity of the body and of the displaced fluid are both in the 
same vertical line. 

A ship floating at rest in still water, a submarine boat wholly 
immersed and at rest, and a balloon preserving constant elevation 
are examples of this principle. In each case it will be noted that 
the resultant pressure of the surrounding fluid is equal to the weight 
of the body, and acts vertically upwards through the centre of 
gravity of the body. 



570 MACHINES AND HYDRAULICS 

A body wholly immersed will experience a resultant upward fluid 
pressure equal to the weight of the fluid displaced ; it follows that, 
to maintain the equilibrium of the body, an upward or a downward 
force will be required, depending on whether the weight of the body 
or the weight of the fluid displaced is the greater. Fig. 629 (a) 
illustrates the former case ; W is the weight of the body, B is the 

buoyancy, and, W being greater than B, 
an upward force P is required given by 
P + B-W. 




r \ i Fig. 629 (b) shows the case of B being 

^*"T* - '^ T"~ greater than W, when a downward force 

P is required, given by 
fW 

f 



FIG. 629. Equilibrium of immersed The specific gravity of a Substance IS 

defined as its weight in air as compared 

with the weight of an equal volume of pure water, usually taken at a 
temperature of 60 F. 

Let W 8 = weight of a given substance in air. 

W w = weight of an equal volume of water. 

W 

Then Specific gravity p =^ 

W w 

and W w = '-. 

P 

It therefore follows that we may calculate the buoyancy of a solid 
body wholly immersed in pure water by dividing the weight of the 
body by the specific gravity of its material. This principle may be 
applied to find the specific gravity of a given substance which is 
heavier than water. In Fig. 629 (a), let P be measured by suspending 
the body by means of a fine wire or cord from a balance ; also 
weigh the body in air to find W s . Then 



Also, 



Since (W s - P) is the apparent loss of weight of the body when 
immersed in water, we may state that the specific gravity of a body is 



CENTRE OE PRESSURE 571 

equal to the weight of the body in air divided by its apparent loss of weight 
when immersed in water. 

Centre of pressure. In Fig. 630 (a) is shown a flat vertical plate 
immersed in a liquid. R is the resultant pressure acting on one 
side of the plate and passes through a point C, which is denned as 
the centre of pressure. The position vertically of C may be found 




J.' 

if 

-5 
R 

fa) 

FIG. 630. Centre of pressure. 

by taking moments about OX, the line in which the plate produced 
cuts the surface of the liquid. Considering a small area a at a depth 
y, we have Pressure on a = way, 

Moment of this pressure = way 1 . 
Integration of this will give the total moment. Thus, 
Total moment = w^ay 2 . 

Now 2#y 2 is the second moment of area or moment of inertia 
(p. 145) of the surface of the plate about OX and may be written 
lox or A/ 2 , where A is the area of the plate and k is the radius of 
gyration about OX. Hence, 

Total moment = whk 2 (i) 

Again, if D be the depth of the centre of pressure, 

R = ze/A Y 

and Moment of R = z#AYD (2) 

Hence, from (i) and (2), 



a/A YD 

D = f... ..-(3) 

Both k and Y should be taken in foot units, when D will be in the 
same units. 

The case of an inclined surface is shown in Fig. 630 (/>). If $ is 
the angle of inclination to the horizontal, it may be shown that 

D = sin 2 , (4) 



572 



MACHINES AND HYDRAULICS 



where k is the radius of gyration about OX, the line in which the 
plate cuts the surface of the water, and Y is the vertical depth of 
the centre of area. 

In practical examples, usually the position of C horizontally may 
be easily determined from the symmetry of the plate. 

EXAMPLE. A dock gate is 60 feet wide and has water on one side to a 
depth of 24 feet. Find the centre of pressure. 

Let b = the breadth of the wetted surface. 

</=the depth 

Then I ox = J- = bd '- ; :. & = . 



Also. 



= 5 x 24= 1 6 feet. 

The centre of pressure is therefore at a depth of 16 feet, and lies in 
the central vertical of the gate. 

Stability of a floating body. A body floating at rest in a still 
liquid will be in stable equilibrium when, if rotated through a small 
vertical angle, it experiences a resultant couple tending to return it to 




(a) 



R-wT 

FIG. 631. Stability of a floating body. 



the original position ; the equilibrium will be unstable if the resultant 
couple has a moment tending to increase the angle of rotation. In 
Figs. 631 (a) and (/^).are shown floating bodies which have been 
disturbed slightly from their positions of equilibrium : the weight, in 
each case, is a vertical force W, acting through the centre of gravity G ; 
the buoyancy in each case is a vertical force R = W, acting through 
the centre of buoyancy B. It will be observed that in Fig. 631 (a) a 
couple is formed by R and W tending to restore the body to its 
original position ; the equilibrium in the original position is therefore 



STABILITY OF A FLOATING BODY 



573 



stable. In Fig. 63 r (/>) the couple tends to increase the angle through 
which the body has been turned, and the equilibrium in the original 
position is therefore unstable. A floating ball would be in neutral 
equilibrium. 

It will be noticed that the line of R cuts the original vertical 
through G in a point M, which lies above G in Fig. 631 (a) and below 
G in Fig. 631 (l>). Clearly the sense of rotation of the couple formed 
by R and W is determined by consideration of the position of M above 
or below G ; the couple will be of righting moment if M is above G, 
and of upsetting moment if M is below G. The point M is called 
the metacentre. The metacentre is of importance in calculations 
regarding the stability of ships ; generally the naval architect finds 
the metacentre for transverse angles of displacement, which affects 
questions of the ship rolling, and also the metacentre for longitudinal 
angles of displacement, which affects questions of the ship pitching. 

In Fig. 632, G is the centre of gravity and B is the centre of 
buoyancy of a body floating at 
rest in still water. G and B 



in 

must fall in the same vertical, 
and the conditions of equili- 
brium are satisfied by the re- 
sultant water pressure R being 
equal to W, the weight of the 
body, both forces falling in 
the same straight line BG. 
To test for stability, the body 
is rotated through a very 
small angle 6, which, in order 
to avoid complication in the 
figure, has been secured by 
rotating the water plane from 
its original position ab into 
the position db ' . G will remain 
unaltered in position, and B 
will move to B' in consequence 
of the body now being immersed 
deeper on the right-hand side. 
The weight of the body is now 
W' = W and acts through G in 
a direction perpendicular to 
ab' ; the resultant pressure of 
the water will be R' = W = W, 
acting through B' and also perpendicular to ab '. 
BG produced in the metacentre M. 



"* 




Plan 



FIG. 632. Metacentric height for a floating vessel. 



R' produced cuts 



574 MACHINES AND HYDRAULICS 

As we assume that W and W are equal, it follows that the weight 
of the wedge l>Cb\ which has been added to the volume of water 
displaced, must be equal to that of the wedge aCa, which has been 
taken away. In the plan of the plane of flotation ab (Fig. 632), 
small areas a and a trace out arcs / and /' as seen in the elevation. 

Volume swept by a = al. 

NOW, l -=e-, 



Hence, Volume swept by a = axO. 

Weight of this = waxO, ............. ^ ........... (i) 

w being the weight of the water per cubic unit. 

The total weight of both wedges must be zero from what has been 
said, and may be obtained by integrating (i) over the whole plane 
of flotation. 

Total weight of wedges = ivB^ax = 0. 

Hence, ^ax = 0. 

This shows that the axis CZ in the plan must pass through the 
centre of area of the plane of flotation. 

The resultant effect of the altered distribution of displacement 
will be found by calculating the total moment of weight of both 
wedges about CZ. 

From (i), weight of the small volume = wOax. 
Moment of this about CZ = 
Total moment of both wedges = 

= U>OICZ ................ (2) 

In this result, I cz is the second moment of area of the plane of 
flotation about CZ. 

Now, if R' be brought back to its original position R, we see that 
the effect of the altered distribution of displacement will be the 
couple, of moment R x BB', which must be supplied in consequence 
of the shift. 

Moment of couple = R x BB' = W x BB'. 

Now, 



:. BB' = 6>xBM. 
.*. moment 01 couple = Wx 6 x BM ........................ (3) 

Hence, from (2) and (3), 



(4) 



STABILITY OF A FLOATING BODY 575 

Let V = volume of water displaced by the body. 
Then W = 70 V, 

or V = 

w 

Substituting in (4), we have 

BM = ^ Z . ....(5) 

Writing AJt* cz for I cz , we obtain a well-known equation for BM, 

EM = ^' ..-(6) 

From Fig. 632, we have 



GM will be positive if M falls above G, in which case we have 
stable equilibrium ; the equilibrium will be unstable if G falls below 
M, leading to a negative value of GM. 

It will be noticed that the completion of the calculation depends 
on a knowledge of the position of the centre of gravity of the body. 
In the case of a body of simple outline and homogeneous in structure, 
this point is determined easily, but, in the case of a ship, is obtained 
only by long and laborious calculation. The calculations for 
A/&CZ and for V required in equation (6), and also for the position of 
B, are carried out easily for a ship-shape body, and the result may be 
applied to the finished ship, in an experimental determination of the 
centre of gravity. This is effected by moving weights on board so as 
to produce a small angle of heel, which is measured carefully by 
means of long plumb lines suspended in the holds. From a 
knowledge of the positions of M and B, together with the moment of 
the weights which have been moved and the angle of heel produced 
by this movement, the position of G is calculated easily. Thus, 
referring to Fig. 632, let the line of W cut BB' in N and draw GQ 
perpendicular to B'M. 

Let w = the weight moved, in tons. 

d= the distance through which the weight is moved, in feet. 
6 the angle of heel produced by moving w, in radians- 
Then Capsizing moment due to moving w = wd ton-feet. 
Righting moment = R' x B'N 
= WxGQ 
= W x GM x B. 

Hence, W x GM x 6 = wd, 

wd 






57 



MACHINES AND HYDRAULICS 



The author is indebted to Mr. E. L. Attwood, Member of the 
Royal Corps of Naval Constructors, for the following example of a 
recent inclining experiment on a large ship. 

EXAMPLE. Draught of ship, forward, 24' 4". 

Draught of ship, aft, 26' 2\". 

These dimensions correspond to a displacement of 15,357 tons, and a 
position of the transverse metacentre of 6-76 feet above the load water- 
line of the ship. 100 tons of ballast was used, arranged on the upper 
deck, in four lots of 25 tons each. The following measurements were 
taken by means of pendulums 20 feet in length, one forward, one aft : 



Weight moved, 


Distance 
moved , 


Direction of movement. 


Deflections of pendulums, 
in inches. 




feet. 




Forward . 


Aft. 


25 


62 


Port to starboard 


7-9 


8-1 


50 


62 


55 


157 


15-8 


25 


62 


Starboard to port 


8-0 


8-0 


50 


62 


j 


15-6 


15-6 



Taking the mean of these gives a deflection of 15-84" for a shift of 
50 tons through 62 feet. Hence, 

GM wd 



50X62 
V->C7X 15 ' 84 

K ~^o~ 
= 3-06 feet. 

The centre of gravity of the ship is therefore (676 -3-06) =3-7 feet 
above the load water-line. 

Retaining wall for water. Referring to Fig. 633, ABCD is the 
section of a wall subjected to water pressure on its vertical face AB. 
In considering the stability of the wall, a portion one foot in length 
may be taken. For the simple trapezoidal section of wall illustrated, 
the weight may be calculated easily. Thus, 



W 



, /AD + BC\ 1U 
= w ( ]H lb., 



where w' is the weight of the material in lb. per cubic foot. H is the 
height of the wall, and AD and BC are the thicknesses at the top and 
bottom respectively, all in foot units. 

The centre of gravity of the wall section may be found by applica- 
tion of the following graphical method, Bisect AD in a and also 



RESERVOIR WALLS 



577 



BC in b ; then G lies in ab. Make Kc and G/ equal to BC and AD 
respectively, and join cd cutting ab in G. 

If the reservoir is empty, the point m, in which the line of W cuts 
the base BC, will be the centre of pressure of the base of the wall, 



D a A 




FIG. 633. Stability of a reservoir wall. 

and the pressure on the base will be owing to W only, and hence 
will be entirely normal to the base. To find the pressure on the 
base and the centre of pressure when the reservoir is full, proceed as 
follows : 

Total water pressure on the wall = P = z#AY (p. 567) 



where w is the weight of the water in Ib. per cubic foot. P will act 
at JH feet from B (p. 572), and will meet the line of W &tf. Con- 
struct the parallelogram of forces fgkh for P and W acting at /, thus 
finding the resultant pressure R on the wall base. R intersects BC 
at n, thus giving the centre of pressure of the base for the case of the 
reservoir being full. It is taken usually that the wall will be safe if 
both m and n fall within the middle third of the base. 

Every horizontal section of the wall will have a centre of pressure 
for the reservoir empty and another for reservoir full. If these 
centres be found, curves joining them may be drawn and give the 
lines of pressure for the wall. Fig. 634 shows how the construction 
may be carried out for sections 22', 33' and 44'. P x is the total 
water pressure on the whole wall, P 2 , P 3 and P 4 are the pressures 
respectively for the portions lying above 22', 33' and 44'. W^ is the 
P.M, 2 o 



578 



MACHINES AND HYDRAULICS 



total weight, and W 2 , W 3 and W 4 are the weights corresponding to 
P 2 , P 3 and P 4 . The centres of gravity G I} G 2 , G 3 and G 4 are found 

D 




B 
FIG. 634. Lines of pressure for a reservoir wall. 

as before, and the lines of weight passing vertically through them 
give m lt m 2 , m B and m 4 on the line of pressure for reservoir empty. 
D u a s A Rj of Pj and W l is found by 
means of the triangle of 
forces, and a line drawn from 
the point of intersection of 
Pj and Wj parallel to R x , and 
cutting BC in j gives a point 
on the line of pressure for 
reservoir full. The triangles 
of forces for the remaining 
forces are shown, and enable 
points ;z 2 , n^ and n 4 to be 
found similarly. The lines 
of pressure have been drawn 
separately in Fig. 635 for the 
sake of clearness. In Fig. 635 
uv and st inclose the middle 
thirds of all sections, and the 

FIG. 63S.-Lines of pressure, reservoir full and empty. lines of pressure an^ and am^ 




WORK DONE BY FLUID PRESSURE 



579 



fall throughout within the middle thirds. The student will note that 
the upper ends of the lines of pressure bisect AD in a. 

Work done by a fluid under pressure. Work may be done by a 
fluid, either liquid or gaseous, by allowing it to exert pressure on a 
piston which may move in a cylinder. In 
Fig. 636, 

Let D = the diameter of the cylinder, 

in feet. 

L = the length of the stroke, in feet. 
P = the pressure of the fluid, in 
Ib. per square foot. 



-* 




Z 1 ^ 


** 




i ! r ' 

__* . i 



B 



FIG. 636. Work done by a fluid. 



Then, if a liquid be employed, owing to the absence of any 
expansive property, the pressure P must be maintained by continuous 
admission of liquid to the cylinder. The work done while the piston 
moves from A to B will be 



-D 2 



Work done = P x ----- x L foot-lb. 
4 



Now L is the volume swept by the piston, and also represents 

4 
the volume of liquid admitted in cubic feet ; writing this volume V, 

we have Work done = P V foot-lb. 

This expression also applies to the case of a gas supplied under 

constant pressure throughout the stroke. 

Fig. 637 shows in outline a hydraulic engine using water as the 

working fluid. There are three cylinders, A, B and C, arranged 

at angles of 120; the water 
pressure acts on one side of 
the pistons only, and all the 
pistons are connected to a 
single crank DE. The arrange- 
ment produces a fairly uniform 
turning moment. In engines 
of this type, as the cylinders 
must be filled completely with 
water during each stroke, the 
efficiency will fall very rapidly 
unless the demand for power 
is maintained steadily at its 




FIG. 637. Three cylinder hydiaulic engine. 



maximum amount. Otherwise, devices may be applied by means 
of which the capacity of the engine may be reduced when a 
diminished demand for power occurs. These devices usually take 



MACHINES AND HYDRAULICS 




the form of having the crank of variable throw ; the strokes of the 
pistons will then vary to correspond. The crank adjustment may be 
effected either by means of a governor or by means of an automatic 
spring coupling between the engine and the machine to be driven. 

If a gas is used in the cylinder in Fig. 636, advantage may be 
taken of its expansive property by cutting off the supply after the 
piston has moved a short distance and allowing the remainder of the 
stroke to be completed under the continually diminishing pressure of 

the gas. In Fig. 638 is plotted 
a curve AB, showing the relation 
of pressure (vertical) and volume 
(horizontal) while a gas is ex- 
panding and doing work. Usually 
the law of the curve AB takes 
the form 
_ v _ I x P V n = a constant, 

V . v" *^- -J where P is the pressure of the gas 

2 in Ib. per square foot measured 

FIG. 638. Work done by an expanding gas. 

from zero. On this basis the 

pressure of the atmosphere is about 14-7 Ib. per square inch or 
2116 Ib. per square foot. V is the volume in cubic feet, n is an 
index which depends on the conditions under which the expansion is 
performed. If the temperature is preserved constant, then Boyle's 
law is being followed, n is unity, and the expansion law will be 

PV = a constant \ 
n usually lies between i and 1-5. 

The work done may be found from the area of the diagram under 
AB in Fig. 638. Thus, assuming Boyle's law to be followed and 
taking a narrow strip EF, for which the pressure is P, the volume V 
and the increase in volume represented by the breadth of the strip is 
8V, we have 

Area of the strip = P . 8V. 

Now, from Boyle's law, PV = P 1 V 1 ; 



SV 



Hence, Area of the strip = PjV a 

Total area under AB = P l V l I -y- j 

,". work done = PjVj log,, ^ foot-lb. 



HYDRAULIC TRANSMISSION OF ENERGY 581 

If the expansion law is PV n = a constant, we have, in the same 
manner : A rea of the strip = P . 8V. 

Also, PV = P 1 V 1 ' 1 , 

P V 
p_ r l v l 

v* 

8V 
Hence, Area of the strip = PjV^ 1 



Total area under AB 



PWV 
= PjV^ ^~ 




Remembering that P 1 V 1 n = P 2 V 2 n , the above becomes, by multi- 
plication, p V - P V 

Work done = l YI 2 2 foot-lb. 

n - i 

Hydraulic transmission of energy. In Fig. 639 A and C are two 

cylinders charged fully with water, and connected by a pipe E. 
B and D are plungers or rams fitted to cylinders, 
and carrying loads P and W. Owing to the 
practical incompressibility of water, any descent 
of B will produce an ascent of D, and hence work 
done on P may be transmitted by the medium of 
the moving water under pressure, and be given 
out in the form of work done on W. The con- 
necting pipe E may be of any length. In practice, 
A represents a set of power-driven pumps, which supply water under 
a pressure of 700 to 1000 Ib. per square inch. A pipe system distri- 
butes the water over the district to be supplied, and D may be taken 
to represent one of the machines to be operated. The principal 
appliances required in a hydraulic power distribution plant are 
shown diagrammatically in Fig. 640. A is one of the power-driven 
pumps supplying water to the pipe line BC. A safety valve is placed 
at D. E is an accumulator consisting of a large cylinder fitted with 
a loaded ram, and connected to the pipe line; its function is to 
absorb energy by raising the weight if the machines are stopped and 
the pumps are still working ; it also assists in preserving a steady 
pressure of water. A stop valve F is under the control of the 
consumer, and another safety valve is placed at G in order to guard 
against damage to his pipes and machines. H, H represent two of 
the machines being driven; each is fitted with a control valve K, 



582 



MACHINES AND HYDRAULICS 



K, for the use of the operator. Usually the exhaust water from the 
machines is collected and passed through a meter, where it is 
measured for the purposes of charging for power. 




FIG. 640. Diagram of a hydraulic installation. 

Referring again to Fig. 639, let d 1 and d^ be the diameter of B and 
D respectively in inches, and let p be the water pressure in Ib. per 
square inch ; also let W and P be measured in Ib. Then, neglecting 

friction : ^d 2 

P = l -p, 



Hence, 5~ == ;^2* 

A (t-i 

This gives the mechanical advantage of the arrangement neglecting 
friction. If P descends one inch, then the volume of water delivered 



from A into C will be 



cubic inches. To accommodate this 



volume in C, D will rise a height h inches say, and the additional 

7 

volume in D will be h cubic inches. 
4 



Hence, 



rc , tr-i 
*- h = Li 

4 4 



Now i -h is the velocity ratio of the arrangement. Hence, 

d^ 
Velocity ratio = -h' 



HYDRAULIC MACHINERY 



583 



Comparison of these results shows that in this hydraulic arrange- 
ment, as in other machines, the mechanical advantage when friction 
is neglected is equal to the velocity ratio 
(p. 328). The resistance W, which may be 
overcome by the ram T) in this arrangement, 
may be very large if the ram is made of 
sufficient diameter. For example, a ram 
10 inches in diameter, and supplied with 
water at 700 Ib. per square inch, will exert a 
total force of about 24^ tons. The principle 
is made use of in hydraulic presses, forging 
and other machines. 

Some examples of hydraulic machinery. 
The cylinder for a hydraulic lift is shown in 
some detail in Fig. 641. The ram passes 
through a stuffing box in the lower end of 
the cylinder, and carries two pulleys mounted 
on its end and both running on the same 
spindle. Another pulley is placed on the top 
end of the cylinder. The wire rope used for 
hoisting the cage is attached to a fixed point 
at A, and is led round the pulleys, as shown, 
before being taken away at B to the cage. 
The object is to multiply the comparatively 
small movement of the ram into the larger 
travel required for the cage. The same type 
of cylinder is made use of in hydraulic cranes. 
Some types of leather 

packing are shown in 

Fig. 642. (a) is a U- 

leather, used for keeping 

water-tight rams of fairly 

large diameter ; the 

water may enter the 

hollow interior of the U, 

and presses the leather outwards against the 

wall of the recess and also against the ram. 
FIG. 642. Types of leather In (b] is shown a hat leather, used for sliding 

packing. i i / \ ' 11 

plungers and rods ; (c) is a cup leather, used 
for pistons in cases where the water acts on one side of the piston 
only. 





FIG. 641. Cylinder for a 
hydraulic lift. 



MACHINES AND HYDRAULICS 



A simple hydraulic accumulator is illustrated in Fig. 643. The ram 
A is fixed to the base plate, and the cylinder B is loaded with a 
number of cast-iron plates and may move vertically. A tail rod C is 
fixed to the cylinder, and serves as a guide. Water enters the 
cylinder by way of an axial hole bored through the ram. When the 
cylinder is nearing the top of its lift, it raises the end U of the lever 
DE ; the movement of this lever is transmitted to the belt-striking 

n. 




FIG. 643. Hydraulic accumulator. 

gear on the pump, or to the throttle of the pump engine, and so stops 
the pump. A spring F pulls the levers back to working position 
when permitted by the descent of the accumulator, and so starts the 
pump again. The following simple calculations may be made 
regarding hydraulic accumulators : 

Let //=the diameter of the ram, in inches. 

/ = the water pressure, in Ib. per square inch. 

W = the total accumulator load, in Ib. 

H = the height of lift, in inches. 

72 

Then W=/ x - - Ib., neglecting friction. 

4 



HYDRAULIC MACHINERY 



585 



When the accumulator is " up," the volume of water stored will be 

70 

Volume stored = H cubic inches. 

4 

Also, Energy stored = WH inch-lb. 

Occasionally it occurs that a hydraulic machine requires a greater 
pressure of water than that supplied in the mains, and an intensifier 
is used in order to secure this. In Fig. 644 a cylinder A has a 
hollow ram B which passes through its right-hand end. A fixed 




FIG. 644. Hydraulic intensifier. 

hollow ram C passes into the interior of B as shown. Low-pressure 
water is supplied at D, and water of a higher pressure is discharged 
at E. 

Let /j = the lower pressure, in Ib. per square inch. 

p. 2 = higher 

d l = the external diameter of B, in inches. 
d^ the external diameter of C, in inches. 
Then, neglecting friction, 



or 



A 4 2 

With a ratio of diameters of 2 to i, the supply pressure of 700 Ib. 
per square inch may be intensified to 2800 Ib. per square inch, 
neglecting friction. Valve arrangements are provided for enabling 
the lower pressure to be used in the machine, and at the moment 
when the higher-pressure water is required, low-pressure water is 
admitted to A in the intensifier, and at the same time the machine is 
connected to E. 

Pumps. Fig. 645 shows a hydraulic pump suitable for supplying 
water for operating hydraulic machines. A cylinder A is fitted with 



586 



MACHINES AND HYDRAULICS 



a piston B operated by a plunger rod C. Water enters the cylinder 
at D, and is prevented from flowing back by the suction valve E. 
G is a discharge valve opening to the discharge branch H, and F is a 
passage connecting the right-hand side of the piston to the discharge. 
The valves E and G are cushioned on lifting against rubber discs, 
separated by metal washers ; the piston packing consists of two cup 
leathers. The action is as follows : Suppose the piston to be moving 




FIG. 645. Hydraulic pump. 



towards the right as shown ; E is open and G is closed. W T ater will 
flow into the pump through E, and will fill the space vacated by 
the receding piston ; at the same time, the water on the right-hand 
side of the piston is being forced into the discharge pipe through F. 
If the diameters of the piston and of the plunger rod are d^ and d^ 
respectively, and if the stroke is L inches, then the volume discharged 
from the right-hand side of the piston during this stroke will be 

f--t__~a-)L cubic inches. 
\ 4 4 / 

Now let the piston be moving towards the left ; E will be closed 
and G will open, and the water on the left-hand side of the piston 
will be discharged through G. The volume so discharged will be 

7 2 

L cubic inches, but a portion of this only will be sent into the 

discharge pipe, the remainder finding its way through F to the right- 
hand side of the piston ; the amount so passing through F will be 

( ) L cubic inches ; hence the volume discharged from the 

\ 4 4 / 

pump during this stroke will be 



\ 4 4 / 



L cubic inches. 



RECIPROCATING PUMPS 



587 



The pump is thus double-acting, i.e. water is discharged during both 
strokes. For equality of discharge, we have 



V 4 



or 



Ln t*-o 
= i. 

4 / 4 

^l 2 ~ 4 2 = ^2 2 ' 



This result may be expressed also by stating that the sectional area 
of the plunger rod should be half that of the piston. 

Fig. 646 illustrates a type of bucket pump used in raising water from 
a lower to a higher level. The piston or bucket is shown ascending, 
and water is passing into the cylinder A 
through B and the suction valve C. The 
water already on the top of the bucket is 
being discharged through the discharge valve 
F and the passage G. During this stroke, the 
bucket valve is closed. On the downward 
stroke, the suction and discharge valves C and 
F both close, and the bucket valve opens, per- 
mitting water to pass from the lower to the 
upper side of the bucket. It is not absolutely 
necessary to have a discharge valve F in this 
type of pump, but, if fitted, it serves as a 
check on the suction valve during the down- 
ward stroke of the bucket. This pump is 
single-acting. 

In Fig. 647 is shown a single-acting plunger 
pump. On the upward stroke of the plunger 
B, water enters the pump through the suction 
valve C, and is delivered, during the downward 
stroke, through the discharge valve D. E is an air vessel, the 
function of which is to get rid of shocks. The water coming from 
the pump flows partly into the air vessel, during the early part of the 
discharge stroke, and compresses the air contained therein; during 
the later part of the discharge stroke, and also possibly during part of 
the suction stroke, the pressure of the compressed air drives some 
of the water out of the air vessel into the discharge pipe. The accele- 
ration required to be given to the column of water in the discharge 
pipe in starting it into motion is lowered by the action of the air 
vessel, and hence the force required is also lowered, and shock is 




FIG. 646. Bucket pump. 



588 



MACHINES AND HYDRAULICS 



avoided entirely. The cushion of compressed air is also beneficial 
in quietly closing the discharge valve at the end of the stroke without 
depending on any backward movement of the mass of water in the 
discharge pipe; thus hammering of the valve is avoided. The air 





FIG. 647. Boiler feed pump. 



FIG. 648. Combined plunger and 
bucket pump. 



vessel should be situated always as close as possible to the discharge 
valve. The type of pump illustrated is much used for forcing the 
feed water into steam boilers. 

Fig. 648 illustrates a combined plunger and bucket pump. During the 
downward stroke, the suction valve C is closed, and the bucket valve 
D is open; the plunger E is thus operating in discharging water 



EXERCISES ON CHAPTER XXII. 589 

through F. During the upward stroke, the bucket valve D is closed 
and the suction valve C opens. Fresh water thus flows into the 
cylinder A from B, and the water already on the top of the bucket is 
discharged through F. As is the case in the pump shown in 
Fig. 645, the area of the plunger should be one-half that of the 
bucket for equality of discharge on the two strokes. 

Pumps may be placed at some height above the supply water, and 
in this case depend on the pressure of the atmosphere acting on the 
supply water and forcing it up the suction pipe into the partial 
vacuum created by the action of the pump bucket or plunger. The 
maximum possible height through which the atmospheric pressure 
will raise water thus is about 34 feet; from 25 to 30 feet is the 
greatest practical height. 

EXERCISES ON CHAPTER XXII. 

1. A rectangular tank is 4 feet long, 3 feet wide and 2 feet deep. 
Find the total pressures on the bottom, on one side and on one end when 
the tank is full of oil which weighs 50 Ib. per cubic foot. 

2. A tank 10 feet long has a horizontal bottom 4 feet wide. The ends 
of the tank are vertical, and both the sides are inclined at 45 to the 
horizontal. Water is contained to a depth of 6 feet. Find the total 
pressures on the bottom, on one side and on one end. Take 2^ = 62-5 Ib. 
per cubic foot. 

3. A dock gate is 80 feet wide and has sea water to a depth of 30 feet 
on one side and 9 feet on the other side. Find the total pressure on each 
side of the gate, and show the lines of action. Find also the resultant 
force on the gate, and show its position. Take 1^ = 64 Ib. per cubic foot. 

4. A tank is in the form of an inverted cone, 6 feet diameter at the 
top and 4 feet vertical depth. When full of oil having a specific gravity 
0-8, find the weight of the contained oil and the total pressure on the 
curved surface of the tank. 

5. A rectangular opening in a reservoir wall is 4 feet high and 3 feet 
wide, and has its top edge 20 feet below the water level. Find the total 
water pressure on the door or gate closing the opening, and find also the 
centre of pressure. 

6. A rectangular pontoon 100 feet long and 30 feet wide has a draught 
in fresh water of 8 feet (i.e. the bottom of the pontoon is 8 feet below the 
surface of the water). Find the weight of the pontoon. Supposing the 
weight to remain unaltered^ and the pontoon to be floating in sea water, 
what will be the draught? For fresh water 2/ = 62'5, and for sea water 
ze/=64 Ib. per cubic foot. 

7. The weight of a submarine is 200 tons, and it lies damaged and 
full of water at the bottom of the sea. Supposing the specific gravity of 
its material to be 7-8, find what total pull must be exerted by the lifting 
chains in order to raise the vessel from the bottom. Take 1^ = 64 Ib. per 
cubic foot. 



$90 MACHINES AND HYDRAULICS 

8. For the pontoon in Question 6, when floating in fresh water, find 
the heights of the transverse and longitudinal metacentres above the 
centre of buoyancy. 

9. The pontoon in Question 8 carries a crane, and is hoisting a load 
which produces a transverse capsizing moment of 200 ton-feet. Calculate 
the angle of heel. It may be assumed that the centre of gravity of the 
complete pontoon is 0-5 foot below the surface level of the fresh water. 

10. A retaining wall for water is triangular in section and has the 
wetted surface vertical. The height is 30 feet and the breadth of the 
base is 25 feet. Fresh water has its surface level 3 feet below the top of 
the wall. The weight of the material is 140 Ib. per cubic foot. Take one 
foot length of wall and find the resultant force acting on the base. 
Answer the same if the reservoir is empty. Do these forces fall within 
the middle third of the base ? 

11. Answer Question 10 for sections at 3 feet, 10 feet and 20 feet from 
the top of the wall, using graphical methods so far as is possible. Plot 
the lines of pressure for the reservoir full and empty. 

12. Water is supplied by a hydraulic company at a pressure of 700 Ib. 
per square inch, and is charged at the rate of 18 pence per thousand 
gallons. How much water must be used in an hour to obtain one horse- 
power, and what would be the cost ? Neglect waste. 

13. A single-acting hydraulic engine has three rams, each 3^ inches 
diameter by 6 inches stroke. The effective mean water pressure on the 
rams is 120 Ib. per square inch, and the engine runs at 90 revolutions per 
minute. Neglect all sources of waste and calculate the horse-power. If 
the efficiency is 65 per cent., what is the useful horse-power ? 

14. 2 cubic feet of air at an absolute pressure of 80 Ib. per square inch 
are expanded in a cylinder until the volume is 5 cubic feet. Assuming 
that the law PV = a constant is obeyed, calculate what work is done. 

15. Answer Question 14 if the law of expansion is PV 1-41 = a constant. 

16. A hydraulic accumulator has a ram 7 inches in diameter and the 
lift is 12 feet. If the water pressure is to be 700 Ib. per square inch, find 
the weight required. How much water is stored when the accumulator is 
up ? Find also the energy stored. 

17. In the hydraulic lift cylinder shown in Fig. 641, find the velocity 
ratio if there are three rope pulleys on the ram end and two pulleys on 
the cylinder top. Suppose the ram to be 4 inches diameter and that the 
water pressure is 700 Ib. per square inch, and calculate the pull on the 
cage rope, neglecting frictional waste. What is the pull if the total 
efficiency is 65 per cent. ? What stroke of ram is required for a total cage 
lift of 60 feet ? 

18. A hydraulic pump, similar to that shown in Fig. 645, has a piston 
4-25 inches and a plunger rod of 3 inches in diameter ; the stroke is 
1 8 inches. If the pump makes 60 double strokes per minute, how much 
water will be delivered, neglecting waste ? If the water pressure is 750 Ib. 
per square inch, find the force which must be applied to the rod (a] when 
the piston is moving towards the valves, (&) when the piston is moving in 
the contrary direction, assuming the pressure on the suction side to be 
15 Ib. per square inch. Neglect friction. 



EXERCISES ON CHAPTER XXII. 591 

19. A bucket pump (Fig. 646) has to raise 400 gallons of water per 
minute to a height of 30 feet. If the pump makes 30 double strokes (one 
up and one down) per minute, and if the length of the stroke is 1-5 times 
ftie diameter of the bucket, find the stroke and the bucket diameter, 
neglecting waste. Calculate the useful work done per minute, and, if the 
efficiency is 60 per cent., find the horse-power required. 

20. A boiler feed pump has a plunger 3 inches in diameter. The 
delivery pipe leading to the boiler is 40 feet in length and 3 inches in 
diameter. The pressure in the boiler is 100 Ib. per square inch. There 
is no air vessel. Supposing that the acceleration of the plunger at the 
beginning of the stroke (the water in the delivery pipe being then at rest) 
to be 90 feet per second per second, what total force must be exerted by 
the plunger in order to start the water into motion ? If a perfect-acting 
air vessel were fitted, what total force would be required ? 

21. In finding the total force in the axial direction which a fluid 
exercises upon a piston or ram, we calculate from the cross section of the 
cylinder or ram ; why is the actual shape of the face of the piston or end 
of the ram of no importance ? (B.E.) 

22. A vertical flap closes the end of a pipe 2 feet in diameter ; the 
pressure at the centre of the pipe is equal to a head of 10 feet of water. 
Find the total pressure on the valve in pounds. (You may neglect the 
atmospheric pressure.) (B.E.) 

23. The ram of a vertical accumulator is 4 inches in diameter ; the 
cylinder is 6 inches in internal diameter and 50 feet high. The ram 
carries a total load of 5 tons. Find the water pressure, in Ib. per square 
inch, at the top and bottom of the cylinder. (B.E.) 

24. Determine the depth from the surface of the centre of pressure on 
a rectangular sluice valve, 6 feet long and 3 feet wide. The centre of the 
valve is at a depth of 8 feet below the surface of the water, and the valve 
lies in a plane inclined at an angle of 30 degrees to the horizontal, with 
one of the long edges of the valve parallel to the surface of the water. 

(B.E.) 

25. A horizontal channel of V section, whose sides are inclined at 45, 
is closed at the end by a vertical partition. The water-surface has a 
width of 4 feet, and consequently a maximum depth of 2 feet. Calculate 
the total hydrostatic pressure upon the partition and the height of the 
centre of pressure. (I.C.E.) 

26. Define metacentric height. A vessel has a length of 150 feet 
between perpendiculars and a beam of 28 feet. The mean load draft in 
sea water is 1 1 feet, and the coefficient of fineness, or ratio between the 
product of length, breadth and draft and the displacement volume is 0-47. 
The second moment of the load water-plane about its fore and aft axis is 
63 per cent, of the moment of the circumscribing rectangle about the 
same axis. The centre of buoyancy is situated 3-95 feet below the water- 
line. If the transverse metacentric height is to be limited to 3-62 feet, 
determine the distance from the centre of gravity to the water-line. 

' (L.U.) 

27. A masonry dam with vertical water face is 20 feet high and 13 feet 
wide at the bottom, sloping gradually till it is 6 feet wide at the top. The 
water reaches 2 feet from the top. Draw the line of thrust throughout the 
dam. Specific gravity of masonry, 2-25. (L.U.) 



CHAPTER XXIII. 

HYDRAULICS. FLOW OF FLUIDS. 

Fluid friction. We have seen already that there can be no friction 
in any fluid at rest ; considerable frictional resistances exist, however, 
when the fluid is in motion. For liquids, the laws of fluid friction, as 
deduced from experimental evidence, have been mentioned in 
Chap. XV., and are stated again for reference as follows : 

(a) The resistance is proportional to the extent of the surface 
wetted by the liquid. 

(b) The resistance is independent of the material of which the 
boundary is made, but depends on the roughness of its surface. 

(c) The resistance is independent of the pressure to which the 
liquid is subjected. 

(d) Rise of temperature of the liquid diminishes the resistance. 

(e) At slow speeds the resistance is very small. 

(f) Below a certain critical speed, the resistance is proportional to 
the speed ; at speeds above this, the resistance is proportional to some 
power, approximately the square, of the speed. 

The critical speed depends on the liquid used and its temperature. 
Below this speed the motion of the liquid is steady, the particles 
moving in stream lines; above it, the liquid breaks up into eddies. 
As the flow of water is the most important case in practice, we will 
confine attention to this liquid. 

Kinds of energy of flowing water. Neglecting effects due to 
changes of temperature and of volume, we may state that the total 
energy of a particle of water is made up of (a) potential energy, 
(b) pressure energy, (c) kinetic energy. The potential energy will 
be proportional to the elevation of the particle above some datum 
level ; the kinetic energy will be proportional to the square of the 
velocity of the particle. The pressure energy requires some fuller 
explanation. 




ENERGY OF FLOWING WATER 593 

In Fig. 649 is shown a cylinder fitted with a piston and supplied 
with water from an overhead tank, in which the level is maintained 
constant. If the piston is allowed to move outwards slowly, work 
will be done by the water pressure on the 
piston overcoming the external resistance 
acting on the other side of the piston or on 
the piston rod. 

Let P = fluid stress on piston, in Ib. per 

square foot. 

A = area of piston, in square feet. 
L = the distance piston is moved, in 

feet. FIG. 649. Pressure energy of 

Then Work done = PAL foot-lb. 

In performing this work, a volume AL cubic feet of water has been 
admitted to the cylinder, and the work has been done at the expense 
of the energy of this water. The work done per cubic foot of water 
may be found by dividing the above result by AL, giving 
Work done per cubic foot of water = P foot-lb. 

Let w = weight of one cubic foot of water in Ib. 

Then = volume of one Ib. of water. 

w 

P 
Hence, Work done per Ib. of water = foot-lb. 

w 

It has been assumed that there has been no waste of energy; 

p 
therefore represents the whole energy available in one pound of 

water due to its pressure. We may say therefore that water at rest 

and under pressure possesses energy due to its pressure to the 

p 
amount of foot-lb. per pound of water. 

w 

Transformations of energy in flowing water. In Fig. 650 is 
shown two tanks at different levels, and connected by a pipe so 
that water may flow from the upper into the lower tank. OX is 
an arbitrary datum level. Considering a pound of water at A and 
assuming it to be at rest, there will be no kinetic energy; it will, 
however, possess H A foot-lb. of potential energy owing to its elevation 
H A feet above OX. The water, being exposed to atmospheric 

pressure P a Ib. per square foot, will also possess pressure energy to 

p 
the amount of foot-lb. per pound. Hence, 

w 

Total energy at A = ^H A + - j ft.-lb. per Ib. of water. 
D.M. 2 P 



594 



MACHINES AND HYDRAULICS 



It is reasonable to suppose that the bulk of the water in the upper 
tank is at rest, only a portion near the pipe entrance, mapped out 
by the dotted curve abc (Fig. 650), will possess any considerable 

velocity. Hence, at B, a pound of water will have potential energy 

p 
H B foot-lb., together with pressure energy -- foot-lb. owing to its 

absolute pressure P B Ib. per square foot. Therefore, 

(P \ 
H B -r- J ft.-lb. per Ib. of water. 




FIG. 650. Transformation of energy of flowing water. 

Consider now a pound of water at C, having acquired a velocity of 
# c feet per second, under pressure P c Ib. per square foot and at an 
elevation H c above OX. The total energy will be given by 

Total energy at C = (H C + ^ + ~] ft.-lb. per Ib. of water. 

In the same way, a pound of water at D will have a total energy 
given by 

Total energy at D = (H D + + ft.-lb. per Ib. of water. 



At the surface level E in the lower, tank, the water may be assumed 
to be at rest again, and also exposed to atmospheric pressure. The 
total energy here will be 



Total energy at E = H E + ft.-lb. per Ib. of water. 

We may now trace the transformations of energy which have taken 
place during the passage of the water from A to E. It may be 
assumed that a pound of water moves from A to B very slowly, and 



BERNOULLI'S LAW 595 



arrives without any appreciable diminution of energy, since the 
frictional resistances will be very small. Hence, 

Total energy at A = total energy at B, 

or H A + = H B + ^. 

w w 

The water has given up potential energy represented by (H A - H B ) 
ft.-lb. per pound, and has acquired an equal amount of pressure energy 

(P P \ 
^___^\ ft.-lb. per pound. 

During the passage from B to C, considerable velocity has been 
acquired, and hence the frictional resistance will produce correspond- 
ing waste of energy. In passing along the pipe from C to D there 
will be further frictional waste of energy. If these sources of waste 
be disregarded we may apply the principle of the conservation of 
energy in asserting that the total energies at B, C and D are equal. 
Hence, 

Total energy at C = total energy at D, 



c + + 

W 2g W 2g 

This equation is the algebraic expression of Bernoulli's law, which 
asserts that if there be no waste of energy, the total energy of water flowing 
from one place to another remains constant. Calculations may be made 
on this assumption, and then corrections can be applied in order to 
account for known sources of waste. 

Referring again to Fig. 650, the water leaving the pipe and entering 
the lower tank will produce surging of the water in this tank, accom- 
panied by a considerable waste of energy. The total waste of energy 
in the complete passage from A to E may be estimated by taking the 
difference in total energies at these places. Thus, 

Total waste of energy = (H A + ~) - (H E + ^ 

= (H A - H E ) ft.-lb. per lb. of water. 

It will be noted that (H A - H E ) is simply the difference in surface 
levels of the water in the two tanks, H feet say (Fig. 650). Hence, in 
the case before us, the total waste of energy per pound of water is 
represented by H foot-lb. 

Venturi water meter. In Fig. 651 is shown a straight horizontal 
pipe, which converges from A to B and then enlarges again between 
B and C. As the pipe is horizontal, there will be no change in the 
potential energy of the water flowing through it ; there will, however, 



596 



MACHINES AND HYDRAULICS 



be interchanges of pressure and kinetic energies, and if pressure 
gauges be fitted as shown so that the pressure heads may be measured, 
it is possible to calculate the velocity of flow, and hence the quantity 
of water flowing, from a knowledge of the pipe diameters. 




A C 

FIG. 651. Principle of the Venturi meter. 

The same quantity of water, Q cubic feet, will pass all sections of 
the pipe per second. If the sectional areas be A l , A 2 and A 3 square 
feet at A, B and C respectively, and if the velocities v l9 v 2 and v 3 be 
measured in feet per second, we have 

Q = v^ Aj = z/ 2 A 2 = # 3 A 3 (i) 

Applying Bernoulli's law and neglecting any frictional waste, we 
have ,,2 -.2 ,7, 2 



2g 2g 2g 

H 1? H 2 and H 3 being the pressure heads in feet. 

If the pipe diameters at A and C are equal, as is usually the case, 
v l and v z will be equal, and H l and H 3 will also be equal, neglecting 
friction. Using the first two terms of (2), 



From (i), 



Hence, 



or 



H 1~ H 2= 2 2g l ' 




AI 




V^jrV^ 

A 2 




(^-^ 

TT TT V^2 / 


/A T 2 \ 
W" 1 ) 


*g 


2^ 


A^-Ao 2 9 




= 1 A 9 2 zv*, 
2^A 2 2 




v 2_ 2 o-/u _ n\ 


A 2 2 


u \ - -& V L L 1 A X 2/ / A 


A, 



Now 




\/2 < -(H 1 - H 2 ) / ' 2 2 cubic feet per sec. ... 

V Aj - A 2 " 



(3) 



STREAM LINE MOTION 597 

Practically, the quantity flowing differs somewhat from the result 
calculated from equation (3). A coefficient, the value of which is 
approximately 0-98, may be used for multiplying the right-hand 
side of (3). Small Venturi meters used in laboratories for testing 
purposes usually require calibration, especially for low heads and 
velocities. 

Steady motion. Steady motion of a fluid may be denned as that 
state of motion when all particles passing through any fixed point 
arrive at the point with the same velocity, both as regards magnitude 
and direction. Thus, in steady motion, the particles will be 
travelling in lines or filaments either straight or curved, these 
filaments being called stream lines. For example, if a fine jet of 
coloured water be injected into a mass of water moving with steady 
motion, the coloured water will follow the stream line which passes 
through the point of injection, and will 
move unbroken through the mass of 
water, giving a coloured band which 
will be straight or curved depending on 
the circumstances of the flow, but will 
appear to remain fixed in position. 

A fluid can only move in straight 
stream lines provided there is no resul- 
tant force acting on the boundary of 
the filament in a direction perpendicular 

tO that of the motion of the filament. FIG. 652. Transverse pressures on 
. i r -11 j i curved stream lines. 

Any such force will produce a change in 

the direction of the motion, and the path of the filament will be 
curved, the resultant force being found on the convex side of the 
filament (Fig. 652). 

In a mass of fluid moving in curved stream lines, each stream 
line communicates pressures to the adjacent stream lines and is 
itself reacted on. As the concave side of any stream line is in 
contact with the convex side of the adjacent stream line, the pressure 
on the concave side ab of the first will be equal to that on the convex 
side ab of the second; let this pressure be p (Fig. 652 (a)). The 
pressure on the concave side cd will be less than / by an amount S/, 
and that on ef will be greater than / by another small amount /. 
Applying the same reasoning to all stream lines in a body of fluid 
moving steadily in a curved path (Fig. 652 (&)), we see that the 
pressure p^ on the convex boundary ab will diminish gradually across 
the stream, attaining a lower value / 2 at the concave boundary cd. 




598 



MACHINES AND HYDRAULICS 



Discharge from an orifice. One of the simplest cases of the flow of 
water is found in a jet discharged through a small sharp-edged circular 

hole in a thin plate. In Fig. 653 
such a hole is formed in the 
vertical side of a tank, WL being 
the free surface 
steady head H 
orifice de. OX 




level, giving a 
feet over the 
may be taken 
as a datum level. At A, a pound 
of water being at rest will have 
a total energy given by 



FIG. 653. Discharge through 
orifice. 



Pa being, as before, the atmo- 
spheric pressure in Ib. per square 
a sharp-edged foot, and w the weight of the 

water in Ib. per cubic foot. 
Passing to a point B on the same level as the centre of the orifice, 
some of the potential energy possessed at A will have been converted 
into pressure energy, giving a total energy at B of 



T? TT , r B 

^B = -tl B + - 



(2) 



As the motion of a particle passing from A to B will be very 
slow, it is reasonable to suppose that frictional losses may be 
disregarded. Hence, 



or 



Again, 



E A =E B , 

Pa 

w 



w 



w 



or 



H = H A -H B ......................... (3) 

This equation simply expresses the fact that the superatmospheric 
pressure head at B is H. 

Assuming that the motion inside the region of important velocity, 
abC) is stream line, we may state that a particle situated at />, level 
with the centre of the orifice, will move along a straight horizontal 
stream line and so pass out ; particles crossing the boundary abc at 
other points will approach the orifice in curved stream lines. Clearly 
the sharp edges of the orifice de cannot produce a sudden change in 



DISCHARGE FROM ORIFICES 599 

the direction of any stream line ; hence the curvature will be main- 
tained for some distance after the plane of the orifice has been 
passed. This leads to contraction of the issuing jet, and such 
contraction will not be complete until a section CD has been 
reached ; this section is called the contracted vein. 

In the body of water between de and CD, the stream lines are 
convex towards the axis of the jet ; hence there must be resultant 
fluid pressures acting transversely to each stream line and directed 
outwards towards the boundary of the jet. As the boundary is 
exposed to atmospheric pressure / a , it follows that superatmospheric 
pressure, of values gradually increasing towards the axis of the jet, 
will be found in the interior of the jet, the maximum pressure 
occurring at the axis. From CD onwards the stream lines will be 
parallel ; hence the water in the jet beyond CD will be under 

uniform pressure equal to P a , and will possess pressure energy 

p 
given by ft.-lb. per pound. 

w 

The velocity of any particle has been increased gradually in passing 
from the boundary abc to the section CD, and hence the particle has 
been acquiring kinetic energy gradually, this being obtained at the 
expense of its other kinds of energy. For example, a pound of water 
at b has had its superatmospheric pressure energy, H foot-lb, changed 
into an equal quantity of kinetic energy (neglecting frictional losses) 
while passing from b to CD. The conversion is completed on arriving 
at CD, and hence we find the maximum velocity at this section. 
Supposing V feet per second to be the velocity of the jet at CD, 
then the total energy per pound of water at CD will be 

E CD = H CD + + ...................... (4) 



Applying Bernoulli's law and neglecting frictional effects, the total 
energies at A and CD will be equal. Hence, 



v 2 

.'. H A -H CD = , 

H=5 ..................................... (5) 

This equation may be written 




or V = v/^H ............................... (6) 



600 MACHINES AND HYDRAULICS 

The actual velocity at CD will be somewhat less than V, due to 
waste of energy in overcoming frictional resistances in the flow 
between abc and CD. 

Experimentally it is found that the actual velocity V a is about 
o-97V, this number being called the coefficient of velocity, written c v . 
Hence, 

V = W^H ............................... (7) 

The quantity of water discharged can be obtained, provided we 
know the area of the section CD. For a small round orifice this will 
be about 0-64 of the area of the orifice ; this number is called the 
coefficient of contraction, written c c . 

Let Q = the quantity discharged per second, in cubic feet. 

A = the area of the orifice, in square feet. 

H = the head over the centre of the orifice, in feet. 
Then 



(8) 

In this result, ca c^v is called the coefficient of discharge. For a 
round orifice its value will be 

^ = 0-64 x 0-97 

= 0-62. 

The discharge from a small round sharp- edged orifice therefore will 
be given by Q ^ . 62 A^/^H cubic feet per second. 

If the orifice is situated in the tank bottom so that the jet dis- 
charges vertically downwards (Fig. 654), contraction does not cease 
at CD. This is owing to the potential 
d & energy of the water in the falling jet con- 

tinually diminishing; hence the kinetic energy, 
and therefore the velocity, must be increasing 
continually. In a steady jet (prior to its 
breaking up into drops) the same quantity of 
water passes each section per second, and there- 

FlG ' 6 inTtl h nk rp bot d t5m orifice fore the area of the J et must be diminishing 
as the jet recedes from the orifice. The 
approximate velocity at any section may be estimated from 



v <r v \2^H, 

where H is the head measured from the free surface level in the tank 
to the section considered. 




DISCHARGE FROM ORIFICES 601 

Contraction of the jet after passing the orifice may be got rid of by 
means of a trumpet orifice (Fig. 655). In this case the discharge is 
estimated by applying a coefficient of velocity only. 
Thus, Q = ^A\/^H. 

Flow of a gas through an orifice. Assuming that 
the pressure in the reservoir containing the gas is 
only slightly greater than the pressure in the space 
into which the jet of gas is discharged, and that 
there is no change in temperature, there will be FIG. 655. Trumpet 
very little change in the weight of the gas per cubic 
foot, and the flow through the orifice may be estimated in the same 
manner as for a liquid. 

Let ij l = the velocity in the reservoir. 

#2 = the maximum velocity of the jet, in feet per second. 
p l = the pressure in the reservoir, in Ib. per square foot. 
/ 2 = the pressure in the space which the jet enters, in Ib. 

per square foot. 
w = the weight of a cubic foot of the gas, in Ib., under the 

conditions existing in the reservoir. 

A = the area of the orifice, in square feet. 

Then A,n_A 




Hence, applying a coefficient of discharge C<z, we have 




cubic feet per sec. 

w 

Experiments show that for circular sharp-edged orifices discharging 

air, the value of ca is in the neighbourhood of 0-6. 

Reaction of a jet. In Fig. 656 is shown a tank mounted on 
wheels and discharging water through a trumpet 
orifice in one side. The issuing water has 
acquired momentum in passing out of the 
orifice, and a resultant force acting towards 
the right on the water in the mouthpiece is 
required in order to produce this change of 

FIG. 6 5 6. Reaction of a jet. momentum. There must also be an equal 




602 MACHINES AND HYDRAULICS 

opposite reaction, and hence there will be a tendency to move the 
tank towards the left when the jet is flowing. The magnitude of 
this force may be found by estimating the change of momentum 
per second. 

Let H = the head over the centre of the orifice, in feet. 

v velocity of jet, in feet per second. 
A = area of jet, in square feet. 
w = mass in pounds of a cubic foot of water. 
Then Quantity flowing per second = Avw pounds ; 

.'. momentum acquired per second = Avw . v. 



Force required = - Ib. 
Neglecting the coefficient of velocity, we have 



,, . , Aw . 2gYL 

Hence, Force required = -- 



Ib. 

If the orifice be closed by a plate, the pressure on the plate would 
be AwH Ib. ; hence the reaction of the jet is double the pressure on 
a plate closing the orifice. 

In the Borda mouthpiece, a short tube projects into the interior of 
the tank, and has its inner edge sharpened (Fig. 657). This orifice 
I produces an effect differing considerably from a 

trumpet orifice or from a simple hole in the tank 
C' ' side. In the latter cases, owing to the curvature of 

-- + v the stream lines in the vicinity of the orifice, the 
walls of the tank there are somewhat relieved of 
pressure, the pressure diminishing from a maximum 
FIG. 6 57 . Borda at the axis of the orifice to a minimum at the 
mouthpiece. boundary. In the Borda mouthpiece, the curved 
portions of the stream lines are removed sufficiently from the tank 
side as not to modify the pressures on the sides. Hence, the force 
producing change of momentum in the issuing water will be simply 
that which would exist on a plate closing the orifice. 
L 61 A = area of orifice, in square feet. 

H = head of water, in feet. 
a = area of jet, in square feet. 
v = velocity of jet, in feet per sec. 



PRINCIPLE OF SIMILAR FLOW 603 

Then Force producing change of momentum = wH A Ib. 

Quantity flowing per second = wav pounds ; 
Change of momentum per second = 

wav 2 ., 
. . reaction of jet = Ib. 

o 

__ wav 2 
Hence, wH A = 

Neglecting the coefficient of velocity, we have 



or A = 20, 

= JA. 

This mouthpiece has therefore a coefficient of contraction of 0-5. 
Thomson's principle of similar flow. Prof. James Thomson's 
principle of similarity is of importance in dealing with the flow 
through orifices and over weirs