.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 c