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Full text of "Mechanics applied to engineering"

BOUGHT WITH THE INCOME 
FROM THE 

SAGE ENDOWMENT FUND 

THE GIFT OF 

fletirg W, Sage 

1S91 



ll;.f^MaqH^ iLimif... 



3777 



Cornell University Library 
TA 350.G65 1914 



Mechanics applied to engineering. 




3 1924 004 025 338 




Cornell University 
Library 



The original of tliis book is in 
tine Cornell University Library. 

There are no known copyright restrictions in 
the United States on the use of the text. 



http://www.archive.org/details/cu31924004025338 



MECHANICS APPLIED TO 

ENGINEERING 



MECHANICS APPLIED TO 
ENGINEERING 



JOHN GOODMAN 

Wh. Sch., M.I.C.E., M.I.M.E. 

PROFESSOR OF ENGINEERING IN THE UNIVERSITY OF LEEDS 




With 741 Illustrations and Numerous Examples 



EIGHTH EDITION 



LONGMANS, GREEN AND CO. 

39 PATERNOSTER ROW, LONDON 
FOURTH AVENUE & 30th STREET, NEW YORK 

BOMBAY, CALCUTTA, AND MADRAS 

I9I4 

All rights reserved 



PREFACE 



This book has been written especially for Engineers and 
Students who already possess a fair knowledge of Elementary 
Mathematics and Theoretical Mechanics ; it is intended to 
assist them to apply their knowledge to practical engineering 
problems. 

Considerable pains have been taken to make each point 
clear without being unduly diffuse. However, while always 
aiming at conciseness, the short-cut methods in common use 
have often — ^and intentionally — been avoided, because they 
appeal less forcibly to the student, and do not bring home to 
him the principles involved so well as do the methods here 
adopted. 

Some of the critics of the first edition expressed the opinion 
that Chapters I., II., III. might have been omitted or else con- 
siderably curtailed ; others, however, commended the innovation 
of introducing Mensuration and Moment work into a book on 
Applied Mechanics, and this opinion has been endorsed by 
readers both in this country and in the United States. In 
addition to the value of the tables in these chapters for reference 
purposes, the worked-out results afford the student an oppor» 
cunity of reviewing the methods adopted. 

The Calculus has been introduced but sparingly, and then 
only in its most elementary form. That its application does 
not demand high mathematical skill is evident from the 
working out of the examples in the Mensuration and Moment 
chapters. For the benefit of the beginner, a very elementary 
sketch of the subject has been given in the Appendix ; it is 
hoped that he will follow up this introduction by studying such 
works as those by Barker, Perry, Smith, Wansbrough, or others. 

For the assistance of the occasional reader, all the symbols 
employed in the book have been separately indexed, with the 
exception of certain ones which only refer to the illustrations 
in their respective accompanying paragraphs. 



vi Preface. 

In this (fourth) edition, some chapters have been con- 
siderably enlarged, viz. Mechanics ; Dynamics of Machinery ; 
Friction; Stress, Strain, and Elasticity; Hydraulic Motors and 
Machines ; and Pumps. Several pages have also been added 
to many of the other chapters. 

A most gratifying feature in connection with the publication 
of this book has been the number of complimentary letters 
received from all parts of the world, expressive of the help it 
has been to the writers ; this opportunity is taken of thanking 
all correspondents both for their kind words and also for their 
trouble in pointing out errors and misprints. It is believed that 
the book is now fairly free from such imperfections, but the 
author will always be glad to have any pointed out that have 
escaped his notice, also to receive further suggestions. While 
remarking that the sale of the book has been very gratifying, 
he would particularly express his pleasure at its reception in the 
United States, where its success has been a matter of agreeable 
surprise. 

The author would again express his indebtedness to all 
who kindly rendered him assistance with the earlier editions, 
notably Professor Hele-Shaw, F.R.S., Mr. A. H. Barker, B.Sc, 
Mr. Aiidrew Forbes, Mr. E. R. Verity, and Mr. J. W. Jukes. 
In preparing this edition, the author wishes to thank his old 
friend Mr. H. Rolfe for many suggestions and much help ; also 
his assistant, Mr. R. H. Duncan, for the great care and pains 
he has taken in reading the proofs ; and, lastly, the numerous 
correspondents (most of them personally unknown to him) who 
have sent in useful suggestions, but especially would he thank 
Professor Oliver B. Zimmerman, M.E., of the University of 
Wisconsin, for the " gearing " conception employed in the 
treatment of certain velocity problems in the chapter on 
" Mechanisms." 

JOHN GOODMAN. 



Thk University of Leeds, 
August, I904' 



PREFACE TO EIGHTH EDITION 



New Chapters on " Vibration " and " Gyroscopic Action " 
have been added to this Edition. Over a hundred new 
figures and many new paragraphs have been inserted. The 
sections dealing with the following subjects have been added 
or much enlarged — Cams, Toothed Gearing, Flywheels, 
Governors, Ball Bearings, Roller Bearings, Lubrication. 
Strength of Flat Plates, Guest's Law, Effect of Longitudinal 
Forces on Pipes under pressure. Reinforced Concrete Beams, 
Deflection of Beams due to Shear, Deflection of Tapered 
Beams, Whirling of Shafts, Hooks, Struts, Repeated Loading. 
Flow of Water down Steep Slopes, Flooding of Culverts, Time 
of Emptying Irregular Shaped Vessels, Continuous and Sinuous 
flow in Pipes, Water Hammer in Pipes, Cavitation in Centri- 
fugal Pumps. 

The mode of treatment continues on the same lines as 
before ; simple, straightforward, easily remembered methods 
have been used as far as possible. A more elegant treatment 
might have been adopted in many instances, but unfortunately 
such a treatment often requires more mathematical knowledge 
than many readers possess, hence it is a "closed book" to the 
majority of engineers and draughtsmen, and even to many 
who have had a good mathematical training in their student 
days. 

There are comparatively few Engineering problems in 
which the data are known to within, say, s per cent., hence it 
is a sheer waste of time for the Engineer in practice to use 
long, complex methods when simple, close approximations 
can be used iii a fraction of the time. For higher branches 
of research work exact, rigid niethods of treatment may be, 
and usually are, essential, but the number of Engineers who 
require to make use of such methods is very small. 

Much of the work involved in writing and revising this 



viii Preface to the Eighth Edition. 

Edition has been performed under very great difficulties, in 
odd moments snatched from a very strenuous life, and but for 
the kind and highly valued assistance of Mr. R. H. Duncan 
in correcting proofs and indexing, this Edition could not 
have been completed in time for this Autumn's publication. 



JOHN GOODMAN. 



The University of Leeds, 
August, 1914. 



CONTENTS 



CHAP. JAGE 

I. Introductory i 

11. Mensuration .20 

III. Moments ... 50 

IV. Resolution of Forces . . . , < 106 

V. Mechanisms . .... ...... iig 

VI. Dynamics of the Steam-engine .... -179 

VII. Vibration . . . 259 

VIII. Gyroscopic Action . . 277 

IK. Friction . . . ... 284 

X. Stress, Strain, and Elasticity ... ... 360 

XI. Beams ... 429 

XII. Bending Moments and Shear Forces . . 474 

XIII. Deflection of Beams 506 

XIV. Combined Bending and Direct Stresses . . . 538 
XV. Struts 550 

XVI. Torsion. General Theory 571 

XVII. Structures S93 

XVIII. Hydraulics 637 

XIX. Hydraulic Motors and Machines . . . .691 

XX. Pumps 738 

Appendix 781 

Examples 794 

Index ... 846 



ERRATA. 



Pages 34 and 35, bottom line, " + " should be ^' - ." 
Page 79, top of page, " IX." should be " XI." 

„ 102, the quantity in brackets should be multiplied by " — ." 

„ 203, middle of page, "IX." should be "XI." 

St 0-2' 
bottom, " 45° " should be " 90°." 
top, " sin Ra " should be " sin 20." 
top, " A " should be " Ao." 
top, " h* " should be " h." 
top, " /i " should be " //„.' 
top, " L " is the length of the suction pipe in feel. 



247, line 12 from top, should be ' 
16 



395. . 


, 10 


663, 


, 6 


»» J 


. 7 


676, , 


M 


747. . 


5 



MECHANICS APPLIED TO 
ENGINEERING 



CHAPTER I. 

INTRODUCTOR Y. 

The province of science is to ascertain truth from sources far 
and wide, to classify the observations made, and finally to 
embody the whole in some brief statement or formula. If 
some branches of truth have been left untouched or unclassi- 
fied, the formula will only represent a part of the truth ; such 
is the cause of discrepancies between theory and practice. 

A scientific treatment of a subject is only possible when 
our statements with regard to the facts and observations are 
made in definite terms ; hence, in an attempt to treat such a 
subject as Applied Mechanics from a scientific standpoint, we 
must at the outset have some means of making definite state- 
ments as to quantity. This we shall do by simply stating how 
many arbitrarily chosen units are required to make up the 
quantity in question. 

Units. 

Mass (M). — Unit, one pound. 

I pound (lb.) = 0'454 kilogramme. 

I kilogramme = 2"2046 lbs. 

I hundredweight (cwt.) = SO'8 kilos. 

I ton = 1016 ,, (tonneau or Millier). 

I tonneau or Millier = 0'984 ton. 

Space {s). — Unit, one foot. 

t foot = 0-305 metre. i mile = l6o9'3 metres. 

I metre = 3'28 feet. i kilometre = I093'63 yards. 

[ inch = 25'4 millimetres. = 0'62I mile. 

I millimetre = o'0394 inch. i sq. foot = 0-0929 sq. metre. 

I yard = 0'9I4 metre. I sq. metre = 10764 sq. feet. 

I metre = l'094 yards. I sq. inch = 6'45I sq. cms. 

B 



Mechanics applied to Engineers 



I sq. 
I sq. 



mm, 
cm. 



r sq. metre 
I atmosphere 

I lb. per sq. inch 



= O'00l55 sq. inch. 

= O'ISS sq. inch. 

= o'ooio76 sq. feet. 

= 10764 sq. feet. 

= I'igS sq. yards. 

= 760 mm. of mercury. 

= 29-92 inches of mercury. 

= 33'9o feet of water. 

= I4'7 lbs. per. sq. inch. 

= I '033 kg. per sq. cm. 

= 0-0703 kg. per sq. cm. 

= 2-307 feet of water. 

= 2-036 inches of mercury. 

= 68970 dynes per sq. cm. 
I lb. per sq. foot = 479 dynes per sq. cm. 
I kilo, per sq. cm. = 14-223 lbs. per sq. inch. 
I cubic inch = 16-387 c. cms. 

I cubic foot = 0-0283 cubic metre. 

I cubic yard = 0-7646 c. metre. 

I c. cm. = 0-06103 c. inch. 

I c. metre = 35-31 c. feet. 

(See also pp. 4, 9, 10, 11, 19.) 

Dimensions. — The relation which exists between any 

given complex unit 

and the fundamental 

units is termed the 

dimensions of the 

unit. As an example, 

see p. 20, Chapter 

II. 

Speed. — ^When a 

body changes its 

position relatively to 

surrounding objects, 

it is said to be in 

motion. The rate at 

which a body changes 

its position when 

moving in a straight 

line is termed the 

speed of the body. 

Uniform Speed. — A body is said to have uniform speed 

when it traverses equal spaces in equal intervals of time. The 

body is said to have unit speed when it traverses unit space in 

unit time. 

„,,.,, ,, space traversed (feet) s 

Speed (m feet per second) = — — : ; f-r — ~ = - 

time (seconds) t 




X 3 * 

Tim& "in seconds 
Umforiwsp. 

Fig. 



Introductory. 3 

Varying Speed. — When a body does not traverse equal 
spaces in equal intervals of time, it is said to have a varying 
speed. The speed at any instant is the space traversed in an 
exceedingly short interval of time divided by that interval; 
the shorter the interval taken, the more nearly will the true 
speed be arrived at. 

In Fig. I we have a diagram representing the distance 
travelled by a body moving with uniform speed, and in 
Fig. 2, varying speed. The speed at any instant, a, can be 
found by drawing a tangent to the curve as shown. From 
the slope of this tangent we see that, if the speed had been 




1 i. 3 4- 5 

Tune in seconds 

Varying sjieett 

Fig. a. 

uniform, a space of 4*9 — 1 "4 = 3*5 ft. would have been 

traversed in 2 sees., hence the speed at a is — = 175 ft. per 

2 

second. Similarly, at h we find that 9 ft. would have been 

traversed in 5-2 — 2*3 = 2-9 sees., or the speed at 3 is -^ = 

3"i ft. per second. The same result will be obtained by taking 
any point on the tangent. For a fuller discussion of variable 
quantities, the reader is referred to either Perry's or Barker's 
Calculus. 

Velocity (z/). — The velocity of a body is the magnitude of 
its speed in any given direction ; thus the velocity of a body 
may be changed by altering the speed with which it is moving, 
or by altering the direction in which it is moving. It does not 



4 Mechanics applied to Engineering. 

follow that if the speed of a body be uniform the velocity will 
be also. The idea of velocity embodies direction of motion, 
that of speed does not. 

The speed of a point on a uniformly revolving wheel is 
constant, but the velocity is changing at every instant. Velocity 
and speed, however, have the same dimensions. The unit of 
velocity is usually taken as i foot per second. 

Velocity in feet 1 _ space (feet) traversed in a given direction 
per second ) "~ time (seconds) 

s . 

V = -J OT s — vt 

I ft. per second = o"3o5 metre per second 

„ „ = o"682 mile per hour 

„ „ = IT kilometre per hou: 

I metre per second = 3'28 ft. per second 

J ( = o'o^28 ft. per second 
I cm. per second < i u 

^ ( = 0*0224 miles per hour 

., u f = I '467 ft. per second 

I mile per hour < ^ ' f , 

'^ ( = 0-447 metre per second 

I kilometre " \ = °'^'l ^^- , 

( = 0-278 metre „ 

Angular Velocity (u), or Velodty of Spin. — Suppose a 
body to be spinning about an axis. The rate at which an 
angle is described by any line perpendicular to the axis is 
termed the angular velocity of the line or body, or the velocity 
of spin J the direction of spin must also be specified. When 
a body spins round in the direction of the hands of a watch, 
it is termed a + or positive spin ; and in the reverse direction, 
a — or negative spin. 

As in the case of linear velocity, angular velocity may be 
uniform or varying. 

The unit of angular measure is a " radian ; " that is, an angle 

subtending an arc equal in length to the radius, The length of 

6° 
a circular arc subtending an angle 6° is 2irr X -^-5, where ir 

360 

is the ratio of the circumference to the diameter {2r) of a circle 

and 6 is the angle subtended (see p. 22). 

Then, when the arc is equal to the radius, we have — • 

2irrO n ^60 ,□ 

—T- = >' e=i_ = 57-296° 

360 2ir >" ' 



Introductory. 5 

Thus, if a body be spinning in such a manner that a radius 
describes 100 degrees per second, its angular velocity is — 

0) = = i*7S radians per second 

57-3 

It is frequently convenient to convert angular into linear 
velocities, and the converse. When one radian is described 
per second, the extremity of the radius vector describes every 
second a space equal to the radius, hence the space described 

in one second is wr = v, ox <a = —. 

r 

Angular velocity in radians per sec. = "near velocity (ft. per sec.) 

radius (ft.) 

The radius is a space quantity, hence — 

_ J _ I 
"^ ~ Js~ 1 

Thus an angular velocity is not affected by the unit of space 
adopted, and only depends on the time unit, but the time unit 
is one second in all systems of measurement, hence all angular 
measurements are the same for all systems of units — an important 
point in favour of using angular measure. 

Acceleration (/,) is the rate at which the velocity of a 
body increases in unit time — that is, if we take feet and 
seconds units, the acceleration is the number of feet per second 
that the velocity increases in one second ; thus, unit acceleration 
is an increase of velocity of one foot per second per second. It 
should be noted that acceleration is the rate of change of 
velocity, and not merely change of speed. The speed of a body 
in certain cases does not change, yet there is an acceleration 
due to the change of direction (see p. 18). 

As in the case of speed and velocity, acceleration may be 
either uniform or varying. 

Uniform ac-^ 

celeration I _ increase of velocity in ft. per sec, in a given time 
in feet perj ~ time in seconds 

sec. per sec.J 

f _ ^2 - ^1 _ V 

^'~ t ~1 
hence v =fj, 0XVi-v^=fJ . . , . (j.) 



Mechanics applied to Engineering. 




where »a is the velocity at the end of the interval of time, 
and »! at the beginning, and v is the increase of velocity. In 

Fig. 3, the vertical distance of 
any point on any line ab from 
the base line shows the velo- 
city of a body at the corre- 
sponding instant : it is straight 
because the acceleration is as- 
sumed constant, and therefore 
the velocity increases directly 
as the time. If the body start 
from rest, when v-i is zero, the 
mean velocity over any inter- 
val of time will be — , and the 

2 

spate traversed in the interval will be the mean velocity 
X time, or — 

s = —t = •'-5— (see equation i.) 
and/. = - 

Acceleration in feet per sec. per sec. = constant X space (in>/) 

(time)^ (in seconds) 
When the body has an initial velocity v^, the mean velocity 
during the time t is represented by the mean height of the figure 
oabc. 



t a- 3 

Time irv s0conds 

Fig. j. 



Mean velocity = ■ ' = —^ — 1 = z-^ 4--ii 

2 2 2 

(see equation i.) 
The space traversed in the time t — 



. = (.+4^. 



(ii.) 



aii.) 



which is represented in the diagram by the area of the diagram 
oabc. From equations i. and ii., we get — 



v^ 



Substituting from iii., we get — 



^" ■(;-)/•'=/•• 



'-* = 2/,J 



or v^ = z/,2 -f 2/.J 



Introductory. 7 

When a body falls freely due to gravity,/. = g = 32-2 ft. 
per second per second, it is then usual to use the lei'ter A, the 
height through which the body has fallen, instead of s. 

When the body starts from rest, we have Vi = o, and z'j = » ; 
then by substitution from above, we have — 

V = ij 2gh = 8'o2 ij h .... (iv.) 

Momentum (Mo). — If a body of mass M * move with a 
velocity v, the moving mass is said to possess momentum, or 
quantity of motion, = Mv. 

Unit momentum is that of unit mass moving with unit 
velocity — 

Mo = Mv = — - 

Impulse. — Consider a ball of mass M travelling through 
space with a velocity z/j, and let it receive a fair blow in the line 
of motion (without causing it to spin) as it travels along, in such 
a manner that its velocity is suddenly increased from v^ to V2- 

The momentum before the blow = M»i 
„ after „ = Mw^ 

The change of momentum due to the blow = M{vz — »i) 

The effect of the blow is termed an impulse, and is measured 
by the change of momentum. 

Impulse = change of momentum = M(Vi — v^) 

Force (F). — If the ball in the paragraph above had received 
a very large number of very small impulses instead of a single 
blow, its velocity would have been gradually changed, and wq 
should have had — 

The whole impulse per second = the change of momentum 

per second 

When the impulses become infinitely rapid, the whole impulse 
per second is termed \!ae. force acting on the body. Hence the 
momentum may be changed gradually from M.-ffl\ to MaZ/j by a 
force acting for t seconds. Then — 

' For a rational definition of mass, the reader is referred to Prof. Kar 
Pearson's " Grammar of Science," p. 357. 



8 Mechanics applied to Engineering. 

Yt = M(z/si - »,) 
, „ _ total change of momentum 
time 

But ^' ~ ^' =/, (acceleration) (see p. 5) 

hence F = M/, = -r- 
Hence the dimensions of this unit are — 

Force = mass X acceleration 
Unit force = unit mass X unit acceleration 

Thus unit force is that force which, when acting on a mass 

of one jP"'™ \ for one second, will change its velocity by 

°"^ (Simetre) P" '^^°"'^' ^"'^ ^' *^™^*^ °°^ {d?ne!^^'' 

We are now in a position to appreciate the words of 
Newton — 

Change of momentum is proportional to the impressed force, 
and takes place in t/ie direction of the force ; . . . zho, a body will 
remain at rest, or, if in motion, will move with a uniform velocity 
in a straight line unless acted tipon by some extei-nal force. 

Force simply describes how motion takes place, not why it 
takes place. 

It does not follow, because the velocity of a body is not 
changing, or because it is at rest, that no forces are acting 
upon it ; for suppose the ball mentioned above had been acted 
upon by two equal and opposite forces at the same instant, 
the one would have tended to accelerate the body backwards 
(termed a negative acceleration, or retardation) just as much as 
the other tended to accelerate it forwards, with the result that 
the one would have just neutralized the other, and the velocity, 
and consequently the momentum, would have remained un- 
changed. We say then, in this case, that the positive acceleration 
is equal and opposite to the negative acceleration. 

If a railway train be running at a constant velocity, it must 
not be imagined that no force is required to draw it ; the force 
exerted by the engine produces a positive acceleration, while 

' The poundal unit is nevei used by engineers. 



Introductory. 5 

the friction on the axles, tyres, etc., produces an equal and 
opposite negative acceleration. If the velocity of the train be 
constant, the whole effort exerted by the engine is expended in 
overcoming the frictional resistance, or the negative accelera- 
tion. If the positive acceleration at any time exceeds the 
negative acceleration due to the friction, the positive or forward 
force exerted by the engine will still be equal to the negative 
or backward force or the total resistance overcome ; but the 
resistance now consists partly of the frictional resistance, and 
partly the resistance of the train to having its velocity increased. 
The work done by the engine over and above that expended in 
overcoming friction is stored up in the moving mass of the 
train as energy of motion, or kinetic energy (see p. 14). 

Units of Force. 

Force. Mass. Acceleration. 

Poundal. • One pound. One foot per second per second. 
Dyne. One gram. One centimetre per second per second. 

I poundal = 13,825 dynes. 
I pound = 445,000 dynes. 

Weight (W). — The weight pf a body is the force that 
gravity exerts on that body. It depends (i) on the mass of the 
body ; (2) on the acceleration of gravity (£), which varies 
inversely as the square of the distance from the centre of the 
earth, hence the weight of a body depends upon its position as 
regards the centre of the earth. The distance, however, of all 
inhabited places on the earth from the centre is so nearly 
constant, that for all practical purposes we assume that the 
acceleration of gravity is constant (the extreme variation is 
about one-third of one per cent.). Consequently for practical 
purposes we compare masses by their weights. 

Weight = mass X acceleration of gravity 
W = M^ 

We have shown above that — 

Force = mass X acceleration ' 

' Expressing this in absolute units, we have — 

Weight or force (poundals) = mass (pounds) x acceleration (feet pei 

second per second) 
Then- 
Force of gravity on a mass of one pound = i x 32*2 = 32 '2 poundals 
But, as poundals are exceedingly inconvenient units to use for practical 



lo Mechanics applied to Engineering. 

hence we speak of forces as being equal to the weight of so 
many pounds; but for convenience of expression we shall 
speak of forces of so many pounds, or of so many tons, as the 
case may be. 

Values of g-.' 

In centimetre- 
In foot-pounds, sees. grammes, sees. 

The equator 32'09i ... gyS'io 

London 32'i9l ••. 9^i'i7 

The pole 3Z'2SS — Q^S"" 

Work. — When a body is moved so as to overcome a resist- 
ance, we know that it must have been acted upon by a force 
acting in the direction of the displacement. The force is then 
said to perform work, and the measure of the work done is the 
product of the force and the displacement. The absolute unif 
of work is unit force (one poundal) acting through unit dis- 
placement (foot), or one foot-poundal. Such a unit of work is, 
however, never used by engineers ; the unit nearly always used 
in England is the "foot-pound," i.e. one pound weight lifted 
one foot high. 

Work = force X displacement 
= FS 

The dimensions of the unit of work are therefore —5- . 

purposes, we shall adopt the engineer's unit of one pound weight, i.e. a 
unit 32-2 times as great ; then, in order that the fundamental equation may 
hold for this unit, viz. — 

Weight or force (pounds) = mass X acceleration 

we must divide our weight or force expressed in poundals by 32'2, and 

we get — 

Weight or force (pounds)= weight or force (poundals) _ mass X acceleration 

or — 

, , , , mass in pounds , ... 

weight or force (pounds) = -— x acceleration in ft. -sec. per sec. 

32 2 

Thus we must take our new unit of mass as 32*2 times as great as the 
absolute unit of mass. 

Readers who do not see the point in the above had better leave il 
alone — at any rate, for the present, as it will not affect any question we 
shall have to deal with. As a matter of fact, engineers always do 
(probably unconsciously) make the assumption, but do not explicitly 
state it. 

' Hicks's " Elementary Dynamics," p. 45. 



Introductory. 1 1 

Frequently we shall have to deal with a variable force 
acting through a given displacement; the work done is then 
the average ' force multiplied by the displacement. Methods 
of finding such averages will be discussed later on. In certain 
cases it will be convenient to remember that the work done in 
lifting a body is the weight of the body multiplied by the 
height through which the centre of gravity of the body is lifted. 

Units of Work. 

Force. Displacement. Unit of work. 

Pound. Foot. Foot-pound. 

Kilogiam, Metre. Kilogrammetre. 

Dyne. Centimetre. Erg. 

I foot-pound = 32*2 foot-poundals. 
„ = 13,560,000 ergs. 

Power. — Power is the rate of doing work. Unit power 
is unit work done in unit time, or one foot-pound per second. 

„ total work done Ff 

Power = -. i — 5 — r- = ■— 

time taken to do it / 

The dimensions of the unit of power are therefore -—. 

The unit of power commonly used by engineers i^ an 
arbitrary unit established by James Watt, viz. a horse-power, 
which is 33,000 foot-pounds of work done per minute. 

Horse-power 

_ foot-pounds of work done in a given time 

~ time (in minutes) occupied in doing the work X 33,000 

I horse-power = 33jOoo foot-pounds per minute 

= 7*46 X 10° ergs per second. 

I French horse-power = 32,500 foot-pounds per minute 
= 736 X 10^ ergs per second. 

I horse-power = 746 watts 

I watt =10' ergs per second. 

Couples. — When forces act upon a body in such a manner 
as to tend to give it a spin or a rotation about an axis without 
any tendency to shift its c. of g., the body is said to be acted 

' Space-average. 




12 Mechanics applied to Engineering. 

upon by a couple. Thus, in the figure the force F tends 
to turn the body round about the point O. If, however, 
this were the only force acting on the body, it would have a 
motion of translation in the direction of the force as well as 

a spin round the axis j in order 
to prevent this motion of trans- 
lation, another force, Fu equal 
and parallel but opposite in direc- 
tion to F, must be applied to the 
body in the same plane. Thus, a 
couple is said to consist of two 
parallel forces of equal magnitude 
acting in opposite directions, but 
not in the same straight line. 
P,Q ^ The perpendicular distance x 

between the forces is termed the 
arm of the couple. The tendency of a couple is to turn 
the body to which it is applied in the plane of the couple. 
When it tends to turn it in the direction of the hands of a 
watch, it is termed a clockwise, or positive (-)-) couple, and in 
the contrary direction, a contra-clockwise, or negative (— ) 
couple. 

It is readily proved ^ that not only may a couple be shifted 
anywhere in its own plane, but its arm may be altered (as long 
as its moment is kept the same) without affecting the equili- 
brium of the body. 

Moments. — The moment of a couple is the product of 
one of the forces and the length of the arm. It is usual to 
speak of the moment of a force about a given point — that is, 
the product of the force and the perpendicular distance from 
its line of action to the point in question. 

As in the case of couples, moments are spoken of as clock- 
wise and contra-clockwise. 

If a rigid body be in equilibrium under any given system 
of moments, the algebraic sum of all the moments in any given 
plane must be zero, or the clockwise moments must be equal 
to the contra-clockwise moments in any given plane. 

Moment = force X arm 
= F« 

The dimensions of a moment are therefore — ^. 

C' 

' See Hicks's " Elementary Mechanics." 



Introductory, 13 

Centre of Gravity (c. of g.). — The gravitation forces 
acting on the several particles of a body may be considered to 
act parallel to one another. 

If a point be so chosen in a body that the sum of the 
moments of all the gravitation forces acting on the several 
particles about the one side of any straight line passing through 
that point be equal to the sum of the moments on the other 
side of the line, that point is termed the centre of gravity of the 
body. 

Thus, the resultant of all the gravitation forces acting on a 
body passes through its centre of gravity, however the body 
may be tilted about. 

Centroid. — The corresponding point in a geometrical 
surface which has no weight is frequently termed the centroid ; 
such cases are fully dealt with in Chapter III. 

Suergy. — Capacity for doing work is termed energy. 

Conservation of Energy. — Experience shows us that 
energy cannot be created or destroyed ; it may be dissipated, 
or it may be transformed from any one form to any other, hence 
the whole of the work supplied to any machine must be equal 
to the work got out of the machine, together with the work 
converted into heat,i either by the friction or the impact of the 
parts one on the other. 

Mechanical Equivalent of Heat. — It was experiment- 
ally shown by Joule that in the conversion of mechanical into 
heat energy,* 772 foot-lbs. of work have to be expended in 
order to generate one thermal unit. 

Efficiency of a Machine. — The efificiency of a machine 
is the ratio of the useful work got out of the machine to the 
gross work supplied to the machine. 

_„. . work got out of the machine 

Efificiency = — = — 2 — — 

work supplied to the machine 

This ratio is necessarily less than unity. 
The counter-efficiency is the reciprocal of the efficiency, 
and is always greater than unity. 

_ ^ „ . work supplied to the machine 

Counter-efficiency = -, — &£ ^-^ -^. — 

work got out of the machine 

' To be strictly accurate, we should also say light, sound, electricity, 
etc. 

' By far the most accurate determination is that recently made by Pro- 
fessor Osborne Reynolds and Mr. W. H. Moorby, who obtained the value 
776-94 (see Phil. Trans., vol. igo, pp. 301-422) from 32° F. to 212° F., 
which is equivalent to about 773 at 39° F. and 778 at 60° F. 



14 Mechanics applied to Engineering. 

Kinetic Energy.— From the principle of the conservation 
of energy, we know that when a body falls freely by gravity, the 
work done on the falling body must be equal to the energy of 
motion stored in the body (neglecting friction). 

The work done by gravity on a weight of W pounds m 
falling through a height h ft. = WA foot-lbs. But we have 

shown above that h = —, where v is the velocity after falling 
through a height h ; whence — 

W/4 = — , or 

2g 2 



This quantity, , is known as the kinetic energy of the 

body, or the energy due to its motion. 

Inertia. — Since energy has to be expended when the 
velocity of a body is increased, a body may be said to offer a 
resistance to having its velocity increased, this resistance is 
known as the inertia of the body. Inertia is sometimes defined 
as the " deadness of matter." 

Moment of Inertia (I). — We may define inertia as the 
capacity of a body to possess momentum, and momentum as 
the product of mass and velocity {Mv). If we have a very 

small body of mass M 
rotating about an axis 
at a radius r, with an 
angular velocity ui, the 
linear velocity of the 
body will be z/ = ar, 
and the momentum will 
beMz/. But if the body 
be shifted further from 
the axis of rotation, 
and r be thereby in- 
creased, the momen- 
tum will also be in- 
creased in the same 
ratio. Hence, when we are dealing with a rotating body, we 
have not only to deal with its mass, but with the arrangement 
of the body about the axis of rotation, i.e. with its moment 
about the axis. 

Let the body be acted upon by a twisting moment, Yr = T, 



M 



GrooveAjUiUey 
considered^ 0£ 



-*/» 



Fig. 5. 



Introductory. 1 5 

then, as the force P acts at the same radius as that of the body, 
it may be regarded as acting on the body itself. The force 
P acting at a radius r will produce the same effect as a 

r 

force n? acting at a radius . The force P actmg on the 

mass M gives it a linear acceleration /„ where P = M^, or 

P • I 

/, = -—. The angular velocity (o is - times the hnear velocity, 

M T 

hence the angular acceleration is - times the linear accelera- 
tion. Let A = the angular acceleration ; then — 

r Mr M/-2 M^ 

, , . . twisting moment ^ 

or angular acceleration = 5__ — _ — _ 

mass X (radius)" 

In the case we have just dealt with, the mass M is supposed to 
be exceedingly small, and every part of it at a distance r from 
the axis. When the body is great, it may be considered to be 
made up of a large number of small masses. Mi, M^, etc., at radii 
»-i, ^2, etc., respectively ; then the above expression becomes — 



A = 



(Min' + M^Ta" + Mar,^ +, etc.) 



The quantity in the denominator is termed the "moment of 
inertia " of the body. 

We stated above that the capacity of a body to possess 
momentum is termed the " inertia of the body." Now, in a 
case in which the capacity of the body to possess angular 
momentum depends upon the moment of the several portions 
of the body about a given axis, we see why the capacity of a 
rotating body to possess momentum should be termed the 
" moment of inertia." 

Let M = mass of the whole body, then M = M1+M2+M3, 
etc. ; then the moment of inertia of the body, I, = Mk^ 
= (Miz-i" + M^r^^ etc.). 

Radius of Gyration (k). — The k in the paragraph above 
is known as the radius of gyration of the body. Thus, if we 
could condense the whole body into a single particle at a 
distance k from the axis of rotation, the body would still have 

' The reader is advised to turn back to the paragraph on " couples," 
so that he may not lose sight of the fact that a couple involves tuio forces. 



i6 



Mechanics applied to Engineering. 



the same capacity for possessing energy, due to rotation about 
that axis. 

Representation of Displacements, Velocities/ 
Accelerations, Forces by Straight Lines. — Any 

I displacement] 
,' ■ I is fully represented when we state its magni- 
force J 

tude and its direction, and, in the case of force, its point of 
application. 

Hence a straight line may be used to represent any 

Idisplacemenfj 
velocity r ^^ length of which represents its magni- 
force j 

tude, and the direction of the line the direction in which the 
force, etc., acts. 

I displacements! 
Velocities 1 • • 

accelerations ' "^^^^^ ^' ^ P°''''' ""^^ 
forces / 

be replaced by one force, etc., passing through the same point, 
which is termed the resultant force, etc. 
(■displacements! 

If two P^l°"'ies. not in the same straight line, 

1 accelerations , 6 > 



I forces 




Fig. 6. 



meeting at a point a, be represented 
, by two straight lines, ab, ac, and if 
two other straight lines, dc, hd, be 
drawn parallel to them from their 
extremities to form a parallelogram, 
abdc, the diagonal of the parallelogram 
ad which passes through that point 

I displacement \ 
acceleration I ^ magnitude 
force ) 

and direction. 

Hence, if a force equal and opposite to ad act on the point 
in the same plane, the point will be in equilibrium. 

It is evident from the figure that bd is equal in every 



Including angular velocities or spins. 



Introductory. 17 

respect to ac; then the three forces are represented by the three 
sides of the triangle ai, bd, ad. Hence we may say that if three 
forces act upon a point in such a manner that they are equal 
and parallel to the sides of a triangle, the point is in equi- 
librium under the action of those forces. This is known as 
the theorem of the " triangle of forces." 

Many special applications of this method will be dealt with 
in future chapters. 

The proof of the above statements will be found in all 
elementary books on Mechanics. 

Hodograph. — The motion of a body moving in a curved 
path may be very conveniently analyzed by means of a curve 
called a "hodograph." In Fig. 7, suppose a point moving 
along the path P, Pj, Pa, with varying velocity. If a line, op, 
known as a "radius vector," be drawn so that its length 
represents on any given scale the speed of the point at P, 
and the direction of the radius vector the direction in 
which P is moving, the line op completely represents the 
velocity of the point P. If other radii are drawn in the same 
manner, the curve traced out 
by their extremities is known 
as the "hodograph" of the 
point P. The change of ve- 
locity of the point P in pass- 
ing from P to Pi is represented 
on the hodograph by the 
distance ppi, consisting of a 
change in the length of the 
line, viz. q-^p-^ representing the 
change in speed of the point 
P, and/^i the change of velo- 
city due to change of direction, Fig. 7. 
if a radius vector be drawn 

each second ; then //i will represent the average change of 
velocity per second, or in the limit the rate of change of 
velocity of the point P, or, in other words, the acceleration 
(see p. s) of the point P ; thus the velocity of / represents the 
acceleration of the point P. • 

If the speed of the point P remained constant, then the 
length of the line op would also be constant, and the hodo- 
graph would become the arc of a circle, and the only change 
in the velocity would be the change in direction pq-^. 

Centrifugal Force. — If a heavy body be attached to the 
end of a piece of string, and the body be caused to move round 




1 8 Mechanics applied to Engineering. 

in a circular path, the string will be put into tension,the amount 
of which will depend upon (i) the mass of the body, (2) the 
length of the string, and (3) the velocity with which the body 
moves. The tension in the string is equal to the centrifugal 
force. We will now show how the exact value of this force may 
be calculated in any given instance.' 

Let the speed with which the body describes the circle be 
constant; then the radius vector of the hodograph will be 
of constant length, and the hodograph it- 
self will be a circle. Let the body describe 
the outer of the two circles shown in the 
figure, with a velocity v, and let its velocity 
at A be represented by the radius OP, the 
inner circle being the hodograph of A. 
Now let A move through an extremely 
small space to Ai, and the corresponding 
radius vector to OPj; then the line PPj 
p,e J represents the change in velocity of A 

while it was moving to Ai. (The reader 
should never lose sight of the fact that change of velocity 
involves change of direction as well as change of speed, and 
as the speed is constant in this case, the change of velocity is 
wholly a change of direction.) 

As the distance AA, becomes smaller, PPj becomes more 
nearly perpendicular to OP, and in the limit it does become 
perpendicular, and parallel to OA ; thus the change of velocity 
is radial and towards the centre. 

We have shown on p. 17 that the velocity of P represents 
the acceleration of the point A ; then, as both circles are de- 
scribed in the same time — 

velocity of P _ OP 
velocity of A ~ OA 

lad 
of 
OA = R; then— 




But OP was made equal to the velocity of A, viz. v, and 
OA is the radius of the circle described by the body. Let 



velocity of P v 
V = R 



or velocity of P = 



R 



' For another method of treatment, see Barker's " Graphic Methods o( 
Engine Pesi{rn." 



Introductory. 19 

and acceleration of A = ^ 

and since force = mass x acceleration 

we have centrifugal force C = ^- 

. . , . ^ W»2 
or in gravitational units, C = — „- 

This force acts radially outwards from the centre. 

Sometimes it is convenient to have the centrifugal force 
expressed in terms of the angular velocity of the body. We 
have — 

V = <dR 
hence C = Mw^R 
W<o=R 



or C = 



g 



Change of Units. — It frequently happens that we wish 
to change the units in a given expression to some other units 
more convenient for our immediate purpose ; such an alteration 
in units is very simple, provided we set about it in systematic 
fashion. The expression must first be reduced to its funda- 
mental units; then each unit must be multiplied by the 
required constant to convert it into the new unit. For 
example, suppose we wish to convert foot-pounds of work to 
ergs, then — 

The dimensions of. work are — 5- 

r 

, . „ J , pounds X (feet)" 

work in ft.-poundals = - — ; ,^,„ '' 

(seconds)^ 

work in ergs = g^ams X (centimetres)^ 
(seconds)^ 
I pound = 453"6 grams 
I foot = 3o"48 centimetres 

Hence — 

I foot-poundal = 4S3'6 X 3o"48' = 421,390 ergs 
and I foot-pound = 32-2 foot-poundals 

= 32-2 X 421,390 = 13,560,000 ergs 



CHAPTER II. 

MENSURATION. 

Mensuration consists of the measurement of lengths, areas, 
and volumes, and the expression of such measurements in 
terms of a simple unit of length. 

Length. — If a point be shifted through any given distance, 
it traces out a line in space, and the length of the line is the 
distance the point has been shifted. A simple statement in 
units of length of this one shift completely expresses its only 
dimension, length ; hence a line is said to have but one dimension, 
and when we speak of a line of length /, we mean a line con- 
taining / length units. 

Area. — If a straight line be given a side shift in any given 
plane, the line sweeps out a sraface in space. The area of the 
surface swept out is dependent upon two distinct shifts of the 
generating point : (i) on the length of the original shift of 
the point, i.e. on the length of the gene- 
^T I rating line (J); (2) on the length of the 
U i side shift of the generating line (d). 
_X_ 1 Thus a statement of the area of a given 



— I > surface must involve two length quantities, 

Fio. 9. / and d, both expressed in the same units 

of length. Hence a surface is said to have 

two dimensions, and the area of a surface Id must always be 

expressed as the product of two lengths, each containing so 

many length units, viz. — 

Area = length units x length units 
= (length units)' 

Volume. — If a plane surface be given a side shift to bring 
it into another plane, the surface sweeps out a volume in space. 



Mensuration. 



21 



The volume of the space swept out is dependent upon three 
distinct shifts of the generating point : (i) on the length of the 
original shift of the generating point, i.e. on the length of the 
generating line /; (2) on the length 
of the side shift of the generating 
line d; (3) on the side shift of the 
generating surface /. Thus the state- 
ment of the volume of a given body 
or space must involve three length 
quantities, /, d, t, all expressed in 
the same units of length. 

Hence a volume is said to have three dimensions, and the 
volume of a body must always be expressed as the product of 
three lengths, each containing so many length units, viz. — 

Volume = length units X length units >< length units 
= (length units)' 



/i 




f 


1 


,'' 


/ 


< 1-- 

FlG. 10. 


— * 





22 Mechanics applied to Engineering. 

Lengths. 

Straight line. 



Circumference of circle. 

Length of circumference = ird 

/^ ^\ = 3 •14161/ 



/ \ 



6*2832r 



Y / The last two decimals above may usually 

V^ _...^ be neglected ; the error will be less than \ in. 
■* dj * on a lo-ft. circle. 



Fis. II. 




Length of arc = —7- 



2irr0 rO 
or =• ^ 



360 57-3 
For an arc less than a semicircle — 

8C — C 

Fioril Length = — ^ ° approximately 



Arc of ellipse. 



d \^ Length of circumference approx. 




/^ A 4D - d) 

= ir^+ 2(D -d)- ^^ '- 



r,a. x3. ^ V(D+rf)(D + 2rf) 



Mensuration. 23 

The length of lines can be measured to within -^ in. with 
a scale divided into either tenths or twentieths of an inch. 
With special appliances lengths can be measured to within 

' in. if necessary. 



1000000 



The mathematical process by which the value of ir is deter- 
mined is too long for insertion here. One method consists of 
calculating the perimeter of a many-sided polygon described 
about a circle, also of one bscribed in a circle. The perimeter 
of the outer polygon is greater, and that of the inner less, than 
the perimetpr of the circle. The greater the number of sides 
the smaller is the difference. The value of ir has been found 
to 750 places of decimals, but it is rarely required for practical 
purposes beyond three or four places. For a simple method 
of finding the value of tt, see " Longmans' School Mensura- 
tion," p. 48. 



The length of the arc is less than the length of the 

Q 

circumference in the ratio —^. 
360 

Length of arc = -ira X -— - = -pr- 
360 360 

The approximate formula given is extremely near when A is 
not great compared with C„-; even for a semicircle the error is 
only about i in 80. The proof is given in Lodge's " Mensura- 
tion for Senior Students " (Longmans). 



No simple expression for the exact value of the length of 
an elliptic arc can be given, the value opposite is due to Mr. 
M. Arnold Pears, of New South Wales, see Trautwine's 
" Pocket-book," 18th edition, p. 189. 



24 Mechanics applied to Engineering 

Arc of parabola. 




Length of arc = 2 



(approximately) 



Fig. 14. 



Irregular curved line abc. 

Set ofT tangent cd. With pair 
of dividers start from a, making 
small steps till the point c is 
reached, or nearly so. Count 
number of steps, and step off 
-— . d same number along tangent. 




FrG. 15. 



Areas. 




Area of figure = Ih 



Triangles. 




Area of figure = 



bh 



Equilateral triangle. 




Area of figure 



tbv^ = 0-4., 



433^ 



Fio. 18. 



Mensuration 

No simple expression can be given for the 
parabolic arc — a common approximation is that e 
opposite page. The error is negligible when h is 
pared with b, but when h is equal to b the error 
about 8J per cent. 


25 

length of a 
;iven on the 
small corn- 
amounts to 


The stepping should be commenced at the end remote 
from the tangent ; then if the last step does not exactly coincide 
with c, the backward Stepping can be commenced from the 
last point without causing any appreciable error. The greater 
the accuracy required, the greater must be the number of 
steps. 


Areas. 

See Euc. I. 35, 




See Euc. I. 41. 








2 4 



26 Mechanics applied to Engineering. 

Triangle. 




Let s = 



a ■\-h -\- c 



A^i Area of 



figure = ,Js(s — d){s-b)(s—c) 



FrG, ig. 



Quadrilateral, 




Area of figure = 



bh 



Fic. 9a 



Trapezium, 

-A. 




Area of figure = ( j li 



Fig. «. 



Irrtgular straight-lined figure. , 
6 




Area of figure = area ahdef — area 3frf 
or area of triangles (acb-^acf-\-cfe->rced) 



Fig. »9* 



Mensuration. 27 



The proof is somewhat lengthy, but perfectly simple (see 
" Longmans' Mensuration," p. 18). 



Area of upper triangle = — -' 
2 

„ lower triangle = — - 

2 

both triangles = b( ^l±Jh \ = 



bh 
2 



Aiea. of parallelogram = 61/1 

Area of triangle = \ ~ " 



2 



Area of whole figure = (^ - '^1 + ^'^■)^ = iA±m 



2 2 



Simple case of addition and subtraction of areas. 



28 Mechanics applied to Engineering. 



Area of figure = izr^ = 3-1416^' 
or — = o-i^SAiP 




Sector of circle. 




Area of figure = — ^ 
360 



Fig. 34. 



Segment of circle. 




Area of figure = f C^^ when h is small 
= A(6C. + 8C,) nearly 



Fig. 25. 



Hollow circle. 




Area of figure = area of outer circle — 
area of inner circle 

= TrTj" — wTi* 

= T(r,» - r^) 

= irr^ 

or = cySSifj' 

or = '' ^V , i.e. mean cir- 
2 

cumf. X thickness 



Fig. 36. 



Mensuration. 29 



The circle may be conceived to be made up of a great 

number of tiny triangles, such as the one shown, the base of 

each little triangle being b units, then the area of each triangle 

. br 

IS — J but the sum of all the bases equals the circumference, or 

%b = 2irr, hence the area of all the triangles put together, 

. , 2irr . r , 

ue. the area of the circle, = = '^r'- 



The area of the sector is less than the area of the circle in 

6 Trr^d 

the ratio —r-, hence the area of the sector = — j- ; if fl be the 
300 360 '^ 

angle expressed in circular measure, then the above ratio 
becomes — • 

The area = — ^ 
2 



When k is less than — ^, the arc of the circle very nearly 

coincides with a parabolic arc (see p. 31). For proof of second 
formula, see Lodge's " Mensuration for Senior Students " 
(Longmans). 



Simple Case of Subtraction of Areas. — The substitution of r^ 
for r^ — r^ follows from the properties of the right-angled 
triangle (Euc. I. 47). 

The mean circumference X thickness is a very convenient 
form of expression ; it is arrived at thus — 

Ttid^ + (fi) 

Mean circumference = — ' ■ 

2 

thickness = — ^ 

2 

product = ^-^^^ X ^ = ^(4= - d^) 



30 Mechanics applied to Engineering. 

Ellipse. 

Area of figure = Trr^r^ 

or = —d-id^ 
4 

Fig. 37. 




Parabolic segments. 




'^ Area of figure = |BH 

I i.e. f (area of circumscribing rectangle) 




Fig. 29. 



Area of figure = f area of A aie 




Make de = ^e 
area of figure = area of A aid 



Mensuration. 



31 



An ellipse may be regarded as a flattened or an elon- 
gated circle j hence the area of an ellipse is |^®* \ than 

igreaterJ 

the area of a circle whose diameter is the /™?J°'^1 a^is of an 

Immorf 



ellipse {J}, in the ratio } J ;° J- 



Area = ^\ 4' = -'^i<^2, or !^' X ^ = ""-M^ 
4 "2 4 4 t^i 4 

From the properties of the parabola, we have — 
H= B 



V B B* 

area of strip = h . db = ( — - ) 
V B* ^ 



r- 


rfS-.^-^ 






X 


k 


¥ 


k 


-«— . 


i 


s/ 



db 



Tig. 28a. 



b = B 



area of whole figure 






= |HB 



l>\db = ^X?r 



B* t 



The area ac^^has been shown 
to be fHB. Take from each the 
area of the A «^^, then the re- 
mainder abt: = 5 the A i^be ; but, 
from the properties of the para- 
bola, we have ed = \eb, hence 
the area abc = § area of the cir- 




FlG. 29a. 



cumscribing A abd. 

From the properties of the parabola, we also have the height 
of the A abd = 2 (height of the A abc) ; hence the area of the 
A abd = 2 (area of A abc), and the area of the parabolic segment 
= 2X1 area A abc = | area A abc. 



By increasing the height of the A abc to g its original 
height, we increase its area in the same ratio, and consequently 
make it equal to the area of the parabolic segment. 



32 Mechanics applied to Engineering. 

d 



Area of shaded figure = \ area of A o.bd 




Surfaces bounded by an irregular curve. 




Area of figure = areas of para- 
bolic segments (a — b-\rC-\-d-\-e) 
+ areas of triangles {g + K). 



Fig. 32. 



Mean ordinate method. 




Area of figure = (/« + Aj + ^ + 
hi +, etc.)a: 



Fio. 33. 



Mensuration. 33 



The area abc = f area of triangle abd, hence the remainder 
\ of triangle abd. 



Simply a case of addition and subtraction of areas. It is 
a somewhat clumsy and tedious method, and is not recom- 
mended for general work. One of the following methods is 
considered to be better. 



This is a fairly accurate method if a large number of ordi- 
nates are taken. The value of ^ + ^1 -)- ^2 + h^, etc., is most 



^ ' ^1 I K^ /ij I and so on. 



Fig. 33fl. 

easily found by marking them off continuously on a strip of 
paper. 

The value of x must be accurately found ; thus, If n be 

the number of ordinates, then x= -. 

n 

The method assumes that the areas a, a cut off are equal to 
the areas a^, a^ put on. 

D 



34 



Mechanics applied to Engineering. 



Simpson's Method. 

Area of figure = -{k-\- 4^1 ■\-2hs, 

+ 4/^3 + 2hi + 4/^6 + 2-4g 

+ 4/4, + 2,48 + A^h +^a) 

The end ordinates should be 
obtained by drawing the mean 
lines (shown broken). If they 
Fig. 34. are taken as zero the expres- 

sion gives too low a result. 
Any odd number of ordinates may be taken ; the greater 
the number the greater will be the accuracy. 




Curved surface of a spherical indentation. 
I Curved surface 



Fig. 34*. 



Mensuration. 



35 



This is by far the most accurate and useful of all methods 
of measuring such areas. The proof is as follows : — 
The curve gfedc is assumed to be a parabolic arc. 

Area aieg = j/ '' "*" ' -' j . . . . (i,) 

„ .3« = 4^) .... (ii.) 

„ abceg= ^(Ai + 2^^ + ^^) . (i.+ii.) ^/f-' 
„ abcjg= 2:c(^L±i5)=;c(/5, + >^3)(iii.) I*' 



Area of A g'^ = (i-) + ("•) — (iii.) 

X 
2 

2 




= -{K-V'iK-'rh^ — x{h^-\-h^ Fio. 34a. 

(2/^ - h^- h^ (iv.) 

Area of parabolic ) _ 4/- v 2X, 

segment gcdef \ " 3^'^-/ " y (2'^2 - fh- 4) ■ (v.) 

Whole figure = (iii.) + (v.) = x{hy, + h^) ->r^{2fh - fh- h^ 

= ^(/i, + 4/i, + /g 

If two more slices were added to the figure, the added area 

X 

would be as above = -{h^ + 4/^4 + h^, and when the two are 

X 

added they become = -(/^i + \li^ + 2^3 + 4/^4 + h^. 



The curved surface of the slice = ^irr^s 
By similar triangles we have 

S _ R 

ly- r^ 

Substituting the value of S we have 

Curved surface of slice = ziiRZy 
„ ,, indentation = zttRY 

Expressing Y in terms of R and d we have 



H-Y ^ 



Fig. 34^. 



tRY = 2!rR(^R +/^R^ - -) 



36 



Mechanics applied to Engineering. 



Surfaces of revolution. 




Pappus^ or Guldiwis' Method. — 

Area of surface swept out by ^ 

the revolution of the line > = L X zirp 

defaboMt the axis ab ) 

Length of line =• L 
Radius of c. of g. of line defi _ 

considered as a fine wire y ~ ^ 

This method also holds for any part of 
a revolution as well as for a complete 
revolution. The area of such figures as 
circles, hollow circles, sectors, parallelo- 
grams (p = cc ), can also be found by this 
method. 



Surface of sphere. 




Area of surface of sphere = 47rr^ 

The surface of a sphere is the same as 
the curved surface of a cylinder of same 
diameter and length = d. 



Fig. 36. 



Surface of cone. 

I 
.1 




Area of curved surface of cone = wrh 



Fig. 37- 



Mensuration. 37 

The area of the surface traced out by a narrow strip of 

i<.„„i.i, /4 and radius po = 2t/|,po\ , 
length \, f^" T"], and so on. 

Area of whole surface 1 

= 2t(/oPo + kp\ +, etc.) -^ 1 ^ 

= 2ff(each elemental length of u/^ I ^x 

wire X its distance from axis 'F^' y-~-'-'y-~---~-'-\ 

of revolution) | ° _a ^ 

= 2'7r(total length of revolving wire I '"* 

X distance of c. of g. from j 

axis of revolution) (see p. | 

58) Fig. 35a. 

= (total length of revolving wire 

X length of path described by its centre of gravity) 
= Lzirp 

N.B. — The revolving wire must lie wholly on one side of 
the axis of revolution and in the same plane. 



The distance of the c. of g. of any circular arc, or wire bent 

re 

to a circular arc, from the centre of the circle is y = — = Pi 

a 

where r = radius of circle, t: chord of arc, a length of arc (see 

p. 64). 

In the spherical surface a = L = irr, c = 2r, p = — = — 

2r 
Surface of sphere = irr . 2t . — = 47ir* 



Length of revolving wire = \, = h 
radius of c. of g. „ „ = p = - 

hlirr 
surface of cone = = m-h 



38 



Mechanics applied to Engineering. 



Hyperbola. 




Area of figure = XY log.r 

log, = 2-31 X ordinary log 




Fig. 3» 



Area of figure = 



XY - X.Yi 



Mensuration. 



In the hyperbola we have — 

XY = X,Yi = xy 

u XY 

hence v = — 

X 

dx 
area of strip = y . dx = XY — 



Area 
whole 
figure 



°n r 

lie > = XY 
re j j 



« = Xi 



dx 

X 



x= X 

= XY (log. X, - log. X) I 



= XY log. 



Xi 




Using the figure above, in this case we have — 



YX» = YiXi" = yxT 

YX» 
hence y = — rr- 

YX- 



area of strip = y .dx = —^dx = YX"x-"dx 

J* = Xi ,-v 1 - n _ Vl - «\ 
x-"dx = YXH \_„ ) 
jc = X • 
_ YX"Xi'-"- YX 
I — n 

But Y,Xi" = YX» 

Multiply both sides by Xj'"" 

then Y,X, = YX"X,'-- 

Substituting, we have — 

Area of whole figure = iii^=i lii 

I — « 

YX - YiX, 
or = 1 — 1. 



40 



Mechanics applied to Engineering. 



Irregular areas. 

Irregular areas of every description are most easily and 
accurately measured by a planimeter, such as Amsler's or 
Goodman's. 

A very convenient method is to cut out a piece of thin 
cardboard or sheet metal to the exact dimensions of the area ; 
weigh it, and compare with a known area (such as a circle or 
square) cut from the same cardboard or metal. A convenient 
method of weighing is shown on the opposite page, and gives 
very accurate results if reasonable care be taken. 



Prisms. 



^ I 

Fin. II. 



Ur 



Volumes. 

Let A = area of the end of prism ; 
/ = length of prism. 
Volume = /A 
Parallelopiped. 

Volume = Idt 




Hexagonal prism. 

Volume = 2'598.fV 



Cylinder. 

Volume = "^ — = o-yS^aTV 
4 ^ 



Mensuration. 



41 





Suspend a knitting-needle or a straight piece of wire or 
wood by a piece of cotton, 
and accurately balance by 
shifting the cotton. Then 
suspend the two pieces of 
cardboard by pieces of 
cotton or silk ; shift them 
tilltheybalance ; then mea- 
sure the distances x and^. 

Then A^ = '&y 

or B=: — 

y 

The area of A should not differ very greatly from the area 
of B, or one arm becomes very short, and error is more likely 
to occur. 



Area of end = td 
volume = ltd 



Fig. 40. 



Area of hexagon = area of six equilateral triangles 
= 6 X o*433S'' (see Fig. 18) 
volume = 2'598SV 

or say 2"6SV 



Area of circular end = -d^ 
4 

■KdH 

volume = 

4 




42 Mechanics applied to Engineering. 



Prismoid. 
Simpsoris Method. — 



Volume = '(A1+4A2+2A8+4A44-AJ) 
3 

and so on for any odd number of sections. 

Contoured volume. 

N.B. — Each area is to be taken as in- 
cluding those within it, not the area between 
the two contours. A3 is shaded over to 
make this clear. 



&ddntorCber of 
£yuidCstant slices 




FjR. 45- 



Solids 0/ revolution. 




Method of Pappus or Guldinus.— 

Let A = area of full-lined surface ; 
p = radius of c. of g. of surface. 

Volume of solid of revolution = 2irpA 

N.B. — The surface must lie wholly on 
one side of the axis of revolution, and in the 
same plane. 

This method is applicable to a great 
number of problems, spheres, cones, rings, 
etc. 



Fig. 47. 



Mensuration. 43 



Area of end (or side) = -(h^ 4. 4^^ + 2,^3 +, etc.) (see p. 34), 
where h^^ k^, etc., are the heights of the sections. 

X 

Volume = -{hJ-\- iji4 -V 2/^3/+, etc.) 

= -(Ai + 4A2 + 2A3 +, etc.) 
3 

The above proof assumes that the sections are parallelo- 
grams, i.e. the solid is flat-topped along its length. We shall 
later on show that the formula is accurate for many solids 
having surfaces curved in all directions, such as a sphere, 
ellipsoid, paraboloid, hyperboloid. 

If the number of sections be even, calculate the volume of 
the greater portion by this method, and treat the volume of the 
remainder as a paraboloid of revolution or as a prism. 



Let the area be revolved around the axis ; then — 
The volume swept out by an"! ^^-" 

elemental area «„, when re- 1 _ „ ^^o ( 
volving round the axis at aj " / f* 

distance p, ' \ ^' 

Ditto ditto a^ and pi = fli X 'iirp^ \ < 

and so on. V 

Whole volume swept out by all^ 
the elemental areas, a^, a^, etc., 
when revolving round the axis 
at their respective distances, poj 
Pi, etc. 

= 2ir(each elemental area, a^, a^, etc. x their respective 

distances, po, Pi, from the axis of revolution) 
= 2ir(sum of elemental areas, or whole area X distance of 
c. of g. of whole area from the axis of revolution) 
(see p. 58) 
= A X 2irp = 27rpA 
But 27rp is the distance the c. of g. has moved through, or the 
length of the path of the c, of g. ; hence — 
Whole volume = area of generating surface X the length of the 
path of the c. of g. of the area 
This proof holds for any part of a revolution, and for any 
value of p; when p becomes infinite, the path becomes a 
straight line, in such a case as a prism. 



Fn. 46a. 

= 27r(a„po + aipi +, etc.) 



44 



Mechanics applied to Engineering. 



Sphere. 




Volume of sphere = — , or ^irr^ 

Volume of sphere = | volume of circum- 
scribing cylinder 

Hollow sphere. 

Volume of ■> _ f volume of outer sphere — 
External diameter = a. hoUow Sphere/ " \ volume of inner sphere 



Internal diameter :: 
Fig. 48. 



_ -n-d.^ _ ltd? 
6 6 



= JW 



d?) 



Slice of sphere. 



/T 


"^ 


f 


"^ 




..^ 


— 




'■■^., 






,/■' 



Volume of slice = -{3R(Yj,''-Yi'')-Y,»-l-Y,n 
3 
N.B. — The slice must be taken wholly 
on one side of the diameter; if the slice 
includes the diameter, it must be treated as 
two of the following slices. 



Fig. 49. 



I 



.--ttT---,. 
V ; i -^^ 

! A^ 



Special case in which Y, = R. 



/ Volume of slice = -(2R» - 3RY1' + Yi') 



Fig. 50. 



Mensuration. 



45 



Sphere. — The revolving area is a semicircle of area 

The distance of the c. of e. 1 Ar , , ,, 

from the diameter f = '' = ^ ('^^ P- ^^) 



Ar 



■K(P 



volume swept out = 27r X ^L, x — - = \iti^ = _ 
3T 2 ^ 6 

or by Simpson's rule — ' 

Volume of sphere = - (o + 47r;i + o) == iwr" 



Volume of elemental slice = icc^dy 
= 7r{R2 - (R2 +/ - '2.^y))dy 
— ir(zRy — y'^)dy 



Volume of 
whole 1- = 
slice 



ry=~-i 

(2Rj 



Ky-f')dy 



-t 



2R/ _/ 



y=Y, 

y=Y, 



_s_ 



±rJi_ 



= J zRC^iri^) _ Y,3-Y.n 



Fic. 4g<z. 



The same result can be obtained by Simpson's method — 

Volume = -(irCj2 + /[ttC' + ■rCi') 



\4i 



For X substitute 

(^ „ (2 R;/ —y) with the proper suflSxes. 
The algebraic work is long, but the results by the two 
methods will be found to be identical. 



46 



Mechanics applied to Engineering. 



*- /? 



Special case in which Yj = o. 
Volume of slice = -(sRY^^ - Y/) 

When Ya = R, and Yj = o, the slice 
becomes a hemisphere, and the^ 



Volume of hemisphere = -(2R') 



Fig. 51. 



which is one-half the volume of the sphere found by the other 
method. 



Paraboloid. 




Volume of\_^„2„ ^a 
paraboloid /- 2^ "■"'^ 8° " 

= iSyR'H, or o-39D''H 
's \ volume circumscrib- 
ing cylinder 



Fig. 51. 



Cone. 




Volume of cone = -R'H 
3 



=J!:d»h 

12 
= ^ volume circumscrib- 
ing cylinder 



Fig. sj. 



Mensuration. 
{Continued from page 45.) 

For the hemisphere it comes out very 
easily, thus— 



R 



R 



«»= R»- 



R' 



Volume = ^^{o+,r(4R2 - R^) + ,rR^} 



6 

= |7rR^ 



Fio. 51, 



47 




From the properties of the parabola, we have — 

R" H 
H 



Volume of slice = 
volume of solid = 



-J, irRV/ 




r 
R 



Volume of slice = irT^dh = 
volume of cone = 




# 




48 Mechanics applied to Engineering. 

Pyramid. 

., B,BH 
Volume of pyramid = — — 

= — , whenB,=B=H 

3 
= i volume circumscrib- 

«^--— ^f- ->'■' ing solid 

F:g. 54- 

Slightly tapered body. 

'a' 

'I™ Mean Areas Method. — 

^ ''?..V.V.:^/;:.':1| volume of body=(^^t^^^t^)/(approx.) 

' ill = (mean area)/ 

SI' 

Fig. ss- 

Ring. 



wd' 





Volume of ring = — X tD = 2'^i(PY) 
4 



Fig. s6. 





Weight or 


Materials. 






Aluminium 






.. 0'093 lb. per 


cubic inch, | 


Brass and bronze 






• o'3o ,, 




i 


Copper 






• 0-32 








Iron — cast 






., 0-26 „ 








„ wrougtit 






■ 0-278 ,, 








Steel 






.. 0-283 








Lead 






.. 0-412 „ 








Brickwork 






.. 100 to 140 lbs 


.per 


cubic foot. 


Stone 






.. 150 to 180 


n 


i> 



Mensuration. 49 

This may be proved in precisely the 
same manner as the cone, or thus by 
Simpson's method — 

Volume=^jO+4(?X?^)+BxB. 




This method is only approximately true when the taper is 
very slight. For such a body as a pyramid it would be 
seriously in error ; the volume obtained by this method would 
be T^HMnstead of ,^H3. 



The diameter D is measured from centre to centre of the 
sections of the ring, i.e. their centres of gravity — 

Volume = area of surface of revolution x length of path of 
c. of g. of section 



Weight of Materials. 



Concrete 
Pine and larch 
Pitch pine and oak 

Teak 

Greenheart ... 



130 to 150 lbs. per cubic foot. 

301040 „ 

40 to 60 ,, ,, 

4StoS5 

65 to 75 



CHAPTER III. 



MOMENTS. 

That branch of applied mechanics which deals with moments 
is of the utmost importance to the engineer, and yet perhaps 
it gives the beginner more trouble than any other part of the 
subject. The following simple illustrations may possibly help 
to make the matter clear. We have already (see p. 12) 
explained the meaning of the terms " clockwise " and " contra- 
clockwise " moments. 

In the figures that follow, the two pulleys of radii R and Rj 
are attached to the same shaft, so that they rotate together. 
We shall assume that there is no friction on the axle. 




Fio. 57. 





n^ 



-R. — ' 



J 



Fig. 59. 



Let a cord be wound round each pulley in such a manner 
that when a force P is applied to one cord, the weight W will 
be lifted by the other. 

Now let the cord be pulled through a sufficient distance to 
cause the pulleys to make one complete revolution j we shall 
then have — ■ 



Moments. 5 1 

The work done by pulling the cord = P x 2irR 
„ „ in lifting the weight = W X zttRj 

These must be equal, as it is assumed that no work is wasted 
m friction; hence — 

PairR = W2irR, 

or PR = WRi 
or the contra-clockwise moment = the clockwise moment 

It is clear that this relation will hold for any portion of a 
revolution, however small ; also for any size of pulleys. 

The levers shown in the same figures may be regarded as 
small portions of the pulleys ; hence the same relations hold in 
their case. 

It may be stated as a general principle that if a rigid body 
De in equilibrium under any given system of moments, the 
algebraic sum of all the moments in any given plane must be 
zero, or the clockwise moments must be equal to the contra- 
clockwise moments. 

r force (/) \ 

rirst Moments.— The product oi & < mass («;) f 

\ volume {v) ) 

the length of its arm /, viz. <^ ^/ ^> is termed ihe first moment 




force "^ 

of the < „ >>, or sometimes simply the moment. 

volume \ 

i force 

A statement of the first moment of a -s „__„ \- must 

I area 

\ volume 

f force units X length units. 

consist of the product of \ "^^^^ "'?[*« ></^'^g* "'?''^- 
'^ I area units X length units. 

\ volume units X length units. 

In speaking of moments, we shall always put the units of 
force, etc., first, and the length units afterwards. For example, 
we shall speak of a moment as so many pounds-feet or tons- 
inches, to avoid confusion with work units. 



52 



Mechanics applied to Engineering. 



/force (/) \ 

Second Moments. — The product of a -; ^g^ (j^ \ ^^ 

(^volume (v)) 
the squarq or second power of the length {I) of its arm, viz. 
(fl\~\ (force I 

^^f ( , is termed the second moment of the } ^^^* \ . The 

^vP J (volume) 

second moment of a volume or an area is sometimes termed 
the "moment of inertia" (see p. 78) of the volume or area. 
Strictly, this term should only be used when dealing with 
questions involving the inertia of bodies ; but in other cases, 
where the second moment has nothing whatever to do with 
inertia, the term " second moment " is preferable. 

C force \ 

A statement of the second moment of a < ™*®^ > must 

1 area ( 

I volume I 
( force units X (length units)'! 
,1 mass units X (length units)*, 
consist of the product of < ^rea units X (length units)". 

\ volume imits x (length units)'. 



First Moments. 



Levers. 

<r-ljr->^ — ^- ■ 



*S «5 -"i 



Fig. 60. 



T' 



Fig. St. 



Cloclcwise moments 
about the point a. 



Contra-clockwise moments 
about the point a. 






lUjt + wj. 



= a'iA 



= a/,4 



Moments. 



53 



Reactidh R at fulcrum tf, 

z.f. the resultant of all 

the forces acting 

on lever. 


Remarks. 


o'l+w.+a'a+a'. 


To save confusion in the diagrams, the / has in 
some cases been omitted. In every case the sufBx 
of / indicates the distance of the weight w bearing 
the same suffix from the fulcrum. 


w^—w^—w. 





i. (_ 



54 



Mechanics applied to Engineering. 



1(3 ITS 



rr 



Ai^ 



Fig. 62. 



<■—- ij"»* -- ? -» 



-«? 



i^-f-.: 



li'lG. 63. 



rSnnnnnoo 



Clockwise moments 
about the point a. 



W-or 

2 2 

If w = dis- 
tributed load 
per unit length, 
w/= W 



W-or — 
2 2 



W/ 

W = weight of 
long arm of 
lever 

W, = weight of 



Contra-clockwise moments 



r 



about the point a. 



=wJi+w.J.i+wJi 



= w^l^ 




= Wi^ + wA 

or — i — I- Wi/i 

2 



= W,/,+w/„-|-a:'3/, 

/= distance of c. of 

g. of long arm 

from a 
/i = distance of c. 

of g. of short 

arm from a 



= P/ 

/i = distance of c. 
of g. of lever 
from a 



orl 
P^J_ 

is 



Reaction R at fulcrum a, 

i.e. the resultant of all 

the forces acting 

on lever. 


Moments. 55 
Remarks. 


TO, + K'j — JOj + TOj — 01/5 




H/, + W 


N.B.— The unit length for w must be of the 
same kind as the length units of the lever. 


Wi+W.+W 




Wa + W.+a/j+W 


This is the arrangement of the lever of the 
Buckton testing machine. Instead of using a huge 
balance weight on the short arm, the travelling 
weight OTj has a contra-clockwise moment when the 
lever is balanced, and the load on the specimen, 
viz. KI3, is zero. As w, moves along its moment 
is decreased, and consequently the load w, is 
increased. When a/, passes over the fulcrum, its 
moment is clockwise ; then we have W/ x wj„ 
= W,/, + w,l,. 


W+a;,-P 


This is the arrangement of an ordinary lever 
safety-valve, where P is the pressure on the valve. 




The weight of the levers may be taken into 
account by the method already shown. 



S6 



Medianics applied to Engineering. 




Fic. 68. 



Clockwise moments 
about the point a. 



ii'Ji 



Contra- clockwise 
moni«?nL5 



r 



about the point a. 



= Wji + Wji 



■w4i + w/s 



= Wj/i + w/. 




W/ + w4^ 

W = weight 
of horizon- 
tal arm 



/ = distance of c. 
of g. from a 



Fig. 70. 






tia 
a 




W2= weight of 
long curved 
arm of lever 

Wi = ditto 
short arm 



= W.A + uuU 

12= distance of c. of 
g. of long arm 
from a 
li = ditto short arm 
/<= perpendicular 
distance of the 
line of w^ from a 



W = weight of 
body 



I = perpendicular 
distance of force 
W from a 



Moments. 



Reaction R. 



Remarks. 



57 



■ In all 
these cases 
it must be 
found by 
the paral- 
lelogram 
of forces. 



It should be noticed that the direction of the 
resultant R varies with the position of the weights ; 
hence, if a bell-crank lever be fitted with a knife- 
edge, and the weights travel along, as in some 
types of testing-machines, the resultant passes 
through the knife-edge, but not always normal to the 
seating, thus causing it to chimble away, or to 
damage its fine edge. 




Fig. 68a. 



The shape of the lever makes no difference whatever to the 
! leverage. 

Consider each force as acting through a cord wrapped round 
the pulleys as shown, then it will be seen that the moment of 
each force is the product of the force and the radius of the 
pulley from which the cord proceeds, i.e. the perpendicular 
distance of the line of action of the force from the fulcrum. 



58 



Mechanics applied to Engineering. 



Centres of Gravity, and Centroids. — We have already 
given the following definition of the centre of gravity (see p. 13). 
If a point be so chosen in a body that the sum of the moments 
of all the gravitational forces acting on the several particles 
about the one side of any straight line passing through that 
point, be equal to the sum of the moments on the other side of 
the line, that point is termed the centre of gravity ; or if the 
moments on the one side of the line be termed positive ( + ), 
and the moments on the other side of the line be termed 
negative ( — ), the sum of the moments will be zero. 

From this definition it will be seen that, as the particles of 
any body are acted upon by a system of parallel forces, viz. 



.1. 



"■^S- 



■Jiof. 



w. 



<> 



gravity acting upon each, the 
algebraic sum of the moments 
of these forces about a line 
must be zero when that line 
passes through the c. of g. of 
the body. 

Let the weights Wi, Wj, 

^'G' 72- be attached, as shown, to a 

balanced rod — we need not consider the rod itself, as it is 

balanced — then, by our definition of the c. of g., we have 

W,L, = W,Lj. 

In finding the position of the c. of g., it will be more 
convenient to take moments about another point, say x, 
distant /j and ^ from Wj and Wj respectively, and distant /, 
(at present unknown) from the c. of g. 

Wi4 + WaLa= R/, 

= (W, + W,)/, 
. _ W/i + Wa4 
' W, + W, 

If we are dealing with a thin sheet of uniform thickness 
and weighing K pounds per unit of area, the weight of any 
given portion will be K« pounds. Then we may put Wi = Kai, 
and W, = Kfflj ; 

J / _ K(gi/i + aa4) _ g/i + aj^ 
'"'^ - K(«. + a,) A— 

or, expressed in words — 

distance of c. of g. from the point x 

the sum of the moments of all elemental su rfaces about x 
area of surface 



or = 



Moments. 59 

the moment of surface about x 
area of surface 



where A = «i + a^ = whole area. 

In an actual case there will, of course, be a great number 
of elemental areas, a, a^, a^, a^, etc., with their corresponding 
arms, /, /j, 4> 4> etc. Only two have been taken above, they 
being sufficient to show the principle involved. 

When dealing with a body at rest, we may consider its 
whole mass as being concentrated at its centre of gravity. 

When speaking of the c. of g. of a thin weightless lamina 
or a geometrical surface, it is better to use the term " centroid " 
instead of centre of gravity. 



Position of Centre of Gravity, or Centroid. 
Parallelograms. 
Intersection of diagonals. 

Height above base ab = — 
2 



In a symmetrical figure it is evident that the c. of g. lies on 
the axis of symmetry. A parallelogram has two axes of 
symmetry, viz. the diagonals ; hence the c. of g. lies on each, 
and therefore at their intersection, and as they bisect one 

another, the intersection is at a height — from the base. 







F 


a- i 

IG. 73. 



6o 



Mechanics applied to Engineering. 



Triangle. 




H I 



Fig. 74. 



Intersection of ae and bd, where d 
and e are the middle points of ac and 
be respectively. 

Height above base be = — 

,. • . V. 2H 

height above apex a = ^ — 



Triangle. 
a 




Distance of c. of g. from b 

« + S 



= ^/=: 



Ditto from e = ef = 



3 

z + S 



Fig, 74a. 



Trapezium. 



<?<-;:v5r;;.::^ * 




Intersection of a3 and ed, 
where a and b are the middle 
points of S and Si, and ed = Si, 
/^ = S. 

J Height above) _ H(2S + SQ 
base Hi 5 3(5 + sj 

depth below) _ H(S + 2S1) 
top H, ] 3(s + Si) 



'I 
Moments. 6i 

Conceive the triangle divided up into a great number of 
very narrow strips parallel to one of the sides, viz. be. It is 
evident that the c. of g. of each strip will be at the middle 
points of each, and therefore will lie on a line drawn from the 
opposite angle point a to the middle point of the side e, i.e. on 
ae; likewise it will lie on bd; therefore the c. of g. is at the 
intersection of ae and bd, viz. g. 

Join de. Then by construction ad= dc = — , and be — ec 

2 

= - ; hence the triangles acb and dee are similar, and therefore 

2 

de = —. The triangles agb and dge are also similar, hence 

ag ae 
eg= ^=—. 

^. 3 

Since g is situated at 5 height from the base, we have 

2/S 



jr = - t.e. = -( x] 

3 3\2 / 

t/+x = b/= 



Draw the dotted line parallel to the sloping side of the 
trapezium in Fig. 75*. 

Height of c. of g. of figure from base 

_TT _ area of parallg. x ht. of its c. of g. + area of A x ht. of its c. of g. 
' area of whole figure 

SHxH , ,„ „>H H 

/S + S,w ■ 3(S + S,) 

SH X H . /„ (j,H 2H 

-^— + (b. - S)- X — ^ ^^g ^ ^g^ ^ 



(S±S)„ 



3(S + S,) 



^1 = gS + Si ^ I 

H, 2S, + S s, +§ 



< — s 


\k 


H 


V 


■ ' \ 


V 



or^=— ' *-^'-'- 

ga ad ■ ^'o- 7S«. 



62 Mechanics applied to Engineering. 

Position of Centre of Gravity, or Ckntroid. 



Ti \i 



,..->Ij:;^-:;;-- 




Area dbe = Ai 
ice = Aj 
c^e = A, 

c. of g. of area a&e Q 

J) JJ i^CC U2 

„ „ whole fig- 
ure C1.J.S. 



Trapezium and triangles. 

Intersection of line joining c. of g. 

„.(; of triangle and c of g. of trapeziunij 
viz. ab and cd, where ac = area of 
trapezium, and db area of triangle, ac 

\ is parallel to bd. 



Fig. 77. 



r 



Lamina with hole. 



Let A = area abcde; 

H = height of its c. of g. from 

ed; 

Hi = height of its c. of g. from 

<Ci'' drawn at right angles 

to ed; 

a = area of hole g/i; 

h = height of its c. of g. from 

ed; 

hi = height of its c. of g. from 

d/. 

Then— 

Hf = height of c. of g. of whole figure from ed 

K'c = height of c. of g. of whole figure from df 

Tjr AH — ah 

Kc=—r- 

A — a 

HV = • '^■^1 ~ ^^ 




Fig. 78. 



Moments. 



63 



The principle of these graphic methods is as follows : — 
Let the centres of gravity of two areas, Aj and Aj, be 

situated at points Q and Ca ^ 

respectively, and let the common 

centre of gravity be situated at 

c, distant «, from Ci, and X2 from 



•■^■■ 



'A/ 
Caj then we shall have Ai*, 

=Aa«a- From Ca set off a line 
cj)^, whose length represents on 
some given scale the area Aj, 
and from Ci a line c^b^^ parallel 
to it, whose length represents 
on the same scale the area Aj. 
Join bi, b^ Then the intersec- 
tion of bj>i and QCa is the 
common centre of gravity c. 
The two triangles are simikr, therefore — 

~r = ~Z^) '^^ "-i^i ~ ■'Vs^a < X, 

Ai *i ^ X, >c, ' 

N.B. — The lines C,*, arid C^b, \^ /J' 
are set off on opposite sides of ^ / ^ 

C„ Cj, and at opposite ends to their 
respective areas, at any convenient 
angle ; but it is undesirable to have 
a very acute angle at c, otherwise 
the point will not be well defined. 
When one of the areas, say Aj, is 
negative, i.e. is the area of a hole 
or a part cut out of a lamina, then 
the lines f ,i, and c^b, must be set off 
on the same side of the line, thus — 

Then— 



'Az 




Fig. 76a. 






ai" 


■^ 


'S 


<—a>,- — ■, 


A, 

f 




i 



k'^^ 



-r-' = ^, or AiJCi = A^j 



Fig. yti. 




64 Mechanics applied to Engineering. 

Position of Centre of Gravity, or Centroid. 

Graphical method. 

Lamina with hole. j^ ^^ ^^ ^^^ ^_ ^f g_ „f ^3^^, . 

^2 >. ). SJ^- . 

Join ^1, ^2, and produce; 
set off C2K2 and fjKi parallel 
to one another and equal to 
A and a respectively; through 
the end points K5, Kj draw a 
line to meet the line through 
^1, (Tj in "^1.21 which is the c. of 



Fig. 70. 



■■■^'■-s 



g. of the whole figure. 



Note.— The lines c^Ks, f,K, need not be at right angles to the line 
^,fj, but the line KjK, should not cut it at a very acute angle. 



Portion of a regular polygon or an arc of a circle, considered as 
a thin wire. 

Let A = length of the sides of the poly- 
gon, or the length of the arc 
in the case of a circle ; 
R = radius of a circle inscribed in 
the polygon, or the radius of 
the circle itself; 
C„ = chord of the arc of the polygon 

or circle ; 
Y = distance of the c. of g. from 
the centre of the circle. 



Then A : R 
R Y 





or Y = 



RC. 



N.B. — The same expression holds for an arc greater than a semicircle. 



Moments. 



65 



See p. 63. 




Fig. 8a>>. 



Regard each side of the polygon as a piece of wire of 
length I; the c. of g. of each side will be at the middle point, 
and distant _j'i,j'2,_)'3, etc., from 
the diameter of the inscribed 
circle; and let the projected f>y 

length of each side on the ^j/'' 
diameter be c-i, c,, c,, etc. // 

The triangles def and Oba Ui;^ . 
are similar ; .' /{ 

,^ , fd Oa / R M-- 
therefore-V = -^, or - = — ^■ 
Je ab C-, y^ 

and J'l = — 

likewise y^ = -j, and so on 

Let Y = distance of c. of g. of portion of polygon from the 
centre O ; 
■w = weight of each side of the polygon. 
Then — 

y ^ wyi + wy.i+, etc. 
wn 
where n = number of sides. The w Cancels top and bottom. 
Substituting the values oiyi,yi, etc., found above, we have— 

Y = -k, + c,+, etc.) 
nl 

(Proof concluded on p. i>i^ 




66 Mechanics applied to Engineering. 

Position of Centre of Gravity or Centroid. 
Semicircular arc or wire. 

9\ \ ^ 

i I 
-VLL 

cs^i^ ' 

Fig. 8z. 

Circular sector considered as a thin sheet. 
..A 

! ^^^ 

^\c.(fq.of/^ V — 'RC„ 

Fig. 82. 

Semicircular lamina or sheet 

"-■ofQ- ofs^etr « 4R 2D 

_^ \ 3ir 3T 

^--CgZR > 

Fig. 83. 

Parabolic segment. 

ri \i V = fH 

where Y = distance of c. of g. from apex. 
'\c.fy. }A The figure being symmetrical, the c. of 

•,1 i \ S- li^s °^ ^^^ 3.xis. 

Fig. 84. 






Moments. 



67 



but Ci + ^2 +1 etc. = the whole chord subtended by the sides of 
the polygon 
= C, 
and nl = A. 

RC, 
A 

When n becomes infinitely great, the polygon becomes a 
circle. 

The axis of symmetry on which the c. of g. lies is a line 
drawn from the centre of the circle at right angles to the chord. 



Y = 



In the case of the sector of a polygon or a circle, we have 
to find the c. of g. of a series of triangles, instead of their 
bases, which, as we have shown before, is situated at a distance 
equal to two-thirds of their height from the apex ; hence the 
c. of g. of a sector is situated at a distance = |Y from the 
centre of the inscribed circle. 



From the properties of the parabola, we hav 



h_ 
H 



' = ¥* 



B>% 



Area of strip = b .dk = — -j dh 
rl 




{Continued on p. 69.) 



68 Mechanics applied to Engineering. 

Position of Centre of Gravity or Centroid, 



Parabolic segment. 




Y = f H (see above) 
where Y, = distance of c. of g. from axis. 



Fig. 8j. 



Moments. 
moment of strip about apex = 



69 



h.b .dh = — idh 



moment of whole figure about apex = — f^^fi — — r X -«- 

H* J o H' 7 



= fBH2 



the area of the figure = |BH (see p. 30) 



the dist. of the c. of g. from the apex : 



|BIP 
fBH 



= |H 



From the properties of the parabola, we have- 





h 


p 






H 


B" 




orM 


H 


b^ 
B^ 




B"(H 


-K) 


= ^''H 






B^y^o 


= B2H- 


tm 




h. 


B»^ 


-b-^) 



Apea 




area of strip = h^db 
moment of strip about axis = b . h^b = ^^ (S'b — P)db 

= 5 J (B'6-i')db 

H rBV _ ^n B 

B^L 2 Jo 

= H/'B* _ B^\ ^ HI 
B\ 2 4 / 4 



moment of whole area) 
about axis ) 



the area of the figure = #BH 



a-" 
HB^ 



the distance of the c. of g. )_ 4 _ 3 „ 
from the axis ) 2r>-H ' 



^o 



Mechanics applied to Engineering. 




^Apeoc 



Position of Centre of Gravity or Centroid. 
Ex-parabolic segment. 

< B 

Y = AH 

where Y, = distance of c. of g. from 
axis; 
Y = distance of c. of g. from 
apex. 

Fig. 86. 



Irregular figure. 




Y = 



Mhy +3^+5 >^8+7^4+, etc.) 
2('4i+'^2+'4,+.«i+, etc.) 



where Y = distance of c. of g. from 
line AB ; 
111 = width of strips ; 
A = mean height of the 
strips. 



Y. = '^{ ^'i+3^„+5^'b+7A\+, etc. x 
2 V li\+k\+A',+/i',+, etc. J 

where Y„ = distance of c. of g. from 
Une CD ; 
lel = width of strips ; 
A' = mean height of the strips. 



Moments. 



71 



By the principle of moments, we have — 
Distance of c. of g. of figure from axis 

f area of rect. X (^ist of its c. of g.| _|area ofpara.| >< fdist. of its c. of g. \ 
_\ I. from axis J I segment ) \ from axis / 



Yi = 



Likewise — 



area of figure 

BH X ? - |BH X fB 
H 



fB 



BH X - - |BH X f H 

Y = ^ = -5-H 

iBH 



This is a simple case of moments, in which we have — 
Distance of c. of g.^ _ moment of each strip about AB 

area of whole figure 



from line AB 



Moment of first strip= w^, x — 

2 

„ second „ ^wh^y.^ 

2 

,, third „ =a/^x^- 

2 



: wh-i, + whi + w^s +, etc. 



The area of the first strip = wht, 
„ „ second „ =whi 

„ „ third „ =wht 

and so on. 

Area of whole figure 
Distance of c. j w/^i X - + wA, X ^ + wA, X 5^ +,etc. 

of g. frninl_Y_ ^ ' , 2 

line AB J whx + wh^ + wh^ +, etc. 

one w cancels out top and bottom, and we have — 
Y ^-w / K + 3^2 + 5^'3 + ih +, etc. 
2 V h-i-{-K-\- hi-\- hi-V, etc. 



and similarly with Yj. 
The division of 
the figure may be 
done thus : Draw a 
Une, xy, at any angle, 
and set oif equal parts 
as shown; project the 
first, third, fifth, etc., 
on to xz drawn normal 
to AB. 




Fig. 87a. 




Mechanics applied to Engineering. 



Position of Centre of Gravity 
OR Centroid. 

Wedge. 

On a plane midway between the ends, 

and at a height — from base. 
3 
For frustum of wedge, see Trapezium. 



Pyramid or cone. 




On a line drawn from the middle. point of 
the base to the apex, and at a distance f H 
from the apex. 



Fig. Bg. 



Frusitim of pyramid or cone. 
t .f,4P«« 




On a line drawn from the middle point 

of the base to the apex, and at a height 

/I — f&\ H 

|H( ; 1 from the apex, where « = _- 1 

* V I - ;;V H 



or ^ 



Moments. 



73 



A wedge may be considered as a large number of triangular 
laminae placed side by side, the c. of g. of each being situated 

at a height — from .the base. 



Volume of layer = b'^ . dh 
moment of layer about apex = b'^ . k . dh 



But ^-=1 
h B. 



b = 



h. B 

H 
B^/JV 



moment of layer about apex = -^^dh 




Yia. igtt. 



moment of the whole pyramid! _ B^ I ,3 ., B'H^ B^H^ 
about apex J" iP J o 4H» " 4 



volume of pyramid 



B'H 

3 
B^H^ 



distance of c. of g. from apex = „^„ - 3 ' 



In the case above, instead of integrating between the limits 
of H and o for the moment about the apex, we must integrate 
between the limits H and Hi ; thus — 

Moment of frustum of pyramid) _B^ f ^ .3 

about the (imaginary) apex S H'' „ 

J Hi 



_ F /- ff HiM 
- ffV 4 ~ 4 ) • 

volume of frustum = 



Bi 



4 4 
B^H Bi'Hi 

3 3 

Hi 



(i.) 
(ii.) 



substituting the value -^=^ = n 

then the distance of the c. of g-\ ilj ^ ag /' '"^'^ ~\ 
from the apex / (ii.) * Vi-«^> 



Fio. 91, 



74 Mechanics applied to Engineering. 

Position of Centre of Gravity or Centroid. 
Locomotive or other symmetrical body. 

The height of the c. of g. 
above the rails can be found 
graphically, after calculating 
a;,byerecting a perpendicular 
to cut' the centre line. 

W, the weight of the 
engine, is found by weighing 
it in the ordinary way. Wa 
is found by tilting the engine 
as shown, with one set of 
wheels resting on blocks on 
the platform of a weighing 
machine, and the other set 
resting on the ground. 

Let hi be the height of 
the c. of g. above the rails. 

Fig. 92. 




Irregular surfaces. 

Also see Barker's " Graphical Calculus," p. 179, for a 
graphical integration of irregular surfaces. 



Moments. 
By taking moments about the lower rail, we have- 

But - = I- 
x^ G 

whence ~%^ = Wjj- 



75 



«i =■ 



W,G 
W 



y = — - ^1 

2 



G 

2 



By similar triangles — 
y A 



K 



W,G 
W 



^ 
^ 



^ 



/4, = 



h 



(These symbols refer to Fig. 92 only.) 




The c. of g. is easily found by balancing methods ; thus, 
if the c. of g. of an irregular surface be required, cut out the 
required figure in thin sheet metal 
or cardboard, and balance on the 
edge of a steel straight-edge, thus : 
The points a, a and b, b are 
marked and afterwards joined: 
the point where they cut is the 
c. of g. As a check on the 
result, it is well to balance about 
a third line cc; the three lines 
should intersect at one point, and not form a small triangle. 

The c. of g. of many solids can also be found in a similar 
manner, or by suspending them by means of a wire, and 
dropping a perpendicular through the points of suspension. 

Second Moments — Moments of Inertia. 

A definition of a second moment has been given on p. 52. 
In every case we shall find the second moment by summing up 



Fig. 93. 



76 



Mechanics applied to Engineering. 



or integrating the product of every element of the body or 
surface by the square of its distance from the axis in question. 
In some cases we shall find it convenient to make use of the 
following theorems : — 

Let lo = the second moment, or moment of inertia, of any 
surface (treated as a thin lamina) or body about 
a given axis ; 
I = the second moment, or moment of inertia, of any 
surface (treated as a thin lamina) or body about 
a parallel axis passing through the c. of g. ; 
M = mass of the body ; 
A = area of the surface ; 
Ro = the perpendicular distance between the two axes. 

Then I„ = I + MR„2, or 1 + ARo" 

Let xy be the axis passing through the c. of g. 

Let «;yi be the axis of 
revolution, parallel to xy and 
in the plane of the surface or 
lamina. 

Let the elemental areas, 
<hi (h., thi etc., be situated at 
distances r„ r,, r„ etc., from 
xy. Then we have — 
I„ = ai(R„ + ri)2 + «,(R, 
+ ;j)-''+,etc. 

= c^(R,' + r,' + 2R,rO 

+, etc. 
= ajTi" + a.f.^ +, etc. 
+ R„'(a,+a,+,etc.) 
+ 2R|,(airi + tVa 
+ , etc.) 

But as xy passes through the c. of^. of the section, we 
have (hr■^ + cv^ +i etc. = o (see p. 58), for some r's 
are positive and some negative ; hence the latter term 
vanishes. 

The second term, ffj + <?, +, etc. = the whole area (A) ; 
whence it becomes Ro'^A. 

In the first term, we have simply the second moment, ot 
moment of inertia, about the axis passing through the c. of g. 
= I ; hence we get — 

I„ = 1 + R.2A 




Moments. 



77 



We may, of course, substitute Wj, m^, etc., for the elemental 
masses, and M for the mass of the body instead of A. 

When a body or surface (treated as a thin lamina) revolves 
about an axis or pole perpendicular to its plane of re/olution, 
the second moment, 
or moment of in- 
ertia, is termed the 
second polar mo- 
ment, or polar mo- 
ment of inertia. 

The second polar 
moment of any sur- 
face is the sum of 
the second moments 
about any two rect- 
angular axes in its 
own plane passing 
through the axis of 
revolution, or \^ 

Ip = I. + Iv 

Consider any ele- 
mental area a, distant 
r from the pole. 

I, about ox = ay'' 

I, » oy = ax^ 

Tp „ pole = ar- 

But r'=«2+y 

and ar'^=ax'^+ay 

hence I,= I,-f I, 

In a similar way, it 

may be proved for every element of the surface. 

When finding the position of the c. of g., we had the following 
relation : — 




Distance of c. of g. from 
the axis xy (k) 



first moment of surface about x j 
) area of surface 

_ first moment of body about xy 
volume of body 

lar 



78 



Mechanics applied to Engineering. 



Now, when dealing with the second moment, we have a 
corresponding centre, termed the centre of gyration, at which 
the whole of a moving body or surface may be considered to 
be concentrated ; the distance of the centre of gyration from 
the axis of revolution is termed the " radius of gyration." 
When finding its value, we have the following relation : — 

Radius of gyrationV second moment of surface about xy 
about the axis xy (k?) J = area of surface 

second moment of body about xy 

volume of body 
2^ I 
A 



or = 



^ = ' 



- = "X. or I = Ak' 



Second Moment, or Moment of Inertia (I). 

Parcdklogram treated as a thin lamina about 
its extreme end. 



O 



I.= 



BIP 



K M 

o 

o 




If the figure be a 
square — 

B = H = S 
we have I,, = — 



Fio. g6. 



Radius of gyration 



H 



Moments, 



79 



For other cases of moments of inertia or second moments, 
see Chapter IX., Beams. 



Area of elemental strip = "& . dh 
second moment of strip = 'R.h'' .dh 



second moment 
of whole surface 



!-// 



Ifl.dh = 



BH^ 



-/I. 



5! 

3 



area of whole | _ -dtt 
surface ) ~ 
square of radius 1 _ BH^ 
of gyration ) ~ 3BH ' 

radius of gyration = —p^ 

It will be seen that the above reasoning holds, however 
the parallelogram may be distorted sideways, as shown. 



<^ 



-M 



Fig. 96a, 



8o Mechanics applied to Engineering. 

Second Moment, or Moment of 
Inertia (I). 





o 


< 


-//■■:> 




O 



\ Parallelogram treated as a 
jj thin lamina about its 
; central axis. 




J BH» S* 
I = or — 

12 12 



Radius of gyration 



H 

\/l2 



< // -> 



\c.cfp.B 



Parallelogram treated as 
a thin lamina about an 
axis distant 'R^from its 
c. ofg. 



Ri;-i 



H 



■^oi 



Io = Bh(^' + R„') 



o 



Fig. gS. 






Moments. &i 



This is simply a case of two parallelograms such as the 

XT 

above put together axis to axis, each of length — ; 

\2 ) BH^ 
Then the second moment of each = — 



3 8X3 

then the second moment "i _ /BH°\ BH'' 

of the two together / \ 24 / 12 
area of whole surface = BH 



radius of gyration = k/ - 



BH' H 



2BH- V77 



From the theorem given above (p. 76), we have- 
I, = [ + Ro'A I = ^ 



^^+R/BH A = BH 

12 " 



= BH(!i + R/j 
when Ro = o, lo = I 



L L- 



82 Mechanics applied to Engineering. 

Secoxd Moment, or Moment of Inertia (I). 







Fig. gg. 



Hollow parallelogi-am. 

Let le = second moment of 
external figure ; 

\i = second moment of 
internal figure ; 

lo = second moment of 
hollow figure ; 

lo = I. - I.. 



Radius of gyration 



Triangle about an axis parallel to the base 
passing through the apex. 

O 




Io = 



BH3 



H 



Triangle about an axis parallel to the base 
passing through the c. of g. 




I = 



BH3 
36 



H 
Vil 



Moments. 



83 



The lo for the hollow parallelogram is simply the difference 
between the I, for the external, and the Ij for the internal 
parallelogram. 



Area of strip = b.dh; but * = ^ 


„ =\h.dh 

second moment of strip = — /z^ . dh 
rl 


^ 3-^ 

<:^'"l ^ 


A 

i? 


B f^ 
„ triangle = g h^-dh 

J 


<- ^ -> 

< M- 


> 


r 

V 


BH* BH^ 
" - 4H 4 

area of triangle = 




Fig. Tooa. 


/BH» 

radius of gyration = / —2— = \y — = -^ 
/ BH 2 V 2 


From the theorem on p. 76, we'have — 


I„ = I + R.^'A I^ _ BH^" 

I-I„_R„.A R„._('Hy=4H^ 

J _ BH3 _ 4H2 ^ BH ;^ _ BH ' 
492 2 
_ BH3 

36 

/bh» 

radius of gyration = / _2£«=x/— s= -^= 
/ BH ^ 18 ^,8 


V 



84 Mechanics applied to Engineering. 



Second Moment, or Moment of Inertia (I). 




O Triangle about an axis 
at the base. 



_BH' 

12 



FtG. T02. 



Radius of gyration 



H 
^6 



Trapezium about an axis coinciding with its 
short base. 
O 



4 



<--- H 



Io= 



(3B + BQH' 



Let Bi= «B. 



V 6(« + 



3_ 
6(« + i) 



^ 



Fig. 103. 



Trapezium about an axis coinciding with its 
long base, 





-H 






I.,= 



(3B, + B)H ' 
12 

or-77(3«+ i) 



Fig. ro4. 



H 



/ 3 "+ I 
V 6(« 4- i) 



Moments. 



85 



From the theorem quoted above, we have- 



I„ = 1 + Ro'A 
BIT W 
36 + 9 
I -BH' 



R„^ = 



H" 



I« = 



BH 



Radius of gyration obtained as in the last case. 



This figure may be treated as a parallelogram and a triangle 
about an axis passing through the apex. 

BiH' • 
. For parallelogram, lo = — — 



for triangle, I,, = 



(B - BQH' 



B.ff (B-B.)H ' 
for trapezmm, lo — • — - — + 7 



I„ 



(3E + BQff 




Fig. 103a. 



When the axis coincides with the long base, the I for the 

/■D "D \TT3 

triangle = ^^ — ; then, adding the I for the parallelogram 

as above, we get the result as given. 

When « = r, the figures become parallelograms, and 

I = , as found above, 

^ , r BH' 

When « = o, the figures become triangles, and the I = — — 

for the first case, as found for the triangle about its apex ; and 

I = for the second case, as found for the triangle about 

12 

its base. 



86 Mechanics applied to Engineering. 

Second Moment, or Moment of Inertia (I). 

Radius ol gyration 

ma parallel wtm cne oase. 
O 




, _ (B.' + 4B.B + B°)H3 
36(B. + B) 
^^BIP/«Mm«jLi\ 
36 \ « + I ) 
Bi 

For a close approximation, 
see next figure. 



Approximate method for trapezium about axis 
passing through c. of g. 

The I for dotted rectangle 

about an axis passing through 

its c. of g., is approximately 

C^ the same as the I for trapezium. 

For dotted rectangle — 

J _ (B + B.)H3 
orif B, = «B 




Fig. 106. 



I = (« + i) 

24 



Moments. 



87 



From the theorem on p. 76, we have- 
I = I„ - R,2A 



l^ ^ (3B1 + B)ff ^^^^^^ i^^g 
^'^ base) 



Substituting the values in the above equation and simplify- 
ing, we get the result as given. The working out is simple 
algebra, but too lengthy to give here. 



■D'XJ3 

The I for a rectangle is (see p. 80). Putting in the 



value 



5-+^ = B', we get- 



I = 



_ B + B. ^ H3 _ (B + B,)H3 



24 



The following table shows the error involved in the above 
assumption ; it will be seen that the error becomes serious 
when « < o"S : — 



Value 
of «. 


Approx. method, 


the correct 


value being i. 


°"9 


I 001 


0'8 


i'oo5 


07 


l-OII 


06 


I -021 


o-S 


I -039 


0-4 


I '065 


0-3 


no? 


0-2 


■ •174 



The approximate method always gives too high results. 



88 Mechanics applied to Engineering. 

Second Moment, or Moment of Inertia (I). 



Square about its diagonal. 




t2 



Radius of gyration 



s 

^12 




Circle about a diameter. 



I =: 



64 




Hollow circle about a dia- 
meter. 



D 
4 



VD.^+D.^ 



Moments. 



89 



This may' be taken as two triangles about their bases fsee 
p. 84). 



In this case, B = v' 2S 







--i, 


/\ 


-^ 




/ -JIS 


•■i|. 


12 


vS 


*/:' 


12 







area of figure = S^ 


Fig. ztyjct. 


/ S'' S 
radius of gyration — a/ j^S^ "~ J~ 





From the theorem on p. 77, we have I, = 1, + I,; in the 
circle, Ij = L. 



Then L= 2L 



irD* 



rD* 



and I, = f = 73^, = -^ (see Fig. 118). 



The I for the hollow circle is simply the difference between 
the I for the outer and inner circles. 



90 Mechanics applied to Engineering. 

Second Moment, or Moment of Inertia (I). 



Hollow eccentric circle about a line normal to 
the line joining the two centres, and passing 
through the c. of g. of the figure. 



where x is the eccentricity. 

Note. — When the eccentricity 
is zero, i.e. when the outer and 
inner circles are concentric, the 
latter term in the above expression 
vanishes, and the value of I is the 
y same as in the case given above 
for the hollow circle. 




Radius of gyration 



Ellipse about minor axis. 
o 




64 




Moments. 



91 



The axis 00 passes through the c. of g. of the figure, and 
is at a distance b from the centre of the outer circle, and a from 
the centre of the inner circle. 

From the principle of moments, we have — 



--QHb + «) = -V).^b 
4 4 



whence b — 



D,^ - W 



also--D,2(a-^) = -D,2a 
4 4 



whence a = 



BJ'x 



From the theorem on p. 76, we 




have- 

I', = I, + A/^ for the outer circle 
about the c. of g. of figure 

64 4 

also F, = I, + A,fl^ for the inner circle about the c. of g. of 
figure 

64 4 
and I = r. — I', for the whole figure. 

Substituting the values given above, and reducing, we get 
the expression given on the opposite page. 



The second moment, or moment of inertia, of a figure 
varies directly as its breadth taken parallel to the axis of 
revolution ; hence the I for an ellipse about its minor axis is 

simply the I for a circle of diameter Da reduced in the ratio =:-i 

D2 



or-D^xg' 
64 D, 



64 



92 Meclianics applied to Engineering. 


Second Moment, or Moment of Inertia (I). 


^ 


^ Ellipse about major axis. 


Radius of gyration 


f 

c- 


tA I - '^D/D' 


i^ 


V 


1 / ^^ 


4 


V 







Fig. 112. 




Parabola about its axis. 









V^ y^ 


ApeXf 




A 


1 \ 1 = ,^HB' 


B 
^5 


< -aS- > 











Fio. nj. 


1 



Moments. 



93 



And for an ellipse about its major axis, the I is that for 
a circle of diameter Di increased in the ratio — - 

or^*X°? = 3!^^ 
64 D, 64 



h = ^{\ - -^ (see p. 69). 



area of strip =: h .db 
second moment of strip = b"^ .h. db 



second moment of whole 
figure 



.3 SB^j o 
"F _ B'l 
.3 sj 



= H 

2HB 



15 



second moment for double"! 
figure shown on opposite > = AHB' 
page 







94 Mechanics applied to Engineering. 

Second Moment, or Moment of Inertia (I). 

Radius of gyration 



Parabola about its base. 





■jApejc I = Ji^BH^ 



Fig. Z14. 



V^H 



Irregular figures. 




4 
49*1 +. etc.) 



Fig. T15. 



(See opposite 
page.) 



b = 



Moments. 
B .h* 



95 



H* 



area of strip = b . dh 
second moment J ^^jj_ ^^,3 _^^ 



of strip 




{Yi.-hfBh\dh 



YC 



second moment of whole ■! 
figure / 



Fig. ii4ff. 

B fH 

= ^ {Wh* + h^ - 2h^B)dh 



B_ \ 2Wk^ 
H* }_ 3 






2K 



= B(|ff + f H» - 4H') 

18 T)XJ3 

for the double figure shown^ _ _32_t)tt8 
on opposite page ) ' "* 

Divide the figure up as shown in Fig. 87a. 

Let the areas of the strips be a^, a^, a^, a^, etc., respectively ; 
and their mean distances from the axis be r^, r^, r,, r^, etc., 
respectively. 

Then Iq = a-^r^ + a^i + a^^ +, etc. 
But fli = wb^, and a^ = wb^, and so on 

and n =— , ^2 = -^, ^3 = ^— , and so on 
2 2 2 

hencel. = «.{^0 + <fj + .3(?j+,et.c.} 



4 



{bi + 9*a + 25*3 + 49^4 +, etc.) 



Also k' 



— (^ + 9*2 + 25^3 + 49^4 +, etc.) 

ii>{bi + b^ + bs + bi+, etc. 
^ ^ / ^1 + 9/^2 + 25<^3 +, etcT 



2 V ^j 4. ^^ + ^3 +^ etc. 

This expression should be compared with that obtained for 

finding the position of the c. of g. on p. 71. Tke comparison 

helps one to realize the relation between the first and second 

moments. 



96 Mechanics applied to Engineering. 


Second Moment, or Moment of Inertia (I). 




^ 1 Graphic method. 


Radius 0^ gyration 


c 


<- Af -- 

\ ^ 

<- Y^ 
9 

Fig. ii( 


1 

5J 

i Let A = shaded area ; 

N^..,-.-J»/ Y = distance of c. 

imv ' ofg. of shaded 

\ ; S areafromOO; 

1 |i|^ H = extreme di- 

; 1 j ® mension of 

1 J figure mea- 

ilii '^ sured normal 

\^\ to 00 ; 

° \ ' Then I„ = AYH 




Second 

Paraildog 
c. ofg. 


Polar Moments of Surfaces or T 
rram about a pole passing through its 

I. = ??(H^ + B^) 


hin Laminae. 






,/«•+- 




/ 


n 


/ 


/ 1 


M square of side S — 


S 


■< 


-H- 

Fig 




117. 


i=si 

' 6 


V^ 


Circle about a pole passing through the c. of g. 





^ 


y I, = — . or 


D 

a/8" 
or ^ 




Fig. ii8. 





Moments, 



97 



Divi.le the figure up into a number of strips, as shown in 
Fig. ii6j project each on to the base-line, e.g. ab projected to 
fli^ij join «i and b^ to c, some convenient point on 00, cutting 
ab in a^^, and so on with the other lines, which when joined 
up give the boundary of the shaded figure. Find the c. of g. of 
shaded figure (by cutting out in cardboard and balancing). The 
principle of this construction is fully explained in Chap. IX., 
p. 360. 

See also Barker's "Graphical Calculus," p. 184, and Line- 
ham's " Text-book of Mechanical Engineering," Appendix. 



From the theorem on p. 77, we have — 



I, = I, + I» = ^+4f 



BH" , HB' 
12 12 



\y 



12 






Fig. Tiya. 



I 
4 



Thickness of ring = dr 

area of ring = zirr . dr 
second moment of ring = 2irr .r^ . dr 

fR 

circle = 2ir \ r^ . dr 



= 2ir I f^ . 



(«R-e) 



2«-R^_ 
4 

2 X 16 



2 




98 Mechanics applied to Engineering. 

Second Polar Moment, or Polar Moment of Inertia. 

ff Hollow circle about a pole 

passing through the c. 
of g. a7td normal to the 
plane. 




I, = ^(D.* - D<^) 



Radius of gyration 



V 8~" 



or 



/ R.' + R.' 



Second (Polar) Moments of Solids. 

Gravitational Units. 

Bar of rectangular section about a pole passing 
through its c. of g. 



, ^1~~ 



O 

)f IG. I20. 



B"! For a circular bar of 
radius R — 

12 ^ 



L'' + B^ 



V'^ 



si- 



L' + 3R' 

12 



Cylinder about its axis, 
o 



■Di- 



<- -R--* 



: T ttD^HW irR^HW 

H I»= — > or 

■ 32 ^ 2 g 





Fig. 191. 




Moments, 99 



The Ij, for the hollow circle is simply the difference between 
the L for the outer and the L for the inner circles. 



The bar may be regarded as being made up of a great 
number of thin laminae of rectangular form, of length L and 
breadth B, revolving about their polar axis, the radius of 

/\? + B^ 
gyration of each being K = a/ — — — (see Fig. 1 1 7), which 

is the radius of gyration of the bar. The second moment of the 

LBH W 

bar will then be K^ (weight of bar), or -^fl{U + ^l- 

Where W is the weight of 1 cubic inch or foot of the 
material, according to the units chosen. 



The cylinder may be regarded as being made up of a great 
number of thin circular lamins revolving about a pole passing 
through their centre, the radius of gyration of each being 

The second moment of cylinder = K^ (weight of cylinder) 

D^ irD^HW ttD^HW 

= - X = 

8 4 ^ 32 ^ 



lOO 



Mechanics applied to Engineering. 



<■■ R. 



Second Polar Moment, or Polar 
Moment of Inertia. 



g— V-/Pf--> ^ Hollow cylinder about Radius of gyration 




M 



I. 



I a pole passing ihi-ottgh 

I its axis. 
H 



or —(R/ - R/)- 



or 



Disc flywheel. 



i 



Treat each part sepa- 
rately as hollow cylinders, 
and add the results. 







1 


1 







o 

Fio. IB3. 



Moments. 



lOl 



The Ij, for the hollow cylinder is simply the difference 
between the I, for the outer and the inner cylinders. 



It must be particularly noticed that the radius of gyration 
of a solid body, such as a cylinder, flywheel, etc., is not the 
radius of gyration of a . 




plane section ; the radius , 

of gyration of a plane P V 

section is that of a thin 
lamina of uniform thick- 
ness, while the radius of 
gyration of a solid is that 
of a thin wedge. The 
radius of gyration of a 
solid may be found by 
correcting the section in 

this manner, and finding [^ Jffe >- 

the I for the shaded figure Fig. 123a. 

treated as a plane surface. 

The construction simply reduces the width of the solid 
section at each point proportional to its distance from OO ; it 
is, in fact, the " modulus figure " (see Chap. IX.) of the section. 

Let y^ = the distance of the c. of g. of the second modulus 
figure from the axis 00 shown black ; 
A = the area of the black figure. 
Ai = the area of the modulus figure 
then Ii = AiK^ = hyj (see page 96) 

Ax 



K^ = : 



The moment of inertia 
of the wheel 



V- 



Weight of the wheel X k-ycy 



102 



Mechanics applied to Engineering. 



Second Polar Moment, or Moment of 

Flyw heel with arms. 

Treat the rim and boss 
separately as hollow cylin- 
ders, and each arm thus 
(assumed parallel) — 

For each arm (see Fig. 
120) — 
L (sectional area),, a p ga 
12 

F'°-"4. + I2R„") 

where R, = the radius of the c. of g. of the arm. 
For most practical purposes the rim only is 
considered, and the arms and boss neglected. 




Inertia. 

Radius of gyration 



Sphere about its diameter. 





W 



or i,rD» 



w 



RVf 

D 

or-^= 

V 10 



Moments. 



103 



The arms are assumed to be of rectangular section; if they 
are not, the error involved will be exceedingly small. 



The sphere may be regarded as being made up of a great 
number of thin circular layers of radius rj, and radius of 

gyration -T=(see p. 96). 

+2RJ' 

= 2Rj-/ 

volume of thin layer = irr^dy 
second moment of \ „ r^ 

layer about 00 / = '^''1 '0' X T 



= -Ai^y-ffdy 




= j(4Ry+y-4R/Kj»' 



second moment of hemi- 1 
sphere, i.e. of all layers | 
on one side of the I 
diameter dd J 






(4Ry+/-4R/K^ 



_^r4Ry /_4R/1-y = ^ 

. 2L 3 ^5 4 \y=o 

IT r4R° R^ 4Rn 



second moment of sphere = iV^R" 



L-3 



4 J 



I04 Mechanics applied to Engineering. 

Second Polar Moment, or Polar Moment of Inertia. 

Radius of gyration 



Sphere about an external axis. 








w 

When the axis becomes a 
tangent, Ro = R ; 



I = 



rR=— 



Fig. 126. 



Cone about its axis. 
O 




I,=^R^H^. 
10 ^ 

or ^D^H^ 
160 g 



I i.e. I of the I for the cir- 
■ cumscribing cylinder. 



^ 10 

Dv'X 

^ 40 



Moments. 



105 



From the theorem on p. 7 6, we have- 



K = 



= (^R» + I^R'Ro") 



V = |,rR» 



The cone may also be regarded as being 
a great number of thin layers. 

Volume of thin layer = irr^dk 

second moment of \ «,, ^ r' '^t*Ji. 

layer about axis OOj 

7rR*/J< 



second moment 
of cone 



2H' 



dh 



v%\y'''i 



rR*H 




CHAPTER IV. 



RESOLUTION OF FORCES. 

We have already explained how two forces acting on a point 
may be replaced by one which will have precisely the same 
efifect on the point as the two. We must now see how to apply 
the principle involved to more complex systems of forces. 

Polygon of Forces. — If we require to find the resultant 

of more than two forces which act on a point, we can do so by 

finding the resultant of any two by means of the parallelogram 

of forces, and then take the resultant of this resultant and the 

, next force, and so on, as shown 

in the diagram. The resultant 

of I and 2 is marked Ri.2., 

!;.,R,i and so on. Then we finally 

get the resultant Ri. 2.3.4. for 

the whole system. 

Such a method is, however, 
J clumsy. The following will be 
found much more direct and 
convenient : Start from any 
point O, and draw the line i 
parallel and equal on a given 
scale to the force 1 ; from the 
extremity of 1 draw the line 
2 equal and parallel to the 
force 2 ; then, by the triangle 
of forces, it will be seen that 
the line R1.2. is the resultant 
of the forces 1 and 2. From 
the extremity of 2 draw 3 in a similar manner, and so on with 
all the forces; then it will be seen that the line Ri,2.3.4. 
represents the resultant of the forces. In using this con- 
struction, there is no need to put in the lines R1.2., etc. ■ 
in the figure they have been inserted in order to make it 




Fig. 128. 



Resolution of Forces. 



107 




clear. Hence, if any number of forces act upon a point in 
such a manner that Unas drawn parallel and equal on some 
given scale to them form a closed polygon, the point is in 
equilibrium under the action of those forces. This is known 
as the theorem of the polygon of forces. 

Method of lettering Force Diagrams. — In order to 
keep force diagrams clear, it is essential that the forces be 
lettered in each diagram to prevent con- 
fusion. Instead of lettering the force 
itself, it is very much better to letter the 
spaces, and to designate the force by the 
letters corresponding to the spaces on each 
side, thus : The force separating a from b 
is termed the force ab ; likewise the force 
separating d from b, db. fig- "9. 

This method of notation is usually attributed to Bow; 
several writers, however, claim to have been the first to 
use it. 

Funicular or Link Polygons. — When. forces in equi- 
librium act at the corners of a series of links jointed together 
at their extremities, 
the force acting 
along each link can 
be readily found by 
a special application 
of the triangle of 
forces. 

Consider the 
links ag and bg. 
There are three 
forces in equili- 
brium, viz. ab^ ag, bg, acting at the joint. The magnitude of ab 
is known, therefore the magnitude of the other two acting on the 
links may be obtained from the triangle of forces shown on 
the right-hand side, viz. abg. Similarly consider all the other 
joints. It will be found that each triangle of forces contains 
a line equal in every respect to a line in the preceding 
triangle, hence all the triangles may be brought together to 
form one diagram, as shown to the extreme right hand. It 
should . be noticed that the external forces form a closed 
polygon, and the forces in the bars are represented by radial 
lines meeting in the point or pole g. 

It will be evident that the form taken up by the polygon 
depends on the magnitude of the forces acting at each joint. 




Fig. 130. 



io8 



Mechanics applied to Engineering. 



Saspecsiou Bridge. — ^Another special application of the 
triangle of forces in a funicular polygon is that of finding the 
forces in the chain of a suspension bridge. The platform on 
which the roadway is carried is supported from the chain by 
means of vertical ties. We will assume that the weight sup- 
ported by each tie is known. The force acting on each 
portion of the chain can be found by constructing a triangle of 
forces at each joint of a vertical tie to the chain, as shown in 
the figure above the chain. But bo occurs in both triangles; 
hence the two triangles may be fitted together, bo being com- 
mon to each. Likewise all the triangles of forces for all the 
joints may be fitted together. Such a figure is shown at 
the side, and is known as a ray or vector polygon. Instead, 




Fig. 131. 

however, of constructing each triangle separately and fitting 
them together, we simply set oiF all the vertical loads ab, be, 
etc., on a straight line, and from them draw lines parallel to 
each link of the suspension chain ; if correctly drawn, all the 
rays will meet in a point. The force then acting on each 
segment of the chain is measured off the vector polygon, to 
the same scale as the vertical loads. In the figure the vertical 
loads are drawn to a scale of 1" = 10 tons ; hence, for example, 
the tension in the segment ao is 9*8 tons. 

The downward pressure on the piers and the tension in the 
outer portion of the chain is given by the triangle aol. 

If a chain (or rope) hangs freely without any platform 
suspended below, the vertical load will be simply that due to 
the weight of the chain itself. If the weight per foot of hori- 
zontal span were constant, it is easy to show that the curve 
taken up by the chain is a parabola (see p. 493). In the same 
chapter, the link and vector polygon construction is employed 



Resolution of Forces, 



109 



to deteimine the bending moment due to an evenly distributed 
load. The bending moment M^ at x is there shown to be the 
depth D, multiplied by the polar distance OH (Fig. 132) ; ix. 




Fig. X33. 

the dip of the chain at x multiplied by the horizontal com- 
ponent of the forces acting on the links, viz. the force acting 
on the link at x, or M^ = D,, X OH. In the same chapter, it 

is also shown that with an evenly distributed load M^^ = -5-, 

o 

where w is the load per foot run, and / is the span in feet 



Hence 



8 



= D, . OH = D,, 



or .^ = 



8D« 



where h is the tension at the bottom of the chain, viz. at x. 

It will, however, be seen that the load on a freely hanging 
chain is not evenly distributed per foot of horizontal run, 
because the inclination of the chain varies from point to point. 
Therefore the curve is not parabolic; it is, in reality, a 
catenary curve. For nearly all practical purposes, however, 
when the dip is not great compared with the span, it is suffi- 
ciently accurate to take the curve as being parabolic. 

Then, assuming the curve to be parabolic, the tension at 
any other point, y, is given by the length of the corresponding 
line on the vector polygon, which is readily seen to be — 



T, = ^h^ + w'l^ 
The true value of the tension obtained from the catenary 



is — 



T, = >4 + wD, 



(see Unwin's " Machine Design," p. 421), which will be found 
to agree closely with the approximate value given above. 



no 



Mechanics applied to Engineering. 



Data for Force Polygons. — Sometimes it is impossible 
to construct a polygon of forces on accoimt of the incomplete 
ness of the data. 

In the case of the triangle and polygon of forces, the follow- 
ing data must be given in order that the triangle or polygon can 
be constructed. If there are « conditions in the completed 
polygon, « — 2 conditions must be given ; thus, in the triangle 
of forces there are six conditions, three magnitudes and three 
directions: then at least four must be supplied before the 
triangle can be constructed, such as — 

3 magnitude(s) and i direction(s) 



Likewise in a five-sided polygon, there are ten conditions, eight 
of which must be known before the polygon can be constructed. 
When the two imknown conditions refer to the same or 
adjacent sides, the construction is perfectly simple, but when 
the unknown conditions refer to non-adjacent sides, a special 
construction is necessary. Thus, for example, suppose when 
dealing with five forces, the forces i, 2, and 4 are completely 
known, but only the directions, not the magnitudes, of 3 and 5 
are known. We proceed thus : 

Draw lines r and 2 in the polygon of forces. Fig. 133, in the 
usual way. From the extremity of 2 draw a line of indefinite 




Fig. 133. 




Fig. 134. 



length parallel to the force 3 ; its length cannot yet be fixed, 
because we do not know its value. From the origin of i draw 
a line of indefinite length parallel to 5 ; its leng& is also not 
yet known. From the extremity of 4 in the diagram of forces, 
Fig. 134, drop a perpendicular ah on to 3, and in the polygon 
of forces, Fig. 133, draw a line parallel to 3, at a distance ab 



Resolution of Forces. 



Ill 



from it. The point where this line cuts the line S is the 
extremity of 5. From this point draw a line parallel to 4 ; then 
by construction it will be seen that its extremity falls on the 
line 3, giving us the length of 3. 

The order in which the forces are taken is of no import- 
ance. 

Forces in the Members of a Jib Crane. Case I. 
The weight W simply suspended from the end of the jib. — There 
is no need to construct a separate dia- 
gram of forces. Set off be = W, or BC 
on some convenient scale,^ and draw ca 
parallel to the tie CA ; then the triangle 
bac is the triangle of forces acting on the 
point b. On measuring the force dia- 
gram, we find there is a compressive 
force of 15 "2 tons along AB, and a 
tension force of 9-8 tons along AC. 

The pressure on the bottom pivot is 
W (neglecting the weight of the crane 
itself). The horizontal pull at the top 
of the crane-post is ad, or 7-9 tons ; and 
the force (tension) acting on the post 
between the junction of the jib and the 
tie is cd, or 6 tons. 

The bending moment at y will be 
ad X h, or W X /. For determining the bending stress at y, 
see Chap. IX. 

Taking moments about the pivot bearing, we have — 





rM/> 


1 


D M-^ 


f 

1 

[ 


...^£..-1....- 


c 


W^//MM 


i. 

''"W 


m 



d, 



Fig. 135. 



p^x=p^ = '^l 

W/ 

or A =/, = — 



The sections ot the various parts of the structure must be 
determined by methods to be described later on. 

The weight of the structure itself should be taken into 
account, which can only be arrived at by a process of approxi- 
mation ; the dimensions and weight may be roughly arrived at by 
neglecting the weight of the structure in the first instance. Then, 
as the centre of gravity of each portion will be approximately 
at the middle of each length, the load W must be increased to 



' In this case the scale is o'l inch = 2 tons, and W = 7 tons. 



112 



Mechanics applied to Engineering. 



W + ^(weight of jib and tie). The downward pressure on the 
pivot will be W + weight of structure. 

The dimensions of the structure must then be increased 
accordingly. In a large structure the forces should be again 
determined, to allow for the increased dimensions. 

The bending moment on the crane-post at y may be very 
much reduced by placing a balance weight Wj on the crane, as 
shown. The forces acting on the balance-weight members are 
found in a similar manner to that described above, and, neglect- 
ing the weight of the structure, are found to be 8" i tons on the 
tie, and 4*4 tons on the horizontal strut. 

The balance weight produces a compression in the upper 
part of the post of 6*8 tons ; but, due to the tie ac, we had a 
tension of 6"o tons, therefore there is a compression of o'8 ton 





Fig. 136. 



Fig. 137. 



in the upper part of the crane-post The pressure on the lower 
part of the crane-post and pivot is W -1- Wj -f weight of 
structure. 

Then, neglecting the weight of the structure, the bending 
moment on the post at y will be — 

W/ - Wi/, 

W/ 
The moment Wi*^ should be made equal to — , then the 

2 

post will never be subjected to a bending moment of more 

than one-half that due to the lifted load, and the pressure/. 

and /a will be correspondingly reduced. 

Case II. The weight W suspended from a chain passing to a 

barrel on the crane-post. — As both poitions of the chain are 



Resolution of Forces 



"3 



subjected to a pull W, the resultant R is readily determined. 
From c a line ac is drawn parallel to the tie ; then the force 
acting down the jib is ab = i6'4 tons j down the tie ac = 4*4 
tons. The bending moments on the post, etc., are determined 
in precisely the same manner as in Case I. 

When pulley blocks are used for lifting the load, the pull 
in the chain between the jib pulley and the barrel will be less 
than W in the proportion of the velocity ratio. 

The general effect of the pull on the chain is to increase 
the thrust on the jib, and to reduce the tension in the tie. In 
designing a crane, the members should be made strong enough 
to resist the greater of the two, as it is quite possible that a 
link of the chain may catch in the jib pulley, and the conditions 
of Case I. be realized. 

Forces in the Members of Sheer Legs.— In the type 
of crane known as sheer legs the crane-post is dispensed with, 
and lateral stability is given by using two jibs or sheer legs 
spread out at the foot ; the tie is usually brought down to the 
level of the ground, and is attached to a nut working in guides. 
By means of a horizontal screw, the sheer legs can be tilted or 
" derricked " at will : 
the end thrust on the 
screw is taken by a 
thrust block ; the up- 
ward pull on the nut 
and guides is taken by 
bolts passing down to 
massive foundations 
below. The forces are 
readily determined by 
the triangle of forces. 

The line ^^ris drawn 
parallel to the tie, and 
represents the force 
acting on it; then ac 
represents the force 
acting down the middle 
line of the two sheer 
legs. This is shown 
more clearly on the 
projected view of the 

sheer legs, cd is then ^'°- '38- 

drawn parallel to the sheer leg ae; then dc represents the force 
acting down the sheer leg ae ; likewise ad down the leg of, and 

I 




114 



Mechanics applied to Engineering. 



dg the force acting at the bottom of the sheer legs tending to 
make them spread; ch represents the thrust of the screw on 
the thrust block and the force on the screw, and bh the upward 
pull which has to be resisted by the nut guides and the founda- 
tion bolts. 

The members of this type of structure are necessarily very 




Fig. i33«. 



heavy and long, consequently the bending stress due to their 
own weight is very considerable, and has to be carefully con- 
sidered in the design. The problem of combined bending and 
compression is dealt with in Chapter XII. 

Forces in a Tripod or Three Legs.— Let the lengths 



Resolution of Forces. 



"5 



of the legs be measured from a horizontal plane. The vertical 
height of the apex O from the plane, also the horizontal 
distances, AB, EC, CA, must be known. 

In the plan set out the triangle ABC from the known 
lengths of the sides; from A as centre describe an arc of 
radius equal to the length of the pole A, likewise from B 
describe an arc of radius equal to the length of the pole B. 
They cut in the point o-^. From o^ drop a perpendicular on 
AB and produce ; similarly, by describing arcs from B and C 
of their respective radii, find the point o-a, and from it drop a 
perpendicular on BC, and produce to meet the perpendicular 
from Ox in O, which is the apex of the tripod. The plan of 
the three legs can now be filled in, viz. AO, BO, CO. Produce 
AO to meet CB in D. From O set off the height of the apex 
above the plane, viz. O^uj, at right angles to AO ; join Af^m. 
This should be measured to see that it checks with the length 
of the pole A. Join DiPm. From o-a\ set off a length to a con- 
venient scale to represent W, complete the parallelogram of 
forces, then Oxa^ gives the force acting down the leg A, and 
o-a.\d the force acting down the imaginary leg D, shown in 
broken line, which lies in the plane of the triangle OBC ; resolve 
this force down OC and OB by setting off o-ad along OuD equal 
to o-a.\d, found by the preceding parallelogram, then the force 
acting down the leg B is Oyj), and that down C is o-aC 

The horizontal force tending to spread the legs AOm and 
DOui is given by fd. This is set off at A/,'and is resolved along 
AC and AB. The force acting on an imaginary tie AC is Ke, 
and on AB is A^, similarly with the remaining tie. 

When the three legs are of equal length and are symmetric- 
ally placed, the forces can be obtained thus — 




l^J-V - ^ 



Three equal fe^s 
Fig. 138*. 

where / is the force acting down each leg. 



ii6 



Mechanics applied to Engineering. 



Forces in the Members of a Roof Truss.— Let the 

roof truss be loaded with equal weights at the joints, as shown ; 
the reactions at each support will be each equal to half the 
total load on the structure. We shall for the present neglect 
the weight of the structure itself. 

.t 




The forces acting on each member can be readily found by 
a special application of the polygon of forces. 

Consider the joint at the left-hand support BJA or Rj. We 
have three forces meeting at a point ; the magnitude of one, 
viz. Rj or ba, and the direction of all are known ; hence we can 
determine the other two magnitudes by the triangle of forces. 
This we have done in the triangle ajh. 



Resolution of Forces 1 1 7 

Consider the joint BJIC Here we have four forces 
meeting at a point ; the magnitude of one is given, viz. be, and 
the direction of all the others ; but this is not sufficient — we 
must have at least six conditions known (see p. no). On 
referring back to the triangle of forces just constructed, we iind 
that the force bj is known ; hence we can proceed to draw our 
. polygon of forces cbji by taking the length of bj from the tri- 
angle previously constructed. By proceeding in a similar 
manner with every joint, we can determine all the forces 
acting on the structure. 

On examination, we find that each polygon contains one 
side which has occurred in the previous polygon ; hence, if these 
similar and equal sides be brought together, each polygon can 
be tacked on to the last, and so made to form one figure con- 
taining all the sides. Such a figure is shown below the 
structure, and is known as a " reciprocal diagram." 

When determining the forces acting on the various members 
of a structure, we invariably use the reciprocal diagram without 
going through the construction of the separate polygons. We 
have only done so in this case in order to show that the reciprocal 
diagram is nothing more nor less than the polygon of forces. 

We must now determine the nature of the forces, whether 
tensile or compressive, acting on the various members. In 
order to do this, we shall put 
arrows on the bars to indicate the 
direction in which the bars resist 
the external forces. 

The illustration represents a 
man's arm stretched out, resisting 
certain forces. The arrows indi- 
cate the direction in which he is 
exerting himself, from which it ^^^ 

will be seen that when the arrows ' ^^' 

on his arms point outwards his arms are in compression, and 
when in the reverse direction, as in the chains, they are in 
tension ; hence, when we ascertain the directions in which a 
bar is resisting the external forces acting on it, we can at once 
say whether the bar is in tension or compression, or, in other 
words, whether it is a tie or a strut. 

We know, from the triangle and polygon of forces, that the 
arrows indicating the directions in which the forces act follow 
round in the same rotary direction ; hence, knowing the direc- 
tion of one of the forces in the polygon, we can immediately 
find the direction of the others. Thus at the joint BJA we 




ii8 Mechanics applied to Engineering. 

know that the arrow points upwards from b Xa a; then, con- 
tinuing round the triangle, we get the arrow-heads as sliown. 
Transfer these arrows to the bars themselves at the joint in 
question ; then, if an arrow points outwards at one end of a bar, 
the arrow at the other end must also point outwards ; hence 
we can at once put in the arrow at the other end of the bar, 
and determine whether it is a strut or tie. When the arrows 
point outwards the bar is a strut, and when inwards a tie. 
Each separate polygon has been thus treated, and the arrow- 
heads transferred to the structure. But arrow-heads must not 
be put on the reciprocal diagram ; if they are they will cause 
hopeless confusion. With a very little practice, however, one 
can run round the various sections of the reciprocal diagram 
by eye, and put the arrow-heads on the structure without 
making a single mark on the diagram. If a mistake has been 
made anywhere, it is certain to be detected before all the bars 
have been marked. If the beginner experiences any difficulty, 
he should make separate rough sketches for each polygon of 
forces, and mark the arrow-heads ori each side. At some 
joints, where there are no external forces, the direction of the 
arrows will not be evident at first; they must not be taken 
from other polygons, but from the arrow-heads on the structure 
itself at the joint in question. For example, the arrows at the 
joints ABJ and BJIC are perfectly readily obtained, the direc- 
tion being started by the forces AB and BC, but at the joint 
JIA the direction of the arrow on the bars JI and JA are 
known at the joint ; either of these gives the direction for 
starting round the polygon ahij. 

The following bars are struts : BJ, IC, GD, FE, JI, GF. 

The following bars are ties : JA, IH, HA, HG, FA. 

Some more examples of reciprocal diagrams will be given 
in the chapter on " Framework St.Tictures." 



CHAPTER V. 

MECHANISMS. 

Professor Kennedy '^ defines a machine as " a combination 
of resistant bodies, whose relative motions are completely 
constrained, and by means of which the natural energies at our 
disposal may be transformed into any special form of work." 
Whereas a mechanism consists of a combination of simple 
links, arranged so as to give the same relative motions as the 
machine, but not necessarily possessing the resistant qualities 
of the machine parts ; thus a mechanism may be regarded as a 
skeleton form of a machine. 

Constrained and Free Motion. — Motion may be 
either constrained or free. A body which is free to move in 
any direction relatively to another body is said to have free 
motion, but a body which is constrained to move in a definite 
path is said to have constrained motion. Of course, in both 
cases the body moves in the direction of the resultant of all 
the forces acting upon it ; but in the latter case, if any of the 
forces do not act in the 
direction of the desired 
path, they automatically 
bring into play constraining 
forces in the shape of 
stresses in the machine 
parts. Thus, in the figure, 
let a^ be a crank which 
revolves about a, and let 
the force be in the direc- 
tion of the connecting-rod act on the pin at b. Then, if b 
were free, it would move off in the direction of the dotted line, 
but as b must move in a circular path, a force must act along 
the crank in order to prevent it following the dotted line. This 
force acting along the crank is readily found by resolving be in 

' " Mechanics of Machinery,'' p. 2. 




120 Mec/ianics applied to Engineering. 

a direction normal to the crank, viz. bd, i.e. in the direction in 
which b is moving, and along the crank, viz. dc, which in this 
instance is a compression. Hence the path of 6 is determined 
by the force acting along the connecting-rod and the force 
acting along the crank. 

The constraining forces always have to be supplied by the 
pdrts of the machine itself. Machine design consists in 
arranging suitable materials in suitable form to supply these 
constraining forces. 

The various forms of constrained motion we shall now 
consider. 

Plane Motion. — When a body moves in such a manner 
that any point of it continues to move in one plane, such as 
in revolving shafts, wheels, connecting-rods, cross-heads, links, 
etc., such motion is known as plane motion. In plane motion 
a body may have either a motion of translation in any 
direction in a given plane or a motion of rotation about an 
axis. 

Screw Motion. — ^When a body has both a motion of 
rotation and a translation perpendicular to the plane of rota- 
tion, a point on its surface is said to have a screw motion, and 
when the velocity of the rotation and translation are kept 
constant, the point is said to describe a helix, and the amount 
of translation corresponding to one complete rotation is termed 
t\i& pitch of the helix or screw. 

Spheric Motion. — When a body moves in such a manner 
that every point in it remains at a constant distance from a 
fixed point, such as when a body slides about on the surface of 
a sphere, the motion is said to be spheric. When the sphere 
becomes infinitely great, spheric motion becomes plane 
motion. 

Relative Motion. — When we speak of a body being in 
motion, we mean that it is shifting its position relatively to 
some other body. This, indeed, is the only conception we can 
have of motion. Generally we speak of bodies as being in 
motion relatively to the earth, and, although the earth is going 
through a very complex series of movements, it in nowise 
affects our using it as a standard to which to refer the motions 
of bodies ; it is evident that the relative motion of two bodies 
is not affected by any motions which they may have in common. 
Thus, when two bodies have a common motion, and at 
the same time are moving relatively to one another, we may 
treat the one as being stationary, and the other as moving 
relatively to it : that is to say, we may subtract their common 



Mecltan isms. 121 

motion from each, and then regard the one as being at rest. 
Similarly, we may add a common motion to two moving bodies 
without affecting their relative motion. We shall find that such 
a treatment will be a great convenience in solving many 
problems in which we have two bodies, both of which are 
moving relatively to one another and to a third. As an 
example of this, suppose we are studying the action of a valve 
gear on a marine engine; it is a perfectly simple matter to 
construct a diagram showing the relative positions of the valve 
and piston. Precisely the same relations will hold, as regards 
the valve and piston, whether the ship be moving forwards or 
backwards, or rolling. In this case we, in effect, add or 
subtract the motion of the ship to the motion of both the valve 
and the piston. 

Velocity. — Our remarks in the above paragraph, as regards 
relative motion, hold equally well for relative 
velocity. 

Many problems in mechanisms resolve 
themselves into finding the velocity of one 
part of a mechanism relatively to that of 
another. The method to be adopted will 
depend upon the very simple principle that 
the linear velocity of any point in a rotating * /^ 
body varies directly as the distance of that 'a^^"<^'^ 
point from the axis or centre of rotation. ^■°- '*^" 

Thus, when the link OA rotates about O, we have — 

velocity of A V„ r^ 
velocity of B ~ Vj ~ rj 

If the link be rotating with an angular velocity w radians 
per second (see p. 4), then the linear velocity of a, viz. 
V. = uir„ and of b, Vj = eo^j, but the angular velocity of every 
point in the link is the same. 

As the link rotates, every point in it moves at any given 
instant in a direction normal to the line drawn to the centre of 
rotation, hence at each instant the point is moving in the 
direction of the tangent to the path of the point, and the centre 
about which the point is rotating lies on a line drawn normal 
to the tangent of the curve at that point. This property will 
enable us to find the centre about which a body having plane 
motion is rotating. The plane motion of a body is completely 
known when we know the motion of any two points in the 
body. , If the paths of the points be circular and concentric, 
then the centre of rotation will be the same for all positions of 




122 Mechanics applied to Engineering. 

the body. Such a centre is termed a " permanent " or " fixed " 
centre ; but when the centre shifts as the body shifts, its centre 
at any given instant is termed its " instantaneous " or " virtual " 
centre. 

Instantaneous or Virtual Centre. — Complex plane 
motions of a body can always be reduced to one very simply 
expressed by utilizing the principle of the virtual centre. For 
example, let the link ab be part of a mechanism having a 
complex motion. The paths of the two end points, a and b, 
are known, and are shown dotted. In order to find the relative 
velocities of the two points, we draw tangents to the paths at 
a and b, which give us the directions in which each is moving 
at the instant. From the points a, b draw normals aa' and bl/ 
to the tangents ; then the centre about which a is moving at the 
instant lies somewhere on the Une ad, likewise with bb' ; hence 
the centre about which both points are revolving at the instant, 
must be at the intersection of the two lines, viz. at O. This 





Fig. 143. 

point is termed the virtual or instantaneous centre, and the 
whole motion of the link at the instant is the same as if it were 
attached by rods to the centre O. As the link has thickness 
normal to the plane of the paper, it would be more correct to 
speak of O as the plan of the virtual axis. If the bar had an 
arm projecting as shown in Fig. 144, the path of the point 
C could easily be determined, for every point in the body, at 
the instant, is describing an arc of a circle round the centre O ; 
thus, in order to determine the path of the point C, all we have 
to do is to describe a small arc of a circle passing through C, 
struck from the centre O with the radius OC. 

The radii OA, OB, OC are known as the virtual radii of 
the several points. 

If the tangents to the point-paths at A and B had been 
parallel, the radii would not meet, except at infinity. In that 




Mechanisms. \ 23 

case, the points may be considered to be describing arcs of 
circles of infinite radius, i.e. their point-paths are straight 
parallel lines. 

If the link AB had yet another arm projecting as shown in 
the figure, the end point of 
which coincided with the virtual 
centre O, it would, at the in- 
stant, have no motion at all 
relatively to the plane, i.e. it is 
a fixed point. Hence there is 
no reason why we should not 
regard the virtual centre as a 
point in the moving body itself. 

It is evident that there can- 
not be more than one of such 
fixed points, or the bar as a whole would be fixed, and then it 
could not rotate about the centre O. 

It is clear, from what we have said on relative motion, that if 
we fixed the bar, which we will term m (Fig. 146), and move 
the plane, which we will term n, the relative motion of the two 
would be precisely the same. We shall term the virtual centre 
of the bar m relatively to the plane n, Omn. 

Oentrode and Axode. — As the link m moves in such a 
manner that its end joints a and i follow the point-paths, the 
virtual centre Omn also shifts relatively to the plane, and traces 
out the curve as shown in Fig. 146. This curve is simply the 
point-path of the virtual centre, or the virtual axis. This curve 
is known as the centrode, or axode. 

Now, if we fix the link m, and move the plane n relatively to 
it, we shall, at any instant, obtain the same relative motion, 
therefore the position of the virtual centre will be the same in 
both cases. The centrodes, however, will not be the same, 
but as they have one point in common, viz. the virtual centre, 
they will always touch at this point, and as the motions of 
the two bodies continue, the two centrodes will roll on one 
another. 

This rolling action can be very clearly seen in the simple 
four-bar mechanism shown in Fig. 147. The point A moves 
in the arc of a circle struck from the centre D, hence AD is 
normal to the tangent to the point-path of A ; hence the virtual 
centre lies somewhere on the line AD. For a similar reason, it 
lies somewhere on the line BC ; the only point common to tiie 
two is their intersection O, which is therefore their virtual 
centre. If the virtual centre, i.e. the intersection of the two 



124 



Mechanics applied to Engineering. 



bars, be found for several positions of the mechanism, the 
centrodes will be found to be ellipses. 

As the mechanism revolves, the two ellipses will be found to 




r<?»n. 



«?■-• 



Fio. 146. 

roll on one another, because A, B and C, D are the foci of the 
two ellipses. That such is the case can easily be proved 
experimentally, by a model consisting of two ellipses cut out of 
suitable material and joined by cross-bars AD and BC ; it will 
be found that they will roll on one another perfectly. 

Hence we see that, if we have given a pair of centrodes for 
two bodies, we can, by making the one centrode roll on the other, 
completely determine the relative motion of the two bodies. 

Position of Virtual Centre. — ^We have shown above 
that when two point-paths of any body are known, we can 



Mechanisms. 



125 






readily find the position of the virtual centre. In the case of 
most mechanisms, however, we can determine the virtual 
centres without first constructing 
the point-paths. We will show 
this by taking one or two simple 
cases. In the four-bar mechanism 
shown in Fig. 148, it is evident 
that if we consider d as stationary, / 
the virtual centre Odd will be at 
the joint of a and d, and the\ 
velocity of any point in a relatively 
to any point in d will be propor- 
tional to the distance from this 
joint ; likewise with Ode. Then, 
if we consider b as fixed, the 
virtual centre of a and b will also 
be at their joint. By similar , 
reasoning, we have the virtual! 
centre Obc. Again, let d be I 
fixed, and consider the motion 
of b relatively to d. The point- 
path of one end of b, viz. Oab, 
describes the arc of a circle 
about Oad, therefore the virtual 
centre lies on a produced j for a 
similar reason, the virtual centre lies on c produced hence it 
must be at Obd, the meet of the two lines. 




0^0 



Oaxs 




Fig. 14S. 



Odo 



In a similar manner, consider the link e as fixed ; then, for 



126 



Mechanics applied to Engineering. 



the same reason as was given above for b and d, the virtual centre 
of a and c lies at the meet of the two lines b and d, viz. Oac. 

If the mechanism be slightly altered, as shown in Fig. 149, 
we shall get one of the virtual centres at infinity, viz. Ocd. 

oOcd 




Oad, 



Obc 



Fig. 150. 



The mechanism shown in Fig. 149 is kinematically similar 
to the mechanism in Fig. 150. Instead of c sliding to and fro 
in guides, a link of any length may be substituted, and the 
fixed link d may be carried round in order to provide a centre 
from which c shall swing. Then it is evident that the joint 
Obc moves in the arc of a circle, and if c be infinitely long it 
moves in a straight line in precisely the same manner as the 
sliding link c in Fig. 149. 

The only virtual centre that may present any difficulty in 
finding is Oac. Consider the link c as fixed, then the bar d 
swings about the centre Ocd; hence every point in it moves 
in a path at right angles to a line drawn from that point to 
Ocd. Hence the virtual centre lies on the line Ocd, Oad; also, 
for reasons given below, it lies on the prolongation of the bar 
b, viz. Oac. 

Three Virtual Centres on a Line. — By referring to the 
figures above, it will be seen that there are always three virtual 
centres on each line. In Figs. 149, 150, it must be remembered 
that the three virtual centres Ocui, Oac, Ocd are on one line ; 
also Obc, Obd, Ocd. 

The proof that the three virtual centres corresponding to the 
three contiguous links must lie on one line is quite simple, and as 
this property is of very great value in determining the positions 
of the virtual centres for complex mechanisms, we will give it 
here. Let b (Fig. 151) be a body moving relatively to a, an J 
let the virtual centre of its motion relative to a be O^^ ; likewise 
let Oac be the virtual centre of c's motion relative to a. If we 



Mechanisms. 



X'Z'J 




want to find the velocity dof a point in b relatively to a point ii 
€, we must find the virtual centre, Obc. Let it be at O : then, 
considering it as a point of 
b, it will move in the arc 
i.i struck from the centre 
Oab; but considering it as 
a point in c, it will move in 
the arc 2.2 struck from the 
centre Oac. But the tangents 
of these arcs intersect at 0, 
therefore the point O has a 
motion in two directions at 
the same time, which is im- fig. 151. 

possible. In the same manner, 

it may be shown that the virtual centre Obc cannot lie any- 
where but on the line joining Oab, Oac, for at that point only 
will the tangents to the point-paths at O coincide ; therefore 
the three virtual centres must lie on one straight line. 

Relative Linear Velocities of Points in Mechan- 
isms. — Once having found the virtual centre of any two bars 
of a mechanism, the finding of the 
velocity of any point in one bar 
relatively to that of any other point 
is a very simple matter, for their 
velocities vary directly as their ^ 
virtual radii. 

In the mechanism shown, let 
the bar d be fixed; to find the 
relative velocities of the points i 
and 2, we have — 

velocity i _ Obd i _ ^1 
velocity 2 Obd 2 r^ 

Similarly — 

velocity i _ r-^ 
velocity 3 r, 

^^^ vdocity 3 ^ /3 
velocity 4 rt, 




The relative velocities are not affected in the slightest 
degree by the skape of the bars. 

When finding the velocity of a point on one bar relatively to 
the velocity of a point on another bar, it must be remembered 



128 



Mechanics applied to Engineering, 



that at any instant the two bars move' as though they had one 
point in common, viz. their virtual centre. 

As an instance of points on non-adjacent bars, we will pro- 
ceed to find the velocity of the point 8 (Fig. 153) relatively to 
that of point 9. By the method already explained, the virtual 
centre Oac is found, which may be regarded as a point in the 
bar c pivoted at Ocd; likewise as a point in the bar a pivoted 




Fig. 153- 



at Oad. As an aid in getting a clear conception of the action, 
imagine the line Ocd . Oac, also the bar c, to be arms of a toothed 
wheel of radius /04, and the line Oad . Oac, also the bar a, to be 
arms of an annular toothed wheel of radius pa, the two wheels 
are supposed to be in gear, and to have the common point Oac, 
therefore their peripheral velocities are the same. Denoting 
the angular velocity of a as o)„ and the linear velocity of the 
point 8 as Vg, etc., we have — 

<^ap3 = <^cpiy and <o„ = -^ also Vg = oi^ps Vg = 01^ 

Substituting the value of (i)„, we have — 

Vg _ MqPjPs _ piPs ^ i'29 X 076 _ ^ .^^ 

Vg _ OM.S 

Y,~ Odd.g~ 2-i6 



or 



paPs 

2'20 



2'29 X 0'42 



= I'02 



Mechanisms. 1 29 

Similarly, if we require the velocity of the point 6 relatively 
to that of point 5 — 

Vg _ Pb V - ^8P6 

iT » 6 — 

Vs Pa Pi 

^j — P} V = X5P? 

V9 P9 P9 

whence — = — ' .^5^= P^'P'^" = M.' 
Vs Vj pePn PsPsPePi PsPe 



Voac Ps Vfl^ p4 

Vg p4pe 1-29 X 0-3 



= 0-36 



V5 psPs 2-29 X 0-47 

Likewise — 

velocity 7 _ R? _ 2'3i _ 
velocity 8 Rs 2-20 ~ 
velocity 6 p^ x velocity 8 X R 



hence 

velocity 7 Ps X K.7 X velocity 8 

= 0-38 



Ps X R7 076 X 2-31 



Fig. 154. 

This can be arrived at much more readily by a graphical 
process; thus (Fig. 154): With Oad as centre, and p, as 

K 



130 Mechanics applied to Engineering. 

radius, set off Oad.h = p, along the line joining Oad to 8 ; set 
off a line Ai to a convenient scale in any direction to represent 
the velocity of 6. From Oad draw a line through i, and from 
8 draw a line Be parallel to /it ; this line will then represent the 
velocity of the point 8 to the same scale as 6, for the two 
triangles Oad.8.e and Oad.h.i are similar ; therefore — 

8(? Oad • 8 _ Pa _ velocity 8 _ o'8o _ 
hi Oad.h p^ velocity 6 0-32 

From the centre Odd and radius Rj, set off Odd./ = Rj ; 
6ia.vf/.g parallel to <?. 8, and from Odd draw a line through e to 
meet this line in g", then,/^ = V,, for the two triangles Obd.f.g 
and Obd.i.e are similar; therefore — 

fg _ Obd.f _ R7 _ velocity 7 _ 2-3 

8^ ~ Obd.B ~ Rs ~ velocity 8 ~ 2^ ^ ^'°^ 

, velocity 6 ih o'^z 
and —. — H^ = T = —5- = o'38 
velocity 7 ^ o'84 "^ 

The same graphical process can be readily applied to all cases 
of velocities in mechanisms. 

Relative Angular Velocities of Bars in Mechan- 
isms. — Every point in a rotating bar has the same angular 
velocity. Let a bar be turning about a point O in the bar 
with an angular velocity m; then the linear velocity V„ of a 
point A situated at a radius r„ is — 

V 
V„ = u)r„ and m = -^ 

In order to find the relative angular velocities of any two 
links, let the point A (Fig. 155) be first regarded as a point 
in the bar a, and let its radius about Oad be r^. When the 
point A is regarded as a point in the bar b, we shall term it 
B, and its radius about Obd, r^. Let the linear velocity of A 
be Vju and that of B be Vb, and the angular velocity of A be 
0)4, and of B be Mg. Then V^ = Wjj-^, and Vi = m^rj^. 

But Va = Vb as A and B are the same point ; hence — 

<"ii ''a {Oad)A 



Mechanisms. 



131 



This may be very easily obtained graphically thus : Set off 
a line he in any direction from A, whose length on some given 
scale is equal to (Oij join e.Obd; from Oad draw 0«(^/ parallel 

b 



Obd 




Fig. iss. 



Fig. 156. 



X.0 e.Obd. Then A/= 0)3, because the two triangles k.f.Oad 
and A.e.Obd are similar. Hence — 

he ^ (Obd)B _ (0^ 
A/ (Oad)A (ub 

In Fig. 156, the distance A^ has been made equal to h.Oad, 
and gf is drawn parallel to e.Obd. The proof is the same as in 
the last case. When a is parallel to e, the virtual centre is at 
infinity, and the angular velocity of b becomes zero. 

When finding the relative angular velocity of two non- 
adjacent links, such as a and e, we proceed thus : For con- 
venience we have numbered the various points instead of using 
the more cumbersome virtual centre nomenclature (Figs. 157 
and 158). The radius 1.6 we shall term r^^, and so on. 

Then, considering points i and 2 as points of the bar b, we 
have — 



V ■ 



''2.6 



132 Mechanics applied to Engineering. 

Then, regarding point i as a point in bar c, and regarding 
point 2 as a point in bar a — 



V, = w„ X n.. 



Vo = (o.ra.; 



J:: _ 

s 



■?-•"" 





Fig. 158. 



Then, substituting the values of Vj and Vj in the equation 
above, we have — 



Mechanisms. 133 

Draw 4.7 parallel to 2.3 ; then, by the similar triangles 
1.2. 6 and 1.7.4 — 

4-7 ri.4 

and ^a., X i-i., = ?-i,, X 4.7 
Substituting this value above, we have — 



^c 1.6 X ^as 



'1.8 



= -M, or = iiP by similar triangles 



X 4-7 4-7 5-4 



Thus, if the length ^2.3 represents the angular velocity of c, 
and a line be drawn from 4 to meet the opposite side in 7, 
4.7 represents on the same scale the angular velocity of a. Or 
it may conveniently be done graphically thus : Set off from 3 a 
line in any direction whose length 3.8 represents the angular 
velocity off; from 4 draw a line parallel to 3.8; from 5 draw 
a line through 8 to meet the line from 4 in 9. Then 4.9 repre- 
sents the angular velocity of a, the proof of which will be 
perfectly obvious from what has been shown above. 

When b is parallel to d, the virtual centre Oac is at infinity, 
and the angular velocity of a is then equal to the angular 
velocity of c. 

Steam-engine Mechanism. — On p. 126 we showed 
how a four-bar mechanism may be developed into the ordinary 
steam-engine mechanism, which is then often called the 
" slider-crank chain." 

This mechanism appears in many forms in practice, but 
some of them are so 
disguised that they are 
not readily recognized. 
We will proceed to 
examine it first in its // 
most familiar form, viz. 
the ordinary steam- 
engine mechanism. ^^^^ y 

Having given the ' fig. 139. 

speed of the engine in 

revolutions per minute, and the radius of the crank, the velocity 
of the crank-pin is known, and the velocity of the cross-head 
at any instant is readily found by means of the principles laid 
down above. We have shown that — 

velocity P Obd.V 
velocity X ~ OW.X 



■XM 




1 34 Mechanics applied to Engineering. 

From O draw a line parallel to the connecting-rod, and 
from P drop a perpendicular to meet it in e. Then the triangle 
QI2e is similar to the triangle P.OW.X ; hence — 

OP Obd.V velocity of pin 

"p7 ~ Obd.^ ~ velocity of cross-head 




But the velocity of the crank-pin may be taken to be constant. 
Let it be represented by the radius of the crank-circle OP; 
then to the same scale Te represents the velocity of the cross- 
head. Set up fg = P<? at several positions of the crank-pin, 

and draw a curve 
through them; then 
the ordinates of this 
curve represent the 
velocity of the cross- 
head at every point 
in the stroke, where 
the radius of the 
crank - circle repre- 
sents the velocity of 
the crank-pin. 

When the connecting-rod is of infinite length, or in the case 
of such a mechanism as that shown in Fig, i6o, the line gc 
(Fig. 159) is always parallel to the axis, and consequently the 
crosshead- velocity diagram becomes a semicircle. 

An analytical treatment of these problems will be found in 
the early part of the next chapter. 

Another problem of considerable interest in connection 
with the steam-engine is that of finding the journal velocity, or 
the velocity with which the various journals or pins riib on 
their brasses. The object of making such an investigation will 
be more apparent after reading the chapter on friction. 

Let it be required to find the velocity of rubbing of (i) the 
crank-shaft in its main bearings ; (2) the crank-pin in the big 
end brasses of the connecting-rod ; (3) the gudgeon-pin in the 
small end brasses. 

Let the radius of the crank-shaft journal be r„ that of the 
crank-pin be r^, and the gudgeon-pin r,. 

Let the number of revolutions per minute (N) be 160. 
Let the radius of the crank be i'25 feet. 
Let the radii of all the journals be 0*25 foot. We have 
taken them all to be of the same size for the sake of comparing 



Mechanisms. 



135 



the velocities, although the gudgeon-pin would usually be 
considerably smaller. 

(i) V, = 2irr„N 

or u)^r„ =250 feet per minute in round numbers 




(2) We must solve this part of the problem by finding the 
relative angular velocity of the connecting-rod and the crank. 
Knowing the angular velocity of a relatively to d, we obtain the 
angular velocity of b relatively to d thus : The virtual centre 
Oab may be regarded as a part of the bar a pivoted at Oad^ 
also as a part of the bar b rotating for the instant about the 
virtual centre Obd; then, by the gearing conception already 
explained, we have — 

il' = ]^», or <o. = '^ = Yi. 
«. Ri' R. R» 

When the crank-arm and the connecting-rod are rotating in 
the opposite sense, the rubbing velocity — 

This has its maximum value when ^ is greatest, i.e. when 

R» 



136 



Mechanics applied to Engineering. 



Rj is least and is equal to the length of the connecting-rod, i.e. 
at the extreme " in " end of the stroke. Let the connecting-rod 
be n cranks long ; then this expression becomes — 



v, = .,4+i) 



which gives for the example taken — 

= 314 feet per minute 

taking « = 4. 

But when the crank-arm and the connecting-rod are rotating 
in the same sense, the rubbing velocity becomes — 



V =.,«.„(.- 1) 



The polar diagram shows how the rubbing velocity varies at 
the several parts of the stroke. 



Conn£ctui'9 rod b? 
O \ •' tnol/iolder 



Gegrwilh 



'"^'•e Off- 



^■e'd 




Fig. i6xa. 



(3) Since the gudgeon-pin itself does not rotate, the rubbing 
velocity is simply due to the angular velocity of the connecting- 
rod. 



Mechanisms. 



117 



which has its maximum value when R, is least, viz. at the 
extreme end of the " in " stroke, and is then 63 feet per minute 
in the example we have taken. 

By taking the same mechanism, and by fixing the link b 
instead of d, we get another familiar form, viz. the oscillating 
cylinder engine mechanism. On rotating the crank the link d 
becomes the connecting-rod, in reality the piston rod in this 
case, and the link c oscillates about its centre, which was the 
gudgeon-pin in the ordinary steam-engine, but in this case it is 
the cylinder trunnion, and the link c now becomes the cylinder 
of the engine. Another slight modification of the same 
inversion of the mechanism is one form of a quick-return 
motion used on shaping machines. In Fig. i6ia we show the 
two side by side, and in the case of the engine we give a polar 
diagram to show the angular velocity of the link c at all parts 
of the stroke when a rotates uniformly. 

We shall again make use of the gearing conception in the 
solution of this problem, whence we have — 



o)„R„ = oj^Ra, and <*<« = t> 



R. 




Taking the circle mn to represent the constant angular 
velocity of the crank, the polar curves op, qr 
represent to the same scale the angular 
velocity of the oscillating link c for corre- 
sponding positions of the crank. From 
these diagrams it will be seen that the 
swing to and fro of the cylinder is not 
accomplished in equal times. The in- 
equality so apparent to an observer of the 
oscillating engine is usefully applied as a 
quick-return motion on shaping machines. 
The cutting stroke takes place during the 
slow swing of c, i.e. when the crank-pin is 
traversing the upper portion of its arc, and 
the return stroke is quickly effected while 
the pin is in its lower position. The ratio 
of the mean time occupied in the cutting 
stroke to that of the return stroke is termed 
the " ratio of the gear," which is readily 
determined. The link c is in its extreme 
position when the link « is at right angles to it ; the cutting 
angle is 360 — B, and the return angle B (Fig. 161J). 




Fig. i6ij. 



•38 



Mechanics applied to Engineering. 



The ratio of the gear R = 



360 -g 




or 0(R + i) = 360° 

and a = b cos — 
2 

Let R = 2 ; B = 1 20° J b = 2a. 

^ = 3 J ^ — 9°° > ^ — 'i'42a. 
Another form of quick-return motion is obtained by fixing 
the link a. When the link d is driven at a constant velocity, 
the link b rotates rapidly during one part of its revolution and 
slowly during the other part. The exact 
speed at any instant can be found by 
the method already given for the oscil- 
lating cylinder engine. 

The ratio R has the same value as 
before, but in this case we have 

Q 

Fig. i6k. a — d COS — : therefore d must be made 

2 

equal to 2a for a ratio of 2, and i'42a for a ratio of 3. This 
mechanism has also been used for a steam-engine, but it is best 
known as Rigg's hydraulic engine (Fig. i6i«:). This special 
form was adopted on account of its lending itself readily to a 
variation of the stroke of the piston as may be required for 

various powers. This varia- 
tion is accomplished by shift- 
ing the point Oad to or from 
the centre of the fl5rwheel Oab. 
A very curious develop- 
ment of the steam-engine 
mechanism is found in Stan- 
nah's pendulum pump (Fig. 
161/^. The link C is fixed, 
and the link d simply rocks 
to and fro ; the link a is the 
flywheel, and the pin Oab is 
attached to the rim and works 
in brasses fitted in an eye in 
the piston-rod. 

The velocity of the point 




Fig. i6irf. 



Oab is the same as that of any other point in the link b. 
When C is fixed, the velocity of b relatively to C is the same 
as the velocity of C relatively to b when b is fixed, whence from 
p. 133 we have — 



Mechanisms. 



139 



^ Oad 
'Oab 



R. 
«tfjRj 
R. 



",R, 



= 0)^ 



The Principle of Virtual Velocities applied to 
Mechanisms. — If a force acts on any point of a mechanism 
and overcomes a resistance at any other point, the work done 
at the two points must be equal if friction be neglected. 

In Fig. 162, let the force P act on the link a at the 
point 2. Find the magnitude of the force R at the point 4 
required to keep the mechanism in equilibrium. 

If the bar a be given a small shift, the path of the point 2 




will be normal to the link a, and the path of the point 4 will 
be normal to the radius 5.4. 

Resolve P along Fr parallel to a, which component, of 
course, has no turning effort on the bar ; also along the normal 
P« in the direction of motion of the point 2. 

Likewise resolve R along the radius Rr, and normal to the 
radius R«. 

Now we must find the relative velocity of the points 2 and 
4 by methods previously explained, and shown by the construc- 
tion on the diagram. Then, as no work is wasted in friction, 
we have — ^ 

' The small simultaneous displacements are proportional to the velocities. 



140 Mechanics applied to Engineering. 







^H/- 


e-S 










J?^ 


4 


-~L--^ 




'^3 


,^ 


■^^%: 




e'-^<r\ 




«-r X 




"I ^ 


\ 


\a d N 


kg 

\ 




"^ ^ / 


A 




/'~. 




V 




>io -- 





o 

■a 



13 V 



eed°^°'>^_ 




1 Seconds 



V) o / 3 3 4 5 6 7 e a lo II IS 13 /^ /s /e 



/il Seconds 




Seconds 



\ \ \ • Seconds 



Tio. i6j. 



Mechanisms^ 141 

P„V3 = R„V4 and Rn = ^ 

' 4 

Velocity and Acceleration Curves. — Let the link a 
revolve with a constant angular velocity. Curves are con- 
structed to show the velocity of the point / relatively to the 
constant velocity of the point e. 

Divide the circle that e describes into any convenient 
number of equal parts (in this case 16). The point / will 
move in the arc of a circle struck from the centre g; then, by 
means of a pair of compasses opened an amount equal to ef, 
from each position of e set off the corresponding position of/ 
on the arc struck from the centre g. By joining up he, ef,fg, 
we can thus get every position of the mechanism ; but only one 
position is shown in the figure for clearness. Then, in order to 
find the relative velocity of e and /, we produce the links a 
and c to obtain the virtual centre Obd. This will often come 
off the paper. We can, however, very easily get the relative 
velocities by drawing a line ^'parallel to c. Then the triangles 
hej and Obd.e.f are similar, therefore — 

he _ Obd.e _ velocity of e 
hj ~ Obd.f velocity of/ 

The velocity of e is constant ; let the constant length of the 
link a, viz. he, represent it; then, from the relation above, hj 
will represent on the same scale the velocity of/. 

Set off on a straight line the distances on the e curve o. i, 
1.2, 2.3, etc., as the base of the speed curve. At each point 
set up ordinates equal to 4/' for each position of the mechanism. 
On drawing a curve through the tops of these ordinates, we get 
a complete speed curve for the point / when the crank a 
revolves uniformly. The speed curve for the point if is a 
straight line parallel to the base. 

In constructing the change of speed curve, each of 
the divisions o.i, 1.2, 2.3, etc., represents an interval of one 
second, and if horizontals be drawn from the speed curve as 
shown, the height X represents the increase in the speed during 
the interval o.i, i.e. in this case one second; then on the 
change of speed diagram the height X is set up in the middle 
of each space to show the mean change in speed that the 
point / has undergone during the interval 0.1, and so on 
for each space. A curve drawn through the points so 
obtained is the rate of change of speed curve for the point /. 



142 



Mechanics applied to Engineering. 



The velocity curve is obtained in the same way as the speed 

curve, but it indicates the direction of the motion as well 

as its speed, and similarly in the case of the acceleration 

curve. 

Let the curve (Fig. 164) represent the velocity of any 

point as it moves through space. Let the time-interval 

between the two dotted lines be dt, and the change of velocity 

of the point while passing through that space be dv. Then 

, . ,, • , . . dv . change of velocity 
the acceleration during the interval is — , t.e.—. — - — ; — — -: — -, 

dt interval of time 

or the change of velocity in the given interval. 



By similar triangles, we have — • = — . 
■' ° ' 'f* xy 



dt 




When dt= 1 second, dv = mean acceleration. 
Hence, make xy^ = i on the time scale, then z^y^, the 
subnormal, gives us the acceleration measured on the same 

scale as the velocity. 

The sub-normals to the 
curve above have been plotted 
in this way to give the accelera- 
tion curve from 7 to 16. The 
scale of the acceleration curve 
will be the same as that of the 
velocity curve. 

The reader is recommended 

to refer to Barker's " Graphical 

-r. ' ■ Calculus " and Duncan's 

/ / "Practical Curve Tracing" 

Fig. 164, /T \ ° 

(Longmans). 

Velocity Diagrams for Mechanisms. — Force and 
reciprocal diagrams are in common use by engineers for find- 
ing the forces acting on the various members of a structure, 
but it is rare to find such diagrams used for finding the velocities 
of points and bars in mechanisms. We are indebted to Pro- 
fessor R. H, Smith for the method. (For fuller details, readers 
should refer to his own treatise on the subject.') 

Let ABC represent a rigid body having motion parallel to 
the plane of the paper ; the point A of which is moving with a 
known velocity V^ as shown by the arrow ; the angular velocity 
(It of the body must also be known. If oi be zero, then every 

• " Graphics," by R. H. Smith, Bk. I. chap. ix. ; or " Kinematics of 
Machines," by R. J. Durley (J. Wiley & Sons), 



Mechanisms. 



H3 




point in the body will move with the same velocity as V^. 
From A draw a line at right angles to the direction of motion 
as indicated by Va, then the body is moving about a centre 
situated somewhere on this line, but since we know V^ and 
0), we can find the virtual centre P, since coRi = Va. Join 
PB and PC, which are virtual 
radii, and from which we know the 
direction and velocities of B and C, ' 
because each point moves in a path 
at right angles to its radius, and its 
velocity is proportional to the length 
of the radius ; thus — 

Vb = «).PB, and Vo = u-PC 

The same result can be arrived at 
by a purely graphical process; thus — 

From any pole p draw (i) the 
ray pa to represent Va ; (2) a ray 
at right angles to PB ; (3) a ray at 
right angles to PC. These rays 
give the directions in which the 
points are moving. 

We must now proceed to find 
the magnitude of the velocities. 

From a draw ab at right angles to AB ; then pb gives the 
velocity of the point B. Likewise from b draw be at right angles 
to BC, or from a draw ac at right angles to AC ; then/^ gives 
the velocity of the point C. 

The reason for this construction is that* the rays pa, pb, pc 
are drawn respectively at right angles to PA, PB, and PC, i.e. 
at right angles to the virtual radii ; therefore, the rays indicate 
the directions in which the several points move. The ray- 
lengths, too, are proportional to their several velocities, since 
the motion of B may be regarded as being compounded of a 
translation in the direction of Va and a spin, in virtue of which 
the point moves in a direction at right angles to AB ; the com- 
ponent pa represents its motion in the direction Va, and ab its 
motion at right angles to AB, whence /5 represents the velocity 
of B in magnitude and direction. Similarly, the point C par- 
takes of the general motion /a in the direction Va, and, due to 
the spin of ttie body, it moves in a direction at right angles to 
AC, viz. ac, whence pc represents the velocity of C. The point 
c can be equally well obtained by drawing be at right angles 
toBC. 



144 



Mechanics applied to Engineering. 



The triangular connecting-rod of the Musgrave engine can 
be readily treated by this construction. A is the crank-pin, 
whose velocity and direction of motion are known. The 
pistons are attached to the corners of the triangle by means of 
short connecting-rods, a suspension link DE serves to keep the 
connecting-rod in position, the direction in which D moves is 
at right angles to DE. Produce DE and AF to meet in P, 
which is the virtual centre of AD and FE. Join PB and PC. 
From the pole/ draw the ray /a to represent the velocity of 
the point A, also draw rays at right angles to PB, PC, PD. 
From a draw a line at right angles to AD, to meet the ray at 



Prr^,, 




R-9 



Fig. i66. 



right angles to PD in d, also a line from a at right angles to AB, 
to meet the corresponding ray in b. Similarly, a line from a at 
right angles to AC, to meet the ray in c. Then pb, pc, pd give 
the velocities of the points B, C, D respectively. The velocity 
of the pistons themselves is obtained in the same manner. 

As a check, we will proceed to find the velocities by another 
method. The mechanism ADEF is simply the four-bar 
mechanism previously treated ; find the virtual centre of DE 
and AF, viz. Q; then (see Fig. 153) — 

V^ _ AF . EQ 
Vb ED . QF 



Mechanisms. 



145 



Tlie points C and B may be regarded as points on the 
bar AD, whence — 

PC PB RG 

Vc = Vi>p^, and Vb = V^p^, and V = V^^ 

Let the radius of the crank be 16', and the revolutions 
per minute be 80 ; then — 



V,= 



_ 3'i4 X 2 X 16 X 80 



= 670 feet per minute 



Then we' get — 



Vg = 250 feet per minute 

Vc = 790 

Vd = 515 

Vg = 246 „ „ 

Vh = 792 .. .. 



Cnmkpbi 

•, 



Valve rod 




• Fig. 167. 






By taking one more example, we shall probably cover most 
of the points that are likely to arise in practice. We have selected 
that of a link-motion. Having given the speed of the engine 
and the dimensions of the valve-gear, we proceed to find the 
velocity of the slide-valve. 

In the diagram A and B represent the centres of the 
eccentrics ; the link is suspended from S. The velocity of the 
points A and B is known from the speed- of the engine, and 
the direction of motion is also known of A, B, and S. Choose 
a pole /, and draw a ray pa parallel to the direction of the 
motion of A, and make its length equal on some given scale to 

L 



146 Mechanics applied to Engineering. 

the velocity of A : likewise draw pb for the motion of B. Draw 
a ray from p parallel to the direction of motion of S, i.e. at 
right angles to US ; through a draw a line at right angles to 
AS, to cut this ray in the point j. From s draw a line at right 
angles to ST ; from b draw a line at right angles to BT ; where 
this line cuts the last gives us the point /. join//,/ s, which 
give respectively the velocities of T and S. From / draw a line 
at right angles to TV, and from S a line at right angles to SV ; 
they meet in z', : then pvx is the velocity of a point on the link 
in the position of V. But since V is guided to move in a 
straight line, from v^ draw a line parallel to a tangent at V, and 
from / a line parallel to the valve-rod, meeting in v ; then pv is 
the required velocity of the valve, and vv-^ is the velocity of 
" slip " of the die in the link. 

Cams. — When designing automatic and other machinery 
it often happens that the desired motion of a certain portion 
of the machine cannot readily be secured by the use of ordinary 
mechanisms such as cranks, links, wheels, etc.; cams must 
then be resorted to, but unless they are carefully constructed 
they often give trouble. 

A rotating cam usually consists of a non-circular disc 
formed in such a manner that it imparts the desired recipro- 
cating motion, in its own plane of rotation, to a body or 
follower which is kept in contact with the periphery of the 
disc. 

Another form of cam consists of a cylindrical surface which 
rotates about its own axis, and has one or both edges of its 
curved surface specially formed to give a predetermined 
motion to a body which is kept in contact with them thus, 
causing it to slide to and fro in a direction parallel, or nearly 
so, to the axis of the shaft. 

Generally speaking, it is a simple matter to design a cam 
to give the desired motion, but it is a mistake to assume that 
any conceivable motion whatever can be obtained by means 
of a cam. Many cams which work quite satisfactorily at low 
speeds entirely fail at high speeds on account of the inertia of 
the follower and its attachments. The hammering action 
often experienced on the valve stems of internal combustion 
engines is a familiar example of the trouble which sometimes 
arises from this cause. 

Design of Cams. — (i) Constant Velocity Cams. — In dealing 
with the design of cams it will be convenient to take definite 
numerical examples. In all cases we shall assume that they 
rotate at a constant angular velocity. Let it be required to 



Mechanisms. 



147 



design a cam to impart a reciprocating motion of 2 inches 
stroke to a follower moving at uniform speed {a) in a radial 
path, (5) in a segment of a circle. 

In Fig. 168 three cams of different dimensions are shown, 
each of which fulfils condition {a), but it will shortly be shown 
that the largest, will give more satisfactory results than the 
smaller cams. The method of construction is as follows : — 
Take the base circle abed, say 3 inches diameter. Make ce = 2 
inches, that is, the stroke of the follower. Draw the semicircle 
efgsxA divide it into a convenient number of equal parts — say 




Scale : f ths full size. 



Fig. 168. 



6 — and draw the radii. Divide ag, the path of the follower, 
into the same number of equal parts, and from each division 
draw circles to meet the respective radii as shown, then draw 
the profile of the cam through these points. It will be obvious 
from the construction that the follower rises and falls equal 
amounts for equal angles passed through by the cam, and since 
the latter rotates at a constant angular speed, the follower 
therefore rises and falls at a uniform speed, the total lift being 
ce or ag. The two inner cams are constructed in a similar 
manner, but with base circles of t inch and 5 inch respectively. 
The form of the cam is the well-known Archimedian spiral. 



148 



Mechanics applied to Engineering. 



The dotted profile shows the shape of the cam when the 
follower is fitted with a roller. The diameter of the roller 
must never be altered for any given cam or the timing will be 
upset. 

In Fig. 169 a constant velocity cam is shown which raises 
its follower through 2 inches in g of a revolution a to e, keeps 



Jrd full size. 




Fig. 169. 

it there for 5 of a revolution e Xaf, and then lowers it at a con- 
stant velocity in g of a revolution /to g where it rests for the 
remaining \, g to a. The construction will be readily followed 
from the figure. 

When the follower is attached to the end of a radius bar 
the point in contact with the cam moves in a circular arc ag. 
Fig. 170. In constructing the cam the curved path is 
reproduced at each interval and the points of intersection of 
these paths and the circles give points on the cam. The 
profile shown in broken lines is the form of the cam when the 
radius bar is provided with a roller. 

The cams already dealt with raise and lower the follower 
at a constant velocity. At two points, a and if, Figs. 168 and 
170, and at four points a, e,f, g, Fig. 169, the velocity of the 
follower, if it could be kept in contact with the cam, would be 
instantaneously changed and would thereby require an infinitely 
great force, which is obviously impossible ; hence such cams 



Mechanisms. 



149 



cannot be used in practice unless modified by "easing off" 
at the above-mentioned points, but even then the acceleration 
of the follower may be so great at high speeds that the cam 
face soon wears irregularly and causes the follower to run in 
an unsatisfactory manner. By still further easing the cam may 
be made to work well, but by the time all this easing has been 




Fig. 170. 

accomplished the cam practically becomes a simple harmonic 
cam. 

(ii) Constant acceleration or gravity cam. — This cam, as its 
name suggests, accelerates the follower in exactly the same 
manner as a body falling freely under gravity, hence there is 
a constant pressure on the face of the cam when lifting the 
follower, but none when falling. The space through which a 
falling body moves is given by the well-known relation — 

S = \gfi 

The total spaces fallen through in the given times are propor- 
tional to the values of S given in the table, that is, proportional 
to the squares of the angular displacements of the radius vectors. 



Time, / . . . . 


I 


2 




4 


S 


6 


Space, S . . . . 


I 


4 


9 


16 


-' 2S 


36 


Velocity, v . . . 


I 


2 


3 


4 


s 


6 



ISO 



Mechanics applied to Engineering. 



Acceleration diagram 
for the follower. 




Fig. 17X. 



' ' ' 'y/ 




Fig. 172. 



Mechanisms. 



151 



In Fig. 171 the length of the radius vectors which fall 
outside the circle abed are set off proportional to the values of 
S given in the table. For the sake of comparison a " constant 
velocity " cam profile is shown in broken line. A cam of this 
design must also be " eased " off at e or the follower will leave 
the cam face at this point. As a matter of fact a true gravity 
cam is useless in practice; for this reason designers will do 
well to leave it severely alone. 

In Fig. 172 the cam possesses the same properties, but 
the follower is idle during one half the time — dab— and is then 
accelerated at a constant rate during be for a quarter of a 
revolution and finally is allowed to fall with a constant 
retardation during ed, the last quarter of a revolution. 

(iii) Simple harmonic cam. — In a simple harmonic cam 
the motion of the follower is precisely the same as that of a 
crosshead which is moved by a crank and infinitely long con- 
necting rod, or its equivalent — the slotted crosshead, see 
Fig. 160. There would be no reason for using such a cam 
rather than a crank or eccentric if the motion were required 
to take place in one complete revolution of the cam, but in 

A 




Fig. 173. 



the majority of cases the s. h. m. is required to take place 
during a portion only of the revolution and the follower is 
stationary during the remainder of the time. 



152 Mechanics applied to Engineering. 

In the cam abed shown in Fig. 173 the follower is at 
rest for one half the time, during dab, and has s. h. m. for the re- 
mainder of the time, bed. The circle at the top of the figure 
represents the path of an imaginary crank pin, the diameter 
of the circle being equal to the stroke of the follower. To con- 
struct such a cam the semi-circumference of the equivalent crank 
circle is divided up into a number of equal parts, in this case six, 
to represent the positions of the imaginary crank pin at equal 
intervals of time. These points are projected on to the line ag, 
the path of the follower, and give its corresponding positions. 
From the latter points circles are drawn to cut the corre- 
sponding radii of the cam circle. If a cam be required to give 
s. h. m. to the follower without any 
period of rest the same construction 
may be used, the cam itself (only one 
half of which, afe, is shown) then be- 
coming approximately a circle with its 
centre at h. The distance ho being 
equal to the radius of the imaginary 

as!' 
crank viz. — . If the cam were made 
2 

truly circular, as in Fig. 174, it is 
evident that oh, the radius of the 
crank, and ih, the equivalent connect- 
ing rod, would be of constant Jength, 
hence the rotation of the cam imparts 
the same motion to the roller as a 
Fig 174. crank and connecting rod of the same 

proportions. 
Size of Cams. — The radius of a cam does not in any 
way affect the form of motion it imparts to the follower, but in 
many instances it greatly affects the sweetness of running. A 
cam of large diameter will, as a rule, run much more smoothly 
than one of smaller diameter which is designed to give pre- 
cisely the same motion to the follower. The first case to be 
considered is that of a follower moving in a radial path in 
which guide-bar friction is neglected. 

In Fig. 168 suppose the cam to be turned round until the 
follower is in contact at the point i (for convenience the 
follower has been tilted while the cam remains stationary). 
Draw a tangent to the cam profile at i and let the angle 
between the tangent and the radius be 6. The follower is, for 
the instant, acted upon by the equivalent of an inclined plane 
on the cam face whose angle of inclination is a = 90 - 6. 




Mechanisms. 153 

The force acting normally to the radius, / = W tan (a + ^) 
(seep. 291), where ^ is the friction angle for the cam face 
and follower, and W is the radial pressure exerted by the 
follower. When a + <^ = 90° the follower will jamb in the 
guides and consequently the cam will no longer be able to lift 
it, and as a matter of fact in practice unless a + S^" is much less 
than 90° the cam will not work smoothly. Let the cam rotate 
through a small angle 8^. Then a point on the cam at a 
radius p will move through a small arc of llength 8S = p80. 
During this interval the follower will move through a radial 
distance hh. 

Hence, tan a = — ^ 

Thus for any given movement of the follower, the angle a, which 
varies nearly as the tangent for small angles, varies inversely 
as the radius of the cam p at the point of contact. By referring 
to Fig. 168 it will be seen that the angle a is much greater 
for the two smaller cams than for the largest. It has already 
been shown that the tendency to jamb increases as a increases, 
whence it follows that a cam of small radius has a greater 
tendency to jamb its follower than has a cam of larger radius. 

The friction angle ^ is usually reduced by fitting a roller 
to the follower. 

When the friction between the guides and follower is 
considered, we have, approximately, 

/ = (W +/ tan <^i) (tan {a. + <^\) 

where ^1 is the friction angle between the guide and follower. 

The best practical way of reducing the guide friction of 
the follower is to attach it to a radius arm. 

When the follower moves in a radial path the cam, if 
symmetrical in profile, will run equally well, or badly, in both 
directions of rotation, but it will work better in one direction 
and worse in the other when the follower is attached to a 
radius arm. 

In Fig. 170 let the cam rotate until the follower is in 
contact at the point j, draw a tangent to the profile at this 
point, also a line making an angle <^, the friction angle, with it. 
The resultant force acting on the follower is ib where id is the 
pressure acting normally to the radius of the follower arm iy 
and ic is drawn normal to the tangent at the point i. The line 
di is drawn normal to id. The force ib has a clockwise 
moment ib X r about the pin y tending to lift the follower, if 
the slope of the cam is such that bi passes through y the cam 



1 54 



Mechanics applied to Engineering. 



will jamb, likewise if in this particular case the moment of the 
force about y is contra-clockwise, no amount of turning effort 
on the cam will lift the follower. 

Velocity ratio of Cams and Followers. — The cam 
A shown 'in Fig. 175 rotates about the centre Oab. Since 
the follower c moves in a straight guide, the virtual centre Obc 
is on a line drawn at right angles to the direction of ihotion of 
c and at infinity. Draw a tangent to the cam profile at the 
point of contact y which gives the direction in which sliding 
is taking place at the instant. The virtual centre of A and C 
therefore lies on a line passing through the point of contact y 
and normal .to the tangent, it also lies on the line Oab Obc, 




Fig. 175. 



Fig. 176. 



therefore it is on the intersection, viz. Oac. The virtual radius 
of the cam is Oab . x, i.e. the perpendicular distance of Oab from 
the liormal at the point of contact. Let the angular velocity 
of the cam be ua, then 0)^ x Oab . x = V where V is the com- 
ponent of C's velocity in the direction yx. In the triangle of 
velocities, V„ represents the velocity of c in the guides, and 
V, the velocity of sliding. This triangle is similar to the 
triangle Oab ■ Oac , x . hence 



V 



Oab . Oac 
Oab .X 



Mechanisms. 1 5 S 

Substituting the value of V we have 

V„ _ Oah . Oac 

(i)j^ . Oab . X Oab . x 

V 
and — = Oab . Oac 

("a 

In the case in which the follower is attached to an arm as 
shown in the accompanying figure, Fig. 176, the virtual centres 
are obtained in a similar manner, and 

toSibc . Oac = wjdab . Oac 
u„ _ Oab . Oac 
0)^ Obc . Oac 

we also have the velocity V^ of the point of the follower in 
contact with the cam, 

Y, = ,^j:)bc.y 



ft), Oab . Oac V, 



0), Obc. Oac Obey X &)« 

V„ Oab. Oac. Obey ^ , ^ 

— = - — „, „ = Oab . Oac 

o)« Obc . Oac 



f Obey \ 
\0bc.0ac) 



When the follower moves in a radial path the virtual 
centre is at infinity and the quantity in the bracket becomes 
unity; the expression then becomes that already found for 
this particular case. 

In Figs. 171, 172, 173, polar velocity diagrams are given 
for the follower ; the velocity has been found by the method 
given above, a tangent has been drawn to the cam profile 
at y, the normal to which cuts a radial line at right angles 
to the radius Oy in the point z, the length Oz is then trans- 
ferred to the polar velocity diagram, viz. OV, in some cases 
enlarged for the sake of clearness. 

In Figs. 168, 170, the velocity of the follower is zero 
at a and e, and immediately after passing these points the 
velocity is finite ; hence if the follower actually moved as 
required by such cams, the change of velocity and the accele- 
ration would be infinite at a and e, which is impossible ; hence 
if such cams be used they must be eased off at the above- 
mentioned points. 

In Fig. 169 there are four such points; the polar diagram 



156 Mechanics applied to Engineering. 

indicates the manner in which the velocity changes ; the 
dotted lines show the effect of easing oif the cam at the said 
points. 

The cams shown in Figs. 171, 172, are constructed to 
give a constant acceleration to the. follower, or to increase the 
velocity of the follower by an equal amount in each succeed- 
ing interval of time ; hence the polar velocity diagram for 
such cams is of the same form as a constant velocity cam, viz. 
an Archimedian spiral. 

Having constructed the polar velocity diagram, it is a 
simple matter to obtain the acceleration diagram from it, 
since the acceleration is the change of velocity per second. 
Assume, in the first place, that the time interval between any 
two adjacent radius vectors on the polar velocity diagram is 
one second, then the acceleration of the follower — to a scale 
shortly to be determined — is given by the difference in length 
between adjacent radius vectors as shown by thickened lines 
in Figs. 171, r72, 173. 

The construction of the acceleration diagram for the 
simple harmonic cam is given in Fig. 173. The radius 
vector differences on the polar diagram give the mean accele- 
rations during each interval ; they are therefore set off at the 
middle of each space on the base line ij, and at right angles 
to it. It should be noted that these spaces are of equal length 
and, moreover, that the same interval of time is taken for the 
radius vector differences in both cams. In the case of the 
cam which keeps the follower at rest for one half the time, 
the lifting also has to be accomplished in one half the time, or 
the velocity is doubled ; hence since the radial acceleration 
varies as the square of the velocity of the imaginary crank 
pin, the acceleration for the half-time cam is four times as 
great as that for the full time cam. Similar acceleration 
diagrams are obtained by a different process of reasoning in 
Chapter VI., p. 180. Analytical methods are also given for 
arriving at the acceleration in such a case as that now under 
consideration. 

The particular case shown in Fig. 173 is chosen because 
the results obtained by the graphic process can be readily 
checked by a simple algebraic expression shortly to be given. 
In the above-mentioned chapter it is shown that when a crank 
makes N revolutions per minute, and the stroke of the cross- 
head, which is equivalent to the travel of the follower in the 
case of a simple harmonic cam, is 2R (measured in feet), the 
maximum force in pounds weight, acting along the centre line 



Mechanisms. 157 

of the mechanism, required to accelerate a follower of W 
pounds is o'ooo34WRN'^ for the case in which the follower is 
lifted to its full extent in half a revolution of the cam, such as 
afe (only one half of the cam is shown). But when there is an 
idle period, such as occurs with the cam abe. Fig. 173, the 
velocity with which the follower is lifted is greater than before 

in the ratio where a. is the angle passed through by the 

a 

cam while lifting the follower. Since the radial acceleration 

varies as the square of the velocity, we have for such a case— 

The maximum force 'j 
in pounds weight I _ o'ooo34WRN2 x i8o2 _ ii'osWRN^ 
required to accele- j ~ ^2 ~ ^,2 

rate the follower ) 

The pressure existing between the follower and the face of 

a simple harmonic cam, when the follower falls by its own 

weight is — 

,-. , II-03WRN2 
maximum pressure, W -\ ^^ ■ 

a." 

. . „, iro3WRN2 

mmimum pressure, W =^=-r 

a'' 

When the follower is just about to leave the cam face — 

iro3WRN2 

w ^ 

a" 

and the speed at which this occurs is — 

Vr 

Hence a simple harmonic cam must never run at a higher 
speed than is given by this expression, unless some special 
provision is made to prevent separation. A spring attached 
either directly to the follower or by means of a lever is often 
used for this purpose. When directly attached to the follower, 
the spring should be so arranged that it exerts its maximum 
effort when the follower is just about to leave the cam face, 

II-03WRN2 
which should therefore be not less than 5 W, 

when the follower simply rests on the cam, and 

n-o-!WRN2 

^^—5 + W, when the follower is below the cam. 



158 



Mechanics applied to Engineering. 



The methods for finding the dimensions of such springs 
are given in Chapter XIV. 

Instead of a spring, a grooved cam (see Fig. 177), which 
is capable of general application, or a cam with double rollers 





Fig. 177. Fig. 178. 

on the follower (see Fig. 178), can be used to prevent sepa- 
ration, but when wear takes place the mechanism is apt to be 
noisy, and the latter is only applicable to cases where the 
requisite movement can be obtained in 180° movement of the 
cam. When the follower is attached to an arm, the weight 
of the arm and its attachments Wa must be reduced to the 
equivalent weight acting on the follower. 

Let K = the radius of gyration of the arm about the pivot. 
r„ = the radius of the c. of g. of the arm about the 

pivot. 
r = the radius of the arm itself from the follower to 
the pivot. 
We = the equivalent weight acting on the follower when 

considering inertia effects. 
W„ = Ditto, when considering the dead weight of the arm. 

W K W r 

Then W, = -^-^ and W„ = ^^^^ 



Then 



II-03WRN2 iro3W,KRN2 



and W becomes 



a." 

W„/-. 



in the expressions above for the case 



when the follower is attached to a radius arm. 



Mechanisms, 



159 



General Case of Cam Acceleration. — In Fig. 179, 
velocity and acceleration diagrams have been worked out for 
such a cam as that shown in Fig. 175. The polar velocity 
diagram has been plotted abonj a zero velocity circle, in 
which the radius vectors outside the circle represent velocities 
of the follower when it is rising, and those inside the circle 
when it is falling ; they have afterwards been transferred to a 
straight base, and by means of the method given on p. 141, 
the acceleration diagram has also been constructed. It will 
be seen that even with such a simple-looking cam the accele- 



i 6 z 




ration of the follower varies considerably in amount and 
rapidly, and the sign not unfrequently changes. 

Scale of Acceleration Diagrams.— Let the drawing of 

the cam be -th full size,- and let the length of the radius 
n 

vectors on the polar velocity diagram be m times the corre- 
sponding length obtained from the cam drawing. In Fig. 
171, /« = 3. In Fig. 172, m = X. In Fig. 173, m = 2. 

27rN 

Let the cam make N revolutions per mmute, or -g— 

= radians per second. 

9"55 
The velocity of the follower at any point where the length 

of the radius vector is OV (see Fig. 173) measured in feet, is 
V, = feet per second. 



i6o 



Mechanics applied to Engineering. 



One foot length on the polar diagram represents 



inch 



9"5S»2| 



feet 

per 

second 



Let the angle between the successive radius vectors be 6 
degrees, and let the change of velocity during one of the 
intervals be SV. Then each portion of the circle subtending 

9 represents an interval of time 8i = ^ ^^ = ,-tt- seconds. 
^ 360N 6N 

The acceleration of the follower is — 



8V _ 8/ X N2 X « X 6 
8/ ^ 



'X N2x« 



feet per second 
per second. 



9"55 Xfnxd 1-59 XmX6) 

Where 8/ is the difference in length of two adjacent radius 

vectors, measured in feet. The numeral in the denominator 

becomes ig-i if 8/ is measured in inches. 

The force due to the acceleration of the follower is in all 

W 
cases found by multiplying the above expressions by — or 

•§" 
the equivalent when a rocking arm is used. 

Useful information on the designing and cutting of cams 
will be found in pamphlet No. 9 of " Machinery's " Reference 
Series, published by " Machinery," 27, Chancery Lane ; "A 
Method of Designing Cams," by Frederick Grover, A.M.LC.E. ; 
Proceedings I.C.E., Vol. cxcii. p. 258. 

The author wishes to acknowledge his indebtedness to 
Mr. Frederick Grover, of Leeds, for many valuable sugges- 
tions on cams. 

the figure let A and B be the 
A rotates at a given speed ; it is 
required to drive B at some 
predetermined speed of rota- 
tion. If the shafts be pro- 
vided with circular discs a 
and b of suitable diameters, 
and whose peripheries are 
kept in contact, the shaft A 
will drive B as desired so 
long as there is sufficient 
friction at the line of contact, 
but the latter condition is 
the twisting moment is small, and no 
In order to prevent slipping each wheel 



Toothed Gearing. — In 

end elevations of two shafts. 




Fig. 180. 



only realized when 
slipping takes place. 



Mechanisms. 1 6 1 

must be provided with teeth which gear with one another, and 
the form of which is such that the relative speeds of the two 
shafts are maintained at every instant. 

The circles representing the two discs in Fig. i8o are 
known as the pitch circles in toothed gearing. The names 
given to the various portions of the teeth are given in Fig. i8i. 



Clearance — ' /f\ —-.^^^^ 

fioot circle \^ 

Fig. i8i. 

The pitch P is measured on the pitch circle. The usual 
proportions are A = o'sP, B = 0*47 to o'48P, C = 0-53 to 
0-52P. The clearance at the bottom of the teeth is o'lP, thus 
the total depth of the tooth is 07 P. The width is chosen to 
suit the load which comes on each tooth — for light wheels it is 
often as small as o'sP and for heavy gearing it gets up to 5 P. 
Most books on machine design assume that the whole load is 
concentrated on the corner of the tooth, and the breadth B 
calculated accordingly, but gearing calculated on this basis is 
far heavier than necessary. 

Velocity Ratio. — Referring to Fig. 180, let there be 
sufificient friction at the line of contact to make the one wheel 
revolve without slipping when the other is rotated, if this b.e so 
the linear velocity of each rim will be the same. 
Let the radius of a be r„, of bhe. r^; 

the angular velocity of a be a)„, of ^ be wj ; 

N^ = the number of revolutions of « in a unit of 

time; 
N} = the number of revolutions of ^ in a unit of time. 

Then m^ = — ^— ? = 27rN„, and wj = 27rNj 

'a 

„, '■»_"«_ ^irN^ _ N„ 

^^— — — sr ^irr 

7\ u)j 2TrNi, Nj 

M 



1 62 Mechanics applied to Engineering. 

or the revolutions of the wheels are inversely proportional to 
their respective radii. 

The virtual centres of a and c, and of b and c, are evidently 
at their permanent centres, and as the three virtual centres 
must lie in one line (see p. 126), the virtual centre Oab must 
lie on the line joining the centres of a and b, and must be a 
point (or axis) common to each. The only point which fulfils 
these conditions is Oab, the point of contact of the two discs. 
To insure that the velocity ratio at every instant shall be 
constant, the virtual centre Oab must always retain its present 
position. We have shown that the direction of motion of any 
point in a body moving relatively to another body is normal 
to the virtual radius; hence, if we make a projection or a 
tooth, say on A (Fig. 182), the direction of motion of any 
point d relatively to B, will be a normal to the line drawn from 
d through the virtual centre Oab. 

Likewise with any point in B relatively to A. Hence, if a 
projection on the one wheel is required to fit into a recess in 

the other, a normal to their 

^ (aW\ surfaces at the point of con- 

^5~\^I^Wj \. tact must pass through the 

/^\~7/w.^/\v virtual centre Oah. If such 

/A \/ '^^^^^^\p\ ^ normal do not pass through 

_ ^v.^_^ccE^ _\.J^- Oa*, the velocity ratio will 

^ — r ^^ be altered, and if Oab shifts 

J\ about as the one wheel 

^ moves relatively to the 

Fig. 182. Other, the motion will be 

jerky. 
Hence, in designing the teeth of wheels, we must so form 
them that they fulfil the condition that the normal to their 
profiles at the point of contact must pass through the virtual 
centre of the one wheel relatively to the other, i.e. the point 
where the two pitch circles touch one another, or the point 
where the pitch circles cut the line joining their centres. An 
infinite number of forms might be designed to fulfil this con- 
dition; but some forms are more easily constructed than 
others, and for this reason they are chosen. 

The forms usually adopted for the teeth of wheels are the 
cycloid and the involute, both of which are easily constructed 
and fulfil the necessary conditions. 

If the circle e rolls on either the straight line or the arc of 
a circle/, it is evident that the virtual centre is at their point 
of contact, viz. Oef\ and the path of any point d in the circle 



Mechanisms. 



163 



moves in a direction normal to the line joining d to Oef, or 
normal to the virtual radius. When the circle rolls on a 
straight line, the curve traced out is termed a cycloid (Fig. 
183) ; when on the outside of a circle, the curve traced out is 




Fig. 184. 

termed an epicycloid (Fig. 184) ; when on the inside of a circle, 
the curve traced out is termed a hypocycloid (Fig. 185). 

If a straight line / (Fig. 
186) be rolled without slipping 
on the arc of a circle, it is 
evident that the virtual centre 
• is at their point of contact, 
viz. Oef, and the path of any 
point d in the line moves in 
a direction normal to the line 
/, or normal to the virtual 
radius. The curve traced out 
by d is an involute. It may 
be described by wrapping a 
piece of string round a circular disc and attaching a pencil 
at ^ ; as the string is unwound d moves in an involute. 

When setting out cycloidal teeth, only small portions of the 
cycloids are actually used. 
The cycloidal portions can d .7^ 
be obtained by construc- 
tion or by rolling a cir- 
cular disc on the pitch 
circle. By reference to 
Fig. 187, which represents 
a model used to demonstrate the theory of cycloidal teeth, the 
reason why such teeth gear together smoothly will be evident. 





Fig. 186. 



164 



Mechanics applied to Engineering. 



A and B are parts of two circular discs of the same diameter 
as the pitch circles ; they are arranged on spindles, so that 
when the one revolves the other turns by the friction at the line 
of contact. Two small discs or rolling circles are provided with 




Fig. 187. 



double-pointed pencils attached to their rims ; they are pressed 
against the large discs, and turn as they turn. Each of the 
large discs, A and B, is provided with a flange as shown. 
Then, when these discs and the rolling circles all turn together, 



Mechavisms. i65 

the pencil-point i traces an epicycloid on the inside of the 
flange of A, due to the rolling of the rolling circle on A, in 
exactly the same manner as in Fig. 184 ; at the same time the 
pencil-point 2 traces a hypocycloid on the side of the disc B, 
as in Fig. 185. Then, if these two curves be used for the pro- 
files of teeth on the two wheels, the teeth will work smoothly 
together, for both curves have been drawn by the same pencil 
when the wheels have been revolving smoothly. The curve 
traced on the flange of A by the point i is shown on the lower 
figure, viz. i.i.i ; likewise that traced on the disc B by the 
point 2 is shown, viz. 2.2.2. In a similar manner, the curves 
3.3.3, 4.4.4, have been obtained. The full-lined curves are those 
actually drawn by the pencils, the remainder of the teeth are 
dotted in by copying the full-lined curves. 

In the model, when the curves have been drawn, the discs 
are taken apart and the flanges pushed down flush with the 
inner faces of the discs, then the upper and lower parts of the 
curves fit together, viz. the curve drawn by 3 joins the part 
drawn by 2 ; likewise with i and 4. 

From this figure it will also be clear that the point of 
contact of the teeth always lies on the rolling circles, and that 
contact begins at C and ends at D. The double arc from C 
to D is termed the " arc of contact " of the teeth. In order 
that two pairs of teeth may always be in contact at any one 
time, the arc CD must not be less than the pitch. The 
direction of pressure between the teeth when friction is 
neglected is evidently in the direction of a tangent to this arc 
at the point of contact. Hence, the greater the angle the 
tangent makes to a line EF (drawn normal to the line joining 
the centres of the wheels), the greater will be the pressure 
pushing the two wheels apart, and the greater the friction on 
the bearings ; for this reason the angle is rarely allowed to be 
more than 30°. The effect of friction is to increase this angle 
during the arc of approach by an amount equal to the friction 
angle between the surfaces of the teeth in contact, and to 
diminish it by the same amount during the arc of recess. The 
effect of friction in reducing the efficiency is consequently 
more marked during approach than during recess, for this 
reason the teeth of the wheels used in watches and clocks are 
usually made of such a shape that they do not rub during the 
arc of approach. In order to keep this angle small, a large 
rolling circle must be used. 

In many instances the size of the rolling circle has to be 
carefully considered. A large rolling circle increases the path 



1 66 Mechanics applied to Engineering. 

of contact and tends to make the gear run smoothly with a 
small amount of outward thrust, but as the diameter of the 
rolling circle is increased the thickness of the tooth at the 
flank is decreased and consequently weakened. If the 
diameter of the rolling circle be one half that of the pitch 
circle the flanks become radial, and if larger than that the 
flanks are undercut. Hence, in cases where the strength of 
the teeth is the ruling factor, the diameter of the rolling circle 
is never made less than one half the diameter of the smallest 
wheel in the train. 

Generally speaking the same rolling circle is used for all 
the wheels required to gear together, but in special cases, 
where such a practice might lead to undercut flanks in the 




Fig. i88. 

small wheels of a train, one rolling circle may be used for 
generating the faces of the teeth on a wheel A, and the flanks 
of the teeth they gear with on a wheel B, and another size of 
rolling circle may be used for the flanks of A and the faces of 
B. In some instances the teeth are made shorter than the 
standard proportions in order to increase their strength. 

In setting out the teeth of wheels in practice, it is usual 
to make use of wooden templates. The template A (Fig. i88) 
is made with its inner and outer edges of the same radius as 
the. pitch circle, and the edge of the template B is of the same 
radius as the rolling circle. A piece of lead pencil is attached 
by means of a clip to the edge of the template, and having its 
point exactly on the circumference of the circle. The template 



Mechanisms. 



167 



A is kept in place on the drawing paper 
weight or screws, a 
pencil run round the 
convex edge gives the 
pitch circle. The tem- 
plate B is placed with 
the pencil point just 
touching the pitch circle; 
it is then rolled, without 
slipping, on the edge of 
the template A and the 
pencil traces out the 
epicycloid required for 
the face of a tooth. The 
template A is then 
shifted until its concave 
edge coincides with the 
pitch circle. The tem- 
plate B is then placed 
with the pencil point on 
the pitch line, and co- 
inciding with the first 
point of the epicycloid, 
then when rolled upon 
the inside edge of the 
template A, the pencil 
traces out the hypocy- 
cloid which gives the 
profile of the flank of 
the tooth. A metal tem- 
plate is then carefully 
made to exactly fit 
the profile of one side 
of the tooth thus ob- 
tained, and the rest of 
the teeth are set out by 
means of it. It is well 
known that cycloidal 
teeth do not work well 
in practice in cases 
where it is difficult to 
ensure ideal conditions 
as regards constancy of 
shaft centres, perfection 



by means of a heavy 





1 68 Mechanics applied to Engineering. 

of workmanship, freedom from grit, etc. For this reason 
involute teeth are almost universally used for engineering 
purposes. 

The model for illustrating the principle of involute teeth is 
shown in Fig. 189. Here again A and B are parts of two 
circular discs connected together with a thin cross-band which 
rolls off one disc on to the other, and as the one disc turns it 
makes the other revolve in the opposite direction. The band 
is provided with a double-pointed pencil, which is pressed 
against two flanges on the discs ; then, when the discs turn, the 
pencil-points describe involutes on the two flanges, in exactly 
the same manner as that described on p. 164. 

Then, from what has been said on cycloidal teeth, it is 
evident that if such curves be used as profiles for teeth, the 
two wheels will gear smoothly together, for they have been 
drawn by the same pencil as the wheels revolved smoothly 
together. 

The point of contact of the teeth in this case always lies 
on the band ; contact begins at C, and ends at D. The arc 
of contact here becomes the straight line CD. In order to 
prevent too great pressure on the axles of the wheels, the 
angle DEF seldom exceeds is|° ; this gives a base circle |f of 
the pitch circle. In special cases where pinions are required 
with a small number of teeth, this angle is sometimes increased 
to 20°. 

Involute teeth can be set off by templates similar to those 
shown in Fig. 188, but instead of the template A being made 
to fit the pitch circle, it is curved to fit the base circle, and 
the template B is simply a straight-edge with a pencil attached 
and having its point on the edge itself. 

If for any reason the distance between the centres of two 
involute gear wheels be altered by a small amount, the teeth 
will still work perfectly, provided the path of contact is not 
less than the pitch. By reference to Fig. 189 it will be evident 
that altering the wheel centres only alters the diameters of the 
pitch circles, but does not affect the diameters of the base 
circles upon which the velocity ratio entirely depends. This 
is a very valuable property of involute teeth, and enables them 
to be used in many places where the wheel centres cannot for 
many reasons be kept constant. If the same angle DEF, 
Fig. 189, be used in setting out the teeth, all involute wheels 
of the same pitch will gear with one another. 

The portions of the flanks inside the base circles are made 
radial. 



Mechanisms. 



169 



Readers interested in mechanical devices for drawing the 
teeth of wheels and the pistons for rotary pumps and blowers 
should refer to a paper by Dr. Hele-Shaw, F.R.S., in the 
British Association Report for 1898, an extract of which will 
be found in Dunkerley's " Mechanism " (Longmans). The 
general question of the design of toothed gearing will also 
be found in standard books on machine design. Readers 
should also refer to Anthony's " Essentials of Gearing " (D. C. 
Heath & Co., Boston, U.S.A.), and the series of pamphlets 
published by " Machinery," 27, Chancery Lane. No. i,iWorra 
Gearing; No. 15, Spur Gearing; No. 20, Spiral Gearing. 

Velocity Ratio of Wheel Trains. — In most cases 
the problem of finding the velocity ratio of wheel trains is 




Fig. 190. 



Fig. 191. 



easily solved, but there are special cases in which difficulties 
may arise. The velocity ratio may have a positive or a 
negative value, according to Hob. form of the wheels used; thus 
if a in Fig. 190 have a clockwise or + rotation, h will have an 
anti-clockwise or — rotation; but in Fig. 191 both wheels 
rotate in the same sense, since an annular wheel, i.e. one with 
internal teeth, rotates in the reverse direction to that of a 
wheel with external teeth. In both cases the velocity ratio is — 

V = ?:» = I-' = ^ai-^bc ^ N, 
' R, T„ Qab.Oac N^ 

where T,, is the number of teeth in a, and T5 in b, and N„ is the 
number of revolutions per minute of a and N5 of b. 

In the case of the three simple wheels in Fig. 192, we have 
the same peripheral velocity for all of them ; hence — 

o)„R„ = — wjRj = co„R„ 

<«5 ^ R„ ^ Tj ^ N^ 

°' a,, R,. T„ N„ 



17 o 



Mechanics applied to Engineering. 



and the first and last wheels rotate in the same sense. The 
same velocity ratio could be obtained with two wheels only, 
but then we should have the sense of rotation reversed, since — 



or - = - — 

<«s R. 




Fig. 192. 



Thus the second or " idle " wheel simply reverses the sense 
of rotation, and does not affect the velocity ratio. The velocity 

ratio is the same in Figs. 
191, 192, 193, but in the 
second and third cases the 
sense of the last wheel is 
the same as that of the 
first. When the radius line 
R„ of the last wheel falls on 
the same side of the axle 
frame as that of the first 
wheel, the two rotate in the 
same sense; but if they fall on opposite sides, the wheels 
rotate in opposite senses. When the second wheel is com- 
pound, i.e. when two 
wheels of different sizes 
are fixed to one another 
and revolve together, it 
is no longer an idle 
wheel, but the sense of 
rotation is not altered. 
If it is desired to get the 
same velocity ratio with 
an idle wheel in the train 
the wheel c must be altered 




as with 



Fig. 193. 

a compound wheel, 
b 



in the proportion r„ where b is the driven and V the driver. 
The velocity ratio V, of this train is obtained thus — 

and - (OiRy = <o„R„ 
Substituting the value of — wj, or eoj, we get — 
<«.R„R» 



<iR» = 



R»' 



andV, = '!^« = |4^: 

0)„ RaRj- 



T,T, 



N2 



Mechanisms. 



171 



Thus, taking a as the driving wheel, we have for the velocity 
ratio — 

Revolutions of driving wheel 



Revolutions of the last wheel in train 

_ product of the radii or number of teeth in driven wheels 
product of the radii or number of teeth in driving wheels 

The same relation may be proved for any number of wheels 
in a train. 

If C were an annular wheel, the virtual centres would be 
as shown. R„ is on the opposite side of d to R„ ; therefore 
the wheel c rotates in the opposite sense to a (Fig. 194). 

In some instances the wheel C rotates on the same axle 
as a ; such a case as this is often met with in the feed arrange- 
ments of a drilling-machine (Fig. 
195). The wheel A fits loosely on 
the outside of the threads of the 
screwed spindle S, and is driven by 
means of a feather which slides in 
a sunk keyway, the wheels B and 
B' are both keyed to the same 
shaft ; C, however, is a nut which 
works freely on the screw S. 
Now, if A and C make the same 
number of revolutions per minute, 
the screw will not advance, but if 
C runs faster than A, the screw will advance ; the number of 
teeth in the several wheels are so arranged that C shall do so. 

For example — Let A have 30 

teeth, B, 20, B', 21, C, 29, 
and the screw have four threads 
per inch : find the linear ad- 
vance of the screw per revo- 
lution of S. For one revolu- 
tion of A the wheel C makes 

30 X 21 „„o 

= 5f2 = i'o86 revo- 

20 X 29 ^*° 

lution. Thus, C makes o-o86 

revolution relatively to A per 

revolution of the spindle, or it 

advances the screw H:^ 




Fig. 194. 




Jig. ips. 



o"o2i inch per revolution of S. 
1- 
Change Speed Gears. — The old-fashioned back gear 



1/2 



Mechanics applied to Engineering. 



so common on machine tools is often replaced by more con- 
venient methods of changing the speed. The advent of 
motor-cars has also been responsible for many very ingenious 
devices for rapidly changing speed gears. The sliding key 
arrangement of Lang is largely used in many change-speed 
gears ; in this arrangement all the wheels are kept continuously 
in mesh, and the two which are required to transmit the power 
are thrown into gear by means of a sliding cotter or key. In 
Fig. 196 the wheels on the shaft A are keyed, whereas the 
wheels on the hollow shaft B are loose, the latter wheels are 
each provided with six keyways, a sliding key or cotter which 
passes through slots in the shaft engages with two of these key- 
ways in one of the desired wheels. The wheel to be driven is 




Fig. 196. 

determined by the position of the key, which is shifted to and 
fro by means of a rod which slides freely in the hollow shaft. 
The bosses of the wheels are counterbored to such an extent that 
when the key is shifted from the one wheel to the next both 
keyways are clear of the key, and consequently both wheels are 
free. The sliding rod is held in position by a suitable lever 
and locking gear, which holds it in any desired position. 

Epicyclic Trains. — In all the cases that we have con- 
sidered up to the present, the axle frame on which the wheels 
are mounted is stationary, but when the frame itself moves, its 
own rotation has to be added to that of the wheels. In the 
mechanism of Fig. 197, if the bar the fixed, and the wheel a 
be rotated in clockwise fashion, the point x would approach 
c, and the wheel b would rotate in contra-clockwise fashion. 



Mechanisms. 



173 




If a be fixed, the bar c must be moved in contra-clockwise 
fashion to cause c and x to approach, but b will still continue 
to move in contra-clockwise fashion. Let c be rotated through 
one complete revolution in contra-clockwise fashion; then h 
will make — N revolutions due to 

the teeth, where N = 7^, and at the 
lb 

same time it will make — 1 revolu- 
tion due to its bodily rotation round 
a, or the total revolutions of h will 
be — N — I or — (N -f i) revolu- 
tions relative to a, the fixed wheel. '^'°' '*'" 
The — sign is used because both the arm and the wheel rotate 
in a contra-clockwise sense. But if b had been an annular 
wheel, as shown by a broken line, its rotation would have been 
of the opposite sense to that of c; consequently, in that case, 
b would make N — i 
revolutions to one of c. 
If either idle' or com- 
pound wheels be intro- 
duced, as in Fig. 198, 
we get the revolutions of 
each wheel as shown for 
each revolution of the 

T 
arm e, where v^j, = ~, 



and v.^ = : 



when there 




is an idle wheel between, 

or z/,e = V '^ ° when 

there is a compound 
wheel. 

In the last figure the 
wheel c is mounted loosely on the same axle as a and d. In this 
arrangement neither the velocity ratio nor the sense is altered. 

The general action of all epicyclic trains may be summed 
up thus : The number of revolutions of any wheel of the train 
for one revolution of the arm is the number of revolutions 
that the wheel would make if the arm were fixed, and the first 
wheel were turned through one revolution, -|- i for wheels that 
rotate in the same sense as the arm, and - i for wheels that 
rotate in the opposite sense to the arm. 



174 



Mechanics applied to Engineering. 



In the case of simple, i.e. not bevil trains, it should be 
remembered that wheels on the «th axle rotate in the same 
sense as the arm when n is an even number, and in the opposite 
sense to that of the arm when n is an odd number, counting 
the axle of the fixed wheel as " one." Hence we have — 



« 


Sense of rotation of »th wheel. 


Even 


Same as arm. 


Odd 


Same as arm if V, is less than i. 


Opposite to arm if V, is greater than I. 



Epicyclic Bevil Trains. — When dealing with bevil 
trains the sense of rotation of each wheel must be carefully 




Fig. igg. 

considered, and apparently no simple rule can be framed to 
cover every case; in Fig. 199 the several wheels are marked 
S for same and O for opposite senses of rotation — arrows on 
the wheels are of considerable assistance in ascertaining the 
sense of rotation. The larger arrows indicate the direction in 
which the observer is looking. 



Mechanisms. 



I7S 



JEEM 

11!.""^ . Jl 



Qlueruer 
U m/cUui tllli 



wcu/ 




Fig. 



The bevil train shown in Fig. 200 is readily dealt with, 
thus let T„, T„, etc., represent the number of teeth in a and c 
respectively — b is an idle 
wheel, and consequently 
Tj does not affect the 
velocity ratio. Fix D, and 
rotate a contra-clockwise 
through one revolution, 

T 

then c makes ~ revolu- 

tions in a clockwise 

direction. Fix a and 

rotate the arm D through 

one clockwise revolution. 

Then since the tooth of 

b, which meshes with the stationary tooth of a, may be regarded 

as the fulcrum of a lever ; hence the tooth of b, which meshes 

with c, moves in the same direction as D, and at twice the speed. 

T 

Hence the number of revolutions of cis 7=r + i for one revolution 

of D. The problem may also be dealt with in the same manner 
as the simple epicyclic train. The number of revolutions of any 
wheel of the train for one revolution of the arm is the number 
of revolutions that the wheel would make if the arm were 
fixed and the first wheel were turned through one revolution, 
+ I for wheels that rotate in the same sense as the arm, and 
— I for wheels that rotate in the opposite sense to the arm. 

For elementary bevil trains, such as that shown in Fig. 
200, wheels on the «th axle rotate in the opposite sense to the 
arm when n is an even number, and in the same sense when 
n is an odd number. Hence we have for elementary bevil 
trains — 



n 


Sense of rotation of «th wheel. 


Odd 


Same as arm. 


Even 


Same as arm if V, is less than i. 


Opposite to arm if V, is greater than i . 



1/6 



Mechanics applied to Engineering. 



In all cases the actual number of revolutions per minute of 
the several wheels calculated for one revolution of the arm 
must be multiplied by the number of revolutions per minute 
of the arm N^, also, if the wheel a rotates, its revolutions per 
minute must be added, with due regard to the sign, to the 
revolutions per minute of each wheel calculated for a stationary 
wheel. 

The following table may be of assistance in this connection. 







T„ 


T. 


N» 


N* 


N„ when ~- = i. 


N„ when -^ = 1-5. 




V„+. = 2. 


V„+. = =-5. 


lOO 





-100 


-ISO 


lOO 


5 


- 100 + 2 X 5=- 90 


-I50+2-SX 5 = -i37"S 


— lOO 


5 


100-J-2X s= no 


150+2-5 X 5= i62-s 


100 


-s 


-100— 2X 5=-IIO 


- 150-2-5 X 5= -162-5 


lOO 


20 


- 100+2 X 20=— 60 


- 150+2-5 X 20= — 100 


lOO 


50 


- 100+2 X 50= 


- 150+2-5 X 50=- 25 


100 


60 


- 100+2 X 60= 20 


- 150+2-5 X 60= 


lOO 


70 


- 100+2 X 70= 40 


- 150+2-5 X 70= 25 


lOO 


100 


-100+2X100= 100 


-150+2-5x100= 100 


— 100 


100 


IOO+2XIOO= 300 


150+2-5x100= 400 


100 


— 100 


-100— 2X100= -300 


— 150 — 2-5 X 100= —400 


40 


30 


- 40+2 X 30= 20 


- 60+2-5 X 30= 15 


40 





-40 


-60 





S 


2X5 = 10 


2-5x5=12-5 





-100 


2(- 100) =-200 


2-5(-I00)=-2SO 



Humpage's Gear. — This compound epicyclic bevil train 
is used by Messrs. Humpage, Jacques, and Pedersen, of Bristol, 
as a variable-speed gear for machine tools (see The Engineer, 
December 30, 1898). 

The number of teeth in the wheels are : A = 46, B = 40, 
Bi = 16, C = 12, E = 34. The wheels A and C are loose on 
the shaft F, but E is keyed. The wheel A is rigidly attached 
to the frame of the machine, and C is driven by a stepped 
pulley ; the arm d rotates on the shaft F ; the two wheels B 
and Bi are fixed together. Let d make one complete clock- 
wise revolution ; then the other wheels will make — 



Revs, of B on own axle = — -," z= — |£ 

1 h 



= -I'lS 



C absolute = -? 4. 1 = || + i =: ^.gj 



Mechanisms. 



177 



Revs. E = revs, of B x t^-' + i = - i-i'S X M + i 
= -0-541 + I = 0-459 



Whence for one revolution of E, C makes 



4-83 
°'459 



= io"53 



revolutions. 

The -f- sign in the expressions for the speed of C and E is 




Fig. 20I. 

on account of these wheels of the epicyclic train rotating in the 
same sense as that of the arm d. 

As stated above the «th wheel in a bevil train of this type 
rotates in the same sense as the arm when n is odd, and in the 
opposite sense when n is even ; hence the sign is + for odd 
axes, and — for even axes, always counting the first as " one." 

It may help some readers to grasp the solution of this 
problem more clearly if we work it out by another method. 
Let A be free, and let d be prevented from rotating ; turn A 
through one —revolution] then — 



Revs.l product of teeth in drivers 

of E j product of teeth in driven wheels 
Revs.l Ta 



T,XT,^ 

't:xt: 



-0-541 



ofc}=T:=3-«3 

Hence, when A is fixed by clamping the split bearing G, and 
d is rotated, the train becomes epicyclic, and since C and E are 
on odd axes of bevil trains, they rotate in the same sense as the 
arm ; consequently, for reasons already given, we have — 

Revs. E _ Ne _ — 0-541 + I _ 1 
Revs. C Ne 3-83 + I 10-53 



178 Mechanics applied to Engineeritig, 

Particular attention must be paid to the sense of rotation. 
Bevil gears are more troublesome to follow than plain gears ; 
hence it is well to put an arrow on the drawing, showing the 
direction in which the observer is supposed to be looking. 

This mechanism can also be used as a simple reduction 

gear — in which case the shaft Fj is not continuous with Fj. 

The wheel C is then keyed to Fj, or drives it through the set 

„ „. „ . , J , T. , Revs. E Revs. Fj 

screw S. Since E is keyed to F, we have =; ;; = =; -. 

Revs. C Revs. F2 

Thus, if F2 is coupled to a motor running at 1053 revs, per 

min., the low speed shaft Fi will run at 100 revs, per min. for 

the set of wheels mentioned above. 



CHAPTER VI. 

lOYNAMICS OF THE STEAM-ENGINE. 

Reciprocating Farts. — Oh p. 133 we gave the construction 
for a diagram to show the velocity of the piston at each 
part of the stroke when the velocity of the crank-pin was 
assumed to be constant. We there showed that, for an infinitely 
long connecting-rod or a slotted cross-head (see Fig. 160), such 
a diagram is a semicircle when the ordinates represent the 
velocity of the piston, and the abscissae the distance it has 
moved through. The radius of the semicircle represents the 
constant velocity of the crank-pin. We see from such a dia- 
gram that the velocity of the reciprocating parts is zero at each 
end of the stroke, and is a maximum at the middle ; hence 
during the first half of the stroke the velocity is increased, or 
the reciprocating parts are accelerated, for which purpose 
energy has to be expended ; and during the second half of the 
stroke the velocity is decreased, or the reciprocating parts are 
retarded, and the energy expended during the first half of the 
stroke is given back. This alternate expenditure and paying 
back of energy very materially affects the smoothness of run- 
ning of high-speed engines, unless some means are adopted for 
counteracting these disturbing effects. 

We will first consider the case of an infinitely long con- 
necting-rod, and see how to calculate 
the pressure at any part of the stroke 
required to accelerate and retard the 
reciprocating parts. 

The velocity diagram for this case 

is given in Fig. 202. Let R represent 

V, the linear velocity of the crank-pin, 

assumed constant ; then the ordinates 

Yi, Ya represent to the same scale the velocity of the piston 

V Y 
when it is at the positions Ai, Aj respectively, and -^ = ~ 




i8o Mechanics applied to Engineering. 

Let the total weight of the reciprocating parts = W. 
Then— 

The kinetic energy of the V _ WVi'' 
reciprocating parts at Aj \ ~ 2g 



Likewise at A. = 



2^-R^ 



The increase of kinetic energy^ _ WV^ , ^ 

during the interval A1A2 • j ~ 2°^^' ~^^) 

This energy must have come from the steam or other motive 
fluid in the cylinder. 

Let P = the pressure on the piston required to accelerate 
the moving parts. 

Work done on the piston in accelerating the) _ -pi _ \ 
moving parts during the interval 3 ~ '^^ ■*!' 

But V + Y,^ = xi + Ya^ = R 
hence Y^ — Yi = x^ — x^ 

and — 

Increase of kinetic energy of the] -nnn 

reciprocating parts during the > = (x^ — x^) 

interval J ^S^ 

then P(*, - X,) = ^,(«/ - *.») 

WV 



where * is the mean distance ' '*^ of the piston from the 



Dynamics of the Steam-Engine. 



iSi 



middle of the stroke ; and when ;«: = R at the beginning and 
end of the stroke, we have — 



P = 









^ 



-^ 



We shall term P the " total acceleration pressure." Thus 
with an infinitely long connecting-rod the pressure at the end 
of the stroke required to accelerate or retard the reciprocating 
parts is equal to the centrifugal force (see p. 19), assuming the 
parts to be concentrated at the crank-pin, and at any other 
part of the stroke distant x from the middle the pressure is less 

00 

in the ratio — . 
K. 

Another simple way of arriving at the result given above is 

as follows : If the connecting-rod be infinitely long, then it 

always remains parallel 

to the centre line of 

the engine ; hence the 

action is the same as 

if the connecting-rod 

were rigidly attached • ,/ 

to the cross-head and '■- - ' "~- •'' 

piston, and the whole Fig. 203. 

rotated together as one solid body, then each point in the body 

would describe the arc of a circle, and would be subjected to 

the centrifugal force C = — 5-, but we are only concerned with 

the component along the centre line of the piston, marked P 
in the diagram. It will be seen that P vanishes in the middle 
of the stroke, and increases directly as the distance from the 
middle, becoming equal to C at the ends of the stroke. 

When the piston is travelling towards the middle of the 
stroke the pressure P is positive, 
and when travelling away from 
the middle it is negative. Thus, 
in constructing a diagram to 
show the pressure exerted at all 
parts of the stroke, we put the 
first half above, and the second 
half below the base-line. We 
show such a diagram in Fig. 2 04. 
The height of any point in the 
sloping line ab above the base-line 




Fig. 204. 

represents the pressure 



182 



Mechanics applied to Engineering. 



at that part of the stroke required to accelerate or retard the 
moving parts. It is generally more convenient to express the 
pressure in pounds per square inch, /, rather than the total 

p 
pressure P ; then -7- =/> where A = the area of the piston. We 

W 
will also put w = — , where w is the weight of the reciprocatmg 

parts per square inch of piston. It is more usual to speak of 
the speed of an engine in revolutions per minute N, than of 
the velocity of the crank-pin V in feet per second. 

27rRN ■ 



and/ '■ 



60 
6o=^R 



= o'ooo34ze'RN'' 



N.B. — -The radius of the crank R is measured in feet. 

In arriving at the value of w it is usual to take as reciprocating 
parts — the piston-head, piston-rod, tail-rod (if any), cross-head, 
small end of connecting-rod and half the plain part of the rod. 
A more accurate method of finding the portion of the connect- 
ing-rod to be included in the reciprocating parts is to place the 
rod in a horizontal position with the small end resting on the plat- 
form of a weighing machine or suspended from a spring balance, 
the reading gives the amount to be included in the reciprocating 
parts. When air-pumps or other connections are attached to 
the cross-head, they may approximately be taken into account 
in calculating the weight of the reciprocating parts j thus — 




Fig. 205. 



weight of \ piston -)- piston and ail-rods -f- both cross-heads -f small 



area of 



end of con. rod -(- ' ' 1- air- 

2 

piston 



ir-pumpplungerf ^j \^.i^^l\ \ 



Dynamics of the Steam- Engine. 183 

The kinetic energy of the parts varies as the square of the 
velocity; hence the \j\ 

Values of w in pounds per square inch of piston : — 

Steam engines with no air pump or other attachments 2 to 4 
„ ' „ attachments 3 to 6 

In compound and triple expansion engines the reciproca- 
ting parts are frequently made of approximately the same total 
weight in each cylinder for balancing purposes, in such cases 
w is often as high as 6 lbs. for the H.P. cylinder and as low as 
I lb. for the L.P. 

Influence of Short Connecting-rods. — In Fig. 159 
a velocity diagram is given for a short connecting-rod ; repro- 
ducing a part of the figure in Fig. 206, we have — 

Short rod — 

cross-head velocity _ OX 
crank-pin velocity OCj 

Infinite rod- — , 

cross-head velocity _ OXj 

crank-pin velocity OCi 

velocity of cross-head with short rod _ OX 

velocity of cross-head with long rod OXj 

But at the " in " end of the stroke .-2J = ^ = li+^ 

and at the " out " end of the stroke = i — — 

Thus if the connecting-rod is n cranks long, the pressure at 

the "in" end is - greater, and at the "out" end - less, than 
n n ' 

if the rod were infinitely long. 

The value of/ at each end of the stroke then becomes — 

p = o'ooo34wRNV I -|- - 1 for the " in " end 
p = o'ooo34wRN-( 1 — - ) for the " out " end 



184 Mechanics applied to Engineering. 





Si 
\ 0) 

\ 




a) 




Si 


^° 





• 






1 


^B 


\ 




XL 

% 


^^p^ 






^ 










% 


^ 




^ 


1 




1- 

T3 \ 


\ 


\ 












^8 


^p 


^ 




1 f' rS 




\ 

\ 














\ 






















1 








nc/v* 


^^/ — 


cc 











W iL,,^ 




N<^o 











'\\ 




/^ 








~-> 




---5 


3 









\ 



Dynamics of the Steam-Engine. 185 

The line ab is found by the method described on p. 133. 

Set off aa„ = —, also bl>„ = — , and cc. = — , ^ is the position 
n n n 

of the cross-head when the crank is at right angles to the centre 

line, i.e. vertical in this case. The acceleration is zero where 

the slope of the velocity curve is zero, i.e. where a tangent to it 

is horizontal. Draw a horizontal line to touch the curve, viz. 

at/. As a check on the accuracy of the work, it should be 

noticed that this point very nearly indeed corresponds to the 

position in which the connecting-rod is at right angles to the 

crank; 'the crosshead is then at a distance R(V'«^-f- i — n), 
"P 

or very nearly — , from the middle of the stroke. The point/ 

having been found, the corresponding position of the cross- 
head g is then put in. At the instant when the slope of the 
short-rod velocity curve is the same as that of the long-rod 
velocity curve, viz. the semicircle (see page 134), the accelera- 
tions will be the same in both cases. In order to find 
where the accelerations are the same, draw arcs of circles 
from C as centre to touch the short-rod curve, and from the 
points where they touch draw perpendiculars to cut the circle 
at the points M and i. The corresponding positions of the 
cross-head are shown at h and i respectively. In these 
positions the acceleration curves cross one another, viz. at h„ 
and 4. It will shortly be shown that when the crank has 
passed through 45° and 135° the acceleration pressure with 
the short rod is equal to that with the infinitely long rod. 
From ]i drop a perpendicular to k„ set off kji„ = L, also 
k'k = L, and from the point where the perpendicular from 
k'„ cuts ab draw a horizontal to meet the perpendicular from 
k, where they cut is another point on the acceleration curve 
for the short rod ; proceed similarly with /. ~ We now have 
eight points on the short-rod acceleration curve through which 
a smooth curve may be drawn, but for ordinary purposes three 
are sufBcient, say a„, c„ b„. 

The acceleration pressure at each instant may also be 
arrived at thus — 

Let 6 = the angle turned through by the crank starting 
from the " in " end ; 

V = the linear velocity of the crank-pin, assumed 

constant and represented by R ; 

V = the linear velocity of the cross-head. 

' Engineering, }\x\y 15, 1892, p. S3 ; also June 2, 1899. 



1 86 Mechanics applied to Engineering. 

Then— 

v_ si n (g + g ) 

V sin (90- a) ^^^^^'^•^°'''' 

^,/sin 6 cos a + cos B sin a\ 
» = V( ) 

\ cos a / 

In all cases in practice the angle a is small, consequently 
cos a is very nearly equal to unity; even with a very short 
rod the average error in the final result is well within one per 
cent. Hence— 

z* = V (sin Q + cos sin a) nearly 

We also have — 







L 
R 


« = 


sin 6 
sin u. 






sin a = 


sin 6 
n 


1 




Substituting 


this value- 


— 










V = 


v(sin 


e + 


cos 6 sin 
n 


ti 




or z) = 


V^sin 


6 + 


sin 26\ 
in ) 






dB~ 


v(cos 


1^ + 


cos 20\ 
n J 





The acceleration of the cross-head /„ = — = —„• — 

■" dt de dt 

Let B = the angle in radians turned through in the 
time t. 

Then e = ■ ^'■*' 



radius 



N.t ^dB V 

, . dv Y 

whence/. = -^._ 

Substituting the value of — found above, we have — 
, Vy . , cos2(9\ 



Dynamics of the Steam- Engine. 187 

and the acceleration pressure, when the crank has passed 
through the angle Q from the " in " end of the stroke, is— 

wVY cos 2ff\ 

or/ = o-ooo34ze/RN=('cos B + E2if_^^ 

When 6 = 45° and 135°, cos 26 = and the expression 
becomes the same as that for a rod of infinite length. When 

6 = and 180° the quantity in brackets becomes 1 + i and 



I 
1 . 



Correction of Indicator Diagram for Acceleration 
Pressure.— An indicator diagram only shows the pressure of 
the working fluid in the cylin- 
der; it does not show the 
real pressure transmitted to 
the crank-pin because some 
of the energy is absorbed in 
accelerating the reciprocating 
parts during the first part of _ 

the stroke, and is therefore " ~— ™™™'-' rr^:. — > 

not available for driving the 

crank, whereas, during the latter part of the stroke, energy is 
given back from the reciprocating parts, and there is excess 
energy over that supplied from the working fluid. But, apart 
from these effects, a single indicator diagram does not show 
the impelling pressure on a piston at every portion of the 
stroke. The impelling pressure is really the difference be- 
tween the two pressures on both sides of the piston at any 
one instant, hence the impelling pressure must be measured 
between the top line of one diagram and the bottom line of 
the other, as shown in full lines in Fig. 207. 

The diagram for the return stroke is obtained in the same 
manner. The two diagrams are set out to a straight base in 
Fig. 208, the one above the line and the other below. On 
the same base line the acceleration diagram is also given to 
the same scale. The real pressure transmitted along the 
centre line of the engine is given by the vertical height of the 
shaded figures. In the case of a vertical engine the accelera- 
tion line is shifted to increase the pressure on the downward 
stroke and decrease it on the upward stroke by an amount zc, 



i88 



Mechanics applied to Engineering. 



see Fig. 209. The area of these figures is not altered in any 
way by the transformations they have passed through, but it 
should be checked with the area of the indicator diagrams in 
order to see that no error has crept in. 

When dealing with engines having more than one cylinder, 
the question of scales must be carefully attended to ; that is, 
the heights of the diagrams must be corrected in such a 
manner that the mean height of each shall be proportional to 
the total effort exerted on the piston. 




■^'- III! I l|ll I, , 



Fig. Z08. 



Fig. 2og, 



Let the original indicator diagrams be taken with springs of 

the following scales, H.P. -, I.P. -, L.P.-. Let the areas of 

oc z 

the pistons (allowing for rods) be H.P. X, LP. Y, L.P. Z. Let 
all the pistons have the same stroke. Suppose we find that 
the H.P. diagram is of a convenient size, we then reduce all the 
others to correspond with it. If, say, the intermediate piston 
were of the same size as the high-pressure piston, we should 
simply have to alter the height of the intermediate diagram in 
the ratio of the springs ; thus — 



Corrected height of LP. diagram) ^ f iiSltll x f 
if pistons were of same size 1 | diagram i i 



= actual height X — 

X 



Dynamics of the Steam-Engine. 



189 



But as the cylinders are not of the same size, the height of 
the diagram must be multiplied by the ratio of the two areas ; 
thus — 



Height of intermediate diagram j j-actual height of ^ ^ 

corrected for scales of springs \ — \ intermediate V X ^ X - 
and for areas of pistons J ( diagram j « X 



-J'v^ 



diagram j 

= actual height X ^— 

Similarly for the L.P. diagram — 
Height of L.P. diagram corrected . ^^^^^j ^^^.^^ ^^. ^^ 

It is probably best to make this correction for scale and 
area after having reduced the diagrams to the form given in 
Fig. 208. 

Pressure on the Crank-pin. — The diagram given in 





/^ 


'~y<^ 


"^•',^ 









/ -^ 


'' 


N^^^^ ~~^ 


^ 




/ 


/ /■ 






1;*-^— . 


■~'\ 


J— \— -v 


/ 


1^ 

4 










\\ 


. 






w 



Fig. 210. 



-/ 



Fig. 208 represents the pressure transmitted to the crank-pin 
at all parts of the stroke. The ideal diagram would be one in 
which the pressure gradually fell to zero at each end of the 
stroke, and was constant during the rest of the stroke, such as 
a, Fig. 210. 

The curve 3 shows that there is too much compression 
resulting in a negative pressure — / at the end of the stroke ; at 
the point x the pressure on the pin would be reversed, and, if 
there were any " slack " in the rod-ends, there would be a 
knock at that point, and again at the end of the stroke, when 
the pressure on the pin is suddenly changed from — / to +p. 
These defects could be remedied by reducing the amount of 
compression and the initial pressure, or by running the engine 
at a higher speed. 

The curve (c) shows that there is a deficiency of pressure at 



1 90 Mechanics applied to Engineering. 

the beginning of the stroke, and an excess at the end. The 
defects could be remedied by increasing the initial pressure 
and the compression, or by running the engine at a lower 
speed. 

For many interesting examples of these diagrams, as applied 
to steam engines, the reader is referred to Rigg's " Practical 
Treatise on the Steam Engine ; " also a paper by the same 
author, read before the Society of Engineers ; and to Haeder 
and Huskisson's " Handbook on the Gas Engine," Crosby 
Lockwood, for the application of them to gas and oil engines. 

Cushioning for Acceleration Pressures. — In order 
to counteract the effects due to the acceleration pressure, it is 
usual in steam-engines to close the exhaust port before the 
end of the stroke, and thus cause the piston to compress the 
exhaust steam that remains in the cylinder. By choosing 
the point at which the exhaust port closes, the desired amount 
of compression can be obtained which will just counteract the 
acceleration pressure. In certain types of vertical single- 
acting high-speed engines, the steam is only admitted on the 
downstroke ; hence on the upstroke some other method of 
cushioning the reciprocating parts has to be adopted. In the 
well-known Willans engine an air-cushion cylinder is used ; 
the required amount of cushion at the top of the cylinder is 
obtained by carefully regulating the volume of the clearance 
space. The pistons of such cylinders are usually, of the trunk 
form ; the outside pressure of the atmosphere, therefore, acts 
on the full area of the underside, and the compressed air 
cushion on the annular top side. 

Let A = area of the underside of the piston in square inches j 
A„ = area of the annular top side in square inches ; 
W = total weight of the reciprocating parts in lbs. ; 
c = clearance in feet at top of stroke. 

At the top, i.e. at the " in end," of the stroke we have — 

Pi = o-ooo34WRN2(^i -f i^ - W -f i4-7A 

Assuming isothermal compression of the air, and taking 
the pressure to be atmospheric at the bottom of the stroke, we 
have — 

I47A„(2R -t- f) = P,f 

whence c = ^ 

Pi - i4-7Aa 



Dynamics of the Steam-Engine. 191 



Or for adiabatic compression — 

I4-7A„(2R + <r)'"" = Pi<;^'" 



2R 



Vi47Aa/ 

in the expression for c given above. 

The problem of balancing the reciprocating parts of gas and 
oil engines is one that presents much greater difiSculties than in 
the steam-engine, partly because the ordinary cushioning 
method cannot be adopted, and further because the eifective 
pressure on the piston is different for each stroke in the cycle. 
Such engines can, however, be partially balanced by means of 
helical springs attached either to the cross-head or to a tail-rod, 
arranged in such a manner that they are under no stress when 
the piston is at the middle of the stroke, and are under their 
maximum compression at the ends of the stroke. The weight 
of such springs is, however, a great drawback ; in one instance 
known to the author the reciprocating parts weighed about 
1000 lbs. and the springs 800 lbs. 

Polar Twisting-Moment Diagrams. ^ From the 
diagrams of real pressures transmitted to the crank-pin that 



Fig. 211. 

we have just constructed, we can readily determine the twisting 
moment on the crank-shaft at each part of the revolution. 

In Fig. 2ir,let/ be the horizontal pressure taken from such 
a diagram as. Fig. 209. Then /i is the pressure transmitted 
along the rod to the crank-pin. This may be resolved in a 
direction parallel to the crank and normal to it (/„) ; we need 
not here concern ourselves with the pressure acting along the 
crank, as that will have no turning effect. The twisting moment 
on the shaft is then /„R ; R, however, is constant, therefore the 
twisting moment is proportional to/„. By setting off values of 
p„ radially from the crank-circle we get a diagram showing the 



192 Mechanics applied to Engineering. 

twisting moment at each part of the revolution. /„ is measured 

on the same scale, say -, as the indicator diagram ; then, if A 

be the area of the piston in square inches, the twisting moment 
in pounds feet =/„a:AR, where /„ is measured in inches, and 
the radius of the crank R is expressed in feet. 

When the curve falls inside the circle it simply indi- 






FiG. 2ia. 



Gates that there is a deficiency of driving effort at that 
place, or, in other words, that the crank-shaft is driving the 
piston. 

In Fig. 212 indicator diagrams are given, which have been 
set down in the manner shown in Fig. 208, and after correct- 
ing for inertia pressure they have been utilized for construct- 
ing the twisting moment diagram shown in Fig. 213. The 
diagrams were taken from a vertical triple-expansion engine 
made by Messrs. McLaren of Leeds, and by whose courtesy 
the author now gives them. 

The dimensions of the engine were as follows : — 



Dynamics of the Steam-Engine. 



193 



Diameter of cylinders — 

High pressure 

Intermediate 

Low pressure 
Stroke 



9*01 inches, 
I4"2S .1 
22-47 >. 
2 feet 





Fig. 182*. 



194 Mechanics applied to Engineering. 

The details of reducing the indicator diagrams have been 
omitted for the sake of clearness ; the method of reducing them 
has been fully described. 

Twisting Moment on a Crankshaft. — In some 
instances it is more convenient to calculate the twisting 

moment on the crank- 
shaft when the crank 
has passed through the 
angle 6 from the inner 
dead centre than to 
construct a diagram. 
Let P, = the effort on the piston-rod due to the working 
fluid and to the inertia of the moving parts ; 
Pi = the component of the eflfort acting along the 
connecting-rod ; 

, . sin 9 

P, = Pi cos o, and sm a = ^- 

from which a can be obtained, since 6 and n are given. 
The tangential component — 

T = Pi cos </) 
and </) = 90 — (^ ^- a) 

whence T= ^^cos {9o-(e + a)] = ^' ^'" ^^ + "^ 
cos a <■' ^ cos a 

Flywheels. — The twisting-moment diagram we have just 
constructed shows very clearly that the turning effort on the 
crank-shaft is far from being constant ; hence, if the moment of 
resistance be constant, the angplar velocity cannot be constant. 
In fact, the irregularity is so great in a single-cylinder engine, 
that if it were not for the flywheel the engine would come to a 
standstill at the dead centre. 

A flywheel is put on a crank-shaft with the object of storing 
energy while the turning effort is greater than the mean, and 
giving it back when the effort sinks below the mean, thus 
making the combined effort, due to both the steam and the 
flywheel, much more constant than it would otherwise be, and 
thereby making the velocity of rotation more nearly constant. 
But, however large a flywheel may be, there must always be 
some variation in the velocity ; but it may be reduced to as 
small an amount as we please by using a suitable flywheel. 

In order to find the dimensions of a flywheel necessary for 
keeping the cyclical velocity within certain limits, we shall make 
use of the twisting-moment diagram, plotted for convenience 



Dynamics of the Steam-Engine. 



195 



to a straight instead of a circular base-line, the length of the 
base being equal to the semicircumference of the crank-pin 
circle. Such a diagram we give in Fig. 215. The resistance 
line, which for the present we shall assume to be straight, is 
shown dotted; the diagonally shaded portions below the mean 
line are together equal to the horizontally shaded area above. 

During the period AC the effort acting on the crank-pin is 
less than the mean, and the velocity of rotation of the crank- 
pin is consequently reduced, becoming a minimum at C. 
During the period CE the effort is greater than the mean, and 
the velocity of rotation is consequently increased, becoming a 
maximum at E. 




Fig. 215. 

Let V = mean velocity of a point on the rim at a radius equal 
to the radius of gyration of the wheel, in feet per 
second — usually taken for practical purposes as the 
velocity of the outside edge of the rim : 

V„ = minimum velocity at C (Fig. 215); 

V^ = maximum „ E ; 

W = weight of the flywheel in pounds, usually taken for 
practical purposes as the weight of the rim ; 

R„ = radius of gyration in feet of the flywheel rim, usually 
taken as the external radius for practical purposes. 

For most practical purposes it is sheer waste of time cal- 
culating the moments of inertia and radii of gyration for all 
the rotating -parts, since the problem is not one that permits 
of great accuracy of treatment, the form of the indicator 
diagrams does not remain constant if any of the conditions 
are altered even to a small extent, then again the coefficient 
-of fluctuation k is not a definitely known quantity, since 
different authorities give values varying to the extent of two 
or three hundred per cent. The error involved in using the 
above approximations is not often greater than five per cent, 
which is negligible as compared with the other variations, and 
by adopting them a large amount of time is thereby saved. 

' Figures 207, 208, 210, 215 are all constructed from the same indicator 
diagram. 



196 Mechanics applied to Engineering. 
Then the energy stored in the flywheel at C = - 



E = 



^g 



W 
The increase of energy during CE = — (V/— V„^) (i.) 

This increase of energy is derived from the steam or other 
source of energy ; therefore it must be equal to the work 
represented by the horizontally shaded area CDE = E„ (Fig. 

215)- 

Let E„ = OT X average work done per stroke. 

Then the area CDE is m times the work done per stroke, 
or m times the whole area BCDEF. Or — 

p, _ m X indicated horse-power of engine x 33000 ... . 

where N is the number of revolutions of the engine per minute 
in a double-acting engine. 

Whence, from (i.) and (ii.), we have — 

W/^a 3. _ OT X I.H.P. X 33000 , 

Tg- '~ °' 2N ■ • ^ ' 

But '"' ° = V (approximately) 
2 

or V. + V. = 2V 

V — V 
also — i^r= — 5 = K, " the coefficient of speed 

fluctuation " 

and V. - V. = KV 

v." - V/ = 2KV 

Substituting this value in (iii.) — 

-(2KV=) = *^ ^ ^•^■^- ^ 33000 ^ E 
2/ 2N " 

„„A w - 48,5o°.o°o/« X I.H.P. 

KWRJ ' • ' ^^^-^ 

The proportional fluctuation of velocity K is the fluctuation 
of velocity on either side of the mean ; thus, when K = 0*02 
it is a fluctuation of i per cent, on either side of the mean. 
The following are suitable values for K : — 



Dynamics of the Steani-Engine. 



197 



K = o'oi to o'oa for ordinary electric-lighting engines, but for 
public lighting and traction stations it often gets as low 
as o"ooi6 to 0-0025 to allow for very sudden and large 
changes in the load ; the weight of all the rotating 
parts, each multiplied by its own radius of gyration, is 
to be included in the flywheel ; 

= o'o2 to o'o4 for factory engines; 

= o'o6 to o'i6 for rough engines. 

When designing flywheels for public lighting and traction 
stations where great variations in the load may occur, it is 
common to allow from 2-4 to 4-5 foot-tons (including rotor) 
of energy stored per I.H.P. 

The calculations necessary for arriving at the value of E„ 
for any proposed flywheel are somewhat long, and the result 
when obtained has an element of uncertainty about it, because 
the indicator diagram must be assumed, since the engine so 
far only exists on paper. The errors involved in the diagram 
may not be serious, but the desired result may be arrived at 
within the same limits of error by the following simpler process. 
The table of constants given below has been arrived at by 
constructing such diagrams as that given in Fig. 215 for a 
large number of cases. They must be taken as fair average 
values. The length of the connecting-rod, and the amount of 
pressure required to accelerate and retard the moving parts, 
affect the result. 

The following table gives approximate values of m. In 
arriving at these figures it was found that if « = number of 

cranks, then m varies as -^ approximately. 



Approximate Values of m for Double-acting Steam-Engines.' 



Cut-off. 


Single cylinder. 


Two cylinders. 
Cranks at right angles. 


Tiiree cylinders. 
Cranks at 120° 


O'l 


0'35 


0-088 


0-040 


0'2 


o'33 


0-082 


0-037 


0-4 


0-31 


0-078 


0-034 


0-6 


0-29 


0-072 


0-032 


0-8 


0-28 


0-070 


0-031 


End of stroke 


0-27 


0-068 


0-030 



' The values of m vary much more in the case of two- and three- 
cylinder engines than in single-cylinder engines. Sometimes the value of 
m is twice as great as those given, which are fair averages. 



198 



Mechanics applied to Engineering. 



m FOK Gas 


- AND OiL-ENGINES. 








Otto cycle. 


Double acting. 




Number of cylinders. 






X 


2 


4 


I 


3 


Exploding at every 
cycle .... 


Single 


37 to 
4-5 


— 


— 


2-3 to 
2-8 


— 


Twin or 
tandem 


— 


i-S to 
I -8 


0-3 to 
0-4 


— 


0-3 to 
0-4 


When missing every 
alternate charge. 


Single 


8-5 to 
9-8 


2-5 to 
3-0 


— 


— 


— 



Gas and oil engines, single acting (Otto) w = i"5 + >Jd 
„ „ double „ 1X1= 2'5 + 'I'd'Jd 

d 2 
High speed petrol engines, a' = 5 + 3 

o d 

Tandem engines, per line, from i'8 to I'g times the above 
values. 

Where d is the diameter of the cylinder in inches. 

Relation betvireen the Work stored in a Flywheel 
and the Work done per Stroke. — For many purposes it 
is convenient to express the work stored in the flywheel in 
terms of the work done per stroke. 

WV* 
The energy stored m the wheel = 

Then from equation (iv.), we also have — 

The energy stored in the^ _ E„ 
wheel / ~ 2K 

and the average work done'! _ E„ 
per stroke / ^ 

_ I.H.P. X 33000 ^ ffor a double- 
2N )\ acting engine 

the number of average* _ 2K _ m . 

strokes stored in flywheel/ ~ E„ ~ 2K ^'' 



Dynamics of the Steam- Engine. 



199 



In the following table we give the number of strokes that 
must be stored in the flywheel in order to allow a total fluctua- 
tion of speed of i per cent., i.e. \ per cent, on either side of 
the mean. If a greater variation be permissible in any given 
case, the number of strokes must be divided by the per- 
missible percentage of fluctuation. Thus, if 4 per cent., i.e. 
K = o'o4, be permitted, the numbers given below must be 
divided by 4. 



Number of Strokes stored in a Flywheel for Double-acting 
Steam-Engines.' 



Cut-off. 


Single cylinder. 


Two cylinders. 
Cranks at right angles. 


Three cylinders. 
Cranks at 120°. 


o-i 


18 


4'4 


2'0 


0-2 


17 


4" I 


I "9 


0-4 


16 


3'9 


I "8 


06 


IS 


3-6 


'7 


0-8 


14 


3-S 


1-6 


End of stroke 


«3 


3-4 


i-S 



Gas-Engines (Mean Strokes). 





Otto cycle. 


Double acting. 




Number of cylinders. 






I 


2 


4 


z 


2 


When exploding at 


Single 


185 to 
225 


— 


■ — 


115 to 
140 


— 


every cycle. 


Twin or 
tandem 


— 


75 to 
90 


15 to 
20 


— 


15 to 

20 


When missing every 
alternate charge . 


Single 


425 to 

490 


125 to 
ISO 


— 


— 


— 



Shearing, Punching, and Slotting Machines (K not 
known). — It is usual to store energy in the flywheel equal 
to the gross work done in two working strokes of the shear, 
punch, or slotter, amounting to about 15 inch-tons per square 
inch of metal sheared or punched through. 

' See note at foot of p. 197. 



200 



Mechanics applied to Engineering. 



Gas-Engine Flywheels. — The value of m for a gas- 
engine can be roughly arrived at by the following method. 




Fig. 2i6. 



The work done in one explosion is spread over four strokes 
when the mixture explodes at every cycle. Hence the mean 
effort is only one-fourth of the explosion-stroke effort, and 
the excess energy is therefore approximately three-fourths 
of the whole explosion-stroke effort, or three times the mean : 
hence m = 2,. 

Similarly, when every alternate explosion is missed, m = j. 

By referring to the table, it will be seen that both of these 
values are too low. 

The diagram for a 4-stroke case is given in Fig. 216. It has 
been constructed in precisely the same manner as Figs. 207, 
208, and 215. When oil- or gas-engines are used for driving 
dynamos, a small flywheel is often attached to the dynamo 
direct, and runs at a very much higher peripheral speed than 
the engine flywheel. Hence, for a given weight of metal, the 
small high-speed flywheel stores a much larger amount of 
energy than the same weight of metal in the engine flywheel. 

The peripheral speed of large cast-iron flywheels has to be 
kept below a mile a minute (see p. 201) on account of their 



Dynamics of the S team-Engine. 



201 



danger of bursting. The small disc flywheels, such as are used 
on dynamos, are hooped with a steel ring, shrunk on the rim, 
which allows them to be safely run at much higher speeds than 
the flywheel on the engine. The flywheel power of such an 
arrangement is then the sum of the energy stored in the two 
wheels. 

There is no perceptible flicker in the lights when about forty 
impulse strokes, or i6o average strokes (when exploding at 
every cycle, and twice this number when missing alternate 
cycles), are stored in the flywheels. 

Case in which the Resistance varies. — In all the 
above cases we have assumed that the resistance overcome by 
the engine is constant. This, however, is not always the case ; 
when the resistance varies, the value of E„ is found thus : 




Fig. 217. 

The line aaa is the engine curve as described above, the line 
bbb the resistance to be overcome, the horizontal shading 
indicates excess energy, and the vertical deficiency of energy. 
The excess areas are, of course, equal to the deficiency areas 
over any complete cycle. The resistance cycle may extend over 
several engine cycles; an inspection or a measurement will 
reveal the points of maximum and minimum velocity. The 
value of m is the ratio of the horizontal shaded areas to the 
whole area under the line aaa described during the complete 
cycle of operations. See The Engineer, January 9, 1885. 

Stress in Flywheel Rims. — If we neglect the effects of 
the arms, the stress in the rim of 
a flywheel may be treated in the 
same manner as the stresses in 
a boiler-shell or, more strictly, 
a thick cylinder (see p. 421), in 
which we have the relation — 

or P,R„ =/, when t = -i inch 

The P, in this instance is the 
pressure on each unit length of 
rim due to centifugal force. We 




Fig. 2i3. 



202 Mechanics applied to Engineering. 

shall find it convenient to take the unit of length as i foot, 
because we take the velocity of the rim in feet per second. 
Then— 

WV ^ WV " 

where W, = the weight of i foot length of rim, i square inch 
in section 
= 3 "I lbs. for cast iron 

We take i sq. inch in section, because the stress is expressed in 
pounds per square inch. Then substituting the value of W, 
in the above equations, we have — 

32"2 

/= 0-096 V„' 

V " / 
or/= -^ (very nearly) 
10 

In English practice V„ is rarely allowed to exceed 100 feet 
per second, but in American practice much higher speeds are 
often used, probably due to the fact that American cast iron is 
much tougher and stronger than the average metal used in 
England. An old millwright's rule was to limit the speed to a 
mile a minute, i.e. 88 feet per second, corresponding to a stress 
of about 800 lbs. per square inch. 

The above expression gives the tensile stress set up in a 
thin plain rotating ring, due to centrifugal force ; but it is not 
the only or even the most important stress which occurs in 
many flywheel and pulley rims. The direct stress in the 
material causes the rim to stretch and to increase in diameter, 
but owing to its attachment to the arms, it is unable to do so 
beyond the amount permitted by the stretch of the arms, 
with the result that the rim sections bend outwards between 
the arms, and behave as beams which are constrained in 
direction at the ends. If the arms stretched sufficiently to 
allow the rim to remain circular when under centrifugal stress 
there would be no bending action, and, on the other hand, 
if the arms were quite rigid the bending stress in the rim 
sections between the arms could be calculated by treating 
them as beams built in at each end and supporting an evenly 
distributed load equal in intensity to the centrifugal force 
acting on the several portions of the rim ; neither of these 
conditions actually hold and the real state of the beam 



Dynamics of the Steam-Engine. 203 

is intermediate between that due to the above-mentioned 
assumptions, the exact amount of bending depending largely 
upon the stretch of the arms. 

A rigid solution of the problem is almost impossible, the 
results obtained by different authorities are not in agree- 
ment, owing to arbitrary assumptions being made. In all 
cases it is assumed that there are no cooling stresses exist- 
ing in the arms of the wheel, which every practical man 
knows is not always correct. However, the results obtained 
by the more complete reasoning are unquestionably nearer 
the truth than those obtained by the elementary treatment 
given above. 

For a more complete treatment the reader is referred to 
Unwin's "Elements of Machine Design," Part II. (Longmans). 
The following approximate treatment may be of service to 
those who have not the opportunity of following the more 
complete theory. 

The maximum bending moment in pound-inches on an 

initially straight beam built in at both ends is — (see p. 529), 

where w is the evenly distributed load per inch run, and / is 
the length between supports in inches. In the case of the 
flywheel, w is the centrifugal force acting on the various 

portions of the rim, and is ° ^ " — , where 0-26 is the weight 

of a cubic inch of cast iron, V„ is the rim velocity in feet per 
second, A the area of the section of the rim in square inches, 
g the acceleration of gravity, R„ the radius of the rim in feet, 
Z the tension modulus of the section (see Chapter IX.). 

Then the bending stress in the rim due to centrifugal force 

o-26V„^A/' 
'^ ■'" 12 X 32-2 X R„ X Z 

The rim section, however, is not initially straight, hence 
the ordinary beam formula does not rigidly hold. / is taken 
as the chord of the arc between the arms. As already ex- 
plained, the stress due to bending is really less than the above 
expression gives, but by introducing a constant obtained by 
comparing this treatment with one more complete, we can 
bring the results into approximate agreement : this constant is 
about 2'2. Hence 

o-26V„^A/' ^ V^'A/' 

■^^ ~ 2-2 X 12 X 32'2R„Z ~ 327oR„Z 



204 Mechanics applied to Engineering. 

and the resulting tensile stress in the rim is — 

All who have had any experience in the foundry will be 
familiar with the serious nature of the internal cooling stresses 
in flywheel and pulley arms. The foundry novice not unfre- 
quently finds that one or more of the arms of his pulleys are 
broken when taken out of the sand, due to unequal cooling ; 
by the exercise of due care the moulder can prevent such 
catastrophes, but it is a matter of common knowledge that, in 
spite of the most skilful treatment it is almost impossible to 
ensure that a wheel is free from cooling stresses. Hence only 
low-working stresses should be permitted. 

When a wheel gets overheated through the use of a friction 
brake, the risk of bursting is still greater; there are many 
cases on record in which wheels have burst, in some instances 
with fatal results, through such overheating. In a case known 
to the Author of a steam-engine fitted with two flywheels, 5 feet 
in diameter, and running at 160 revolutions per minute, one 
of the wheels broke during an engine test. The other wheel 
was removed with the object of cutting through the boss in 
order to relieve the cooling stresses; but as soon as the cut 
was started the wheel broke into several pieces, with a loud 
report like a cannon, thus proving that it was previously sub- 
jected to very serious cooling stresses. In order to reduce 
the risk of the bursting of large flywheels, when made with 
solid rims, they should always be provided with split bosses. 
In some cases the boss is made in several sections, each being 
attached to a single arm, which effectually prevents the arms 
from being subjected to initial tension due to cooling. Built-up 
rims are in general much weaker than solid rims ; but when 
they cannot be avoided, their design should be most carefully 
considered. The question is discussed in Lanza's " Dynamics 
of Machinery " (Chapman & Hall). 

Large wheels are not infrequently built up entirely of 
wrought iron or steel sections and plates ; in some instances 
a channel rim has been used into which hard-drawn steel wire 
is wound under tension. Such wheels are of necessity more 
costly than cast-iron wheels of the same weight, but since the 
material is safe under much higher stresses than cast iron, the 
permissible peripheral speed may be much higher, and conse- 
quently the same amount of energy may be stored in a much 
lighter wheel, with the result that for a given speed fluctuation 



Dynamics of the Steam-Engine. 



205 



the cost of the built-up wheel may be even less than that of a 
cast-iron wheel. 

In the place of arms thin plate webs are often used with 
great success ; such webs support the rim far better than arms, 
and moreover they have the additional advantage that they 
materially reduce the air resistance, which is much more im- 
portant than many are inclined to believe. 

Experimental Determination of the Bursting Speed 
of Flywheels. — Professor C. H. Benjamin, of the Case School, 
Cleveland, Ohio, has done some excellent research work on 
the actual bursting speed of flywheels, which well corroborates 
the general accuracy of the theory. The results he obtained are 
given below, but the original paper read by him before the 
American Society of Mechanical Engineers in 1899 should be 
consulted by those interested in the matter. 



Bursting speed 
in feet per sec. 


TO 


Thickness 
of rim. 


Remarks, 




lbs. sq. in. 


inch. 




430 


18,500 


068 


Solid rim, 6 arms, 15 ins. diam. 


388 


15,000 


0-56 


. ) » » » » 


192 


3.700 




Jointed rim, ,, „ 


38" 


14,500 


0-65 


Solid rim, 3 arms, „ 


363 


13,200 


0-38 


.. .» ». 


38s 


14,800 


IS 


„ 6 arms, 24 ins. diam. 
Two internal 


igo 


3.610 


075 


flanged joints, ,, „ 


305 


9.300 




Linked joints „ ,, 



From these and other tests. Professor Benjamin con- 
cludes that solid rims are by far the safest for wheels of 
moderate size. The strength is not much affected by bolting 
the arms to the rim, but joints in the rims are the chief sources 
of weakness, especially when the joints are near the arms. 
Thin rims, due to the bending action between the arms, are 
somewhat weaker than thick rims. 

Some interesting work on the bending of rims has been 
done by Mr. Barraclough (see I.C.E. Proceedings, vol. cl.). 

For practical details of the construction of flywheels, 
readers are referred to ; — Sharpe on " Flywheels," Manchester 
Association of Engineers. Haeder and Huskisson's " Hand- 
book on the Gas-Engine." " Flywheels," " Machinery " 
Reference books. 



2o6 Mechanics applied to Engineering. 

Arms of Flywheels. — In addition to the unknown 
tensile stresses in wheels with solid bosses and rims, the arms 
are under tension due to the centrifugal force acting (i) on the 
arm itself, (ii) on a portion of the rim, amounting to approxi- 
mately one-fourth of the length of rim between the arms. In 
addition to these stresses the arms are subjected to bending 
due (iii) to a change in the speed of the wheel, (iv) to the 
power transmitted through the wheel when it is used for 
driving purposes. 

The tension due to (i) is arrived at thus — 

Let Aa = Sectional area of the arm in square inches as- 
sumed to be the same throughout its length. 
r = Radius from centre of wheel to any element of 

the arm in feet. 
(1) = Angular velocity of the arm. 
/"i = Radius of inside of the rim of the wheel in feet, 
fa = Radius of boss of the wheel in feet. 
w = Weight of a cubic inch of the material. 

The centrifugal force acting on the element of the arm is 

g 
and on the whole arm 



g 



Jvi g \ 2 / 






6' 

The tension in the arm due to the centrifugal force acting 
on one-fourth of the rim between the arms is 

_ ■w\NJ' 
Hence the tensile stress in the arm at the boss due to both is 

7r Ir --T— + T>PP'°'- 

Let the 'acceleration of the rim, arising from a change of 

speed of the shaft, be 8V feet per sec. per sec. Let W be the 

weight of the rim in pounds, then, since the arms are built in 

at both ends, the bending moment on them (see p. 504) is 

6WR„.8V . ^ J 1 ,. ^ .. 

mch-pounds, approximately ; the bendmg stress is 



Dynamics of the Steam-Engine. 207 

6WR .8V 

'^ — , where n is the number of arms and Z the modulus 

gnT. 

of the section of the arms in bending. 

If the flywheel be also used for transmitting power, and P 

be the effective force acting on the rim of the wheel, the 

bending stress in the arms is —?, on the assumption that 

the stress in the most strained arm is twice the mean, which 
experiments show is a reasonable assumption. 

Hence the bending stress in the arms due to both causes 

The strength of flywheel and pulley arms should always be 
checked as regards bending. 

Bending Stresses in Locomotive Coupling-rods. — 
Each point in the rod describes a circle (relatively to the 




Fig. 219. 

engine) as the wheels revolve ; hence each particle of the rod 
is subjected to centrifugal force, which bends the rod upwards 
when it is at the top of its path and downwards when at the 
bottom. Since the stress in the rod is given in pounds per 
square inch, the bending moment on the rod must be ex- 
pressed in pound-inches ; and the length of the rod / in 
inches. In the expression for centrifugal force we have 
foot units, hence the radius of the coupling crank must be in 
feet. 

The centrifugal force acting ) ^ ^ o-ooo34«/R„N^ 
on the rod per men run ^ jt 

where w is the weight of the rod in pounds per inch run, 
ox w = o'28A pounds, where A is the sectional area of 
the rod. 

The centrifugal force is an evenly distributed load all along 
the rod if it be parallel. 



2o8 Mechanics applied to Engineering. 

The maximum bending moment ) _ C/^ _ /At' 
in the middle of the rod \~ i> ~ y 

(see Chapters IX. and X.) 

where k^ = the square of the radius of gyration (inch units) 
about a horizontal axis through the c. of g. ; 
y ■= the half-depth of the section (inches). 

Then, substituting the value of C, we have — 

000034 X o'zS X A X R, X N° X /° X J' 

~ 8 X A X K^ ' 

o-oooor2R„Ny;/ _ R^N^ 

^~ k" ' ""^ ~ 84,0001^ 

The value of k' can be obtained from Chapter III. For a 

rectangular section, k' = — ; and for an I section, k' = 

BH° - bh? 
I2(BH - bh) 

It should be noticed that the stress is independent of the 
sectional area of the rod, but that it varies inversely as the 
square of the radius of gyration of the section ; hence the im- 
portance of making rods of I section, in which the metal is 
placed as far from the neutral axis as possible. If the stress 
be calculated for a rectangular rod, and then for the same rod 
which has been fluted by milling out the sides, it will be found 
that the fluting very materially reduces the bending stress. 

The bending stress can be still further reduced by 
removing superfluous metal from the ends of the rod, i.e. by 
proportioning each section to the corresponding bending 
moment, which is a maximum in the middle and diminishes 
towards the ends. The " bellying '' of rods in this manner is a 
common practice on' many railways. 

In addition to the bending stress in a vertical plane, there 
is also a direct stress of nearly uniform intensity acting over 
the section of the rod, sometimes in tension and sometimes in 
compression. This stress is due to the driving effort trans- 
mitted through the rod from the driving to the coupled wheel, 
but it is impossible to say what this eiFort may amount to. 
It is usual to assume that it amounts to one-half of the total 
pressure on the piston, but a safer method is to calculate it 
from the maximum adhesion of the coupled wheels. The 
coefficient of friction between the wheels and rails may be 
taken at o'3. 



Dynamics of the Steam-Engine. 



209 



On account of the bending moment on these rods a certain 
amount of deflection occurs, which reaches its maximum value 
when the rod is at the top and bottom of its traverse. When 
the rod is transmitting a compressive stress it becomes a strut 
loaded out of the centre, and the direct stress is no longer 
distributed uniformly. The deflection due to the bending 
moment already considered is 



8 = 



384EI 8o7,oooEI 



if T be the thrust on the rod in pounds. 

The bending stress due to the eccentric loading is 

TS_ TR.NV* 
Z ~ Soy.oooEK'^Z 
and the maximum stress due to all causes is 

RoN'/'j)' ( . T/^ 



84,oook^ 



1 + 



9-6iEI 



- A 



The + sign refers to the maximum compressive stress and 
the — sign to the tensile stress. The second term in the 
brackets is usually very small and negligible. 

A more exact treatment will be found in Morley's " Strength 
of Materials," page 263. 

In all cases coupling rods should be checked to see that 
they are safe against buckling sideways as struts ; many break- 
downs occur through weakness in this direction. 

Bending Stress in Connecting-rods. — In the case of 
a coupling-rod of uniform section, in which each particle 
describes a circle of the same radius as the coupling-crank pin, 
the centrifugal force produces an evenly distributed load ; but 
in the case of a connecting-rod the swing, and therefore the 
centrifugal force at any sec- 
tion, varies from a maximum 
at the crank-pin to zero 
at the gudgeon-pin. The 
centrifugal force acting on 
any element distant x from 

the gudgeon-pin is —, 

where C is the centrifugal ^"^ "°- 

force acting on it if rotating in a circular path of radius R, i.e. 

the radius of the crank, and /is the length of the connecting-rod. 

Let the rod be in its extreme upper or lower position, and 

p 







org. C 


• x. — 


^^^^ I — 





2IO Mechanics applied to Engineering. 

let the reaction at the gudgeon-pin, due to the centrifugal 
force acting on the rod, be R,. Then, since the centrifugal 
force varies directly as the distance x from the gudgeon-pin, 
the load distribution diagram is a triangle, and — 

•n , C/ / , „ C/ 

RJ = — X - whence R„ = -r 

"^2 3 "6 

The shear at a section distant x')_^_ fCx x\ 
from the gudgeon-pin ) 6 V / 2/ 

C/ C:x^ I 

The shear changes sign when -^ = — ^ or when x = — ;=: 

o 2/ V3 

But the bending moment is a maximum at the section 
where the shear changes sign (see p. 482). The bending 
moment at a section y distant x from R^ — 

M^ = R^ 5- X - X - = R^a; - -^ 

"^ / 2 3 " 6/ 

The position of the maximum bending moment may also 
be obtained thus — 



for a maximum value 





dx ~ 


0.1 
6 


6/ 


c/ 

6 


3C^ 
6/ 


and 


/ 
^=V-3 



By substitution of the values of x and R, and by reduction, 
we have — 

_ _c^ _ c/" 

■M-inai. — /- — -■ 

9V3 150 

^ ^ ^ A- . X RNV^ 

and the bendmg stress/ = —z ^ 

° "^ i64000K^ 

which is about one-half as great as the stress in a coupling-rod 
working under the same conditions. 



Dynamics of the Steam- Engine. 211 

The rod is also subjected to a direct stress and to a very 
small bending stress due to the deflection of the rod, which 
can be treated by the method given for coupling rods. 

Readers who wish to go very thoroughly into this question 
should refer to a series of articles in the Engineer, March, 
1903. 

Balancing Revolving Axles. 

Case I. " Standing Balancer — If an unbalanced pulley or 
wheel be mounted on a shaft and the shaft be laid across two 
levelled straight-edges, the shaft will roll until the heavy side of 
the wheel comes to the bottom. 

If the same shaft and wheel are mounted in bearings and 
rotated rapidly, the centrifugal force acting on the unbalanced 
portion would cause a pressure on the bearings acting always 
in the direction of the unbalanced portion ; if the bearings were 
very slack and the shaft light, it would lift bodily at every 
revolution. In order to prevent this action, a balance weight 
or weights must be attached to the wheel in its own plane of 
rotation, with the centre of gravity diametrically opposite to the 
unbalanced portion. 

Let W = the weight of the unbalanced portion ; 
Wj = „ „ balance weight ; 

R = the radius of the c. of g. of the unbalanced 

portion ; 
Ri = the radius of the c. of g. of the balance weight. 

Then, in order that the centrifugal force acting on the balance 
weight may exactly counteract the centrifugal force acting on 
the unbalanced portion, we must have — 

0-00034WRN2 = o-ooo34WiRiN* 

or WR = WiRi 
or WR - WiR, = o 

that is to say, the algebraic sum of the moments of the rotating 
weights about the axis of rotation must be zero, which is 
equivalent to saying that the centre of gravity of all the rotating 
weights must coincide with the axis of rotation. When this is 
the case, the shaft will not tend to roll on levelled straight- 
edges, and therefore the shaft is said to have "standing 
balance." 



212 



Mechanics applied to Engineering. 



When a shaft has standing balance, it will also be perfectly 
balanced at all speeds, //-mafei/ that all the weights rotate in the 
same plane. 

We must now consider the case in which all the weights do 
not rotate in the same plane. 

Case II. Running Balance. — If we have two or more 
weights attached to a shaft which fulfil the conditions for 

standing balance, but yet do not 
/(iC rotate in the same plane, the 

shaft will no longer tend to lift 
bodily at each revolution ; but it 
will tend to wobble, that is, it 
I will tend to turn about an axis 
I perpendicular to its own when it 
^ rotates rapidly. If the bearings 
were very slack, it would trace out 
the surface of a double cone in 
space as indicated by the dotted 
Fig. 221. lines, and the axis would be con- 

stantly shifting its position, i.e. it 
would not be permanent. The reason for this is, that the 
two centrifugal forces c and c-^ form a couple, tending to turn 
the shaft about some point A between them. In order to 




Fig, ! 



counteract this turning action, an equal and opposite couple 
must be introduced by placing balance weights diametrically 
opposite, which fulfil the conditions for " standing balance." and 



Dynamics of the Steam-Engine. 213 

moreover their centrifugal moments about any point in the 
axis of rotation must be equal and opposite in effect to those 
of the original weights. Then, of course, the algebraic sum of 
all the centrifugal moments is zero, and the shaft will have no 
tendency to wobble, and the axis of rotation will be permanent. 
In the figure, let the weights W and W, be the original 
weights, balanced as regards " standing balance," but when 
rotating they exert a centrifugal couple tending to alter the 
direction of the axis of rotation. Let the balance weights 
Wa and W3 be attached to the shaft in the same plane as 
Wi and W, i.e. diametrically opposite to them, also having 
"standing laalance." Then, in order that the axis may be 
permanent, the following condition must be fulfilled : — 

<y + c^y-i = c^y^i + hy 

o-ooo34N^(WRy+W,R,ji'i) = o'ooo34N'(W,R2j/,+W3R3jFs) 
or WRj/ + WjRi^i - W,R,j/s, - WsR^^a = o 

The point A, about which the moments are taken, may be 
chosen anywhere along the axis of the shaft without affecting 
the results in the slightest degree. Great care must be taken 
with the signs, viz. a + sign for a clockwise moment, and a — 
sign for a contra-clockwise moment. 

The condition for standing balance in this case is — 

WR - WiR, - W^Ra + WaRa = o 

So far we have only dealt with the case in which the 
balance weights are placed diametrically opposite to the 
weight to be balanced. In some cases this may lead to more 
than one balance weight in a plane of rotation ; the reduction 
to one equivalent weight is a simple matter, and will be dealt 
with shortly. Then, remembering this condition, the only other 
conditions for securing a permanent axis of rotation, oy a 
" running balance," are — 

SWR = o 
and SWR,)- = o 

where 2WR is the algebraic sum of the moments of all 
the rotating weights about the axis of rotation, and y is the 
distance, measured parallel to the shaft, of the plane of rotation 
of each weight from some given point in the axis of rotation. 
Thus the c. of g. of all the weights must lie in the axis of 
rotation. 



214 



Mechanics applied to Engineering. 



Graphic Treatment of Balance Weights. — Such a 
problem as the one just dealt vpith can be very readily treated 
graphically. For the sake, however, of giving a more general 
application of the method, we will take a case in which the 




'J3 






Fig. 223. 



weights are not placed diametrically opposite, but are as shown 
in the figure. 

Let all the quantities be given except the position and 
weight of W4, and the arm yi, which we shall proceed to find 
by construction. 



Standing balance. 

There must be no tendency for 
the axis to lift bodily ; hence the 
vector sum of the forces C,, Cj, C„ 
Ci, must be zero, i.e. they must form 
a closed polygon. Since C is pro- 
portional to WR, set- oflf W,R„ 



WR, 




WjRj, W,R„ to some suitable scale 
and in their respective directions ; 
then the closing line of the force 
polygon gives us W^R, in direction, 
magnitude, and sense. The radius 
R, IS given, whence W, is found by 
dividing by R,. 



Runfiing balance. 

There must be no tendency for 
the axis to wobble ; hence the vector 
sum of the moments C,_j'„ etc., 
about a. given plane must be zero, 
i.e. they, like the forces, must form 
a closed polygon. We adopt Pro- 
fessor D^by's method of taking 
the plane of one of the rotating 
masses, viz. W, for our plane 
of reference ; then 
the force C, has no 
moment about the 
plane. Constructing 
the triangle of mo- 
ments, we get the 
value of W,R,^4 from the closing 
line of the triangle. Then dividing 
by WjR,, we get the value of V4. 



W,R,», 



Dynamics of the S team-Engine. 



215 



/to 



A 



B 



Tcrr 



Fig. 224. 



Provided the above-mentioned conditions are fulfilled, the 
axle will be perfectly balanced at all speeds. It should be 
noted' that the second condition cannot' be fulfilled if the 
number of rotating masses be less than four. 

Balancing of Stationary Steam-Engines. — Let the 
sketch represent the scheme of a two- cylinder vertical steam- 
engine with cranks at right angles. Consider the moments of 
the unbalanced forces/, and/j about the point O. When the 
piston A is at the bottom of' its stroke, 
there is a contra-clockwise mome.n\.,p^y„ 
due to the acceleration pressure p^ tend- 
ing to turn the whole engine round in a- 
contra-clockwise direction about the point 
O. The force /j is zero in this position 
(neglecting the effect of the obliquity of 
the rod). When, however, A gets to 
the top of its stroke, there is a moment, 
^oJ«i tending to turn the whole engine 
in a contrary direction about the point 
O. Likewise with B ; hence there is a 
constant tendency for the engine to lift 
first at O, then at P, which has to be counteracted by the 
holding-down bolts, and may give rise to very serious vibra- 
tions unless the foundations be very massive. It must be 
clearly understood that the cushioning of the steam men- 
tioned on p. 190 in no way tends to reduce this effect; balance 
weights on the cranks will partially remedy the evil, but it is 
quite possible to entirely eliminate it in such an engine as 
this. 

A two-cylinder engine can, however, be arranged so that 
the balance is perfect in every 
respect. Such a one is found in 
the Barker engine. In this engine 
the two cylinders are in line, and 
the cranks are immediately op- 
posite and of equal throw. The 
connecting-rod of the A piston is 
forked, while that of the B piston is coupled to a central crank ; 
thus any forces that may act on either of the two rods are 
equally distributed between the two main bearings of the 
bed-plate, and consequently no disturbing moments are set 
up. Then if the mass of A and its attachments is equal to 
that of B, also if the moments of inertia of the two con- 
necting-rods about the gudgeon-pins are the same, the 




2i6 Mechanics applied td Engineering. 

disturbing effect of the obliquity of the rods will be entirely 
eliminated. 

A single cylinder engine can be balanced by a similar 
device. Let B represent the piston, cross-head, and connecting 
rod of a single-cylinder engine, and let a " bob-weight " be 
substituted for the piston and cross-head of the cylinder A. 
Then, provided the " bob-weight " slides to and fro in a 
similar manner and fulfils the conditions mentioned above, a 
perfect balance will be established. In certain cases it may 
be more convenient to use two " bob-weights," Aj and A^, each 
attached to a separate connecting rod. Let the distances of 
the planes of the connecting rods from a plane taken through 
B be Oil and x^, and the "bob-weights" be Wj and Wa, and 
the radii of the cranks r^ and r.^, respectively. Let the weight 
of the reciprocating parts of B be'W and the radius of the 
crank r. Then using connecting rods of equal moment of 
inertia about the gudgeon pin for Aj and Aj, and whose com- 
bined moments of inertia are equal to that of B, we must have — 

Wi/-i(a:i -f x^ = WrjCj, also '^ir4^Xi + x^= Vfrx^. 

The same arrangement can be used for three-cylinder 
engines,^the "bob-weights" then become the weights of the 
reciprocating parts of the two cylinders Aj and Aj. 

Any three-cylinder engine can be balanced in a similar man- 
ner by the addition of two extra cranks or eccentrics to drive 
suitable "bob-weights," the calculations for arriving at the neces- 
sary weights, radii of cranks and their positions along the shaft 
can be readily made by the methods shortly to be discussed. 

If slotted cross-heads are used in order to secure simple 
harmonic motion for the reciprocating parts the matter is 
simplified in that no connecting rods are used and consequently 
there are no moments of inertia to be considered. 

A three-cylinder vertical engine having cranks at r2o°, and 
having equal reciprocating and rotating masses for each cylinder, 
can be entirely balanced along the centre line of the engine. 
The truth of this statement can be readily demonstrated by 
inserting the angles 6, 6 + 120°, and B + 240° in the equation 
on p, 186 ; the sum of the inertia forces will be found to be 
zero. The proof was first given by M. Normand of Havre, 
and will be found in " Ripper's Steam Engine Theory and 
Practice." 

There will, however, be small unbalanced forces acting at 
right angles to the centre line, tending to make the engine 
rock about an axis parallel to the centre line of the crank-shaft. 



Dynamics of the Steam- Engine. 



217 



A four-cylinder engine, apart from a small error due to the 
obliquity of the rods, can be perfectly balanced j thus — 




Fig. 226. 



JtBi, 



"msf 




Let the reciprocating masses be 
Wi, Wa, etc. ; 

the radii of the cranks be Rj, Rj, etc.; 
the distance from the plane of refer- 
ence taken through the first crank 
bejj'a.ji'a, etc. 

Then the acceleration pressure, neglecting the obliquity 
of the rods at each end of the stroke, will be o"ooo34WRN^, 
with the corresponding suffixes for each cylinder. Since the 
speed of all of them is the same, the acceleration pressure will 
be proportional to WR. It will be convenient to tabulate the 
various quantities, thus — 



t5' 


.Weight of 

reciprocating 

parts. 

lbs. 


Radius of 
crank. 


Proportional 
acceleration force. 


Distance of 

centre line 

from plane of 

reference. 


Proportional 

acceleration force 

moment. 


I 
2 

3 

4 


W,= 7.?o 
W,= iooo 

W5=I20O 

W.=i230 


R,= I2" 

R,= i4" 
R3 = i4" 

R, = I2" 


■W]R,= 9,000 

■Wi,Rj= 14,000 
W3R3= 16,800 

W<Ri = 14,800 




y^= 40" 
y,= 80 
;/<=li2" 




"^^0^= 560,000 

W3R;jy3= 1,344,000 
W<R4j/4= 1,653,000 



2i8 Mechanics applied to Engineering. 

The vector sum of both the forces and the moments of the 
forces must be zero to secure perfect balance, i.e. they must 
form closed polygons ; such polygons are drawn to show how 
the cranks must be arranged and the weights distributed. 

The method is due to Professor Dalby, who treats the 
whole question of balancing very thoroughly in his " Balancing 
of Engines." The reader is recommended to consult this book 
for further details. 

Balancing Locomotives. — In order that a locomotive 
may run steadily at high speeds, the rotating and reciprocating 
parts must be very carefully balanced. If the rotating parts 
be left unbalanced, there will be a serious blow on the rails 
every time the unbalanced portion gets to the bottom; this 
is known as the " hammer blow." If the reciprocating parts 
be left unbalanced, the engine will oscillate to and fro at every 
revolution about a vertical axis situated near the middle of 
the crank-shaft ; this is known as the " elbowing action." 

By balancing the rotating parts, the hammer blow may 
be overcome, but then the engine will elbow j if, in addition, 
the reciprocating parts be entirely balanced, the engine will be 
overbalanced vertically ; hence we have to compromise matters 
by only partially balancing the reciprocating parts. Then, 
again, the obliquity of the connecting-rod causes the pressure 
due to the inertia of the reciprocating parts to be greater at 
one end of the stroke than at the other, a variation which 
cannot be compensated for by balance weights rotating at a 
constant radius. 

Thus we see that it is absolutely impossible to perfectly 
balance a locomotive of ordinary design, and the compromise 
we adopt must be based on experience. 

The following symbols will be used in the paragraphs on 
locomotive balancing : — 

W„ for rotating weights (pounds) to be balanced. 

Wp, for reciprocating weights (pounds) to be balanced. 

Wb, for balance weights ; if with a suffix p, as Wsp, it will 
indicate the balance weight for the reciprocating parts, 
and so on with other suffixes. 

R, for radius of crank (feet). 

R„ „ „ coupling-crank. 

Rb, „ „ balance weights. 

Rotating Parts of Locomotive. — ^The balancing of 
the rotating parts is effected in the manner described in the 
paragraph on standing balance, p. 2 1 1, which gives us — 



Dynamics of the Steam-Engine, 219 

W,R = W,,R3 
and W,, = ^ 

■Kb 

The weights included in the W, vary in different types of 
engines ; we shall consider each case as it arises. 

Reciprocating Parts of Locomotive. — We have 
already shown (p. 181) that the acceleration pressure at the 




Fig. 227. 

end of the stroke due to the reciprocating parts is equal to 
the centrifugal force, assuming them to be concentrated at the 
crank-pin, and neglecting the obliquity of the connecting-rod. 

Then, for the present, assuming the balance weight to 
rotate in the plane of the crank-pin, in order that the recipro- 
cating parts may be balanced, we must have — 

C = C 

o-ooo34Wbp . Rb . N^ = o-ooo34Wp .R.N' 
Wbp . Rb = Wp . R 

andWBp=^%^ (i.) 

•Kb 

On comparing this with the result obtained for rotating 
parts, we see that reciprocating parts, when the obliquity of 
the connecting-rod is neglected, may for every purpose be 
regarded as though their weight were concentrated in a heavy 
ring round the crank-pin. 

Now we come to a much-discussed point. We showed 
above that with a short connecting-rod of n cranks long, the 

acceleration pressure was - greater at one end and - less at 

the other end of the stroke than the pressure with an infinitely 

long rod : hence if we make Wsp - greater to allow for the 



220 Mechanics applied to Engineering. 

2 , 

obliquity of the rod at one end, it will be - too. great at the 

other end of the stroke. Thus we really do mischief by 
attempting to compensate for the obliquity of the rod at either 
end; we shall therefore proceed as though the rod were of 
infinite length. 

If the reader wishes to follow the effect of the obliquity 
of the rod at all parts of the stroke, he should consult a paper 
by Mr. Hill, in the Proceedings of the Itistitute of Civil Engineers , 
vol. civ. ; or Barker's " Graphic Methods of Engine Design ; " 
also Dalby's " Balancing of Engines." 

The portion of the connecting-rod which may be regarded 
as rotating with the crank-pin and the portion as reciprocating 
with the cross-head may be most readily obtained by find- 
ing the centre of gravity of the whole rod — let its distance 
from the crank-pin centre be x, the length of the rod centres /, 
then the portion to be included in the reciprocating parts is 

—j-, where W is the weight of the whole rod. The remainder 

(i— j-Jistobe included in the rotating parts. If the rod 

be placed horizontally with the small end on a weighing 
machine or be suspended from a spring balance the reading 
will give the weight to be included in the reciprocating parts. 
For most purposes it is sufficiently near to take the weight 
of the small end together with one-half the plain part as 
reciprocating, and the big end with one-half the plain part as 
rotating. 

Inside-cylinder Engine (uncoupled).- -In this case 
we have — 

Wp = weight of (piston -|- piston-rod -f cross-head -|- small 
end of connecting-rod -|- \ plain part of rod) ; 

W, = weight of (crank-pin -1- crank- webs ' -f- big end of 
connecting-rod -P \ plain part of rod). 

If we arrange balance weights so that their c. of g. rotates 
in the same plane as the crank-pins, their combined weight 
would be Wg, -|- Wup, placed at the radius Rb, and if we only 
counterbalance two-thirds of the reciprocating parts, we get — 



W 



R(|Wp + W,) 
See p. 229, 



_ J-V^3»vp T T',y ... , 

»'B0 — t5 V"'J 



Dynamics of the Steam-Engine. 



221 



Balance weights are not usually placed opposite the crank- 
webs as shown, but are distributed over the wheels in such a 
manner that their centrifugal moments about the plane of 
rotation of the crank-pin is zero. If W be the balance weight 
on one w^heel, and Wj the other, distant y' and yl from the 
plane of the crank, then — 

or Wy = W,;/i' 

which is equivalent to saying that the centre of gravity of the 
two weights lies in the plane of rotation of the crank. The 
object of this particular arrangement is to keep the axis of 




WboX 

y 

i°off'K/heel 
opposite 'off "crank 



Cofe.ifCr^/rmi^/llS. 



WboZ 
y 

On "near" wheel 

opposite "near^Qrank 




Flc. 328. 



rotation permanent. Then, considering the vertical crank 
shown in Fig. 228, by taking moments, we get the equivalent 
weights at the wheel centres as given in the figure. 
We have, from the figure — 



X = ■ 



z=y-^i 



J 



2 

_y ■\-c 



222 Mechanics applied to Engineering. 

Substituting these values, we get — 

^»/ y — c) = Wbi, as the proportion of the balance weight 
on the " off" wheel opposite the far crank 

and — "(v + ^) = Wb2, as the proportion of the balance weight 
on the " near " wheel opposite near crank 

Exactly similar balance weights are required for the other 
crank. Thus on each wheel we get 
one large balance weight W^a at N 
(Fig. 229), opposite the near crank, 
and one small one Wei at F, opposite 
the far crank. Such an arrangement 
\f would, however, be very clumsy, so we 
shall combine the two balance weights 
by the parallelogram of forces as 
shown, and for them substitute the 
large weight Wb at M. . 




ThenWB= VWbi' + Wb 



On substituting the values given above for Wei and Wj, 
we have, when simplified — 



W„ 



V2Wb 

2y 



V/ + ^ 



In English practice^ = 2'^c (approximately) 
On substitution, we get — 

Wb = o-76Wb, 

Substituting from ii., we have — 

o-76R(|W, + W,) 
Rb 

Let the angle between the final balance weight and the 
near crank be a, and the far crank Q + 90. 
Then a = 180 - 6 

andtane = S^=^^ 
Wbs ;» + <■ 



Dynamics of the Steam- Engine. 223 

Substituting the value oi y for English practice, we get — 

tan B = — —= o"42Q 
3-5 ^ ^ 

Now, 6 = ^ very nearly ; hence, for English practice, if 
the quadrant opposite the crank quadrant be divided into 




W„ = Wb 



■ nearly 



Fig. 230. 

four equal parts, the balance weight must be placed on the first 
of these, counting from the line opposite the near crank. 

Outside-eylinder Engine (uncoupled). — Wp and W, 
are the same as in the last paragraph. If the plane of rotation 
of the crank-pin nearly coincides, as it frequently does, with the 
plane of rotation of the balance weight, we have — 

Rb 

and the balance weight is placed diametrically opposite the 
crank. 

When the planes do not approximately coincide — 

Let J/ = the distance between the wheel centres ; 

cylinder centres ; 
cylinder centre line and 

the " near " wheel ; 
cylinder centre line and 

the " off" wheel. 

2 



c = 

X = 



* = 



c-y 



The balance weight required! 

on the "off" wheel opposite? 

the " far " crank ' 

The balance weight required. 



Wbo^: 



2y 



on the "near" wheel oppo-|= —'^- = ^^(c+y) = W^ 
site the "near" crdnk ) ^ '-^ 



224 



Mechanics applied to Engineering. 



Then W, = 



^2W„ 

2y 



'Jf + c" 



which is precisely the same expression as we obtained for 
inside-cylinder engines, but in this casejc = o'8f to o'<)c. On 
substitution, we get Wb = i'I3Wb„ to i-osW^o, and fl = 6° to 3°. 

The same reasoning applies to the coupling-rod balance 
weights Wbo in the next paragraphs. 

Inside-cylinder Engine (coupled). — In this case we 
have Wp the same as in the previous cases. 

Wo = the weight of coupling crank-web and pin ^ -f coupling 
rod from a to b, or c to d, otbto c (Fig. 195), as the 
case may be ; 

Wbo = the weight of the balance weight required to counter- 
balance the coupling attachments ; 
Ro = the radius of the coupling crank. 




Fio. 231. 



In the case of the driving-wheel of the four-wheel coupled 
engine, we have Wb arrived at in precisely the same manner 
as in the case of the inside-cylinder uncoupled engine, and 

^^ Rb • 

The portion of the coupling rod included in the Wo is, in 
this case, one-half the whole rod. The balance weight Wbo is 
placed diametrically opposite the coupling crank-pin. After 
finding Wb and Wbo. tliey are combined in one weight Wbf by 
the parallelogram of forces, as already described. 

With this type of engine the balance weight is usually 
small. Sometimes the weights of the rods are so adjusted 
that a balance weight may be dispensed with on the driving- 
wheel. 

' See p. 229. 



Dynamics of the Sieam- Engine. 



225 



It frequently happens, however, that W^ is larger than Wpo ; 
in that case Wgp is placed much nearer Wj than is shown in 
the figure. 

On the coupled wheel the balance weight Wjo is of the 
same value as that given above, and is placed diametrically 
opposite the coupling crank-pin. 




Fig. 232. 

In the six-wheel coupled engine the method of treatment is 
precisely the same, but one or two points require notice. 

RcWc 



W'b„ = . 



Rn 



The portion of the coupling rod included in the Wo is from 
b\.o e; whereas in the Wbo the portion is from a to ^ or ^ to d. 

Coupling cranks * have been placed with the crank-pins ; 
the balance weights then become very much greater. They are 
treated in precisely the same way. 

Some locomotive-builders evenly distribute the balance 
weights on coupled engines over all the wheels : most authorities 
strongly condemn this practice. Space will not allow of this 
point being discussed here. 

Outside-cylinder Engine (coupled). 

W is the same as before ; 

W, is the weight of crank-web ' and pin -|- coupling rod 

from a\.ob ■\- big end of connecting-rod + half plain 

part of rod ; 
Wc is the same as in the last paragraph; 

Ro = R 
VV3 = \Vn„ = 

' See Proc. Inst. C.£., vol. Uxxi. p. 



m 


w, + w,) 

Rb 




122. 


' See 


p. 229 
Q 



226 



Mechanics applied to Engineering. 



The six-wheel coupled engine is treated in a similar way ; 
the remarks in the last paragraph also apply here. 




The above treatment only holds when the planes of the 
crank-pins and wheels nearly coincide, as already explained 
when dealing with the uncoupled outside-cylinder engine. 

On some narrow-gauge railways, in which the wheels are 
placed inside the frames, the crank and coupling pins are often 

at a considerable distance from 
the plane of the wheels. Let 
the coupling rods be on the 
outer pins. It will be convenient, 
| .,. . when dealing with this case, to 

l^'i jt ~ ^jp^^^^la^i^^;^^ ' iJij', find the distance between the 
1^ planes containing the centres of 

I '«--t}-i — ^— y ' ' gravity of the coupling and 

r* g connecting rods, viz. C,. 

„ _ W. Q + (W, -I- fW,)C 



c 



I 



Fig. 234. 



W, -f W, + fWp 



The W, must include the weight of the crank-webs and 
pins all reduced to the radius of the pin and to the distance C. 
For all practical purposes, C, may be taken as the distance 
between the insides of the collars on the crank-pin. Then, by 
precisely similar reasoning to that given above — 

where W.^'^^tyVp + W.-fW.) 

In some cases _)> is only osC, ; then — 

Wb = i-s8Wbo, and 0= 18° 



Dynamics of the Steam- Engine. 227 

Hammer Blow. — If the rotating parts only of a loco- 
motive are fully balanced there is no variation of load on 
the rail due to their centrifugal force, but when, in addition, 
the reciprocating parts are partially or fully Ijalanced the 
vertical component of the centrifugal force of the excess 
balance weight over and above that required to balance the 
rotating parts causes a considerable variation of the load on 
the rail, tending to lift the wheel off the rail when the balance 
weight is on top and causing a very rapid increase of rail load 
— almost amounting to a blow — when the balance weight is 
at the bottom. The hammer blow is rather severe on the 
cross girders of bridges and on the permanent way generally. 
At very high speeds the upward force will actually lift the 
wheel off the rail if it exceeds the dead weight on the wheel. 
On some American railroads the proportion of the recipro- 
cating parts to be balanced is settled by the maximum speed 
the engine is likely to reach. 

In arriving at this speed, the balance weight required to 
completely balance the rotating parts is first determined, then 
the balance weight for both rotating and reciprocating parts is 
found, the difference between the two is the unbalanced portion 
which is responsible for the hammer blow and the tendency to 
lift the wheel off the rail. 

Let the difference be W^' then the centrifugal force tending 
to lift the wheel is 

o"ooo34VV^RbN* 

and when this exceeds the dead weight W„ on the rail the 
wheel will lift, hence the speed of lifting is 



-sj 



W 

N 



o-ooo34W^Rb 



When calculating the speed of the train from N it must not 
be forgotten that the radius of the wheel is greater than Rb. 

Centre of Gravity of Balance Weights and Crank- 
webs. — The usual methods adopted for finding the position 
and weight of balance weights are long and tedious ; the follow- 
ing method will be found more convenient. The effective 
balance weight is the whole weight minus the weight of the 
spokes embedded. 

Let Figs. 235, 236, 237 represent sections through a part of 
the balance weight and a spoke ; then, instead of dealing first 
with the balance weight as a whole, and afterwards deducting 



228 



Mechanics applied to Engineering. 



the spokes, we shall deduct the spokes first. Draw the centre 
lines of the spokes x, x, and from them set off a width w on 




Fig. ass- 
each side as shown in Fig. 236, where wi = half the area of 
the spoke section ; in the case 



of the elliptical spoke, wf = 



o'785Di^ 



o'392Di/ 



of the rectangular spoke, wi = — 

2 



o'sD^ 



By doing this we have not altered either the weight or the 
position of the centre of gravity of the section of the balance 
weight, but we have reduced it to a much simpler form to 
deal with. If a centre line yy (Fig. 235) be drawn through 
the balance weight, it is only necessary to dealt with the 
segments on one side of it. 

Measure the area of the segments when thus treated. 



Dynamics of the Steam-Engine. 



229 



Let them be Aj, A^, A3 ; then the weight of the whole balance 
weight is the sum of these segments — 

Wb = 2twJ,k^ + A, + A3) 




where zc„ = the weight per cubic inch of the metal. 
For a cast-iron weight — 

Wb = o-52/(Ai + Aj, + A3) 

For a wrought-iron or cast-steel weight — 

Wb = o-56/(Ai + A, + A3) 

all dimensions being in inches. 

The centre of gravity of each section can be calculated, but 
it is far less trouble to cut out pieces of cardboard to the shape 
of each segment, and then find the position of the centre of 
gravity by balancing, as described on p. 75. Measure the 
distance of each centre of gravity from the line AB drawn 
through the centre of the wheel. 

Let them be ri, r^, r^ respectively ; then the radius of the 
centre of gravity of the whole weight (see Fig. 235) — 

Rb- A, + A,-t-A. (^^«P-S8) 



md WgRj = \ or 
(0-56; 



i?(Airi + AaZ-a + Aa^s) 



?30 



Mechanics applied to Engineering. 



If there were more segments than those shown, we should 
get further similar terms in the brackets. 





Tig. 237- 



Fig. 238. 



AVhen dealing with cranks, precisely the same method may 
be adopted for finding their weight and the position of the 
centre of gravity. 

In the figures, the weight of the crank = zto™ X shaded 
areas. The position of the centre of gravity is found as before, 
but no material error will be introduced by assuming it to be 
at the crank-pin. 

Governors. — The function of a flywheel is to keep the 
speed of an engine approximately constant during one revolu- 
tion or one cycle of its operations, but the function of a 
governor is to regulate the number of revolutions or cycles 
that the engine makes per minute. In order to regulate the 
speed, the supply of energy must be varied proportionately to 
the resistance overcome ; this is usually achieved automatically 
by a governor consisting essentially of a rotating weight 
suspended in such a manner that its position relatively to the 
axis of rotation varies as the centrifugal force acting upon it, 
and therefore as the speed. As the position of the weight varies, 
it either directly or indirectly opens and closes the valve 
through which the energy is supplied, closing it when the speed 
rises, opening it when it falls. 

The governor weight shifts its position on account of a 
change in speed ; hence some variation of speed must always 
take place when the resistance is varied, but the change in 



Dynamics of the Steam-Engine. 



231 



speed can be reduced to a very small amount by suitably 
arranging the governor. q 

Simple Watt Governor. — Let the 
ball shown in the figure be suspended by 
an arm pivoted at O, and let it rotate round 
the axis OOi at a constant rate. The ball 
is kept in equilibrium by the three forces 
W, the weight of the ball acting vertically 
downwards (we shall for the present neglect 
the weight of the arm and its attachments, 
also friction on the joints) ; C, the centri- 
fugal force acting horizontally; T, the tension 
in the supporting arm. 




f IG. 239. 



Let H = height of the governor in feet ; 
h = „ „ „ inches; 

R = radius of the ball path in feet ; 
N, = number of revolutions made by the governor per 

second ; 
N = number of revolutions made by the governor per 

minute ; 
V = velocity (linear) in feet per second of the balls. 

By taking moments about the pin 0, we have— 

WV^H 



CH = WR, 
hence H = 



i;R 



-=WR 



g^^ 



o'8i6 



47r^R^N/ N 1 



Expressing the height in inches, and the speed in revolu- 
tions per minute, we get — 

3513? 



h = 



Thus we see that the height at which a simple Watt governor 
will run is entirely dependent upon the number of revolutions 
per minute at which it runs. The size of the balls and length 
of arms make no difference whatever as regards the height 
when the balls are " floating." 

The following table gives the height of a simple Watt 
<• governor for various speeds : — 



232 



Mechanics applied to Engineering. 







Change of height 


Revolutions per 
minute («). 


Height of governor 
in inciics (A). 


corresponding to 
a change of speed 
of 10 revolutions 
per minute. 






Inches. 


50 


14-09 


— 


(54-2) 


(I2-00) 


— 


60 


979 


4-30 


70 


7-19 


260 


80 


S'Si 


1-68 


90 


4-3S 


i-i6 


100 


3-52 


083 


no 


2-91 


o'6i 


120 


2 -45 


0-46 



These figures show very clearly that the change of height 
corresponding to a given change of speed falls off very rapidly 
as the height of the governor decreases or as the apex angle 
Q increases ; but as the governing is done entirely by a cliange 
in the height of the governor in opening or closing a throttle 
or other valve, it will be seen that the regulating of the motor 
is much more rapid when the height of the governor is great 
than when it is small ; hence, if we desire to keep the speed 
within narrow limits, we must keep the height of the governor 
as great as possible, or the apex angle 6 as small as possible, 
within reasonable limits. 

Suppose, for instance, that a change of height of 2 inches 
were required to fully open or close the throttle or other valve ; 
then, if the governor were running at 60 revolutions per minute, 
the 2-inch movement would correspond to about 7 per cent, 
change of speed; at 80, 15 per cent.; at 100, 24 per cent; 
at 120, 36 per cent. 

The greater the change of height corresponding to a given 
change of speed, the greater is said to be the sensitive?KSs of 
the governor. 

A simple Watt governor can be made as sensitive as we 
please by running it with a very small apex angle, but it then 
becomes very cumbersome, and, moreover, it then possesses 
very little " power " to overcome external resistances. 

Loaded Governor. — In order to illustrate the principle 
of the loaded governor, suppose a simple Watt governor to be 
loaded as shown. The broken lines show the position of the 
governor when unloaded. 

When the load W„ is placed on the balls, the " equivalent 



Dynamics of the Steam- Engine. 



233 



height of the simple Watt governor " is increased from H to 

H^. Then, constructing the triangle of forces, or, by taking 

W 
moments about the pin, and remembering that — - acts ver- 

2 

tically downwards through the centre of the ball, we have — 




Fig. 240. 



\V 

R. r — 



Then, by precisely the same reasoning as in the case given 
above, we have — 



H.= 



W ^ 
W 4-—!° 
o-8i6/ ^ 2 



N/ 



W 



or h, ■ 



W + — 

3523°! L 

N" \ W 



If W„ be m times the weight of one ball, we have — 

the value of m usually varies from 10 to 50. 

This expression must, however, be used with caution. 
Consider the case of a simple Watt governor . both when 
unloaded and when loaded as shown in Figs. 239 and 240. 
If the same governor be taken in both instances, it is evident 
that its maximum height, i^. when it just begins to lift, also its 



234 



Mechanics applied to Engineering. 



minimuni apex angle, will be the same whether loaded or 
unloaded, and cannot in any case be greater than the length 
of the suspension arm measured to the centre of the ball. 
The speed of the loaded governor corresponding to any given 
height will, however, be greater than that of the unloaded 

governor in the ratio » /i + — to i, and if the engine runs at 

the same speed in both cases, the governor must be geared up 
in this ratio, but the alteration in height for any given alteration 
in the speed of the engine will be the same in both cases, or, in 
other words, the proportional sensitiveness will be the same 
whether loaded or unloaded. We shall later on show, however, 
that the loaded governor is better on account of its greater power. 

In the author's opinion most writers on this subject are in 
error ; they compare the sensitiveness of a loaded governor at 
heights which are physically impossible (because greater even 
than the length of the suspension arms), with the much smaller, 
but possible, heights of an unloaded governor. If the reader 
wishes to appeal to experiment he can easily do so, and will 
find that the sensitiveness actually is the same in both cases. 

The following table may help to make this point clear. 



m has been chosen as i6 : then 



/ m 



On comparing the last column of this table with that for 
the unloaded governor, it will be seen that they are identical, 
or the sensitiveness is the same in the two cases. 







Change of height 






corresponding to 






a change of speed 


Revolutions per 


Height of loaded 


of 30 revolutions 


minute of 


governor in 


per minute of 


governor. 


inches. 


govemorp or 10 
revolutions per 
minute of engine. 






Inches. 


150 


14-09 




180 


979 


4 "3° 


210 


7-19 


2 -60 


240 


551 


1-68 


270 


4'35 


116 


300 


3"5z 


0-83 


330 


2'9I 


0'6i 


360 


2-45 


0-46 



Dynamics of the Steam- Engine. 235 

If, by any system of leverage, the weight W„ moves up and 
down .t: times as fast as the balls, theabove expression becomes — 






Porter and other Loaded Governors. — The method 
of loading shown in Fig. 240 is not convenient, and is rarely 
adopted in practice. The usual method is that shown in 
Fig. 241, viz. the Porter governor, in which the links are 
usually of equal length, thus making x = 2; but this proportion 
is not always adhered to. Then — - 



u _ 35230/W + W„N 



^' = ^(^ + *") 

Occasionally, governors of this type are loaded by means of 
a spring, as shown in Fig. 243, instead of a central weight. The 
arrangement is, however, bad, since the central load increases 
as the governor rises, and consequently makes it far too sluggish 
in its action. 

Let the length of the spring be such that it is free from 
load when the balls are right in, «'.*. when their centres coincide 
with the axis of rotation, or when the apex angle is zero. The 
reason for making this stipulation will be apparent when we 
have dealt with other forms of spring governors. 

Let the pressure on the spring = Pa:, lbs. when the spring 
is compressed x„ feet ; 

/ = the length of each link in feet ; 
H, = the equivalent height of the governor in feet ; 
^V = the weight of each ball in lbs. ; 
X. = 2(1 - H.) ; 

N = revolutions of governor per minute ; 
Then, as in the Porter- governor, or, by taking moments 
about the apex pin, and remembering that the force Px^ acts 
parallel to the spindle, we have — 

H. _ W + Pa:. _ W + gP/ - zPH. 
R, ~ C ~ o-ooo34WR,N!' 
,HXo-ooo34WN2 + 2P) = W + 2P/ 
„ W + 2PI . , 

"• = 0-00034WN' + 2P ^°' ** ^"''*=^^ SovernoT 

zP/ 
H, = 'iTrNT2~i — T> >i horizontal „ 



236 



Mechanics applied to Engineering. 



In the type of governor shown in Fig. 244, which is 
frequently met with, springs are often used instead of a dead 
weight. The value of * is usually a small fraction, consequently 
a huge weight would be required to give the same results as 
a Porter or similar type of governor. But it has other inherent 
defects which will shortly be apparent. 

Isochronous Governors. — A perfectly isochronous 
governor will go through its whole range with the slightest 





Fig. 241. 



Fig. 242. 



variation in speed ; but such a governor is practically useless 
for governing an engine, for reasons shortly to be ^cussed. 




Fig. 243. ' 




But when designing a governor which is required to be very 
sensitive, we sail as near the wind as we dare, and make it 
very nearly isochronous. In the governors we have considered. 



Dynamics of the Steam-Engine. 



237 



the height of the governor has to be altered in order to alter 
the throttle or other valve opening. If this could be accom- 
plished without altering the height of the governor, it could 
also be accomplished without altering the speed, and we should 
have an isochronous governor. Such a governor can be con- 
structed by causing the balls to move in the arc of a parabola, 
the axis being the axis of rotation. Then, from the pro- 
perties of the parabola, we know that the height of the' 
governor, i.e. the subnormal to the path of the balls, is constant 
for all positions of the balls ; therefore the sleeve which 
actuates the governing valve moves through its entire range 
for the smallest increase in speed. We shall only consider 
an approximate form which is very commonly used, viz. the 
crossed-arm governor. 

The curve abc is a parabolic arc ; the axis of the parabola 
is O^; then, if normals be drawn to the curve at the highest 
and lowest positions of the 
ball, they intersect at some 
point d on the other side of 
the axis. Then, if the balls 
be suspended from this point, 
they will move in an approxi- 
mately parabolic arc, and the 
governor will therefore be 
approximately isochronous — ■ 
and probably useless because 
too sensitive. If it be de- < 
sired to make the governor 
more stable, the points d, d 
are brought in nearer the 
axis. The virtual centre of 
the arms is at their inter- 
section ; hence the height of 
the governor is H, which is approximately constant. The 
equivalent height can be raised by adding a central weight as 
in a Porter governor. It, of course, does not affect the 
sensitiveness, but it increases the power of the governor to 
overcome resistances. The speed at which a crossed-arm 
governor lifts depends upon the height in precisely the same 
manner as in the simple Watt governor. 

It can also be arrived at thus. By taking moments about 
the pin d, W is the weight of the ball, / the length of the arm 
ad. 




Fig. 24s. 



238 Mechanics applied to Engineering. 



VV(R ■\-x) = CH„ = 0-00034WRNV/'' - (R + xf 



^=s/- ^^+^ 



0-00034 kV/'' - (R + xf 

X H„ - H , „ RH„ o-8i6 x 60^ 
or thus: - = -^^ and H =^^-^ = - ^, — 



--^- 



2937(jc + R) 



KH„ 



when the dimensions of the governor are taken in feet and 
the speed in revolutions per minute. 

In some instances Watt governors are made with the arms 
suspended at some distance, say x, from the axis of rotation, 
as shown in Fig. 249, but without the central weight. Then 
the above expression becomes — 






N- / R-* 



o-ooo34RV/'-(R-a:)^ 

and when the expression for H is used (p. 231), the height H is 
measured from the level of the ball centres to the point where 
the two arms cross (see Fig. 249). 

Astronomical Clock Governor. — A beautiful applica- 
tion of the crossed-arm principle as applied to isochronous 
governors is found in the governors used on the astronomical 
clocks made by Messrs. Warner and Swazey of Cleveland, 
Ohio. Such a clock is used for turning an equatorial telescope 
on its axis at such a speed that the telescope shall keep exactly 
focussed on a star for many hours together, usually for the 
purpose of taking a photograph of that portion of the heavens 
immediately surrounding the star. If the telescope failed to 
move in the desired direction, and at the exact apparent 
speed of the star, the relative motion of the telescope and star 
would not be zero, and a blurred image would be produced ; 
hence an extreme degree of accuracy in driving is required. 
The results obtained with this governor are so perfect that no 
ordinary means of measuring time are sufficiently accurate to 
detect any error. 

The spindle A and cradle are driven by the clock, whose 



Dynamics of the Steam- Engine. 



239 




speed has to be controlled, A short link, c, is pivoted to arms 
on the driving spindle at b ; the 
governor weights are suspended 
by links from the point d; a brake 
shoe, e, covered with some soft 
material, is attached to a lug on 
the link c, and, as the governor 
rotates, presses on the fixed drum 
f. The point of suspension, d, is 
so chosen that the governor is , 
practically isochronous. The ; 
weights rest in their cradle until '^V 
the speed of the governor is 
sufficiently high to cause them to 
lift ; when in their lowest position, ^'°' ''*^- 

the centre line of the weight arm passes through b, and con- 
sequently the pull along the arm has no moment about this 
point, but, as soon as the speed rises sufficiently to lift the 
weights, the centre line of the weight arm no longer coincides 
with b, and the pull acting along the weight arm now has a 
moment about b, and thus sets up a pressure between the 
rotating brake shoe e and the fixed drum /. The friction 
between the two acts as a brake, and thus checks the speed 
of the clock. 

It will be seen that this is an extremely sensitive arrange- 
ment, since the moment of the force acting along the ball link, 
and with it the pressure on the brake shoe, varies rapidly as the 
ball rises ; but since the governor is practically isochronous, 
only an extremely small variation in speed is possible. It 
should be noted that the driving effort should be slightly in 
excess of that required to drive the clock and telescope, apart 
from the friction on the governor drum, in order to ensure that 
there is always some pressure between the brake shoe and 
the drum. 

Wilson Hartnell Governor. — Another well-known and 
highly successful isochronous governor is the " Wilson Hartnell " 
governor. 

In the diagram, c is the centrifugal force acting on the ball, 
and / the pressure due to the spring, i.e. one-half the total 
pressure. As the balls fly out the spring is compressed, and since 
the pressure increases directly as the compression, the pressure 
p increases directly (or very nearly so) as the radius r of the 
balls ; hence we nn.ay write / = Kr, where K is a constant 
depending on the stiffness of the spring. 



240 



Mechanics applied to Engineering. 



Let r^ = nr, frequently n =. \. 

Then Wj — pr 
and o'ooo34W>-''«N^ = YJ-' 

and N'' = 



K 



o"ooo34Wn 



For any given governor the weight W of the ball is con- 
stant; hence the denominator of the fraction is constant, 

whence N^, and therefore N, 
is constant ; i.e. there is only 
one speed at which the 
governor will float, and any 
increase or decrease in the 
speed will cause the balls to 
fly right out or in, or, in 
other words, will close or 
fully open the governing 
valve ; therefore the governor 
is isochronous. 

There are one or two 
small points that slightly 
affect the isochronous cha- 




FlG. 247. 



racter of the governor. For example, the weight of the ball, 
except when its arm is vertical, has a moment about the 
pivot. Then, except when the spring arm is horizontal, the 
centrifugal force acting on the spring arm tends to make 
the ball fly in or out according as the arm is above or below 
the horizontal. 

We ■ shall shortly show how the sensitiveness can be varied 
by altering the compression on the spring. 

Weight of Governor Arms. — Up to the present we have 
neglected the weight of the governor arms and links, and have 
simply dealt with the weight of the balls themselves ; but with 
some forms of governors such an approximate treatment would 
give results very far from the truth. 

Dealing first with the case of the arm, and afterwards with 
the ball- 
Let the vertical section of the arm be a square inches, and 
all other dimensions be in inches. The centrifugal moment 
acting on the element is — 

wadr.rurh _ wau^r^dr 
1 2p 1 2"- tan Q 



Dynamics of the Steam-Engine. 
and on the whole arm 



241 



J 



r^dr 



1 2g tan 0. 

12X3^ tan 9 

which may be written 

„ R' «-' 

waR X — X : — a 

3 J2g tan 

The quantity wa^ is the 
weight of the whole arm VV„, 

R2 
and — IS the square of the 

.3 
radius of gyration of the arm 
about the axis of rotation, and 
the product is the moment of 




Fio. 248. 



inertia of the arm in pounds weight and inch ^ units. 

lo)' W„RV 



The centrifugal moment) 
acting on the arm ) 



i2g tan 36^ tan 6 



In the case of the ball we have the centrifugal moment for 
the elemental slice. 



■waidf\r' i(i?ki 



12g 

and for the whole ball^ 



woyr 



Via?/ 



(77i)«"^=MVn 



The quantity outside the brackets is the sum of the 
moments of the weight of each vertical slice of the ball about 
the axis of rotation, which is the product of the weight of 
the ball and the distance of its centre of gravity from the axis. 
The centrifugal moment of the ball is — • 

We X Rb X <o' X K 



This is the value usually taken for the centrifugal moment 

R 



242 Mechanics applied to Engineering. 

of the governor, but it of course neglects the arms. A common 
way of taking the arms into account is to assume that the 
centrifugal force acts at the centre of gravity of the arm; but 
it is incorrect. By this assumption we get for the centrifugal 
moment — 

wd^ R H „ M.aRa)^R= 

X X — X «)' = — ^ 

g 2X12 2 48^ tan B 

W„RV 

48^ tan Q 

Hence the ratio of the true centrifugal moment of the 
arm to the approximation commonly used is f, thus the error 
involved in the assumption is 33 per cent, of the centrifugal 
moment of the arm. In cases in which the weight of the arm 
is small compared with the weight of the ball the error is not 
serious, but in the case of some governors in which thick 
'stumpy arms are used, commonly found where the balls and 
arms are of cast or malleable iron, the error may amount to 
as much as 20 per cent., which represents about 1 1 per cent, 
error in the speed. The centre of gravity assumption is a 
convenient one, and may be adhered to without error by 
assuming that the weight of the arm is f of its real weight. 
When the arm is pivoted at a point which is not on the axis 
of rotation, the corresponding moment of inertia of the arm 
should be taken. The error involved in assuming it to be on 
the axis is quite small in nearly all cases. 

In all cases the centripetal moment in a gravity governor 
is found by taking the moment of the arm and ball about the 
point of suspension. The weight of the 
arm and ball is considered as concen- 
trated at the centre of gravity ; but in 
the case of the lower links, the bottom 
joint rises approximately twice as fast 
as the upper joint, hence its centre of 
gravity rises 1-5 times as fast as the 
top joint, and its weight must be taken 
as I 'S times its real weight; this re- 
mark also applies to the centrifugal 
moment, in that case the lower link 
is taken as twice its real weight. 

The height of this governor is H„ 
not h^ ; i.e. the height is measured from the virtual centre at 
the apex. 




Dynamics of the Steam-Engine. 



243 



H = 



n - '-a 



A governor having arms suspended in this manner is very 
much more stable and sluggish than when the arms are sus- 
pended from a central pin, and still more so than when the 
arms are crossed. 

In Fig. 250 we show the governor used on the De Laval 
steam turbine. The ball weights in this case consist of two halves 
of a hollow cylinder mounted on knife-edges to reduce the 
friction. The speed of these governors is usually calculated 
by assuming that the mass of each arm is concentrated at its 
centre of gravity, or that it is a governor having weightless 
arms carrying equivalent balls, as shown in broken lines in 
the figure. It can be readily shown by such reasoning as that 
given above that such an assumption is correct provided the 
arms are parallel with the spindle, and that the error is small 
provided that the arms only move through a small angle. 
These governors work exceedingly well, and keep the speed 
within very narrow Hmits. The figure is not drawn to scale. 




Fig. 250. 

Crank-shaft Governors. — The governing of steam- 
engines is often effected by varying the point at which the 
steam is cut off in the cylinder. Any of the forms of governor 
that we have considered can be adapted to this method, but 
the one which lends itself most readily to it is the crank-shaft 
governor, which alters the cut-off by altering the throw of the 
eccentric. We will consider one typical instance only, the 
Hartnell- McLaren governor, chosen because it contains many 
good points, and, moreover, has a great reputation for govern- 
ing within extremely fine limits (Figs. 251 and 252). 

The eccentric E is attached to a plate pivoted at A, and 
suspended by spherical-ended rods at B and C. A curved cam. 



244 Mechanics applied to Engineering. 

DD, attached to this plate, fits in a groove in the governor 
weight W in such a manner that, as the weight flies outwards 
due to centrifugal force, it causes the eccentric plate to tilt, 
and so bring the centre of the eccentric nearer to the centre of 
the shaft, or, in other words, to reduce its eccentricity, and 
consequently the travel of the valve, thus causing the steam 
to be cut off earlier in the stroke. The cam DD is so arranged 
that when the weight W is right in, the cut-off is as late as the 
slide-valve will allow it to be. Then, when the weight is right 
out, the travel of the valve is so reduced that no steam is 
admitted to the cylinder. A spring, SS, is attached to the 
weight arm to supply the necessary centripetal force. The 
speed of the engine is regulated by the tension on this spring. 
In order to alter the speed while the engine is running, the 
lower end of the spring is attached to a screwed hook, F. The 
nut G is in the form of a worm wheel ; the worm spindle is 
provided with a small milled wheel, H. If it be desired to 
alter the speed when running, a leather-covered lever is pu.shed 
into gear, so that the rim of the wheel H comes in contact with 
it at each revolution, and is thereby turned through a small 
amount, thus tightening or loosening the spring as the case may 
be. If the lever bears on the one edge of the wheel H, the 
spring is tightened and the qpeed of the engine increased, and 
if on the other edge the reverse. The spring S is attached 
to the weight arm as near its centre of gravity as possible, in 
order to eliminate friction on the pin J when the engine is 
running. 

The governor is designed to be extremely sensitive, and, 
in order to prevent hunting, a dashpot K is attached to the 
weight arm. 

In the actual governor two weights are used, coupled 
together by rods running across the wheel. The figure must 
be regarded as purely diagrammatic. 

It will be seen that this governor is practically isochronous, 
for the load on the spring increases as the radius of the weight, 
and therefore, as explained in the Hartnell governor, as the 
centrifugal force. 

The sensitiveness can be varied by altering the position of 
suspension, J. In order to be isochronous, the path of the 
weight must as nearly as possible coincide with a radial line 
drawn from O, and the direction of S must be parallel to this 
radial line. 

A later form of the same governor is shown in Fig. 252, 
an inertia weight I is attached to the eccentric. The speeding 



Dynamics of the Steam-Engine. 



24S 



up is accomplished 
by the differential 
bevil gear shown. 
The outer pulley A is 
attached to the inner 
bevil wheel A, and the 
inner pulley B to the 
adjoining wheel; by 
applying a brake to 
one pulley the bevil 
wheels turn in one 
direction, and when 
the brake is applied 
to the other pulley 
the wheels turn in the 
opposite direction, 
and so tighten or 
slacken the springs 
SS. 

Inertia Effects 
on Governors. — 
Many governors rely 





246 Mechanics applied to Engineering. 

entirely on the inertia of their weights or balls for regulating 
the supply of steam to the engine when a change of speed 
occurs, while in other cases the inertia effect on the weights is 
so small that it is often neglected ; it is, however, well when 
designing a governor to arrange the mechanism in such a 
manner that the inertia effects shall act with rather than against 
the centrifugal effects. 

In all cases of governors the weights or balls tend to fly out 
radially under the action of the centrifugal force, but in the 
case of crank-shaft governors, in which the balls rotate in one 
plane, they are subjected to another foi^ce, acting at right angles 
to the centrifugal, whenever a change of speed takes place; the 
latter force, therefore, acts tangentially, and is due to the 
tangential acceleration of the weights. For convenience of 
expression we shall term the latter the " inertia force." 

The precise effect of this inertia force on the governor 
entirely depends upon the sign of its moment about the point 
of suspension of the ball arm : if the moment of the inertia force 
be of the same sign as that of the centrifugal force about the 
pivot, the inertia effects will assist the governor in causing it 
to act more promptly ; but if the two be of opposite sign, tiiey 
will tend to neutralize one another, and will make the governor 
sluggish in its action. 

Since the inertia of a body is the resistance it offers to 
having its velocity increased, it will be evident that the inertia 
force acts in an opposite sense to that of the rotation. In the 
figures and table given below we have only stated the case in 
which the speed of rotation is increased ; when it is decreased 
the effect on the governor is the same as before, since the 
moments act together or against one another. 

In the case of a governor in which the inertia moment 
assists the centrifugal, if the speed be suddenly increased, both 
the centrifugal and the inertia moments tend to make the balls 
fly out, and thereby to partially or wholly shut off the supply 
of steam, — the resulting moment is therefore the j«/« of the two, 
and a prompt action is secured ; but if, on the other hand, the 
inertia moment aqts against the centrifugal, the resulting moment 
is the difference of the two, and a sluggish action results. If, 
as is quite possible, the inertia moment were greater than the 
centrifugal, and of opposite sign, a sudden increase of speed 
would cause the governor balls to close in and to admit more 
steam, thus producing serious disturbances. The table given 
below will serve to show the effect of the two moments on the 
governor shown in Fig. 253. 



Dynamics of the Steam- Engine. 



247 



In every 
inertia force. 



case c is the centrifrugal force, and T the 

dv W dv 

T = M— 7, or = — ■ — , where W is the weight 
dv _dt ^ g^'" 



df 



of the ball, and ^ the acceleration in feet per second per 

second. For example, let the centre of gravity of a crank- 
shaft governor arm and weight be at a radius of 4 inches when 
the governor is running at 300 revolutions per minute, and let 
it be at a radius of 7 inches when the governor is running at 
312 revolutions per minute, and let the change take place in 
o'2 second. The weight of the arm and weight is 25 pounds. 
The change of velocity is — 

2 X ^'14 

— (7 X 312 — 4 X 300) = 8-59 feet per sec^ond. 



X 60 



SV8-SQ 



and the acceleration -kt —7-^ = 42*95 feet per sec. per sec, 
and the force T = n_='-' ^ — ? = 23-4 pounds. 

22'2 



Sense of 
rotation. 


Position 
of ball. 


Centrifugal 
moment. 

A B 


Inertia moment for an 
increase of speed. 

A B 


Effect of inertia on 
governor. 


+ 

+ 
+ 


I 
2 

3 

I 

2 

' 3 


-c^3 


CxXi 











Ketards its action 
No effect 
Assists its action 

»i II 
No effect 
Retards its action 




Sensitiveness of Governors. — The sensitiveness and 
behaviour of a governor when running can be very conveniently 
studied by means of a diagram showing the rate of increase of 



248 



Mechanics applied to Engineering. 



Scale 
■ 0-2? feet 



the centrifugal and centripetal moments as the governor balls 
fly outwards. These diagrams are the invention of Mr. Wilson 
Hartnell, who first described them in a paper read before the 
Institute of Mechanical Engineers in 1882. 

In Fig. 254 we give such a diagram for a simple Watt 
governor, neglecting the weight of the arms. The axis 
OOi is the axis of rotation. The ball is shown in its two 
extreme positions. The ball is under the action of two 
moments — the centrifugal moment CH and the centripetal 
moment W„R, which are equal for all positions of the 

ball, unless the ball is 
being accelerated or 
retarded. The centri- 
fugal moment tends 
to carry the ball out- 
wards, the centripetal to 
bring it back. The four 
numbered curves show 
the relation between the 
moment tendingto make 
the balls fly out (ordi- 
nates) and the position 
of the balls. The centri- 
petal moment line shows 
the relation between the 
moment tending to 
bring the balls back and 
the position of the balls, 
which is independent of 
the speed. 
We have — 

CH = o-ooo34WRN2H 
= o-ooo34WN''(RH) 
= KRH 

The quantity O'ooo34 
WN^ is constant for any 
given ball running at any 
given speed. Values of 
KRH have been calcu- 
lated for various posi- 
tions and speeds, and 
the curves plotted, 
directly as the radius ; 




The 



Fig. 254. 

value of W.R 



Dynamics of the Steam-Engine. 249 

hence the centripetal line is straight, and passes through the 
origin O. From this we see that the governor begins to lift 
at a speed of about 82 revolutions per minute, but gets to a 
speed of about 94 before the governor lifts to its extreme 
position. Hence, if it were intended to run at a mean speed 
of 88 revolutions per minute, it would, if free from friction, 
vary about 9 per cent, on either side of the mean, and when 
retarded by friction it will vary to a greater extent. 

For the centrifugal moment, W = weight of (ball 
+ 1 arm + | link). The resultant acts at the centre of 
gravity of W. For the centripetal moment, W„ = weight of 
(ball + sleeve + arm + i link). The resultant acts at the 
centre of gravity of W„ , the weight of the sleeve being re- 
garded as concentrated at the top joint of the link. It is 
here assumed that the sleeve rises twice as fast as the top pin 
of the link. 

If the centrifugal and centripetal curves coincided, the 
governor would be isochronous. If the slope of the centrifugal 
curve be less than that of the centripetal, the governor is 
too stable ; but if, on the other hand, the slope of the 
centrifugal curve be greater than that of the centripetal, 
the governor is too sensitive, for as soon as the governor 
begins to lift, the centrifugal moment, tending to make the 
balls fly out, increases more rapidly than the centripetal 
moment, tending to keep the balls in — consequently the balls 
are accelerated, and fly out to their extreme position, com- 
pletely closing the governing valve, which immediately causes 
the engine to slow down. But as soon as this occurs, the 
balls close right in and fully open the governing valve, thus 
causing the engine to race and the balls to fly out again, and 
so on. This alternate racing and slowing down is known as 
hunting, and is the most common defect of governors intended 
to be sensitive. 

It will be seen that this action cannot possibly occur 
with a simple Watt governor unless there is some disturbing 
action. 

When designing a governor which is intended to regulate 
the speed within narrow limits, it is important to so arrange 
it that any given change in the speed of the engine shall be 
constant for any given change in the height throughout its 
range. Thus if a rise of 1 inch in the sleeve corresponds to a 
difference of 5 revolutions per minute in the speed, then each 
|th of an inch rise should produce a difference of i revolution 
per minute of the engine in whatever position the governor 



250 



Mechanics applied to Engineering. 



may be. This condition can be much more readily 
realized in automatic expansion governors than in throttling 

governors. 
Friction of Governors. 
— So far, we have neglected 
the effect of friction . on the 
sensitiveness, but it is in reality 
one of the most important 
factors to be considered in 
connection with sensitive 
governors. Many a governor 
is practically perfect on paper 
— friction neglected — but is to 
all intents and purposes -useless 
in the material form on an 
engine, on account of retarda- 
tion due to friction. The 
friction is not merely due to 
the pins, etc., of the governor 
itself, but to the moving of 
the governing valve or its 
equivalent and its connections. 
In Fig. 255 we show how 
friction affects the sensitiveness 
of a governor. The vertical 
height of the shaded portion represents the friction moment 
that the governor has to overcome. Instead of the governor 
lifting at 8q revolutions per minute, the speed at which it 
should lift if there were no friction, it does not lift till the 
speed gets to about 92 revolutions per minute ; likewise on fall- 
ing, the speed falls to 64 revolutions per minute. Thus with 
friction the speed varies about 22 per cent, above and below 
the mean. Unfortunately, very little experimental data exists 
on the friction of governors and their attachments ; ' but a 
designer cannot err by doing his utmost to reduce it even to 
the extent of fitting all joints, etc., with ball-bearings or with 
knife-edges (see Fig. 250). 

The effect of friction is to increase the height of the 
governor when it is rising, and to reduce it when falling. The 
exact difference in height can be calculated if the frictional 

' See Paper by Ransome, Proc. Inst. C.£., vol. cxiii. j the question 
was investigated some years ago by one of the author's students, Mr. 
Eurich, wJio found that when oiled the Watt governor tested lagged 
behind to the extent of 7'S per cent., and when unoiled I7"5 per cent. 




Fig. 255. 



Dynamics of the Steam-Engine. 



251 



resistance referred to the sleeve is known ; it is equivalent 
to increasing the weight on the sleeve when rising and re- 
ducing it when falling. 

In the well-known Pickering governor, the friction of the 
governoritself is reduced 

to a minimum by mount- 

ing the ballson a number 
of thin band springs in- 
stead of arms moving on 
pins. The attachment 
of the spring at the c. 
of g. of the weight and 
arm, as in the McLaren 
governor, is a point also ^ 
worthy of attention. We 
will now examine in de- 
tail several types of go- 
vernor by the method 
just described. 

Porter Governor 
Diagram. — In this case 
the centripetal force is 
greatly increased while 
the centrifugal is un- 
affected by the central 
weight W^, which rises 
twice as fast as the balls 
(Fig. 256) when the 
links are of equal 
length. Resolve W„ in 
the directions of the 
two arms as shown : it 
is evident that the com- 
ponent ab, acting along 
the upper arm, has no 
moment about O, but 
l)d = Wo has a centri- 
petal moment, WoR,, ; then we have — 

CH = WR -f W„R„ 

Values of each have been calculated and plotted as in 
Fig. 254. In the central spring governor W„ varies as the balls 
liltj in other respects the construction is the same. 

It should be noticed that the centripetal and centrifugal 




Fig. 256. 



252 Mechanics applied to Engineering. 

moment curves coincide much more closely as the height of 
the governor increases j thus the sensitiveness increases with 
the height — a conclusion we have already come to by another 
process of reasoning. 

Crossed-arm Governor Diagram. — In this governor H 
is constant, and as C varies directly as, the radius for any given 
speed, it is evident that the centripetal and centrifugal lines 
are both straight and comcident, hence the governor is 
isochronous. 

Wilson Hartnell Governor Diagram (Fig. 257). — 
When constructing the curves a, b, c, d, e, the moment of the 
weight of the ball on either side of the suspension pin, also 
the other disturbing causes, have been neglected. 

We have shown that cr„ = pr, also that c and p vary as R, 
hence the centrifugal moment lines (shown in full) and the 
centripetal moment line Oa both pass through the origin, 
under these conditions the governor is isochronous. A com- 
mon method of varying the speed of such governors is to alter 
the load on the spring by the lock nuts at the top ; this has 
the effect of bodily shifting the centripetal moment line up or 
down, but it does not alter the slope, such as db, ec, both of 
which are parallel to Oa. But such an alteration also affects 
the sensitiveness ; if the centripetal line was db, the governor 
would hunt, and if ec, it would be too stable. These defects 
can, however, be remedied by altering the stiffness of the 
spring, by throwing more or less coils out of action by the 
corkscrew nut shown in section, by means of which db can be 
altered to dc and ec to eb. For fine governing both of these 
adjustments are necessary. 

When the moment of the weight of the ball and other 
disturbing causes are taken into account, the curves / and g 
are obtained. 

Instead of altering the spring for adjusting the speed, some 
makers leave a hollow space in the balls for the insertion of 
lead until the exact weight and speed are obtained. It is 
usually accomplished by making the hollow spaces on the 
inside edge of the ball, then the centrifugal force tends to 
keep the lead in position. 

The sensitiveness of the Wilson Hartnell governor may 
also be varied at will by a simple method devised by the author 
some years ago, which has been successfully applied to several 
forms of governor. In general, if a governor tends to hunt, 
it can be corrected by making the centripetal moment increase 
more rapidly, or, if it be too sluggish, by making it increase less 



Dynamics of the Steam- Engine. 
Oi 



253 




Fro. 257. 



254 Mechanics applied to Engineering. 

rapidly as the centrifugal moment of the balls increases. The 
governor, which is shown in Fig. 258, is of the four-ball 
horizontal type ; it originally hunted very badly, and in order 
to correct it the conical washer A was fitted to the ball path, 
which was previously flat. It will be seen that as the balls fly 
out the inclined ball path causes the spring to be compressed 
more rapidly than if the path were flat, and consequently the 
rate of increase of the centripetal moment is increased, and 
with it the stability of the governor. 

Id constructing the diagram it was found convenient to 
make use of the virtual centre of the ball arm in each position ; 
after finding it, the method of procedure is similar to that 
already given for other cases. In order to show the efiect of 
the conical washer, a second centripetal curve is shown by_ a 
broken line for a flat plate. With the conical washer, neglect- 
ing friction, the diagram shows that the governor lifts at 430 
revolutions, and reaches 490 revolutions at its extreme range ; 
by experiment it was found that it began to lift at 440, and 
rose to 500, when the balls were lifting, and it began to fall at 
480, getting down to 415 before the balls finally closed in. 
The conical washer A in t"his case is rather too steep for 
accurate governing. 

The centrifugal moment at any instant is — 

4 X o-ooo34WRN=H 

where W is the weight of one ball. 
And the centripetal moment is — 

Load on spring x R. 

See Fig. 258 for the meaning of R„ viz. the distance of 
the virtual centre from the point of suspension of the arm. 

Taking position 4, we have for the centrifugal moment at 
450 revolutions per minute — 

4 X 0-00034 X 2'S X n^ X 450^* X I'sS = 260 pound-inches 

and for the centripetal moment — 

The load on the spring =111 lbs. ; and R, = 278 

centripetal moment = in x 278 = 310 pound-inches 

McLaren's Crank-shaft Governor. — In this governor 
we have CR, = SR. ; but C varies as R, hence if there be no 
tension on the spring when R is zero, it will be evident that S 
will vary directly as R ; but C also varies in the same manner, 
hence the centrifugal and centripetal moment lines are nearly 



256 



Mechanics applied to Engineering. 



straight and coincident, The centrifugal lines are not abso- 
lutely straight, because the weight does not move exactly on a 
radial line from the centre of the crank-shaft. 




Fig. 359. 



Governor Dashpots. — ^A dashpot consists essentially of 
a cylinder with a leaky piston, around which oil, air, or other 
fluid has to leak. An extremely small force will move the piston 
slowly, but very great resistance is offered by the fluid if a 
rapid movement be attempted. 

Very sensitive governors are therefore generally fitted with 
dashpots, to prevent them from suddenly flying in or out, and 
thus causing the engine to hunt. 

If a governor be required to work over a very wide range 
of power, such as all the load suddenly thrown off, a sensi- 
tive, almost isochronous governor with dashpot gives the best 
result ; but if very fine governing be required over small 
variations of load, a slightly less sensitive governor without a 
dashpot will be the best. 

However good a governor may be, it cannot possibly 
govern well unless the engine be provided with sufficient fly- 
wheel power. If an engine have, say, a 2-per-cent cyclical 



Dynamics of the Steam- Engine. 257 

variation and a very sensitive governor, the balls will be 
constantly fluctuating in and out during every stroke. 

Power of Governors. — The "power" of a governor is its 
capacity for overcoming external resistances. The greater the 
poweiTj the greater the external resistance it will overcome with 
a given alteration in speed. 

Nearly all governor failures are due to their lack of power. 

The useful energy stored in a governor is readily found 
thus, approximately : — 

Simple Watt governor, crossed-arm and others of a 
similar type — 

Energy = weight of both balls X vertical rise of balls 

Porter and other loaded governors — 

Energy = weight of both balls X vertical rise of balls + weight 
of central weight X its vertical rise 

Spring governors — 

Energy = weight of both balls X vertical rise (if any) of balls 
. ^ max. load on spring + min. load on sp ring \ 

X the stretch or compression of spring 
where n = the number of springs employed j express weights 
in poimdSj and distances in feet. 

The following may be taken as a rough guide as to the 
energy that should be stored in a governor to get good results : 
it is always better to store too much rather than too little 
energy jn a governor : — 

Foot-pounds of energy 
Type of governor. stored per inch diametei 

of cylinder. 
For trip gears and where small resistances have to 

be overcome ... ... ... .-• ... o'5-o*7S 

For fairly well balanced throttle-valves 075-1 

In the earlier editions of this book values were given for 
automatic expansion gears, which were bfsed on the only data 
available to the author at the time; but since collecting a 
considerable amount of information, he fears that no definite 
values can be given in this form. For example, in the case of 
governors acting through reversible mechanisms on well- 
balanced slide-valves, about 100 foot-pounds of energy per inch 
diameter of the high-pressure cylinder is found to give good 
results ; but in other cases, with unbalanced slide-valves, five 

s 



258 Mechanics applied to Engineering. 

times that amount of energy stored is found to be insufficient. 
If tiie driving mechanism of the governor be non-reversible, 
only about one-half of this amount of energy will be required. 

A better method of dealing with this question is to calculate, 
by such diagrams as those given in the " Mechanisms " chapter, 
the actual effort that the governor is capable of exerting on the 
valve rod, and ensuring that this effort shall be greatly in excess 
of that required to drive the slide-valve. Experiments show 
that the latter amounts to about one-fifth to one-sixth of the 
total pressure on the back of a slide-valve (j.e. the whole area 
of tha back x the steam pressure) in the case of unbalanced 
valves. The effort a governor is capable of exerting can also 
be arrived at approximately by finding the energy stored in the 
springs, and dividing it by the distance the slide-valve moves 
while the springs move through their extreme range. 

Generally speaking, it is better to so design the governor 
that the valve-gear cannot react upon it, then no amount of 
pressure on the valve-gear will after the height of the governor ; 
that is to say, the reversed efficiency of the mechanism which 
alters the cut-off must be negative, or the efficiency of the 
mechanism must be less than 50 per cent. On referring to the 
McLaren governor, it will be seen that no amount of pressure on 
the eccentric will cause the main weight W to move in or out. 

Readers who wish to go more fully into the question of 
governors will find detailed information in " Governors and 
Governing Mechanism," by H. R. Hall, The Technical 
Publishing Co., Manchester ; " Dynamics of Machinery," by 
Lauza, Chapman and Hall ; " Shaft Governors," Trinks and 
Housum, Van Nostrand & Co., New York. 



CHAPTER VII. 

VIBRA TION. 

Simple Harmonic Motion (S.H.M.).— When a crank 
rotates at a uniform velocity, a slotted cross-head,, such as is 
shown in Fig. i6o, moves to and fro with simple harmonic 
motion. In Chapter, VI. it is shown that the force required 
to accelerate the cross-head is 

T, WV^ * /■^ 

^^ = 7r^r •. ■ ■ • • «• 

where W = weight of the cross-head in pounds. 

V = velocity of the crank-pin in feet per sec. 
R = radius of the crank in feet. 

X = displacement of cross-head in feet from the 

central position. 
g = acceleration of gravity. 
K = radius of gyration in feet. 

Then the acceleration of the crosshead in this position 

'\'^x _ Y! V i ''* displacement from the middle 
■ "rs" ~ r2 ^ 1 of the stroke, 

V _ /acceleration _ /^jS_ ,..> 
°' R - V displacement " V W^ • • • . (n^ 

These expressions show that the acceleration of the cross-head 
is proportional to the displacement x, and since the force 
tending to make it slide is zero at the middle of the stroke, 
and varies directly as x, it is clear that the direction of the 
acceleration is always towards the centre O. 

The time t taken by the cross-head in making one complete 
journey to and fro is the same as that taken by the crank in 
making one complete revolution. 



W " 



26o 



Mechanics applied to Engineering 



Hence t ■■ 



27rR 



= 2TrsJ' 



displacement 
acceleration 









(iii) 



On referring to the argument leading up to expression (i) 
in Chapter VI. it will be seen that Pj is the force acting along 
the centre line of the cross-head when it is displaced an 
amount x from its middle or zero-force position. When 
dealing with the vibration of elastic bodies W is the weight of 
the vibrating body in pounds, and Pj is the force in pounds 
weight required to strain the body through a distance x feet. 

When dealing with angular oscillations we substitute thus — 



Angular. 


Linear. 


The moment of inertia of the body 
I = in pound-fdot units. 


Mass of the body ( — ). where W is 
in pounds. 


The couple acting on the body 
C = lA in pound-foot units. 


W 
Force acting on the body P, = —/ 

in pounds. 


The angular displacement (9) in 
radians. 


Linear displacement {x) in feet. 


The angular velocity (oj) in radians 
per second. 


The linear velocity (v) in feet per 
second. 


The angular acceleration (A) in 
radians per second per second. . 


The linear acceleration (/) in feet 
per second per second. 


Kinetic energy J Ia>' 


Kinetic energy - ~v' 



Hence, when a body is making angular oscillations under the 
influence of a couple which varies as the angular displacement, 
the time of a complete oscillation is — 



"•V 



le 
c 



(iv) 



These expressions enable a large number of vibration 
problems to be readily solved. 



Vibration. 



261 



Simple Pendulum. — In the case of a simple pendulum 
the weight of the suspension wire is regarded as negligible 
when compared with the weight of the bob, and 
the displacement x is small compared to /. The 
tension in the wire is normal to the path, hence 
the only accelerating force is the component of 
W, the weight of the bob, tangential to the path, 



hence Pj = W sin 6 = — r- 



/Wxl / 



g 



(v) 



where / is the length of the pendulum in feet rm. ^go. 

measured from the point of suspension to the 

c. of g. of the bob; t is the time in seconds taken by the 

pendulum in making one complete oscillation through a small 

arc. 

The above expression may be obtained direct from (iv) by 

substituting — for I, and "Wx or W/5 for C. 



g 



Then 






VfBg 

Compound Pendulum. — When the weight of the arm is 
not a negligible quantity, the pendulum is termed compound. 

Let W be the weight of the bob and 
arm, and /, b^ the distance of their 
c. of g. from the point of suspension. 
Then if ;«; be a small horizontal dis- 
placement of the c. of g. 



C = W^c = W49 (nearly), and 



W/' 



(vi) 



CofG 



4? 




Fig. 261. 



where Ko is the radius of gyration of the body about the point 
of suspension. If K be the radius of gyration of the body 
about an axis through the centre of gravity, then (see p. 76) 



262 



and 



Mechanics applied to Engineering. 

J ^ VVK vv^ ^ w , 
S g g 



If / be the length of a simple pendulum which has the same 
period of oscillation, then 






g 
1 = 



K= + // 



and/„(/-4) = K^ 



or OG.GOi = K°, and since the position, of G will be the same 
whether the point of suspension be O or Oj the time of oscil- 
lation will be the same for each. 

Oscillation of Springs. — Let a weight W be suspended 
from a helical spring as shown, and let it stretch an amount 
8 (inches) when supporting the load W, where 

8 = -^^ (see page 587). 

D is the mean diameter of the coils in inches 

d „ diameter of the wire in inches 

n ,, number of free coils 

G „ modulus of rigidity in pounds per 

square inch 
W „ weight in pounds, 

Then, neglecting the weight of the spring itself, we have, for 
the time of one complete oscillation of the spring (from iii) 

/W^ / W8 / D'Ww 

t = 2-rsJ s— = 2'r\/ rr— = 2-K\J -— 

^ ^ig ^ i2W^ V v^gGd" 

t = o-904>/ -^^ (vm) 

t = 2TT\/ ■- , where A is the deflection in feet. For a 

simple pendulum t = 2ir\J - , hence the time of one complete 
oscillation of a spring is the same as that of a simple pendulum 




' Vibration. 263 

whose length is equal to the static deflection of the spring due 
to the weight W. 

The weight of the spring itself does not usually affect the 
problem to any material extent, but it can be taken into 
account thus : the coil at the free end of the spring oscillates 
at the same velocity as the weight, but the remainder of the 
coils move at a velocity proportional to their distance from the 
free end. 

Let the weight per cubic inch of the spring = w, the area 
of the section = a. Consider a short length of spring dl, distant 
/from the fixed end of the spring, L being the total length of 
the wire in the spring. 



Weight of short length = wa . dl 
V/ 
L 



V/ 
Velocity of element = -j- 



w , ci , V^ 
Kinetic energy of element = ' Ml 



2g\I 



waVV'=^„ „ waV\} 



Kmetic energy of sprmg = ^ I I dl = 



3 X 2g\j 



3 X 2^ ~ 3\ 2g ) 

where W, is the weight of the spring. Thus the kinetic energy 
stored in the spring at any instant is equal to that stored in an 
oscillating body of one-third the weight of the spring. Thus 
the W in the expression for the time of vibration should 
include one-third the weight of the spring in addition to that 
of the weight itself. 

Expression (viii.) now becomes — 

Example. — D = 2 inches d = 0-2 inch 

W = 80 pounds « = 8 
G = 12,000,000 pounds per sq. inch. 
Find the lime of one complete oscillation, (a) neglecting 
the weight of the spring, (d) taking it into consideration. 



264 Mechanics applied to Engineering. 



/ 2= X 80 X 8 / — 7- 

/=o'Q04x/ - = o'Q04v 0-267 

V 12,000,000 X o'ooib 

= 0-4668 sec, say 0*47 second. 

In some instances the deflection x is given in terms of the 
force Pi ; if the deflection S be woiked out by the expression 
given on page 587, it will be found to be 1-33" or o'li foot for 
every 50 lbs. Then- 



/. 

'■ = 2W/y/ 



80 X o-ii 

= 0-47 sec. 



50 X 32-2 

The weight of the spring = o-69«^D 

= 0-69 X 8 X 0-04 X 2 
Wj = 0-44 lb. 

Then allowing for the weight of the spring — 



/ 8 X 80-15 X 8 ^ , 

t = o-oo4x/ 2 = o"4074 second 

'V 12,000,000x0-0010 

say 0-47 sefcond. 

Thus it is only when extreme accuracy is required that the 
weight of the spring need be taken into account. 

In the case of a weight W pounds on the end of a cantilever 
L inches long, the time of one complete oscillation is found 
by inserting the value for 8 (see p. 587) in equation (iii), which 
gives — 

The value of 8 can be inserted in the general equation for any 

cases that may arise. 

Oscillation of 
Spring-controlled 
Governor Arms. — In 
the design of governors it 
is often of importance to 
know the period of oscil- 
lation of the governor 
arms. This is a case of 
angular oscillation, the 

time of which has already been given in equation (iv), viz. 




■= 27r^ 



10 
C 



Vibration. 265 

The I is the moment of inertia of the weight and arm about 
the pivot in foot and pound units, and is obtained thus (see 
page 76) 

g S^ 12 4) 

where K is the radius of gyration of the weight about an axis 
parallel to the pivot and passing through the centre of gravity. 

For a cylindrical weight K'' = — , and for a spherical weight 

o 

K^ = — . The weight of the arm, which is assumed to be of 
10 

uniform section is w. More complex forms of arms and weights 

must be dealt with by the graphic methods given on page loi. 

Example. — The cylindrical weight W is 48 pounds, and is 

5 inches diameter. The weight of the arm =12 pounds, 

Zj = 18 inches, 4=12 inches, /j = 8 inches, h = 2 inches. 

The spring stretches o'3 inch per 100 pounds. If the 

dimensions be kept in inch units, the value for the moment of 

inertia must be divided by 144 to bring it to foot units. 

T = 48 /25 \ 12 y/3 24 + \ \ 3^1 

32-2x144X8 ^V"^32-2 X i44^V 12 / 43 

= 1-8 pounds-feet units. 

0-3 100 X 8 
e = —^ when C = 



hence t= 2 X 2,'^^\/ a J ,Z. ^ a ~ °'^ ^^"^• 



X o"3 X 12 
8 X 100 X 8 



Vertical Tension Rod. — For a rod of length / supported 
at top with a weight W at the lovyer end^ Inserting the values 

// - / w/ 

X =-<g and Pi = A/ we get t = ztt^ g^ 

Torsional Oscillations of Shafts. — If an elastic shaft 
be held at one end, and a body be attached at the other as 
shown, it will oscillate with simple harmonic motion if it be 
turned through a small arc, and then released. The time of 
one complete oscillation has already been shown (iv) to be 



/Te 
= ^W c 



266 Mechanics applied to Engineering. 

on page 579 it is shown that 

= — ^ 
GI„ 




6^ 



Fig 264. 



radians, where M, is the twisting moment 
in pounds inches, and I^ is the polar 
moment of inertia of the shaft section 
in inch' units. G is the modulus of 
rigidity in pounds per square inch, / is 
the length in inches. 

Substituting this value, we get — 






I2lM^ 

M.GI„ 



/12I/ 

2,r^ GT 



Note. — C is in pounds-feet, and M, in pounds-inches 
units. Writing L for the length in feet, we get — 



/i44lL / J,L 



where Ij is the moment of inertia of the oscillating body in 
pound and • inch units. If all the quantities be expressed in 
inch units, and the constants be reduced, we get 



-Wc^. 



If greater accuracy is required, one-third of the polar 
moment of inertia of the rod should be added to I^, but in 
general it is quite a negligible quantity. 

If there are two rotating bodies or wheels on a shaft, there 
will be a node somewhere between them, and the time of a 
complete oscillation will be 



and 




It not infrequently happens in practice that engine crank 









^3 


— 










1. 


, 1 




I, 


M 


'2 


H 



Vibration. 267 

shafts fracture although the torsional stresses calculated from 
static conditions are well within safe limits. On looking more 
closely into Ihe matter, it may often be found that the frequency 
of the crank effort fluctuations agrees very 
nearly with the frequency of torsional 
vibration of the shaft, and when this' is 
the case the strain energy of the system 
may be so great as to cause fracture. 
An investigation of this character shows 
why some engine crank shafts are much 
more liable to fracture when running 
at certain speeds if fitted with two fly- 
wheels, one at each end of the shaft, than 
if fitted with one wheel of approximately 
twice the weight and moment of inertia. f,^. ^cs. *^ 

In both cases the effective length / is 

roughly one half the distance between the two wheels, or it 
may be the distance from the centre of the crank pin to the 
flywheel ; but the moment of inertia I in the one case is only 
one half as great as in the other, and consequently the natural 
period of torsional vibration with the two flywheels is only 

—7= = 07 of the period with the one flywheel. If the lower 

speed ha'pperis to synchronise, or nearly so, with the crank 
effort fluctuations, fracture is liable to occur, which would not 
happen with the higher period of torsional vibration. 

Vibration and Whirling of Shafts. — If the experi- 
ment be made of gradually increasing the speed of rotation 
of a long, thin horizontal shaft, freely supported at each end, 
it will run true up to a certain speed, apart from a little 
wobbling at first due to the shaft being not quite true, and 
will then quite suddenly start to vibrate violently, and will 
whirl into a single bow with a node at each bearing. As the 
speed increases the shaft straightens and becomes quite rigid 
till a much higher speed is reached, when it suddenly whirls 
into a double bow with a node in the middle. It afterwards 
straightens and whirls into three bows, and so on. High speed 
shafts in practice are liable to behave in this way, and may 
cause serious disasters, hence it is of great importance to avoid 
running at or near the speed at which whirling is likely to 
occur. An exact solution of many of the cases' which occur 
in practice is a very complex matter (see a paper by Dunker- 
ley, FMl. Trans., Vol. 185), but the following approximate 



268 Mechanics applied to Engineering. 

treatment for certain simple cases gives results sufficiently 
accurate for practical purposes. 

The energy stored at any instant in a vibrating body 
consists partly of kinetic energy and partly of potential or strain 
energy, the total energy at any instant being constant, but the 
relation between the two kinds of energy changes at every 
instant. In the case of a simple pendulum the whole energy 
stored in the bob is potential when it is at its extreme position, 
but it is wholly kinetic in its central position, and in inter- 
mediate positions it is partly potential and partly kinetic. 

If a periodic force of the same frequency act on the 
pendulum bob it will increase the energy at each application, 
and unless there are other disturbing causes the energy will 
continue to increase until some disaster occurs. In an elastic 
vibrating structure the energy increases imtil the elastic limit 
is passed, or possibly even until rupture takes place. When a 
horizontal shaft rests on its hearings the weight of the shaft 
sets up bending stresses, consequently some of the fibres are 
subjected to tension and some to compression, depending upon 
whether they are above or below the neutral axis, hence when 
the shaft rotates each fibre is alternately brought into tension 
and compression at every revolution. The same distribution 
and intensity of stress can be produced in a stationary shaft 
by causing it to vibrate laterally. Hence if a shaft revolve at 
such a speed that the period of rotation of the shaft agrees 
with the natural period of vibration, the energy will be increased 
at each revolution, and if this particular speed be persistent 
the bending of the shaft will continue to increase until the 
elastic limit is reached, or the shaft collapses. The speed at 
which this occurs is known as the whirling speed of the shaft. 
The problem of calculating this speed thus resolves itself into 
finding the natural period of lateral vibration of the shaft, if 
this be expressed in vibrations per minute, the same number 
gives the revolutions per minute at which whirling occurs. 

The fundamental equation is the one already arrived 

For the present purpose it is more convenient to use revolutions 
per minute N rather than the time in seconds of one vibration, 
also to express the deflection of the shaft 8 in inches. 

t 27rV vva 



at. viz.- /^ 



Vibration. 269 

In this case the displacing force Pi is the load W which 
tends to bend the shaft, and x is the deflection. Reducing the 
constants we get — 

In the expression for the deflection of a beam it is usual to 

take the length (/) in inches, but in Professor Dunkerley's 

classical papers on the whirling of shafts he takes the length L 

in feet, and for the sake of comparing our approximations with 

his exact values, the length in the following expressions is 

taken in feet. The diameter of the shaft {d) is in inches. The 

weight per cubic inch = o'zS lb. E is taken at 30,000,000 lbs. 

. ird* 
per square inch. The value of I is -7— . For the values of S 

see Chapter XIII. 

Parallel Shaft Supported freely at each end. — 



- _ ^wl^ _ 270WL* _ 



384EI EI 25000^' 



O-ziiird^ , 

where w = pounds 

4 



i8y 29S90i/ y First critical speed 



■^ "~ ' — ^- iJ \ ■ — ^single bow 

25000;^'' 



munkerley gets ^^ / ) 

At the second critical speed the virtual length is — , hence 

N,= ^^=il^ double bow 
. (9 

,„d N3=^ = ?%^^reblebow. 

The following experimental results by the author will show 
to what extent the calculations may be trusted. Steel shaft — 
short bearings, d = 0-372 inch, L = 5-44 feet. 



270 



Mechanics applied to Engineering. 



Condition. 


Whirling speed R. p.m. 


Calculated. 


Experiment. 


Single bow .... 


370 


350 to 400 


Double bow 


1480 


1460 to 1550 


Treble bow. . . . 


3340 


3200 to 3500 



It is not always easy to determine the exact speed at which 
whirHng occurs, the experimental results show to what degree 
of accuracy the speed can be determined with an ordinary 
speed indicator. 

Parallel Shaft, supported freely at each end, 
central load W. — 

W/^ _ 36WL° 
48EI ~ EI 

""^-^^ (-»^-«"-^) 

Experimental results — </= o'-gg^ inch, L = ys-i feet 





Whirling speed R. p.m. 




Calculated. 


Experiment. 


Unloaded .... 


2940 


3100 


W = 33 pounds . . 


1080 


1 100 



The weight of the shaft itself may be taken into account 
thus — the deflection coefficient for the weight of the shaft is 
jfj = say ^, and for the added load, ^, hence 4f of the weight 
of the shaft should be added to W, in general the weight of 
the shaft is a negligible quantity. 

Tapered Shaft, supported freely at each end, cen- 
tral load. — If the taper varies in such a manner that — 

is constant except close to the ends where it must be enlarged 
on account of the shear, but such a departure from the 



Vibration. 27 1 

assumed conditions does not appreciably aifect the deflection, 

■py 

then since p = —- = constant, the beam bends to an arc of a 
circle, and — 

s _ ^ = 54WL° 

32EI~ EI 
^^ 31050^ 

Note. — The d is the diameter at the largest section 
If tlie taper varies £ 
stant, and we have — 



If tlie taper varies so that the stress is constant, then — is con- 



W/s _6sWL' 



N = 



26-5EI EI 
281001^ 

Vwi? 



Shafts usually conform roughly to one of the above con- 
ditions but it is improbable that the taper will be such as to 
give a lower whirling speed than the latter value. For calcu- 
lating a shaft of any profile the following method is convenient. 

In Chapter XIII., we show that 
I 



I 



S = :^ (2 m .X .dxioi z. parallel shaft. 
In the case of a tapered shaft this becomes 

S JmxBx 






_ Moment of the area of half the be nding moment diagram 

~ Moment of the area of half the b.ni. diagram after altering 

its depth at each section in the ratio of the moment of 

inertia in the middle to the moment of inertia at that 

section. 

W/= area e/gA x h ,„. .. \ 

^' = i8EI^ • 1 <^'S.^66.) ^ 

area ehg X - 
3 



2/2 



Mechanics applied to Engineering. 



where—, = =^ = ^r a similar correction being made at every 
ctb 1-^ a" 

section. 

The I in the above expressions is for the largest section, 

. ttD 

VIZ. —. — . 

64 




Fig. 266. 



For some purposes the following method is more convenient. 
Let the bending moment diagram egh be divided into a num- 
ber of vertical strips of equal width, it will be convenient to 

take 10 strips, each — wide. Start numbering from the 

support (see page 508). 



Mean bending moment at the 
middle of the 


The distance of each 
strip from the support. 


W / 

1st Strip = — X — 

"^ 2 40 


l_ 
40 


W 3/ 

2nd „ = — X ^- 

2 40 


3/ 
40 


.3-"=?x| 


Si 
40 


and so on. 


and so on. 



Vibration. 



273 



The respective moments of inertia are Ij, I2, I3, etc., that at 
the middle of the shaft being I. 
Then 



EIVSoIi 20 40 80I2 20 40 

+ V-r X — X 3- + etc. I 
80I3 20 40 / 

S = W/^ ^i^9_)_^5^49 ,81 ^121 ^169 ^ 225 

64oooE\Ii I2 I3 I4 Ib le ~ J7 Is 

289 , 36i \ 



8==J^a + J-^ + ii + l? + etc. + f,^) 
3i42E\a'f 4* (/s* fl'i (/lo / 

„ WL' / I , , 36i\ 



+ I + I 



Experimental Results. — d = 0-995 inch, parallel in 
middle, tapering to 0*5 inch diameter at ends. L = 3"i7 
feet. 





Whirling speed R. p.m. 




Calculated. 


Experiment. 


Parallel for 4 ins. in 
middle, unloaded 


2606 


2750 


Ditto for I in. . . . 


2320 


2425 to 2500 


Ditto for 4 ins. 

Loaded, W = 33 lbs. 


900 


900 to 910 


Ditto for 1 in. 
Loaded, W = 33 lbs. 


800 


750 to 770 



When the middle of a shaft is rigidly gripped by a 
wheel boss, or its equivalent, of length 4, the virtual length 
of shaft for deilection and whirling, purposes is/— 4 instead 
of/. 

If the framing, on which the- shaft- bearings are- mounted is^ 

T 



274 



Mechanics applied to Engineering. 



not stifF, the natural period of vibration of the frame must 
be considered. An insufficiently rigid frame may cause the 
shaft to whirl at speeds far below those calculated. 

Vertical shafts, if perfectly balanced, do not whirl, but if 
thrown even a small amount out of balance they will whirl at 
the same speeds as horizontal shafts of similar dimensions. 

The following table gives a brief summary of the results 
obtained by Dunkerley. See "The Whirling and Vibration 
of Shafts" read before the Liverpool Engineering Society, 
Session 1894-5, also by the same author in " The Trans- 
actions of the Royal Society," Vol. 185A, p. 299. 

N = the number of revolutions per minute of the shaft 

when whirling takes place 
d = the diameter of the shaft in inches 
L = the length of the shaft in feet 



Description of shaft. 


Remarks on /. 


N. 


L. 


U nloaded : overhanging from 
a long bearing which fixes 
its direction 


Length of over- 
hang in feet 


n645j;. 


-v^ 


Unloaded : resting in short 
or swivelled bearings at 
each end 


Distance between 
centres of bear- 
ings in feet 


3^364^, 


■VI * 


Unloaded : supported as in 
last case, but one end 
overhanging c feet 


Ditto 
c 


d 
For values 


ofa see Table I. 


Unloaded : supported in a 
long bearing at one end, 
and a short or swivelled 
bearing at the other 


Distance between 
inner edge of 
long bearing 
and centre of 
short bearing 


5'34°i 


"Vl 


Unloaded : shaft supported 
in three short or swivelled 
bearings, one at each end 

-I 


L, = shorter span 

Lj = longer span 

in feet 


d 

For 
see 


values of <j 
Table II. 



Vibration. 



275 



Description of shaft. 


Remarks on /. 


N. 


L. 


Unloaded : shaft supported 
in long bearings which fix 
its direction at each end 


Clear span be- 
tween inner 
edges of bearings 


d 
7497 'l. 


VI 


Unloaded : long continuous 
shaft supported on short 
or swivelled bearings equi- 
distant 


Distance between 
centres of bear- 
ings in feet 


3286^ 


-Vi 


Loaded : shaft supported in 
short or swivelled bear- 
ings, single pulley of 
weight fV pounds cen- 
trally placed between 
the bearings 


Distance between 
the centres of 
the bearings in 
feet 


N - 37,35o^^j i 


Loaded ; long bearings 
which fix its direction at 
each end 


Clear span be- 
tween inner 
edgesof bearings 


^ ^'''^"VwL' 



276 



Mechanics applied to Engineering 
TABLE I. 



Value of »■ = ?_. 


... 


va« 


I -00 


7.554 


87 


075 


12,044 


no 


0-50 


20,931 


HS 


0-33 


28.095 


168 


0-25 to O'lO 


31,000 


176 


Very small 


31.590 


178 


Zero 


32,864 


iSi 



TABLE IL 



Value of r = i 
h 


0> 


V5. 


i-oo too7S 


32,864 


18: 


0-5 to 07s 


36.884 


192 


o'33 


41,026 


203 


0-25 


43,289 


208 


0'20 to 0-14 


44.312 


211 


0"I25 to O'lO 


47.I2S 


217 


Very small 


50.654 


225 



For hollow shafts in which — 

<^, = the external diameter in inches 
d^ = „ internal „ „ 

substitute for d in the expressions given above the value — 

fjd^ + di for unloaded shafts, 

and substitute for d^ the value — 

Vi^i* - d^ for loaded shafts. 



CHAPTER VIII. 



GYROSCOPIC ACTIO X. 



When a wheel or other body is rotating at a high speed a 
considerable resistance is experienced if an attempt be made 
to rapidly change the direction of its 
plane of rotation. This statement can 
be verified by a simple experiment on 
the front wheel of a bicycle. Take 
the front wheel and axle out of the 
fork, suspend the axle from one end 
only by means of a piece of string. 
Hold the axle horizontal and spin the 
wheel rapidly in a vertical plane, on 
releasing the free end of the axle it 
will be found that the wheel retains 
its vertical position so long as it con- 
tinues to spin at a high speed. The 
external couple required to keep the 
wheel and axle in this position is 
known as the gyroscopic couple. 

The gyroscopic top is also a 
famiUar and striking instance of gyro- 
scopic action, and moreover affords an 
excellent illustration for demonstrating 
the principle involved, which is simply 
the rotational analogue of Newton's 
second law of motion, and may be stated thus — 

When a < til ^^^^ ^"^ ^ body, or self-contained system 
the change of J^^g^^j^^ momentum^ 
is proportional to the ii^pressed j , |. 

Hence, if an external couple act for a given time on the 
wheel of a gyroscope the angular momentum about the axis of 
the couple will be increased in proportion to the couple. 




Fig. 267. 

From Worthington's "Dyna- 
mics of Rotation." 



momentum ) generated in a given time 



278 



Mechanics applied to Engineering. 



If a second external couple be impressed on the system 
for the same length of time about the same or another axis 
the angular momentum will be proportionately increased about 
that axis. Since angular momenta are vector quantities they 
can be combined by the method used in the parallelogram of 
forces. 

Precession of Gyroscope. — A general view of a gyro- 
scopic top is shown in Fig. 267, a plan and part sectional 
elevation in Figs. 268 and 269. 




Fig. 269. 



If the wheel were not rotating and a weight W were 
suspended from the pivot x as shown, the wheel and gimbal 
ring would naturally rotate about the axis yy., but when the 
wheel is rotating at a high speed the action of the weight at 
first sight appears to be entirely different. 

Let the axis 00 be vertical, and let the gimbal ring be placed 
horizontal to start with. Looking at the top of the wheel (Fig. 
268), let the angular momentum of a particle on the rim be 
represented by Qm in a horizontal plane, and let 0« I'epresent 
the angular momentum in a direction at right angles to Om 
generated in the same time by the moment of W about yy. 
The resultant angular momentum will be represented by Or, 
that is to say, the plane of rotation of the wheel will alter 
or precess into the direction shown by Or; thus, to the 
observer, it.precesses in a contra-clockwise sense at a rate 
shortly to be determined. The precession will be reversed 
in sense if the wheel rotates in the opposite direction or if the 
external couple tends to lift the right-hand pivot x. 

If any other particle on the rim be considered, similar 



Gyroscopic Action. 



279 



results, as regards the precession of the system will be 
obtained. 

Consider next the case of a horizontal force P applied to 
the right-hand pivot. 

Let ym (Fig. 270) represent the angular momentum of a 
particle on the rim of the wheel, which at the instant is moving 

in a vertical plane. Let yn 
represent the angular mo- 
mentum generated by the 
external couple of P about 
the axis 00. The resultant 
angular momentum is repre- 
sented by yr, thus indicating 
that the plane of rotation 
which was vertical before the 
application of the external 
moment has now tilted over 
or precessed to the inclined 
position indicated by yr. 
Thus an observer at y-^ look- 
ing horizontally at the edge 
of the wheel sees it precess- 
ing about y-^^y in a contra- 
clockwise sense. If the sense 
of rotation of the wheel be 
reversed, or if P act in the 
opposite direction, the sense of the precession will also be 
reversed. 

We thus get the very curious result with a spinning gyro- 
scope that when it is acted on by an external couple or twist- 
ing moment the resulting rotation does not occur about the 
axis of the couple but about an axis at right angles to it. 

An easily remembered method of determining the sense in 
which the gyroscope tends to precess is to consider the motion 
of a point on the rim of the wheel just as it is penetrating the 
plane containing the axis of the wheel and the external couple. 
In Fig. 268 this point is on the axis 00 on either the top or 
bottom of the rim. In Fig. 270 it is on the axis yy. The 
diagram of velocity' is constructed on a plane tangential 

IV 




Angular momentum = In? ; 



K 



Since I and K are constant, the angular momentum generated is pro- 
portional to V, hence velocity diagrams may be used instead of angular 
momentum diagrams. 



28o 



Mechanics applied to Engineering. 



to the rim of the wheel at the point in question and parallel 
to the axis of rotation, by drawing lines to represent 

(i) the velocity of the point owing to the rotation of the 
wheel ; 

(ii) the velocity of the point due to the extemal coupje ; 

(iii) the resultant velocity, the direction of which indicates 
the plane towards which the wheel tends to precess. 

In the case of rotating bodies in which the precession is 
forced the sense and direction of lines (i) and (iii) are known ; 
hence the sense and direction of (ii) are readily obtained, thus 
supplying all the data for finding the sense of the external 
couple. 

When only the sense of the precession is required, it is 
not necessary to regard the magnitude of the velocities or the 
corresponding lengths of the lines. 

Rate of Precession. — When the wheel of a gyroscope is 
rotating at a given speed, the rate at which precession occurs 



Ae 






p* 



Fig. 271. 




is entirely dependent upon the applied twisting moment, the 
magnitude of which can be readily found by a process of 
reasoning similar to that used for finding centrifugal force. 
See the paragraph on the Hodograph on p. 17. 

The close connection between the two phenomena is 
shown by the following statements : — 

(Gyroscopic couple) : —3 ((wheel)] "((spinning)) 

at a constant < /,\ > speed and be constrained in such a 

.1. ^-A moves ) , . Ca centre O in the plane of 
manner that it J(precesses)r^°"* l(an axis 00 at right angles 

motion I 

to the axis of rotation) j* 

Let OP, represent the [(^'^^^,^^^^] velocity of the \i^^^l^^ 



Gyroscopic Action. 281 

when in position r and let a[(/^J^f^^| act on the ^ "J, J 

in such a manner as to bring it into position 2. 

Then OP2 represents its corresponding \ , '"^ j"^ ^ > velocity, 

and PiP„ represents, to the same scale, the change of 
r velocity j 

I (angular velocity)/- 

'^he{(eouple)|^^1"'-d '° ^^^^ ^^^ change ofjjj^^tr 

velocity)] '^g'^^^^y 

W _ WVS2 \ 

i/'^^^ WVKS2 WK^^ ^^ \| 
(^CK = = ii(o = IQo) j 

or Centrifugal force 

= mass of body x velocity x angular velocity. 

Gyroscopic couple 



{moment of inertia of 
wheel about the axis 
of rotation 



angular ) 



velocity X S^T'^'' ^^^°"'y 
of wheel) I of precession. 



In the above expressions — 

W = Weight of body or the wheel in pounds. 

V = Velocity in feet per second of the c. of g. of the 
body or of a point on the wheel at a radius equal 
to the radius of gyration K (in feet). 

i2 = The angular velocity of the body or the wheel in 
radians per second. 

to = The rate of precession in radians per second. 

N = Revolutions per minute of the wheel. 

n = Revolutions per minute of the precession. 

For engineering purposes it is generally more convenient 
to express the speeds in revolutions per minute. Inserting 
the value of ^ and reducing we get — 

WK^N« 
Gyroscopic couple = = o'ooo34WK^N« in pounds- 
feet. 



282 



Mechanics applied to Engineering. 



The following examples of gyroscopic effects may serve to 
show the magnitude of the forces to be dealt with. 

The wheels of a locomotive weigh 2000 pounds each, the 
diameter on the tread is 6 feet 6 inches, and the radius of 
gyration is z'8 feet. Find the gyroscopic couple acting on 
the wheels when running round a curve at 50 miles per hour. 
Radius of curve 400 feet. Find the vertical load on the 
outer and inner rails when there is no super-elevation, and 
when the dead load on each wheel is 10,000 pounds. Rail 
centres 5 feet. 



N = 215-5 n = r75 
Gyroscopic couple = 



2000 X 2'8^ X 2is'5 X f75 



2937 

= 2015 pounds-feet. 
Arm of couple, 5 feet. Force, 403 pounds. 

The figure will assist in determining the sense in which 
the external couple acts. Since the wheels precess about a 




vertical axis passing through the centre of the curve, we know 
that the external forces act in a vertical direction. As the 
wheels traverse the curve their direction is changed from oni 
to or, hence the external couple tends to move the top of the 
wheel in the direction on. The force, which is the reaction of 
the outer rail, therefore acts upwards, which is equivalent to 
saying that the vertical pressure on the rail is greater by the 



Gyroscopic Action. 283 

amount of the gyroscopic force than the dead weight on the 
wheel. Since the total pressure due to both wheels is equal 
to the dead weight upon them, it follows that the vertical 
pressure on the inner rail is less by the amount of the gyro- 
scopic force than the dead weight on the wheel. The vertical 
pressure on the outer rail is 10,009 + 403 = iOi403 pounds, 
a«id on the inner rail 10,000 — 403 = 9597 pounds. (No 
super-elevation.) 

Thus it will be seen that the gyroscopic effect intensifies 
the centrifugal effect as far as the overturning moment is con- 
cerned. 

If the figure represented a pair of spinning gyroscopic 
wheels which are not resting upon rails, but with the frame 
supported at the pivot O, the gyroscope would precess in the 
direction of the arrow if an upward force were applied to 
the left-hand pivot x, or if a weight were suspended from the 
right-hand pivot x. 

In the case of a motor-car, in which the plane of the fly- 
wheel is parallel to that of the driving wheels, the angular 
momentum of the flywheel must be added to that of the road 
wheels, when the sense of rotation of the flywheel is the same 
as that of the road wheels, and subtracted when rotating in 
the opposite sense. 

When the flywheel rotates in a plane at right angles to 
that of the road wheels, and its sense of rotation is clockwise 
when viewed from the front of the car, the weight on the 
steering wheels will be increased and that on the driving 
wheels will be diminished when the car steers to the chauffeur's 
left hand, and vice versa when steering to the right ; the 
actual amount, however, is very small. When a car turns 
very suddenly, as when it skids in turning a corner on a 
greasy road, the gyroscopic effect may be so serious as to 
bend the crank shaft. Lanchester strengthens the crank web 
and the neck of the crank-shaft next the flywheel, in order 
to provide against such an accident. 



CHAPTER IX. 



FRICTION. 



W=iV 



When one body, whether solid, liquid, or gaseous, is caused to 

slide over the surface of another, a resistance to sliding is 

experienced, which is termed the " friction " between the two 

bodies. 

Many theories have been advanced to account for the 

friction between sliding bodies, but none are quite satisfactory. 

To attribute it merely to the roughness between the surfaces is 

but a very crude and incomplete statement j the theory that the 

surfaces somewhat resemble a short-bristled brush or velvet 

pile much more nearly accounts for known phenomena, but 

still is far from being satisfactory. 

However, our province is not to account for what happens, 

but simply to observe and 

classify, and, if possible, to sum 

up our whole experience in a 

brief statement or formula. 

Frictional Resistance. 

(F). — If a block of weight W be 

placed on a horizontal plane, as 

shown, and the force F applied 

parallel to the surface be required 

to make it slide, the force F 

is termed the frictional resistance of the block. The normal 

pressure between the surfaces is N. 

Coefficient of Friction (/i). — Referring to the figure 

F F . 

above, the ratio^ or ^ = /*, and is termed the coefficient of 

friction. It is, in more popular terms, the ratio the friction 
bears to the normal pressure between the surfaces. It may be 
found by dragging a block along a plane surface and measuring 
F and N, or it may be found by tilting the surface as in Fig. 274. 
The plane is tilted till the block just begins to slide. The vertical 



Fig. 273. 



Friction, 



285 



weight W may be resolved normal N and parallel to the plane 
F. The friction is due to the normal pressure N, and the 





Fig. 274. 



Fig. 27s. 



force required to make the body slide is F ; then the coefficient 

F F 

of friction ^ — ^3& before. But ^ = tan <^, where ^ is the 

angle the plane makes to the horizontal when sliding just 
commences. 

The angle ^ is termed the " friction angle," or " angle of 
repose." The body will not slide if the plane be tilted at an 
angle less than the friction angle, a force Fo (Fig. 275) will 



-r-^/r.. 





Fig. 276. 



Fig. 277. 



then have to be applied parallel to the plane in order to make 
it slide. Whereas, if the angle be greater than ^, the body will 
be accelerated due to the force Fj (Fig. 276). 

There is yet another way of looking at this problem. Let 
the body rest on a horizontal plane, and let a force P be 
applied at an angle to the normal ; the body will not begin to 
slide until the angle becomes equal to the angle ^, the angle of 
friction. If the line representing P be revolved round the 
normal, it will describe the surface of a cone in space, the apex 
angle being 2^; this cone is known as the "friction cone." 



286 Mechanics applied to Erigineering. 

If the angle with the normal be less than <^, the block will not 
slide, and if greater the block will be accelerated, due to the 
force Fj, In this case the weight of the block is neglected. 

If P be very great compared with the area of the surfaces 
in contact, the surfaces will seize or cling to one another, and 
if continued the surfaces will be torn or abradec!. 

Friction of Dry Surfaces. — The experiments usually 
quoted on the friction of dry surfaces are those made by Morin 
and Coulomb ; they were made under very small pressures and 
speeds, hence the laws deduced from them only hold very 
imperfectly for the pressures and speeds usually met with in 
practice. They are as follows : — 

1. The friction is directly proportional to the normal 
pressure between the two surfaces. 

2. The friction is independent of the area of the surfaces 
in contact for any given normal pressure, i.e. it is independent 
of the intensity of the normal pressure. 

3. The friction is independent of the velocity of 
rubbing. 

4. The friction between two surfaces at rest is greater than 
when they are in motion, or the friction of rest ' is greater 
than the friction of motion. 

5. The friction depends upon the nature of the surfaces in 
contact. 

We will now see how the above laws agree with experiments 
made on a larger scale. 

The first two laws are based on the assumption that the 
coefficient of friction is constant for all pressures; this, 
however, is not the case. 

The cmves in Fig. 278 show approximately the diflFerence 
between Coulomb's law and actual experiments carried to high 
pressures. At the low pressure at which the early workers 
worked, the two curves practically agree, but at higher 
pressures the friction falls off, and then rises until seizing 
takes place. 

Instead of the frictional resistance being 

F =/x,N 

it is more nearly given by F = f<,N"''', or F = /h^'I'v'N 

The variation is really in /a and not in N, but the ex- 
pression, which is empirical, assumes its simplest form as given 
above. 

For dry surfaces ft has the following values :■ — 

' The friction of rest has been very aptly termed the " sticktion." 



Friction. 



287 



Wood on wood 
Metal „ 
Melal on metal 



o'2S to 0-50 
020 to o'6o 
0'I5 to o'30 



Leather on wood 
,, metal 

Stone on stone 



0-25 to o'5o 
o'3o to o'6o 
040 to o"6s 



These coefficients must always be taken with caution ; they 
vary very greatly 
indeed with the 
state of the sur- 
faces in contact. 

The third law 
given above is far 
from representing 
facts ,: in the limit 
the fourth law 
becomes a special 
case of the third. '^ 
If the surfaces 
were perfectly '^ 
clean, and there ij; 
were no film of 
air between, this 
law would pro- 
bably be strictly 
accurate, but all 
experiments show that the friction decreases with velocity of 
rubbing. 




IntensUy of jn-essuro 

Fig. 278. 




+ Morin-. 
O Ri'iLtiCe. 
• Westinghouse &■ Ga/ton 



30 40 SO 60 

Speed in feet per second. 

Fig. 279. 



288 



Mechanics applied to Engineering. 



The following empirical formula fairly well agrees with 
experiments : — 

Let ft = coefficient of friction ; 

K = a constant to be determined by experiment ; 
V = the velocity of sliding. 

Ihen u = 7= 

2-4VV 

The results obtained with dry surfaces by various experi- 
menters are shown in Fig. 279. 

The fourth law has been observed by nearly every experi- 
menter on friction. The following figures by Morin and others 
will suffice to make this clear : — 





Coefficient of friction. 


Materials. 


Rest. 


Velocity 3 to 
5 ft.-sec. 


Wood on wood 

»» »» 

Metal on metal 

Stone on stone 

Leather on iron 


0-S4 
0-69 

034 
074 
0-S9 


0'46 

0-43 
0-26 
063 
0-52 



The figures already quoted quite clearly demonstrate the 
truth of the fifth law given above. 

Special Cases of Sliding Bodies. — In the cases we are 

f ^ ^ about to consider, we shall for 

sake of simplicity, assume that 
Coulomb's laws hold good. 

Oblique Force re- 
squired to make a Body 
slide on a Horizontal 
Plane. — If an oblique force 
P act upon a block of weight 
W, making an angle Q with 
the direction of sliding, we 
can find the magnitude of 
P required to make the block 
slide; the total normal pres- 
sure on the plane is the 
normal component of P, viz. n, together with W. From a draw a 
line making an angle <^ (the friction angle) with W, cutting P in 




Friction. 289 

the point b ; then be, measured to the same scale as W, is the 
magnitude of the force P required to make the body slide.. 
The frictional resistance is F, and the total normal pressure 
« 4- W; hence F = /^(« + W). When 6 = o, P„ = fa? = /tW. 

When 6 is negative, it simply indicates that Pj is pulling 
away from the plane : the magnitude is given by ce. From the 
figure it is clear that the least value of P is when its direction 
is normal to ab, i.e. when 6 = 4>; then — 

P,„in. = Po cos <j> = fxW cos (t> 
= tan "^ W cos <l> 
= W sin 4> 

It will be seen from the figure that P is infinitely great 
when ab is parallel to be — that is, when P is just on the edge 
of the friction cone, or when go — 6 = <j>. When 6 = — 90°, P 
acts vertically upwards and is equal to W. 

A general expression for P is found thus — 

;? = P sin e 

F = P cos e = /x(W + n) 
F = //.(W + P sin ff) 
and P(cos 6 + i>. s\n 6) = fiW 

/X.W tan </. W 



hence P = 
P = 



cos ^ + /x sin 6 cos 6 + /u, sin 
sin <f> W 



cos <^ cos + sin <^ sin 6 
_ W sin <^ 
~ cos (^ + 6) 

When P acts upwards away from the plane, the — sign is 
to be used in the denominator ; and for the minimum value of P, 
<f> = —6; then the denominator is unity, and P = W sin <j>, the 
result given above, but arrived at by a different process. 

Thus, in order to drag a load, whether sliding or on wheels, 
along a plane, the line of pull should be upwards, making an 
angle with the plane equal to the friction angle. 

Force required to make a Body slide on an Inclined 
Plane. — A special case of the above is that in which the plane 
is inclined to the horizontal at an angle a. Let the block of 
weight W rest on the inclined plane as shown. In order to 
make it slide up the plane, work must be done in lifting the 
block as well as overcoming the friction. The pull required 
to raise the block is readily obtained thus : Set down a line be 
tc represent the weight W, and from e draw a line ed, making 

V 



290 



Mechanics applied to Engineering. 




Fig. aSi. 



an angle a with it ; then, if from b a. line be drawn parallel to 
the direction of pull Pi, the line M^ represents to the same 

scale as W the required 
pull if there were no 
friction. An examination 
of the diagram will at 
once show that id^c is 
simply the triangle of 
forces acting on the 
block ; the line cdi is, of 
course, normal to the 
plane. 

When friction is 

taken into account, draw 

the line ce, making an 

angle <f> (the friction 

angle) with cd; then ie^ 

gives the pull Pj required 

to drag the block up the 

plane including friction. 

For it will be seen that the normal pressure on the plane is 

cdo, and that the friction parallel to the direction of sliding, 

viz. normal to cd, is — 

fx, . cdQ ^= tan (^ . cd^ 
= d^^ 

Then resolving d^^ in the direction of the pull, we get the pull 

lequired to overcome the 
friction dye-^ ; hence the total 
pull required to both raise 
the block and overcome the 
friction is be^. 

Least Pull.— The least 
pull required to pull the block 
up the plane is when be has 
its least value, i.e. when be is 
normal to ce ; the direction of 
pull then makes an angle <^ 
with the plane, ox B = ^, for 
cd is normal to the plane, and 
F,o. 28s. <■* makes an angle <^ with cd. 




Then ?„,„ = W sin (<A + a) 



(i-) 



Friction. 



291 



Horizontal Pull.— When the body is raised by a hori- 
zontal pull, we have (Fig. 283) — 



Pj = W tan (^ + a) 



(ii.) 




Fig. 283. 




Fig. 284. 




Thus, in all cases, the effect of friction is equivalent to 
making the slope steeper by , -» 

an amount equal to the friction I - 

angle. 

Parallel Pull. — When the '^ 

body is raised by a pull parallel ^ /?--- 

to the plane, we have (Fig. 284) — fig^s. 

V^^^ + db 
But ed = dc tan ^ = /idc 
and dc = W , cos a 
therefore^ = /t . W . cos a 
and d3 = W sin o 
hence P, = W(fj. . cos a -f sin a) , . 

This may be expressed thus (see Fig. 285) — 



(Hi) 



p, = w(.g-fg) 

or P,L = W//,B + WH 



or, Work done in 
dragging a body 
of weight W up a . 
plane, by a force / 
acting parallel to 
the plane / 



'work done in dragging^ 
the body through the 
same distance on a 
horizontal plane 
against friction. 



/work done 
in lifting 
I the body 



292 Mechanics applied to Engineering. 

General Case. — When the body is raised by a pull making 
an angle Q with the plane — 




P = 



P = 



of 



(iv.) 



COS (e - ^) 
Substituting the value 
P„,„. from equation (i.) — 
W sin («^ + a ) 
COS (6 - <^) 
When the line of action of 
P is towards the plane, as in 
Po, the B becomes minus, and 
we get — 

W sin (<^ + a) 
cos ( — ^ — <^) 

F,o. 286. All the above expressions 

may be obtained from this. 

When the direction of pull, Po, is parallel to ec, it will 
only meet .ec at infinity — that is, an infinitely great force would 
be required to make it slide ; but this is impossible, hence the 
direction of pull must make an angle to the plane 6 < (90 — <j>) 
in order that sliding may take place. 

We must now consider the case in which a body is dragged 
down a plane, or simply' allowed to slide down. If the angle a 
be less than <f>, the body must be dragged down, and if a be 



Po = 





Note. — The friction now assists the lowering, hence « is set off to the 
right olcti. 



Friction. 



293 



greater, a force must be applied to prevent it from being 
accelerated. 

Least pull when body is lowered, <^ < a (Fig. 287). 

Pmin. =. be = W sin'(a - <^) and 6 = <^ . . . (v.) 

When <^>a, be^ is the least force required to make the body 
slide down the plane. 

P,».„. = ■^i = W sin (<^ - a) . . . . (vi.) 

when <^ = a, P„|„, is of course zero. 

The remaining cases are arrived at in a similar manner ; we 
will therefore simply state them. 





*<«. 


0>O. 


Least pull 

Parallel pull 
Horizontal pull 

(General case 


W sin (b — <))) 
W (sin — |U cos 0) 
W tan (a - ^) 
W sin (0 — <f) 
cos (9 + <p) 


\V sin (f — a) 
W (jit cos — sin a) 
W tan (((> - a) 
W sin (<^ - 0) 
cos (fl - <p) 



Note.— If the line of pull comes below the plane, the angle 9 takes 
the — sign. 

In the case of the parallel pull, it is worth noting that when 
t^ < a, we have — 

Total work done = work done in lowering the body — work 
done in dragging the body through the 
horizontal distance against friction 

and when <^>a we have the same relation, but the work done 
is negative, that is, the body has to be retarded. 

It should be noticed that the effect of friction on an inclined 
plane is to increase the steepness when the block is being 
hauled up the plane, and to decrease it when hauling it down 
the plane by an amount equal to the friction angle. 

EflSciency of Inclined Planes. — If an inclined plane be 
used as a machine for raising or lowering weights, we have — 

■ccc ■ useful work done (t.e. without friction) 

EtSciency = ; ; — ^-^^ — , . , ^ . . — r — i- 

actual work done (with friction) 

Inclined Plane when raising a Load. — The maximum 
efficiency occurs when the pull is least, i.e. when 6 = tji. The 
useful work done without friction is when = o ; then — 



294 Mechanics applied to Engineering. 

The work done without friction = -^ — from (iv.) 

cos 6 

„ „ with „ = LW sin (<^ + a) from (i.) 

where L = the distance through 'which the bpdy is dragged ; 
a = the inclination of the plane to the horizon ; 
B = the inclination of the force to the plane ; 
^ = the friction angle. 

LW sin a 



(vii.) 



Then maximuml _ cos 6 _ sin a 

efficiency ) LW sin (<^ + a) cos B sin (^ + a) 

When the pull is horizontal, ^ = a, and — 

„^ . sin o tan a. / ■■■ \ 

Efficiency = — rr-r — ^ = - — jj—. — •■ (viii.) 

cos a . tan (<;* + a) tan (^ + a) 

when the pull is parallel, 6 = o, and cos 6 = i ; 

■,71V- • sin a . cos <i /• \ 

Efficiency = ^ — -. — -— f .... (ix.) 
sin (a + (ji) 

General case, when the line of pull makes an angle 6 with 
the direction of sliding^ 



sin a . cos (0 — ^) / > 
o — ■„ / 1 I — \ • • • (^•) 



Efficiency = -^ — -■ — 7-7— i — ( 

' cos 6 . sin (0 + a) 




Friction of Wedge. — This is simply a special case of the 
inclined plane in which the pull 
is horizontal, or when it acts 
normal to W. We then have 
from equation (ii.) P = W tan 
(<^ + a) for a single inclined 
plane; but here we have two 
inclined planes, each at an angle 
F,Q 'j5g, a, hence W moves twice as far 

for any given movement of P 
as in the single inclined plane ; hence — 

P = zW tan (<j> + a) for a wedge 

The wedge will not hold itself in position, but will spring 
back, if the angle a be greater than the friction angle </!>. 

From the table on p. 293 we have the pull required to 
withdraw the wedge — 

- P = 2W tan (a - ,^) 



Friction. 295 

The efficiency of the wedge is the same as that of the 
inclined plane, viz. — 

Efficiency = : — , ° , ^ when overcoming a resistance (xi.) 

reversed | ^ tan (a - ^) ^^^^ withdrawing from a resistance 
efficiency J tan a (^^^ p 335) 

Efficiency of Screws and Worms— Square Thread. 
— A screw thread is in effect a narrow inclined plane wound 
round a cylinder; hence the efficiency is the same as that of 
an inclined plane. We shall, however, work it out by another 
method. 



Fig. 290. 

Let/ = the pitch of the screw ; 

^0 = the mean diameter of the threads ; 
W = the weight lifted. 

The useful work done per revolution! _ -^p _ -^^^ ^^^ 

on the nut without friction I " 

The force applied at the mean radius ofl ^ fP = Wtan(a+^) 
the thread required to raise nut I I (see equation ii.) 

The work done in turning the nut) _ p^^^ 
through one complete revolution f 

= 'Wirda tan (a + 4>) 
Wx^ o tan g 
Efficiency, when raising the weight, = YV^^^laiToSTT^) 

tan a sin a cos (a + 0) 



tan (a + </)) cos a sin (a + 4>) 

_ sinJ2a_+ « ^) - sin ^ ^ ^ 2 sin (^ 

'" sin"(2a"+ <^) + sin <^ sin(2a + ^) + sin <^ 

This has its maximum value when the fractional part is 
least, or when sin (2a + ^) = i. 



296 Mechanics applied to Engineering. 

Since the sine of an angle cannot be greater than i, then 
2a + c^ = 90 and u = 45 . Inserting this value, 

maximum efficiency = | ^ I = i — 2/i (nearly). 




In addition to the friction on the threads, the friction on 
the thrust collar of the screw must be taken into account. The 
thrust collar may be assumed to be of the same diameter as 
the thread ; then the 

efficiency of screw thread 1 _ tan a . 

and thrust collar ] " tan (a + 2^) WP"""^-) 

In the case of a nut the radius at which the friction acts 
will be about i^ times that of the threads ; we may then say — 

efficiency of screw thread and nut ) _ tan o 
bedding on a flat surface j ~ tan (a + 2*5 A) 

If the angle of the thread be very steep, the screw will be 
reversible, that is, the nut will drive the screw. By similar 
reasoning to that given above, we have — 

reversed efficiency = ^ — (see p. 335) 

Such an instance is found in the Archimedian drill brace, 
and another in the twisted hydraulic crane-post used largely on 
board ship. By raising and lowering the twisted crane-post, 
the crane, which is in reality a part of a huge nut, is slewed 
round as desired. 

Triangular Thread. — In the triangular thread the normal 
pressure on the nut is greater than in the square-threaded 

Wn I W 

screw, in the ratio oi ~ = -, and Wq = — , where 6 is 

W p (7 

cos - cos — 

2 2 

the angle of the thread. In the Whitworth thread the angle 6 
is 55°, hence Wo = 1-13 W, In the Sellers thread 6 = 60° and 
W„ = i'i5 W J then, taking a mean value of Wo = i'i4 W, we 
have — 

efficiency = *^" " 



tan (a -|- I "141^) 



Friction. 



297 



In tlie case of an ordinary bolt and nut, the radius at which 
the friction acts between the nut and the 
washer is about i^ times that of the thread, 
and, taking the same coefficient of friction 
for both, we have — 




efficiency \ 

of a bolt = tan a 

and nut I tan (a + 2-6+.^) 



(approx.) 



The following table may be useful in 
showing roughly the efficiency of screws. 
In several cases they have been checked 
Fig. 291. by experiments, and found to be fair 

average values ; the efficiency varies greatly 
with the amount of lubrication : — 



Table of Approximate Efficiencies of Screw Threads. 





ElHciency per cent, when 


Ef&cieocy^er cent, allow- 


Angle of 


no friction between nut 


ing for friction between 


thread a. 


and washer or a thrust 


nut and washer or a 




collar. 




thrust collar. 




Sq. thread. 


V-thread. 


Sq. thread. 


V-thread. 


2» 


19 


17 


II 


8 


->o 


26 


23 


14 


12 


4° 


32 


28 


17 


16 


5° 


36 


33 


21 


20 


10° 


ss 


52 


36 


29 


20° 


67 


63- 


48 


42 


45-1 


79 


75 


52 


44 



In the above table <p has been taken as 8*5°, 01 fi = o'i5. 

For the efficiency of a worm and wheel see page 344. 

Rolling Friction. — When a wheel rolls on a yielding 
material that readily takes a permanent deformation, the 
resistance is due to the fact that the wheel sinks in and makes 
a rut. The greater the weight W carried by the wheel, the 
deeper will be the rut, and consequently the greater will be the 
resistance to rolling. 

When the wheel is pulled along, it is equivalent to con- 
stantly mounting an obstacle at A ; then we have — 



298 



Mechanics applied to Engineering. 



P . BA = W . AC 
W.AC 



orP = 
LetAC = K; 
Then P = 



BA 



W.K 



J 


B 


///// 


1 



w 



But h is usually small compared 

with R ; hence we may write — ^'°- "^^ 

P = ^ (nearly) 

P and W, also K and R, must be measured in the same 
units, or the value of K corrected accordingly. The above 
treatment is very rough, but the relation holds fairly well in 
practice. There is much need for further research in this 
branch of friction. 



Values of K. 

Iron or steel wheels on iron or steel rails ... 

I. » wood 

,, ,, macadam 

„ ,, soft ground 

Pneumatic tyres on good road or asphalte ... 

„ „ heavy mud 

Solid indiarubber tyres on good road or asphalte 
„ „ heavy mud 



K (inches). 
O'oo7-o"oi5 
O"o6-o'io 

OOS-0'20 

3-S 

0"02-0'022 

0'04-o-o6 

0-04 

o-og-o'ii 



Some years ago Professor Osborne Reynolds investigated 
the action of rollers passing over elastic materials, and showed 
clearly that when a wheel 
rolls on, say, an indiarubber 
road, it sinks in and com- 
presses the rubber imme- 
diately under it, but forces 
out the rubber in front and 
behind it, as shown in the 
sketch. That forced up in 
the front slides on the surface 
of the wheel in just the 
reverse direction to the mo- 
tion of the wheel, and so hinders its progress. Likewise, as the 
wheel leaves the heap behind it, the rubber returns to its original 




Fig. 293. 



Friction. 



299 



. place, and again slips on the wheel in the reverse direction to 
its motion. Thus the resistance to rolling is in reality due 
to the sliding of the two surfaces. On account of the stretch- 
ing of the path over which 
the wheel rolls, the actual 
length of path rolled over 
is greater than the hori- 
zontal distance travelled 
by the wheel, hence it 
does not travel its geo- 
metrical distance ; the 
amount it falls short of it 
or the " slip " depends 
upon the hardness of the 
surfaces in contact. Even Fig. 294. 

with the balls in ball bear- 
ings the " slip " is quite appreciable. 

Antifriction Wheels. — In order to reduce the friction 
on an axle it is sometimes mounted on antifriction wheels, as 
shown. A is the axle in question, B and C are the anti- 
friction wheels. If W be the load on the axle, the load on 
each antifriction wheel bearing will be — 




W„=- 



W W 

-1, and the load on both ^ 

^' cos d 



2 cos 

Let Ra = the radius of the main axle ; 

R= „ „ antifriction wheel ; 

r — „ ,, axle of the antifriction wheel. 

The rolling resistance on the surface j _ WK 
of the wheels J "~ R^ cos d 

The frictional resistance referred toj ^ 

the surface of the antifriction wheels, V = ~- 

or the surface of the main axle ) ^ • ^°^ " 

W /K , u,A 

The total resistance = ^^1 ^ + -5- ^ 

cos ^ \R„ R / 

If the main axle were running in plain bearings, the 
resistance would be /aW ; hence — 

/<.R cos B 



frictio n with plain bearings 
friction with antifriction wheels 



k| + ^. 



300 



Mechanics applied to Engitieering. 



In some instances a single antifriction wheel i? used, the 
axle A being kept vertically over the axle of the wheel by 
means of guides. The main trouble with all such devices is 
that the axle travels in an endwise direction unless prevented 
by some form of thrust bearing. One British Railway Com- 
pany has had large numbers of waggons fitted with a single 
antifriction disc on each axle bearing ; the roUing resistance is 
materially less than when fitted with ordinary bearings, but the 
discs are liable to get " cross cornered," and to give trouble 
in other ways. 

Roller Bearings. — There are many forms of roller bear- 
ings in common use, but unfortunately few of them give really 
satisfactory results. The friction of even the worst of them is 
considerably lower than that of bearings provided with ordinary 
lubrication. If the rollers are not perfectly parallel in them- 
selves (in a cylindrical bearing), and are not kept absolutely 
parallel with the shaft, they tend to roll in a helical path, but 
since the cage and casing prevent them from doing so, they 
press the cage against the flange of the casing and set up what 




Lcfmibuduwl Sectiorv. 



If aW Cross Section, 
onljine A-B 
Fig. 295. 



I/aJf Cross Section- 
an C0rUreLvi£ 



is known as " end-thrust," which thereby gives rise to a large 
amount of friction between the end of the cage and the flange 
of the bearing, and in other ways disturbs the smoothness of 
running. Few, if any, roller bearings are entirely free from 
this defect ; it is moreover liable to vary greatly from time to 
time both as regards its amount and direction. In general, it 
is less at high speeds than at low, and it increases with the load 
on the bearing ; it does not appear to be greatly affected by 
lubrication. Bearings in which the end-thrust is high nearly 
always show a high coefficient of friction, and vice versa. High 
friction is always accompanied with a large amount of wear 
and vibration. In order tO- reduce wear and to ensure smooth 



Friction. 



301 



running, the rollers, sleeve, and liner should be of the hardest 
steel, very accurately ground and finished. In many of the 
cheaper forms of roller bearing no sleeve is used, hence the 
rollers are in direct contact with the shaft. The outer casing is 
usually split to allow of a bearing on a long line of shafting 
being replaced when necessary without removing the pulleys 
and couplings, or without taking the shafting down. This is an 
undoubted advantage which is not possessed by ball bearings. 
The reason why ball bearings will not run with split or jointed 
races will be obvious after reading the paragraphs devoted to 
ball bearings. 

The commonest form of roller bearing is that shown 
in Fig. 295. There is no sleeve on the shaft and no liner in 
the casing ; the steel rollers are 
kept in position by means of a 
gun-metal cage, which is split 
to allow of the rollers being 
readily removed. 

Another cheap form of roller 
bearing which is extensively 
used is the Hyatt, in which the 
rollers take the form of helical 
springs ; they are more flexible 
than solid rollers, and conse- 
quently accommodate them- 
selves to imperfections of align- 
ment arid workmanship. — 

In the Hoffmann short roller 
bearing. Fig. 296, the length 
of the rollers is equal to their diameter; the roller paths and 
rollers are of the hardest steel ground to a great degree of 
accuracy. The end-thrust is almost negligible and the co- 
efficient of friction is low ; the bearings will run under a con- 
siderably higher load than a ball bearing of similar dimensions. 
The following table gives fair average results for a friction test 
of an ordinary roller bearing : — 




Centre of Shaft 



1 



Fig. 296. 



Total load 


40 revolutions per minute. 


400 revolutions per minute. 


in lbs. 


V- 


End thrust in lbs. 


V- 


End thrust in lb:i. 


2000 
4000 
6000 
8000 
10,000 


o'oi3i 
o'oo94 
0-0082 
00076 

0'0072 


82 

147 
212 
276 


0-0053 
0-0035 
0-0029 
0-0026 
0-0024 


51 

89 

128 

166 

205 



303 Mechanics applied to Engineering. 

A test of a Hoffmann short roller bearing gave the following 
results : — 



Total load in lbs. 


C 


End thrust in lbs. 


Temperature air at 
62° F. 


2000 


0-OOI2 


None 





4000 


O'OOIO 




■ — 


6000 


00008 


7 


76° F. 


8000 


O'OOIO 


84 


89° F. 


10,000 


o-ooii 


2X0 


96° F. 


12,000 


O'OOIO 


270 


94° F. 


14,000 


O'OOIO 


20 


95° F. 


14,500 


O'OOIO 





95° F. 



Diameter of sleeve ... 


.. 4' 7 5 inches 


Diameter of rollers . . . 


.. I'ooinch 


Length of rollers ... 


.. I'ooinch 


Number of rollers . . . 


.. 14 


Revolutions per minute 


.. 400 



Hardened steel 
highly finished. 



For details of other types of roller bearings and results of 
tests, the reader should refer to a paper by the author on 
" Roller and Ball Bearings," Froceedings of the Institution of 
Civil Engineers, vol. clxxxix. 

Ball Bearings.— The "end-thrust" troubles that are 
experienced with roller bearings can be entirely avoided by 
the substitution of balls for rollers. The form of the ball path, 
however, requires careful consideration. 

Various types of ball races are shown in diagrammatic form 
in Fig. 297. A is known as a four-point radial bearing, the 
outer cones screw into the casing, and thereby permit of adjust- 
ment as the bearing wears. B is a three-point bearing capable 
of similar adjustment. C is a three-point bearing ; the inner 
coned rings are screwed on to the shaft, and can be tightened 
up as desired. None of these forms of adjustment are satis- 
factory for heavy loads. A four-point thrust bearing is shown 
at F; the races are ground to an angle of 45° with the axle, 
but since the circumferential speed of the race at a is greater 
than at b, the circle aa on the ball tends to rotate at a higher 
speed than the circle bb, but since this cannot occur, grinding 
and scratching of the ball take place. In order to avoid this 
defect, races were made as shown at G; the circle aa was 

greater than bb in the ratio ~. By this means it was expected 

that a true rolling motion would occur, but the bearing was 



Friction. 



303 



not a success. The three-point bearing H was designed on 
similar lines, but a considerable amount of grinding of the 
balls took place. 

To return to the radial bearings. A two-point bearing is 
shown at D; the balls rolled between tfro plain cylindrical 
surfaces. A cage was usually provided for holding the balls 
in their relative position. In E the balls ran in grooved races. 
In these bearings a true rolling motion is secured, the balls do 




not grind or scratch, and the friction is considerably less than in 
A, B, C. Thrust bearings designed on similar lines are shown 
at I and J. A more detailed view of such bearings is shown 
in Fig. 298. The lower race is made with a spherical seat 
to allow it to swivel in case the shaft gets out of line with its 
housing, a very wise precaution which greatly increases the 
life of the bearing. 

When designing bearings for very heavy loads, the difficulty 
is often experienced of placing in one row a sufficient number 



304 



Mechanics applied to Engineering. 



of balls of the required diameter. In that case two or more 
rows or rings may be arranged concentrically, but it is almost 

impossible to get the 
workmanship sufificiently 
accurate, and to reduce 
the elastic strain on the 
housings to such an ex- 
tent as to evenly dis- 
tribute the load on both 
rings. The one set 
should therefore be 
backed with a sheet of 
linoleum or other soft 
material which will yield 
to a sufficient extent to 
equalize approximately the load on each ring. Such a bearing 
is shown in Fig. 299. The sheet-iron casing dips into an oil 
channel for the purpose of excluding dust. The lower half 
of the housing is made with a spherical seat. 




Fig. 258. 




Fig. 299. 

Modern cylindrical or radial bearings are almost always 
made of the two-point type; for special purposes plain 
cylindrical races may be used, but balls running in grooved 
races will safely carry much higher loads than when they run 
on plain cylinders. With plain raceS there is no difficulty in 
inserting the full number of balls in the bearing, but when 
grooved races are used, only about one-half the number can 
be inserted unless some special device be resorted to. But 



Friction. 



305 



since the load-carrying capacity of a bearing depends upon 
the number of balls it contains, it is evidently important to 
get the bearing as nearly filled as possible. After packing in 
and spacing as many balls as possible, the remaining balls are 
inserted through a transverse slot in one side of the race ; the 
depth of this slot is slightly less than that of the groove in 
which the balls run, hence it does not in any way affect the 
smoothness of running. See Fig. 300. 

When two or more rows of balls are used in a cylindrical 
bearing, each row must be provided with a separate ring. 
The inner ring or sleeve must be rigidly attached to the shaft, 
and the outer ring should be backed with linoleum in order 




Fio. 300. 



to evenly distribute the load on each row of balls. The 
housing itself should be provided' with a spherical seating to 
allow for any want of alignment. A design for such a bearing 
is shown in the Author's paper referred to above. A special 
form of bearing has been designed by the Hoffmann Co. with 
the same object. 

It is of great importance to attach the sleeve, or inner ring, 
of the bearing rigidly to the shaft. It is sometimes accom- 
plished by shrinking ; in that case the shrinkage must not be more 

, diameter of shaft ^r .1 ■ ^- ■ j j ■ 

than . If this proportion is exceeded the 

2000 

ring is liable to crack on cooling, or to expand to such an 

extent as to jamb the balls. Where shrinking is not resorted 



3o6 



Mechanics applied to Engineering. 



to the ring is sometimes made taper in the bore, and is 
tightened on to the shaft by means of a nut, or by a clamping 
sleeve, as shown in Fig. 300. 

In the Skefco ball bearing shown in Fig. 301 the outer 
ball race is ground to a spherical surface, and the sleeve is 
provided with grooved races to receive two rows of balls. By 
this arrangement the full number of balls can be packed in 
by tilting the inner race, but the great feature of the bearing 
is the spherical ball race which enables the bearing to be 
used on a shaft which does not run true, or on a machine in 
which the frame springs considerably relative to the shaft. 




Centr e of Sha ft 
Fig. 301. 




Fig. 302. 



Approach of the Balls Race when under Load. — 

When an elastic ball is placed between two elastic surfaces 
which are pressed together, the ball yields under the pressure 
and the surfaces become hollow ; the theory of the subject was 
first enunciated by Hertz, and afterwards by Heerwagen. The 
results obtained by the two theories are not identical, but there 
is no material diiference between them. Experimental re- 
searches on the strain which steel balls undergo when loaded 
show that the theories are trustworthy within narrow limits. 
The following expression is due to Hertz. 

Let 8 = the amount, in inches, the plates approach one 
another when loaded. 
P = the load on the ball in pounds. 
d = the diameter of the ball in inches. 



Friction. 



307 



Then 



;V 



/pa 



32,000 ■*■ d 

Distribution of the Load on the Balls of a Radial 
Bearing. — The load on the respective balls in a radial 
bearing may be arrived at by Stribeck's method. Thus — 

When the bearing is loaded the inner race approaches 
the outer race by an amount S„. The load on the ball a 
immediately in the line of loading is p^. The load on the 
adjacent balls b is less, because they are compressed a smaller 
amount than a, namely, 8j = 8„ cos aj. Similarly, 8„ = S^ cos a^. 
Let m be the number of balls in the bearing, then — 

360° , 2 X ■?6o° 

a„ = •s and a„ = -^ 

m m 



3 /pa 



Then 8„ = - 

32,000 
p 2 p 2 p s 
hence ^ = jT = Tl ~ ^^'^•' when d does not vary. 



P* = 



PA^ 



P„8jcosi(^°) 



8«^ 



' = P„ cos'^ 



\ m / 



and P„ = P, cos!' 2 






The total load on the bearing W is — 
W = P„ + 2P, + 2P, + etc. 

= P.[. + 4c«.i(f) + c.,i<3^) + e,c.}] 

Exapiples — 



m 


10 


IS 


20 


360 


36° 


24° 


18° 


W 

Pa 


2'2S 


3'44 


4-58 




4-38 


4-37 


4' 36 



Thus P„ = 



4-37W 



3oS 



Mechanics applied to Engineering. 



This expression is only true when there is no initial " shake " 
or " bind " in the bearing and no distortion of the ball races. 
To allow for such deficiencies Stribeck proposes — 

P„ = 5^and W = ^ 
m 5 

Necessity for Accuracy of Workmanship. — The 

expression for the approach of the plates shows how very 
small is the amount for ordinary working conditions. In a 
one-inch ball the approach is about j^ of an inch ; hence any 
combined errors, such as hills on the races or balls to the 
extent of j^ of an inch, will increase the load on the ball 
by 50 per cent. Extreme accuracy in finishing the races and 
balls is therefore absolutely essential for success. 

Friction of Ball Bearings. — The results of experiments 
tend to show that the friction of a ball bearing — 

(i) Varies directly as the load; 

(2) Is independent of the speed ; 

(3) Is independent of the temperature ; 

(4) The friction of rest is but very slightly greater than the 
friction of motion ; 

(5) Is not reduced by lubrication in a clean well-designed 
bearing. 

The following results, which may be regarded as typical, 
lend support to the statements (i), (2), (3) : — 



Load, in lbs. 
Friction moment, 
inch-lbs. 



1000 
S-o 



2000 
6-0 



3000 
8-4 



4000 

12-8 



Sooo 
lyo 



5ooo 

20'4 



7000 

25-2 



8030 

30-4 



9000 

36-0 



10,000 
40'o 



Speed, in revs, per 
min. (approx.)... 

Friction moment, 
inch-lbs. 



S 

20'0 



5° 
19-8 



100 

19-7 



500 
19-8 



800 

20'I 



1000 
20'0 



Temperature, Fahrenheit .. . 
Friction moment, inch-lbs. 



58° 
37-6 



65° 
38-1 



77° 
39-0 



39"o 



98° 
38-7 



The curves given in Fig. 303 were obtained by an auto- 
graphic recorder in the author's laboratory. They show clearly 



Friction. 



309 



lo'Ol 




L 


^earing 


001 

1. 


Same loads in all cases. 


Mail Searings 


/ 


0. 1 


.234 



SevobtUotu ofSlia/t 

Fig. 303. 



that the friction of rest in the case of a ball bearing is practically 
the same as the friction of motion, and that it is very much less 
than that of an ordinary bearing. 

Although the friction of a ball 
bearing is not reduced by lubri- 
cation, yet a small amount of 
lubrication is necessary in order 
to prevent rust and corrosion of 
the balls and races. Thick grease 
resembling vaseline, which has 
been freed from all traces of cor- 
rosive agents, is used by many 
makers ; others find that the best 
lard oil is preferable, but in any 
case great care must be taken to 
get a lubricant which will not 
set up corrosion of the balls and 
races. 

Cost of Ball Bearings.— 
The cost of a ball bearing or a 
first-class roller bearing is considerably greater than that of an 
ordinary bearing, but owing to the fact that they are more 
compact, and that the mechanical efficiency of a machine fitted 
with ball bearings is much higher than when fitted with ordinary 
bearings, a considerable saving in metal may be efiected by 
their use ; with the result that the first cost of some machines, 
such as electric motors, is actually less when fitted with ball 
bearings than with ordinary bearings. The quantity of lubricant 
required by a ball bearing is practically nil, and they moreover 
require practically no attention. Provided ball bearings are 
suitably proportioned for the load and speed, and are intelli- 
gently fitted and used, they possess great advantages over 
other types of bearings. 

Safe Loads and Speeds for Ball Bearings. — The 
following expressions are based on the results of a large 
number of experiments by the author : — 

W = The maximum load which may be placed on a bear- 
ing in pounds. 

m = The number of balls in the bearing. 

d = The diameter of the balls in inches. 

N = The number of revolutions per minute made by the 
shaft. 

D = The diameter of the ball race, measured to the 
middle of the balls, in inches. 



3IO Mechanics applied to Engineering. 

ND + C^ 
where K and C have the following values :— 



Type of bearing 


K 


C 


Cylindrical — no grooves 


1 ,000,000 


2000 


Cylindrical — grooved races 


2,000,000 to 
2,500,000 


2000 


Thrust*— no grooves 


500,000 


200 


Thrust — grooved races 


1 ,000,000 
1,250,000 


200 



Information on ball bearings can also be found in the fol- 
lowing publications : — Engineering, April 12, 1901 ; December 
26, 1902; February 20, 1903. Proceedings of the Institution 
of Civil Engineers, vols. Ixxxix. and clxxxix. " Machinery " 
Handbooks—" Ball Bearings." 

Friction of Lubricated Surfaces.— The laws which 
appear to express the behaviour of well-lubricated surfaces 
are almost the reverse of those of dry surfaces. For the sake 
of comparison, we tabulate them below side by side — 



Dry Surfaces, 

I. The frictional resistance is 
nearly proportional to the normal 
pressure between the two surfaces. 



2. The frictional resistance is 
nearly independent of the speed for 
low pressures. For high pressures 
it tends to decrease as the speed 
increases. 



Lubricated Surf cues. 

1. The frictional resistance is 
almost independent of the pressure 
with bath lubrication so long as 
the oil film is not ruptured, and 
approaches the behaviour of dry 
surfaces as the lubrication becomes 
meagre. 

2. The frictional resistance of a 
flooded bearing, when the tempera- 
ture is artificially controlled, in- 
creases (except at very low speeds) 
nearly as the speed, but when the 
temperature is not controlled the 
friction does not appear to follow 
any definite law. It is high at low 
speeds of rubbing, decreases as the 
speed increases, reaches a minimum 
at a speed dependent upon the tem- 
perature and the intensity of pres- 
sure ; at higher speeds it appears to 
increase as the square root of the 
speed ; and finally, at speeds of over 
3000 feet per minute, some believe 
that it remains constant. 



Friction. 



311 



3. The frictional resistance is not 
greatly affected by the temperature. 



4. The frictional resistance de- 
pends largely upon the nature of 
the material of which the rubbing 
surfaces are composed. 



5. The friction of rest is slightly 
greater than the friction of motion. 



6. When the pressures between 
the surfaces become excessive, seizing 
occurs. 



7. The frictional resistance is 
greatest at first, and rapidly de- 
creases with the time after the two 
surfaces are brought together, pro- 
bably due to the polishing of the 
surfaces. 

8. The frictional resistance is 
always greater immediately after 
reversal of direction of sliding. 



3. The frictional resistance de- 
pends largely upon the temperature 
of the bearing, partly due to the 
variation in the viscosity of the oil, 
and partly to the fact that the 
diameter of the bearing increases 
with a rise of temperature more 
rapidly than the diameter of the 
shaft, and thereby relieves the bear- 
ing of side pressure. 

4. The frictional resistance with 
a flooded bearing depends but 
slightly upon the nature of the 
material of which the surfaces are 
composed, but as the lubrication 
becomes meagre, the friction follows 
much the same laws as in the case 
of dry surfaces. 

5. The friction of rest is enor- 
mously greater than the friction of 
motion, especially if thin lubricants 
be used, probably due to their being 
squeezed out when standing. 

6. When the pressures between 
the surfaces become excessive, 
which is at a much higher pressure 
than with dry surfaces, the lubri- 
cant is squeezed out and seizing 
occurs. The pressure at which this 
occurs depends upon the viscosity 
of the lubricant. 

7. The frictional resistance is 
least at first, and rapidly increases 
with the time after the two surfaces 
are brought together, probably due 
to the partial squeezing out of the 
lubricant. 

8. Same as in the case of dry 
surfaces. 



The following instances will serve to show the nature of 
the experimental evidence upon which the above laws are 
based. 

I. The frictional resistance is independent of the pressure 
with bath lubrication. 



312 Mechanics applied to Engineering. 

Tower's Experiment 



[jubricant. 



Pressure in pounds per square inch. 



153 205 310 415 S20 625 



Frictional resistance in pounds. 



Olive oil ... 
Lard oil 
Sperm oil ... 
Mineral grease 



0-89 
0-90 
0*64 



0-87 


0-82 


0-84 


_ 


0-87 


o-8o ! 0-86 


090 


0-87 


0-87 


0-57 


o-SS 


0-S9 


— 


— 


1-27 


I -.35 


1-24 


I-I2 


I-I4 



1-25 



The results shown in Fig. 304 were, obtained from the 
author's friction testing machine. In the case of the '"oil 




4600 490P fZOO 1000 8O0 OOP fOV ZOO O 

LOfI DS IN POUNDS SO INCH 

Fig. 304, 

bath" the film was ruptured at a pressure of about 400 lbs. 
per square inch, after which the friction varied in the same 
manner as a poorly lubricated bearing. It is of interest to 
note that the friction of a dry bearing is actually less than that 
of a flooded bearing when the intensity of pressure is low. 



Friction. 



313 



2. The manner in which the friction of a flooded bearing 
varies with the velocity of rubbing is shown in Fig. 305. 
Curves A and B were obtained from a solid bush bearing 
such as a' lathe neck by Heiman {Zeitschrift des Vereines Deut- 
scher Ingenieure, Bd. 49, p. n6i). Curves C and D were 



1 










Ui/mf 


%'l 


A 


Hc,»*/,n 


*3 


20 


a 


HeiKAua 


43 


SO 


c 


Stmidcck 


57 


ss 


D 


STniaecK 


213 


25 


E 


GOOOIHAH 


75 


40 


, F 


GaaoMAH 


ISO 


40 




Uubbing Speeet: (Feet per TrUnuteJ. 
Fig. 305. 

obtained by Stribeck (Z des V. D. Ing., September 6, 7902) 
with double ring lubrication. Curves E and F were obtained 
by the Author with bath lubrication {Proceedings I. C. E. 
vol. clxxxix.). 

The erratic fashion in which 
the friction varies is due to many 
complex actions, which have not 
as yet been reduced to rigid 
mathematical treatment, although 
Osborne Reynolds, Sommerfeld, 
Petrofif, and others have done 
much excellent work in this direc- 
tion. An examination of the problem on the assumption that 
the friction is due to the shearing of a viscous film of oil of 
uniform thickness is of interest, although it does not give 
results entirely in accord with experiments. 




Fig. 306. 



314 Mechanics applied to Engineering. 

In the theory of the shear of an elastic body we have the 
relation (see page 376) — 

/ G AG 

where/, is the intensity of shear stress. 
F, is the total resistance to shear. 
A is the cross-sectional area of the element sub- 
jected to shear. 
G is the modulus of rigidity. 

But in the case of viscous fluids in which the resistance to flow 
varies directly as the speed S, we have — 

S_/. _ F. AKS 

7=K-AK ^"'^ ^'^—r 

where K is the coefficient of viscosity. 

When a journal runs in a solid cylindrical bush of diameter 
d and length L with a film of oil of uniform thickness inter- 
posed, the friction of the journal is — 

If W be the load on the journal, and /x be the coefficient of 
friction, then 

_F, _x^LKS 
'* ~ W ~ W/ 

lip be the nominal intensity of pressure— 

W 

hence M=— = - 

From this relation we should expect that the coefficient of 
friction in an oil-borne brass would vary directly as the speed, 
as the coefficient of viscosity, also inversely as the intensity of 
pressure and as the thickness of the film. But owing to dis- 
turbing factors this relation is not found to hold in actual 
bearings. Osborne Reynolds and Sommerfeld have pointed 
out that the thickness of the oil film on the " on " side of a 
brass is greater than on the "off" side. The author has 



Friction. 3 1 g 

experimentally proved that this is the case by direct measure- 
ment, and indirectly by showing that the wear on the " off" 
side is greater than on the " on " side. The above-mentioned 
writers have also shown that the difference in the thickness on 
the two sides depends on the speed of rotation, the eccentricity 
being greatest at low speeds. Reynolds has shown that the 
friction increases with the eccentricity ; hence at low velocities 
the effect of the eccentricity is predominant, but as the speed 
increases it diminishes with a corresponding reduction in the 
friction until the minimum value is reached (see Fig. 305). 
After the minimum is passed the effect of the eccentricity 
becomes less important, and if the temperature of the oil film 
remained constant the friction would vary very nearly as the 
speed, but owing to the fact that more heat is generated in 
shearing the oil film at high speeds than at low the tempera- 
ture of the film increases, and thereby reduces the viscosity of 
■the oil, also the friction with the result that it increases less 
rapidly than a direct ratio of the speed. Experiments show 
that when the temperature is not controlled the friction varies 
more nearly as the square root of the speed. This square 
root relation appears to hold between the minimum point and 
about 500 feet per minute ; above that speed it increases less 
rapidly than the square root, and when it exceeds about 3000 
feet per minute the friction appears to remain constant at all 
speeds. For a flooded bearing in which the temperature is 
not controlled the friction appears to follow the law — 

between the above-mentioned limits. 

The fact must not be overlooked that the deviation from 
the straight line law of friction and speed after the minimum 
is passed is largely due to the fact that the temperature of the 
film does not remain constant. In some tests made by the 
author in 1885 {Proceedings Inst. C. Engineers, vol. Ixxxix., 
page 449) the friction was found to vary directly as the speed 
when the temperature was controlled by circulating cooling 
water through the shaft. The speeds varied from 4 to 200 
feet per minute with both bath and pad lubrication, and with 
brasses embracing arcs from 180° to 30°. Tests of a similar 
character made on white metals on a large testing machine 
also showed the straight line law to hold between about 15 
and 1000 feet per minute with both bath and pad lubrication. 



3i6 Mechanics applied to Engineering. 

Where the temperature of the bearing is controlled, the 
friction-speed relation appears to closely agree with that de- 
duced above from viscosity considerations, viz. — 

<rS 
3. The curves given in Fig. 307 show the relation between 



8» 







5 50 



(3 
u 

i" 30 



• Specimen A. Temp&rautLu-a conjtroU£d.a£ 120°F. 

C Specimen A. Tempsralure ctSaiuettto vafi/. 

+ Specimen fl. Temperatur-e controUeoLat t20'*F. 

4 Speoimen B. TemperaUi-re aJiojuedt to vary . 




SCfFt 



-L. 



2.000 4.000 6,000 a,ooo fO,00O /z,ooo /C,OOCf 

Lo(z<3^ 07V Meexj^iTxg : Pounds ■ 
Fig. 307. 

the friction and the temperature. When other conditions are 
kept constant the relation between the coefficient of friction 



Friction. 



317 



ju. and the temperature t may be represented approximately 
by the empirical expression — 

constant 

where /a, is the coefficient of friction at the temperature t F. 
Thus, if the coefficient at 60° F. is o'oi76 the constant is 0-332, 
and the coefficient at 120° F. would be 0-005 r. 
Tower showed that the relation 

constant 

^' = — T~ 

closely held for many of his tests. 

When making comparative friction tests of bearing metals 
it is of great importance to control the temperature of the 
bearing at a predetermined point for all the metals under test. 

In all friction testing machines provision should be made 
for circulating water through the shaft or the bearing for coii • 
trolling the temperature. 

The curves in Fig. 307- shows the effect of controlling the 




TemperaUtre in. Degrees Fakrenheit ft) 
Fig. 3070:. 



temperature when testing white antifriction metals, also of 
allowing it to vary as the test proceeds. 

4. Mr. Tower and others have shown that in the case of a 



3i8 



Mechanics applied to Engineering. 



flooded bearing there is no metallic contact between the shaft 
and bearing ; it is therefore quite evident that under such 
circumstances the material of which the bearing is composed 
makes no difference to the friction. When the author first 
began to experiment on the relative friction of antifriction 
metals, he used profuse lubrication, and was quite unable to 
detect the slightest difference in the friction ; but on using the 
smallest amount of oil consistent with security against seizing, 
he was able to detect a very great deal of difference in the 
friction. In the table below, the two metals A and B only 
differed in composition by changing one ingredient, amounting 
to 0-23 per cent, of the whole. 



Load in lbs. sq. inch 

Coefficient of (A 

friction tB 



150 
0-0143 
0-0083 


250 

0-OII2 
0-0062 


3S0 
0-0091 
0-0054 


450 

0-0082 
o;ooso 


750 
0-0075 
0-0045 



950 

0-0083 
0-0047 



5. The following tests by Thurston will show how much 
greater is the friction of rest than of motion : — 



Load in lbs. sq. inch 

Coefficient of j A^°inst^t"'\ 
fr"="°" I of starting/ 



SO 
0-013 


100 
o-oo8 


250 
0-005 


500 
0-004 


750 
0-0043 


0-07 


0-I3S 


0-14 


0-15 


0-185 



1000 
0-009 

0-18 



Oil used, sperm. 

The ratio between the starting and the running coefficients 
depends largely upon the viscosity of the oil, as shown by the 
following tests by the author. See Proceedings Inst. C. E., 
vol. Ixxxix. p. 433. 





Coefficient of friction. 






Running. 


At starting. 




Machinery oil 

Thick valve oil |g 
Grease 


0-0084 
0-0329 
0-0252 
0-0350 


0-192 
O-171 
0-147 
0-090 


22-9 
S-2 

S-8 
2-6 



Friction. 319 

6. Experiments by Tower and others show that a steel 
shaft in a gun-metal bearing seizes at about 600 lbs. square 
inch under steady running, whereas when dry the same materials 



ft 

9 


No load. 


§ 1^ IdOCt<i' SOO lh& so inch' 


s 


■> ■/' 




Time , 


6 M, Second « / inAih, 




Fin. 308. 

seize at about 80 lbs.' square inch. The author finds that the 
seizing pressure increases as the viscosity of the oil increases. 

7. Fig. 308 is one of many drawn autographically on 
the author's machine. The lever which applies the load 
on the bearing was lifted, and the machine allowed to run 
with only the weight of the bearing 
itself upon it ; the lever was then 
suddenly dropped, the friction 
being recorded automatically. 

An indirect proof of this state- 
ment is to be found in the case of 
connecting-rod ends, and on pins 
on which the load is constantly 
reversed ; at each stroke the oil is 
squeezed away from the pressure fig. 309. 

side of the pin to the other side. 

Then, when the pressure is reversed, there is a large supply of 
oil between the bearing and the pin, which gradually flows to 
the other side. Hence at first the bearing is floating on oil, and 
the friction is consequently very small ; as the oil flows away, the 
friction increases. This is the reason why a much higher 
bearing pressure may be allowed in the case of a connecting- 
rod end than in a constantly revolving bearing. 

8. In friction-testing machines it is always found that the 
temperature and the friction of a bearing is higher after reversal 
of direction, but in the course of a few hours it gets back to 
the normal again. Some metals, however, appear to have a 

-grain, as the friction is always much greater when running one 
way than when running the other way. 

Nominal Area of Bearing. — The pressure on a cylin- 
drical bearing varies from point to point; when the lubrication 



320 



Mechanics applied to Engineering. 




is very meagre or with a dry bearing it is a maximum at the 
crown, and is least at the two sides. When the bearing is 
flooded with oil the distribution of 
pressure can be calculated from hydro- 
dynamical principles, an account of 
which will be found in Dr. Nicolson's 
paper, " Friction and Lubrication," 
read before the Manchester Associa- 
tion of Engineers, November, 1907. 
Fig. 310. For the purpose of comparing 

roughly the intensity of pressure on 
two bearings, the pressure is assumed to be evenly distri- 
buted over the projected area of the bearing. Thus, if w be 
the width of the bearing across the chord, and / the length 
of the bearing, the nominal area is wl, and the nominal pressure 

W 
per square inch is — „ where W is the total load on the bearing. 

Beauchamp Tower's Experiments. — These experi- 
ments were carried out for a research committee of the 



Centre 




Fig. 311. 



Institution of Mechanical Engineers, and deservedly hold the 
highest place amongst friction experiments as regards accuracy. 
The reader is referred to the Reports for full details in the 
Institution Proceedings, 1885. 

Most of the experiments were carried out with oil-bath 
lubrication, on account of the difficulty of getting regular 
lubrication by any other system. It was found that the 
bearing was completely oil-borne, and that the oil pressure 



Friction. 



321 



varied as shown in Fig. 3CI, the pressure being greatest on 
the "off" side. In this connection Mr. Tower shows that 
it is useless — worse than useless — to drill an oil-hole on the 
resultant line of pressure of a bearing, for not only is it 
irnpossible for oil to be fed to the bearing by such means, but 
oil is also collected from other sources and forced out of the 
hole (Fig. 312), thus robbing the bearings of oil at exactly the 
spot where it is most required. If oil-holes are used, they must 
communicate with a part of the bearing where there is little 
or rio pressure (Fig. 313). The position of the point of 
minimum pressure depends somewhat on the speed of 
rotation. 





Fig. 312. 



Fig. 313. 



A general summary of the results obtained by Mr. Tower 
are given in the following table. The oil used was rape ; the 
speed of rubbing 150 feet per minute; and the temperature 
about 90° F. : — 



Form of bear- 
ing. 



o 



Load at which 
seizing occur- 
red, in lbs. 
sq. inch. 

Coefficient of 
friction 



150 



370 
0'Oo6 



55° 
0'Oo6 



600 




90 
0-035 



Other of Mr. Tower's experiments are referred to in 
preceding and succeeding paragraphs. 

Professor Osborne Reynolds' Investigations, — A 
theoretical treatment of the friction of a flooded bearing has 
been investigated by Professor Osborne Reynolds, a full 

Y 



322 



Mechanics applied to Engineering. 



account of which will be found in the Philosophical Transac- 
tions, Part I, 1886; see also his published papers, Vol. II. 
p. 228. In this investigation, he has shown a complete agree- 
ment between theory and experiment as regards the total 
frictional resistance of a flooded bearing, the distribution of 
oil pressure, and the thickness of the oil film, besides many 
other points of the greatest interest. Professor Petroff, of 
St. Petersburg and Sommerfeld have also done very similar 
work. A convenient summary of the work done by the 
above-mentioned writers- will be found in Archbutt and Deeley's 
" Lubrication and Lubricants." 

The theory of the distribution of oil pressure in a 
flooded collar bearing has been recently investigated by 
Michell, who has successfully applied it to the more difficult 
problem of producing an oil-borne thrust bearing. See a 
paper by Newbigin in the Proceedings of the Institution of 
Civil Engineers, Session 1913-1914; also Zeitschrift fiir 
Mathematik und Physik, Vol. 52, 1905. 

Goodman's Experiments. — The author, shortly after 
the results of Mr. Tower's experiments were published, repeated 
his experiments on a much larger machine belonging to the 
L. B. & S, C. Railway Company ; he further found that the 
oil pressure could only be registered when the bearing was 
flooded ; if a sponge saturated with oil were applied to the 
bearing, the pressure was immediately shown on the gauge, but 
as the oil ran away and the supply fell oflf, so the pressure fell. 
In another case a bearing was provided with an oil-hole 
on the resultant line of pressure, to which a screw-down valve 
was attached. When the oil-hole was open the friction on the 
bearing was very nearly 25 per cent, 
greater than when it was closed 
and the oil thereby prevented from 
escaping. 

Another bearing was fitted with 
a micrometer screw for the purpose 
of measuring the thickness of the oil 
film ; in one instance, in which the 
conditions were similar to those 
assumed by Professor Reynolds, the 
thickness by measurement was found 
to be 0-0004 inch, and by his calcu- 
lation o'ooo6 inch. By the same appliance the author found 
that the thickness was greater on the " on " side than on the 
" off" side of the bearing. The wear always takes place where 
the film is thinnest, i.e. on the " off" side of the bearing 




Fig. 314. 



Friction. 



323 



exactly the reverse of what would be expected if the shaft were 
regarded as a roller, and the bearing as being rolled forwards. 
When white metal bearings are tested to destruction, the metal 
always begins to fuse on the " off " side first. 

The side on which the wear takes place depends, however, 
upon the arc of the bearing in contact with the shaft. When 
the arc subtends an angle greater than about 90° (with white 
metal bearings this angle is nearer 60°) the wear is on the off 
side ; if less than 90°, on the "on" side. This wear was measured 
thus : The four screws, a, a, a, a, were fitted to an overhanging lip 




- J?iarrL o^ 




Fig. 315. 



O-ZS O'J 0-7J t-O 

Cfwrcts irv cemtact 

Fig. 316. 



on the bearing as shown. They were composed of soft brass. 
Before commencing a run, they were all tightened up to just 
touch the shaft; on removing the bearing after some weeks' 
running, it was seen at once which screws had been bearing and 
which were free. 

Another set of experiments were made in 1885, to ascertam 
the effect of cutting away the sides of a bearing. The bearings 
experimented upon were semicircular to begin with, and the 
sides were afterwards cut away step by step till the width of 
the bearing was-only i d. The effect of removing the sides is 
shown in Fig. 316. 



324 



Mechanics applied to Engineering. 



The relation may be expressed by the following empirical 
formula : — 

Let R = frictional resistance ; 

_ width of chord in contact 
diameter of journal 
K and N are constants for any given bearing. Then — 
R = K + NC 

Methods of Lubricating. — In some instances a small 
force-pump is used to force the oil into the bearing ; it then 
becomes equivalent to bath lubrication. Many high-speed 




Fig. 318.— Collar bearing. 



Fig. 319.— Pivot or footstep. 



engines and turbines are now lubricated in this way. The oil 
is forced into every bearing, and the surplus runs back into 



Friction. 325 

a receiver, where it is filtered and cooled. When forced 
lubrication is adopted with solid bushes, or in bearings in 
which the load constantly changes in direction, as in connect- 
ing rods, the clearance must be considerably greater than when 
meagre lubrication is supplied. The clearances usually adopted 
are about one thousandth of the diameter of the shaft for 
ordinary lubrication, and rather more for flooded bearings. 

Relaiive Friction of Different Systems of Luhrication. 



Mode of lubrication. 


Tower. 


Goodman. 


Bath 

Saturated pad 
Ordinary pad 
Syphon 


I '00 

6'48 
7-06 


I '00 

1-32 
2-21 
4'20 



Seizing of Bearings. — It is well known that when a 
bearing is excessively loaded, the lubricant is squeezed out, 
and the friction takes place between metal and metal ; the two 
surfaces then appear to weld themselves together, and, if the 
bearing be forced round, small pieces are torn out of both 
surfaces. The load at which this occurs depends much upon 
the initial smoothness of the surfaces and upon the nature of the 
material, but chiefly upon the viscosity of the oil. If only the 



Fig. 320. 



viscosity can be kept up by artificially keeping the bearing cool, 
by water-circulation or otherwise, the surfaces will not seize 
until the pressure becomes enormous. The author has had a 
bearing running for weeks under a load ol two tons per square 
inch at a surface velocity of 230 feet per minute with pad 
lubrication, temperature being artificially kept at 110° Fahr. by 
circulating water through the axle.' 

Seizing not unfrequently occurs through unintentional high 
pressures on the edges of bearings. A very small amount of 

' In another instance nearly four tons per square inch for several 
hours. 



326 



Mechanics applied to Engineering. 



spring will cause a shaft to bear on practically the edge of the 
bearing (Fig. 320), and thereby to set up a very intense pressure. 

This can be readily avoided by 
using spherical-seated bearings. 
For several examples of such 
bearings the reader is referred to 
books on " Machine Design." 

Seizing is very rare indeed with 
soft white metal bearings ; this is 
probably due to the metal flowing 
and adjusting itself when any un- 
even pressure conies upon it. 
This flowing action is seen clearly 
in Fig. 321, which is from photo- 
graphs. The lower portion shows 
the bearing before it was tested in 
a friction-testing machine, and the 
upper portion after it was tested. 
The metal began to flow at a 
temperature of 370° Fahr., under a 
pressure of 2000 lbs. per square 
inch : surface speed, 2094 feet per 
minute. The conditions under 
which the oil film ruptures in a 
flooded bearing will be discussed 
shortly. 

Bearing Metals. — If a 
bearing can be kept completely 
oil-borne, as in Tower's oil-bath 
experiments, the quality of the 
bearing metal is of very little 
importance, because the shaft is not in contact with the bear- 
ing; but unfortunately, such ideal conditions can rarely be 
ensured, hence the nature of the bearing metal is one of great 
importance, and the designer must very carefully consider the 
conditions under which the bearing will work before deciding 
upon what metal he will use in any given case. Before going 
further, it will be well to point out that in the case of bearings 
running at a moderate speed under moderate loads, practically 
any material will answer perfectly ; but these are not the cases 
that cause anxiety and give trouble to all concerned : the really 
troublesome bearings are those that have to run under extremely 
heavy loads or at very high speeds, and perhaps both. 

The first question to be considered is whether the bearing 




Fig. 32X. 



Friction. 327 

will be subjected to blows or not; if so, a hard tough metal must 
be used, but if not, a soft white metal will give far better 
frictional and wearing results than a harder metal. It would be 
extremely foolish to put such a metal into the connecting-rod 
ends of a gas or oil engine, unless it was thoroughly encased to 
prevent spreading, but for a steadily running journal nothing 
could be better. 

The following is believed to be a fair statement of the 
relative advantages and disadvantages of soft white metal for 
bearings : — • 

Soft White Metals for Bearings. 
Advantages. Disadvantages. 

The friction is much lower than Will not stand the hammering 

with hard bronzes, cast-iron, etc., action that some shafts are sub- 
hence it is less liable to heat. jected to. 

The wear is very small indeed The wear is very rapid at first 

after the bearing has once got well if the shaft is at all rough ; the 
bedded (see disadvantages). action resembles that of a new file 

on lead. At first the file cuts 
rapidly, but it soon clogs, and then 
ceases to act as a file. 

It rarely scores the shaft, even if It is liable to melt out if the 

the bearing heats. bearing runs hot. 

It absorbs any grit that may get If made of unsuitable material 

into the bearing, instead of allowing it is liable to corrode, 
it to churn round and round, and so 
cause damage. 

As far as the author's tests go, amounting to over one 
hundred different metals on a 6-inch axle up to loads of 
10 tons, and speeds up to 1500 revolutions per minute, he finds 
that ordinary commercial lead gives excellent results under 
moderate pressure : the friction is lower than that of any other 
metal he has tested, and, provided the pressure does not greatly 
exceed 300 lbs. per square inch, the wear is not excessive. 

A series of tests made by the author for the purpose of 
ascertaining the effect of adding to antifriction alloys small' 
quantities of metals whose atomic volume differed from that 
of the bulk, yielded very interesting results. The bulk metal 
under test consisted of lead, 80; antimony, 15 ; tin, 5 ; and 
the added metal, 0*25. With the exception of one or two 
metals, which for other reasons gave anomalous results, it was 
found that the addition of a metal whose atomic volume was 
greater than that of the bulk caused a diminution in the friction, 
whereas the addition of a metal whose atomic volume was less 
than that of the bulk caused an increase in the friction, and 



328 



Mechanics applied to Engineering. 



metals of the same atomic volume had apparently no effect on 
the friction. 

All white metals are improved if thoroughly cleaned by 
stirrjng in sal ammoniac and plumbago when in a molten state. 
Area of Bearing Surfaces. — From our remarks on 
seizing it will be evident that the safe working pressure for 
revolving bearings largely depends upon their temperature and 
the lubricant that is used. If the temperature rise abnormally, 
the viscosity of the oil is so reduced that it gets squeezed out. 
The temperature that a bearing attains to depends (i) on the 
heat generated; (2) on the means for conducting away the 
heat. 

Let S = surface speed in feet per minute ; 
W = load on the bearing in pounds ; 
/„ = number of thermal units conducted away per 
square inch of bearing per minute in order to 
keep the temperature down to the desired limit. 

u,WS 
The thermal units generated per minute = - — 

773 
The nominal area of bearing surfacel _ A^WS 
in square inches, viz. dh J "~ 7734 

As a first approximation the following values of /a, and 4 
may be assumed : — 

VaLDES of /i AND tu. 
Method of lubrication. Value of i^ 

Bath o'004 

Pad 0'0i2 

Syphon 0020 





Values of tu. 


Conditions of running. 


Crank and 


Continuous 


Crank and 


Continuous 




other pins. 


running 
bearing. 


other pins. 


running 
bearing. 




Maximum temperature of bearing. 




140° F. 


140° F. 


100° F. 


100° F. 


Exposed to currents of cold air 










or other means of cooling, 










as in locomotive or car axles 


4-7 


«-is 


2-3 -S 


0-5-0-75 


In tolerably cool places, as in 










marine and stationary en- 










gines 


0-75-I 


0-3-0-5 


0-4-O-S 


0-15-0-25 


In hot places and where heat is 










not readily conducted away 


0-4-o-S 


0-I-0-3 


0-2-0-25 


~ 



Friction. 



329 



After arriving at the area by the method given above, it 
should be checked to see that the pressure is not excessive. 



Bearing. 
Crank-pins. — Locomotive ... 

Marine and stationary 

Shearing machines 

Gudgeon pins. — Locomotive 

Marine and stationary 
Railway car axles 
Ordinary pedestals, — Gun-metal 

Good white metal 
Collar and thrust bearings, — Gun-metal 

Good white metal 
Lignum vilse ... 
Slide blocks. — Cast-iron or gun-metal 

Good white metal 
Chain and rope pulleys for cranes, — Gun-metal bush 



aximum permis- 
sible pressure in 
'bs. per sq. inch. 
1500 
600 
3000 
ZOOO 
800 

200 

500 

80 

200 

SO 

80 

250 

1000 



Work absorbed in Revolving Bearings. 

Let W = total load on bearing in pounds ; 
D = diameter of bearing in inches ; 
N = number of revolutions per minute ; 
L = length of journal in inches. 

For Cylindrical Bearings. — 

Work done per minute"! _ /uWttDN 
in foot-pounds j ~ 

horse-power absorbed = 



12 
WttDN/* 



_ /itWDN 

12 X 33)000 126,000 



A convenient rough-and-ready estimate of the work absorbed 
by a bearing can be made by assuming that the frictional 
resistance F on the surface of a bearing is 3 lbs. per square 
inch for ordinary lubrication, 2 lbs. for pad, r lb. for bath, the 
surface being reckoned on the nominal area. 

Work done in overcoming the friction \ _ ttD^LFN 
per minute in foot-pounds / "~ i^ 

Flat Pivot. — If the thrust be evenly distributed over the 
whole surface, the intensity of pressure is — 

' 7rR» 



330 



Mechanics applied to Engineering. 



pressure on an elementary ring = 2irrp . dr 
moment of friction on an elementary ring = iirr^iip . dr 
moment of friction on whole surface = 2irfjipfr' . dr 

3 

Substituting the value of/ from above — 

M, =|/x,WR 

work done per minute in foot-pounds = 

5 73 
ftWDN 
189,000 




horse-power absorbed: 



This result might have been arrived at 
thus : Assuming the load evenly distributed, 
the triangle (Fig. 323) shows the distribution of 
pressure, and consequently the distribution of 
the friction. The centre of gravity of the 
triangle is then the position of the resultant 
friction, which therefore acts at a radius equal 
to f radius of the pivot. 

If it be assumed that the unequal wear of 
the pivot causes the pressure to be unevenly 
distributed in such a manner that the product 
of the normal pressure / and the velocity of 
rubbing V be a constant, we get a different 
Fig. 322. value for M,; the f becomes \. It is very 

uncertain, however, which is the true value. The same remark 
also applies to the two following paragraphs. 

Collar Bearing (Fig. 325).— By similar 
reasoning to that given above, we get — 




Fig. 323. 



Moment of friction 1 _ 
on collar J ~ "^^^ 

M _ 2/^W(R,^ - R33) 
3(Ri" - R2=) 



jr=R, 



dr 



Conical Pivot. — ^The intensity of pressure p all over the 
surface is the same, whatever may be the angle a. 

Let Po be the pressure acting on one half of the cone — 

VV 
a sm a 



Friction. 



331 



The area of half the surface of the cone is — 

■rRL_ ttR' 
2 2 sin a 



A = : 



. _ Po _ W . 2 sin a _ W _ weight 
A 2sina.irR^ jtR^ projected area 




Fig. 324. 




Fig. : 



Total normal pressure on any elementary ring = zirrp . dl 
moment of friction on elementary ring = zttz-V/ ■ dl 
/, ,, dr \ iirr^apdr 

\ sin a) 



sm a 

Sirixp i 



moment of friction on whole surface = • fr^ . dr 

sm a 



M,= 



27r/t/R° 

3 sin a 



Substituting the value of/, we have — 

2/iWR 



M,= 



J sin o 



The angle a becomes 90°, and sin » = i when the pivot 
becomes flat. 

By similar reasoning, we get for a truncated conical pivot 
(Fig. 326)— 



2/.W(R,^ - R,°) 
*^^'- 3 sin a(R,2 - R,') 



332 



Mechanics applied to Engineering. 



Schiele's Pivot and Onion Bearing (Figs. 327, 328). — 
Conical and flat pivots often give trouble through heating, pro- 
baBly due to the fact that the wear is uneven, and therefore the 
contact between the pivot and step is imperfect, thereby giving 
rise to intense local pressure. The 
object sought in the Schiele pivot is 
to secure even wear all over the pivot. 
As the footstep wears, every point 
in the pivot will sink a vertical dis- 
tance h, and the point a sinks to «i, 
where aa^ = h. Draw ab normal to 
the curve at a, and ac normal to the 
axis. Also draw ba^ tangential to 
the dotted curve at b, and ad to the 
full-lined curve at a ; then, if h be 
taken as very small, ba-^ will be 
practically parallel to ad, and the 
two triangles aba-^ and acd will be 
practically similar, and — 




Fig. 326. 



ad aa-, , ac Y. aa, 

— =— 2, 01 ad = -' 

ac ba ba 



or ad = 



ba 



But ba is the wear of the footstep nornial to the pivot, which is 
usually assumed to be proportional to the friction F between 
the surfaces, and to the velocity V of rubbing ; hence — 

ba 00 FV 00 f>.p . 27rrN 
or ba = Vi-iJ-pr 

where K is a constant for any given speed and rate of wear ; 
hence — 



ad : 



K/u//- K/t/ 



But h is constant by hypothesis, and /* is assumed to be constant 
all over the pivot; / we_have already proved to be constant 
(last paragraph) ; hence ad, the length of the tangent to the 
curve, is constant; thus, if the profile of a pivot be so con- 
structed that the length of the tangent ad = the constant, the 
wear will be (nearly) even all over the pivot. Although our 



Friction. 



333 



assumptions are not entirely justified, experience shows that 
such pivots do work very smoothly and well. The calculation 
of the friction moment is very similar to that of the conical 
pivot. 





Fig. 327. Fig. 328. 

The normal pressure at every point is — 



weight _ 



W 



projected area 7r(Ri'' — R^) 

By similar reasoning to that given for the conical pivot, we 
have — 

Moment of friction on an elementary^ _ 2-irr'ii.pdr 
ring of radius r ) iuTo" 

(but -J^ = /) = 2TTtu.prdr 

\ sm a / '^ 

and moment of friction for whole pivot = 2Ttt)x.p r .dr 



Mf = 2-jr ffip 



^1' - R2' 



Substituting the value of/, M, = Wju/" 



334 Mecltanics applied to Engineering, 

The onion bearing shown in the figure is simply a Schiele 
pivot with the load suspended from below. 

Friction of Cup Leathers. — The resistance of a 
hydraulic plunger sliding through a cup leather has been 
investigated by Hick, Tuit, and others. The formula proposed 
by Hick for liie friction of cup leathers does not agree well 
with experiments ; the author has therefore recently tabulated 
the results of published experiments and others made in his 
laboratory, and finds that tiie following formulae much more 
nearly agree with experiment : — 

Let F = frictional resistance of a leather in pounds per 
square inch of water-pressure ; 
d = diameter of plunger in inches ; 
p = water-pressure in pounds per square inch. 

Then F = o"o8/ -I — -j- when in good condition 

F = 0-08/ + ^ „ bad 

Efficiency of Machines. — In all cases of machines, the 
work supplied is expended in overcoming the useful resistances 
for which the machine is intended, in addition to the useless or 
frictional resistances. Hence the work supplied must always 
be greater than the useful work done by the machine. 

Let the work supplied to the machine be equivalent to 
lowering a weight W through a height h ; 

the useful work done by the machine be equivalent to 
raising a weight W„ through a height /«„ ; 

the work done in overcoming friction be equivalent to 
raising a weight W, through a height A^. 

Then, if there were no friction — 

Supply of energy = useful work done 
W/4 = W„/5„ 

or mechanical advantage = velocity ratio 

When there is friction, we have — 

Supply of energy = usefiil work done -f- work wasted in friction 
W/t = WA + W/, 



Friction. 335 

and — 

, L ■ 1 a: ■ useful work done 

the mechanical efticiency = — — , 7—= 

total work done 

the work sot out 

or = — i ;-e — 

the work put in 

Let t] = the mechanical eflficiency ; then — 

„ _ WA ^^ WA 



' -wh ' WA + w/, 

Tj is, of course, always less than unity. The " counter- 
efficiency " is -, and is always greater than unity. 

Reversed Efficiency. — When a machine is reversed, for 
example, when a load is being lowered by lifting-tackle, the 
original resistance becomes the driver, and the original driver 
becomes the resistance ; then — 

_ A cc • useful work done in lifting W through h 

Reversed efficiency = — = =-^i r-- . °„, , °, , 

total work done in lowering W„ through h„ 

W/^ _ WA-W/^ 

''' W„/5„ WA 

When W acts in the same direction as W„, i.e. when the 

machine has to be assisted to lower its load, i;, takes the 

negative sign. In an experiment with a two-sheaved pulley 

block, the pull on the rope was 170 lbs. when lifting a weight 

J. 

of 500 lbs.; the velocity ratio in this case R = -^ = ^. 

TV,»„ - ^"'''« - 5°° X I _„.-,. 

1 hen 17 = ,^r- = =07^5 

' W/4 T70 X 4 •^ 

The friction work in this case y^/h, was 170 x 4 — 500 X 1 

= 180 foot-lbs. Hence the reversed efficiency w, = 

500 

= o'64, and in order to lower the 500 lbs. weight gently, the 

backward pull on the rope must be — 

-—- X 0*64 = 80 lbs. 
4 

If the 80 lbs. had been found by experiments, the reversed 
efficiency would have been found thus — 

80 X 4 ./- 

m, = 3l = 0-64 

500 X I 



336 Mechanics applied to Engineering. 

The reversed efficiency must always be less than unity, and 
may even become negative when the frictional resistance of 
the machine is greater than the useful resistance. In order to 
lower the load with such a machine, an additional force acting 
in the same sense as the load has to be applied ; hence such a 
machine is self-sustaining, i.e. it will not run back when left to 
itself. The least frictional resistance necessary to ensure that it 
shall be self-sustaining is when W/i, = W„^„ ; then, substituting 
this value in the efficiency expression for forward motion, we 
have — 

Thus, in order that a machine which is not fitted with a 
non-return mechanism may be self-supporting its etficiency 
cannot be over 50 per cent. This statement is not strictly 
accurate, because the frictional resistance varies somewhat 
with the forces transmitted, and consequently is smaller when 
lowering than when raising the load ; the error is, however, 
rarely taken into account in practical considerations of 
efficiency. 

This self-supporting property of a machine is, for many 
purposes, highly convenient, especially in hand-lifting tackle, 
such as screw-jacks, Weston pulley blocks, etc. 

Combined Efficiency of a Series of Mechanisms. — 
If in any machine the power is transmitted through a series 
of simple mechanisms, the efficiency of each being jj„ j/j, %, 
etc., the efficiency of the whole machine will be — 



>SJ^ 17 = iji X r;a X %, etc. 



If the power be transmitted through n mechan- 
isms of the same kind, each having an efficiency iji, 
the efficiency of the whole series will be approxi- 
mately — ■ 

p Hence, knowing the efficiency of various simple 
' mechanisms, it becomes a simple matter to calculate 

with a fair degree of accuracy the efficiency of any 

complex machine. 

Efficiency of Various Machine Elements. 
Fra. 329. Pulleys.— In the case of a rope or chain pass- 

ing over a simple pulley, the frictional resistances 
are due to (i) the resistance of the rope or chain to bending ; 
(2) the friction on the axle. The first varies with the make. 



W 



Friction. 



337 



size, and newness of rope ; the second with the lubrication. 
The following table gives a fairly good idea of the total 
eflficiency at or near full load of single pulleys j it includes 
both resistances i and 2 : — 



Diameter of rope 

,, . f Clean and well oiled 
Maximum U; 

efficiency! Clean and well oiled, 1 
per cent, y ^j^j^ gjiij- ne,,. jopg ] 



i in. 
96 
94 



Jin. 
93 
9> 

91 



91 
89 



i\ in.r 
88 
86 



chain. 
95-97 
93-96 



These figures are fair averages of a large number of 
experiments. The diameter of the pulley varied from 8 to 1 2 
times the diameter of the rope, and the diameter of the pins 
from \ inch to i^ inch. 

It is useless to attempt to calculate the efficiency with any 
great degree of accuracy. 

Pulley Blocks.' — When a number of pulleys are combined 

for hoisting tackle, the ,^\\\\^^\^^^^^^^^\\^^Kx\\^^^^^^^ 

efficiency of the whole 

maybe calculatedapproxi- 

mately from the known 

efficiency of the single 

pulley. The efficiency of 

a single pulley does not 

vary greatly with the load 

uqless it is absurdly low ; 

hence we may assume that 

the efficiency of each is 

the same. Then, if the 

rope passes over n pulleys, 

each having an efficiency 

1J1, we have the efficiency 

of the whole — 

The following table 
will serve to show how 
the efficiency varies in different pulley blocks. 




338 



Mechanics applied to Engineering. 





Single 


pulley. 


Two-sheaved. 


Three-sheaved. 














pounds. 


Old l-in. 


New J-in. 


Old i-in. 


New i-in. 


Old }-in. 


New J-in. 




rope. 


rope. 


rope. 


rope. 


rope. 


rope. 


«4 


94 


90 








_ 


_ 


28 


94-S 


90-5 


80 


75 


30 


24 


56 


95 


91 


84 


78-S 


5° 


35 


112 


96 


92 


86 


91-5 


60 


41 


i6g 






87-S 


93 


65 


44 


224 


— 


— 


89 


93 


69 


47 


280 


— 


— 


90 


94 


72 


50 


336 


— 


— 


— 


— 


74 


53 


448 








— 


78 


56 



Weston Pulley Block. — This is a modification of the 
old Chinese windlass ; the two upper pulleys are rigidly 
attached ; the radius of the smaller one is r, 
and of the larger R. Then, neglecting fric- 
tion for the present, and taking moments 
about the axle of the pulleys, we have — 

2 2 

w 
-(R - ^) = PR 

2 




and the velocity ratio- 

W 
P 



v, = — = 



2R 
R- 



The pulleys are so chosen that the velocity 
ratio is from 30 to 40. The efficiency of 
these blocks is always under 50 per cent., 
consequently they will not run back when 
left alone. 

From a knowledge of the efficiency of a 
single-chain pulley, one can make a rough 
estimate of the relative sizes of pulleys required to prevent 
such blocks from running back. Taking the efficiency of each 
pulley as 97 per cent, when the weight is just on the point of 
running back, the tension in the right-hand chain will be 
97 per cent, of that in the left-hand chain due to the friction 



Friction. 



339 



on the lower pulley; but due to the friction on the upper 
pulley only 97 per cent, of the effort on the right-hand chain 
can be transmitted to the left-hand chain, whence for equi- 
librium, when P = o, we have — 

W W„ 

~r = o'97 X o'97 x — R 



ox r = o'94R 

2R 
and the velocity ratio = 



R - 0-94R 



= 33 



which is about the value commonly adopted. The above 
treatment is only approximate, but it will serve to show the 
relation between the efficiency and the ratio between the 
pulleys. 

Morris High-efficiency Self-sustaining Pulley 
Block. — In pulley blocks of the Weston type the efficiency 
rarely exceeds 45 per cent., but in geared self-sustaining blocks 
it may reach nearly 90 per cent. 

The self-sustaining mechanism is shown in Fig. 332. 
When hoisting the load the sprocket wheel A together with 



Ratchet 
\ 
Back *f4s^e/-l__J Brake | 



Driver in hoisting 




Drivers irfien hoisting. 



Fig. 33J. 
By kind permission of Messrs. Herbert Morris, Ltd.; Lougliborough. 

the nut N are rotated by means of an endless hand chain 
running in a clockwise sense of rotation. The nut traverses 
the quick running thread until the leather brake ring presses 
on the face of the ratchet wheel, the friction between these 
surfaces^ also between the back of the ratchet wheel and the 
back washer, becomes sufficiently great to lock them altogether. 
The back washer is keyed to the pinion shaft, the pinion gears 



340 Mechanics applied to Engineering. 

into a toothed wheel provided with a pocketed groove for the 
lifting chain. 

Let P,, = the pull on the hand-chain. 

D = the diameter of the hand-chain sprocket wheel A. 
d„ = the mean diameter of the screw thread (see 

page 295). _ 
P = the circumferential force acting at the mean 

diameter of the screw thread when lifting. 
W = the axial pressure exerted by the screw when 
lifting. 
e = the mechanical efficiency of the gear from the 

lifting hook to the brake, 
a = the angle of the screw thread. 
<^ = the friction angle for the threads and hut which 

is always less than a. 
/ij = the coefficient of friction between the brake ring 

and the ratchet wheel. 
/i,» = ditto, back washer and ratchet wheel. 
Di = mean diameter of brake ring. 
D,„ = mean diameter of back washer bearing surface. 

Then P = ?i^ = W tan (a + <^) . . . . (i.) 

(see page 295) 
^^= W^. tan (a -J- <^) (ii.) 

The axial pressure must be at least sufficient to produce 
enough friction on the brake ring and on the back washer to 
prevent the load on the hook from running down when the 
hand chain is released. 

Hence 

P D 

W^iisDj -f j«.,„D,„) must be greater than -^^— (iii.) 

In order to provide a margin of safety against the load 
running back, the friction on the back washer may be neg- 
lected ; then from (ii.) and (iii.) 

W^i.D, = Wrf„ tan (a -f <^) 

Dj tan (a -f- <^) ,. ^ 

and —r — * (iv.) 



Friction. 



341 



When these conditions are fulfilled the brake automatically 
locks on releasing the hand chain. The overall mechanical 
efficiency of the pulley block can be calculated from the 
mechanical efficiency of the toothed gearing and the friction 
of the chain in the pocketed grooves. 

When lowering the load the hand-chain wheel revolves in 
the opposite direction, thus tending to relieve the pressure of 
the brake. At the same time the sleeve on which the hand 
wheel is mounted bears against the washer and nut at the end 
of the pinion shaft, so that a drive in the lowering direction 
can be obtained through the gears. 

General Efficiency Law. — A simple law can be found 
to represent tolerably accurately the friction of any machine 
when working under any load it may be capable of dealing 
with. It can be stated thus : " The total effort F that must 
be exerted on a machine is a constant quantity K, plus a 
simple function of the resistance W to be overcome by the 
machine." 

The quantity K is the effort required to overcome the 
friction of the machine itself apart from any useful work. 
The law may be expressed thus — 

F = K + Wa; 

The value of K depends upon the type of machine under 
consideration, and the value of si upon the velocity ratio v^ of 




the machine. From Fig. 333 it will be seen how largely the 
efficiency is dependent upon the value of K. The broken and 



342 



Mechanics applied to Engineering. 



the full-line efficiency curves are for the same machine, with a 
large and a small initial resistance. 



The mechanical efficiency i; = 



VV 



Yvr (K + Wx)v, 



Thus we see that the efficiency increases as the load W 

K. 
increases. Under very heavy loads ^^ may become negligible ; 

hence the efficiency may approach, but can never exceed — 



Vmar 



The following 
experiments : — 


values give 


results 


agreeing well with 




X 


K 


n 


Rope pulley blocks 

Chain blocks of the\ 
Weston type / 

Self-sustaining geared\ 
blocks J 


I + 0-052/, 

■Vr 

1 + O-Wr 

V, 

I + oo09», 

Vr 


2v,dVas. 
3 lbs. 
i-S lbs. 


W 


W{i+o-osz',.) + Kz', 
W 


W(l+0-IZ/r) + K»r 

W 


W(i + ooogz/r) + Kz-, 



d = diam. of rope in inches. 

Levers. — ^The efficiency of a simple lever (when used at 
any other than very low loads) with two pin joints varies from 
94 to 97 per cent., the lower value for a short and the higher 
for a long lever. 

When mounted on well-formed knife-edges, the efficiency is 
practically 100 per cent. 

Toothed Gearing. — The efficiency of toothed gearing 
depends on the smoothness and form of the teeth, and whether 
lubricated or not. Knowing the pressure on the teeth and the 
distance through which rubbing takes place (see p. 165), also the 
fi, the efficiency is readily arrived at ; but the latter varies so 
much, even in the same pair of wheels, that it is very difficult to 
repeat experiments within 2 or 3 per cent. ; hence calculated 
values depending on an arbitrary choice of //, cannot have any 



Friction, 



343 



pretence to accuracy. The following empirical formula fairly 
well represents average values of experiments : — 

For one pair of machine-cut toothed wheels, including the 
friction on the axles — 

t\ = o'g6 — 



for rough unfinished teeth — 

1/ = o'go — 



2-5N 



2-5N 



Where N is the number of teeth in the smallest wheel. 
When there are several wheels in one train, let n = the 
number of pairs of wheels in gear ; 

Efficiency of train 17, = 17" 

The efficiency increases slightly with the velocity of the 
pitch lines (see Engineering, vol. xli. pp. 285, 363, 581; also 
Kennedy's " Mechanics of Machinery," p. 579). 



Velocity of pitch line in \ 
feet per minute ...J 
Efficiency 



10 
0*940 



5° 
o"972 



100 
o'gSo 



150 
o'9S4 



200 
0-986 



Screw and Worm Gearing.— We have already shown 




Fig. 334- 



how to arrive at the efficiency of screws and worms when the 
coefficient of friction is known. The following table is taken 
from the source mentioned above : — 



344 



Mechanics applied to Engineering. 



Velocity of pitch line in feet per) 
minute .... ) 


10 


50 


100 


ISO 


200 




Efficiency per cent. 


Angle of thread o, 45° 


87 


94 


95 


96 


97 


30° 


82 


90 


93 


94 


95 


20° 


7S 


86 


90 


92 


92 


■5° 


70 


82 


87 


89 


90 


10° 


62 


76 


82 


«S 


86 


7° 


S3 


69 


76 


80 


81 


s° 


4S 


62 


70 


74 


76 



The figure shows an ordinary single worm and wheel. As 
the angle a increases, the worm is made with more than one 
thread ; the worm and wheel is then known as screw gearing. 
For details, the reader should refer to books on machine 
design. 

Friction of Slides. — A slide is generally proportioned so 
that its area bears some relation to the load ; hence when the 
load and coefficient of friction are unknown, the resistance to 
sliding may be assumed to be proportional to the area ; when 
not unduly tightened, the resistance may be taken as about 
3 lbs. per square inch. 

Friction of Shafting. — A 2-inch diameter shaft running at 
100 revolutions per minute requires about i horse-power per 
100 feet when all the belts are on the pulleys. The horse- 
power increases directly as the speed and approximately as the 
cube of the diameter. 

This may be expressed thus — 

Let D = diameter of the shafting in inches ; 
N = number of revolutions per minute j 
L = length of the shafting in feet ; 
F = the friction horse-power of the shafting. 



Then F = 



NLD« 
80,000 



The horse-power that can be transmitted by a shaft is— 
H.P. 



_ to — - (see p. 580) 



according to the working stress. 



Friction. 345 

)f line shafting on which the« 

_ horse-power transmitted - friction horse-power 



Hence the efficiency of line shafting on which there are 
numerous pulleys is — 



horse-power transmitted 



1\ = 


64 


' LND=» 

80,000 




ND^ 


= 


I — 


64 

L r 

for a 

1250 


and 


I — 


L 
2000 




r — 


L 

2960 



for a working stress of 5000 lbs. sq. inch 
,, 8000 



i> i> 



Thus it will be seen that ordinary line shafting may be 
extremely wasteful in power transmission. The author knows 
of several instances in which more than one-half the power of 
the engine is wasted in driving the shafting in engineers' shops ; 
but it must not be assumed from this that shafting is necessarily 
a wasteful method of transmitting power. Most of the losses 
in line shafting are due to bending the belts to and fro over the 
pulleys (see p. 349), and to the extra pressure on the bearings 
due to the pull on the belts and the weight of the pulleys. 

In an ordinary machine shop one may assume that there 
is, on an average, a pulley and a 3-inch belt at every 5 feet. 
The load on the bearings due to this belt, together with the 
weight of the shaft and pulley, will be in the neighbourhood 
of 500 lbs. The load on the countershaft bearings may be 
taken at about the same amount or, say, a load on the bearings 
of 1000 lbs. in all. Let the diameter of the shafting be 
3 inches J the S feet length will weigh about 120 lbs., hence 
the load on the bearings due to the pulleys, belts, etc., will 
be about eight times as great as that of the shaft itself — and 
considering the poor lubrication that shafting usually gets, one 
may take the relative friction in the two cases as being roughly 
in this proportion. Over and above this, there is considerable 
loss due to the work done in bending the belt to and fro. 

We shall now proceed to find the efficiency of shafting, 
which receives its power at one end and transmits it to a 
distant point at its other end, i.e. without any intermediate 
pulleys. 



346 Mechanics applied to Engineering. 

Consider first the case of a shaft of the same diameter 
throughout its entire length. 

Let L = the length of the shaft in feet; 
R = the radius of the shaft in inches ; 
W = the weight of the shaft i square inch in section 

and I foot long ; 
/A = the coefficient of friction ; 
■q = the efficiency of transmission ; 
/ = the torsional, skin stress on the shaft per square 

inch; 

Weight of the) „. r,2T ik * 
shaft f=W,rR=Llbs. 

moment of the 1 ., t,3t • u iu 
friction |=/^W,rR3Lmch-lbs. 

the maximum] 

twisting mo- , ^3 

ment at the )=-'-i inch-lbs. (see p. 576) 

motoi end I ^ 

of the shaft I 

^^^ f ??"^"*^^ ) the effective twisting moment at the far end 

of the trans- \ = — -r — r-r—. 2 — 

mission n I *^ twisting moment at the motor end 

_ maximum twisting moment — friction moment 

maximum twisting moment 

_ _ friction moment 

maximum twisting moment 

/aWttR^L _ 2j^WL 

" ' ~ /3rR3 * - /. 

2 

For a hollow shaft in which the inner radius is - of the 
outer, this becomes — 

2«WL/ «" \ 

Now consider •^he case in which the shaft is reduced in 
diameter in order to keep the skin stress constant throughout 
its length. 

Let the maximum twisting moment at the motor end of the 
shaft = T, ; 



Friction. 347 

Let the useful twisting moment at the far end of the 
shaft = Tj. 

Then the increase of twisting moment dt due to the friction 
on an elemental length dl = fjiW-n-R^d/ = dt. 

For the twisting moment / we may substitute — 

'=-^(seep. 576) 

or 7rR3 = ?;? 
J* 

by substitution, we get — 

dt= ^Jt^^^ 

and^^=?^^ 
i /. 

Integrating — 




where e =■ 372, the base of the system of natural logarithms. 
The efficiency ,' = J" =.^ 

_ 2>iWL 

and for a hollow shaft, such as a series of drawn tubes, which 
are reduced in size at convenient intervals — 

_ 2) jWLg' 

The following table shows the distance L to which power 
may be transmitted with an efficiency of 80 per cent. For 
or(Unary bearings we have assumed a high coefficient of 
friction, viz. 0*04, to allow for poor lubrication and want of 
accurate alignment of the bearings. For ball bearings we also 



348 



Mechanics applied to Engineering. 



take a high value, viz. o'ooz. Let the skin stress y^ on the 
shaft be 8000 lbs. sq. inch, and let « = i"25. 



Form of shafting 


Parallel. 


Taper. 


Kind of bearings 


Ordinary. 


Ball. 


Oidinary, 


Ball 


Solid shaft mth belts ' 

,, „ without belts 
Hollo* „ 


Feet. 

400 

6000 

9840 


Feet. 

120,000 
197,000 


Feet. 

76,600 
J2S.SSO 


Feet. 

1,530,000 
2,505,000 



These figures at first sight appear to be extraordinarily high, 
and every engineer will be tempted to say at once that they 
are absurd. The author would be the last to contend that 
power can practically be transmitted through such distances 
with such an efficiency, mainly on account of the impossibility 
of getting perfectly straight lines of shafting for such distances, 
and the prohibitive costs ; but at least the figures show that 
very economical transmission may, under, convenient circum- 
stances, be accomplished by shafting — and when straight 
lengths of shafting could be put in they would unquestionably 



Driver 




B 



n=/ 



Fig. 335. 



71=3 



-fV=^ 



be far more economical in transmitting power than could 
be accomplished by converting the mechanical energy into 
electrical by means of a dynamo, losing a certain amount of 
the energy in the mains, and finally reconverting the electrical 
energy into mechanical by means of a motor ; but, of course, 
in most cases the latter method is the most convenient and the 
cheapest, on account of the ease of carrying the mains as 
against that of shafting. The possibility of transmitting power 
very economically by shafting was first pointed out by Professor 

' A part from the loss in bending rh? belts to and fro as they pass over 
the pulleys. 



Friction. 



349 



Osborne Reynolds, F.R.S., in a series of Cantor Lectures on 
the transmission of power. 

Belt and Rope Transmission. — The efficiency of belt 
and rope transmission for each pair of pulleys is from 95 to 
96 per cent., including the friction on the bearings ; hence, if 
there are n sets of ropes or belts each having an efficiency rj, 
the efficiency of the whole will be, approximately — 

% = ij" 

Experiments by the author on a large number of belts 
show that the work wasted by belts due to resistance to bending 
over pulleys, creeping, etc., varies from 16 to zi foot-lbs. per 
square foot of belt passed over the pulleys. 

Mechanical EfSciency of Steam-engines. — The 
work absorbed in overcoming the friction of a steam-engine is 
roughly constant at all powers ; it increases slightly as the power 
increases. A full investigation of the question has been made 
by Professor Thurston, who finds that the friction is distributed 
as follows : — 

Main bearings 3S~47 P^f cent. 

Piston and rod 21-33 

Ciank-pin 5-7 

Cross-head and gudgeon-pin 4-5 

Valve and rod ... 2°5 balanced, 22 unbalanced 

Eccentric strap 4-S 

Link and eccentric 9 

The following instances may be of interest in illustrating 
the approximate constancy of the friction at all powers : — 



Experimental Engine, Univeestty College, Lot 
Syphon Lubrication. 



London. 



LH.P 

B.H.P 

Friction H.P. ... 


2-75 

O'O 

27s 


9 "25 
5-63 
3-62 


1023 
7-50 
273 


11-14. 
7-66 
3-48 


12-34 
^•09 

3-25 


13-95 

II '09 

2-86 


14-29 

11-25 

3-04 



Experimental Engine, The University, Leeds. 
Syphon and Pad Lubrication. 



LH.P. 
B.H.P. 
Friction H.P. 


2-48 

00 

248 


S-i6 

2-35 
2-81 


6-83 

3-94 
2-89 


8-30 
5-61 
2-69 


11-50 
8-70 
2-8o 


13-84 

1082 
302 


17-02 

13-89 

3-«3 


22-30 
19-09 

3-21 



3S0 



Mechanics applied to Engineering. 



Belliss Engine, Bath (Forced) Lubrication. 
(See Proc. J.M.E., 1897.) 



I.H.P. 
B.H.P. 
Friction H.P. 



49-8 


102 '7 


147 I 


193-6 


44'S 


97-0 


140*6 


i860 


5-3 


S-7 


6'S 


7-6 



217-5 
209-5 

8-0 



Friction Pressure. — The friction of an engine can be 
conveniently expressed by stating the pressure in the working 
cylinder required to drive the engine when running light. 
Under the best conditions it may be as low as i lb. per square 
inch {%e& Engineer, May 30, 1913, p. 574). In ordinary 
steam engines in good condition the friction pressure amounts 
to 2J to 3^ lbs. square inch, but in certain bad cases it may 
amount to 5 lbs. square inch. It has about the same value 
in gas and oil engines per stroke, or say from 10 to 14 lbs. 
square inch, reckoned on the impulse strokes when exploding 
at every cycle, or twice that amount when missing every 
alternate explosion. 

Thus, if the mean effective pressure in a steam-engine 
cylinder were 50 lbs. square inch, and the friction pressure 
3 lbs. square inch, the mechanical efficiency of the engine 

would be = 94 per cent, if double-acting, and - — '^— 

5° S° 

= 88 per cent, if single-acting. 

The mean effective pressure in a gas-engine cylinder seldom 
exceeds 75 lbs. square inch. Thus the mechanical efficiency 
' is from 81 to 87 per cent. 

The friction horse-power, as given in the above tables, can 
also be obtained in this manner. 



Mechanical Efficiency per Cent, of Various Machines. 
(From experiments in all cases with more than quarter full load.) 



Weston pulley block (J ton) 

„ „ „ (larger sizes) 

Epicycloidal pulley block ... 

Morris 

One-ton steam hoists or windlasses 

Hydraulic windlass 

„ jack 

Cranes (steam) 

Travelling overhead cranes 



30-40 
40-47 
40-45 
75-85 
50-70 
60-80 
80-90 
60-70 
30-50 



Friction. 35 1 

T .. draw bar H.P. ,^ „^ 

Locomotives o5~75 

1. ri.r. 

Two-ton testing-machine, worm and wheel, screw and 

nut, slide, two collars ... ... ... 2-3 

Screw displacer— hydraulic pump and testing-machine, 
two cup leathers, toothed-gearing four contacts, three 
shafts (bearing area, 48 sq. inches), area of flat slides, 
18 sq. inches, two screws and nuts 2-3 

(About 1000 H.P. engines, 
spur-gearing, and engine 
friction 74 

Rope drives 70 

Belt , 71 

Direct (400-H.P. engines) ... 76 



Belts. 

Coil Friction. — Let the pulley in Fig. 336 be fixed, and 
a belt or rope pass round a portion of it as shown. The 
weight W produces a tension Tj ; in order to raise the weight 
W, the tension Tg must be greater than T, by the amount of 
friction between the belt and the pulley. 

Let F = frictional resistance of the belt ; 

/ = normal pressure between belt and pulley at any 
point. 

Then, if /* = coefficient of friction — 
F = T, - T, = 2/./ 

Let the angle a embraced by the belt be divided into a 

a 
great number, say «, parts, so that - is very small ; then the 

tension on both sides of this very small angle is nearly the 
same. Let the mean tension be T ; then, expressing a in 
circulav measure, we have — 

/ = T? 

• n 

The friction at any point is (neglecting the stiffness of the 
belt)— 

ft* = mT - = Tj' - T,' 

But we may write - as 8a : also Tj - T,' as ST. Then— 
fiT . 8a = 8T 



352 Mechanics applied to Engineering. 

which in the limit becomes — 

/iT . da = </r 
— = li.da 

We now require the sum of all these small tensions ex- 
pressed in terms of the angle 
embraced by the belt : — 





log, Tj - log. Ti = ixa 



n 



log. 



'1\ 



ixa 



= /^° 



or 



-©= 



±o'4343/^" 



V l— /J^ 



where e =■ 272, the base of the 
system of natural logarithms, 
and log e = 0-4343. 

When W is being raised, the 
+ sign is used in the index, and 
when lowered, the — sign. The 
value of /A for leather, cotton, or 
hemp rope on cast iron is from 
o'2 to o'4, and for wire rope 0*5. 
If a wide belt or plaited 
rope be used as an absorption 
dynamometer, and be thoroughly 
smeared with tallow or other 
thick grease, the resistance will 
be greatly increased, due to the shearing of the film of grease 
between the wheel and the rope. By this means the author 
has frequently obtained an apparent value of /n of over i — a 
result, of course, quite impossible with perfectly clean surfaces. 
Power transmitted by Belts. — Generally speaking, the 
power that can be transmitted by a belt is limited by the 
friction between the belt and the pulley. When excessively 
loaded, a belt usually slips rather than breaks, hence the 




Fig. 336. 



Friction. 353 

friction is a very important factor in deciding upon the power 
that can be transmitted. When the belt is just on the point of 
slipping, we have — 



Horse-power transmittec = = ^— ^ i^— 

33.000 33,000 

33.000 



t/i - ^ 



where the friction F is expressed in pounds, and V = velocity 
in feet per minute. Substituting the value of <f, and putting 
/A = o'4 and o = 3"i4 (180°), we have the tension on the tight 
side 3 '5 times that on the slack side. 

H.P. = ""y^"^'^ 

33,000 

For single-ply belting Tj may be taken as about 80 lbs. per 
inch of width, allowing for the laced joints, etc. 
Let w = width of belt. 



Then T2=8ow 




J tr r. 0-7 2 X 8o7</V 

and H.P.= — ^ = 

33.000 


600 


for single-ply belting; 




andH.P-«'V 
.300 





for double-ply belting. 

The number of square feet of belt passing over the pulleys 

per minute is — • 
^ 12 

Hence the number of square feet of belt required per 
minute per horse-power is — 

vN 

H.z= 50 square feet per minute for single-ply, and 

wV 25 square feet per mtaute for double-ply 

600 

2 A 



354 Mechanics applied to Engineering. 

This will be found to be an extremely convenient expression 
for committal to memory. 

Centrifugal Action on Belts. — In Chapter VI. we 
showed that the two halves of a flywheel rim tended to fly 
apart due to the centrifugal force acting on them ; in precisely 
the same manner a tension is set up in that portion of a belt 
wrapped round a pulley. On p. 202 we showed that the 
stress due to centrifugal force was — 

g 

where W, is the weight of i foot of belting i square inch in 
section. W, = 0*43 lb., and V„ = the velocity in feet per 
second : V = velocity in feet per minute ; hence — 

o-43V„'' ^ v." ^ V^ 

32"2 75 270,000 

and the effective tension for the transmission of power is — 

V= 
T — 



270,000 



The usual thickness of single-ply belting is about 0-22 inch, 
and taking the maximum tension as 80 lbs. per inch of width, 

this gives -; — = 364 lbs. per square inch of belt, and the 

power transmitted per square inch of belt section is — 









P = 


TaV- 


270,000 








d\ 


T,- 


3^= 
270,000 


For 


maximum 


power 












T.= 


270,000 


and V = 


= 5700 


feet 


: per minute. 





Friction. 



355 



The tension in the belt when transmitting the maximum 
power is therefore — 

Ts — r— = 364 — 121 = 243 lbs. per square inch. 

270,000 

and the maximum horse-power transmitted per square inch of 
belt section — 

072 X 243 X 5700 

H.Pma«. = —— = 3° nearly. 

33,000 

For ropes we have taken the weight per foot run as o'35 lb. 




2000 3000 1000 5000 6000 7000 

Velocity in Feeifer Minult, 

Fig. 337. 



per square inch of section, and the maximum permissible 
stress as 200 lbs. per square inch. On this basis we get the 
maximum horse-power transmitted when V = 4700 feet per 
minute, and the maximum horse-power per square inch of 
rope = i7*i. 

The curves in Fig. 337 show how the horse-power trans- 
mitted varies with the speed. 

The accompanying figure (Fig. 338), showing the stretch of 
a belt due to centrifugal tension, is from a photograph of an 
indiarubber belt running at a very high speed ; for comparison 



356 Mechanics applied to Engineering. 

the belt is also shown stationary. The author is indebted to 
his colleague Dr. Stroud for the photograph, taken in the 
Physics laboratory at the Leeds University. 

Creeping of Belts. — ^The material on the tight side of 9 
belt is necessarily stretched more than that on the slack side, 
hence a driving pulley always receives a greater length of belt 
than it gives out ; in order to compensate for this, the belt creeps 
as it passes over the pulley. 

Let / = unstretched length of belt passing over the pulleys 
in feet per minute ; 
Li = stretched length on the Tj side ; 

A ^^ )) )» » ■'■1 »i 

N, = revolutions per minute of driven pulley ; 

N2= , driving „ 

di = diameter of driven pulley) measured to the middle 
(/a = „ driving „ J of the belt ; 

X = stretch of belt in feet ; 
E = Yoimg's modulus ; 
/, Atid/j = stresses corresponding to T, and Tg in lbs. square 
inch. 



Then x -. 


E 






i,=/+x = 


= '(■ 


+® = 


7r</,N, 


A- 


-(■ 


-l) = 


W,N, 










N, 


_ (E ^A)d^ 

(E +/,)-/, 





If there were no creeping, we should have — 

E = from 8,000 to io,ooo lbs. per square inch. Taking 



Friction. 



357 



Ta = 80 lbs. per inch of width, and the thickness as o'2 2 inch, 
we have when a — 3-14 — 

f-i = = 364 lbs. per square inch 

0*22 

Q _ Q _ 

and Ti = Trr^-~T. = — = 23 lbs. per inch width 

2"12 3'5 

/i = — 2_ = 104 lbs. per square inch 

0'22 

Hence %±fy = ^°'°°° + ^°4 = ^.975 

E +/a 10,000 + 364 ^'^ 




Fig. 338. 



or the belt under these conditions creeps or slips 2's per 
cent. 

When a belt transmits power, however small, there must be 
some slip or creep. 



358 Mechanics applied to Engineering. 

When calculating the speed of pulleys the diameter of the 
pulley should always be measured to the centre of the belt ; thus 
the effective diameter of each pulley is D + /, where / is the 
thickness of the belt. In many instances this refinement is of 
little importance, but when small pulleys are used and great 
accuracy is required, it is of importance. For example, the 
driving pulley on an engine is 6 feet diameter, the driven 
pulley on the countershaft is 13 inches, the driving pulley on 
which is 3 feet 7 inches diameter, and the driven pulley on a 
dynamo is 8 inches diameter; the thickness of the belt is 
o"22 inches; the creep of each belt is 2-5 per cent.; the 
engine runs at 140 revolutions per minute : find the speed of the 
dynamo. By the common method of finding the speed of 
the dynamo, we should get — 

— — rr-^ = 4168 revolutions per minute 

13 X 8 

But the true speed would be much more nearly — 

140 X 72-22 X 43'22 X o'Q75 X o"o7s „ 

— i 5j? — ^15 11^ = 3822 revs, per mmute 

13'22 X 8'22 '^ 

Thus the common method is in error in this case to the extent 

of 9 per cent. 

Chain Driving. — In cases in which it is important to 
prevent slip, chain drives should be used. 
They moreover possess many advantages 
over ordinary belt driving if they are 
properly designed. For the scientific 
designing of chains and sprocket wheels, 
the reader is referred to a pamphlet on 
the subject by Mr. Hans Renold, of 
Manchester. 

Fig. 339. Rope Driving. — When a rope does 

not bottom in a grooved pulley, it wedges 

itself in, and the normal pressure is thereby increased to — 

p -JL 

sm - 
2 

The angle 6 is usually about 45°; hence P, = 2 -6?. 

The most convenient way of dealing with this increased 
pressure is to use a false coefficient 2"6 times its true value. 
Taking /i = 0*3 for a rope on cast iron, the false /i for a 
grooved pulley becomes 2-6 x 0-3 = o'78. 




Friction. 359 

The value of ei^' now becomes lo'i when the rope embraces 
half the pulley. The factor of safety on driving-ropes is very 
large, often amounting to about 80, to allow for defective 
splicing, and to prevent undue stretching. The working 
strength in pounds may be taken from loc^ to i6c'^, where c\s 
the circumference in inches. 

Then, by similar reasoning to that given for belts, we get 
for the horse-power that may be transmitted per rope for the 
former value — 

„ „ c^V </-V 
H.P. = , or 

3740 374 

where d = diameter of rope in inches. 

The reader should refer to a paper on rope driving by 
Mr. Coombe, Insf. Mech. Engrs. Proceedings, 1889. 



Coefficients of Friction. 

The following coefficients obtained on large bearings will 
give a fair idea of their friction : — 

Ball bearings with plain| ^ . 

cyhndrical ball races, i ^ ^' 

_, r Flat ball races „ „ o-ooo8 to 0-0012 

Ihrust ■jOneflat,one vrace, 3 ,. „ mean 00018 

"^^""Sslxwoyraces, 4 » .. .. o'o°S5 

Gun-metal bearings r Plain cylindrical journals^ 
tested by Mr. with bath lubrication / 
Beauchamp Tower Plain cyhndrical journals V 
for the Institution' with ordinary lubrication/ 
of Mechanical Thrust or collar bearingj ^.^ 
Engineers ^ well lubricated / ^ 

Good white metal (author) with very meagrej ^.^^ 

lubrication > 

Poor white metal under same conditions o*oo3 



o'ooi 
o'oi 



Reference-books on Friction. 

" Lubrication and Lubricants," Archbutt and Deeley. 

"Friction and Lubrication," Dr. J. T. Nicolson, Man- 
chester Association of Engineers, 1907— 1908. 

" Cantor Lectures on Friction," by Dr. Hele-Shaw, F.R.S. 
Published by the Society of Arts. 



CHAPTER X. 

STSESS, STRAIN, AND ELASTICITY. 

Stress. — If, on any number of sections being made in a body, it 
is found that there is no tendency for any one part of it to move 
relatively to any other part, that body is said to be in a state of 
ease; but when one part tends to move relatively to the other 
parts, we know that the body is acted upon by a system of 
equal and opposite forces, and the body is said to be in a state 
of stress. Thus, if, on making a series of saw-cuts in a plate 
of metal, the cuts were found to open or close before the saw 
had got right through, we should know that the plate was in a 
state of stress, because the one part tends to move relatively. to 
the other. The stress might be due either to external forces 
acting on the plate, or to internal initial stresses in the material, 
such as is often found in badly designed castings. 

Intensity of Stress. — The intensity of direct stress on 
any given section of a body is the total force acting normal to 
the section divided by the area of the section over which it is 
distributed ; or, in other words, it is the amount of force per 
unit area. 

Intensity of stress in) _ the given force in pounds 

pounds per sq. inch J ~ area of the section over which the 

force acts in sq. inches 

For brevity the word " stress " is generally used for the 
term " intensity of stress." 

The conditions which have to be fulfilled in order that the 
intensity of stress may be the same at all parts of the section 
are dealt with in Chapter XV. 

Strain. — The strain of a body is the change of form or 
dimensions that it undergoes when placed in a state of stress. 
No bodies are absolutely rigid ; they all yield, or are strained 
more or less, when subjected to stress, however small in amount. 

The various kinds of stresses and strains that we shall 
consider are given below in tabular form. 



Stress, Strain, and Elasticity. 



361 



s.s. 

O >ig 
H <3 V 

41 3 4-> 

5 «.3 



-H 


-^4 


II 




.a 








bfl 




n 














s 


1 


M 


:& 








Hi-- 



hI--> 



.a 

c 

.!3 



hK 



>2 






,_^ 


(3 


,*<, 










c 






4» 










M 





O ii 



0) u 

o o 



J3 



s 


i 















e ° 






ai 



V 
•-J u 



e53 



c " 
o tuo 
"S a 



•S u »j 
2 S-S 

§■£■3 
S ai 



G 5 ""■ 

.Sou 



^ a 






^g 



g 



a a 






•? 



01s 

,a a 



I 



1 




,r 






■I© 



fVi 



362 Mechanics applied to Engineering. 

Elasticity, — A body is said to be elastic when the strain 
entirely disappears on the removal of the stress that produced 
it. Very few materials can be said to be perfectly elastic except 
for very low stresses, but a great many are approximately so 
over a wide range of stress. 

Fenuanent Set. — That part of the strain that does not 
entirely disappear on the removal of the stress is termed 
" permanent set." 

Elastic Limit. — The stress at wjiich a marked permanent 
set occurs is termed the elastic limit of the material. We use 
the word marked because, if very delicate measuring instruments 
be used, very slight sets can be detected with, much lower stresses 
than those usually associated with the elastic limit. In elastic 
materials the strain is usually proportional to the stress ; but 
this is not the case in all materials that fulfil the conditions of 
elasticity laid down above. Hence there is an objection to the 
definition that the elastic limit is that point at which the strain 
ceases to be proportional to the stress. 

Plasticity, — If none of the strain disappears on the 
removal of the stress, the body is said to be plastic. Such 
bodies as soft clay and wax are almost perfectly plastic. 

Ductility. — If only a small part of the strain be elastic, 
but the greater part be permanent after the removal of the 
stress, the material is said to be ductile. Soft wrought iron, 
mild steel, copper, and other materials, pass through such a 
stage before becoming plastic. 

Brittleness. — When a material breaks with a very low 
stress and deforms but a very small amount before fracture, it 
is termed a brittle material. 

Behaviour of Materials subjected to Tension. 

Ductile Materials. — If a bar of ductile metal, such as wrought 
iron or mild steel, be subjected to a low tensile stress, it will 
stretch a certain amount, depending on the material ; and if the 
stress be doubled, the stretch will also be doubled, or the stretch 
will be proportional to the stress (within very narrow limits). 
Up to this point, if the bar be relieved of stress, it will return 
to its original length, i.e. the bar is elastic ; but if the stress be 
gradually increased, a point will be reached when the stretch 
will increase much more rapidly than the stress ; and if the bar 
be relieved of stress, it will not return to its original length — in 
other words, it has taken a " permanent set." The stress at 
which this occurs is, as will be seen from our definition above, 
the elastic limit of the material. 

Let the stress be still further increased. Very shortly a 



Stress, Strain, and Elasticity. 



363 



point will be reached when the strain will (in good wrought 
iron and mild steel) suddenly increase to 10 or 20 times its 
previous amount. This point is termed \ih& yield point of the 
material, and is always quite near the elastic limit. For all 
commercial purposes, the elastic limit is taken as being the 
same as the yield point. Just before the elastic limit was 
reached, while the bar was still elastic, the stretch would only 
be about xrjo of the length of the bar ; but when the yield 
point is reached, the stretch would amount to y^, or ^ of the 
length of the bar. 

The elastic extensions of specimens cannot be taken by 
direct measurements unless the specimens are very long 
indeed; they are usually measured by some form of exten- 
someter. That shown in Fig. 340 was designed by the author 




Fig. 340. 

some years ago, and gives entirely satisfactory results ; it reads 
to ^q^(,J of an inch ; it is simple in construction, and does 
not get out of order with ordinary use. It consists of suitable 
clips for attachment to the specimen, from which a graduated 
scale is supported ; the relative movement of the clips is read 
on the scale by means of a pointer on the end of a 100 to i 
lever. 

In Fig. 341 several elastic curves are given. In the case 
of wrought iron and steel, the elastic lines are practically 
straight, but they rapidly bend off at the elastic limit. In the 
case of cast iron the elastic line is never straight ; the strains 
always increase more rapidly than the stresses, hence Young's 
modulus is not constant Such a material as copper takes a 
" permanent set " at very low loads ; it is almost impossible to 
say exactly where the elastic limit occurs. 



364 



Mechanics applied to Engineering. 



As the stress is increased beyond the yield point, the strain 
continues to increase much more rapidly than before, and the 
material becomes more and more ductile ; and if the stress be 
now removed, almost the whole of the strain will be found to 
be permanent. But still a careful measurement will show that 
a very small amount of the strain is still elastic. 



OOlB 



fs 







^4 6 8 10 

Stress in. Ions per Sf Inxfi, 

Fig. 341. 



14 



Just before the maximum stress is reached, the material 
appears to be nearly perfectly plastic. It keeps on stretching 
without any increase in the load. Up to this point the strain 
on the bar has been evenly distributed (approximately) along 
its whole length; but very shortly after the plastic state has 
been reached the bar extends locally, and "stricture" com- 
mences, «.<f. a local reduction in the diameter occurs, which is 
followed almost immediately by the fracture of the bar. The 
extension before stricture occurs is termed the " proportional " 
extension, and that after fracture the " final " extension, which 



Stress, Strain, and Elasticity. 



365 



is known simply as the "extension" in commercial testing. 
We shall return to this point later on. 

The stress-strain diagram given in Fig. 342 will illustrate 
clearly the points mentioned above. 

Brittle Materials. — Brittle materials at first behave in a 
similar manner to ductile materials, but have no marked elastic 
limit or yield point. They break oflf short, and have no 
ductile or plastic stage. 

Extension of Ductile Materials. — We pointed out 
above that the final extension of a ductile bar consisted of two 
parts — (i) An extension evenly distributed along the whole 
length of the bar, the total amount of which is consequently 



Stress 



orMiltl Steel- 



Stricture 




tJu-esT 



Fis. 



proportional to the length of the bar ; (2) A local extension at 
fracture, which is very much greater per unit length than the 
distributed or proportional extension, and is independent 
(nearly so) of the length of the bar. Hence, on a short bar the 
local extension is a very much greater proportion of the whole 
than on a long bar. Consequently, if two bars of the same 
material but of different lengths be taken, the percentage of 
extension on the short bar will be much greater than on the 
long bar. 



366 



Mechanics applied to Evgineering. 



The following results were obtained from a bar of Lowmoor 




Flo. 343- 



The local extension in this bar was 54 per cent, on 2 inches. 

The final extensions reckoned on various lengths, each 
including the fracture, were as follows : — 



Length 

Percentage of extension 



10 
22 



24-5 



6" 
34 



4 
41 



2 
54 



(See papers by Mr. Wicksteed \-a Industries, Sept. 26, 1890, and 
by Professor Unwin, I.C.E., vol. civ.) Hence it will be seen 

that the length on which the 
percentage of extension is 
measured must always be 
stated. The simplest way of 
obtaining comparative results 
for specimens of various 
lengths is to always mark 
them out in inches throughout 
their whole length, and state 
the percentage of extension 
on the 2 inches at fracture 
as well as on the total length 
T)f the bar. A better method 
would be to make all test 
specimens of similar form. 







Stress 

Via. 344. 



t.e. the diameter a fixed proportion of the length ; but any one 
acquainted with commercial testing knows how impracticable 
such a suggestion is. 

Load-strain diagrams taken from bars of similar material, but 
of different lengths, are somewhat as shown in Fig. 344. 



Stress, Strain, and Elasticity. 



367 



If L = original length of a test bar between the datum 
points ; 
Li = stretched length of a test bar between the datum 
points ; 

Then Lj — L = x, the extension 

The percentage of extension is — 

L, — L loox 

-^— — X roo = — — 



In Fig. 345 we show some typical fractures of materials 
tested in tension. 



Gun 
metal. 



Hard 
steel. 



Soft Delta 

steel. Copper. metal. 




Fig. 345. 

If specimens are marked out in inches prior to testing, and 
after fracture they are measured up to give the extension on 
various lengths, always including the fracture, they will, on 
plotting, be found to give approximately a law of the form — 

Total extension = K + «L 

where K is a constant depending upon the material and the 
diameter of the bar, and n is some function of the length. 
Several plottings for different materials are given in Fig. 346. 
Beduction in Area of Ductile Materials. — The 
volume of a test bar remains constant within exceedingly small 
limits, however much it may be strained ; hence, as it exterids 



368 



Mechanics applied to Engineering. 



the sectional area of the bar is necessarily reduced. The 
reduction in area is considered by some authorities to be the 
best measure of the ductility of the material. 



Annealed Coppe* 



Mild Steel 
Wroughi Iron 




4- e a 

Length. 

Fig. 346. 



to. tnches 



Let A = the original sectional area of the bar ; 
Ai = the final area at the fracture. 

Then the percentage of reduction in area is — —^ x 100 

A 

If a bar remained parallel right up to the breaking point, 
as some materials approximately do, the reduction in area caii 
be calculated from the extension, thus : 



Stress, Strain, and Elasticity. 369 



The volume of the bar remains constant ; hence — 


LA = LjAi, or Aj 


LA 


and the reduction in area is — 




A- A, 




A 




Then, substituting the value of Aj 


, we have-^ 


LiA - LA L, - 


L * 


L,A Li 


L. 



Thus the reduction in area in the case of a test bar which 
remains parallel is equal to the extension on the bar calculated 
on the stretched length. This method should never be used 
for calculating the reduction in area, but it is often a useful 
check. The published account of some tests of steel bars 
gave the following results : — 

Length of bar, 2 inches ; extension, 6'o per cent. ; reduc- 
tion in area, 4*9 per cent. ; 

6x2 

Then x in this case was '— = o" 1 2 inch 

100 

and Li = 2" 12 inches 

„ , . . o'i2 X 100 

Reduction ui area = \ = S'66 per cent. 

Thus there is probably an error in measurement in getting 
the 4'9 per cent., for the reduction in area could not have been 
less than 5 '66 per cent, unless there had been a hard place in 
the metal, which is improbable in the present instance. 

Real and Nominal Stress in Tension. — It is usual to 
calculate the tensile stress on a test bar by dividing the 
maximum load by the area of the original section. This 
method, though convenient and always adopted for commercial 
purposes, is not strictly accurate, on account of the reduction of 
the area as the bar extends. 

Using the same notation as before for the lengths and 
areas — 

Let W = the load on the bar at any instant ; 

W 
S = the nominal stress on the bar, viz. -r ; 

. W 
S, = the real stress on the bar, viz. -r-- 

2 B 



370 Mechanics applied to Engineering. 

Then, as the volume of the bar remains constant — 

L A 
LA = LiA„ and ^ = . - 

W 

S W Ai L 
~K 

SLn 

OT the real stress Sj = -^r- 

The diagram of real stress may be conveniently constructed 
as in Fig. 347 from the ordinary stress-strain diagram. 

The construction for one point only is given. The length 



Tl^-'" 




A 

Fig. 347. 



i^ e strain % 



L of the specimen is set oflf along the strain axis, and the 
stress ordinate de is projected on to the stress axis, viz. ao. 
The line ba is then drawn to meet ed produced in c, which 
gives us one point on the curve of real stress. For by similar 
triangles we have — 



S, L. 



SLi 



which we have shown above to be the real stress. 

The last part of the diagram, however, cannot be obtained 



Stress, Strain, and Elasticity. 371 

thus, as the above relation only holds as long as the bar 
remains parallel; but points on the real stress diagram between 





^-Cu. 


Mid 
/^teel 


fIfvurM 

\ 


\ (^^ 




\ 



Xoad 

Fig. 348. 

g and / can be obtained by stopping the test at intervals, 
noting the load and the corresponding diameter of the bar in 
the stricture : the load divided by the corresponding stricture 
area gives the real stress at the instant. 




Fig. 349. — Steel containing several percentages of carbon. 

Typical Stress. Strain Curves for Various 
Materials in Tension. — The curves shown in Figs. 348, 



372 



Mechanics applied to Engineering. 



349, were drawn by the author's autographic recorder 
(see Engineering, December 19, 1902), from bars of the same 
length and diameter. 

Some of the curves in Fig. 350 are curiously serrated, i.e. 
the metal does not stretch regu- 
larly (these serrations are not 
due to errors in the recording 
apparatus, such as are obtained 
by recorders which record the 
faults of the operator as well as 
the characteristics of the ma- 
terial). The author finds that 
all alloys containing iron give a 
serrated diagram when cold and 
a smooth diagram when hot, 
whereas steel does the reverse. 
This peculiar effect, which is 
disputed by some, has been 
independently noticed by Mons. 
(«) Le Chatelier. 

Artificial Raising of the 




Fig. 350. — (a) Rolled aluminium . 
roHed copper i^c) rolled "bull "metal, 
temp. 400° Fahr. ; (a!) ditto 60° Kahr. 

N.B.— Bull metal and delta metal ElastiC Limit. The form of 

behave in practically the same way in „t „ 4. ■_ j 1 

the testing-machine. a strcss-strain curve depends 

much upon the physical state 

of the metal, and whether the elastic limit has been artificially 

raised or not. It has been known for many years that if a 

piece of metal be loaded beyond the elastic limit, and the load 

be then released, the next 
time the material is loaded, 
the elastic limit will approxi- 
mately coincide with the pre- 
vious load. In the diagram in 
Fig- 35 1, the metal was loaded 
up to the point c, and then re- 
leased ; on reloading, the 
elastic limit occurred at the 
stress cd, whereas the original 
elastic limit was at the stress 
ab. Now, if in manufacture, 
by cold rolling, drawing, or 
otherwise, the limit had been 
thus artificially raised, the 
stress-strain diagram would have been dee. 

Young's Modulus of Elasticity (E). — We have already 




Stress, Strain, and Elasticity. 



373 



stated that experiments show that the strain of an elastic body 
is proportional to the stress. In some elastic materials the 
strain is much greater than in others for the same intensity of 
stress, hence we need some means of concisely expressing the 
amount of strain that a body undergoes when subjected to a 
given stress. The usual method of doing this is to state the- 
intensity of stress required to strain the bar by an amount 
equal to its own length, asstiming the material to remain 
perfectly elastic. This stress is known as Young's modulus 
of (or measure of) elasticity. We shall give another definition 
of it shortly. 



4 



.[] 



J _^....'Vj; 



~~yiPess~ 




Fig. 352. 

In the diagram in Fig. 352 we have shown a test-bar of 
length / between the datum points. The lower end is supposed 
to be rigidly fixed, and the upper end to be pulled ; let a stress- 
strain diagram be plotted, showing the strain along the vertical 
and the stress along the horizontal. As the test proceeds we 
shall get a diagram abed as shown, similar to the diagrams 
shown on p. 365. Produce the elastic line onward as shown 
(we have had to break it in order to get it on the page) until 
the elastic strain is equal to /; then, if x be the elastic strain 
at any point along the elastic line of the diagram corresponding 
to a stress/, we have by similar triangles — 

I E 



374 Mechanics applied to Engineering. 

The stress E is termed " Young's modulus of elasticity," 
and sometimes briefly " The modulus of elasticity." Thus in 
tension we might have defined the modulus of elasticity as 
The stress required to stretch a bar to twice its ori^nal length, 
assuming the material to remain perfectly elastic. It need hardly 
be pointed out that no constructive materials used by engineers 
do remain perfectly elastic when pulled out to twice their 
original length ; in fact, very few materials will stretch much 
more than the one-thousandth of their length and remain elastic. 
It is of the highest importance that the elastic stretch should 
not be confused with the stretch beyond the elastic limit. It 
will be seen in the diagram above that the part bed has nothing 
whatever to do with the modulus of elasticity. 

We may write the above expression thus : 

E=Z 

X 

/ 



Then, if we reckon the strain per unit length as on p. 361, 
ha' 
thus :- 



we have - = unit strain, and we may write the above relation 



Young's modulus of elasticity = — ^ 

unit stram 

Thus Young's modulus is often defined as the ratio of the unit 
stress to the unit strain while the material is perfectly elastic, 
or we may say that it is that stress at which the strain becomes 
unity, assuming the material to remain perfectly elastic. 

"The first definition we gave above is, however, by far the 
clearest and most easily followed. 

'For compression the diagram must be slightly altered, as in 

Fig- 353- 

In this case the lower part of the specimen is fixed and the 
upper end pushed down ; in other respects the description of 
the tension figure applies to this diagram, and here, as before, 
we have — 

/ E 

For most materials the value of E is the same for both 
tension and compression ; the actual values are given in tabular 
form on p. 427. 



Stress, Strain, and Elasticity. 



375 



Occasionally in structures we find the combination of two 
or more materials having very different coefficients of elasticity ; 
the problem then arises, what proportion of the total load is 



b|:":;a":: 



••^ stress 




Fig. 353. 

borne by each? Take the case of a compound tension 
member. 

Let E, = Young's modulus for material i ; 

^2 ^^ )» J» )) 2 j 

Ai = the sectional area of i ; 
■"2 = )i _ II 2 ; 

/i = the tensile stress in i ; 
/i — >i I. 2 ; 

W = the total load on the bar. 



Then 



/2 Eja'/i Eg 



since the components of the member are attached together at 
both ends, and therefore the proportional strain is the same in 
both; 



376 



Mechanics applied to Engineering. 



and 



A2/2 A.E^ Wj 



which gives us the proportion of the load borne by each of the 
component members ; 

W 
and Wi = 



1 + 



A,E, 



W,= 



A1A2 

w 



1 + 



AjEj 

A2E2 



By similar reasoning the load in each component of a bar 
containing three different materials can be found. 

The Modulus of Transverse Elasticity, or the 
Coefficient of Rigidity (G). — The strain or distortion of an 




Fig. 354. 

element subjected to shear is measured by the slide, x (see 
p. 361). The shear stress required to make the slide x equal 
to the length / is termed the modulus of transverse elasticity, or 
the coefficient of rigidity, G. Assuming, as before, that the 
material remains perfectly elastic, we can also represent this 
graphically by a diagram similar to those given for direct 
elasticity. 

In this case the base of the square element in shear is 
rigidly fixed, and the outer end sheared, as shown. 



Stress, Strain, and Elasticity. 



177 



From similar triangles, we have 
I G 



_ / stress 

G = - = - 

X strain 



Relation between the Moduli of Direct and Trans- 
verse Elasticity. — Let abed be a square element in a perfectly 
elastic material which is to be subjected to — 

(i) Tensile stress equal to the modulus stress; then the 

length / of the line ab will be stretched to 2/, viz. abi, and the 

2/— / 
strain reckoned on unit length will be ■ — - — = i. 

(2) Shearing stress also equal to the modulus stress ; then 

the length / of the line ab will be stretched to is/P + P 

= ^2l= I "41/, and the strain reckoned on unit length 

1-41/-/ 
will be = 0-41. 

Thus, when the modulus stress is reached in shear the 
strain is 0-41 of the strain when the modulus stress is reached 




Fig. 3SS. 

in tension ; but the stress is proportional to the strain, therefore 
the modulus qf transverse elasticity is o'4i, or | nearly, of the 
modulus of direct elasticity. 

The above proof must be regarded rather as a popular 
demonstration of this relation than a scientific treatment. The 
orthodox treatment will be given shortly. 

Strength of Wire. — Surprise is often expressed that the 




378 Mechanics applied to Engineering. 

strength of wire is so much greater than that of the material 
from which it was made; the great difference between the two 
is, however, largely due to the fact that the nominal tensile 
strength of a piece of material is very much less than the real 
strength reckoned on the final area. The process of drawing 
wire is equivalent to producing an elongated stricture in the 
material ; hence we should expect the strength of the wire to 
approximate to that of the real strength of the material from 
which it was made (Fig. 356). That it does so 
will be clear from the following diagrams. In 
addition to this the skin of the wire is under very 
severe tensile stress, due to the punishing action 
of the draw-plate, which causes a compression 
_ g of the core, with the result that the density of 
the wire is slightly increased with a correspond- 
ing increase in strength. Evidence will shortly be given to 
show that the skin is in tension and the core in compression. 

The process of wire-drawing very materially raises the 
elastic limit, and if several passes be made without annealing 
the wire, the elastic limit may be raised right up to the break- 
ing point ; the permanent stretch of the wire is then extremely 
small. If a given material will stretch, say, 50 per cent, in 
the stricture before fracture, and a portion of the material be 
stretched, say, 48 per cent., by continual passes through the 
draw-plate without being annealed, that wire will only stretch 
roughly the remainder, viz. 2 per cent., before fracture. We 
qualify this remark by saying roughly, because there are other 
disturbing factors ; the statement is, however, tolerably accu- 
rate. If a piece of wire be annealed, the strength will be 
reduced to practically that of the original material, and the 
proportional extension before fracture will also approximate 
to that of the undrawn material. 

If a number of wires of various sizes, all made from the 
same material, be taken, it will be found that the real stress on 
the final area is very nearly the same throughout, although the 
nominal strength of a small, hard, i.e. unannealed, wire is 
considerably greater than that of a large wire. 

The above remarks with regard to the properties of wire 
also apply to the case of cold drawn tubes and extruded 
metal bars. 

The curves given in Fig. 357 clearly show the general 
effects of wire-drawing on steel ; it is possible under certain 
conditions to get several " passes " without annealing. 

The range of elastic extension of wires is far greater than 



Stress, Strain, and Elasticity. 379 

that of the material in its untreated state; in the latter case 
he elastic extension is rarely more than ^Ao of the length of 
the bar, but in wires it may 
reach i^Vo "■ " 



The elastic ex 

tension curve for a sample 
of hard steel wire is given 
in Fig. 358. In some in- 
vestigations by the author 
it was found that Young's 
Modulus for wires was con- 
siderably lower than that for 
the undrawn material. On 
considering the matter, he 
concluded that the highly 
stressed skin of the wire 
acted as an elastic tube tightly 
stretched over a core of 
material, which thereby com- 
pressed it transversely, and 
caused it to elongate longi- 
tudinally. If this theory be 
correct, annealing ought to 
increase Young's Modulus for the wire. On appealing to ex- 
periment it was found that such was the case. A further series 




I 2 

NumAer or Passes 
Fig 357. 




30 40 50 60 70 

stress in. tons j)er Sq. IntJi. 

Tig. 3s8. 



90 



of experiments was made on wires of different sizes, all made 



38o 



Mechanics applied to Engineering, 



from the same billet of steel; the results corroborated the 
former experiments. In every case the hard-drawn steel wires 
had a lower modulus than the same wires after annealing. 
The results were — 



o*i6o 
0*160 



0*210 
o'aio 



0-174 
0*174 



0*146 
0*146 



0*115 
0*115 



Elastic 
limit. 



Maxi- 
mum 

stress. 



Pounds per sq. inch. 



181,700 

59,600 



78,000 
63,800 



125,000 
71,600 



z6o,ooo 
7i»3«> 



316,000 
102,400 



126,700 



164,800 
124,800 



189,700 
xxi,ooo 



Extension 
per cent 



on 10 
ins, 



9-6 



3'S 



on 3 
ins. 



19*0 

19*5 



5 n 



34'S 
54'o 



46-8 
5a'3 



31*5 
51 'o 



30*4 
S7'i 



E. 
Pounds 
sq. inch. 



25,430,000 
28,500,000 



37,520,000 
37,800,000 



25,4x0,000 
a7,5oo»ooo 



35,300,000 
27,200,000 



Remarks- 



Hard drawn 

Annealed 



Rolled rod 
Ditto annealed 



Rod after one " pass " 
Ditto annealed 



Rod after two " 
Ditto annealed 



188,000 
72,000 



318,900 
1x3,800 



30*3 






35.330,000 
37,000,000 
31,400,000 



Rod after three " paa 
Ditto annealed 
Ditto after breaking 



Wire Bopes. — The form in which wire is generally used 
for structural purposes is that of wire ropes. The wires are 
suitably twisted into strands, and the strands into ropes, either 
in the same or in the opposite sense as the wires according to 
the purpose for which the rope is required j for details, special 
treatises on wire ropes must be consulted. 

The hauling capacity of a wire rope entirely depends upon 
the strength at its weakest spot, which is usually at the attach- 
ment of the hook or shackle. The terminal attachment, or the 
" capping," as it is generally tended, can be accomplished in 
many ways, but, unfortunately, very few of the methods are 
at all satisfactory. In the table below the average results of 
a large number of tests by the author are given. 

The method adopted for testing purposes in the Leeds 
University Machine is shown in Fig. 359. After binding the 
rope with wire, and tightly " serving " with thick tar band in 
order to keep the strands in position, the ends are frayed out, 



Stress, Strain, and Elasticity. 



381 



cleaned thoroughly with emery cloth, and finally a hard 
white metal 1 end is cast on. With ordinary ropes high 
efficiencies are obtained, but with very hard steel wires, which 
only stretch a very small amount before breaking, the wires 
have not the same chance of adjusting themselves to the 
variable tension in each (due 
to imperfect manufacture and 
capping), and consequently tend 
to break piecemeal at a much 
lower load than they would if 
each bore its full share of the 
load. 

On first loading a new wire 
rope the strain is usually large, 
due to the tightening up of the 
strands and wires on one another 
(see Fig. 359)., but the rope 
shortly settles down to an elastic 
condition, then passes an ill- 
defined elastic limit, and ulti- 
mately fractures. From its be- 
haviour during the elastic stage, 
a value for Young's modulus 
can be obtained which is always 
very much lower than that of 
the wires of which it is com- 
posed. This low value is largely 
due to the tightening of the 
strands, which continues more 
or less even up to the breaking 
load. In old ropes which have 
taken a permanent " set " the 
tightening effect is reduced to a 
minimum, and consequently Young's modulus is greater than 
for the same rope when new. 

The value of E for old ropes varies from 8000 to 10,000 
tons per square inch. The strength is often seriously reduced 
by wear, corrosion, and occasionally by kinks. 




Fig. 35g. — Method of capping wire ropes 



' A mixture of lead 90 per cent., antimony 10 per cent. 



382 



Mechanics applied to Engineering. 











Breaking load. 




Number 


Diameter of 


Section 
of 


E. 
Tons 








Tensile strengtli 


of 








of rope. 


wires. 








Rope. 


Sum of 
wires. 


Ratio. 






ins. 


sq. ios. 










tons sq. in. 


20 


0-0895 


0-126 


6000 


14-0 


14-6 


096 


Ii7\ 


24 


0-085 


0-136 


6870 


18-0 


18-5 


0-97 


85 




30 


0-091 


0-195 


S«40 


23-5 


24-6 


0-96 


127 




V 30 


0-085 


0-187 


SSSo 


8-25 


9-1 


0-91 


49 




3" 


0-136 


0-522 


7200 


47-2 


47-5 


0-99 


91 


Steel 


4? 


0081 


0-220 


5400 


20-0 


20-4 


0-9S 


90 wire 


■^s 


0-023 


0-232 


6330 


309 


31-75 


0-98 


13^ 


ropes 


108 


0-065 


0-358 


6020 


41-0 


42-3 


0-97 


"5 




222 


0-044 


0-337 


5560 


35 '4 


46-3 


0-76 


105 




222 


0039 


0-264 


6530 


24-9 


30-0 


0-83 


95' 


19 


0-231 


0-796 


,3770 


6-95 


8-27 


0S4 


8-7 ^l"- 


7 


0-231 


0-293 


35yo 


225 


3-03 


0-74 


7'7 cable 



Work done in fracturing a Bar. — Along one axis a 
load-strain diagram shows the resistance a bar offers to being 

pulled apart, and along the 
other the distance tliough 
which this resistance is over- 
come; hence the product of 
the two, viz. the area of the dia- 
gram, represents the amount 
of work done in fracturing the 
bar. 

Let a = the area of the dia- 
gram in square 
inches ; 
/ = the length of the bar 
in inches (between 
datum points) \ 
A = the sectional area of 
the bar. 
If the diagram were drawn i inch = i ton, and the strain were 
full size, then a would equal the work done in fracturing the 
bar ; but, correcting for scales, we have — 





N 


M . 

- ^ ^ 


i 

! 



Iioeui TTu tons ■■ 
Fig. 360. 



N 



= work done in inch-tons in fracturing the bar 



and TT-vj = work done in inch-tons per cubic inch in fracturing 
''^'^ the bar 



Stress, Strain, and Elasticity. 383 

Sir Alexander Kennedy has pointed out that the curve 
during the ductile and plastic stages is a very close approxima- 
tion to a parabola. Assuming it to be so, the work done can 
be calculated without the aid of a diagram, thus : 
Let L = the elastic limit in tons per square inch ; 
M = the maximum stress in tons per square inch ; 
X = the extension in inches. 

Then the work done in inch-tons per) _ ^ . 2 ,^ _ , . 
square inch of section of bar )~ "t" 3^ / 

= |(L + 2M) 

work done in inch-tons per cubic inch = — 7(L + 2M) 

^ X , . . 

But Y X 100 = tf, the percentage of extension 

hence the work done in inch-tons per) £_,y , ^> 

cubic inch ) ~ 3oo'' ' 

The work done in inch-tons per cubic inch is certainly by 
far the best method of measuring the capacity of a given 
material for standing shocks and blows. Strictly speaking, in 
order to get comparative results from bars of various lengths, 
that part of the diagram where stricture occurs should be 
omitted, but with our present system of recording tests such a 
procedure would be inconvenient. 

The value of the expression for the '' work done " in fractur- 
ing a bar is evident when one considers the question of bolts 
which are subjected to jars and vibration. It was pointed out 
many years ago that ordinary bolts are liable to break off short 
in the thread when subjected to a severe blow or to long- 
continued vibration, and further, that their life may be greatly 
increased by reducing the sectional area of the shank down to 
that of the area at the bottom of the thread. The reason for this 
is apparent when one calculates the work done in fracturing 
the bolt in the two cases ; it is necessarily very small if the 
section of the shank be much greater than that at the bottom 
of the thread, because the bolt breaks before the shank has 
even passed the elastic limit, consequently all the extension is 
localized in the short length at the bottom of the thread, but 
when the area of the shank is reduced the extension is evenly 
distributed along the bolt. The following tests will serve to 
emphasize this point : — 



384 



Mechanics applied to Engineering. 



Diameter 
of bolt. 


Length. 


Work done in 
fracturing the bolt. 


Remarks. 


I in. 
I „ 


I3'2 ins. 
132 » 


I0'4 inch-tons 
39'9 .. .. 


Ordinary state 
Turned shank 


i:; 


14 ins. 
14 » 


2' 1 5 inch-tons 
lb-8 „ „ 


Ordinary state 
Turned shanlc 


Jin. 

f.. 


14 ins. 
14 .1 


175 inch-tons 
74 » .. 


Ordinary state 
Turned shank 



In this connection it may be useful to remember that the 
sectional area at the bottom of the thread is — 

-^ '- sq. inches (very nearly) 

100 ^ .1 J I 

where d is the diameter of the bolt expressed in eighths of an 
inch. The author is indebted to one of his former students, 
Mr. W. Stevenson, for this very convenient expression. 

Behaviour of Materials subjected to Compression. 







Aluminium. Original 
form. 



Gun 
metal. 



Cast 



Soft 



Cast iron. 



Fia. 361 



Ductile Materials. — In the chapter on columns it is shown 
that the length very materially affects the strength of a piece of 
material when compressed, and for getting the true compressive 
strength, very short specimens have to be used in order to 



Stress, Strain, and Elasticity. 



38s 




Fig. 362. 



prevent buckling. Such short specimens, however, are incon- 
venient, for measuring accurately the relations between the 
stress and the strain. 
(Jp to the elastic limit, 
ductile materials be- 
have in much the 
same way as they do 
in tension, viz. the 
strain is proportional 
to the stress. At the 
yield point the strain 
does not increase so 
suddenly as in tension, 
and when the plastic 
stage is reached, the 
sectional area gradu- 
ally increases and the 
metal spreads. With 
very soft homo- 
geneous materials, this 
spreading goes on 
until the metal is squeezed to a flat disc without fracture. 
Such materials are soft 
copper, or aluminium, 
or lead. 

In fibrous mate- 
rials, such as wrought 
iron and wood, in 
which the strength 
across the grain is 
much lower than with 
the grain, the material 
fails by splittbg side- 
ways, due to the lateral 
tension. 

The . usual form 
of the stress - strain 
curve for a ductile 
material is somewhat 
as shown in Fig. 362. 

If the material 

reached a perfectly 

^, , , . load at any instant (W) , 

plastic stage, the real stress, t.e. — -. ' ' / , ' 

'^ ° sectional area at that mstant (Aj) 

2 c 




Fig? 363. 



386 



Mechanics applied to Engineeiing. 



would be constant, however much the material was compressed ; 
then, using the same notation as before — 

A, - _ 

W 
and from above, — = constant 

Ai 

Substituting the value of Ai from above, we have — 
-— ' = constant ; 

/A. 

But /A, the volume of the bar, is constant ; 
hence W/j = constant 

or the stress-strain curve during the plastic period Is a 

hyperbola. The material 
never is perfectly plastic, 
and therefore never perfectly 
complies with this, but in 
some materials it very nearly 
approaches it. For example, 
copper and aluminium 
(author's recorder) (Fig. 
363). The constancy of the 
real stress will be apparent 
when we draw the real stress 
curves. 

Brittle Materials. — Brittle 
materials in compression, as 
in tension, have no marked 
elastic limit or plastic stage. 
When crushed they either 
split up into prisms or, if of 
cubical form, into pyramids, 
and sometimes by the one- 
half of the specimen shearing 
over the other at an angle of 

about 45°. Such a fracture is shown in Fig. 361 (cast iron). 
The shearing fracture is quite what one might expect from 

purely theoretical reasoning. In Fig. 365 let the sectional 

area = A ; 




Fig. 364.— Asphalte, 



then the stress on the cross-section S = 



W 



Stress, Strain, and Elasticity. 3^7 

and the stress on an oblique section aa, making an angle a 



'»5J5m5^&?;$^J%%M^ 




A 


w / 


tt. 




N\j 








/^ 




a 









Fig. 366. — Portland cement. 

with the cross-section, may be found thus : resolve W into 
components normal N = W cos o, and tangential T = W sin a. 



388 Mechanics applied to Engineering. 

The area of the oblique section aa = Ao= 

^ cos a 

.1. 1 ^ N W cos a W cos* a _ „ , 

the normal stress = — = — - — = ^ a . cos" a 



cos a 



tangential or si ar-) _ T _ W sin a _ W cos a sin a 



f Ao 



ing stress f , Aj A A 

cos a 
= S COS a sin a 

If we take a section at right angles to aa, T becomes the 
normal component, and N tiie tangential, and it makes an 
angle of 90 — o with the cross-section ; then, by similar reason- 
ing to the above, we have — 

A A 

The area of the oblique section = A^,' = ■■ -. .^ = -: — 

cos (90 — a) sm o 

normal stress = S sin* a 

tangential stress or shearing stress = S sin a cos a 

So that the tangential stress is the same on two oblique 
sections at right angles, and is greatest when a = 45°; it is 
then = S X 071 X 0-71 = 0-5 S. 

From this reasoning, we should expect compression speci- 
mens to fail by shearing along planes at 45° to one another, 
and a cylindrical specimen to form two cones top and bottom, 
and a cube to break away at the sides and become six 
pyramids. That this does occur is shown by the illustrations 
in Figs. 364-366. 

Real and Nominal Stress in Compression (Fig. 367). 
— In the paragraph on real and nominal stress in tension, we 
showed how to construct the curve of real stress from the 
ordinary load-strain diagram. Then, assuming, that the volume 
remains constant and that the compression specimen remains 
parallel (which is not quite true, as the specimens always 
become barrel-shaped), the same method of constructing the 
real stress curve serves for compression. As in the tension 
curve, it is evident that (see Fig. 347) — 

S L 
c SL] 



Stress, Strain, and Elasticity. 



389 



Behavia ar of Materials subjected to Shear. Nature 
of Shear Stress. — If an originally square plate or block be 




Sffess 



Fig. 367. 



acted upon by forces P parallel to two of its opposite sides, the 
square will be distorted into a rhombus, as shown in Fig. 368, 




. *^ v 






a- b 




d 




y§ 




d C 


' ' 




t— ^ — 



Fig. 368. 



? 



Fig. 369. 



p 

and the shearing stress will be / = -,, taking it to be of unit 

thickness. This block, however, will spin round due to the 
couple P . arf or P . 6c, unless an equal and opposite couple be 
applied to the block. In order to make the following remarks 



390 



Mechanics applied to Engineering, 



perfectly general, we will take a rectangular plate as shown in 

Fig- 369- 

The plate is acted upon by a clockwise couple, P . ad, or 
f,.ab . ad, and a contra-clockwise couple, P, . a^ ox f, .ad . ab, 
but these must be equal if the plate be in equilibrium ; 

theny^ .ab.ad =fi .ad.ab 

or/. =/; 

t.A the intensity of stress on the two sides of the plate is the 
same. 

Now, for convenience we will return to our square plate. 
The forces acting on the two sides P and Pi may be resolved 
into forces R and E.i acting along the diagonals as shown in 
Fig. 370. The effect of these forces will be to distort the 
square into a rhombus exactly as before. (N.B. — The rhombus 





Fig. 370. 



Fig. 371. 



in Fig. 368 is drawn in a wrong position for simplicity.) These 
two forces act at right angles to one another ; hence we see 
that a shear stress consists of two equal and opposite stresses, 
a tension and a compression, acting at right angles to one 
another. 

In Fig. 371 it will be seen that there is a tensile stress 
acting normal to one diagonal, and a compressive stress normal 
to the other. The one set of resultants, R, tend to pull the two 
triangles abc, acd apart, and the other resultants to push the two 
triangles abd, bdc together. 

Let/o = the stress normal to the diagonal. 

Then/oa<r =/,,»/ lab, Oif^Jibc = R 
But V 2P, or ij 2f,. ab, or ij 2/,. be ='&. 
hence/, =/, =/,' 



Stress, Strain, and Elasticity. 



391 



Thus the intensity of shear stress is equal on all the four 
edges and the tension and compression on the two diagonals 
of a rectangular plate subjected to shear. 

Materials in Shear. — ^When ductile materials are sheared, 
they pass through an elastic stage similar to that in tension and 
compression. If an element be slightly distorted, it will return 
to its original form on the removal of the stress, and during 
this period the strain is proportional to the stress; but after 
the elastic limit has been reached, the plate becomes perma- 
nently deformed, but has not any point of sudden alteration as 
in tension. On continuing to increase the stress, a ductile and 
plastic stage is reached, but as there is no alteration of area 
under shear, there is no stage corresponding with the stricture 
stage in tension. 

The shearing strength of ductile materials, both at the 
elastic limit and at the maximum stress, is about f of their 
tensile strength (see p. 400). 

Ductile Materials in Shear. — The following results, ob- 
tained in a double-shear shearing tackle, will give some idea of 
the relative strengths of the same bars when tested in tension 
and in shear ; they are averages of a large number of tests : — 



' Material. 



Nominal 
tensile 

strength. 



Shearing 
strength. 



Shearing strength. 



Tensile strength. 



Work done 
per sq. in. 
of metal 
sheared 
through. 



Cast iron — hard, close grained 
,, ordinary . . . 

„ soft, open grained 

Best wrought iron .... 

Mild steel 

Hard steel 

Gun-metal 

Copper 

Aluminium 



I4'6 

io'9 

7-9 

22'0 

26-6 

48-0 

13-5 
150 

8-8 



13-5 

I2"9 

107 
i8-i 

20'9 

34'o 
15-2 

II-O 

S7 



0-92 
i-i8 
1-36 
•0-82 
079 
071 
I-I3 
073 
0-65 



From autographic shearing and punching diagrams, it is 
found that the maximum force required occurs when the 
shearing tackle is about \ of the way through the bar, and 
when the punch is about \ of the way through the plate. 

From a series of punching tests it was found that where — 



■'• n\rt 



load on punch 



circumference of hole X thickness of plate 



392 



Meclianics applied to Engineering. 







ft 


Ratio. 


Wrought iron . . 


19-8 


24-8 


o-8o 


Mild steel . . . 


22-2 


28-4 


078 


Copper. . . . 


10-4 


147 


071 



Brittle materials in shear are elastic, although somewhat 
imperfectly in some cases, right up to the point of fracture ; 
they have no marked elastic limit. 

It is generally stated in text-books that the shearing 
strength of brittle materials is much below \ of the tensile 
strength, but this is certainly an error, and has probably come 
about through the use of imperfect shearing tackle, which has 
caused double shear specimens to shear first through one 
section, and then through the other. In a large number of 
tests made in the author's laboratory, the shearing strength of 
cast iron has come out rather higher than the tensile stress in 
the ratio of i"i to i. 

Shear combined with Tension or Compression. — 
We have shown above that when a block or plate, such as abed, 
is subjected to a shear, there will be a direct stress acting 
normally to the diagonal bd. Likewise if the two sides ad, be 
are subjected to a normal stress, there will be a direct stress 
acting normally to the section ef\ but when the block is sub- 
jected to both a direct stress and a shear, there will be a direct 
stress acting normally to a section occupying an intermediate 
position, such as gh. 

Consider the stresses acting on the triangular element shown, 
which is of unit thickness. The intensity of the shear stress 
on the two edges will be equal (see p. 390). Hence — 

The total shear stress on the face gi = f,.gi = Pj 
„ „ „ „ hi =f, . M= F 

„ direct „ „ gi=/,.gi = T 

Let the resultant direct stress on the face gA, which we are 
about to find in terms of the other stresses, be /„. Then the 
total direct stress acting normal to the face^^ =/u-ih = Pj. 



Stress, Strain, and Elasticity. 393 

Now consider the two horizontal forces acting on the 
«,« C't— ^ 




Fig. 372. 



element, viz. T and P, and resolve them normally to the face 
gh as shown, we get T. and P,. 



394 



Mechanics applied to Engineering. 



cos d COS Q 

alsoP„ + T„ = P„=/«i;4 
hence, substituting the above values, we have — 
ff'B +/.. hi =fggh cos e =f„.'gl 

and/.+-^=/« 

Next consider the vertical force acting on the element, 
viz. Pj. 

Pi = P„ sin =fagh_s\n 6 =fjd 
or^.^7=/>'~and ^-^^ = M 

-. = ^ = tan e 

Substituting this value of hi in the 
equation above, we have — 

or/«/. ^ff=f^ 
Jtt ~ Juft —J, 




Fig. 373. 



Solving, we get 



/. 



= J±V^ 



7? 



+^ 



The maximum tensile stress on the) ^i _ft , /ft , ^ 
{a.cegh 5 -'■'■- 2 "^ V ^+-/' 

and the compressive stress on the face) 



at right angles to gh, viz.j'h 



i 



/»=f-vf+^" 



Some materials are more liable to fail by shear than by 
tension, hence it is necessary to find the maximum shear stress. 
Drawy^ at right angles to gh, and// making an angle a with it, 
the value of which will be determined. The compressive 
stress onj'h is/^, and the total stress represented by mn isj'h ./„. 
The maximum shear stress^ occurs on the i&CQJl, and the 
total shear stress on // is /^ . jl. The tensile stress on tiie face kl 
IS fa and the total stress is -4/ .^ which is represented by fq. If 
the tensile stress be + the compressive stress will be — . Since 



Stress, Strain, and Elasticity. 



395 



we want to find the shear stress we must resolve these stresses 
in the direction of the shear. Resolve pg, also mn^ parallel 
and normal to jl. 

pr = pq sin a = kl.fa sin a 

on = tun cos d = —jk.f^ cos a 
f.Ji = pr — on = kl.fa sin a - jk ./„ cos o 

fm. = Jjf^ sm a -J^f^ cos a 

=-f^ cos a sin u, — _/^ cos a sin a = (/„ —f^ cos a sin a 

= sm 2a, 

2 

This is a minimum when 2a = 90° and sin Ra = i. 



Then /^ = 



yirt JK 



/» 



, = \/// 



+ 



f^ 



This is the maximum inten- 
sity of shear stress which occurs 
when a piece of elastic material 
is subjected to a direct stress 
ft and a shear stress /,. The 
direction of the most stressed 
section is inclined at 45° to 
that on which the maximum 
Intensity of tensile stress occurs. Bi^^*— 

Compound Stresses. — 
Let the block ABCD, of uni- 
form thickness be subjected to 
tensile stresses f^ and/, acting 
normal to the mutually perpen- 
dicular faces AD and DC or 
BC and AB. The force 

P^=/^. AD or/«.BC 

P,=/,.ABor/,.DC 

P„=/„.AC 

V,=ft- AC. (/j is the tangential or shear stress). 




Fig. 374. 



It is required to find (i) the intensity and direction of the 



396 Mechanics applied to Engineering' 

normal stress /„ acting on the face AC, the normal to which 
makes an angle Q with the direction of P,. (ii) the in- 
tensity of the shear or tangential stress y^ acting on the face 
AC. 

Resolving perpendicular to AC we have — 

P„ = P„, + P.« = Py cos + P. sin 
/;AC =/,AB cos d +/«BC sin Q 
, ,AB ., ,BC . a 
/» =/»Xc *^°^ ^ "^-^'AC ^'° 

/„ =/, cos" e +/, sin» e 

also P, = P„ - P,. 

P, = P, sin 6 -V, cos 6 
/,A.C =/»AB sin -/,BC cos 6 
. .AB . . ,BC . 
•^' "•^'AC ^'" ~-^'AC *^°^ 
^ =^ cos sin B — /, sin 9 cos 

ft = (/» -/.) cos d sin e =^^^^ sin 2^ 



This is a maximum when B = 45°. The tangential or 

shear stress is then = — — —. Thus the maximum intensity of 

shear stress is equal to one half the difference between the two 
direct stresses. 

The same result was obtained on page 395 by a different 
process. 

FoiBSOn's Ratio. — When a bar is stretched longitudinally, 
it contracts laterally; likewise when it is compressed longi- 
tudinally, it bulges or spreads out laterally. Then, terming 
stretches or spreads as positive (+) strains, and compressions 
or contractions as negative (— ) strains, we may say that when 
the longitudinal strain is positive (+), the lateral strain is 
negative (— ). 

Let the lateral strain be - of the longitudinal strain. The 

fraction - is generally known as Foisson's ratio, although in 

reality Foisson's ratio is but a special value of the fraction, viz. \. 

Strains resulting from Three Direct Stresses 

acting at Bight Angles. — In the following paragraph it will 



Stress, Strain, and Elasticity. 397 

be convenient to use suffixes to denote the directions in which 
the forces act and in which the strains take place. Thus any 
force P which acts, say, 
normal to the face x will 

be termed P,, and the 
■^ 
strain per unit length -~ 

will be termed S„ and the 
stress on the face f^ ; then 

/x 



S,='g(seep. 374). 



Every applied force 
which produces a stretch 
or a + strain in its own 
direction will betermed +, 
and vice versd. 

The strains produced 




Thuiknj&ss 



Fig. 375. 



by forces acting in the various 



directions are shown in tabulated form below. 



Fio. 376. 



Force acting 

oa face of 

cube. 


Strain in' 
direction Xn 


Strain in 
directioilj'. 


Strain id 

direction *, 

Si. 


p« 


E 


E« 


_/x 

E« 


p» 


E» 


E 


E« 


p. 


-A 
'S.n 


E» 


4 



398 



Mechanics applied to Engineering. 



These equations give us the strains in any direction due to 
the stresses/,,^,/, acting alone; if two or more act together, 
the resulting strain can be found by adding the separate strains, 
due attention being paid to the signs. 

Shear. — We showed above (p. 390) that a shear consists of 
two equal stresses of opposite sign acting at right angles to one. 
another. The resulting strain can be obtained by adding the 
strains given in the table above due to the stresses^ andyi, 
which are of opposite sign and act at right angles to one 
another. 

The strains are — 



4 + 4; = T^f 1 + -^ in the direction (i) 
» .. (3) 



E «E 

-4+4=0 



«E ■ «E 

Thus the strain in two directions has been increased by 
- due to the superposition of the two stresses, and has been 
reduced to zero in the third direction. 

LetS=4(r+i). 

If a square abed had been drawn on the side of the element, 
it would have become the rhombus db'dd' after the strain, the 




AT— 



l*ac- 



..a. .;:, 




A":>\d' 



Fig. 378. 

long diagonal of the rhombus being to the diagonal of the 
square as i + S to i. The two superposed are shown in Fig. 378. 
Then we have — 

^ + (/+*)»=(i+S)» 
or 2/» + 2/a: + «» = I + 2S + S» 



Stress, Strain, and Elasticity. 399 

But as the diagonal of the square = i, we have— 

2/»= I 
X 

And let 7 = So ; x = IS^; then by substitution we have — 

a/» + 2/% + /%" =1 + 23 + 3" 

and I + So + — = 1 + 2S + S' 
2 

Both S and S, are exceedingly small fractions, never more than 
about YoVo" ^i^'i their squares will be still smaller, and therefore 
negligible. Hence we may write the above — 

r I + S„ = I + 2S 

orSo=2S = ^(x+i) 
/ /E E/ I \ E« 



^"'""^ ^=:777rr^( TTi )^ ^(«+^) 



hence n = 5 y^, and E = — ^ 

When « = s, G = -j^E = o-42E. 
« = 4, G = |E = o-4oE. 
« = 3. G = f E = 0-38E. 
« = 2, G = |E = 0-33E. 

Some values of n will be given shortly. 
We have shown above that the maximum strain in an 
element subject to shear is — 



=K.-0 



but the maximum strain in an element subject to a direct stress 
in tension is — 

»<- E 



hence — = 



3 iG+3 



3,- / 

E 



= (.^0^ 



orS 



= s,(. + I) 



400 



Mechanics applied to Engineering. 





S 


Safe shear stress 




s; 


safe tensile stress 


When « = S 


il 


1 


„ « = 4 


t 


i 


» » = 3 


« 


\ 


» « = 2 


i 


\ 






J^ 



Taking « = 4, we see that the same material will take a 
permanent set, or will pass the elastic limit in shear with | of 
the stress that it will take in tension j or, in other words, the 
shearing strength of a material is only f of the tensile strength. 
Although this proof only holds while the material 
is elastic, yet the ratio is approximately correct for 
the ultimate strength. 

Bar under Longitudiiial Stress, but -with 
Lateral Strain prevented. in one Direc- 
tion. — Let a bar be subjected to longitudinal 
stress, _/^, in the direction x, and be free in the 
direction y, but be held in the direction z. 

The strain in the direction 2 due to the stresses _;^ and/j. is 
f f . . 

■~f — ^~= = o, because the strain is prevented m this direction. 



r 



Flo. 379. 



Hence ^ = 






The strain in the direction x due to these stresses is — 

Jx Jz Jx Jx _^ Jxi l_\ 15-^ 



E «E E «^E 



E 



Thus the longitudinal strain of a bar held in this manner 
is only y| as great as when the bar is free in both lateral 
directions. 

Bar under Longitudinal Stress with Lateral Strain 
prevented in both Directions. 

«E 

_A_^ _^ _ q\ because the 
E «E «E I strain is pre- 
. —fi _f^ _f»_ ^Q [vented in these 
E «E «E / directions. 



Strain in direction :«; = ^ — =^ ■ 
E «E 



Stress, Strain, and Elasticity. 



401 



Then -=^ 4- — 
E «E^«E 

and /^ = «/, -f^ =fy{n - 1) since /„ =/, 

The strain in direction x is — 

E «E(« - 1) E\ «(« - 1)/ "^ E 



Or the longitudinal strain of a bar 
held in this manner is only f as great as 
when the bar is free. \^J [/| 

Anticlastio Curvature. — When a yecllljt' 
beam is bent into a circular arc some of 
the fibres are stretched and some com- 
pressed (see the chapter on beams), the 
amount depending upon their distance 
from the N.A.; due to the extension 
of the fibres, the tension side of the 
beam section contracts laterally and 
the compression side extends. Then, 




Fig. 380. 



corresponding to the strain at EE on the beam profile, we 
have - as much at E'E' on the section ; also, the proportional 
strain — 

EE - LL _y 
LL p 

E'E' - L'L' _ y 
L'L' p. 

ButZ = i.2 
Pi » P 
hence pi = np = 4p 



and 



This relation only holds when all the fibres are free laterally, 
which is very nearly the case in deep narrow sections ; but if the 
section be shallow and wide, as in a flat plate, the layers which 
would contract sideways are so near to those which would 
extend sideways, that they are to a large extent prevented from 
moving laterally ; hence the material in a flat wide beam is 
nearly in the state of a bar prevented from contracting laterally 
in one direction. Hence the beam is stifTer in the ratio of 

«3 



«!"- 



= yI than if the section were narrow. 



2 D 



402 Mechanics applied to Engineering. 

Boiler Shell. — On p. 421 we show that P. = aP,; 



/. = 2/. ; /. = 



/. 



ii 



I 




Fig. 381. Fig. 382. 

Strains. 
Let « = 4. 

' E E« EV2 J ^E 

' E« E« B.\2n J "E 

" ~ E« ^ E ~ E*. i^/ ~ » E 

Thus the maximum strain is in the direction S^ 

By the thin-cylinder theory we have the maximum strain 

- ^ > thus the real strain is only -j as great, or a cylindrical 

boiler shell will stand f= iT4or 14 per cent, more pressure 
before the elastic limit is reached than is given by the ordinary 
ring theory. 

Thin Sphere subjected to Internal Fluid Pressure. 
— In the case of the sphere, we have P. = P. ; /i = ^. 

Strains. — 

c _ /. _ 21 =-6/'i - i^ = s/' 
^' E E« EV «/ *E 

c A - A = _/'/'i 4- ^^ - _ i/' 

°'~ E« E« E\n^ nj »E 

*•" E^^E EV n)~*E 



Stress, Strain, and Elasticity. 403 

But^ in this dase ='^in the case given above; 

.-. S. in this case = | X ^ = |^ 

Maximum strain in sphere _ -f _ 3. 
maximum strain in boiler shell \ '' 

Hence, in order that the hemispherical ends of boilers should 
enlarge to the same extent as the cylindrical shells when under 
pressure, the plates in the ends should be f thickness of the 
plates in cylindrical portion. If the proportion be not adhered 
to, bending will be set up at the junction of the ends and the 
cylindrical part. 

Cylinder exposed to Longitudinal Stress when 
under Internal Pressure. — When testing pipes under 
pressure it is a common practice to close the ends by flat 
plates held in position by one long bolt passing through the 
pipe and covers, or several long bolts outside the pipe. The 
method may be convenient, but it causes the pipes to burst 
at lower pressures than if flanged covers were used. In the 
case of pipes with no flanges a long rod fitted with two pistons 
and cup leathers can be inserted in the pipe, and the water 
pressure admitted between them through one of the pistons, 
which produces a pure ring stress, with the result that the 
pipes burst at a much higher pressure than when tested as 
described above. 

(i) Strain with simple ring stress = ^ 

zff 
Pressure required to burst a thin cylinder/ = — 

a 



(ii) Strain with flanged covers = 8„ 



2ft 



Pressure required to burst a thin cylinder = p = ^-~ 

(iii) When the covers are held in position by a longitudinal 
bolt, the load on whith is Pj, the longitudinal compression in 

P 
the walls of the pipe is-j =7^. 

The circumferential ring strain due to internal pressure and 
longitudinal compression — 



404 Mechanics applied to Engineering. 

where m =-^ 

Jx 

n 2ft 

and/ = — j — X -^ 
n + m a 

Example. — Let d= 8 ins. / = 05 in. f^ = ro.ooo lbs. 
sq. in. 

Simple ring stress / = 1250 lbs. sq. in. 
With flanged covers/ = 1428 lbs. sq. in. 
With longitudmal bolt and let «? = 2. "j 
This is a high value but is sometimes \ p = 830 lbs. sq. in. 
experienced ' 

Thus if a pipe be tested with longitudinal bolts under the 
extreme condition assumed, the bursting pressure will be only 
58 per cent, of that obtained with flanged covers. 

Alteration of Volume due to Stress. — If a body 
were placed in water or other fluid, and were subjected to 
pressure, its volume would be diminished in proportion to the 
pressure. 

Let V = original volume of body ; 

Sv = change of volume due to change of pressure ; 
Bp „ pressure. 

Sp change of pressure 

~ Sv "" change ofvolume per cubic unit of the body 
~V 

V 

K is termed the coefficient of elasticity of volume. 

The change of volume is the algebraic sum of all the strains 
produced. Then, putting p =f^ =fy =/; for a fluid pressure, 
we have, from the table on p.. 397, the resulting strains — 

E £«"*"£ E«'^E E« E E« E E« 

_/ _ pEn _ E« - _ EG 



3/« — bp 3« — 6 9G — 3E 
ButE = ^-^(^(p.399) 

hence K = '^^^ = i(-^)e and n = 'g + ^^ 
3« - 6 ^\« - 2/ 3K - 2G 



Stress, Strain, and Elasticity. 



405 



The following table gives values of K in tons per sc^uare 
inch also of n : — 



Material. 


K. 


n. 


Water 


140 





Cast iron 


6,000 


3'0 to 47 


Wrought iron 


8,800 


3-6 


Steel 


11,000 


3-6 to 4"6 


Brass 


6,400 


31 to 3-3 


Copper 


10,500 


29 to 30 


Flint glass ... 


2,400 


3'9 


Indiarubber ... 




2-0 



The « given above has not been calculated by the above 
formula, but is the mean of the most reliable published 
experiments. 

Strain Energy Stored in a Plate. — Let the plate be 
subjected to stresses^ and/,,. 

Strain in direction x = ^ — '^ 



y = 



Energy stored 
inch 



per cub 



'1= 



fy_L 

E «E 
strain x stress 



\E «E/2'^\E «E/2 

~2EV' ^■'' n ) 

Strength of Plat Plates. — An attempt 
treatment of the strength of flat 
plates subjected to fluid pressure 
is long and tedious. See Mor- 
ley's "Strength of Materials." 
The following approximate treat- 
ment yields results sufficiently 
accurate for practical purposes, 

Kectangular Plate of 
thickness /. Consider a diagonal section d. 

The moment of the water pressure about the 
diagonal acting on the triangle efg 



exact 




pab 



X 



4o6 Mechanics- applied to Engineering. 

The moment of the reactions of the edges oi\ ^^^ ^ 
the plate when freely supported, the resultant is = — X - 
assumed to act at the middle of each side j 2 2 

The resulting moment is equal to the moment of resistance 
of the plate to bending across the diagonal, hence — 

ab c ab c , dfi ,^ _,, ^_. 

p—X~-p—X- = f-r (See Chapter XI) 

22-^23'^6 "^ ' 

pabc _ , 

^"■^ 

, ab 

But d = y/a^ + 1^ and c = -7=?==^ 



2{cP + ^)/" 



•^^'^^^ „/„8 I »\/a ~ f when freely supported. 



and / 2 1 psM ~'f w^^'' ^^ edges are rigidly held. 
For a square plate of side a this becomes 

■^ = / when freely supported. 
V 

and -Ya = /when the edges are securely bolted. 

From experiments on such plates, Mr. T. H. Bryson, of 
Troy, U.S.A., arrived at the expression 
pj^_ 

When the length of a is very great as compared with b, the 
support from the ends is negligible. When the edges are 
rigidly held, the plate simply becomes a built-in beam of 
breadth a, depth t, span b — 

p.ab.~=f-r 
12 

Pl-r 

For a circular plate of diameter d and radius r 
The moment of the water pressure about a^ trr^p 4^ 

diameter j ~ ~2~ ^ rir 

The moment of the reactions of the edge of) tn^p 2r 

the plate when freely supported \ ^ ~^ ^ '^ 



Siress, Strain, and Elasticity. 
The. !^(-)=/£j! ^=/ 

When the edges are rigidly held ^ = / 



407 



w 






W 

Fig. 384. 



Riveted Joints. 

Strength of a Perforated Strip. — If a perforated strip 
of width w be pulled apart in the testing-machine, it will break 
through the hole, and if the material be only very slightly 
ductile, the breaking load will be (approximateLy)«* 
W =ft{'W — d)t, where _;^ is the tensile strength of 
the metal, and t the thickness of the plate. If the 
metal be very ductile the breaking load will be 
higher than this, due to the fact that the tensile 
strength is always reckoned on the original area of 
the test bar, and not on the final area at the point of 
fracture. The difference between the real and the 
nominal tensile strength, therefore, depends upon the 
reduction in area. If we could prevent the area 
from contracting, we should raise the nominal tensile 
strength. In a perforated bar the minimum area 
of the section — through the hole — is surrounded by 
metal not so highly stressed, hence the reduction in 
area is less and the nominal tensile strength is greater 
than that of a plain bar. This apparent increase in strength 
does not occur until the stress is well past the 
elastic limit, hence we have no need to take it 
into account in the design of riveted joints. 

Strength of an Elementary Fin Joint. 
— If a bar, perforated at both ends as shown, 
were pulled apart through the medium of pins, 
failure might occur through the tearing of the 
plates, as shown at aa or bb, or by the shearing 
of the pins themselves. 

Let d = the diameter of the pins ; 
c = the clearance in the holes. 
Then the diameter of the holes =d + c no. 38s. 

LetyS = the shearing strength of the material in the pins. 

For simplicity at the present stage, we will assume the pins 
to be in single shear. Then, for equal strength of plate and 
pins, we have— -^2 




4o8 



Mechanics applied to Engineering. 



If the holes had been punched instead of drilled, a thin 
ring of metal all round the hole would have been 
damaged by the rough treatment of the punch. This 
damaging action can be very clearly seen by ex- 
amining the plate under a microscope, and its effect 
demonstrated by testing two similar strips, in one of 
which the holes are drilled, and in the other punched, 
the latter breaking at a lower load than the former. 

Let the thickness of this damaged ring be — ; then 

2 

the equivalent diameter of the hole will be <^-f-f +K, 
and for equal strength of plate and pins — 

4 

A riveted joint differs from the pin joint in one important 
respect : the rivets, when closed, completely (or ought to) fill 
the holes, hence the diameter of the rivet \^d-\-c when closed. 
In speaking of the diameter of a rivet, we shall, however, 
always mean the original diameter before closing. 

Then, allowing for the increase in the diameter of the rivet, 
the above expressions become, when the plates are not damaged 
as in drilling — 




Fig. 386. 



{■w-{d^c)W, 



T{d + cy 



/ 



When the plates are punched — 

[w-{d + c+TL)]ff,= 



r(d+cY 



-/. 



The value of c, the clearance of a rivet-hole, may be 
taken at about ^V of the original diameter of the rivet, or 
c = o'o^d. 

The diameter of the rivet is rarely less than | inch or 
greater than i^ inch for boiler work ; hence, when convenient, 
we may write c = o'os inch. 

The equivalent thickness of the ring damaged by punch- 
ing may be taken at ^ of an inch, or K = 0-2 inch. 
This value has been obtained by very carefully examining 
all the most recent published accounts of tests of riveted 
joints. 



Stress, Strain, and Elasticity. 



409 



The relation between the diameter of the rivet and the 
thickness of the plate is largely a matter of practical conve- 
nience. In the best modern practice somewhat smaller rivets 
are used than was the custom a few years ago; taking 
d = i"i ■// and {d + cf = r^T,t appears to agree with present 
practice. 

Instead of using 7^, we may write f^ (see 400), where 
/, is the tensile strength of the rivet material. 

The values oifi,f„ B.nAf, may be taken as follows, but if in 
any special case they differ materially, the actual values should 
be inserted. 



Material. 
Iron ... 
Steel ... 



A' fr 



... Jo 24 21 K . , 

W 28 28 \ ^"^ P" square inch. 



Ways in which a Riveted Joint may fail. — Riveted 
joints are designed to be equally strong in tearing the plate 



o 



o 




Tearing plate 
through rivet- 
holes. 



Shearing of rivet. 



-S- 



J^ 



Bursting 

through edge 

of plate; 

Fig. 387. 



-Q- 



Shearing 
through 
edge of 
plate. 







Crushing of 
rivet. 



and in shearing the rivet ; the design is then checked, to see 
that the plates and rivets are safe against crushing. 

Failure through bursting or shearing of the edge of the 
plate is easily avoided by allowing sufficient margin between 
the edge of the rivet-hole and the edge of the plate ; usually 
this is not less than the diameter of the rivet 



See paragraph on Bearing Pressure. 



4IO 



Mechanics applied to Engineering. 



Lap and Single Cover-plate Riveted Joints. — When 
such joints are pulled, the plates bend, as shown, till the two 





Fig. 388. 



Fig. 389. 



lines of pull coincide. This bending action very considerably 
increases the stress in the material, and consequently weakens 
the joint. It is not usual to take this bending action into 
account, although it is as great or greater than the direct stress, 
and is the cause of the dangerous grooving so often found in 
lap-riveted boilers.' 

' The bending stress can be approximately arrived at thus : The 

. P/ 
maximum bendmg moment on the plates is — (see p. 505), where P is 

the total load on a strip of width w. This bending moment decreases as 
the plates bend. Then, /( being the stress in the metal between the 

rivet-holes, the stress in the metal where there are no holes 'i& M — ^— |> 

\ w ) 

hence— 

and the bending moment = -^^ - 

The plate bends along lines where there are no holes ; hence 

Z = ^ (see Chap. XI.) 
6 

and the skin stress due to bending — 



Stress, Strain, and Elasticity. 



411 



Single Row of Rivets. 
Punched iron — 

{w ~ d — c ■ 



4 5 





Fig. 390. 



Fig. 391. 



Then, substituting 0-05 inch for e on the left-hand side of 
the equation, and putting in the numerical value of K as given 
above, also putting {d + cf = i"33/', we have on reduction — 

/ 
w — d = o-83'^ + o'25 inch 

Jt 
w — d= 1*20 inches 

Thus the space between the rivet-holes (w — d) of all punched 
iron plates with single lap or cover joints is i"20 inch, or say 
I5 inch for all thicknesses of plate. 

By a similar process, we get for — 

Punched steel — 

w — d = I '09 inches 

In the case of drilled plates the constant K disappears, 
hence w — diso'2 inch less than in the case of punched plates ; 
then we have for — 



Drilled iron — 
Drilled steel — 



w — d = I'o inch 

m — d = o"89 inch 
Double Row of Rivets. — In this case two rivets have to 





Fig. 393. 



Fig. 393. 



412 



Mechanics applied to Engineering. 



be sheared through per unit width of plate w; hence we 
have — 



Punched iron— 

{iv — d — 



-Yi)ff.=VL^d^c)^^ 
4 5 

/ 
IV — d = i-gi"^ + 0-2S inch 
Jt 



Punched steel — 
Drilled iron — 
Drilled steel — 



w — d = zt6 inches 
w — d = i-g2 inches 
w — d = vg6 inches 
w — d = V]2 inches 



Double Cover-plate Joints. — In this type of joint there 
is no bending action on the plates. Each 
rivet is in double shear; therefore, with a 
single row of rivets the space between the 
rivet-holes is the same as in the lap joint 
with two rows of rivets. The joint shown 
in Fig. 394 has a single row of rivets («.*. in 
each plate). 

Double Row of Rivets. — In this case 
there are two rivets in double shear, which 
is equivalent to four rivets in single shear 
for each unit width of plate w. 

Punched iron — 

(w-d-c-Y:)tf, = ^(d + cj^ 
4 5 




Fig. 394. 



■w-d= 3-347 + 0-25 
It 



r \ i 


w — d = 4*07 inches 


k- "*"-■% \ 


Punched steel — 


000 


w — d= 3*59 inches 


K) 


Drilled iron — 


1 


w — d= 3-87 inches 


• 


Drilled steel— 


Tig. 395. 


w — d=i 3'39 inches 



Stress, Strain, and Elasticity. 



413 



Diamond Riveting. — In this case there are five rivets in 
double shear, which is equivalent to ten rivets in single shear, 




Fig. 396. 

in each unit width of plate w ; whence we have for drilled steel ' 
plates — 

{JV - d - c)if, = ^^{d ^,cf^ 
4 5 

w — d= 8*4 inches 

Combined Lap and Cover-plate Joint. — This joint 
may fail by — 

(i) Tearing through the outer row of rivet-holes. 
' (2) Tearing through the | \ ^m^ 

inner row of rivet-holes and 
shearing the outer rivet (single 
shear). 

(3) Shearing three rivets in 
single shear (one on outer row, 
and two on inner). 

Making ( i ) = (3), we have 
for drilled steel — 




o ' o 1 o o 

■---I — j 

oo6o<i>ooool 



l^ 



Fio. 397. 

(^-u>-d-c)tf, = ^{d + cf^^' 
4 5 

w — d = 2'7i inches 

If we make (3) = (2), we get 

{w-2d- 2c)tf, + -(^ + cf"^ = ^-{d -f cf-^ 
4 5 4 5 

On reduction, this becomes — 

w — 2d= 2*07 inches 

* This joint is rarely used for other than drilled steel plates. 



414 



Mechanics applied to Engineering. 



If the value of d be supplied^ it will be seen that the joint 
will not fail by (2), hence such joints may be designed by 
making (i) = (3). 

Fitch of Rivets. — The pitch of the rivets, i.e. the distance 
from centre to centre, is simply w ; in certain cases, which are 

IS) 

very readily seen, the pitch is — . The pitch for a number of 

joints is given in the table below. The diameters of the 
rivets to the nearest ^V of an inch are given in brackets. 









Iron plates and rivets. 


Steel plates and rivets. 




Thick- 
ness of 


Diameter 
of rivet. 






Type of joint.' 












plate, t. 


d=i-W7 


Punched. 


Drilled. 


Punched. 


Drilled. 








1'20 + d. 


i+d. 


109 + rf. 


0-89 + d. 








in. 


in. 
















\ 


°-67 GS) 


1-87 


1-67 


176 


1-56 


A 


1 




5 


078 (i) 


1-98 


178 


I '87 


1-67 




' 




% 


0-87 (?) 


2-07 


1-87 


1-96 


1-76 








\ 


0-9S HI) 


2-IS 


1-95 


2-04 


1-84 








2-i6 + d. 


1-96 + d. 


1.92 + d. 


lyi + d. 


1 


i 


0-67 (JU 


(a) 2-63 
283 


(a) Z-43 
2-63 


(a) 2-21 
2-S9 


(a) 2-01 
■2-39 


ji ■ 






\ 


o78(i) 


2-94 


2-74 


(a) 2-51 
270 


(a) 2-31 
2-50 


B 1 


\ 




i 


0-87 (?) 


3-03 


2-S3 


(«) 2-76 
279 


(a) 2-56 
2-59 








; 


0-9S «i) 


3-II 


2-91 


2-87 


2-67 


fttf 








1-03(1) 


319 


2-99 


295 


27s 




I 


I -10(11) 


3-26 


3-06 


302 


2-82 








407 + d. 


3'87 + </. 


3-59 + d. 


3-39 + d. 


1 


\ 


078 (1) 


4-68 
4^ 


4-65 


3-99 


3-79 


1 


t 


0-87 (?) 


4-94 


4-74 


4-40 


4-20 


C 1 


\ 


0-95 (ti) 


5-02 


4-82 


4-54 


4-34 


II 


i 


I -03 (I) 


S-io 


4-90 


4-62 


4-42 




I 


1 -10(11) 


S-I7 


4-97 


4-69 


4-49 




14 


Ji7(ift) 


5-24 


5-04 


476 


4-56 



Stress, Strain, and Elasticity. 



415 









Thick- 
ness of 
plate, t. 


Diameter 
of rivet. 


Steel plates and rivets 
(drilled holes). 




Inner row. 


Outer row. 
8-4 +rf. 


1 


1 




in, 

3 

1 

I 
I 
I 
I 


in. 

0-9S (}■§) 
103 (I) 
i-io(ij) 

1-17(1^) 
1-23 (ii) 

1-30 (ift) 


4-67 
4-71 

4;7S 

4-81 
4-85 


9-3S 

9"43 
9-50 


D i 


.■:•.•) 




1 


........... 

• 


9"57 
9-63 


1 


, 1 




970 
















271 + d. 


^ f 


I 


0-78 (3) 
0-87 (?) 
0-9S (H) 
1-03(1) 

I'IO(lJ) 

i-i7(ij|) 


I -75 
1-79 
i;83- 
1-87 
i-gi 
1-94 


3-49 
3-88 



It is found that if the pitch of the rivets along the caulked 
edge of a plate exceeds six times the diameter of the rivet, the 
plates are liable to pucker up when caulked ; hence in the above 
table all the pitches that exceed this are crossed out with a 
horizontal line, and the greatest permissible pitch inserted. 

Bearing Pressure. — The ultimate, bearing pressure on a 
boiler rivet must on no account exceed 50 tons per square inch, 
or the rivet and plate will crush. It is better to keep it below 
45 tons per square inch. The bearing area of a rivet is dt, 
or (^d -{■ c)t. Let_^ = the bearing pressure. 
The total load on a"] 

group of Y^ets r ^^^ ^^ 

m a strip of plate / •'"^ ^ ' 

of width w j 

theloadonastripof. ^ (^ _ ^ _ .yt, or {w - d - c - Y.)f,t 

plate of width ?e// ^ vt> \ ui 

then (w-d-c)f^ = nf^{d + c)t 



and/s 



_ \w-d- c)f. 



]]^So tons per sq. inch 



n{d -f- c) 
The bearing pressure has been worked out for the joints 



4i6 



Mechanics applied to Engineering. 



given in the table above, and in those instances in which it is 
excessive they have been crossed out with a diagonal line, and 
the greatest permissible pitch has been inserted. 
Efficiency of Joints. — 

^''offSU-^-^tl?" 

joint J strength of plate 

effective width of metal between rivet-holes 



w — d — c 



or 



pitch of rivets 
w — d — c —K. 






The table on the following page gives the efficiency of 

joints corresponding to the table of pitches given above. All 

the values are per cent. 

Zigzag Riveting. — In zigzag riveting, if the two rows are 

placed too near together, the plates tear across in zigzag fashion. 

If the material of the plate were equally 

Q Q strong with and across the grain, then a:,, 

\ ,''' "■■, the zigzag distance between the two holes, 

"0' Q X 

-X-^j ^ should be -. The plate is, however. 

Fig. 398. weaker along than across the grain, con- 

sequently when it tears from one row to 
the other it partially follows the grain, and therefore tears more 
readily. The joint is found to be equally strong in both 

directions when Xi = — . 
3 
Riveted Tie-bar Joints. — When riveting a tie bar, a 

very high efficiency 
can be obtained by 
properly arranging the 
-._^ rivets. 

The arrangement 
shown in Fig. 399 is 
radically wrong for 
tension joints. The 
strength of such a 
joint is, neglecting the 
— » clearance in the holes, 
and damage done by 
punching — 

(w — 4d)/^ at aa 



Fig. 399. 
-ted 




abed 

Fic. 400. 



Stress, Strain, and Elasticity. 
Efficiency of Riveted Joints. 



417 



Type of joint. 



It:: 



Thick- 
ness of 
plate. 



i^ 



Iron plates and 
rivets. 



Punched. Drilled 



51 

48 
46 
44 



[a) 65 
67 

65 



63 


67 


61 


66 


60 


64 


59 


62 



78 

77 
76 

75 
74 
73 



57 
53 
51 
49 



(a) 70 
73 
70 



82 
81 

79 
78 

77 
76 



Steel plates and 
rivets. 



Punched. Drilled. 



48 

43 

4' 



{«) 58. 
64 

(a) 59 
62 

(«)59 
60 

58 
57 
55 



74 
74 
74 
72 

71 
70 



54 
50 
48 
46 



(a) 64 
70 

(0)64 
67 

W64 
64 
62 
6r 
59 



78 
78 
77 
76 

74 
73 



87 
87 
86 



72 
70 
69 
67 
66 
65 



41 8 Mechanics applied to Engineering. 

whereas with the arrangement shown in Fig. 400 the strength 

is — 

(w — i)fit at aa (tearing only) 

(k' - ■iisff + -d^f, at bb (tearing and shearing one rivet) 
4 

{w - 2,<^f + — -^X at cc ( „ „ three rivets) 

4 

(w - i4)f,t 4- -^d^f. at dd ( „ „ six „ ) 

4 

By assuming some dimensions and working out the strength 
at each place, the weakest section may be found, which will be 
far greater than that of the joint shown in Fig. 399. The joint 
in Fig. 400 will be found to be of approximately equal strength 
at all the sections, hence for simplicity of calculation it may be 
taken as being — 

(w - d)ff 

In the above expressions the constants c and K have been 
omitted j they are not usually taken into account in such 
joints. 

The working bearing pressure on the rivets should not 
exceed 8 tons per square inch, and where there is much vibra- 
tion the bearing pressure should not exceed 6 tons per square 
inch. 

Tie bar joints are frequently made with double cover 
plates ; the , bearing stress on the rivets in such joints often 
requires more careful consideration than the shearing stress. 
When ties are built up of several thicknesses of plate they 
should be riveted at intervals in order to keep the plates 
close together, and so prevent rain and moisture from entering. 

For this reason the pitch of the rivets in outside work should 
never exceed 12/, where t is the thickness of the outside plates 
of the joint. Attention to this point is also important in the 
case of all compression members, whether for outside or inside 
work, because the plates tend to open between the rivets. 

GrcSups of Rivets. — In the Chapter on combined bend- 
ing and direct stress, it is shown that the stress may be very 
seriously increased by loading a bar in such a manner that 
the line of pull or thrust does not coincide with the centre of 
gravity of the section of the bar. Hence, in order to get the 
stress evenly distributed over a bar, the centre of gravity of the 



Stress, Strain, and Elasticity. 



419 



group of rivets must lie on the line drawn through the centre 
of gravity of the cross-section of the bar. And when two bars 




Fig. 401. 



not in line are riveted, as in the case of the bracing and the 
boom of a bridge, the centre of gravity of the group must lie 




Fig. 402. 

on the intersection of the lines drawn through the centres of 
gravity of the cross-sections of the two bars. 



420 Mechanics applied to Engineering. 

In other words, the rivets must be arranged symmetrically 
about the two centre lines (Fig. 4°i)- 

Punching E£fects. — Although the strength of a punched 
plate may not be very much less than that of a similar drilled 
plate, yet it must not be imagined that the effect of the punch- 
ing is not evident beyond the imaginary ring of ^ inch in 
thickness. When doing some research work upon this question 
the author found in some instances that a i-inch hole punched 
in a mild steel bar 6 inches wide caused the whole of the 
fracture on both sides of the hole to be crystalline, whereas 
the same material gave a clean silky fracture in the unpunched 
bars. Fig. 402 shows some of the fractures obtained. The 
' punching also very seriously reduces the ductility of the bar ; 
in many instances the reduction of area at fracture in the 
punched bars was not more than one-tenth as great as in the un- 
punched bars. These are not isolated cases, but may be taken 
as the general result of a series of tests on about 150 bars of 
mild steel of various thicknesses, widths, and diameters of hole. 



Strength of Cylinders subjected to Internal 
Pressure. 

Thin Cylinders. — Consider a short cylindrical ring i inch 
in length, subjected to an internal 
pressure of p lbs. square inch. 
Then the total pressure tending 
to burst the cylinder and tear the 
plates at aa and bb is /D, where 
D is the internal diameter in 
inches. This bursting pressure 
has to be resisted by the stress 
in the ring of metal, which is 
>2//, where /, is the tensile stress 
in the material. 

^'"- ■'°^' Then/D = if,t 

or/R =f,t 

When the cylinder is riveted with a joint having an efficiency 
■H, we have — 

/R =/A 

In addition to the cylinder tending to burst by tearing 
the plates longitudinally, there is also a tendency to burst 




Stress, Strain, and Elasticity. 421 

circumferentially. The total bursting pressure in this direction 
Is/ttR'', and the resistance of the metal Is zttR^,, where /j is 
the tensile stress in the material. 

Then/7rR2 = 27rR^ 
or/R = 2tff 

and when riveted — 

/R = 2tfy 

Thus the stress on a circumferential section is one-half as 
great as on a longitudinal section. On p. 402 a method is 
given for combining these two stresses. 

The above relations only hold when the plates are very 
thin; with thick-sided vessels the stress is greater than the 
value obtained by the thin cylinder formula. 

Thick Cylinders. 

Barlow's Theory. — When the cylinder is exposed to an 
internal pressure (/), the radii will be increased, due to the 
stretching of the metal. 

When under pressure — 

Let /; be strained to r, + «jr, = ^^(1 + «<) 
r, ■„ „ r, + V. = ''.(i + «.) 

where « is a small fraction indicating 
the elastic strain of the metal, which 
never exceeds yoVo fo'' ^^^^ working 
stresses (see p. 364). 

The sectional area of the cylinder 
will be the same (to all intents 
and purpose:^) before and after 
the a^iication of the pressure ; 
hence — 

Fig. 404. 

T(r.=' - r?) = ^{r.^{i + „,y - r,\i + «,)'} 

which on reduction becomes — 

ri%n? + 2n^) = r,\n,^ + 2«,) 

« being a very small fraction, its square is still smaller and 
may be neglected, and the expression may be written — 




422 Mechanics applied to Engineering. 

The material being elastic, the stress/ will be proportional 
to the strain n ; hence we may write — 

^'' =4 or/,n^ =/,;-." 

re Ji 

that is, the stress on any thin ring varies inversely as the square 
of the radius of the ring. 

Consider the stress / in any ring of radius r and thickness 
dr and of unit width. 

The total stress on any section of the) _ f j 
elementary ring ) 

=LflLdr 

r\ 
= fir?.r-Hr 

The total stress on the whole section) _ , i\'-2j 
of one side of the cylinder ) ~ •'''''' I ^ 

_ /^«v. '-/<n 
— I 

(Substituting the value of /(;»■,' from above) 

— I 
= /'■«-//'. 

This total stress on the section of the cylinder is due to the 
total pressure//-,; hence — 

pn=/,r,-/^. 
Substituting the value of/„ we have — 







pr,= 


M- 


Mr. 
r^ 


Dividing by r 


and 


reducing, we 


have — 






Pr. 
Pr. 




-U, 



Stress, Strain, and Elasticity. 423 

For a thin cylinder, we have — 
pr, =ft 

Thus a thick cylinder may be dealt with by the same form of 
expression as a thin cylinder, taking the presstire to act on the 
external instead of on the internal radius. 

The diagram (Fig. 404, abed) shows the distribution of 
stress on the section of the cylinder, ad representing the stress 
at the interior, and be at the exterior. The curve dc isfr^ = 
constant. 

Lames Theory .^All theories of thick cylinders indicate 
that the stress on the inner skin is 
greater than that on the outer when the 
cylinder is exposed to an internal pres- 
sure, but they do not all agree as to the 
exact distribution of the stress. 

Consider a thin ring i inch long and 
of internal radius ;-, of thickness hr. 

Let the radial pressure on the inner 
surface of the ring be /, and on the 
outer surface {p — Sp), when the fluid 
pressure is internal. In addition to 
these pressures the ring itself is sub- 
jected to a hoop stress_/, as in the thin 
cylinder. Each element of the ring is 
subjected to the stresses as shown in !"'=■ 4°5- 

the figure. 

The force tending to burst this thin ring = / x 2;- . . (i.) 

„ prevent bursting ={ ^V/t^S^S 

These two must be equal — 




2pr = 2pr + 2/8r — 2r%p -|- 2/Sr 



(iii.) 



We can find another relation between/ and p, which will 
enable us to solve this equation. The radial stress/ tends to 
squeeze the element into a thinner slice, and thereby to cause 
it to spread transversely; the stress / tends to stretch the 
element in its own direction, and causes it to contract in 



424 Mechanics applied to Engineering. 

thickness and normal to the plane of the paper. Consider the 

P 
strain of the element normal to the paper ; due to / it is — - 

/ 
(see p. 397), and due to/ it is — ~, and the total longitudinal 

P f 
strain normal to the paper is ^ p. Both /and/, however, 

diminish towards the outer skin, and the one stress depends 
upon the other. 

Now, as regards the strain in the direction/ both pressures 
/and/ act together, and on the assumption just made — 

/ — / = a constant = 2a 

The 2 is inserted for convenience of integration. 
From (iii.) we have — 

Ip 2p 2a dp . . . 

-^ = — — ^ = -f^ m the hmit 
or r r dr 

Integrating we get — . 

b , 

/ = 7. + « 

, b 

where a and b are constants. 

Let the internal pressure above the atmosphere be /<, and 
the external pressure/, = o, we have the stress on the inner 
skin — 

Reducing Barlow's formula to the same form of expression, we 
get— 

fi=Pi^^_,.^ 

Kxperlmental Determination of the Distribution 
of Stress in Thick Cylinders. — In order to ascertain 
whether there was any great difference between the distribution 
of stress as indicated by Barlow's and by Lamp's theories, the 



Stress, Strain, and Elasticity. 



425 



author, assisted by Messrs. Wales, Day, and Duncan, students, 
made a series of experiments on two cast-iron cylinders with 
open ends. Well-fitting plugs were inserted in each end, and 
the cylinder was filled with paraffin wax ; the plugs were then 
forced in by the 100-ton testing machine, a delicate extenso- 
meter, capable of reading to ^ ,'0 „ of an inch, was fitted on 



/rttenor of 
Cylinder 



per ^ 
iMeril 
pressure 



'US 



2-607!m% 
Sqmck 
imefnal.^ 
jfressare-^ 




033T0I 
SqrinA. 
oderiml 
p» press ure 



By experimenJti. 
Lame's theory . 
Barlow^ theory _ 






W 






\ 






^ 



x\ 



-Cyhnder WaU- 



FlG. 406. 



the outside j a series of readings were then taken at various 
pressures. Two small holes were then drilled diametrically 
into the cylinder to a depth of 0*5 inch; pointed pins were 
loosely inserted, and the extensometer was applied to their 
outer ends, and a second set of observations were taken. The 
holes were then drilled deeper, and similar observations taken 



426 



Mechanics applied to Engineering. 



at various depths. From these observations it is a simple 
matter to deduce the proportional strain and the stress. The 
results obtained are shown in Fig. 333^, and for purposes of 
comparison, Lamp's and Barlow's curves are inserted. 

Built-up Cylinders. — In order to equalize the stress over 
the section of a cylinder or a gun, various devices are adopted. 
In the early days of high pressures, 
cast-iron guns were cast round chills, 
so that the metal at the interior was 
immediately cooled; then when the 
outside hot metal contracted, it 
brought the interior metal into com- 
pression. Thus the initial stress in 
a section of the gun was somewhat 
as shown by the line ah, ag being 
compression, and bh tension. Then, 
when subjected to pressure, the curve 
of stress would have been dc as 
before, but when combined with 
ab the resulting stress on the section is represented by ef, thus 
showing a much more even distribution of stress than before. 

This equalizing process is effected in modem guns by 
either shrinking rings on one another in such a manner that 
the internal rings are initially in compression and the externa] 
rings in tension, or by winding wire round an internal tube to 
produce the same effect. The exact tension on the wire re- 
quired to produce the desired eflfect is regulated by drawing the 
wire through friction dies mounted on a pivoted arm — in 
effect, a friction brake. 




Fig. 407. 



Stress, Strain, and Elasticity. 



427 



Strength and Coefficients of Elasticity of Materials in 
Tons square inch. 





Elastic limit. 


Breaking strength. 


E. 

Tension 


G. 

Shear. 


i 




Material. 




■i 






J 










is 










or com- 


s 







1 


s 


i 


1 

c 


1 


Is 


pression. 




■a 


■« 

3 






6 




V 


u 


i 










Wrought - iron) 
iars 5 


12-15 





10-12 


21-24 





17-19 


( 11,000- 

1 13,000 


5,000) 
6,000 j 


10-30 


15-40 


Plates with grain 


13-15 


— 


— 


20-Z2 


— 


— 




— 


5-10 


7-13 


,, across ,, 


H-13 


— 


— 


18-20 


— 


.— 


— 


~- 


2-6 


3-7 


Best Yorkshire, ) 
with grain ... } 

Best Yorkshire,! 
across grain i 


13-14 


lz-14 


— 


20-23 


— 


17-19 


12,000 


— 


15-30 


40-50 






















13-14 






19-20 










IO-3Z 


13-ao 




















Steel, o-i % C. ... 


13-14 




lO-II 


21-22 


— 


16-17 


( 13.000 
(14,000 


5,0001 
6,000 J 


27-30 


45-50 


„ o-=%C.... 


17-18 


16-17 


13-14 


30-32 


— 


34-26 




„ 


20-23 


27-32 


„ o-s%C.... 


20-21 


— 


16-17 


34-35 


— 


28-29 


„ 


„ 


14-17 


17-20 


„ l-o%C.... 


28-29 


22-24 


21-23 


5^55 


■^ 


42-47 


14,000 


„ 


4-5 


7-8 


Rivet steel 


15-17 


— 


12-14 


26-28 


— 


21-22 


■ — 


.— 


30-35 


30-50 


Steel castings ... 


TO-II 


(not 

neal 


an- 
ed) 


30-25 


• - 




(■12,000 
to 


} ~( 


S-I3 


6-13 


„ ,, 


IS-I7 


(ann 


ealed) 


30-40 


/ — 


^- 


^12,500 


} ~- \ 


ro-zo 


15-35 


Tool steel 


35-45 


40-50 


— 


40-70 


(unh 
en 


ard- 
ed) 


/ 14,000 
to 


^ ~~ f 


1-5 


1-5 


^ 


6g-8o 


— 


— 


60-80 


Chard 


ened) 


USjOoo 


/ - \ 


ni! 


nil 


Nickel steel Ni( 
4% ) 


25-30 


- 


— 


35-40 


- 


- 


- 


- 


30-35 


50-5S 


45-65 


— 


__ 


90-95 


— 


— 


— 


— 


10-12 


24-27 


Manganese steel) 
Mn I % ...; 


15 


- 


- 


30-35 


- 


- 


- 


- 


28-33 


- 


., 2 % 


IS 


— 


— 


50-5S 








— 





5-7 


— 


Chrome steel ( 
Cr 5 % - i 


30-40 


- 


- 


60-75 


- 


- 


- 


- 


10-15 


— 


Chrome Vana- } 


65-85 


wat 


cr 


70-go 


(hard 


ened) 


— 


— 


10-15 


45-55 


ft „ ••• 


60-75 


oil 


oil 


65-80 


(hard 


ened) 


— 


— 


8-18 


45-55 


Chrome nickel ... 


45-50 


— 


~" , 


5S-6o 


(ann 


ealed) 


— 


— 


13-15 


50-55 


.. ......{ 


100- 
i°5 


}air 


air| 


105- 
xzo 


(hard 


ened) 


— 


— 


6-8 


30-35 


ti »i ••• \ 


115- 
130 


^oil 


oil{ 


120- 
130 


(hard 


ened) 


- 


- 


5-8 


8-10 


Cast iron 


■jno 


mar 

lim 


ked) 
it S 


7-X1 


3S-6o 


8-13 


' 6,000 
V 10,000 


2.500 \ 

to 
4,000 J 


pra 
cally 


cti- 
nil 


Copper 


2-4 


10-13 
anne 


aled 


12-15 


20-25 


11-12 


1 7.000 

to 


— 


35-40 


50-60 




10-12 


bard 


drawn 


16-20 


— 





. 7.500 


— 


3-5 


40-55 


Gun-metal 


3-4 


S-6 


2 '5-5 


g-i6 


30-50 


8-12 


i S.ooo- 
l 5.5c" 


2,000 ( 
2,500) 


8-15 


lO-iS 



428 



Mechanics applied to Engineering. 





Elastic limit. 


Breaking strength. 








s 
















E. 






g 


Material. 




J 






1 




Tension 
or com- 


G. 

Shear. 


0.0 

1 = 


d 





B 
o 


S 




g 






pression. 




■K 

C3 


'% 






1* 


1 


*c 


a 


i 








% 




H 


t3 


W 


H 


tS 


<g 






M 


£ 


Brass 


2-4 






7-10 


5-6 




4,000 


2,000 


10-12 


12-15 


Delta, bull metal. 






















etc.— 






















Cast 


5-8 


12--14 


— 


14-20 


60-70 


— 


( 5.50° 
\ '° 


1 — 
^ - 


8-16 


10-22 


Rolled 


IS-2S 


l6-22 





27-34 


45-60 


— 


( 6,000 




17-34 


27-50 


Phosphor bronze 


7-9 








24-26 




— 


6,000 


2,500 






Muntz metal 


20-25 


(roll- 
ed) 

8-10 


— 


=5-30 


— 


— 


— 


— 


10-20 


30-40 


Aluminmm 


2-7 


- 


7-10 


- 


5-6 


J 3,000 
( 5,000 


} 1,700 


4-iS 


30-70 


Duralumin 


4-S 


— 


— 


22-24 


( 


with 


( 4,200 
i 4,500 
\ 


}- 


10-12 


30-38 


Oak 








4-6 


( 


grain 
0*2 
0*07 


500-700 








Soft woods 








1-3 


x-3{ 


to 
0-4 


450-500 









CHAPTER XI. 

BEAMS. 

The beam illustrated in Fig. 408 is an indiarubber model 
used for lecture purposes. Before photographing it for this 
illustration, it was painted black, and some thin paper was stuck 
on evenly with seccotine. When it was thoroughly set the 
paper was slightly damped with a sponge, and a weight was 
placed on the free end, thus causing it to bend ; the paper 
on the upper edge cracked, indicating tension, and that on the 
lower edge buckled, indicating compressioii, whereas between 
the two a strip remained unbroken, thus indicating no longi- 




FlG. 40S. 

tudinal strain or stress. We shall see shortly that such a result 
is exactly what we should expect from the theory of bending. 

General Theory. — The T lever shown in Fig. 409 is 
hinged at tjie centre on a pivot or knife-edge, around which the 
lever can turn. The bracket supporting the pivot simply takes 
the shear. For the lever to be in equilibrium, the two couples 
acting on it must be equal and opposite, viz. W/ = px. 

Replace the T lever by the model shown in Fig. 410. It 



430 



Mechanics applied to Engineering. 



is attached to the abutment by two pieces of any convenient 
material, say indiarubber. The upper one is dovetailed, be- 
cause it is in tension, and the lower is plain, because it is in 




--^- 




FlG. 4og, 



Fig. 410. 



compression. Let the sectional area of each block be a ; then, 
as before, we have "^l = px. But/ =fa, where /= the stress 
in either block in either tension or compression ; 



Or- 



hence W/ = fax 
or = 2fay 



The moment of] fthe moment of the internal forces, or the 
the external I = ! internal moment of resistance of the 
forces I I beam 

= stress on the area a X (moment of the two 
areas {a) about the pivot) 

Hence the resistance of any section to bending — apart alto- 
gether from the strength of the material of the beam — ^varies 
. directly as the area a and as the distance x, or as the moment 
ax. Hence the quantity in brackets is termed the " measure 
of the strength of the section," or the " modulus of the section," 
and is usually denoted by the letter Z. Hence we have — 

W/ = M =/Z, or 
The bending moment^ _ J stress on the) 



(modulus of the 



at any section ) \ material f ( section 

The connection between the T lever, the beam model 01 
Fig. 410, and an actual beam may not be apparent to some 
readers, so in Fig. 411 we show a 
rolled joist or I section, having top and 
bottom flanges, which may be regarded 
as the two indiarubber blocks of the 
model, the thin vertical web serves the 
purpose of a pivot and bracket for 
taking the shear; then the formula that 
we have just deduced for the model 
applies equally well to the joist. We shall have to slightly 




Fio. , 



Beams. 



431 



modify this statement later on, but the form in which we have 
stated the case is so near the truth that it is always taken in 
this way for practical purposes. 

Now take a fresh model with four blocks instead of 
two. When loaded, the 

outer end will sag as ^ (yy) 

shown by the dotted ^ |/o " 
lines, pivoting about the 
point resting on the 
bracket. Then the outer 
blocks will be stretched 
and compressed, or 
strained, more than the 
inner blocks in the ratio 

— or ^ : i.e. the strain is directly proportional to the distance 

«" ^1 

from the point of the pivot. 

The enlarged figure shows this more distinctly, where e, e^ 
show the extensions, and c, Cj show the compressions at the 
distances y, y^ from the pivot. From the similar 




Fl'G, 41a. 



triangles, we have 



also - = ^. But we 



= y. ; also i = y. 

ex y-L h yi 

have previously seen (p. 375) that when a piece 
of material is strained {i.e. stretched or compressed), 
the stress varies directly as the ^'asxa, provided the 
elastic limit has not been passed. Hence, since the 
strain varies directly as the distance from the pivot, 
the stress must also vary in the same manner. 

Let/= stress in outer blocks, and/j = stress 
in inner blocks ; then — 

Z = Zor/,=^ 

/i yx y 

Then, taking moments about the pivot as before, we have™ 
W/ = if ay + z/ioyi 
Substituting the value of/i, we have — 




W/ = 2fay + 



z/ayi' 



If V, = ^. W/ = 2fay + ^* 
2 Ay 

W/ = 2fay(x + \) 



432 



Mechanics applied to Engineering. 



Thus the addition of two inner blocks at one-half the 
distance of the outer blocks from the pivot has only increased 
the strength of the beam by \, or, in other words, the four-block 
model will only support x\ times the load (W) that the two- 
block model will support. 

If we had a model with a very large number of blocks, or a 
beam section supposed to be made up of a large number of 
layers of area a, a^, a^, a,, etc., and situated at distances y, y-^, 
yi, yz, etc., respectively from the pivot, which we shall now 
term the neutral axis, we should have, as above — 



W/ = 2fay + 



2A.yi' 4. zAj-a" + zAjI'a' ^_ gtj._ 



Fig. 414. 



= ■^2((zy= -j- a^yy^ + a-ij/i + a^y^^ -\-, etc.) 



The quantity in brackets, viz. each 
little area (a) multiplied by the square of 
its distance {y'') from a given line (N.A.), 
is termed the second moment or moment 
of inertia of the upper portion of the 
beam section ; and as the two half-sec- 
tions are similar, twice the quantity in 
brackets is the moment of inertia (I) of 
the whole section of the beam. Thus we 
have — 




W/ = 



/I 



or 



The bending moment at any section 

the stress on thei (second moment, or moment of 
_ outermost layer ( ( inertia of the section 

distance of the outermost layer from the neutral axis 

But we have shown above that the stress varies directly as the 
distance from the neutral axis j hence the stress on the outer- 
most layer is the maximum stress on any part of the beam 
section, and we may say — 

The bending moment at any section 

the maximum stress) I second moment, or moment of 
on the section ) 1 inertia of the section 



distance of the outermost layer from the neutral axis 



Beams, 433 

But we have also seen that W/ = /Z, and here we have 

therefore Z = - 

The quantity - is termed the " measure of the strength of the 

section," or, more briefly, the " modulus of the section ; " it is 
usually designated by the letter Z. Thus we get W/ = /Z, or 
M=/Z, or— 

moment' aU = i*^ maximum or skin) (the modulus of 
any section) ' stress on the section) •^ \ the section 

Assumptions of Beam Theory. — To go into the 
question of all the assumptions made in the beam theory 
would occupy far too much space. We will briefly consider 
the most important of them. 

First Assumption. — That originally plane sections of a 
beam remain plane after bending; that is to say, we assume 
that a solid beam acts in a similar way to our beam model 
in Fig. 412, in that the strain does increase directly as the 
distance from the neutral axis. Very delicate experiments 
clearly show that this assumption is true to within exceedingly 
narrow limits, provided the elastic limit of the . material is not 
passed. 

Second Assumption. — That the stress in any layer of a beam 
varies directly as the distance of that layer from the neutral 
axis. That the strain does vary in this way we have just 
seen. Hence the assumption really amounts to assuming that 
the stress is proportional to the strain. Reference to the 
elastic curves on p. 364 will show that the elastic line is 
straight, i.e. that the stress does vary as the strain. In most 
cast materials the line is unquestionably slightly curved, but for 
low (working) stresses the line is sensibly straight. Hence for 
working conditions of beams we are justified in our assumption. 
After the elastic limit has been passed, this relation entirely 
ceases. Hence the beam theory ceases to hold good as soon as the 
elastic limit has been passed. 

Third Assumption. — That the modulus of elasticity in 
tension is equal to the modulus of elasticity in compression. 
Suppose, in the beam model, we had used soft rubber in the 

2 F 




434 Mechanics applied to Engineering. 

tension blocks and hard rubber in the compression blocks, i.e. 
that the modulus of elasticity of the tension blocks was less 
than the modulus of elasticity of the compression blocks ; then 
the stretch on the upper blocks would be greater thjui the 
compression on the lower blocks, with the result that the beam 
would tend to turn about some point other than the pivot, Fig. 
415; and the relations given above entirely cease to hold, for 
the strain and the stress will not vary directly as 
the distance from the pivot or neutral axis, but 
directly as the distance from a, which later on we 
shall see how to calculate. 

For most materials there is no serious error in 
making this assumption ; but in some materials the 
error is appreciable, but still not sufficient to be of 
any practical importance. 

Neither of the above assumptions «t& perfectly 
Fig. 415. true ; but they are so near the truth that for all 
practical purposes they may be considered to be 
perfectly true, but only so long as the elastic limit of the material 
is not passed. In other parts of this book definite experi- 
mental proof will be given of the accuracy of the beam 
theory. 

Graphical Method of finding the Modulns of the 

Section (Z = -). — The modulus of the section of a beam 

y 

might be found by splitting the section up into a great many 

layers and multiplying the area of each by ^, as shown above. 

The process, however, would be very tedious. 

But in the graphic method, instead of dealing with each 
strip separately, we graphically find the magnitude of the 
resultant of all the forces, viz.-^ 

fa +fiai -^fa^ +, etc. 

acting on each side of the neutral axis, also the position or 
distance apart of these resultants. The product of the two 
gives us the moment of the forces on each side of the neutral 
axis, and the sum of the two moments gives us the total amount 
of resistance for the beam section, viz./Z. 

Imagine a beam section divided up into a great number of 
thin layers parallel to the neutral axis, and the stress in each 
layer varying directly as its distance from it. Then if we con- 
struct a figure in which the width, and consequently the area, 



Beams. 43 S 

of each layer is reduced in the ratio of the stress in that layer 
to the stress in the outermost layer, we shall have the intensity 
of stress the same in each. Thus, if the original area of the 
layer be «i, the reduced area of the layer will be — 



<« 



yy- say «! 
whence we havey^Oj =fa^ 

Then the sum of the forces acting over the half-section, viz. — 

fa +/,<?! 4-/2«a +, etc. 
becomes /a +fal ■\-fai +, etc. 
or/(a + «i' + flJa' +, etc 
or /(area of the figure on one side of the neutral axis) 
or (the whole force acting on one side of the neutral axis) 

Then, since the intensity of stress all over i!a& figure is the 
same, the position of the resultant will be at the centroid or 
centre of gravity of the figure. 

Let Ai = the area of the figure below the neutral axis ; 
Aa = „ „ above „ „ 

d-^ = the distance of the centre of gravity of the lower 

figure from the neutral axis ; 
di = the distance of the centre of gravity of the upper 
figure from the neutral axis. 

Then the moment of all the forces acting! _ , . 

on one side of the neutral axis | " -^ ' ^ ' 

Then the moment of all the forces acting j _ fi\ j \ \ j\ 
on both sides of the neutral axis f ~ ■^^^•'^' + ^^'^''> 

= /Z(seep. 354) 
or Z = A,rfi + Aj^/a 

We shall shortly show that Ai = Aj = A (see next para- 
graph). 

Then Z = A(^i + 4) 
Z = AD 

where D is the distance between the two centres of gravity. 
In a section which is symmetrical about the neutral axis 
d = di — d, and (^ + (^a = 2;/- 



436 



Mechanics applied to Engineering. 



Then Z = 2M 
or Z = AD 

The units in which Z is expressed are as follows — 

Z = U (length units)* ^ ^ j^ ^^i^^j3 
y length 

or Z = AD = area X distance 

= (length units)'' X length units 
= (length units)' 

Hence, if a modulus figure be drawn, say, - full size, the 

result obtained must be multiplied by «* to get the true value. 
For example, if a beam section were 
drawn to a scale of 3 inches = i foot, 
i.e. \ full size, the Z obtained on that 
scale must be multiplied by 4' = 64. 

We showed above that in order to 
construct this figure, which we will term 
a "modulus figure," the width of each 
strip of the section had to be reduced in 

the ratio ^, which we have previously seen 

is equal to — . This reduction is easily 

done thus: Let Fig. 416 represent a section through the 
indiarubber blocks of the beam model. Join ao, bo, cutting 
^ ,. ,# the inner block in c and d. Then by 
similar triangles — 




t: 



zszzi 



Fig. 416. 




r 



ab 



or w{ 



' = <y) 



the strip in c and d. 



In the case of a section in which 
the strips are not all of the same width, 
the same construction holds. Project 
the strip ab on to the base-line as 
shown, viz. db'. Join «'<?, l/o, cutting 
By similar triangles we have — 



^^■^ ^1=7=7. °^-'^ = «^(7) 



Beams. 437 

Several fully worked-out sections will be given later on. 

By way of illustration, we will work out the strength of the 
four-block beam model by this method, and see how it agrees 
with the expression found above on p. 432. 

The area A of the modulus figure onl _ ^ 1 ^ ' 

one side of the neutral axis / ' 

The distance d of the c. of g. of the modulusl _^_+J^ 



of the c. of g. of the modulus! _ ay + Oiy-. 
; neutral axis (see p. 58) / a + Oi' 



figure from the 

ButW/=/Z =/X 2Ad 
or W/ = 2/(a + a/) X ~^rff = ^/^^ + 2>i>" 
But a,' = a^, .-. W/ = 2fay + ^^^ 

which is the same expression as we had on p. 431. 

The graphic method of finding Z should only be used 
when a convenient mathematical expression cannot be 
obtained. 

Position of Neutral Axis. — We have stated above 
that the neutral axis in a beam section corresponds with the 
pivot in the beam model; on the one side of the neutral axis 
the material is in tension, and on the other side in compression, 
and at the neutral axis, where the stress changes from tension to 
compression, there is, of course, no stress (except shear, which 
we will treat later on). In all calculations, whether graphic or 
otherwise, the first thing to be determined is the position of the 
neutral axis with regard to the section. 

We have already stated on p. 58 that, if a point be so 
chosen in a body that the sum of the moments of all the 
gravitational forces acting on the several particles about the 
one side of any straight line passing through that point, be 
equal to the sum of the moments on the other side of the 
line, that point is termed the centre of gravity of the body ; or, 
if the moments on the one side of the line be termed positive 
( + )( and the moments on the other side of the line be termed 
negative (— ), the sum of the moments will be zero. We are 
about to show that precisely the same definition may be used 
for stating the position of the neutral axis ; or, in other words, 
we are about to prove that, accepting the assumptions given 
above, the neutral axis invariably passes through the centre of 
gravity of the section of a beam. 



438 



Mechanics applied to Engineering. 



liBt the given section be divided up into a large number of 
strips as shown — 

Let the areas of the strips above 
the neutral axis be fli, (h, ^a> etc. ; 
and below the neutral axis be a/, ^a', 
a,', etc. ; and their respective distances 
above the neutral axis yi, y^, y^, etc. ; 
ditto below j/, y^, yl, etc : and the 
stresses in the several layers above 
the neutral axis be /i, ft, /s, etc. ; ditto 
below the neutral axis be //, fi, fi- 

Then, as the stress in each layer 
varies directly as its distance from 




y',^i 



Fig. 4i3. 



the neutral axis, we have — 



^=^. and/.=^^ 

U y2 yi 



also ^' = ^ 

7i yi 



The total stress in all the layers] =/i«i + fnHi +,etc. 

on one side of the neutral I /,, , , .. \ 

I = -(aiyi + «a;'2 +, etc.) 



axis 1 j/j 

The total stress in all the layers) 



on the Other side of the neutral! =^(fli,'j'i' + a^y^' +, etc.) 

j y. 



axis 



But as the tensions and compressions form a couple, the 
total amount of tension on the one side of the neutral axis 
must be equal to the total amount of compression on the other 
side ; hence — 

ay + a^yi + (jys +. etc. = dy' + a/j/ + aiyl +, etc. 

or, expressed in words, the sum of the moments of all the ele- 
mental areas on the one side of the neutral axis is equal to the 
sum of the moments on the other side of the neutral axis ; but 
this is precisely the definition of a line which passes through 
the centre of gravity of the section. Hence, the neutral axis 
passes through the centre of gravity of the section. 

It should be noticed that not one word has been said in the 
above proof about the material of which the beam is made ; all 
that is taken for granted in the above proof is that the modulus 
of elasticity in tension is equal to the modulus of elasticity in 



Beams. 439 

compression. The position of the neutral axis has nothing 
whatever to do with the relative strengths of the material in 
tension and compression. In a reinforced concrete beam the 
modulus of elasticity of the tension rods differs from that of 
the concrete, which is in compression. Such beams are dealt 
with later on. 

Unsymmetrical Sections, — In a symmetrical section, 
the centre of gravity is equidistant from the skin in tension and 
compression; hence the maximum stress on the material in 
tension is equal to the maximum stress in compression. Now, 
some materials, notably cast iron, are from five to six times as 
strong in compression as in tension ; hence, if we use a 
symmetrical section in cast iron, the material fails on the 
tension side at from ^ to -^ the load that would be required to 
make it fail in compression. In order to make the beam 
equally strong in tension and compression, we make the section 
of cast-iron beams of such s.form that the neutral axis is about 
five or six times ^ as far from the compression flange as from 
the tension flange, so that the stress in compression shall be five 
or six times as great as the stress in tension. It should be 
•particularly noted that the reason why the neutral axis is nearer 
the one flange than the other is entirely due to the form of the 
section, and not to the material ; the neutral axis would be in 
precisely the same place if the material were of wrought iron, 
or lead, or stone, or timber (provided assumption 3, p. 433, is 

true). We have shown above on p. 432 that M =/-, and that 

I . •'' 

Z = -, where VIS the distance of the skin from the neutral axis. 

y . . 

In a symmetrical section y is simply the half depth of the section ; 

but in the unsymmetrical section y may have two values : the 

distance of the tension skin from the neutral axis, or the 

distance of the compression skin from the neutral axis. If 

the maximum tensile stress ft is required, the 7, must be taken 

as the distance of the tension skin from the neutral axis ; and 

likewise when the maximum compressive stress /<, is required, 

the y^ must be measured from the compression skin. Thus we 

have either — 

' We shall show later on that such a great difference as 5 ot 6 is 
undesirable for practical reasons. 



440 



Mechanics applied to Engineering. 



and as - = — , we get precisely the same value for the bending 

yt y<i 

moment whichever we take. We also have two values of 

^ ■ I ^ ,1 r. 

Z, VIZ. - = Z, and - = Z. ; 

andM =y^Z, or/^. 

We shall invariably take /jZ, when dealing with cast-iron 
sections, mainly because such sections are always designed in 
such a manner that they fail in tension. 

The construction of the modulus figures for such sections is 
a simple matter. 



G^mpressioTV 6as& Zma, 




yA / 




Comji ressiorh 
h<xse'tmj& 



Xensio7v 'base Zirve 
Fig. 419. 



Fig. 420. 



Construction for Z^ (Fig. 419). — Find the centre of gravity 
of the section, and through it draw the neutral axis parallel 
with the flanges. Draw a compression base-line touching the 
outside of the compression flange ; set off the tension base-line 
parallel to the neutral axis, at the same distance from it as the 
compression base-line, viz. _j'^; project the parts of the section 
down to each base-line, and join up to the central point which 
gives the shaded figure as shown. Find the centre of gravity 
of each figure (cut out in cardboard and balance). Let 
D = distance between them ; then Zo = shaded area above or 
below the neutral axis X D. 

Construction for Z, (Fig. 420). — Proceed as above, only 
the tension base-line is made to touch the outside of the tension 
flange, and the compression base-line cuts the figure ; the parts 
of the section above the compression base-line have been pro- 
jected down on to it, and the modulus figure beyond it passes 



Beams. 441 

outside the section ; at the base-lines the figure is of the same 
width as the section. The centre of gravity of the two figures 
is found as before, also the Z. 

The reason for setting the base-lines in this manner will be 
evident when it is remembered that the stress varies directly as 
the distance from the neutral axis; hence, the stress on the 
tension flange,/^ is to the stress on the compression flange /„ as 
yt is to y^ 

N.B. — The tension base-line touches the tension flange when the figure 
is being drawn for the tension modulus figure Z,, and similarly for the 
compression. 

As the tensions and compressions form a couple, the total 
amount of tension is equal to the total amount of compression, 
therefore the area of the figure above the neutral axis must be 
equal to the area of the figure below the neutral axis, whether 
the section be symmetrical or otherwise; but the moment of 
the tension is not equal to the moment of the compression 
about the neutral axis in unsymmetrical sections. The 
accuracy of the drawing of mo Julus figures should be tested by 
measuring both areas ; if they only differ slightly (say not more 
than 5 per cent.), the mean of the two may be taken ; but if the 
error be greater than this, the figure should be drawn again. 

If, in any given instance, the Z^ has been found, and the Z, 
is required or vice versA, there is no need to construct the two 
figures, for — 

Z, = Z. X ^-°, or Z„ = Z, X -^ 

.;'. yc 

hence the one can always be obtained from the other. 

Most Economical Sections for Cast-iron Beams.— 
Experiments by Hodgkinson and others show that it is un- 
desirable to adopt so great a difference as 5 or 6 to i between 
the compressive and tensile stresses. This is mainly due to 
the fact that, if sections be made with such a great difference, 
the tension flange would be very thick or very wide com- 
pared with the compression flange ; if a very thick flange were 
used, as the casting cooled the thin compression flange and 
web would cool first, and the thick flange afterwards, and set 
up serious initial cooling stresses in the metal. 

The author, when testing large cast-iron girders with very 
unequal flanges, has seen them break with their own weight 
before any external load was applied, due to this cause. Very 
wide flanges are undesirable, because they bend transversely 
when loaded, as in Fig. 421. 



442 



Mechanics applied to Engineering. 



Experiments appear to show that the most economical section 
for cast iron is obtained when the proportions are roughly those 
given by the figures in Fig. 421. 

"Massing up" Beam Sections.— Thin hollow beam 



CT^f 



/o 



[■■- i- 



/■s 



a 



Fig. 421. 



Fig. 422. 



sections are usually more convenient to deal with graphically if 
they are " massed up " about a centre line to form an equiva- 
lent solid section. " Massing up " consists of sliding in the 
sides of the section parallel to the neutral axis until they meet 
as shown in Fig. 422. 

The dotted lines show the original position of the sides, 
and the full lines the sides after sliding in. The " massing up " 
process in no way affects the Z, as the distance of each section 
from the neutral axis remains unaltered j it is done merely for 
convenience in drawing the modulus figure. In the table of 
sections several instances are given. 



Section. 



Rectangular. 



Square. 



Examples of Modulus Figures. 
< S » 




Fig. 423. 



B = H = S (the side of the square) 



Beams. 



443 



Those who have frequently to solve problems involving 
the strength and other properties of rolled sections will do 
well to get the book of sections issued by The British Standards 
Committee. 



Modulus of the 
section Z. 


Remarks. 


BH' 
6 


BH' 
The moment of iuettia (sec p. 8o) = 




H 




y=2 




BH» 




., 12 BH« 
^ H 6 




2 




Also by graphic method — 


(Square) 
S' 
6 


The area A = — 
4 
^ 2 H H 
-'=3X2=1 




BH H BH» 
Z = ZAJ =2X -—X T=-fi- 
430 



444 Meclianics applied to Engineering. 



Section. 



Hollow rect- 
angles and girder 
sections. 



One corruga- 
tion of a trough 
flooring. 



Examples of Modulus Figures. 




" Corrtjor^c.sston' 



NA- 




^ 






Te.n^torv 



k 




Beams, 



445 



Modulus of the 
section Z. 

BH' - bh' 
6H 



Approximate- 
ly, when the 
web is thin, 
as in a rolled 
joist — 

Bm 

where t is the 
mean thickness 
of the flange. 



BH* 
Moment of inertia for outer rectangle = 



;' = 



inner „ 


_bh' 
12 


hollow „ 
H 

2 

BH» - bh' 


BH' - bh* 

12 


12 BH» - W 


H 

2 


6H 



z = 



This might have been obtained direct from the Z for 
the solid section, thus — 

Z for outer section = —t— 

^ . bhf h bh* 

Z., mner „ =-x- = — 

„ , ,, BH' bh* BH' - bh* 

Z „ hollow ,, = = 

6 6H 6H 

The Z for the inner section was multiplied by the ratio 

-^, because the stress on the interior of the flange is less 

than the stress on the exterior in the ratio of their distances 
from the neutral axis. 

The approximate methods neglect the strength of the 
web, and assume the stress evenly distributed over the two 
flanges. 

For rolled joists B;H is rather nr-«er the truth than 
B^Hg, where Ho is measured to the middle of the flanges, 
and is more readily obtained. For almost all practical 
purposes the approximate method is sufficiently accurate. 

N.B. — ^The safe loads given in makers' lists for their 
rolled joists are usually too high. The author has tested 
some hundreds in the testing-machine on both long and 
short spans, and has rarely found that the strength was 
more than 7$ per cent, of tiiat stated in the list. 

In corrugated floorings or built-up sections, if there are 
rivets in the tension flanges, the area of the rivet-holes 
should be deducted from the BA Thus, if there are « 
rivets of diameter d in any one cross-section, the Z will 
be— 

(B - nd)^ (approx.) 



446 Mechanics applied to Engineering. 

Section. Examples of Modulus Figures. 



Square on 
edge. 




Fig. 426. 



Tees, angles, 
and LI 
sections. 





Fig. 427. 



■ t 

I This figure 

, , H becomes a T or 
ffAj^. \ .\ u when massed 
i up about a ver- 
tical line. 



Fig. 428. 



Beams. 



447 



Modulus of the 
section Z. 



0-II8S' 



The moment of inertia (see p. 88) = — 

S 






§1 

12 VzS' 



= oii8S» 



BiH,'+B.,IV-3A' 



3H, 

H, = 07H approx. 
H, = 0-3H „ 



The moment of inertia for \ _ BiH|* 
the part above the N.A./ ~ 3 

Ditto below = -^— ^ 

3 

Ditto for whole section = ^'^■' + ^'^' ~ ^'^' 

3 
~, , ,, B,H,' + BjHj' - M' 

Z for stress at top = — — * — ■ — ^-2 

3H, 

If the position of the centre of gravity be calculated 
for the form of section usually used, it will be found 
to be approximately 0"3H from the bottom. 

Rolled sections, of course, have not square comers 
as shown in the above sketches, but the error involved 
is not material if a mean thickness be taken. 



448 Mechanics applied to Engineering. 



Section. 



T section 
on edge and 
cruciform sec- 
tions. 



Examples of Modulus Figures. 






2 



ff B > 



Fig. 439. 



Unequal 
flanged sec- 
tions. 







ci,,,::::. 


...i 


f 






f'<' 


,.■ 






c--:.- 


--ja 
.if' 


'"^9 


h 


:■■-::::,:„ 


'y 







Fig. 430. 



Compressitm base'lim 



Tension hose 
line 




Modulus of the section Z, 

iW + B/4» 
6H 

AppTOT. Z 

6 


Beams. 449 

ITTJ 

Moment of inertia for vertical part = 

^ 12 

„ „ horizontal „ = -^ 

iH'+Bi' 
„ ,, whole section = — 

, iH'+BA' 


Zfor „ - gjj 

Or this result may be obtained direct from the 
moduli of the two parts of the section. Thus — 

Z for vertical part = -g- 
„ horizontal „ = _ ^ g (see p. 445) 

" ''l>°l'='=«'=f°°= 6 +6H- 6H 

It should be observed in the figure how 
very little the horizontal part of the section 
adds to the strength. The approximate Z 
neglects this part. 

The strength of the T section when bent in 
this manner is very much less than when bent 
as shown in the previous figure. 


B,H,'+BjH,'-*,V-Vi' 
3H, 


Moment of inertia) B,H,' i,Ai' 
for upper part ) ~ 3 3 

Ditto lower part = ^'"'' ^'^'' 

Moment ^ 
of inertia B,H,»+B,n,»-i,.S,>-*A' 
for whole 3 
section 

Z = Tj- for the stress on the tension flange 

. . 



4SO 



Section, 



Mechanics applied to Engineering. 

Examples of Modulus Figures. 



...JBt.-, 



M/1. 



h 



FlO. 432. 



This 
_ figure be- 
! I comes the' 
i J, same as 
'^{T the last 
i I when 
-f— massed 
Hi about a 
i centre 
line. 



Triangle. 



///« 




Trapezium. 



AfA 




Modulus of the 
section Z. 


Beams. 45 1 

The construction given in Fig. 4.30 is a con- 
venient method of finding the centre of gravity of 
such sections. 

Where ab = area of web, cd = area of top flange, 
ef = area of bottom flange, gh = ab ■\- cd. The 
method is fully described on p. 63. 


For stress at base — 
12 

For stress at apex — 

BH' 

24 


The moment of inertia for al BH' , ^ ^. 
triangle about its c. of g. / 36 ^ ^' 

H 2H 
y for stress at base = — , at apex = — 
J A 

BH» 
Z for stress at base = i^ = 

T 

BH» 
apex= ^^ 


For stress at wide 
side — 

BH'/'«'+4« + i'\ 

12 V, 2« + I J 

For stress at narrow 
side — 

BIP/'«'+4» + l '\ 
12 \ K + 2 ) 

Approximate value 
forZ— 


Moment of ^ BH» ^ »' + 4« + i ^ , ^^ gg. 
inertia / 36 ^ « + 1 j^ ^' ' 

y for stress at wide side = H, = — T ^^^y J 

BH' / «' + 4» + I N 

Z for stress at wide side = 3^ \ « + 1 7 
H / 2« + I N 

3 *^ » + i y 

BH' Z' «' + 4« + I ^ 
~ 12 ^ 2« + I y 

y for stress at narrow side = Hj = — f " j 

and dividing as above, we get the value given in the 
column. 

The approximate method has been described on 
p. 86. It must not be used if the one side is more 
than twice the length of the other. For error in- 
volved, see also p. 87. 



45.2 Mechanics applied to Engineering. 



SectloD. 



Circle. 



Hollow 
circle and 
corrugated 
section. 



Examples of Modulus Figuzvt. 




Fig. 435. 




Beams. 



4S3 



Modulus of the 
section Z. 






10*2 



The moment of inertia of a circle \ _ "'D* 
about a diameter (see p. 88) / ~ 64 

D ' 

D - 32 

2 



ir(D' - D.«) 
32D 



The moment of inertia of a hollow circle \ _ t(D* — Dj*) 
about a diameter (see p. 88) / ~ 64 

D 



^=2 



Z = 



t(D' - D(') 
64 



D 

2 



., _ ir(D' - D««) 
32D 
This may be obtained direct from the Z thus — 



Z for outer circle = 



irD' 
32 

'^^''xg' (see p. 445) 



hollow 



32 

^ t(D' - D/) 
32D 



For corrugated sections in which tbe corrugations are not 
perfectly circular, the error involved is very slight if the 
diameters D and Di are mea- 
sured vertically. The expres- 
sion given is for one corruga- 
tion. It need hardly be pointed 
out that the corrugations must 
not be placed as in Fig. 438. F'°' 438. 

The strength, then, is simply that of a rectangular section of 
height H = thickness of plate. 




454 Mechanics applied to Engineering. 



Irregular sections. 



Bull-headed 
rail. 



Examples of Modulus Figures. 




Fig. 439. 



Flat-bottomed 
rail. 




Fig. 440. 



Tram rail 
(distorted). 




Fig. 441, 



Beams. 



455 



Irregular sections. 



Bulb section. 



Hobson's 
patent floor- 
ing. 



Fireproof 
flooring. 



Examples of Modulus Figures. 




Fic. 442' 




One section. 




Four sections massed up. 
Fig. 443. 




Fig. 444- 



45 6 Mechanics applied to Engineering. 



Irregular sections. 



Fireproof 
flooring. 



Examples oC Modulus Figure 



4- 



FlG. 445. 



Table of 
hydraulic 
press. 







FlQ. 446. 



Beams. 



457 




Fig. 447. 



Shear on Beam Sections. — In the Fig. 447 the rect- 
angular element abed on the unstrained beam becomes aW(^ 
when the beam is bent, and the 
element has undergone a shear. 
The total shear force on any 
vertical section = W, and, assum- 
ing for the present that the shear 
stress is evenly distributed over 
the whole section, the mean shear 

W 
stress = x, where A = the area 

of the section ; or we may write it 

W 
T. TT . But we have shown (p. 

390) that the shear stress along 

any two parallel sides of a rectangular element is equal to 

the shear stress along the other two 

parallel sides, hence the shear stress 

W 
along cd is also equal to x • 

The shear on vertical planes tends 
to make the various parts of the 
beam slide downwards as shown in 
Fig. 448, a, but the shear on the 
horizontal planes tends to make the 
parts of the beam slide as in Fig. 
448, b. This action may be illustrated 
by bending some thin strips of wood, 
when it will be foimd that they slide 
over one another in the manner shown, 
often fail in this manner when tested. 

In the paragraph above 
we assumed that the shear 
stress was evenly distributed 
over the section; this, how- 
ever, is far from being the 
case, for the shearing force 
at any part of a beam section 
is the algebraic sum of the 
shearing forces acting on 
either side of that part of 

the section (see p. 479). We "*' 

will now work out one or ra-449- 

two cases to show the distribution of the shear on a beam 



FlQ. 448. 



Solid timber beams 





By 




z. 




* 




^^. 






\ ; 


Y 


y^:--\ 


y\ ^ 


a; 


NA 


A 




/\ 


A 




y 



4S8 



Mechanics applied to Engineering. 



section by a graphical method, and afterwards find an analytical 
expression for the same. 

In Fig. 449 the distribution of stress is shown by the width 
of the modulus figure. Divide the figure up as shown into strips, 
and construct a figure at the side on the base-line aa, the 
ordinates of which represent the shear at that part of the section, 
i.e. the sum of the forces acting to either side of it, thus — 

The shear at i is zero 

2 is proportional to the area of the strip between 
I and 2 = ft^ on a given scale. 

3 is proportional to the area of the strip between 
I and 3 = 4^ on a given scale. 

4 is proportional to the area of the strip between 
I and 4 = ^A on a given scale. 

5 is proportional to the area of the strip between 
1 and 5 = y on a given scale. 

6 is proportional to the area of the istrip between 
I and 6 = ^/ on a given scale. 

Let the width of the modulus figure at any point distant y 
from the neutral axis = b; then — 

the shear at v = 

2 2 

But^ = ?,and* = l? 
the shear atjF = ?X_^ = ^/_ \{f) 

2 2Y 2Y 

in the figure U =^ ff, = \(j^) 

2 2 1 

Thus the shear curve is a parabola, as the ordinates //,, 
etc., vary as y^ ; hence the maximum ordinate kl = \ (mean 
ordinate) (see p. 30), or the maximum shear on the section 
is f of the mean shear. 

In Figs. 450, 45 1 similar curves are constructed for a circular 
and for an I section. 

It will .be observed that in the I section nearly all the shear 

is taken by the web ; hence it is usual, in designing plate girders 

of this section, to assume that the whole of the shear is taken by 

the web. The outer line in Fig. 45 1 shows the total shear and 

the inner figure the intensity of shear at the different layers. The 

W 
shear at any section (Fig. 452) ab = — , and the intensity of 



Beams. 



459 



W 
shear on the above assumption = —^, where A„ = the sectional 

2 A«, 



But the 



area of the web, or the intensity of shear = 

intensity of shear stress on aa^ = the intensity of shear stress 



2ht 




heights 




Fig. 4SO. 



Fig. 451. 



on ab, hence the intensity of the shear stress between the web 

W 
and flange is also = — ;-. We shall make use of this when 
2ht 

working out the requisite spacing for the rivets in the angles 

between the flanges and the web of a plate girder. 

In all the above cases it should be noticed that the shear 

stress is a maximum at the neutral plane, and the total shear 



JVmJtraZ-pl^zna 



2. 





. r 




«■■-?-■. 


d 

Z 


M 

i 




Fig. 452. 

there is equal to the total direct tension or compression acting 
above or below it. 



460 Mechanics applied to Engineering. 

We will now get out an expression for the shear at any part 
of a beam section. 

We have shown a circular section, but the argument will be 
seen to apply equally to any section. 

Let b = breadth of the section at a distance y from the 
neutral axis ; 
F = stress on the skin of the beam distant Y from the 

neutral axis ; 
/ = stress at the plane b distant y from the neutral axis ; 
M = bending moment on the section cd ; 
I = moment of inertia of the section j 
S = shear force on section cd. 

The area of the strip distant j*) _ i j 

from the neutral axis ) ~ " " 

the total force acting on the strip =f.b.dy 

f V Fy 

But|=^, or/= y- 

Substituting the value of/ in the above, we have — 

■^b.y.dy 

„ „ FI F M , 

But M = Y> ""^ y = Y (see p. 432) 

F 
Substituting the value of y ii the above, we have — 

the total force actmg on the stnp = -jO .y .ay 
But M = S/ (see p. 482) 
Substituting in the equation above, we have — 

-^b.y.dy 

Dividing by the area of the plane, viz. b.I,v{e get — 

S/fY 
The intensity of shearing stress. on that plane = fl> I b.y.dy 

J y 

JY 
^D.y.dy 



Beams. 461 

g 
the mean intensity of shear stress = -r 

nrhcre A = the area of the section. 

S fY 

IB ^-y-^y 
The ratio of the maximum"! _ J o 

intensity to the mean J ~ s 

A 



IBJo 




K = — b.y.dy 



The value of K is easily obtained by this expression for 
geometrical figures, but for such sections as tram-rails a graphic 
solution must be resorted to. 

ri 

The value of I b.y.dyi^ the sum of the 

moments of all the small areas b . dy about the 

N.A., between the limits of _y = o, i.e. starting from 

the N.A., and J = Y, which is the moment of the ^'°' ^^'" 

shaded area Aj about the N.A., viz. AiY„ where Y, is the 

distance of the centre of gravity of the shaded area from 

the N.A. 

Since the neutral axis passes through the centre of gravity 
of the section, the above quantity will be the same whether the 
moments be taken above or below the N.A. 

Deflection due to Shear. — The shear in ieam sections 
increases the deflection over and above that due to the 
bending moment. The shear effect is negligible in solid 
beanie of ordinary proportions, but in the case of beams having 
natrow webs, especially when the length of span is small com- 
pared to the depth of the section, the shear deflection may be 
3 o or 30 per cent, of the total. 

Consider a short cantilever of length /, loaded at the free 
end. The deflection due to shear is x. For the present the 
deflection due to the bending moment is neglected. 

Work done by W in deflecting the beam by shear = — 

2 

Let the force required to deflect the strip of area hdh 
through the distance x be d^, let the shear stress in the strip 
be/„ then — 



462 



Mechanics applied to Engineering. 
X f, dW 



G Gbdh 
V 



(see page 376) 
(fW =ffidh 




Beams. 463 

Work done in deflecting the strip by) _ xd'W _fflbdh 
shear 5 - ~^ - 2G 

Total work done in deflecting the) / V^i -fih^j, _ ^■^ 
beam section by shear j ~ ^j _^ -'• '"^"' ~ ~ 

Let a be the area of the modulus figure between h and Hi. 
Let /i be the skin stress due to bending. The total longi- 
tudinal force acting on the portion of the beam section between 
h and Hi is a/i, and the shear area over which this force is 

distributed is bl, hence /, = -^, since the shear is constant 
throughout the length. 

WG/i_H, 6 ~WG/ 
/■Hi ay/i 
where Ao = / —7—; which is the area of the figure mno^ 

between the limits Hj and — H2, Substituting the value of 

/i in terms of the bending moment M and the modulus of the 

section Zj 

AqM^ _ sAqM^ 

*~WGZl^~2WEZlV 

SAoW/ 
X = " -3 for a cantilever, and the total 

deflection at the free end of a cantilever with an end load, 
due to bending and shear, is — 

^=.S + x = ^ + ^E2? ^^^^ P^^® 5^°^ 
and for a beam of length L = 2/ supporting a central load 



Wi= 2W 



^ W,U 5A„WiL 
48EI''" SEZi^ 
WiL/U; sAA 
\6I "^ Z,= / 



and E = 



8A \6I ' Zi= 

In the case of beams of such sections as rectangles and 
circles the deflection due to shear is very small and is usually 



464 



Mechanics applied to Engineering. 



negligible, especially when the ratio of length to depth is 
great. 

In the case of plate web sections, rolled joists, braced 
girders, etc., the deflection due to the shear is by no means 
negligible. In textbooks on bridge work it is often stated 
that the central deflection of a girder is always greater than 
that calculated by the usual bending formula, on account of 
the "give" in the riveted joints between the web and the 
flanges. It is probable that a structure may take a "permanent 
set " due to this cause after its first loading, but after this has 
once occurred, it is unreasonable to suppose that the riveted 
joints materially affect the deflection ; indeed, if one calculates 
the deflection due to both bending and shear, it will be found 
to agree well with the observed deflection. Bridge engineers 
often use the ordinary deflection formula for bending, and take 
a lower modulus of elasticity (about 9000 to 10,000 tons square 
inch) to allow for the shear. 



Experiments on the Deflection of I Sections. 









E 




Section. 


Span L. 


Depth of 
section H. 
















From 5. 


From A. 


Rolled joist 


28" 


6" 


7,120 


12,300 


j» »» 


3^;; 


6" 


8,760 


12,300 


,) j» 


42" 


6" 


9,290 


12,400 


»» »» 


56" 


6" 


10,750 


12,700 


>i a 


60" 


6" 


11,200 


12,800 


Riveted girder 


60' 


S' 


10,200 


12,200 


»» j» 


60' 


6' 


9,000 


12,200 


)» )» 


75' 


4' 


ir,ooo 


12,600 


tf 1} 


160' 


12' 


8,900 


12,000 



It will be seen that the value of E, as derived from the 
expression for A, is tolerably constant, and what would be 
expected from steel or iron girders, whereas the value derived 
from 8 is very irregular. 

The application of the theory given above often presents 
difficulties, therefore, in order to make it quite clear, the full 
working out of a hard steel tramway rail is given below. The 
section of the rail was drawn full size, but the horizontal width 

of the diagram, i.e^ -7- was drawn 5 of full size, hence the actual 



Beams. 



46s 



area of mnop was afterwards multiplied by 4. The original 
drawing has been reduced to 0-405 of full size to suit the size 
of page. 



a in square inches. 
















«2 


I, 
in inches. 


a' 








b 


Between 


Tolal. 








m and 2 


0-38 


0-38 


0-14 


2*20 


0-07 


2 » 3 


o'6o 


0-98 


0-96 


2-35 


0-41 


3 >. 4 


o'6o 


1-58 


2' 50 


3 "03 


0-82 


4 ., 5 


o"64 


2-22 


4-93 


2-25 


2'19 


5 „ 6 


0-3S 


2-57 


6-6o 


0-72 


9-17 


6 „ 7 


0'12 


269 


7-24 


0-42 


17-2 




O'lO 


2-79 


7-78 


0-42 


I8-S 


8 „ 9 


003 


2-82 


7-95 


0-42 


18-9 


9 „ 10 


— 0-09 


2 '73 


7-45 


0-42 


17-7 


10 „ II 


—0-42 


2-31 


5-34 


0-42 


12-7 


II „ 12 


-0-23 


2-o8 


4-33 


i-oo 


4*33 


12 „ 13 


-0-82 


1-26 


1-59 


5-30 


0-30 


13 .. P 


— I 26 














Depth of section 6-5 ins. 

Modulus of section (Z). In this case the) g 

two moduli are the same 5 

Moment of inertia of section (I) 48T 

Ao (area mnof) . . ' 73'9 sq. ins. 

Span (^ 60 ins. ... 28 ins. 

Load at which the deflection) g ^^^^ _ _ _ ^^ ^^^^ 

IS measured 3 

Mean deflection A for the) (^ j^^ _ _ ^ -^^^ 

above loads ) 

E from -;r^^ tons sq. m. 12,200 . . . 8790 

488! 

Efrom^^(Y4-^°) „ ,, 13.9°° • • • 14,200 

E from a tension specimen) ^^^^ i„ 

cut from the head of rail j ^' J 

. In the case of a rectangular section of breadth B and 
depth H the value of Ao is , by substitution in the expres- 
sion for A for a centrally loaded beam of rectangular section, 
the deflection due to shear x = g^ and the ratio 

2 U 



466 



Mechanics applied to Engineering. 



Deflection due to shear 



32! 



Deflection due to bending 

1^, the shear deflection is about 2 per cent, of 



Taking — as 

the bending deflection. 

Discrepancies between Experiment and Theory. — 
Far too much is usually made of the slight discrepancies 
between experiments and the theory of beams ; it has mainly 
arisen through an improper application of the beam formula, 
and to the use of very imperfect appliances for measuring the 
elastic deflection of beams. 

The discrepancies may be dealt with under three heads — 

(r) Discrepancies below the elastic limit 

(2) » at 

(3) .. after 

(1) The discrepancies below the elastic limit are partly due 
to the fact that the modulus of elasticity (E^) in compression is 

not always the same as in 

Compression- tension (E,). 

For example, suppose a 
piece of material to be tested 
by pure tension for E„ and 
the same piece of material 
to be afterwards tested as a 
beam for Ej (modulus of 
elasticity from a bending 
test) ; then, by the usual beam 
theory, the two results should 
be identical, but in the case 
^"'- ■ts'- of cast materials it will pro- 

bably be found that the bending test will give the higher 
result ; for if, as is often the case, the E„ is greater than the E„ 
the compression area A„ of the modulus figure will be smaller 
than the tension area A,, for the tensions and compressions 
form a couple, and AcE„ = A,E,. The modulus of the section 

will now be A. X D, or IM x y^r^'/TT 
4 V -Ci, + V Jit 




fifA TjuhejtE^' Be 



X D ; thus the Z is 



1 /F 
increased in the ratio . °.„ . If the E, be 10 per cent. 



^E. + ^E,' 



Beams. 



46; 



greater than E„ the Ej found from the beam will be about 
2 "5 per cent, greater than the E, found by pure traction. 

This difference in the elasticity will certainly account for 
considerable discrepancies, and will nearly always tend to 
make the Ej greater than E,. There is also another dis- 
crepancy which has a similar tendency, viz. that some materials 
do r^ot. perfectly obey Hooke's law; the strain increases slightly 
more rapidly than the stress (see p. 364). This tends to 
increase the size of the modulus figures, as shown exaggerated 
in dotted lines, and thereby to increase the value of Z, which 
again tends to increase its strength and stiffness, and con- 
sequently make the E5 greater than E,. , 

On the other hand, the deflection due to the shear (see 
p. 463) is usually neglected in calculating the value Ej, which 
consequently tends to make the deflection 
greater than calculated, and reduces the value 
of Ej. And, again, experimenters often mea- 
sure the deflection between the bottom of 
the beam and the supports as shown (Fig. 
458). The supports slightly indent the beam 
when loaded, and they moreover spring 
slightly, both of which tend to make the 
deflection greater than it should be, and con- 
sequently reduce the value of Ej. 

The discrepancies, however, between 
theory and experiment in the case of beams which are not 
loaded beyond the elastic limit are very, very small, far smaller 
than the errors usually made in estimating the loads on 
beams. 

(2) The discrepancies at the elastic limit are more imaginary 
than real. A beam is usually assumed to pass the elastic limit 
when the rate of increase of the 
deflection per unit increase of 
load increases rapidly, i.e. when 
the slope of the tangent to the 
load-deflection diagram increases 
rapidly, or when a marked per- 
manent set is produced, but the 
load at which- this occurs is far beyond the true elastic 
limit of the material. In the case of a tension bar the 
stress is evenly distributed over any cross-section, hence the 
whole section of the bar passes the elastic limit at the same 
instant; but in the case of a beam the stress is not evenly 
distributed, consequently only a very thin skin of the metal 




Fic. , 




Fig. 458. 



468 



Mechanics applied to Engineering. 




passes the elastic limit at first, while the rest of the section 
remains elastic, hence there cannot possibly be a sudden 
increase in the strain (deflection) such as is experienced in 
tension. When the load is removed, the elastic portion of the 
section restores the beam to very nearly its original form, and 
thus prevents any marked permanent set. Further, in the case 
of a tension specimen, the sudden stretch at the elastic limit 
occurs over the whole length of the bar, but in a beam only 
over a very small part of the length, viz. just where the 
bending moment is a maximum, hence the load at which the 
sudden stretch occurs is much less definitely marked in a beam 
than in a tension bar. . 

In the case of a beam of, say, mild steel, the distribution 

of stress in a section just after passing the elastic limit is 

approximately that shown in the shaded 

modulus figure of Fig. 459, whereas if the 

material had remained perfectly elastic, it 

would have been that indicated by the 

triangles aob. By the methods described 

in the next chapter, the deflection after the 

elastic limit can be calculated, and thereby 

it can be readily shown that the rate of 

increase in the deflection for stresses far 

above the elastic limit is very gradual, hence it is practically 

impossible to detect the true elastic limit of a piece of material 

from an ordinary bending 
test. Results of tests will 
be found in the Appendix. 

(3) The discrepancies 
after the elastic limit have 
occurred. The word "dis- 
crepancy" shouldnotbe used 
in this connection at all, for 
if there is one principle above 
all others that is laid down 
in the beam theory, it is that 
the material is taken to be perfectly elastic, i.e. that it has not 
passed the elastic limit, and yet one is constantly hearing of the 
" error in the beam theory," because it does not hold under 
conditions in which the theory expressly states that it will not 
hold. But, for the sake of those who wish to account for the 
apparent error, they can do it approximately in the following 
way. The beam theory assumes the stress to be proportional 
to the distance from the neutral axis, or to vary as shown by the 



Fig. 459. 




Fig. 460. 



Beams. 



469 



line ab ; under such conditions we get the usual modulus figure. 
When, however, the beam is loaded beyond the elastic limit, 
the distribution of stress in the section is shown by the line 
adb, hence the width of the modulus figure must be increased 
in the ratio of the widths of the two curves as shown, and the 
Z thus corrected is the shaded area X D as before.^ This in 
many instances will entirely account for the so-called error. 
Similar figures corrected in this manner are shown below, 
from which it will be seen that the difference is much greater 
in the circle than in the rolled joist, and, for obvious reasons, 
it will be seen that the difference is greatest in those sections 
in which much material is concentrated about the neutral axis. 




Fig. 461. 



Fig. 462. 



But before leaving this subject the author would warn 
readers against such reasoning as this. The actual breaking 
strength of a beam is very much higher than the breaking 
strength calculated by the beam formula, hence much greater 
stresses may be allowed on beams than in the same material in 
tension and compression. Such reasoning is utterly misleading, 
for the apparent error only occurs afkr the elastic limit has 
been passed. 

Reinforced Concrete Beams. — The tensile strength of 
concrete is from 80 to 250 pounds per square inch, but the 
compressive strength is from 1500 to 5000 pounds per square 
inch. Hence a concrete beam of symmetrical section will always 
fail in tension long before the compressive stress reaches its 

' Readers should refer to Proceedings I.C.E., vol. cxlix. p. 313. It 
should also be remembered that when one speaks of the tensile strength of 
a piece of material, one always refers to the nominal tensile strength, not 
to the real; the difference, of course, is due to the reduction of the section 
as the test proceeds. Now, no such reduction in the Z occurs in the beam, 
hence we must multiply this corrected Z by the ratio of the real to the 
nominal tensile stress at the maximum load. 



470 



Mechanics applied to Engineering. 



ultimate value. In order to strengthen the tension side of the 

beam, iron or steel rods are embedded in the concrete, it is 

then known as a reinforced beam. The position of the neutral 

axis of the section can be obtained thus — 

Let E = the modulus of elasticity of the rods. 

E„ = the „ „ „ concrete. 

E , 
r = :^ = from lo to 12. 

/ = the tensile stress in the rods. 
/, = the maximum compressive stress in the concrete. 
a = the combined sectional area of the rods in square 
inches. 

The tension in the concrete 
may be neglected owing to the 
fact that it generally cracks at 
quite a low stress. It is as- 
sunied that originally plane 
sections of the beam remain 
plane after loading, hence the 
strain varies directly as the 
distance from the neutral axis, 
or 




i^ = - and ■'- = 









The total compressive force acting on the concrete is equal 
to the total tensile force acting on the rods, then for a section 
of breadth b. 



he" 



E. 
bi? 



c E 
or bcx - = a=r X e 
2 E, 



which may be written — =(</■ 
obtained. 



c)ra, from which c can be 



The quantity bcX - is the moment of the concrete area 

about the neutral axis, and a— x « is the moment of the rod 

E. 



Beams. 47 1 

area, increased in the ratio r of the two moduli of elasticity, 
also about the neutral axis. Hence if the sectional area of 
the rods be increased in the ratio of the modulus of elasticity 
of the rods to that of the concrete the neutral axis passes 
through the centre of gravity, or the centroid, of the section 
when thus corrected as shown in Fig. 464. 

The moment of resistance of the section (/Z) is, when 
considering the stress in the concrete, 

/o(7)(^+^) or /„©(§. -f.) 

and when considering the stress in the rods— 
fa{g-\-e) or fa{^c-\-e) 

The working stress in the concrete /, is usually taken 
from 400 lbs. to 600 lbs. per square inch, and / from 10,000 
lbs. to 16,000 lbs. per square inch. 

For the greatest economy in the reinforcements, the moment 
of resistance of the concrete should be equal to that of the 
rods, but if they are not equal in any given case, the lower 
value should be taken. 

The modulus of the section for the concrete side is— 

and the corresponding moment of inertia — 

Similarly, in the case of the reinforced side of the section — 
I = aer^^c + e\ 

Example. — Total depth of section 18 inches; breadth 
6 inches, four rods | inch diameter, distance of centres from 
bottom edge i| inch, the compressive stress in the concrete 
400 lbs. per square inch, r ~ 12. Find the moment of 
resistance of the section, the stress in the rods, and the 
moment of inertia of the section. 

In this case d= 16 '5 inches, a = 0-785 square inch. 

— ^=(i6-5 -^)i2 X 0-785 

c = 5-8 inches. e = 10-7 inches. 



472 Mechanics applied to Engineering. 



Moment of resistance for the concrete — 

For the stress in the rods — 
^2 X 5-8 



/Xo-78s(' 



+ 107 



j= loi, 



400 inch-lbs. 



/= 8860 lbs. per square inch 
The moment of inertia — 



/6 X 5'8'V2 X 5-8 , \ . , 4 ., 

( -^ — 11 2 Y 107 I = 1470 mch -units 



or 



(0785 X 107 




1470 inch*-units 



Fig. 465. 

In reinforced concrete floors, the upper portion is a part 
of the floor; at intervals reinforced beams are arranged as 
shown. 

Example. — ^Total depth = 24 inches, thickness of floor it) 
= 4 inches, breadth {b) = 22 inches, breadth of web (J>') = 7 
inches. Six |" rods, the centres of which are I's" from the 
tension skin. Compressive stress in concrete (/„) 400 lbs. 
per sq. inch. /■= 12. Find the moment of resistance, the 
stress in the rods, and the moment of inertia. 

Here d = 22*5 inches a = 2'6s square inches. 

Position of neutral axis — 

22 X 4{c - 2) -f 7^^ ^ = 2-65 X 12(22-5 - c) 



c=r^s' 



^ = 3-15" 



<?= i5'35 



Beams 473 

Moment of resistance of concrete — 

\ ^(! X 7'i5 + 15-34) - (^^-^J3-i5 

X ^(1 X 3'iS + iS'35)|4oo(i4oo)4oo = 560,000 in.-lbs. 
Stress on rods — 

560,000 =/ X 2-65 X 20'I 

/= 10,500 lbs. square inch. 
Moment of inertia — 

1400 X 7" 1 5 = 10,000 inch*-units 
or 2-65 X 15-35 X 12 X 20-I = 9810 „ „ 

In order to get the two values to agree exactly it would be 
necessary to express c to three places of decimals. 

For further details of the design of reinforced concrete 
beams, books specially devoted to the subject should be 
consulted. 



CHAPTER XII. 



BENDING MOMENTS AND SHEAR FORCES. 

Bending Moments. — When two ^ equal and opposite couples 

are applied at opposite ends 
'y'/'/y ■'■■:■ /i of a bar in suck a manner 

as to tend to rotate it in 
opposite directions, the bar 
is said to be subjected to a 
bending moment. 

Thus, in Fig. 466, the 
bar ab is subjected to the 
two equal and opposite 
couples R . ac and W . be, 
which tend to make the 
two parts of the bar rotate 

in opposite directions round the points; or, in other words, 

they tend to bend the bar. 



i 



R*W 



Fig. 466. 



W-R,+lf2 



Fig. 467. 



Loa/i. 



Svpj ^ort 



\lioaei 



hence the term "bending 
moment." Likewise in Fig. 
467 the couples are RiOf 
and Ra^iT, which have the 
same effect as the couples 
in Fig. 466. The bar in 
Fig. 466 is termed a " canti- 
lever." The couple 
R . af is due to the 
resistance of the wall 
into which it is built. 

The bar in Fig. 467 
is termed a " beam." 

When a cantilever 
or beam is subjected 
to a bending moment 
which tends to bend it 

If there be more than two couples, they can always be reduced to two 



Stj^ tport 




Bending Moments and Shear Forces. 



475 




Fig. 469. 



concave upwards, as in Fig. 468 (a), the bending moment will be 
termed positive (+), and when it tends to bend it the reverse 
way, as in Fig. 468 {b), it will be termed negative (— ). 

Bending-moment Diagrams. — In order to show the 
variation of the bending moments at various parts of a beam, 
we frequently make use of bending-moment diagrams. The 
bending moment at the point c 
in Fig. 469 is W . ^tf ; set down 
from c the ordinate ct^ = 'W .be 
on some given scale. The bend- 
ing moment at d=Vf .bd; set 
down from d, the ordinate 
dd' = W .bd on the same scale ; 
and so on for any number of 
points : then, as the bending 
moment at any point increases directly as the distance of that 
point from W, the points b, d', c, etc., will lie on a straight line. 
Join up these points as shown, then the depth of the diagram 
below any point in the beam represents on the given scale the 
bending moment at that point. This diagram is termed a 
" bending-moment diagram." 

In precisely the same manner the diagram in Fig. 470 is 
obtained. The ordinate 
ddi represents on a given 
scale the bending mo- 
ment Ri«(f, likewise cci 
the bending moment 
Rioc or Ra^c, also ee^ the 
bending moment Rj>e. 

The reactions Ri and 
Rj are easily found by 
the principles of moments 
thus. Taking moments about the point b, we have — 




Fig. 470. 



R^ab = Wbc Ri = 



W.bc 
ab 



R, = W-R, 



In the cantilever in Fig. 469, let W = 800 lbs., be = 675 
feet, bd = 4-5 feet. 

The bending moment ai c = W .be 

= 800 (lbs.) X 6-75 (feet) 
= 5400 (Ibs.-feet) 

Let I inch on the bending-moment diagram = 12,000 (lbs.- 

, /ii. r ,.\ ■ u 12000 Ibs.-feet 
feet), or a scale of 1 2,000 (Ibs.-feet) per inch, or j-. — j-v — . 



476 Mechanics applied to Engineering. 

^, , ■!• S400 (Ibs.-feet) ,. , . 

Then the ordinate ec, = ,,, , \ = 0*45 (inch) 

' 12000 (Ibs.-feet) ^^ ^ ' 

I (inch) 

Measuring the ordinate dd^, we find it to be 0*3 inch. 

Then 0-3 (inch) X "°°° O^s-feet) ^ ^3600 (Ibs.-feet) bending 
•^ ^ ' I (inch) ( moment at a 

In this instance the bending moment could hav£ been 
obtained as readily by direct calculation; but in the great 
majority of cases, the calculation of the bending moment is 
long and tedious, and can be very readily found from a 
diagram. 

In the beam (Fig. 470), let W = 1200 lbs., ac= $ feet, 
dc= 5 feet, ad= 2 feet. 

yv .be 1200 (lbs.) X 3 (feet) 

R>=^r= 8 (feet) =450 lbs. 

the bending moment at ^ = 450 (lbs.) X 5 (feet) 
= 2250 (lbs.-feet) 

Let 1 inch on the bending-moment diagram = 4000 Ibs.- 

4000 (lbs.-feet) 
feet), or a scale of 4000 (lbs. -feet) per mch, or )■ . > — • 

™, , ■,- 2250 (lbs.-feet) , ,. , . 

Then the ordinate cc, = ,,, ^ J. = o'go (mch) 

' 4000 (lbs.-feet ) •' ^ ' 

I (inch) 

Measuring the ordinate ddi, we find it to be 0*225 (inch). 

„, 0-225 (inch) X 4000 (lbs.-feet) ^ 1 900 (lbs.-feet) bending 
I (inch) ( moment at (/ 

General Case of Bending IVEoments. — Tke bending 
moment at any section of a beam is t}u algebraic sum of all the 
moments of the external forces about the section acting either to the 
left or to the right of the section. 

Thus the bending moment at the section/ in Fig. 471 is, 
taking moments to the left of/— 

or, taking moments to the right of/— 



Bendinsr Moments and Shear Forces. 



All 



That the same result is obtained in both cases is easily 
verified by taking a numerical example. 

Let Wj = 30 lbs., W2 = 50 lbs., W3 = 40 lbs.; ac =■ 2 feet, 
cd = 2*5 feet, df = i'8 feet,/^ = 2-2 feet, eb = ^ feet. 



W, 



W2 



w. 



or 



f 



R. 



Fio. 471. 

We must first calculate the values of Ri and Ra. Taking 
moments about b, we have — . 

R,«^ = V^^cb + y^^db + ^^b 
„ W,<r^ + ^^b + Ws^-J 

^' ^ 

_ 3o(lbs.) X 9-5(feet)+5o(lbs.) X 7(feet)+4o(lbs.) X 3(feet ) 
11-5 (feet) 
\ =755 0bs-feet)^ 

11-5 (feet) ^ ^ 

.R,=Wi + Wa + W3-R, 
R, =3o(lbs.)+So(lbs.)+4o(lbs.)-6s-6s (lbs.) = 54-3S (lbs.) 

The bending moment at/, taking moments to the left of/, 
= Ri«/-Wie^-W3^ 

= 65-65 (lbs.) X 6-3 (feet) - 30 (lbs.) X 4"3 (feet) - 50 (lbs.) 
X I -8 (feet) = 194-6 (lbs.-feet) 

The bending moment at/, taking moments to the right oif, 
= R,bf- Wsef 

= 54'3S (lbs.) X 5-2 (feet) — 40 (lbs.) X 22 (feet) 
= 194-6 (lbs.-feet) 

Thus the bending moment at / is the same whether we take 
moments to the right or to the left of the point/ The 
calculation of it by both ways gives an excellent check on the 
accuracy of the working, but generally we shall choose that 
side of the section that involves the least amount of calculation. 
Thus, in the case above, we should have taken moments to the 
right of the section, for that only involves the calculation of two 
moments, whereas if we had taken it to the left it would have 
involved three moments. 



478 



Mechanics applied to Engineering. 



The above method becomes very tedious when dealing 
with many loads. For such cases we shall adopt graphical 
methods. 

Shearing Forces. — When couples are applied to a beam 
in the way described above, the beam is not only subjected to 
a bending moment, but also to a shearing action. In a long 
beam or cantilever, the bending is by far the most important, 
but a short stumpy beam or cantilever will nearly always fail 
by shear. 

Let the cantilever in Fig. 472 be loaded until it fails. It 





0, 


., 1 




i 


i, 


y':-::^ 






11 1 


1 


II 











w 



r/ 




Fia. 472. 



Fig. 473. 



will bend down slightly at the outer end, but that we may 

neglect for the present. The failure will be due to the outer 

part shearing or sliding off bodily from the built-in part of the 

cantilever, as shown in dotted lines. 

The shear on all vertical sections, such as ab or dV, is of 

the same value, and equal to W. 

In the case of the beam in Fig. 473, the middle part will 
shear or slide down relatively to 
the two ends, as shown in dotted 
lines. The shear on all vertical 
sections between h and c is of the 
same value, and equal to Rj, and 
on all vertical sections between a 
and c is equal to Rj. 

We have spoken above of posi- 
tive and negative bending moments. 

We shall also find it convenient to speak of positive and 

negative shears. 



FioC'&d ' 



1 
1 



FlG. 474. 



Bending Moments and Shear Forces. 479 

When the sheared part slides in a I clockwise \ 

'^ t contra-clockwise J 

direction relatively to the fixed part, we term it a 
(positive {A-) shear) 
(negative (— ) shear) 

Shear Diagrams. — In order to show clearly the amount 
of shear at various sections of a beam, we frequently make 
use of shear diagrams. In cases in which the shear is partly 
positive and partly negative, we shall invariably place the 
positive part of the shear diagram above the base-line, and the 
negative part below the base-line. Attention to this point will 
save endless trouble. 

In Fig. 472, the shear is positive and constant at all vertical 
sections, and equal to W. This is very simply represented 
graphically by constructing a diagram immediately under the 
beam or cantilever of the same length, and whose depth is 
equal to W on some given scale, then the depth of this diagram 
at every point represents on the same scale the shear at that 
point. Usually the shear diagram will not be of uniform depth. 
The construction for various cases will be shortly considered. 
It will be found that its use greatly facilitates all calculations of 
the shear in girders, beams, etc. 

In Fig. 47 3, the shear at all sections between a and c is 
constant and equal to Rj. It is also positive (-[-), because the 
slide takes place in a clockwise direction ; and, again, the shear 
at all sections between b and c is constant and equal to Rj, but 
it is of negative ( — ) sign, because the slide takes place in a 
contra-clockwise direction ; hence the shear diagram between 
a and c will be above the base-line, and that between b and c 
below the line, as shown in the diagram. The shear changes 
sign immediately under the load, and the resultant shear at that 
section is Rj — R^. 

General Case of Shear, — The shear at any section of a 
beam or cantilever is the algebraic sum of all the forces acting to 
the right or to the left of that section. 

One example will serve to make this clear. 

In Fig. 475 three forces are shown acting on the canti- 
lever fixed at d, two acting downwards, and one acting 
upwards. 

The shear at any section between a and d = -f W due to W 

„ b „ rf=-W, „ W, 
.. ,. .. c „ d= -t-Wj „ W, 



48o 



Mechanics applied to Engineering. 



Construct the diagrams separately for each shear as shown, 
then combine by superposing the — diagram on the + diagram. 
The unshaded portion shows where the — shear neutralizes the 



«S 



W, 



W 



m> 



m 



iili 


»i 














III 


III 


1 





w 



H'+MJ-Mf" 



iiiiiiiiiiiiwiMiiiiiiir yry_ 

Fig. 475. 



IV 



+ shear ; then bringing the + portions above the base-line and 
the — below, we get the final figure. 



Bending Moments and Shear Forces. 



481 



Resultant shear at any section — 


Between 


To the right. 


To the left. 


a and b 


W 


-W, + W^ - (W + W, - W,) 
= -W 


b and c 


w-w, 


W, - (W + W, - W,) 

= -(w - W.) 


(T and a 


Wj - w, + w 

orW+W,-W, 


-(W + W, - W,) 



In the table above are given the algebraic sum of the forces 
to the right and to the left of various sections. On comparing 
them with the results obtained from the diagram, they will be 
found to be identical. In the case of the shear between the 
sections b and c, the diagram shows the shear as negative. 
The table, in reality, does the same, because Wj in this case is 
greater than W. It should be noticed that when the shear is 
taken to the left of a section, the sign of the shear is just the 
reverse of what it is when taken to the right of the section. 

Connection between Bending-moment and Shear 
Diagrams. — In the construction of shear diagrams, we make 
their depth at any section equal, on some given scale, to the 
shear at the section, i.e. to the algebraic sum of the forces to 
the right or left of that section, and the length of the diagrams 
equal to the distance from that section. 

Let any given beam be loaded thus : Loads Wj, W^, W,, 
— W4, — Wb at distances l^, 4 4i h, h respectively from any given 
section a, as shown in Fig. 476. 

The bending moment at a is = W/a + W3/3 — W4/4 or — Wj^ 

But Wa/j is the area of the shear diagram due to W, between 
W, and the section a, likewise Wg/j is the area between Wj and 
a, also — W4/4 is the area between — W4 and a. The positive 
areas are partly neutralized by the negative areas. The parts 
not neutralized are shown shaded. 

The shaded area = Wa4 + Ws4 — W4/4, but we have shown 
above that this quantity is equal to the bending moment at a. 
In the same manner, it can be shown that the shaded area of 
the shear diagram to the left of the section a is equal to 

2 I 



482 



Mechanics applied to Engineering. 



- W/i + Wj/j, i.e. to the bending moment at a. Hence we 
get this relation — 

The bending moment at any section of a freely supported 
beam is equal to the area of the shear diagram up to that point. 
The bending moment is therefore a maximum where the shear 
changes sign. 



w. 







Wz 


IV3 


' ^s- 


j:1\ ± 


' - h 








h 






Fig. 476. 

Due attention must, of course, be paid to positive and 
negative areas in the shear diagram. 

To make this quite clear, we will work out a numerical 
example. 

In the figure, let W, = 50 lbs. A = i foot 

Wj = 80 lbs. 4=2 feet 

W, = 70 lbs. /, = 4 feet 

By moments we f W4 = ■s.%2-7, lbs. /« = 5 feet 

find (W. = 67-8 lbs. 4 = 4 feet 

The figure is drawn to the following scales — 

Length i inch = 4 feet 
load I inch =160 lbs. 
hence i square inch on the shear diagram = 4 (feet) x 160 (lbs.) 

= 640 (lbs.-feet) 



Bending Moments and Shear Forces. 483 

The area of the negative part of the shear diagram below 
the base-line is — o'4oi sq. inch, and the positive part above the 
base-line is 0*056 sq. inch; thus the area of the shear diagram 
up to the section a is — o'4oi + o"o56 = 0*345 sq. inch. But 
I sq. inch on the shear diagram = 640 (Ibs.-feet) bending 
moment, thus the bending moment at the section a = 
o*345 X 640 = 221 (Ibs.-feet). The area of the shear diagram 
to the left of a = o'345 sq. inch, i.e. the same as the area to the 
right of the section. As a check on the above, we will calculate 
the bending moment at a by the direct method, thus — 

The bending moment at a = W^^ + W/j — W/4 

= 80 (lbs.) X 2 (feet) + 70 (lbs.) X 
4 (feet) - 132-2 (lbs.) X 5 (feet) 
= 221 (Ibs.-feet) 

which is the same result as we obtained above from the shear 
diagram. 

This interesting connection between the two diagrams can 
be shown to hold in all cases from load to slope diagrams. If 
a beam supports a distributed load, it can be represented at 
every part of the beam by means of a diagram whose height is 
proportional to the load at each point ; then the amount of load 
between the abutment and any given point is proportional to 
the area of the load diagram over that portion of the beam. 
But we have shown that the shear at any section is the algebraic 
sum of all the forces acting either to the right or to the left of 
that section, whence the shear at that section is equal to the 
reaction minus the area of the load diagram between the section 
in question and the said reaction. We have already shown the 
connection between the shear and bending moment diagrams, 
and we shall shortly show that the slope between a tangent to 
the bent beam at any point and any other point is proportional 
to the area of the bending-moment diagram enclosed by normals 
to the bent beam drawn through those points. 



484 Mechanics applied to Engineering. 



Cantilever with 

single load at 

free end. 




Fig. 477. 



Cantilever with 
two loads. 



W*w, 




Fig. 478. 



Bending Moments and SJiear Forces. 



485 



Bending moment M in 
ll».-incheft 



= W(lbs.)X/(in.) 
M, = W/, 
M = (/. mn 

at any section where 
d = depth of beriding- 
moment diagram in 
inches 



Depth of bend- 

in^-moment 

dlagnun In 

mches. 

Scale of W, 
m lbs. = 1 indi. 

Scale of/, 

-full size. 



W/ 
mn 



mn 



Remarks. 



The only moment acting to the right 
of X is W/, which is therefore the 
bending moment at x. Likewise at^. 

The complete statement of the units 
for the depth of the bending-moment 
diagram is as follows : — 

xn(lbs.) = i inch on diagram, or -^. — ^ 
^ ' ^ ' l{mch) 

«(in.)=l „ „ 

W (lbs.) /,(inches) W/ 



m (lbs.) 
I inch 



n (inches) mti 



(inches) 



M = 1/ . mn 



mn 



This is a simple case of combining 
two such bending-moment diagrams 
as we had above. The lower one is 
tilted up from the diagram shown in 
dotted lines. 



486 



Mechanics applied to Engineering. 



Cantilever with 
an evenly distri- 
buted load of w 
lbs. per inch run. 




Cfg. 
of loads 

Hendiuruf TTvoTnentff 




apfix- 



Bending Moments and S/iear Forces. 



487 



Bending moment M in 
lbs.-inche8. 



Mx = 



iv? 



w(\hi, ) ^(inches)' 
inches 2 (constant) 
- ^.p (Ibs.-inches) 
constant 



Let W = o)/ 



Depth 3f bend- 

inc-moment 

diagram in 

inches. 

Scale of r/, 

m lbs.=si inch. 

Scale of/, 

- full size. 



2 

M = 1/ , mn 



W/ 
2mn 



Remarks. 



Tn statics any system of forces may 
always be replaced by their resultant, 
which in this case is siluatedat the centre 
of gravity of the loads ; and as the dis- 
tribution of the loading is uniform, the 

resultant acts'^t a distance - from x. 

2 
The total load on the beam is wl, or 
W ; hence the bending moment at * 

/ wl' 

= wl X — = . At any other sec- 

22 



tion, V = wl. X - = — =-, 
■^ '2 2 



Thus the 



bending moment at any section varies 
as the square of the distance of the 
section from the free end of the beam, 
therefore the bending moment dia- 
p;ram is a parabola. As the beam 
is fully covered irith Ioad.s, the sum 
of the forces to the right of any 
section varies directly as the length 
of the beam to the right of the 
section ; therefore the shearing force 
at any section varies directly as the 
distance of that section from the free 
end of the beam, and the depth of the 
shearing-force diagram varies in like 
manner, and is therefore triangular, 
with the apex at the free end as 
shown, and the depth at any point 
distant /, from the free end is wl,, i.e. 
the sum of the loads to the right of /,, 
aird the area of the shear diagram up 

to that point is ^?i2iZ« = !<, ,-.,. 

•^ 2 2 

the bending moment at that point. 



488 Mechanics applied to Engineering. 



Cantilever irregu- 
larly loaded. 




Fig. 480. 



Beam supported 
at both ends, with 
a central load. 




Fig. 481. 



Bending Moments and Shear Forces. 



489 



Bending moment 

Mm 

lbs.-inches. 



M«=W/+W,/, 

"*" 2 
M = </ . mn 



Depth of bending' 

moment diagram 

in inches. 

Scale of W, 

m lbs. ^ X inch. 

Scale of/, 
- full siM. 



to/." 



Remarks. 



This is simply a case of the combina- 
tion of the diagrams in Figs. 478 and 479. 
However complex the loading may 
be, this method can always be adopted, 
although the graphic method to be 
described later on is generally more 
convenient for many loads. 



w/ 


W/ 


M«_ ^ 




^mn 


M = </ . ««« 





Each support or abutment takes one- 

W 

half the weight = — • 

The only moment to the right or left 
. W / W/ 
of the section a: is — x - = — • 
224 

At any other section the bending 
moment varies directly as the distance 
from the abutment ; hence the diagram 
is triangular in form as shown. The 
only force to the right or left of x is 

W 

— ; hence the shear diagram is of 

constant depth as shown, only positive 
on one side of the section x, and negative 
on the other side. 



490 



Mechanics applied to Engineering. 



Beam supported 
at both ends, with 
one load not in the 
middle of the span. 




Fig. 482. 



Beam supported 
at both ends, with 
two symmetrically 
placed loads. 




R'W 



Fio. 483, 



Bending Moments and Shear Forces. 



491 



Bending moment 

Min 

Ibs.-inchcBi 



W 

M.= y(AA) 

M = </ . mn 



Depth of bending- 

moment diagram 

in inches. 

Scale of W, 
It lbs. = I inch. 

Scale of /p 

- full size. 



Imn 



Remarks. 



Taking moments about one support, 
we have R,/ = W/,, or R, = H4. The 
bending moment at x — 



W. 



M. = R,/, = -^(/,4) 



M, = W4 
M„ = W/. 
M = </, mn 



W4 

mn 



The beam being symmetrically 
loaded, each abutment takes one weight 
= W = R. 

The only moment to the right orthe 
right-hand section ;ir is W . 4 ; likewise 
with the left-hand section. 

At any other section y between the 
loads, and distant /v from one of them, 
we have, taking moments to the left of 
y, R{4 + /,) - W . /, = R . 4 + R . /, 
- R . 4 = R . 4 01 W . 4, ».«. the 
bending moment is constant between 
the two loads. 

The sum of the forces to the right or 
left aiy_ = W — R = o, and to the right 
of the right-hand section the sum of the 
forces = R ,= W at every section. 



492 



Mechanics applied to Engineering. 



Beam supported 
at both ends, load 
evenly distributed, 
w lbs. per inch run. 




Fig. 484. 



Bending Moments and Shear Forces. 



493 



Bending 

moment M in 

lbs.-incbes. 



Let W = a// 



Mx = 



8 



N.B.— Be 
very caieful 
to reduce the 
distributed 
load to 
pounds per 
inth tun if 
the dimen- 
sions of the 
beam are in 
inelus. 



Depth of 
bending- 
moment 
diagram 
in inches. 

Scale 

ofW, 

Mlbs. 

s 1 inch. 

Scale of/, 

— full WBt. 






Remarks. 

As in the case of the uniformly loaded cantilever, 
we must replace the system of forces by their 
resultant. 

The load being symmetrically placed, the abut- 
ments each take one-half the load = — • 

2 

Then, taking moments to the left of *, we have — 



wl 



wl I 



2 ^a 2^4- 



wP 
8 



wl 



The — shown midway between x and the abut- 
ment is the resultant of the loads on half the beam, 

acting at the centre of gravity of the load, viz. - 

4 
from X, or the abutment. , 

The bending moment at any other section y, 
distant I, from the abutment, is : taking moments 
to the right oty — 

wl , , ly '"'ly,. 



=f<4««.?(^) 



where I,' = I — /,. 

Thus the bending moment at any section is 
proportional to the product of the segments into 
which the section divides the beam. Hence the 
bending-moment diagram is a parabola, with its 
axis vertical and under the middle of the beam as 
shown. 

The forces acting to the right of the section x = 

=^0 ; i.e. the shear at the middle section 

2 2 

is zero. 

Atthe section>=i»/,- ^ = w (/,--). Hence 

the shear varies inversely as the distance from the 
abutment, and at the abutment, where 4 = o, it is 
wl 



494 



Mechanics applied to Engineering, 



Beam supported 
at two points equi- 
distant from the 
ends, and a load 
of w lbs. per inch 
run evenly distri- 
buted. 



PCTTTTYTXTY^ 
I ^ y f ] 



*■■■ ij >i ^/ -11-. ■-^--> 

J\re^ative B. M. thteto dvorhanffTr^ lo ads 
Positive B. If^eh^e^^centred- ^euv 

ComhmeA B. M. 




SJvear 




Fio. 485. 



Bending Mommts mid Shear Forces. 495 


Bending moment 


Depth of bend- 




inff-moment 
dUsnun 




Ibs.-inches. 






in inches. 






Scale of W 






fff Ibfc. =11 tncn. 






Scale of /, 






i full siie. 






n 


Remarks. 

The bending-moment diagram for the 
loads on the overhanging ends is a com- 
bination of Figs. 479 and 483, and the dia- 
gram for the load on the central span is 
simply Fig. 484. Here we see the im- 
portance of signs for bending moments. 

The beam will be subject to the smallest 


M.= «"'• 




bending moment when M, = M, ; or when 


Mx 




2 


ivl^ _ a//," _ w4» 




mn 
mn 


282 




/, = 283/, 






But /, + 2/, = / 






substituting the value\ _ -, , ,, _ , 
of/, above 1 - 2 83/, + 2/, - / 






or/ =4-83/, 






or say /, = \f for the conditions of maxi- 






mum strength of the beam. 






The shear diagram will be seen to be 
a combination of Figs. 479 and 484. 








496 Mechanics applied to Engineering. 



Beam supported 
at each end and 
irregularly loaded. 




FlQ. 486. 



Bending Moments and Shear Forces. 



497 



Bending 



Min 
Ibs.-inches. 



Depth of bend' 

ing-moment 

diagram 

in inches. 

Scale of W. 
M lbs.=i inch. 

Scale of/, 

— full size. 



Msid , mn 



mn 
or 

mn 
etc 



Remarks, 

The method shown in the upper figure is 
simply that of drawing in the triangular bending- 
moment diagram for each load treated separately, 
as in Fig. 481, then adding the ordinates of each 
to form the final diagram by stepping off with a 
pair of dividers. 

In the lower diagram, the heights ag,gh, etc., 
are set off on the vertical drawn through the abut- 
ment = W,/„ Wj/j, etc., as shown. The sum 
of these, of course, = R,/. From the starting- 
point a draw a sloping line ai, cutting the 
vertical through W, m the point b. Join gb and 
produce to e, join he and produce to d, and so 
on, till the point / is readied ; join fa, which 
completes the bending-moment dis^am abcdcf, 
the depth of which in inches multiplied by mn 
gives the bending moment. The proof of the 
construction is as follows : The bending moment 
at any point * is R,4 — W^^,. 

On the bending-moment diagram -^= j-; or 
R./X4 



K/ = ^ = 



= =R,/. 



0/ 



if X /. W,/, X r. 



= W/. 



and the depth ofl 
the bending-mo- > = KO = K/ - 0/ = R/, - W^r, 
ment diagram ) 

It will be observed that this construction 
does not involve the calculation of R, and R,. 
For the shear diagram R, can be obtained thus : 

Measure off of in inches ; then -^—z = R^, 

where / is the actual length of the beam in 

inches. 



49^ Mechanics applied to Engineering. 



Beam supported 
at the ends and 
irregularly loaded. 




Bending Moments and Shear Forces. 



499 



Bendtne moment 

Min 

IbS'-incbes. 



M = )n.n(Dx 
Ok) 



Remakks. 
Make the height of the load lines on the beam propor- 



tional to the 



loads, Tu. — i. 



W, 



etc., 



inches. Drop 

perpendiculars through each as shown. On a vertical fb 

\V W 

set ofif y!r = — , ed = — ?, etc. Choose any convenient 



point O distant Oh from the vertical. Ok is termed the 
"polar distance." JoinyO, <0, etc. From any pointy 
on the line passing through R, draw a line/m parallel to 

_/0 ; from m draw mK parallel to eO, and so on, till the 
line through R, is reached in g. Join gf, and draw Oa on 
the vector polygon parallel to this last line ; then the 
reaction R, =_/a, and Rj = ia. Then the vertical depth 
of the bending-moment diagram at any given section is 
proportional to the bending moment at that section. 

Proof. — The two triangles jr^wj and Oaf axe similar, for 

Jm is parallel to/D, axA jp to aO, and/w io fa ; also/? 
is drawn at right angles to the base mp. Hence — 

height of A ipm _ base of A iiim 
height of ^ Oaf base of ^ Oaf 

or^ = OA 

mp af 

. h -Ok 

••D.-R-, 

For 7^ = /i and af^ R, ; and let mp = D^, i.e. the depth 
of the bending-moment diagram at the section x, or R,/, 
= D,OA = M, = the bending moment at x. 
By similar reasoning, we have — 

R,4 = rf)t.Oh 
also \V,(4 - /,) = rlC X Oh 
the bending moment at^ = M,, = Rj/j — W,(/j — /,) 
= Oh{,rt - rK) 
= 0/5(K/) 
= OA.D, 

where D, = the depth of the bending-moment diagram at 
the section J/. 

Thus the bending moment at any section is equal to the 
depth of the diagram at that section multiplied by the 
polar distance, both taken to the proper scales, which we 
will now determine. The diagram is drawn so that — 

I inch on the load scale = m lbs, 
I „ „ length „ = » inches. 

Hence the measurements taken from the diagram in inches 
must be multiplied by mn. 

The bending moment expressed \ „ ,t^ ..,,, 

in lbs. .incites at any section ) = M = r« . « . (D . 0-5) 



Soo 



Mechanics applied to Engineering. 




Beam sup- 
ported at two 
points with 
overhanging 
ends and irre- 
gularly loaded. 



Bending Moments and Shear- Forces. 501 



Bending moment 

Min 

lbs. -inches. 



where D is the depth of the diagram in inches at that 
section, and Oh is the polar distance in inches. 

In Chap. IV. we showed that the resultant of such a 
system of parallel forces as we have on the beam passes 
through the meet of the first and last links of the link 
polygon, viz. through u, where /»« cuts gh. Then, as the 
whole system of loads may be replaced by the resultant, 
we have Rj/j = Rj/,. But we have shown above that 

the triangles juw and Oaf are similar ; hence ^ = — ^. 

\jh of 

But jv — /j, therefore af x l, = Oh x uw =: Rj/j, or 

af— R,. Similarly it may be shown that ab = R^. 



M = »j. «(Dx 
0/0 



The loads are set down to the proper scale on the 
vector polygon as in the last case, A pole O is chosen 
as before. The vertical load lines are dotted in order to 
keep them distinct from the reaction lines which are 
shown in full. Starting from the point/ on the reaction 
line Ri, a line jm is drawn parallel to oO on the vector 
polygon, from m a line mk (in the space i) is drawn 
parallel to bO, and so on till the point u is reached, from 
« a line is drawn parallel to fO to meet the reaction line 
R2 in the point g. Join Jg. From the pole O draw a 
line Oj parallel lojg. 

Then ai gives the reaction R„ and if the reaction Rj. 
The bending moment diagram is shown shaded. The 
points where the bending moment is zero are known as 
the points of contrary flexure. 

The construction of the shear diagram will be evident 
when it is remembered that the shear at any section is 
the algebraic sum of the forces to the right or to the left 
of the section. 

The bending moment at any section of a beam loaded 
in this manner can be readily calculated. The reactions 
must first be found by taking moments about one of the 
points of support. The bending moment at any section 
X distant 4 from the load ^is 

Mx = Tdli + 7el, - R,/, + rf/x 
and the distance ig of the point of contrary flexure from 
the load ef is obtained thus 



/« = 



d^h - R.A 



" "" — ^ . -r- 



ef+de- Rj 



502 



MecJianics applied to Engineering. 



Beam supported 
at each end and 
loaded with an 
evenly distributed 
load of TV lbs. per 
inch run over a 
part of its length. 



Beam supported 
at each end with 
a distributed load 
which varies 
directly as the dis- 
tance from one 
end. 




Bending Moments and Shear Forces. 



503 



BendinK moment 

M in 

Ibs.-incbes. 



MnM.= — 



P 



Remarks. 

Remembering that the bending moment at any section 
is equal to the area of the shear diagram up to that section, 

the maximum bending moment will occur at the section 
where the sliear chapges sign. 

R.= 



2a/4' 

/ 


R. 




2/,R, 


_ 2/.R. 


_ 2/,/, 


Kj + R, 


2wL 


I 


V.^x _ 2w4V,« 




2 


p 





^' ' max. — — M 



M,. 



\\P_ 
9V3 



Let ia, be the intensity of loading at any point distant 
/, from the apex of the load diagram. 

p. w H' 

The shear at this point = I^|— I ■U'^ll = Ri — ,- ( /»<// 
Jo 'Jo 





6 


W/.' 
2/ 


Therefore the shear diagram is 


parabolic 


The shear 


is zero when — 






6 


■2.1 




or when /, 


I 



The maximum bending moment occurs at the section 
where the .shear changes sign, and is equal to the area 
of the shear curve ; hence — 

/ ^P 



M„ 






«/3 9^/3 
For another method of arriving at this result, see p. 185 



504 



MecJianics applied to Engineering. 



Beam built in 
at both ends and 
centrally loaded. 



Ditto with 
evenly distributed 
load 



Cantilever 
propped at the 
outer end with 
evenly distributed 
load. 



Beam built in 
at both ends, the 
load applied on 
one of the ends, 
which slides paral- 
lel to the fixed 
end. 



(jgS 



-I \ 



Fig. 491. 



^ 



i*^ o^ ^p 



Fig. 492. 




l^fp. 



Fig. 493. 



■ 



rv 

: | i 



W'/f 



Fig. 494. 



Bending Moments and Shear Forces. 



505 



Bending moment 
Mm 

Ibs.-!nches. 






M,= 



W/ 



M.= 



24 



Mx 



~ 128 

tap 



M„ = 



M,= — 
M, = 



The determination of these bending moments depends 
on the elastic properties of the beams, which are fully 
discussed in Chap. XIII. 

In all these cases the beam is shown built in at both 
ends. The beams are assumed to be free endwise, and 
guided so that the ends shall remain horizontal as the 
beam is bent. If they were rigidly held at both ends, the 
pioblem would be much more complex. 



CHAPTER XIII. 

DEFLECTION OF BEAMS. 

Beam bent to the Arc of a Circle. — Let an elastic beam 
be bent to the arc of a circle, the radius of the neutral axis 

being p. The length of the neutral 

axis will not alter by the bending. 

The distance of the skin from the 

neutral axis = y. 




The original length of ) _ 



the outer skin 



!= 



27rp 



Fig. 495. 



the length of the outer i ^ ^, . 

skin after bending ) ^ ■" 

the strain of the skin ) _ , / i \ 

due to bending C ^^^ ^' 

° ' — 2irp = 27ry 



But we have (see p. 373) the following relation : — 

strain _ stress 

original length modulus of elasticity 
2'^' _ > _ / 



or ■ 



27rp 



But we also have — 



/ = 



M 



Substituting this value in the above equation — 

p EZ 

whence M = — ^ ; or M = — 
P P 

Central Deflection of a Beam bent to the Arc of a 
Circle. — From the figure we have — 



Deflection of Beams. 

,-^ = (p-8)^+(L^y 



S07 



whence 2p8 — 8' = — 
4 

The elastic deflection (S) of a 
beam is rarely more than -p^ of 
the simn (L) ; hence the 8* ■will 

not exceed —^ , which is quite 

360,000 ^ 

negligible ; 

T a 

hence 2pS = — 
4 

But p=^l 



8 = 



8p 



/if V 



hence 8 = 



ML* 
8EI 




\Vc shall shortly give another method for arriving at this result. 

General Statement regarding Deflection. — In 
speaking of the deflection of a cantilever or beam, we always 
mean the deflection measured from 
a line drawn tangential to that 
part of the bent cantilever or 
beam which remains parallel to its 
unstrained position. The deflec- 
tion 8 will be seen by referring to 
the figures shown. 

The point / at which the tan- 
gent touches the beam we shall 
term the " tangent point." When 
dealing with beams, we shall find it 
convenient to speak of the deflec- 
tion at the support, «.& the height 
of the support above the tangent. 

Deflection of a Cantilever. — Let the upper diagram 
(Fig. 498) represent the distribution of bending moment acting 
on the cantilever, the dark line the bent cantilever, and the 
straight dotted line the unstrained position of the cantilever. 
Consider any very small portion ^j/, distant /, from the free end of 
the cantilever. We will suppose the length jy so small that the 
radius of curvature p, is the same at both points, y,y. Let the 
angle subtending yy be 6, (circular measure) ; then the angle 




So8 



Miclianics applied to Engineering. 



between the two tangents ya, yb will also be 6,. Then the 
deflection at the extremity of these tangents due to the bending 
between ^/.j* is— 

o» = ^- ^ 

-yy 



Bute, =-^-^ 
Pi 

and from p. 424, we have — 
'? EI 

where M, is the mean bending 
moment between the points 

y^y- , . . 

Then by substitution, we 
have — 

^'~ EI 

where Q, is the "slope" be- 
tween the two tangents to the 
bent beam at^_j'; 

But Mjj'j' = area (shown shaded) of the bending-moment 
diagram between y, y 

hence ^, = .gr? 




Fig. 498. 



and 8, : 



A/, 
EI 



That is, the deflection at the free end of the cantilever due 
to the bending between the points y, y is numerically equal to 
the moment of the portion of the bending-moment diagram 
over yy about the free end of the cantilever divided by EI. 
The total deflection at the free end is — 

8 = S(8, + 8. -1-, etc.) 

8 = ^5A^, + A.4 -f , etc. 



where the suffix x refers to any other very small portion of the 
cantilever xx. 

Thus the total deflection at the free end of the cantilever is 



Deflection of Beams. 



509 



numerically equal to the sum of the moments of each little 
element of area of the bending-moment diagram about the free 
end of the cantilever divided by EI. But, instead of dealing 
with the moment of each little element of area, we may take 
the moment of the whole bending-moment diagram about the 
free end, i.e. the area of the diagram X the distance of its centre 
of gravity from the free end ; 



or 8 = 



EI 



where A = the area of the bending-moment diagram ; 

Lc = the distance of the centre of gravity of the bending- 
moment diagram from the free end. 
To readers familiar with the integral calculus, it will be seen 
that the length that we have termed yy above, is in calculus 
nomenclature dl in the limit, and the deflection at the free end 
due to the bending over the elementary length dl is — 

M„. /„.<// _ 



8,= 



£1 



and the total deflection between points distant L and o from 
the free end is— 



« = ^I 



Ml.dl 



(ii.) 



where M = the bending moment at the point distant / from 
the free end. 

Another calculus method 
commonly used is as follows. 
The slope of the beam between 
P and Q (the distance PQ is 
supposed to be infinitely small) 

dy 
is denoted by — . This ratio 

dx 

is constant if the beam is 
straight, but in bent beams 
the slope varies from point to 
point, and the change of slope 
in a given length dx is — 




Fig. 499- 



\dx' 
dx 



dx^' 



Sio 



Mechanics applied to Engineering. 



When Q is very small dx becomes equal to the arc subtending 
the angle Q, and p? = Pq = p, then -j- = — -^, in the limit 
PQ = dx, and the change in slope in the length dx is 



<?) . 



dx 



M 



'' ds^ 



(iii.) 



This expression will be utilized shortly for finding the deflection 
in certain cases of loaded beams. 

Case I. — Cantilever with load W on free end. Length L. 
Method (i). 




Fig. 50a 



A = WL X - = 

2 2 



S = 



2 
WL' 
3EI 



EI 



Method (ii.). — By integration 
M = W/ 



W 

hence 8 = — -, 

EI 



'^ = ^' WL' 

/=o 



3EI 



Method (iii.). — Consider a section of the beam distant x 
from the abutment. 

d}v 
The bending moment M = W(L - a;) = El-^ 



L — a; = 



Integrating 



EI^ 
"iN d£- 

L.-^Vc = |f 
2 Yl dx 



where C is the integration constant. When x = o the slope 

dy 



-r is also zero, hence C = o. 

ax 



Deflection of Beams. 



S" 



Integrating again — 

L^= x"^ , ^^ EI 
2 6 W 

When * = o, the deflection y is zero, hence K = o and 

_ WL^ _ ^\ 
•''" E[\ 2 6/ 

which gives the deflection of the beam at any point distant x 
from the abutment. An exactly similar expression can be 
obtained by method (i.). 

The deflection S at the free end of the beam where 
:r = L is — 

'=iEI 

In this particular case the result could be obtained much 
more readily by methods (i.) and (ii.). 

Case II. — Cantilever with load W evenly distributed or w 
per -unit length. Length L. 



Method (i.)— 



XfLx^ 



8 = 



wL* 



8EI 




8EI 
or by integration — 

Method (lu) M = 

hence S = 




/=L 



l\, wJJ 



512 Mechanics applied to Engineering. 

Method (iii.) — Consider a section distant x from the 
abutment. 

rr^, , ,• ,, K'fL — xf „,d^y 
The bending moment M = = El-j-5 

w doc 

r, . x' zL^c" , ^ 2EI dy 

Integratms Ux -i f- C = r 

32 w dx 

For the reason given in Case I, C = o 

. LV , «* L^ , ^ 2EI 

Integrating again f- K = y 

° 2123 w 

as explained above K = o, 



^ = 2lEi^^^'^ + *'-4^^^ 

•which gives the deflection of the beam at any point distant x 
from the abutment. This is an instance in which the value 
of J* is found more readily by method (iii.) than by (i.) or (ii.). 
The deflection 8 at the end of the beam where x='L is — 

_ w\} _ WL° 
8E1 ~ 8EI 

Case III. — Cantilever with load W not at the free end. 



A = WL. X ^ = ^- 

3 3 

L, = L - ^ 



3 
2EI\^ 3 

Fig. 502. •' 

N.B. — The portion db is straight. 



8 = S(l-L') 



Deflection of Beams. 



513 



Case IV. — Cantilever with two loads Wi, Wa, neither oj 
them at the free end. 



(-^) 



2EI \- 3 
, WaL,' 






Wj 




Fig. S03. 

Case V. — Cantilever with load unevenly distributed. 
Length L. 




Let the bending-moment diagram shown above the canti- 
lever be obtained by the method shown on p. 487. 

Then if i inch = m lbs. on the load scale ; 

I inch = n inches on the length scale ; 

D = depth of the bending-moment diagram 

measured in inches ; 

OH = the polar distance in inches ; 
M =fthe bending-moment in inch-lbs.; 

M = w.w.D.OH; 

hence i inch depth on the bending-moment diagram represents 

M . 

=r = m .n . OH mch-lbs. j and i square inch of the bending- 
moment diagram represents tn .rfi . OH inch-inch-lbs. \ hence — 

f area of bending-moment N j ^;^ ^ 
- V diagram in squa re inches y ^ • " ' " ^ ^ ^« 
8= El 

2 L 



514 



Mechanics applied to Engineering. 



The deflection 8 found thus will be somewhere between 
the deflection for a single end load and for an evenly 
distributed load ; generally by inspection it can be seen 
whether it will approach the one or the other condition. 
Such a calculation is useful in preventing great errors from 
creeping in. 

In irregularly loaded beams and cantilevers, the deflection 
cannot conveniently be arrived at by an integration. 

Deflection of a Beam freely supported. — Let the 
lower diagram represent the distribution of bending moment 

on the beam. The 
dark line represents the 
bent beam, and the 
straight dotted line the 
unstrained position of 
the beam. By the same 
process of reasoning as 
in the case of the 
cantilever, it is readily 
shown that the deflec- 
tion of the free end or 
the support is the sum 
of the moments of each 
little area of the bend- 
ing-moment diagram 
between the tangent 
point and the free end 
about the free end ; or, 
as before, instead of 
dealing with the mo- 
ment of each little area, 
we may take the moment of the whole area of the bending- 
moment diagram between the free end and the tangent 
point, about the free end, i.e. the area of the bending-moment 
diagram between the tangent point and the free end X distance 
of the centre of gravity of this area from the free end. Then, 
as before — 




J^eeJBrul 



Fig. 505. 



AL. 

EI 



where A and L, have the slightly modified meanings mentioned 
above. 



Deflection of Beams. 515 

Case VI. — Beam loaded with central load W. Length L. 



A=^xt 



L,=- 



- ...1- 



\ .■? EI 



8 = 



4 
WL3 




Fig. 506. 



48EI 

Or by integration, at any point distant / from the support — 

M = — i 
2 

J 



~ EI I 2 EI V6 z' 

<^ n 



When / = — , we have — 

2 

WLs 



8 = . 



WL» 
2" X 6EI 48EI 

Case VII. — Beam loaded with an evenly distributed load w 
per unit length. Length L. 




8 = ^'xSjxf^ 
16 EI 

S= 5^L^ _ SWL' 
384EI 384EI 



Fig. 507. 



Or by integration, at any point distant / from the support (see 
p. 512) the bending moment is — 

M=^(/L-/^)=^-^ 
2^ 82 

where x is the distance measured from the middle of the beam 
and y the vertical height of the point above the tangent at the 
middle of the beam. Then by method (iii.) — 



d'y wU 



wx 

2 



5i6 



Mechanics applied to Engineering. 



^^^ = "8 6" + ^ 



16 

•w 



(C = o) 

(K = o) 



. so/L* sWL' ^ L 

o = - „, = o „T when * = - 

384EI 384EI 2 

Case VIII. — Beam loaded with two equal weights symmetri- 
cally placed. — By taking the 
*f W moments ofthe triangular area 

._^_ i^ _:^ abc and the rectangle bced, the 

--Z/— ► — i.^—^y—-i---, deflection becomes — 




and when L, = L2 = — , this 
3 



c e 

Fig. 508. 

expression becomes — 

„ 23WL» WL» - , V 
^ = Wl = iSEI <"^"'y> 

or if Wo be the total load — 

Case IX. — Beam loaded with one eccentric concentrated load. 

. L W 

L, 




Fig. 509. 



It should be noted that the point of maximum deflection 
does not coincide with that of the maximum bending moment. 

We have shown that the point of maximum bending moment 
in a beam is the point where the shear changes sign, and we 



Deflection of Beams. 517 

shall also show that the bent form of a beam is obtained by 
constructing a second bending-moment diagram obtained by 
taking the original bending-moment diagram as a load diagram. 

Hence, if we construct a second shear diagram, still treating 
the original bending-moment diagram as a load diagram, we 
shall find the point at which the second shear changes sign, or 
where the second bending moment, i.e. the deflection, is a 
maximum. This is how we propose to find the point of 
maximum deflection in the present instance. 

Referring to p. 483, we have — 

The shear at a section | _ t> _ ^^ 
distant / from Ri j ^ 2L1 



=^.=^(^+^7) 



+ 



ML = 



The shear changes sign and the deflection is a maximum 
when — 

ML/ LiN ML/ M/2 



orwhen/=^^\L.+\)+% = L^ 



3^ 3L ^3 

where Lj = «L and La = L — Lj, i.e. the shorter of the two 
segments ; and the deflection at this point is — 



3EI 3EIL 






The deflection 8 under the load itself can be found thus — 
Let the tangent to the beam at this point o be »«; then 
we have — 

But these are the deflections measured from the tangent vx. 

Let S = slope of the tangent vx; then uv = SLj, and 
xy = SL2, and the actual deflection under the load, or the 
vertical distance of o from the original position of the 
beam, is — 

8 = 81 — «?/, or 8 = 8a -f xy 
whence W_sL^ = 5^V SI. 



5i8 Mechanics applied to Engineering. 

3EI V Lj + La / 



and 8 = 5^' + 



J-ig / ixiJji — K-2i-A2 



3EI ' 3EU L, + L, 



) 



which reduces 




g _ WLi'La' _ WW 
3EI(La + L,) 3EIL 

Case X. — Beam hinged at one end, free at the other, propped 
in the middle. Load at the free end. 

The load on the 
hinge is also equal to W, 
since the prop is central, 
and the load on the prop 
is 2W; hence we may 
treat it as a beam sup- 
ported at each end and 
^^ centrally loaded with a 

load zW. Hence the 

2WL' 
central deflection would be ■ if the two ends were kept 

level ; but the deflection at the free end is twice this amount ; 
or — 

8 = 4WL' ^ WL' 
48EI 12EI 

Case XI. — General case of a learn whose section varies from 
point to point. (See also p. 270.) 

(i.) Let the depth of the section be constant, and let the 
breadth of the section vary directly as the bending moment ; 
then the stress will be constant. We have — 

,, /I EI / 1 
M = ■'^ = — , or ^ = - 

y 9 ^y p 

But, since/, y, and E are constant in any given case, p is also 
constant, whence the beam bends to the arc of a circle. 

(ii.) Let both the depth of the section and the stress vary ; 

then <- = - if V varies directly as /, -t will be constant, and 

y 9 y 

the beam will again bend to the arc of a circle. 



Deflection of Beams. 



519 



From p. 507, we have- 



8EI 

g - 3WL^ 
8EB^ 



32EI 



Let the plate be cut up into strips, and bring the two long 
edges of each together, making a plate with pointed ends of 
the same form as plate i on the plan ; pack all the strips as 
shown into a symmetrical heap. Looked at sideways, we see a 
plate railway spring. 




t: 



1 



> 

Fig. 511 



Let there be n plates, in this case 5, each of breadth b\ 
then B = nb. Substitute in the expression above — 

s _ 3WL» 



8E«,5/» 



If a railway-plate spring be tested for deflection, it may not, 
probably will not, quite agree with the calculated value on 
account of the friction between the plates or leaves. The result 
of a test is shown in Fig. 512. When a spring is very rusty it 
deflects less, and when unloading more than the formula gives, 
but when clean and well oiled it much more closely agrees with 



520 



Medianics applied to Engineering. 



the formula, as shown by the dotted lines. If friction could be 
entirely eliminated, probably experiment and theory would 
agree. 

In calculating the deflection of such springs, E should be 
taken at about 26,000,000, which is rather below the value for 
the steel plates liiemselves. Probably the deflection due to 
shear is partly responsible for the low modulus of elasticity, and 




'V 6 8 10 

Locul on Spring 



12 



from the fact that the small central plate (No. 5) is always 

omitted in springs. 

Case XII. Beam unevenly loaded. — Let the beam be loaded 

as shown. Construct the b ending-moment diagram shown below 

the beam by the method given on p. 498. Then the bending 

moment at any section is M = »«.«. D . OH inch-lbs., using 

the same notation as on p. 499. Then i inch on the vertical 

M 

scale of the bending-mpment diagram = — = »«.«. OH 

inch-lbs., and i inch on the horizontal scale = n inches. 
Hence one square inch on the diagram = m . «'0H inch-lbs. 
Then A = a.m. «^0H, where a = the shaded area measured 
in square inches. 

The centre of gravity must be found by one of the methods 



Deflection of Beams. 



521 



described in Chap. III. Then L„ = « . 4 where /„ is measured 
in inches, and the deflection — 



AL, 
EI 



a.m. «^0H . /, 
EI 



It is evident that the height of the supports above the 
tangent is the same at both 
ends. Hence the moment of 
the areas about the supports 
on either side of the tangent 
point must be the same. The 
point of maximum deflection 
must be found in this way by 
a series of trials and errors, 
which is very clumsy. 

The deflection may be 
more conveniently found by a 
somewhat diflferent process, as 
in Fig. 514. 

We showed above that the deflection is numerically equal 
to the moment of each little element of area of the bending- 
moment diagram about the free end -i- EL The moment of 




Fig. 513- 




Jtefledian, Curve 

Fig. 514. 

each portion of the bending-moment diagram may be found 
readily by a link-and-vector polygon, similar to that employed 
for the bending-moment diagram itself. 

Treat the bending-moment diagram as a load diagram ; split 
it up into narrow strips of width x, as shown by the dotted 
lines ; draw the middle ordinate of each, as shown in full lines : 
then any given ordinate x by « is the area of the strip. Set 
down these ordinates on a vertical line as shown; choose a 



5^2 Mechanics applied to Engineering. 

pole O', and complete both polygons as in previous examples. 
The link polygon thus constructed gives the form of the bent 
beam ; this is then reproduced to a horizontal base-line, and 
gives the bent beam shown in dark lines above. The only 
point remaining to be determined is the scale of the deflection 
curve. 

We have i inch on the load scale of the] 

first bending-moment diagram 
also I inch on the length of the bending-1 _ 

moment diagram j ~ "^ mches 

and the bending moment at any point M = m .n.Ti . OH 



' I = OT lbs. 



where D is the dept h of the bending-moment diagram at the 
point in inches, and OH is the polar distance, also expressed 
in inches. Hence i inch depth of the bending-moment diagram 

represents Y=r-= wz« . OH inch-lbs., and i square inch of the 

bending-moment diagram represents ot«*OH inch-inch-lbs. 
Hence the area xQ represents arDww^OH inch-inch-lbs. ; but 
as this area is represented on the second vector polygon by D, 
its scale is xmn'OH. ; hence — 



s «w«='OHmDiOiHi 

8 = Ei 



EI 

If it be found convenient to reduce the vertical ordinates of 
the bending-moment diagram when constructing the deflection 

vector polygon by say -, then the above expression must be 

multiplied by r. 

The following table of deflection constants k will be 
found very useful for calculating the deflection at any section, 
if the load W is expressed in tons, E must be expressed in 
tons per square inch. The length L and the moment of 
inertia I are both to be expressed in inch units. 

Example. — A beam 20 feet long supports a load of 3 tons 
at a point 4 feet from one support : find the deflection at 
12 feet from the same support. I = 138 inch^-units. E = 
12,000 tons per square inch. 

The position of the load is ^ = o'2. 



Deflection of Beams. 



523 



H 

< 

a 
o 
U 



o 
z 

Q 

a 



Q 

H 
PS 
O 



pEOj JO nopisoj 



b b 



b b 



ON 

b 



i. ^ 



O r^ 



u 



»> =" 
S 01 

< o 
w 9 

N 
O 
o 



Q 

H 
W 
H 

O 

s 

<; 
o 

g 

M 









t^ 


N 


u^ 


iri 


w 


':^ 





r^ 


NO 




ON 

b 




M 


CO 


■ei- 


10 


NO 


NO 


NO 


■<d- 


CNl 






8 


8 


8 






8 


8 


8 


8 


8 








b 


b 


b 


b 


b 


b 


b 


b 


b 








« 


fo 


r^ 


t^ 


CO 








VI 


t^ 




00 




ro 


NO 


00 







M 


M 




-^ 








8 








kH 




•H 


M 



































p 








b 


b 


b 


b 


b 


b 


b 


b 











>f2 


r-* 


■* 





NTl 


10 


f^ 










t^ 




S" 


00 


w 


m 


NO 


NO 


■^ 




NO 































b 































13 






b 


b 


b 


b 


b 


b 


b 


b 


b 


S 
























3 










































a 




























\r\ 


r^ 








NO 


M 


I'l 





Tj- 


f 




LO 





li^ 


00 


On 


On 


VO 


w 


ND 













HH 






M 







m 


b 
















p 











p 


^0 






b 


b 


b 


b 


b 


b 


b 


b 


b 
























'■C 

























u 




•« 


„ 


00 


i/i 


NO 


00 


NO 


U-) 


00 


N* 


VI 





■s 




NO 


ON 





ON 


VO 




NO 













M 











b 





p 

















p 





^ 




•1 


b 


b 


b 


b 


b 


b 


b 


b 


b 


*J 




> 




















■g 






































































2 











vo 


N 


NO 








t^ 


NO 


^ 








« 


NO 


On 


ON 


00 


iri 





10 


Ti* 
















M 







tS 


b 




P 





p 


p 


P 











P 


p: 









b 


b 


b 


b 


b 


b 


b 


b 



















































+j 








































































(2 












r^ 


ITN 


NO 





■^ 


r^ 


NO 






vo 




*:(- 


NO 


NO 


NO 


w 


GO 


ti- 




CO 




8 


hH 








>-4 












b 







p 





3 








p 


p 






b 


b 


b 


b 


b 


b 


b 


b 


b 








r^ 


u^ 


■ 





00 


t^ 


r^ 


fO 


w 










00 




M 







00 


NO 


m 




C4 


















8 

































p 




b 




b 


b 


b 


b 


b 


b 


b 


b 


b 








NO 


t^ 





^ 




<o 


LO 


w 


1^ 








« 




NO 


NO 


NO 


NJ~) 


■^ 


ro 


i-i 

























8 










b 
































b 


b 


b 


b 


b 


b 


b 


b 


b 



puoj JO UOlJtSOJ 



b b 



b 



On 

b 



524 



Mechanics applied to Engineering. 



The position at which the deflection is measured = ^ = 0-6. 
Referring to the table, ^ = 0-0107. 



Then 8 = 



0-0107 X 3 X 240 



= 0-24 inch. 



12,000 X 138 

When a beam supports a number of loads, the deflection 
due to each must be calculated and the results added. When 
the loads are not on the even spaces given in the table, the 
constant can be obtained approximately by interpolation or 
by plotting. 

A very convenient diagram for calculating the deflection of 
beams has been constructed by Mr. Livingstone; it is pub- 
lished by, "The Electrician" Printing and Publishing Co., 
Fleet Street, London. 

Another type of diagram for the same purpose was published 
in Engineering, January 13, 1913, p. 143. 

Example. — A beam 20 feet long, freely supported at each 
end, was loaded as follows : — 



Load in 
tons (W). 


Distance rrom 
end of beam. 


Position of 
load. 


K . 


KW 


3 
S 

2 

4 


3' 6" 

7' 6" 

11' 8" 

IS' a" 


0-175 

0-37S 

059 

0-76 


0-0095 
0-0177 
0-0194 
0-0139 


0-0285 
0-0885 
0-03S8 
0-0556 

0-2114 



Find the deflection under the 2 tons load, 
tons per sq. inch. I = 630. 



E = 12,000 



8 = 



0-2114 X 240' 

I2-000 X 630 



= 0-39 inch. 



By a graphical process 8 = 0-40 inch. 
Deflection of Built-in Beams.— When a beam is built 
in at one end only, it bends down with a convex curvature 



Fig. 515. 



Fig. stS. 



Upwards (Fig. 515); but when it is supported at both ends, 
it bends with a convex curvature downwards (Fig. 516); 



Deflection of Beams. 



525 



and when a beam is built in at both ends (Fig. 517), we get a 
combined curvature, thus — 




Fig. 519. 

Then considering the one kind of curvature as positive and 
the other kind as negative, the curvature will be zero at the 
points XX (Fig. 518), at which it changes sign; such are termed 
"points of contrary flexure." As the beam undergoes no 
bending at these points, the bending moment is zero. Thus 
the beam may be regarded as a short central beam with free 
ends resting on short cantilevers, as shown in Fig. 519. 

Hence, in order to determine the strength and deflection of 
built-in beams, we must calculate first the positions of the 
points *, X. It is evident that they occur at the points at which 
the upward slope of the beam is equal to the downward slope 
of the cantilever. 

We showed above that the slope of a beam or cantilever at 
any point is given by the expression — 

A 



Slope = 



EI 



Case XIII. Beam built in at both ends, with central load. 




■0-2S—* 



Fig. 



A for cantilever 



4 



2 



WT 
A for beam = Iltl^ X i^ = 
2 2 



526 



Mechanics applied to Engineering. 



Hence, as the slope is the same at the point where the beam 
joins the cantilever, we have — 

WL,= _ WL, 



'-, or L, = L2 = — 

4 4 4 

Maximum bending moment in middle of central span^ 

WLs ^ WL 

2 8 

Maximum bending moment on cantilever spans — 
WLi_ WL 



8 



Deflection of central span- 

W(2L, 
48Er 

Deflection of cantilever — 

^L» 
2 ' 

\ = - — 

3EI 



..w@". 



WL« 



48EI 384EI 



3E1 384E1 



WL* 



Total deflection in middle of central span — 

WL» 



8 = 8^ + 8 = 



[92EI 



This problem may be treated by another method, which, 
in some instances, is simpler to apply than the one just given. 




Wlien a beam is built in at both ends, the ends are necessarily 
level, or their slope is zero ; hence the summation of the slope 
taken over the whole beam is zero, if downward slopes be 



Deflection of Beams. 527 

given the opposite sign to that of upward slopes. Since the 
slope between any two sections of a beam is proportional to 
the area of the bending-moment diagram between those 
sections, the net area of the bending-moment diagram for a 
built-in beam must also be zero. 

A built-in beam may be regarded as a free-ended beam 
having overhanging ends, da, Vb, which are loaded in such a 
manner that the negative or pier moments are just sufficient to 
bring those portions of the beam which are over the supports 
to a level position. Then, since the net area must be zero, we 
have the areas — 

feg-adf-gcb^ o 

But in order that this condition may be satisfied, the area 

of the pier-moment diagram adcb must be equal to the area of 

the bending-moment diagram aeb for a freely supported beam, 

or — 

, he 
ad = — 



Whence the bending moment at the middle and ends is -5-, 

and the distance between the points of contrary flexure 

fg = - ; all the other quantities are the same as those found 
2 

by the previous method. 

It will be seen that dc is simply the mean height-line of the 
bending-moment diagram for the free-ended beam. 

Thus when the ends are built in, the maximum bending 
moment is reduced to one-half, and the deflection to one- 
quarter, of what it would have been with free ends. 

Case XIV. — Beam built in at both ends, with a uniformly 
distributed load. 



A. for cantilever- 



A for beam — 



f'wLjL, , a/L, 



+ 



> 



2 ^ 3 

These must be equal, as explained above— 

.^26 



528 Mechanics applied to Engineering. 

Let Iq = wLa. 

Then ^ = ^^^ + ^5^ 
326 

2 = 3«^ + «' 
which on solving gives us « = 0732, 
We also have — 

L, + L, = - 

2 

or 1732L2 = - 
2 

La = o'289L 

and Li = 0732 X o'289L = o'2iiL 



^L. 



Fig. 5sa. 

Maximum bending moment in middle of central span — 

a/La" ^ a; X o-289'L '' ^ wL» 
2 2 24 

Maximum bending moment on cantilever spans — 

w\^ + ^' = «' X 0-289L X o-2iiL+"'^°'"''^' 

2 2 

_ «/L' 

13 

Deflection of central span — 

g ^ 5K/(o-578L)* ^ w\} 
' 384EI 689EI 

Deflection of cantilevers due to distributed load — 




« _ w(o'2iiLy _ wh* 
^ 8El 4038EI 



Deflection of Beams. 

Deflection due to half-load on central part — 

5 _ zfLa X Li' _ w X o'289L x o-2ii'L' 
3EI ~~ ^Ei 

1105EI 
Total deflection in middle of central span — 

K/L* 



529 



= ^ + 8, + 8„ = 



384EI 



This problem may also be treated in a similar manner to 
the last case. The area 
of the parabolic bending- ;v<vj 



moment diagram axbxc 
\bf . ac, and the mean ^ 
height ae = ^bf; whence 
ae, the bending moment at 
the ends, is — 

^~8 TT 

and bg, the bending moment in the middle, is— 

and for the distance xx, we have — 

-(L,L,) = ^ 
2 12 




L,(L - L,) = 



1? 



L] = o'2iiL 

These calculations will be sufficient to show that identical 
results are obtained by both methods. Thus, when the ends 
are built in and free to slide sideways, the maximum bending 
moment on a uniformly loaded beam is reduced to -j-"^ = -f, and 
the deflection to \ of what it would have been with free ends. 

Case XV. — Beam built in at both ends, with an irregularly 
distribtited load. 

Since the ends of the beam are guided horizontally the 
slope of the ends is zero, hence the net area of the bending 



S30 



Mechanics applied to Engineering. 



moment diagram is also zero. The area of the bending 
moment diagram A for a freely supported beam is therefore 
equal to that of the pier moment diagram. 




and Mo = ^- - M. 



Mp = Mo 



x = -\ -Z-. — TTF^ I (See p. 60.) 
3VM<j + Mp/ '^ ^ ' 

Substituting the value of Mq'and reducing — 

Mp = ^(2/- 3a:) 

2A 
also Mq = -^(3^ - I) 

where x = The distance of the centre of gravity of the 
bending moment diagram for a freely supported 
beam from the nearest abutment (the centre of 
gravity of the pier moment diagram is at the 
same distance from the abutment). 
c = The distance of the centre of gravity of the 
bending moment diagram from the middle of 
the beam. 
A = The area of the bending moment diagram for 
a freely supported beam. 
If the beam be regarded as a cantilever fixed at one end, 
say Q, and free at the other. The moment of the external 
system of loading between P and Q causes it to bend down- 
wards, but the pier moment causes it to bend upwards, and 



Deflection of Beams. 531 

since the deflection at P is zero under Ihe two systems of 
loading it is evident that the moment of the banding-moment 
diagram due to the external loads between P and Q is equal 
to the moment of the pier bending-moment diagram, and since 
the areas of the two diagrams are equal the distance of the 
centre of gravity of each is at the same distaijce from Q. 

When the load is symmetrically disposed f = o, and the 
bending moment at the ends of the built-in beam is simply the 
mean bending moment for a freely supported beam, under 
the same system of loading. And the maximum bending 
moment in the middle of the built-in beam is the maximum 
bending moment for the freely supported beam minus the 
mean bending moment. The reader should test the accuracy 
of this statement for the cases already given. 

Beams supported at more than Two Points. — Wlien 
a beam rests on three or more supports, it is termed a 
continuous beam. We shall only treat a few of the simplest 
cases in order to show the principle involved. 

Case XVI. Beam resting on three supports, load evenly 
distributed. — The proportion of the load carried by each 
support entirely depends upon their relative heights. If the 
central support or prop be so low that it only just touches the 
beam, the end supports will take the whole of the load. 
Likewise, if it be so high that the ends of the beam only just 
touch the end supports, the central support will take the whole 
of the load. 

The deflection of an elastic beam is strictly proportional to 
the load. Hence from the deflection we can readily find the 
load. 

The deflection in the middle) _ 3 _ S^L^ 
when not propped j 384EI 

Let Wi be the load on the central prop. 

W,L' 



Then the upward deflection due to W, = Si = 



4SEI 



If the top of the three supports be in one straight line, the 
upward deflection due to Wj must be equal to the downward 
deflection due to W, the distributed load ; then we have — 

5WL3 _ WiL'' 
384EI ~ 48EI 
whence Wi = |W 



532 Mechanics applied to Engineering. 

Thus the central support or prop takes | of the whole 
load ; and as the load is evenly distributed, each of the end 
supports takes one-half of the remainder, viz. ^ of the load. 



Fig. 525. 

The bending moment at any point x distant /j from the 
end support is — 

M, = igwL/i — wliX - 



= z./(^L-9 = ^\3L-84) 



The points of contrary flexure occur at the points where 
the bending moment is zero, i.e. when — 

^'(3L - 8/,) = or when 3L = B/j or /, = fL 

Thus the length of the middle span is — . It is readily shown, 

4 
by the methods used in previous paragraphs, that the maximum 

wP 
bending moment occurs over the middle prop, and is there — , 

32 
or 5 as great as when not propped. 

When the three supports are not level. Let the load on 
the prop be — 

mvL = nW 

Then the upward deflection due to the prop is — 

48EI 

and the difference of level between the central support and the 
end supports is — 

384 EI ~ 48EI ~ 384EI^^ "^ 



Deflection of Beams. 533 

When the result is negative it indicates that the central 

support is higher than the end supports ; if « = i the whole 

WL' 
load is taken by the prop, and its height is — -^=-z above the 

end supports. 

When the load is evenly distributed over the three 
supports « = I the prop is then below the end supports by 
7WL= WL« 

-^^ = —r-v^ nearly. 

11S2EI i6sEI ■' 

Where there are two props symmetrically placed at a 
distance x from the middle of the beam, the downward de- 
flection at these points when freely supported at each end 
is (see page 516) — 

^ ^~384EI 384EI^'4L^ ibx ) 

If the spans are equal x = -^ and 


_ 4-s46wU _ WL' 
^~ 384EI ~ 88-4EI 

The upward deflection due to the two props is — 

o'o309PL' 
El 

where P is the load on each prop ; the constant is taken from 
the table on page 523. 

When all the supports are level — 

WL" _ o-o309PL» 

88-4EI ~ EI 

W 
P = — - = 0-37W 

273 

And the load on each end support is 0T3W. 

Case XVII. Beam with the load unevenly distributed, with 
an uncentral prop. — Construct the bending moment and deflec- 
tion curves for the beam when supported at the ends only 
(Fig. 526). 

Then, retammg the same scales, construct similar curves for 
the beam when supported by the prop only (Fig. 527). If, due 
to the uneven distribution of the load, the beam does not 
balance on its prop, we must find what force must be applied 



534 



Mechanics applied to Engineering. 



at one end of the beam in order to balance it. The unbalanced 
moment is shown by xy (Fig. 527). In order to find the force 
required at z to balance this, join xz and yz, and from the pole 
of the vector polygon draw lines parallel to them ; then the 
intercept x-^y-^, = Wj on the vertical load line gives the required 
force acting upwards (in this case). 




■ A f, y 



Fig. 529. 



In Fig. 528 set off 8 and 80 on a vertical. If too small to be 
conveniently dealt with, increase by the method shown ioj/,ej, 
and construct the rectangle efgh. If the prop be lowered so 
that the beam only just touches it,- the whole load will come 
on the end supports ; the proportion on each is obtained from 
Ri and R3 in Fig. 526. Divide ^/4 in i in this proportion. 

As the prop is pushed up, the two ends keep on the end 



Deflection of Beams. 535 

supports until the deflection becomes 8 + So ; at that instant the 
reaction Rj becomes zero just as the beam end is about to lift 
ofl" the support, but the other reaction Rj supports the un- 
balanced force W]. This is shown in the diagram by ee-^ = W, 
to same scale as Ri and Rj. 

Join ie and ge-^ ; then, if the three supports be level, the prop 
will be at the height/. Draw a horizontal from/ to meet ge-^ in 
gf,; erect a perpendicular. Then the proportion of the load 

taken by the prop is ^^, by the support Rj is |^, by the 
/««« /o«o 

support Rais^. 

Likewise, if the prop be raised to a height corresponding to 
/i, the proportions will be as above, with the altered suffixes 
to the letters. 

In Fig. 528, we have the final bending-moment diagram for 
the propped beam when all the supports are level ; comparing 
it with Fig. 526, it will be seen how greatly a prop assists in 
reducing the bending moment. 

It should be noted that in the above constructions there is 
no need to trouble about the scale of the deflections when the 
supports are level, but it is necessary when the prop is raised 
or lowered above or below the end supports. 

This method, which the author believes to be new, is 
equally applicable to continuous beams of any number of 
spans, but space will not allow of any further cases being 
given. 

Stiffness of Beams.— The ratio '^^^^^^^°" is termed the 

span 
' stiffness " of a beam. This ratio varies from about 



the best English practice for bridge work ; it is often as great 
as 3^ for small girders and rolled joists. 

By comparing the formulas given above for the deflection, 
it will be seen that it may be expressed thus — 

«EI 

where M is the bending moment- and « is a constant depend- 
ing on the method of loading. 

In the above equation we may substitute /Z for M and Zy 
for I ; then — - 

g^/ZL^^/L^ 
nYlLy nEy 



536 Mechanics applied to Engineering. 

Hence for a stiffness of 2^5, we have — 

i = _i_ = fh 

L 2000 wEy 
or 2000/L = wEy 

Let/= 15,000 lbs. square inch ; 
E = 30,000,000 „ „ 

Then «y = L 

But V = - 
2 

where d = depth of section (for symmetrical sections) ; then— 

nd = 2L 

, d 2 

and =r- = - 

L » 

Values of «. 

Beam. Cantilever. 

(a) Central load ... 12 — 

End load ... ... — 3 

{i) Evenly distributed load g-6 4 

{e) Two equal symmetrically placed loads dividing } 

beam into three equal parts ... ... ...\ ^ ^ 

{d) Irregular loading (approx.) 11 3*5 



Values of ~ 






, 


Stifihess. 




iimro 


vm 


^ 


.firaffi, central load 6 


12 


24 


Cantilever, end loa.d 1-5 


3 


6 


.ffMOT, evenly distributed load 4-8 


9-6 


19-2 


Cantilever, „ ,, 2 


4 


8 


Beam, two symmetrically placed loads, as in} ., 
Fig. 423 S"^^ 


9-3 


l8-6 


.ffMOT, irregular loading (approx.) 5-5 


II 


22 


Cantilever, „ „ 175 


3-5 


7 



This table shows tlie relation that must be observed between 
the span and the depth of the section for a given stiffness. 

The stress can be found direct from the deflection of a 
given beam if the modulus of elasticity be known ; as this does 
not vary much for any given material, a fairly accurate estimate 
of the stress can be made. We have above — 

nE.d 



Deflection of Beams. 537 

hence/ = -^^ 

The system of loading being known, the value of n can be 
found from the table above. The value of E must be assumed 
for the material in the beam. The depth of the section d can 
readily be measured, also 8 and L. 

The above method is extremely convenient for finding 
approximately the stress in any given beam. The error cannot 
well exceed lo per cent., and usually will not amount to more 
than 5 per cent. 



CHAPTER XIV. 



COMBINED BENDING AND DIRECT STRESSES. 

In the figure, let a weight W be supported by two bars, i and 2, 
whose sectional areas are respectively Aj and Aj, and the 
corresponding loads on the bars Rj 
and R2; then, in order that the stress 
may be the same in each, W must be 
so placed that Rj and R2 are pro- 
portional to the sectional areas of the 
Ri_Ai 



m 




'/////. 




M 


f?f 


y 

i 




>? 


II ,1 



bars, or 



But Ri« = RjZ, 



M^.■ ■-« - 
Fig. 530. 



or Ai« = Aja ; hence W passes through 
the centre of gravity of the two bars 
when tlie stress is equal on all parts of 
the section. This relation holds, how- 
ever many bars may be taken, even if taken so close together 
as to form a solid section ; hence, in order to obtain a direct 
stress of uniform intensity all over a section, the external force 
musi be so applied that it passes through the centre of gravity of 
the section. 

If W be not placed at the centre of gravity of the section, 
but at a distance x from it, we shall 
have — 

^{u + z) = W(« -f x) 

and when W is at the centre of gravity — 

R2(« -f z) = W« 

Thus when W is not placed at the 
centre of gravity of the section, the 
section is subjected to a bending moment 
Wa: in addition to the direct force W. 
Thus— 

If an external force W acts on a section at a distance xfrom its 
centre of gravity, it will be subjected to a dire J force W acting 



^, 



/f. 



•■^..U-'ii-X- > 



Fio. 531. 



Combined Bending and Direct Stresses. S39 

uniformly all over the section and a bending moment War. Th(^ 
intensity of stress on any part of the section will be the sum 
of the direct stress and the stress due to bending, tension and 
compression being regarded as stresses of opposite sign. 

In the figure let the bar be subjected to both a direct stress 
(+), say tension, and 

bending stresses. The i — '■ ^^^^ ^ — ,leMf ied 

direct stress acting uni- ' ^^^m \juia- 

formly all over the section 

may be represented by 

the diagram aicd, where 

a6 or cd is the intensity 

of the tensile stress (+) ; 

then if the intensity of tensile stress due to bending be 

represented by 6e (+), and the compressive stress ( — ) hy fc, 

we shall have — 

The total tensile stress on the outer skin = ab ■\- be = ae 
„ „ „ inner „ = dc -fc=df 

If the bending moment had been stili greater, as shown in 




side- 




side 



Xtnlocuz0<t- 
sid& 



Fig. 533, the stress (^ would be — , i.e. one side of the bar 
would have been in compression. 

Stresses on Bars loaded out of the Centre. — 
Let W = the load on the bar producing either direct tensile 
or compressive stresses ; 
A = the sectional area of the bar ; 
Z = the modulus of the section in bending ; 
*• = the eccentricity of the load, i.e. the distance of the 
point of application of the load from the centre 
of gravity- of the section ; 
/', = the direct tensile stress acting evenly over the 

section ; 
J\ —■ the direct compressive stress acting evenly over 
the section ; 



540 Mechanics applied to Engineering, 

f, = the tensile stress due to bending ; 

fc = the compressive stress due to bending ; 

M = the bending moment on the section. 

W 
Then j-=/c orf, or/' (direct stress) 

M W* 

"2 = "2" ^-^ ^"^-/^ or/ (bending stress) 

Then the maximum stress on the skinj _ ^ i ^ _ W , Wa: 
of the section on the loaded side /""•' ■'""a "Z^ 

Then the maximum stress on the skin^ _ f _ f _ xv/^ i x\ 
of the section on the unloaded side/ "•' -' ~ \A ~ z) 

In order that the stress on the unloaded side may not be of 
opposite sign to the direct stress, the quantity - must be greater ' 

than -. When they are equal, the stress will be zero on the 

unloaded side, and of twice the intensity of the direct stress on 

J X Z 
the loaded side ; then we have t = v> or — = «. Hence, in 

Jx cj A. 

order that the stress may no": change sign or that there may be 
no reversal of stress in a section, the line of action of the 

7 
external force must not be situated at a greater distance than — 

A 
from the neutral axis. 

Z 
For convenience of reference, we give various values of - 

A 
in the following table : — 



Combined Bending and Direct Stresses. 



541 



i 










/^^ 






/^/^^^ 


i^ 












-«imnil 


iimw 




v5 


.S|| 




1 




t3 


(((®)i)S 






m Mil 


Will 




6 

fa 


C*) 




i lit 


11 




'11111 


P 








g,s.s 
























\^-</ 
































• *U u J9 




























•3^° 
























11 














rt ., 














•Ss 














III 


3 


s 
o 










.5 




<A 










^ 


V 


u 










'% 


^ 


^ 










'd 


ts 










** 














i^ 


s 


s 










' ■§ 








^ 






«M 






3? 


:§ 


5; 


a 


5- 






■3 II 


W|« 


Q|«> 


1 


1 


PQ 
+ 


+ 


+ 


Q 

00 




"S t 






K 


ES. 


a 


s. 


2i 






1 






M 


S 


►*a 


a 












>o 




VO 






H 






. 








•< 










ST- 




1 






:§ 


a 


(5 






S 


K 


Q * 


1 


+ 


1 


■* 




En 


n 


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s 


Q 






8 










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-5 






















_,^ 














, s- 


.~i 




tM 












ST'^ 


Q 

1 

Q 
T, 












-5 
1 

M 


a 


+ 
a 


all 

'mm 


Q 

H 
CO 












P 6 














— ,D 












u 










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ll 


|, 






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K *G 






A 




=i° 























w 

















542 



Mechanics applied to Engineering. 



Af\A. 



i 



General Case of Eccentric Loading. — In the above 

instances we have only dealt 
with sections symmetrical about 
the neutral axis, and we showed 
that the skin stress was much 
greater on the one side than the 
other. In order to equalize the 
skin stress, we frequently use 
unsymmetrical sections. 

Let the skin stress at a due 
to bending and direct stress 



,...^..„_^. 






Fig. 538. 

= /■'„ ; likewise that at i = f\. 



W 



The direct stress all over the section = f = —- 

A 

For bending we have — 

according to the side we are considering; 
or Wa; = — or -r 



hence /a = 



W;cy. 



andA=/+/=T + — ^■- 



also/»=/'-/= 



W 
A 
W 

" A ' 



V^x 



y<. 



I 



When y^ = y^ the expression becomes the same as we 
had above. 

Cranked Tie-bar.— Occasionally tie bars and rods have 




Fig. 539. 



to be cranked in order to give clearance or for other reasons, 
but they are very rarely properly designed, and therefore are a 
source of constant trouble. 



Combined Bending and Direct Stresses. 543 

The normal width of the tie-bar is b ; the width in the cranked 
part must be greater as it is subjected to bending as well as to 
tension. We will calculate the width B to satisfy the condition 
that the maximum intensity of stress in the wide part shall not 
be greater than that due to direct tension in the narrow part. 

Let the thickness of the bar be t. 

Then, using the same notation as before — 

B b 

X = - + u — - 

2 2 

the direct stress on the\ W ^ 
wide part of the bar I Bt ~ ' 

the bending stress on the"! _ Wx \ 2 ^ 2 / 

wide part of the bar / z" ^ BV 

the maximum skm stress) _ W \ 2 2 / 

due to both > Bt '^ Wt 

But as the stress on the wide part of the bar has to be 
made equal to the stress on the narrow part, we have — 

W _ W 6W(B + 2u- b) 
it ~Bt'^ 2BV 

W 
Then dividing both sides of the equation by y, and solvmg, 

we get — 



B = a/66u + 3^ + 26 

Both 6 and u are known for any given case, hence the width 
B is readily arrived at. If a rectangular section be retained, 
the stress on the inner side will be much greater than on the 
outer. The actual values are easily calculated by the methods 
given above, hence there will be a considerable waste of 
material. For economy of material, the section should be 
tapered oif at the back to form a trapezium section. Such a 
section may be assumed, and the stresses calculated by the 
method given in the last paragraph ; if still imequal, the correct 
section can be arrived at by one or two trials. An expression 
can be got out to give the form of the section at once, but it is 
very cumbersome and more trouble in the end to use than the 
trial and error method. 



544 



Mechanics applied to Engineering. 



Bending of Curved Bars. — Let the curved bar in its 
original condition be represented by the full lines, and after 
bending by the broken lines. 



C.A. 




Fig. S40. 

Let the distance of any layer from the central axis which 
passes through the centroid of the section be 4-^ when 
measured towards the extrados, and —y when measured 
towards the intrados. 

Let the area of the cross section be A = S^Sy = 28a 
The original circumferential length of a) _ ^tj , ^r, _ , 
layer distant j/ from the central axis 3 ~ ' +^/''i ~ ^i 

The final length =(R2+j')62 = 4 

The strain on the layer = (R2 + y)^^ — (Rj + y)Q-^^ = x 

a:=RA-RA+J^(^2-<9i) . . . • (i.) 



a: = o when y = — 



6a — Oi 



= h 



(ii.) 



The only layer on which the strain is zero is that at the 
neutral axis, hence h is the distance of the neutral axis from 
the central axis of the section. Rj and ^2 are unknown at this 
stage in the reasoning, hence the expression must be put in 
another form before h can be calculated. 

Substituting the value of h in (i.), 

x= -h{6^-6,)+y{e^-6,) 

x = (y-A){e!,-eO = {y-A)^6 . . (iii.) 



Combined Bending and Direct Stresses. 545 
For elastic materials we have — 

1 = 1=^ and /=^ = E.. . (iv.) 

'-—~h — ^^-x 

Substituting the value of x from (iii.) 

J - j^ = Etf . . . . (vi.) 

Since the total tension on the one side of the neutral axis 
is equal to the total compression on the other, the net force 
on the whole section 2(/Sa) = o 

or/ifli +/2aa + etc. =//«i' +/2V + etc. 

o, e4('J^*>. + elc.} - EA^K-'-i+i),. + ett.} 

^) <^^) ' 4-a->i 

^ ^ R^+y Ri(A - A') . 

A = -^ = -^, '- (see p. 546) . . . (vii.) 

Ri+Jf 
We also have M =/Z =/i^i«i +^^2*2 + etc. 

or M = 27(/. J . S«) = E A6li;|- ^^-^ ~ '^^ 8g| 
Ae= , ^ 

Substituting this value in (vi.) 

2 N 



271 



546 Mechanics applied to Engineering. 

Substituting the value of /i 



f=Z-^ 



M 






But 
Hence 



Ula = o, and Rj -^ — ;— = o 



/ = 



_ M 



Ri+y' 

y — h 



~ hK Ri +y 
The stress at the intrados — 

,_ M{y,-h) 
^' hA{R,-y,) 

The stress at the extrados — 




y:= My. + h) 



hh(K,+y,) 

where yi and ^, are the dis- 
tances of the intrados and 
the extrados respectively from 
the central axis. 
The value of 



■^a-.) 



A' or R: 



can be found graphically 
thus : The section of the 
curved bar is shown in full 
lines ; the centre of curvature 
is at O. The points a and b 
are joined to O ; they cut the 
central axis in d and b'. At 
these points erect perpen- 
diculars to cut the line ab, as 
shown in «„ and b^ which 
are points on the new area A', because 

ab ~Ri+;' 



Combined Bending and Direct Stresses. 547 



Similarly 



cd "~ Ri — _y 



The two areas, A and A', must be accurately measured 
by a planimeter. The author wishes to acknowledge his 
indebtedness to Morley's " Strength of Materials," from which 
the above paragraph is largely drawn. 

Hooks. — In the commonly accepted, but erroneous, theory 
of hooks, a hook is regarded as a special case of a cranked 
tie bar, and the stresses are calcu- 
lated by the expressions given in 
the paragraph on the " general case 
of eccentric loading,'' but such a 
treatment gives too low a value for 
the tensile stress on the intrados, and 
too high a value for the compressive 
stress on the extrados. In spite, 
however, of its inaccuracy, it will 
probably continue in use on account 
of its simplicity; and provided the 
permissible tensile stress be taken 
somewhat lower to allow for the 
error, the method gives quite good 
results in practice. 

The most elaborate and com- 
plete treatment of hooks is that 
by Pearson & Andrews, "On a '°' "^' 

Theory of the Stresses in Crane and Coupling Hooks," 
published by Dulau & Co., 37, Soho Square, W. 

The application of their theory is, however, by no means 
simple when applied to such hook sections as are commonly 
used for cranes. In the graphical treatment the sections must 
be drawn to a very large scale, since very small errors in 
drawing produce large errors in the final result. For a com- 
parison of their theory with tests on large crane hooks, see 
a paper by the author. Proceedings I.C.E., clxvii. 

For a comparison of the ordinary theory with tests on- 
drop forged steel hooks, see a paper by the author. Engineering, 
October 18, 1901. 

The hook section shown in Fig. 541 gave the following 
results — 

A = 14-53 sq. ins. Aj = 15-74 sq. ins. Ri = 5 ins. 
h = 0-384 in. M = 85 in.-tons. yi = 2'46 ins, 

y, = 2-81 ins. Then/, = 12-5 tons sq. in. 




548 



Mechanics applied to Engineering. 



The Andrews- Pearson theory gave i3"9 tons sq. in., the 
common theory 8-8 tons sq. in., and by experiment i3'2 tons 
sq. in. 

By this theory/, = 6*2 tons sq. in., the Andrews-Pearson 
theory gave 4"6, and the common theory 7-6 tons sq. in. 

Inclined Beam. — Many cases of inclined beams occur in 
practice, such as in roofs, etc. ; they 
are in reality members subject to 
combined bending and direct stresses. 
In Fig. 543, resolve W into two com- 
ponents, W, acting normal to the 
beam, and P acting parallel with the 
beam ; then the bending moment at 
the section x = Wj/i. 

But Wi = W sin a 

F.G. 543. ^"'^ ^ = ilHT 

hence M. = W sin ui X -■ 

* sm a 

M, = W/=/Z 

W/ 

^~ Z 

fr.1 ^ . -11 i.u ..• P W cos O 

The tension actmg all over the section = x = a — 




hence y max. = 
and/min. = 



W cos a 
~A 

W cos a 



W/ „, / cos a /\ 



N.B. — The Z is for the section x taken normal to the beam, ftof a 
vertical section. 

Machine Frames sub- 
jected to Bending and Direct 

Stresses. — Many machine frames 
which have a gap, such as punch- 
ing and shearing machines, riveters, 
etc., are subject to both bending 
and direct stresses. Take, for 
example, a punching-machine with 
a box-shaped section through AB. 
Let the load on the punch 
= W, and the distance of the 
punch from the centre of gravity of the section = X. X is at 




Fig. 544. 



Combined Bending and Direct Stresses. 549 

present unknown, unless a section has been assumed, but if 
not a fairly close approximation can be obtained thus : We must 
first of all fix roughly upon the ratio of the compressive to the 
tensile stress due to bending ; the actual ratio 
will be somewhat less, on account of the uni- 
form tension all over the section, which will 
diminish the compression and_ increase the m p 
tension. Let the ratio be, say, 3 to i; then, W^^-i\ 
neglecting the strength of the web, our section 
will be somewhat as follows : — 

Make A. = ^A, 
then X = G, + - approx. '^'* 

4 *''G- S45- 

Z = - = ^4 ^ ^ l±i. (approx.) 

4 
Z :» sAjH (for tension) 
But WX =/Z (/being the tensile stress) 

W (' 



'(g. + 5) = 3A3/ 



wCg.4-5) WrG,4-- 

Ac = Tj ■■ or -— p 

3H/ «H/ 

where n is the ratio of the compressive to the 
tensile stress, 

and A, = «A, 

Having thus approximately obtained the sectional areas of 
the flanges, complete the section as shown in Fig. 546 ; and 
as a check on the work, calculate the stresses by finding the 
centre of gravity, also the Z or the I of the complete section 
by the method given on page 544, or better by the " curved 
bar" method. 




CHAPtER XV. 

STRUTS. 

General Statement. — The manner in which short com- 
pression pieces fail is shown in Chapter X. ; but when their 
length is great in proportion to their diameter, they bend 
laterally, tmless they are initially absolutely straight, exactly 
centrally loaded, and of perfectly uniform material — three 
conditions which are never fulfilled in practice. The nature of 
the stresses occurring in a strut is, therefore, that of a bar 
subjected to both bending and compressive stresses. In 
Chapter XIV. it was shown that if the load deviated but very 
slightly from the centre of gravity of the section, it very greatly 
increased the stress in the material ; thus, in the case of a 
circular section, if the load only deviated by an amount equal 
to one-eighth diameter from the centre, the stress was doubled ; 
hence a very slight initial bend in a compression member very 
seriously affects its strength. 

Effects of Imperfect Loading. — Even it a strut be 
initially straight before loading, it does not follow that it will 



B 

Fig. 547. 

remain so when loaded j either or both of the following causes 
may set up bending : — 

(i) The one side of the strut may be harder and stiffer 
than the other ; and consequently the soft side will yield most, 
and the strut will bend as shown in A, Fig. 547. 



Struts. 55 1 

(2) The load may not be perfectly centrally applied, either 
through the ends not being true as shown in B, or through the 
load acting on one side, as in C. 

Possible Discrepancies between Theory and 
Practice. — We have shown that a very slight amount of 
bending makes a serious difference in the strength of struts ; 
hence such accidental circumstances as we have just mentioned 
may not only make a serious discrepancy between theory and 
experiment, but also between » experiment and experiment. 
Then, again, the theoretical determination of the strength of 
struts does not rest on a very satisfactory basis, as in all the 
theories advanced somewhat questionable assumptions have to 
be made ; but, in spite of it, the calculated buckling loads agree 
fairly well with experiments. 

Bending of Long Struts. — The bending moment at the 
middle of the bent strut shown in Fig. 548 is evidently W8. 

Then WS =/Z, using the same notation as in the 
preceding chapters. 

If we increase the deflection we shall correspondingly 
increase the bending moment, and consequently the 
stress. 

From above we have — 

^ =iz or's'Z, and so on 

O Oj 

But as /varies with 8,"s-= a constant, say K; 

Fig. 548. 

then W = KZ 

But Z for any given strut does not vary whqn the strut bends ; 
hence there is only one value of W that will satisfy the 
equation. 

When the strut is thus loaded, let an external bending 
moment M, indicated by the arrow (Fig. 549), be applied to it 
until the deflection is Sj, and its stress /i ; 

Then W81 + M =/,Z 
But W81 =/iZ 
therefore M = o 

that is to say, that no external bending moment M is required 
to keep the strut in its bent position, or the strut, when thus 
loaded, is in a state of neutral equilibrium, and will remain 



SS2 



Mechanics applied to Engineering. 



when left alone in any position in which it may be placed; 
this condition, of course, only holds so long as the strut is 
elastic, i.e. before the elastic limit is reached. This state of 
neutral equilibrium may be proved experimentally, if a long 
thin piece of elastic material be loaded as shown. 

Now, place a load Wj less than W on the strut, 
say W = Wj + w, and let it again be bent by an 
external bending moment M till its deflection is Sj 
and the stress /i ; then we have, as before — 

WiSi + M =/iZ = W8i = WA + wl^ 
hence M = w\ 
Thus, in order to keep the strut in its bent position 
with a deflection Sj, we must subject it to a + bend- 
ing moment M, i.e. one which tends to bend the 
strut in the same direction as WiSi ; hence, if we 
remove the bending moment M, the deflection will 
become zero, i.e. the strut will straighten itself. 
Now, let a load Wa greater than W be placed on 
the strut, say W s= Wj — a/, and let it again be bent until its 
deflection = Sj, and the stress f^ by an external bending 
moment M ; then we have as before — 

WA + M =/,Z = WA - wS, 
hence M = —w\ 

Thus, in order to keep the strut in its bent position with a 
deflection \, we must subject it to a — bending moment M, i.e. 
one which tends to bend the strut in the opposite direction to 
W281 ; hence, if we remove the bending moment M, the de- 
flection will go on increasing, and ere long the elastic limit will 
be reached when the strain will increase suddenly and much 
more rapidly than the stress, consequently the deflection will 
suddenly increase and the strut will buckle. 

Thus, the strut may be in one of three conditions — 




Fig. 549- 



Condition. 



When slightly bent by an ex- 
ternal bending moment M, 
on being released, the strut 
will- 



When supporting 
a load — 



Remain bent 
Straighten itself 

Bend still further and ultimately 
buckle 



W. 

less than W. 

greater than W. 



Struts. 



SS3 



_ Condition ii. is, of course, the only one in which a strut can 
exist for practical purposes ; how much the working load must 
be less than W is determined by a suitable factor of safety. 

Buckling Load of Long Thin Struts, Euler's 
Formula. — The results arrived at in the paragraph above 
refer only to very long thin struts. 

As a first approximation, mainly for the sake of getting 
the form of expression for the buckling load of a slender strut, 
assume that the strut bends to an arc of a circle. 
Let / = the eifective length of the strut (see Fig. 
S5o); 
E = Young's modulus of elasticity ; 
I = the least moment of inertia of a section 
of the strut (assumed to be of constant 
cross-section). 

Then for a strut loaded thus — 
* = 8Rt(^^«P- 425) =■• 



SEP 
8EI 



8EI 




or W = —7^ (first approximation) 

As the strut is very long and the deflection 
small, the length / remains practically constant, -and 
the other quantities 8, E, I are also constant for j-,<;. j^^^ 
any given strut ; thus, W is equal to a constant, 
which we have previously shown must be the case. 

Once the strut has begun to bend it cannot remain a 
circular arc, because the bending moment no longer remains 
constant at every section, but it will vary directly as the 
distance of any given section from the line of application of 
the load. Under these conditions assume as a second approxi- 
mation that it bends to a parabolic arc, then the deflection — 

<» ^ ^ ■., S I ^r M/^ W8/2 
8=-X-Mx§X--^EI = -7^7 = -7^ 



3 2 

a„dW = 9^ 



' 9-6EI 9-6EI 



The value obtained by Euler was — 
_ g^EI _ 9-87EI _ loEI 



(nearly) 



554 Mechanics applied to Engineering. 

This expression is obtained thus — 
The bending moment M at any point distant x from the middle 
of the strut is 

M = -WS = El^ (see page 510) 

Multiply each side by — 

^.^^ _WS ^ 
dx ds^ EI dx 

Integrating (|J=-S(8^ + c) 

When ^ = o, 8 = A, hence C = - A^ 
dx 






Integrating again — 

^ = A/^sin-'- + K 
V W A 

When a: = o, 8 = 0, therefore K = o 
Hence 8 = A sin \x^ — j 

When a; = - 8 = A 
z 



'CVI,)= 



and sin ( 

■"■ EI> 

The only angles whose sines are = 1 are -, — , etc. We 

2 2 



require the least value of W, hence — 

aV EI ~ 2 
and W = 



EI 

;ei 



Struts. 



555 



It must not be forgotten that this expression is only an 
approximation, since the direct stress on the strut is neglected. 
When the strut is very long and slender the direct compressive 
stress is very small and therefore negligible, but in short struts 
the direct stress is not negligible consequently for such cases 
the above expression gives results very far from the truth. 

Effect of End holding on the Buckling Load. — 
In the case we have j-ust considered the strut was supposed to 
be free or pivoted at the ends, but if the ends are not free the 
stmt behaves in a different manner, as shown in the accompany- 
ing diagram. 



Diagram showing Struts of Equal Strength. 



One end free, the 
other hxed. 



/=3L 







p = 






Both ends pivoted or 
rounded. 



/=L 




Fig. 



W = 



P = 



loEI 
loEp' 



One end rounded or 
pivoted, the other 
end built in or 
fixed. 







P = 



20Ep' 



Both ends fixed or 
built in. 



1=^ 
2 







Each strut is supposed to be of the same section, and loaded 
with the same weight W. 



556 Mechanics applied to Engineering. 

W 
A 



W 
Let P = the buckling stress of the strut, i.e. —, where 



W = the buckling load of the strut ; 
A = the sectional area of the strut. 

We also have r- = p'' (see p. 78), where p is the radius of 

gyration of the section. 

Substituting these values in the above equation, we have — 

' /=> 

The " eflfective " or " virtual "length /, shown in the diagram, 
is found by the methods given in Chapter XIII. for finding the 
virtual length of built-in beams. 

The square-ended struts in the diagrams are shown bolted 
down to emphasize the importance of rigidly fixing the ends ; if 
the ends merely rested on flat flanges without any means of 
fixing, they much more nearly approximate round-ended struts. 

It will be observed that Euler's formula takes no account 
of the compressive stress on the material ; it simply aims at 
giving the load which will produce neutral equilibrium as 
regards bending in a long bar, and even this it only does 
imperfectly, for when a bar is subjected to both direct and 
bending stresses, the neutral axis no longer passes through the 
centre of gravity of the section. We have shown above that 
when the line of application of the load is shifted but one- 
eighth of the diameter from the centre of a round bar, the 
neutral axis shifts to the outermost edge of the bar. In 
the case of a strut subject to bending, the neutral axis shifts 
away from the line of application of the load ; thus the bend- 
ing moment increases more rapidly than Euler's hypothesis 
assumes it to do, consequently his formula gives too high 
results; but in very long columns in which the compressive 
stress is small compared with the stress due to bending, the 
error may not be serious. But if the formula be applied to 
short struts, the result will be absurd. Take, for example, an 
iron strut of circular section, say 4 inches diameter and 40 

mches long; we get P = — z 9000000 1 _ jgj ^^^ jj^^^ 

1000 
per square inch, which is far higher than the crushing strength 
of a short specimen of the material, and obviously absurd. 
If Euler's formula be employed, it must be used exclusively 



Struts. 557 

for long struts, whose length / is not less than 30 dia- 
meters for wrought iron and steel, or 12 for cast iron and 
wood. 

Notwithstanding the unsatisfactory basis on which it rests, 
many high authorities prefer it to Gordon's, which we will 
shortly consider. For a full discussion of the whole question 
of struts, the reader is referred to Todhunter and Pearson's 
" History of the Theory of Elasticity." 

Gordon's Strut Formula rationalized. — Gordon's 
strut formula, as usually given, contains empirical constants 
obtained from experiments by Hodgkinson and others; but 
by making certain assumptions constants can be obtained 
rationally which agree remarkably well with those found 
by experiment. 

Gordon's formula certainly has this advantage, that it agrees 
far better with experiments on the ultimate resistance of 
columns than does the formula propounded by Euler; and, 
moreover, it is applicable to columns of any length, short or 
long, which, we have seen above, is not the case with Euler's 
formula. The elastic conditions assumed by Euler cease to 
hold when the elastic limit is passed, hence a long strut always 
fails at or possibly before that point is reached ; but in the 
case of a short strut, in which the bending stress is small 
compared with the compressive stress, it does not at all follow 
that the strut will fail when the elastic limit in compression is 
reached — indeed, experiments show conclusively that such is 
not the case. A formula for struts of any length must there- 
fore cover both cases, and be equally applicable to short struts 
that fail by crushing and to long struts that fail by bending. 
In constructing this formula we assume that the strut fails 
either by buckling or by crushing,' when the sum of the direct 
compressive stress and the skin stress, due to bending, are 
equal to the crushing strength of the material ; in using the 
term " crushing strength " for ductile materials, we mean the 
stress at which the material becomes plastic. This assumption, 
we know, is not strictly true, but it cannot be far from the 
truth, or the calculated values of the constant (a), shortly to 
be considered, would not agree so well with the experimental 
values. 

' Mons. Considire and others have found that for long columns the 
resistance does not vary directly as the crushing resistance of the material, 
but for short columns, which fail by crushing and not by bending, the re- 
sistance does of course entirely depend upon it, and therefore must appear 
in any formula professing to cover struts of all lengths. 



558 Mechanics applied to Engineering. 

Let S = the crushing (or plastic) strength of a short specimen 
of the material ; 
C = the direct compressive stress on the section of the 
strut; 
then, adopting our former notation, we have — 

C = -and/= — 
then S = C +/ 

S = -- + -=- (the least Z of the section) 

We have shown above that, on Euler's hypothesis, the 
maximum deflection of a strut is — 

g^ M/' _ fZfl 
loEI loEZy 

where y is the distance of the most strained skin from the 
centre of gravity of the section, or from the assumed position 
of the neutral axis. We shall assume that the same expression 

holds in the present case. In symmetrical sections y = -, 

3 

where d is the least diameter of the strut section. 
By substitution, we have — 

8 = 4^. 

, ^ W , W/72 ,. V 

^ SEdZj 



w/ 

_ W/ A/d P\ 
~ A^' + pZ "^ W 



If W be the buckling load, we may replace — by P, 



Then S = p(i + a^^) 



T, S S 



Struts. 559 

where r = ->, which is a modification of " Gordon's Strut 
a 

Formula.'' 

P may be termed the buckling stress of the strut. 

The d in the above formula is the least dimension of the 
section, thus — 




L 



JJ □! 



Fig. 532. 

It now remains to be seen how the values of the constant a 
agree with those found by experiment ; it, of course, depends 
upon the values we choose for / and E. The latter presents 
no difficulty, as it is well known for all materials; but the 
former is not so obvious at first. In equation (i.), the first 
term provides for the crushing resistance of the material 
irrespective of any stress set up by bending ; and the second 
term provides for the bending resistance of the strut. We 
have already shown that the strut buckles when the elastic 
limit is reached, hence we may reasonably take / as the elastic 
limit of the material. 

It will be seen that the formula is only true for the two 
extreme cases, viz. for a very short strut, when W = AS, and 
for a very long strut, in which S =/; then — 

,„ 5E</Z loEI 
W = -j^ or -p- 

which is Euler's formula. It is impossible to get a rational 
formula for intermediate cases, because any expression for 8 
only holds up to the elastic limit, and even then only when the 
neutral axis passes through the centre of gravity of the section, 
i.e. when there is pure bending and no longitudinal stress. 
However, the fact that the rational value for a agrees so well 
with the experimental value is strong evidence that the formula 
is trustworthy. 

Values of S,/, and E are given in the table below j they 



S6o 



Mechanics applied to Engineering. 



must be taken as fair average values, to be used in the absence 
of more precise data. 





Pounds per square in 


ch. 




S 


f 


E 


Soft wrought iron 

Hard „ 

Mild steel 

Hard 

Cast iron 

„ (hard close - grained\ 

metal) / 

Pitchpine and oak 


40,000 

48,000 

67,000 

110,000 

f 80,000 (no 
\marked limit) 

130,000 

8,000 


28,000 
32,000 
45,000 
75,000 

80,000 

130,000 
8,000 


25,000,000 
29,000,000 
30,000,000 
32,000,000 

13,000,000 

22,000,000 
900,000 



Material. 


Form of section. 


sEZ 


a. by experiment. 


Wrought iron ... 


OB 


Tloto,^ 


TJjtOTOJ 




• 


ifetos^ 


^ 







tSntOsi, 


Tsm '0 555 




ML+-LU 


?5iito,J, 


410553 


Mild steel 


■i 


^ 


jfc to m 




• 


^ 


■ ^ 







^ 


A, 




HL + J.U 


^ 


5J11 to 5J, (author) 


Hard steel 


■■ 




ssj to ^ 




• 


BjtOjfe 







^ 


sS,' 




HL + _LU 


^ 


— 



' The discrepancies in these cases may be due to the section being 
thicker or thinner than the one assumed in calculating the value of a. In 
the case of hollow sections, angles, tees, etc., the value of o should be 
worked out. 



Struts. 



561 



Material. 


Formof section. 


sEZ 


a by experiment 


Cast iron ... 


■■ 


raitOTj, 


tI, 




• 


iJlltOT}, 


^ 







iJntOTJs 


,1, (Rankine) jjj 




HL4--LU 


A to A 


A' 


Fitchpine and oak 


■i 


i. 


<^ 




m 


A 


^ (author) ,', 



N.B. — The values of a given in the last column are four times as great 
as those usually given, due to the / used in our formula being taken equal 
to L for rounded ends, whereas some other writers take it for square or 
fixed ends. 

The values of the constant a have been worked out for the 
various materials, and are given in tabulated form above ; also 
values found by experiment as given in Rankine's " Applied 
Mechanics," and by Bovey, " Theory of Structures and Strength 
of Materials " (Wiley, New York). 

Rankine's Strut Formula. — In the above tables it will 
be noticed that the value of a as found by calculation quite 
closely agrees with that found by experiment for the solid 
sections, but the agreement is not so good in the case of 
hollow or rolled sections, largely due to the fact that a varies 
with each form of section and with the thickness of the metal. 
If the value of a be calculated for each section there is no 
objection to the use of the Gordon formula, but if one value 
of a be taken to cover all the cases shown in the tables above 
it is possible that considerable errors may creep in. For such 
cases it is better to use Rankine's modification of Gordon's 
formula. 

Instead of writing in the Gordon formula — 

S = ^(x+^ 
we may write 



) 



AV "^ loEjAKV ~ A\ 



I + 



loE 



KV 



W = 



AS 
i + ^R 



2 o 



562 



Mechanics applied to Engineering. 



where h = 



f 



and 



„ uvw^^ -^ ~ A' '■'■ *^^ equivalent length 

divided by the /^aj/ radius of gyration of the. section about a 
line passing through the centroid- of the section. Values of k 
will be found in Chapters III. and XI. 
The values of b are as follows : — 







Value 


oib. 


Material. 


loE 


By experiment. 


Wrought iron 


5MII 


Sim 


Mild steel 


5755 


7SM 


Hard steel 


?OTn 


SMI 


Cast iron 


TB35 t° -Am 


ims 



The discrepancies are due to the assumed value of/ not 
being suitable for the material experimented upon. The terms 
" mild " and " hard " steel are very vague. If the properties 
of the material in the tested struts were known, the dis- 
crepancies would probably be smaller. It must, however, be 
borne in mind that the strength of struts cannot be calculated 
with the same degree of accuracy as beams, shafts, etc. 

In the case of long cast-iron struts the failure is usually 
due to tension on the convex side, and not to compression on 
the concave side. The expression then becomes — 

T 
P=: 



ar' — \ 



where T = the tensile strength of the material, or rather the 
tension modulus of rupture, i.e. the tensile stress as found from 
a bending experiment. The values of/ and T then vary from 
30,000 to 45,000 lbs. per square inch, and E (at the breaking 
point) varies from i t, 000,000 to 16,000,000 lbs. per square inch. 
The value of a then becomes jg^ for a rectangular section. 
On calculating some values for P, it will be seen that for long 
struts where the fracture might occur through the excessive 



Struts. 



563 



stress on the tension skin, the value given by this formula agrees 
fairly well with the values calculated from the original formula ; 
hence we see that such struts are about as likely to fail by 
tension on the convex side as by compression on the concave side. 
The following tables have been worked out by the formula 
given above to the nearest 100 lbs. per square inch. For those 
who are constantly designing struts, it will be found convenient 
to plot them to a large scale, in the same manner as shown in 
Fig. 553. In order better to compare the results obtained by 
Euler's and by Gordon's formula, curves representing both are 









\ 








■ 




















, 














1 






































s. 










































s 


V 




\ 






































\ 




\ 








































\ 


\ 
\ 


































«? 






\ 




































^ 











\^ 






































\ 


\^ 


• 






































Q 


s; 


■* 






























^ 










^ 


\ 






























j^ 












^ 






























^ 














s 




























Rl 
















'^^ 














































■~- 


;r 


-— 


. ^ 














































— 


— 


^^ 


^^ 


— 


— 



Fio. 553. 



■(4) 



Note. — The two curves in many cases practically coincide after 40 
diameters. In the figure the Gordon curve has been shifted bodily up, to 
better show the relation. 



given, from which it will be seen that they agree fairly well for 
very long struts, but that Euler's is quite out of it for short 
struts. 

The table on the opposite page gives the ultimate or the 
buckling loads ; they must be divided by a suitable factor of 
safety to get the safe working load. 



564 Mechanics applied to Engineering. 

Buckling Load of Struts in Pounds per Square Inch of Section. 











HL 








HL 


r 
or 


■■ 


• 





+ 


■■ 


• 





+ 


I 
d 








XU 








XU 




Wrought iron. 


Mild steel. 


S 


42,600 


42,000 


43,000 


41,700 


64,100 


63,100 


64,700 


62,000 


10 


38,800 


37,200 


39,900 


36,000 


56,600 


56,300 


58,300 


Sijooo 


20 


28,800 


25,600 


30,900 


23,200 


38.500 


33.500 


41,900 


29,800 


30 


20,000 


16,800 


22,200 


14,700 


25,000 


20,600 


28,600 


17,500 


40 


14,000 


11,400 


16,300 


9,600 


17,000 


13,400 


19,700 


11,100 


50 


10,400 


8,000 


12,400 


6,700 


11,900 


9,200 


14,100 


7,800 


60 


7,600 


5,900 


9,100 


4,900 


8,700 


6,700 


10,500 


5.400 


70 


5,800 


4,500 


7,100 


3.700 


6,700 


5,000 


8,100 


4,100 


80 


4,600 


3.500 


S,6oo 


2,900 


5,200 


3.900 


6,300 


3,200 


90 


3.700 


2,800 


4.500 


2,100 


4,200 


3,100 


5,100 


2,500 


100 


3,100 


2,300 


3.800 


1,900 


3.400 


2,500 


4,200 


2,100 




Hard steel. 


Cast iron (soft). 


S 


102,000 


100,300 


104,000 


98,600 


67,100 


64,000 


69,200 


58,900 


10 


85,500 


79,400 


89,600 


74.500 


45,200 


40,000 


49,200 


32.900 


20 


51.500 


43.300 


57.500 


37.900 


19,600 


16,000 


22,900 


11,900 


30 


30,800 


24,600 


36,100 


20,400 


10,100 


8,000 


12,100 


5,800 


40 


19,700 


15,400 


23,700 


12,700 


6,000 


4.700 


7.300 


3.400 


1° 


13.500 


10,300 


16,400 


8,500 


3.900 


3,100 


4,800 


2,200 


60 


9.750 


7,400 


11,900 


6,100 


2,800 


2,200 


3.400 


1,500 


70 


7.300 


5,500 


9,100 


4.500 


2,100 


1,600 


2,500 


1,100 


80 


S.70O 


4.300 


7,100 


3.500 


1,600 


1,200 


2,000 


870 


90 


4,600 


3.400 


5,700 


2,800 


1,300 


980 


1,600 


690 


100 


3.700 


2,800 


4,600 


2,200 


1,000 


790 


1,300 


560 


r 
or 
/ 
d 


Cast iron (hard close-grained). 


Pitchpine 


and oak. 




S 


109,000 


104,000 


112,000 


95.700 


6,300 


5,900 




10 


73.500 


65,000 


80,000 


53.500 


3.800 


3.300 




20 


31,800 


26,000 


37,200 


19,300 


1,500 


1,200 




30 


16,400 


13,000 


19,700 


9.400 


730 


580 




40 


9.700 


7,600 


11,900 


5.500 


430 


340 




50 


6,300 


5,000 


7,800 


3,600 


280 


220 




So 


4,600 


3,600 


5. 500 


2,400 


200 


150 




70 


3.400 


2,600 


4,100 


1,800 


140 


1 10 




80 


2,600 


1,900 


3.300 


1,400 


110 


90 




90 


2,100 


1,600 


2,600 


1,100 


90 


70 




100 


1,600 


1,300 


2,100 


900 


70 


60 





Struts. 



56S 



Factor of Safety for Struts. 



Wrought iron and steel 
Cast iron 
Timber 

350 



Dead loads. 


Live loads. 


... 4 


8 


... 6 


12 


... 5 


10 



300 
250 
200 



-s 
'SISO 



xlOO 



50 















b/ 


/<> 


/ 














/ 


/ 












/ 


v 


/ 


/ 


/ 


• 






/ 


// 


xy 


/ y 


* 










//. 


/ y 


y 










y 


^ 


^ 














^ 

















25 SO 75 100 125 150 175 
Weight in, Pounds per ft. 

Fig. SS4. 



200 225 



In choosing a section for a column, economy in material is 
not the only and often not the most important matter to be 
considered ; every case must be dealt with on its merits. Even 
as regards the cost the lightest column is not always the 
cheapest. In Figs. 554 and 555 we show by means of 
curves how the weight and cost of different sections vary with 
the load to be supported. Judging from the weight only, the 
hollow circle would appear to be the cheapest section, but the 
cost per ton of drawn tubes is far greater than that of rolled 
sections ; hence on taking this into account, we find the hollow 
circle the most expensive form of section. 

The values given in the figures must not be taken as being 
rigidly accurate ; they vary largely with the state of the market. 
Designers, however, will find it extremely useful to plot such 
curves for themselves, not only for struts, but for floorings, 
cross-girders, roof-coverings, roof-trusses, and many other 
details which a designer constantly has to deal with. 

Straight-line Strut Formula. — The more experience 



S66 



Mechanics applied to Engineering. 



one gets in the testing of full-sized struts and columns, the 
more one realizes how futile it is to attempt to calculate the 
buckling load with any great degree of accuracy. If the struts 
are of homogeneous material and have been turned or 
machined all over, and are, moreover, very caitfully tested 
with special holders, which ensure dead accuracy in loading, 
and every possible care be taken, the results may agree within 
5 per cent, of the calculated value; but in the case of com- 
mercial struts, which are not aX^jiSiys perfectly straight, and are 
350r 



300 



250 



1 200 



1l50 
^100 



50 

















A 


// 


















/ 






























^ 






fKJ 








y<- 


^ 


y 


y 








1/ 


V 


^ 














/ ^ 

















£5 ^ £10 £15 

Cost of a 20 foot column. 

Fig. S5S- 



£20 



not always perfectly centrally loaded, the results are frequently 
lo or 15 per cent, out with calculation, even when reasonable 
care has been taken ; hence an approximate expression, such 
as one of the straight-line formulas, is good enough for many 
practical purposes, provided the length does not exceed that 
specified. An expression of this kind is — 

P = M- N-, 
a 

where P = the buckling load in pounds per square inch ; 

M = a constant depending upon the material ; 

N = a constant depending upon the form of the strut 

section ; 

/= the "equivalent length'' of the strut; 

d = the least diameter of the strut. 



Struts. 



567 



Material. 


Form of section. 


M. 


N. 


i not to 

d 

exceed 


Wrought iron ... 


IB 


47,000 


82s 


40 


' 


• 


47,000 


900 


40 







47,000 


775 


40 




Ht-ff-J-U 


47,000 


1070 


30 


Mild steel 


■■ 


71,000 


1570 


30 




# 


71,000 


1700 


30 







73,000 


1430 


30 




HL + XU 


71,000 


1870 


30 


Hard steel 


1^ 


-114,000 


3200 


30 




• 


114,000 


3130 


30 







114,000 


2700 


30 




HL + J.U 


114,000 


3500 


30 


Soft cast iron . . . 


^ 


90,000 


4100 






• 


90,000 


4700 









90,000 


3900 






HL+ J-U 


90,000 


5000 




Hard cast iron ... 


■1 


140,000 


6600 






• 


140,000 


7000 




> 





140,000 


6100 






HL. + -LU 


140,000 


8000 




Pitchpine and oak 


Hi 


8000 


470 


10 




• 


8000 


Soo. 


10 



S68 



Mechanics applied to Engineering. 



XW 



Columns loaded on Side Brackets. — The barbarous 
practice of loading columns on side brackets is 
\Wi unfortunately far too common. As usually carried 
"n..!... out, the I practice reduces the strength of the 
cohimn to one-tenth ' of its strength when cen- 
trally loaded. 

In Fig. 557 the height of the shaded figure 
on the bracket of the column shows the relative 
loads that may be safely placed at the various 
distances from the axis of the column. It will be 
perceived how very rapidly the value of the safe 
load falls as the eccentricity is increased. If a 
designer will take the trouble to go carefully into 
the matter, he will find that it is positively cheaper 
to use two separate centrally loaded columns 
Fig. 556. instead of putting a side bracket on the much 
larger column that is required for equal strength. 
Lety^ = the maximum compressive stress on the material 
due to both direct and bending stresses ; 

ff = the maximum tensile stress on the 
material due to both direct and 
bending stresses ; 
/ = the skin stress due to bending ; 
C = the compressive stress acting all over 

the section due to the weight W ; 
A = the sectional area of the column. 

WX W 
Then/.=/-FC = ^4-^ 




wx_ w 

Z A 
If the column also carries a central 



and /, = 



load Wi, the above become — 
/, = WX W-fWi 
Z ■*" A 



/,= 



WX W + Wi 



Z A 

Columns loaded thus almost invariably 

fail in tension, therefore the strength must 

be calculated on the /, basis. We have 

neglected the deflection due to loading 

• The ten is not used with any special significance here ; may be 
one-tenth or »ven oue-twenlieth. 



Fig. 557- 



Struts. 



569 



(Fig. 558), which makes matters still worse; the tensile stress 
then becomes — 



/.= 



W(X + 8) W_+ W, 



The deflection of a column loaded in this way may be 
obtained in the following manner : — 



The bending moment = WX 
area of bending-moment diagram = WXL 

„ WXL' 



(approximately) 



After the column has bent, the bending moment of course 
is greater than WX, and approximates to W(X + 8), but S is 




"1^ 



if = 0-2 ^ 



S^ff,!^ 



Fig. 558. 



Fig. 559. 



usually small compared with X, therefore no serious error arises 
from taking this approximation. 

A column in a public building was loaded as shown in 
Fig. 559; the deflections given were taken when the gallery 
was empty. The deflections were so serious that when the 
gallery was full, an experienced eye immediately detected 
them on entering the building. 

The building in question has been condemned, the galleries 
have been removed, and larger columns without brackets have 



570 Mechanics applied to Engineering. 

been substituted. Thecolumn, as shown, was tested to destruc- 
tion by the author, with the following results — 

External diameter 4 '95 inches 

Internal diameter ... ... ... 3'7o ,, 

Sectional area ... fi'49sq. inches 

Modulus of the section 8" 18 

Distance of load from centre of column ... 6 inches 

Height of bracket above base ... 8' 6" 

Deflection measured, above base 4' 3" 

Win tons... I 2 I 4 I 6 1 8 I 10 I 12 I 14 I 16 I 18 
5 in inches... | o'035 1 o'o69 1 o'loo 1 0'148 1 o'igs 1 0'245 1 0^300 1 0-355 ' 0'42o 

The column broke at iS'iy tons; the modulus of rupture 
was i3'3 tons per sq. inch. 

Judging from the deflection when the weight of the gallery 
rested on the bracket, it will be seen that the column was in a 
perilously dangerous state. 

Another test of a column by the author will serve to 
emphasize the folly of loading columns on side brackets. 

Estimated Buckling Load if centrally loaded, about 
1000 tons. 

Length 10 feet, end flat, not fixed. 
Sectional area of metal at fracture ... 34*3 sq. inches 

Modulus of section at fracture 75 'O 

Distance of point of application of load 

from centre of column, neglecting slight 

amount of deflection when loaded ... 17 inches 
Breaking load applied at edge of bracket 65'5 tOnS 
Bending moment on section when fracture 

occurred 1114 tons-inches 

Compressive stress all over section when 

fracture occurred I '91 tons per sq. inch 

Skin stress on the material due to bending, 

assuming the bending formula to hold 

up to the breaking point ... ... I4"85 ,, ,, 

Total tensile stress on material due to 

combined bending and compression * ... 12*94 tons per sq. inch 
Total compressive stress on material due 

to combined bending and compression l6'76 ,, ,, 

Tensile strength of material as ascertained 

from subsequent tests 8'45 ,, ,, 

Compressive strength of material as ascer- 
tained from subsequent tests ... ... 30'4 i> i> 

Thus we see that the column failed by tension in the 
material on the off side, i.e. the side remote from the load. 

' The discrepancy between this and the tensile strength is due to the 
bending formula not holding good at the breaking point, as previously 
explained. 



CHAPTER XVI. 

TORSION. GENERAL THEORY. 

Let Fig. 560 represent two pieces of shafting provided with 



y 




Fig. s6o. 

disc couplings as shown, the one being driven from the other 
through the pin P, which is evidently in shear. 

Let S = the shearing resistance of 
the pin. 

Then we have W/ = Sy 

Let the area of the pin = a, and 
the shear stress 'on the pin he/,. 
Then we may write the ab o ve equation — 

W/ = f^y 

Now consider the case in which 
there are two pins, then — 




Fig. 561. 



w/ = Sy + Si^/i =fAy +Aaxyi 

The dotted holes in the figure are supposed to represent the 
pin-holes in the other disc coupUng. Before W was applied 
the pin-holes were exactly opposite one another, but after the 
application of W the yielding or the shear of the pins caused a 



572 Mechanics applied to Engineering. 

slight movement of the one disc relatively to the other, but 
shown very much exaggerated in the figure. It will be seen 
that the yielding or the strain varies directly as the distance 
from the axis of revolution (the centre of the shaft). When 
the material is elastic, the stress varies directly as the strain ; 
hence — 

Substituting this value in the equation above, we have — 



y 
y 



=-^'(a/ + «^>'.') 



Then, if a = Oj, and say y^ 



_y 



4 

Thus the inner pin, as in the beam (see p. 432), has only 
increased the strength by j. Now consider a similar arrange- 
ment with a great number of pins, such a number as to form 
a hollow or a solid section, the areas of each little pin or 
element being a, a^, a^, etc., distant y, y^, y^, etc., respectively 
from the axis of revolution. Then, as before, we have — 

W/=-^(ay^ + a^j^ + 0^,=' +, etc.) 

But the quantity in brackets, viz. each little area multiplied 
by the square of its distance from the axis of revolution, is the 
polar moment of inertia of the section (see p. 77), which we 
will term I,. Then — 

The W/ is termed the twisting moment, M,. /, is the skin 
shear stress on the material furthest from the centre, and is 
therefore the maximum stress on the material, often termed the 
skin stress. 

y is the distance of the skin from the axis of revolution. 



Torsion. General Theory. 



573 



-^ = the modulus of the section = Z^. To prevent confusion, 

we shall use the suffix / to indicate that it is the polar modulus 
of the section, and not the modulus for bending. 

Thus we have M, =/,Zp 
or the twisting moment = the skin stress X the polar modulus 

of the section 

Shafts subject to Torsion. — To return to the shaft 
couplings. When power is transmitted from one disc to the 
other, the pin evidently will be in shear, and will be distorted 




P 



Fig. 562. 

as shown (exaggerated). Likewise, if a small square be marked 
on the surface of a shaft, when the shaft is twisted it will also 
become a rhombus, as shown dotted on the shaft below. 

In Chapter X. we showed that when an element was 
distorted by shear, as shown in Fig. 563 (a), it was 




Fig. 363. 



equivalent to the element being pulled out at two opposite 
corners and pushed in at the others, as shown in Fig. 563, 
(b) and {c), hence all along the diagonal section AB there 



574 



Mechanics applied to Engineering. 





is a tension tending to pull the two triangles ADB, ACB 
apart ; similarly there is a compression along the diagonal CD. 
These diagonals make an angle of 45° with their sides. Thus, 
if two lines be marked on a shaft at an angle of 45° with the 
axis, there will be a tension normal to the one diagonal, and a 

compression normal to the 

J other. That this is the case 

can be shown very clearly 
by getting a piece of thin 
tube and sawing a diagonal 
slot along it at an angle of 
45°. When the outer end is 
p,Q jg^. twisted in the direction of 

the arrow A, there will be 
compression normal to the slot, shown by a full Une, and the 
slot will close ; but if it be twisted in the direction of the arrow 
B, there will be tension normal to the slot, and will cause it 
to open. 

Graphical Method of finding the Polar Modulus 
for a Circular Section. — The method of graphically finding 
the polar modulus of the section is precisely similar in principle 
to that given for bending (see Chap. XL), hence we shall not 
do more than briefly indicate the construction of the modulus 
figure. It is of very limited application, as it is only true for 
circular sections. 

As in the beam modulus figure, we want to construct a 
figure to show the distribution of stress in the section. 

Consider a small piece of a circular 
section as shown, with two blocks equiva- 
lent to the pins we used in th« disc 
couplings above. The stress on the inner 
block = fa, and on the outer block = /, ; 

then -^ = ■^. Then by projecting the 

width of the inner block on to the outer 
circle, and joining down to the centre of 
the circle, it is evident, from similar 
triangles, that we reduce the width and 

area of the inner block in the ratio — , or in the ratio of •^. 

The reduced area of the inner block, shown shaded, we will 

now term ai, where — =-^ =-^, or <j!,|/; = aja- 
<h J, y 




Fig. 565. 



Torsion. General Theory. 



S7S 



Then the magnitude of the resultant force acting on the two 
blocks = af, + «,/,! 

= of, + <f. =/.(« + «i') 

=f, (sHaded area or area of modulus figure) 

And the position of the resultant is distant y,, from the centre, 
where — 

ay + gi>i 

i.e. at the centre of gravity of the blocks. 

Then/.Z, =f,{a + «,') X ^^^^ 

=f.{ay + ch'y,) =^{ay^ + a^y^) 

which is the same result as we had before for W/, thus proving 
the correctness of the graphical method. 

In the figure above we have only taken a small portion of 
a circle ; we will now use the same method to find the Z_ for a 




-no 



Fig. s66. 



circle. For convenience in working, we will set it off on a straight 
base thus : Draw a tangent ab to the circle, making the length 
= irD ; join the ends to the centre O ; draw a series of lines 
parallel to the tangent ; then their lengths intercepted between 
ao and bo are equal to the ciicumference of circles of radii Oi, 
Oa, etc. Thus the triangle Oab represents the circle rolled out 
to a straight base. Project each of these lines on to the tangent, 
and join up to the centre ; then the width of the line I'l', etc., 
represents the stress in the metal at that layer in precisely the 
same manner as in the beam modulus figures. Then — 

The polar modulus of ) S^'^% °^ T^'f^ ^^T ^ f ''f "''^ 
the section Z, = i "^ ^; °^ S- of modulus figtfrefrom 

' ' {_ centre of circle 

or Z, = Ay. 



5/6 Mechanics applied to Engineering. 

The construction for a hollow circle is precisely the same 
as for the solid circle. It is given for the sake of graphically 
illustrating the very small amount that a shaft is weakened by 
making it hollow. 

This construction can be applied to any form of section, 




Fig. 567. 



but the strengths of shafts other than circular do not vary as 
their polar moments of inertia or moduli of their sections; 
serious errors will be involved if they are assumed to be so. 
The calculation of the stresses in irregular figures in torsion 
involves fairly high mathematical work. The results of such 
calculations by St. Venant and Lord Kelvin will be given in 
tabulated form later on in this chapter. 

Strength of Circular Shafts in Torsion.— We have 
shown above that the strength of a cylindrical shaft varies as 

?2 = Z,. In Chapter III., we showed that I. = — , where D 

y 32 

is the diameter, and y in this case = — ; hence — 



2 



which, it will be noticed, is just twice the value of the Z for 
bending. In order to recollect which is which, it should be 
remembered that the material in a circular shaft is in the very 
best form to resist torsion, but in a very bad form to resist 
bending ; hence the torsion modulus will be greater than the 
bending modulus. 

For a hollow shaft — 



Torsion. General Theory. 577 

I„ = — ^^ '-, -where D< = the internal diameter 

32 ' 

hence Z, = .ep = oT96(^^^-j 

If - of the metal be removed from the centre of the shaft, 

n ' 

we have — 

the external area 



The internal area = • 



n 



— ^ = or Dj^ = -a 

4 4» W' 

Z, = o-i96D{i-i) 

The strength of a shaft with a sunk keyway has never 
been arrived at by a mathematical process. Experiments 
show that if the key be made to the usual proportions, viz. 
the width of the key = \ diameter of shaft, and the depth 
of the keyway = \ width of the key, the shaft is thereby 
weakened about 19 per cent. See Engineering, March 3rd 
191 1, page 287. 

Another empirical rule which closely agrees with experi- 
ments is : The strength of a shaft having a sunk keyway is 
equal to that of a shaft whose diameter is less than that of 
the actual shaft by one-half the depth of the keyway j thus, 
the strength of a 2-inch shaft having a sunk keyway 0-25 inch 

deep is equal to a shaft {2 ~j = r87S inches diameter. 

This rule gives practically the same result as that quoted 
above. 



2 p 



578 Mechanics applied to Engineering. 

Strength of Shafts of Various Sections. 





„ . ,/D* - DA 



Fig. 569. 




Fin. 570. 



5 -- 



Fig. 571. 



Z, = -^,or 
Z, = 0-I96D3 



z = 


0-19603(1- 


m^J 


z. 


irTid-^ 





Z, = o-r96D<^ 



Z„ = 0-208S3 



g ^ 584M^ 
GD* 



. _ 584M/ 
G(D* - D<*) 



^^ 292M/(rf» + D") 
GDV 



g _ 4ioM^ 
S*G 



Z„ = 



B<52 



where m = 



3 + I'Sw 
6 



2o5M,/(^ + B') 



^B»G 



S 

Fig. S72. 



B 



Any section not 
containing re-en- 
trant angles (due 
to St. Venant). 



Z.= 



A* 



i^ (aPProx.) 

where A = area of sec 

I, = polar moment of in 

y = distance of furthest cdg 



22901^/ 
^- A*G 
tion; 

ertia of section ; 
e from centre of section, 



Torsion. General Theory. 



579 



Twist of Shafts. — In Chapter X., we showed that 
when an element was sheared, the 
amount of slide x bore the follow- 
ing relation : — 



I G 



(i.) 



where f, is the shear stress on the 

material ; 

G is the coefficient of rigidity. 

In the case of a shaft, the x 

is measured on the curved surface. 

It will be more convenient if we 

express it in terms of the angle of 

twist. 



If Qr be expressed in radians, then — 




Fig. 573- 



e,D 



and Qr — 



2fl 
GD 



If 6 be expressed in degrees — 

irD9 

X = —7- 
360 

Substituting the value of x in equation (i.), we have— 

360/ G irGD 

But M, =/.Z, =f:^, and/. = ^ 
hence 6 - 3^0 X 16 X M/ _ 584M./ 

for solid circular shafts. Substituting the value of Z, for a 
hollow shaft in the above, we get — 



e-. 



584M/ 
G(D^ - D,*) 



for hollow circular shafts. 



N.B. — The stiffness of a hollow shaft is the difference of the stifihess 
of two solid shafts whose diameters are respectively the outer and inner 
diameters of the hollow shaft. 

When it is desired to keep the twist or spring of shafts 
within narrow limits, the stress has to be correspondingly 



580 Mechanics applied to Engineering. 

reduced. Long shafts are frequently made very much stronger 
than they need be in order to reduce the spring. A common 
limit to the amount of spring is 1° in 20 diameters; the stress 
corresponding to this is arrived at thus — 

We have above 6 = ^-dfr 

But when = 1°, / = 20D 

a-GD G_ 

then/. - ^g^ ^ ^^p _ ^^^^ 

For steel, G = 13,000,000; /, = 5670 lbs. per sq, inch 
Wrought iron, G = 11,000,000;^ = 4800 „ „ 

Cast iron, G = 6,000,000;/, = 2620 „ „ 

In the case of short shafts, in which the spring is of no 
importance, the following stresses may be allowed : — 

Steel, ^ = 10,000 lbs. per sq. inch 
Wrought iron,/, = 8000 „ „ 

Cast iron, yi = 3000 „ „ 

Horse-power transmitted by Shafts. — Let a mean 
force of P lbs. act at a distance r inches from the centre of 
a shaft ; then — 
The twisting mo- 1 

ment on the shaft }■ = P (lbs.) X r (inches) 

in Ibs.-inches J 
The work done per ) t, /ii, \ ,, /■ u \ . 

revolution in foot- = P Obs.) X r (mches) X 2^ 

lbs. ) " 

The work done perl _ P (lbs.) x r (inches) x 27rN (revs.) 
minute in foot-lbs. | ^ 

where N = number of revolutions per minute. 

27rPrN 

The horse-power transmitted = = H P 

'^ 12 X 33000 •"' 

then — 

_ 12 X 33000 X H.P. /D» 

^'■~ 2irN ~ 5-1 

_3 _ 12 X 33000 X H.P. X 5-1 _ 321400 H .P. 
2tN/. - N^: 

64-3 H.P . 
N 
taking/, at Sooo lbs. per square inch. 



D = 



Torsion. General Theory. ^^i 

4-^/ — :^ (nearly) for 5000 lbs, per sq. inch 

VHPT 
~ 3"S\/ ~N~ ^'^^ 7500 lbs. per sq. inch 

= 3\/ "1^' ^°^ i2)Ooo lbs. per sq. inch 



In the case of crank shafts the maximum effort is often 
much greater than the mean, hence in arriving at the diameter 
ctf the shaft the maximum twisting moment should be taken 
rather than the mean, and where there is bending as well as 
twisting, it must be allowed for as shown in the next paragraph. 

Combined Torsion and Bending. — In Fig. 574 a shaft 
is shown subjected to torsion only. We have previously seen 




Fig 574. 



(Chapter X.) that in such a case there is a tension acting 




■^y^ Tension ont^ 

Fig. S7S- 

normal to a diagonal drawn at an angle of 45° with the axis of 
the shaft, as shown by the arrows in the figure. In Fig. 575 
a shaft is shown subjected to tension only. In this case the 
tension acts normally to a face at 90° with the axis. In 
Fig. 576 a shaft is shown subjected to both torsion and 



582 



Mechanics applied to Engineering. 



tension ; the face on which the normal tensile stress is a 
maximum will therefore lie between the two faces mentioned 
above, and the intensity of the stress on this face will be 




Fig. 576. 



greater than that on either of the other faces, when subjected 
to torsion or tension only. 

We have shown in Chapter X. that the stress /„ normal to 
the face gh due to combined tension and shear is — 



/. 



If the tension be produced by bending, we have — 

7,-2 
If the shear be produced by twisting— 

^' Z, 2Z 
Substituting these values in the above equation — 






/«Z = M, = 



M+ <JW+Mi 



alsoA X 2Z = /.,Z, = M„ = M + Vm» + M," 



Torsion,. General Theory. 



583 



The M„ is termed the equivalent bending moment, the 
Mjj the equivalent twisting 
moment, that would produce 
the same intensity of stress 
in the material as the com- 
bined bending and twisting. 

The construction shown 
in Fig. 578 is a convenient 
graphical method of finding 

In Chapter X. we also 
showed that — 




Fig. 578. 



f,gi--f^hw[\Q and 
L^ = ^ cos = sin;e 

/»=^Hll=tan0,or 
/« cos d 

—^ = tan 
M« 

From this expression we can find the angle of greatest 
normal tensile stress 6, and therefore the angle at which 
fracture will probably occur, in the case of materials which 
are weaker in tension than in shear, such as cast iron and 
other brittle materials. 

In Fig. 579 we show the fractures of two cast-iron torsion 
test-pieces, the one broken by pure torsion, the other by 
combined torsion and bending. Around each a spiral piece of 
paper, cut to the theoretical angle, has been wrapped in order 
to show how the angle of fracture agreed with the theoretical 
angle 6 ; the agreement is remarkably close. 

The following results of tests made in the author's laboratory 
show the results that are obtained when cast-iron bars are 
tested in combined torsion and bending as compared with pure 
torsion and pure bending tests. The reason why the shear 
stress calculated from the combined tests is greater than when 
obtained from pure torsion or shear, is due to the fact that 
neither of the formulae ought to be used for stresses up to 
rupture ; however, the results are interesting as a comparison. 
The angles of fracture agree well with the calculated values. 



584 



Mechanics applied to Engineering. 



Twisting 


Bending 


Equivalent 


Modulus 


Angle of fracture. 


moment 


moment 
M 


twisting 
moment 


of rupture 
/if tons per 






Mt 






Pounds- 


inches. 


Md 


sq. inch 


Actual. 


Calculated. 


Zero 


2300 


4600 


25-5 


0° 


0° 


777 


1925 


4000 


267 


12° 


11° 


nyo 


2240 


475° 


27-1 


14° 


14° 


1228 


22SS 


4820 


231 


17° 


15" 


1308 


2128 


4628 


24-0 


19° 


i6» 


2606 


137s 


4320 


20-8 


,,0 


31° 


2644 


766 


3520 


i6-2 


38° 


37° 


3084 


Zero 


3084 


i6'o 


43° 


45° Mean of a 


Pure shear ... 




13-0 


0° 


0° large number 


„ tension... 


... 


11-5 


0° 


0° of tests. 



L 





Pure torsion. 



Fig. 579. 



ComHued toision 
and bending. 



In the case of materials which are distinctly weaker in 
shear than in tension it is more important to determine the 
maximum shear stress than the maximum tensile stress, because 
a shaft subjected to combined bending and torsion will fail in 
shear rather than in tension. 



Torsion. General Theory. 5^5 

In Chapter X. it is shown that the maximum shear 
stress — 



/. max. — \/ — + f^ 
4 

Hence the equivalent twisting moment which would produce 
the same intensity of shear stress as the combined bending 
and twisting is — 

and the angle at which fracture occurs is at 45° to the 
face^^ Fig. 577. 

' In the case of ductile materials, in their normal state, the 
angle of fracture, as found by experiment, undoubtedly does 
approximately agree with this theory, but in the case of crank 
shafts broken by repeated stress the fracture more often is in 
accordance with the maximum tension theory. 

The maximum tension theory is generally known as the 
" Rankine Theory," and the maximum shear theory as the 
"Guest Theory," named after the respective originators of 
the two hypotheses. 

Example. — A crank shaft is subjected to a maximum bend- 
ing moment of 300 inch-tons, and a maximum twisting moment 
of 450 inch-tons. The safe intensity of tensile stress for the 
material is 5 tons per sq. inch, and for shear 3 tons per sq. 
inch. Find the diameter of the shaft by the Rankine and the 
Guest methods. 

The equivalent bending moment (Rankine) — 

2 2 ' 

= 420 inch-tons. 

D' 420 _ . , 
= ~ — D = Q'l; mches. 

10-2 5 ^ = 

The equivalent twisting moment (Guest) — 

= Vsoo^ -f 450^ = 540 inch-tons. 

— = ^^-^ D = 07 mches. 

5-13 
Helical Springs. — The wire in a helical spring is, to all 
intents and purposes, subject to pure torsion, hence we can 
readily determine the amount such a spring will stretch or 
compress under a given load, and the load it will safely carry. 



586 



Mechanics applied to Engineering. 



We may regard a helical spring as a long thin shaft coiled 
into a helix, hence we may represent our helical spring thus — 




'xS: 



Fig. s8o. 

In the figure to the left we have the wire of the helical spring 

straightened out into a shaft, and provided with a grooved 

pulley of diameter D, i.e. the mean diameter of the coils in the 

. . WD 
spring; hence the twistmg moment upon it is . That the 

twisting moment on the wire when coiled into a helix is also 

WD 

will be clear from the bottom right-hand figure. The 

length of wire in the spring (not including the ends and hook) 
is equal to /. Let n = the number of coils ; then / = irDn 
nearly, or more accurately / = t/{TrT>nY + L*, Fig. 581, a 
refinement which is quite unnecessary for springs as ordinarily 
made. 

When the load W is applied, the end of the shaft twists, 

so that a point on the surface 
moves through a distance*, and 
a point on the rim of the pulley 
moves through a distance 8, 




where 



-,and8 = ^. 

X f 

But we have -= ^ 

hence S = -pJ^ 



"I 



(i) 



Torsion. General Theory. 587 

AlsoM=/.z,=/i!l^ 
2 16 

_ 16WD _ 2-55WD ,. . 

then 8 = ^SP'^ (from i. and ii.) 

Substituting the value of / = mrD — 

g^ 2-55D'W^,rD _■ SD'Wn 
Gd* Gd* 

D'W« 
G for steel = 12,000,000 8 = 



1,500,000^* 

DHV« 
G for hard brass = 5,000,000 8 = 



Safe Load. — From equation (ii.), W 



625,0001^ 
2-S5D 



Experiments by Mr. Wilson Hartnell show that for steel 
wire the following stresses are permissible : — 

Diameter of wire. Safe stress {/si- 

\ inch . . , 70,000 lbs. per square inch 

I „ ... 60,000 lbs. „ „ 

J „ ... 50,000 lbs. „ „ 

When a spring is required to stretch M times its initial 
length L, let the initial pitch of the coils be/, then L =«/ 

From (i.) 8 = ^ 

8 = L(M - i) = «/(M - 1) = ^~ 

Gd 

pd{M. — 1) _ t/, _ 3-14 X 70,000 _ 1^ 
D^ G 11,000,000 50 

and so(M - 1) = — 

For a close coiled spring, p= d, then — 

Vso(M-i)=? 



588 Mechanics applied to Engineering. 

Work stored in Springs. 

The work done in stretching 1 _ W8 
or compressing a spring \ ~ 2 

f^ X V)lf. ,, . ..... 

= 2 X 2-55D X Qd ^^'""^ "• ^*i ">•) 



Jl^nQ 



(substituting the value of /) = , ^ (inch-lbs.) 



19,500,000 
putting G = 12,000,000 

Weight of Spring. — Taking the weight of i cub. inch of 
steel = o"28 lb., then — 

The weight of the spring w = o"j8^d^I X o'28 = o'22d^/ 
Substituting the value of /, we have — 

w = Q'Sgnd'D 

Height a Steel Spring will lift itself (A). 

work stored in spring 
~ weight of spring 

, /.'^«D /^. , 

'* ~ r62G X o-6gnd^D ' ri2G '"*^^^ 

= -^ — = — feet 

i3'4G 161,000,000 

The value of // is given in the following table corresponding 
to various values of/": — 

f, (lbs. per square inch) ... 30,000 60,000 90,000 120,000 150,000 
A (feet) 5-56 22-4 50-3 89-5 139-8 

h also gives the number of foot-pounds of energy stored 
per pound of spring. 

All the quantities given above are for springs made of wire 
of circular section ; for wire of square section of side S, and 
taking the same value for G as before, we get — 



Torsion. General Theory. 589 

,i35,°°°y ^''""^ 'P""S') 



o = 

2 



890.0008^ (brass springs) 

n4s ) square section' 

W = ^9°^°°^-S brass 
Safe Load for Springs -of Square Section. 

W = ^' — stress 70,000 lbs. per square inch 

24,9608' 
= — ^-^ — » 60,000 lbs. „ „ 

20,8ooS' ,, 

= — i^ „ 50,000 lbs. „ „ 

Taking a mean value, we have — 

W = J. — (square section) 

work stored = ' „ (inch-lbs.) steel 

24,680,000 ^ ' 

weight = o-88«S''D (steel) 

Height a square-section spring) , fl_ ,, . 

will lift itself (steel) ] "■ - 260,600,000 ^ > 

/^ (lbs. per square inch) ... 30,000 60,000 90,000 120,000 150,000 
-iCfeet) 3-45 138 31-1 55-3 86'3 

It will be observed that in no respect is a square-section 
spring so economical in material as a spring of circular 
section. 

Helical Spring in Torsion. — When a helical spring is 
twisted the wire is subjected to a bending moment due to the 
change of curvature of the spring, which is proportional to the 
twisting moment. 



5 go Mechanics applied to Engineering, 

Let /3j and p^ = the mean radii of the spring in inches before 
and after twisting respectively ; 
«, and «2 = the number of free coils before and after 

twisting respectively ; 
^1 and Qi = the angles subtended by the wire in inches 
before and after twisting respectively ; 
= 36o«i and 360/2, respectively ; 
6 = ^1 — ^3, or the angle twisted through by the 
free end of the spring in degrees ; 
M, = the twisting moment in pounds-inches ; 
I = the moment of inertia of the wire section 

about the neutral axis in inch units ; 
d = the diameter or side of the wire in inches ; 
L = the length of frefe wire in the spring in 
inches ; 

E = M,L _ 36oM,L 

2irl(«i — «j) 2irl8 

E = 734opi«iM, j.^j. ^jj.^ ^f circular section 

trO 
E = 4320Pi«iM, f ^^jj.g pf 5 g section 

d*e 

If 6^ be the angle of twist expressed in radians, we have — 

Open-coiled Helical Spring. — In the treatment given 
above for helical springs, we took the case in which the coils 
were close, and assumed that the wire was subjected to torsion 
only ; but if the coils be open, and the angle of the helix be 
considerable, this is no longer an admissible assumption. 
Instead of there being a simple twisting moment WR acting 
on the wire, we have a twisting moment M, = WR cos a, which 
twists the wire about an axis ad. Think of ai as a little shaft 
attached to the spring wire at a, and ei as the side view of a 
circular disc attached to it, then, by twisting this disc, the wire 
will be subjected to a torsional stress. In addition to this, let al> 
represent the side view of half an annular disc, suitably attached 
to the wire at a, and which rotates about an axis ai. Then, by 



Torsion. General Theory. 



591 



twisting this disc, the spring wire can be bent ; thereby its radius 
of curvature will be altered in much the same manner as that 
described in the article on the " Helical Spring in Torsion," 
the bending moment M = WR sin a. The force which pro- 
duces the twisting moment acts in the plane of the disc «, and 




Fig. 582. 



that which produces the bending moment in the plane ab, 
i.e. normal to the respective sides of the triangle of moments 
obi. 

In our expression for the twist of a shaft on p. 579, we 

gave the angle of twist 6c in degrees; but if we take it in 

circular measure, we get * = rd„ where r is the radius of the 
vyire, and — 

I -G 

^_2l_ M, M, 
I- rG- rGZ„ " GI„ 



and 6. = 



AVR cos g 
GL 



Likewise due to bending we have (see p. 590) — 

a, /M /WR sin g 
° ~ EI ~ EI 

We must now find how these straining actions affect the 
axial deflection of the spring. 



592 



Mechanics applied to Engineering. 



The twisting moment about ab produces a strain be = RS^, 
which may be resolved into two components, viz. one, ef = R^, 
sin a, which alters the radius of curvature of the coils, and 
which we are not at present concerned with ; and the other, 
bf = R5„ cos a, which alters the axial extension of the spring 

/WR^cos^a 
by an amount pj 

The bending moment about cd in the plane of the imaginary 

disc ab produces a strain bh = RdJ, of which Ag alters the 

radius of curvature in a horizontal plane, normal to the axis 

of the spring ; and bg- = R6,' sin a alters the axial extension of 

the spring in the same direction as that due to twisting, by an 

/WR'' sin2 o 
amount - 



EI 
whence the total axial ") 



= /WR2 



extension 8 j < w jx. ^ qj^ 

On substitution and reduction, we get — 



cos^ a sm 



EI 



'-) 



8 = 



8«D'W 
Gd* cos a 



cos' o + 



I"2S 



and for the case of springs in which the bending action is 
neglected, we get — 

s wpnv 

~ Gd^ cos a 



Angle. 


Ratio of 


deflection 


allowing for bending and torsion 


deflection allowing for torsion only 


K 






0-998 


IO° 






0-992 


IS" 






0-986 


20° 






0-978 


30° 






0-950 


45° 






0-900 



Thus for helix angles up to 15° there is no serious error 
due to the bending of the coils, and when one remembers how 
many other uncertain factors there are in connection with 
helical springs, such as finding the exact diameter of the wire 
and coils, the number of free coils, the variation in the value 
of G, it will be apparent that such a refinement as allowing for 
the bending of the wire is rarely, if ever, necessary. 



CHAPTER XVII. 

STRUCTURES. 

Wind Pressures. — Nearly all structures at times are exposed 
to wind pressure. In many instances, the pressure of the wind 
is the greatest force a structure ever has to withstand. 
Let V = velocity of the wind in feet per second ; 
V = velocity of the wind in miles per hour ; 
M = mass of air delivered per square foot per second ; 
W = weight of air delivered per square foot per second 

(pounds) ; 
w = weight of 1 cubic foot of air (say o'o8o7 lb.) ; 
P = pressure of wind per square foot of surface 
exposed (poimds). 
Then, when a stream of air of finite cross-section impinges 
normally on a flat surface, whose area is much greater than 
that of the stream, the change of momentum per second per 
square foot of air stream is — 

Mz'= P 

or = P 

g 
But W = wo 

hence — = P 
g 

or expressed in miles per hour by substituting v - i'466V, 
and putting in the value of w, we have — 

p _ o-o8o7 X 1-466'' X V . ^,, 

r = ! = o'ooi;4V* 

32-2 ^^ 

If, however, the section of the air stream is much greatei 
than the area of the flat surface on which it impinges, the 
change of direction of the air stream is not complete, and 
consequently the change of momentum is considerably less 
than (approximately one-half) the value just obtained. Smeaton, 
from experiments by Rouse, obtained the coefficient 0*005, tut 

2 Q 



594 



Mechanics applied to Engineering. 



later experimenters have shown that such a value is probably 
too high. Martin gives 0*004, Kernot o'oo33, Dines 0*0029, 
and the most recent experiments by Stanton (see Proceedings 
I.C.E., vol. clvi.) give 0*0027 for t^^ maximum pressure in the 
middle of the surface on the windward side. In all cases of 
wind pressure the resultant pressure on the surface is composed 
of the positive pressure on the windward side, and a suction 
or negative pressure on the leeward side. Stanton found in the 
case of circular plates that the ratio of the maximum pressure 
on the windward side to the negative' pressure on the leeward 
side was 2*1 to 1 in the case of circular plates, and a mean of 
I "5 to I for various rectangular plates. Hence, for the re- 
sultant pressure on plates, Stanton's experiments give as an 
average P = o*oo36V". 

Recent experiments by Eiffel, Hagen, and others show that 
the total pressure on a surface depends partly on the area of 
the surface exposed to the wind, and partly on the periphery 
of the surface. The total pressure P is fairly well represented 
by an expression of the form — 

P = (a + ^^)SV* 

where S = normal surface exposed to the wind in square feet ; 

/ = periphery of the surface in feet. 
a and b are constants. 

When a wind blowing horizontally impinges on a flat, 
inclined surface, the pressure in horizontal, vertical, and normal 
directions may be arrived at thus — 

the normal pressure P„ = P . sin fl 
the horizontal pressure P, = P„ sin 

= P sin^ 6 
the vertical pressure P, = P„ cos 6 

= P . cos d . sin 

In the above we have neg- 
lected the friction of the air 
moving over the inclined sur- 
face, which will largely ac- 
count for the discrepancy 
between the calculated pres- 
sure and that found by expe- 
riment. The following table 
Fig. 583. will enable a comparison to 

be made. The experimental 
values have been reduced to a horizontal pressure of 40 lbs. per 




Structures. 



595 



square foot of vertical section of air stream acting on a flat 
vertical surface. 





Normal pressure. 


Vertical pressure. 


Horizontal pressure. 


roof. 


Experi- 
ment.' 


P sin 9. 


Experi- 
ment. 


P COS e sin e. 


Experi- 
ment. 


P sin" 9. 


IO° 
20° 

3°: 

50° 
60° 
70° 


97 
i8-i 
26'4 
33-3 
38-1 
40-0 
41-0 


7-0 

137 

20'0 

257 
30-6 
34-6 
37-6 


9-6 
ii7-o 

22-8 
25-5 
24-5 
20'0 

i4"o 


6-9 
12-9 

i7'3 
197 
197 

17-3 
129 


17 

6-2 

13-2 

21-4 
29-2 
34 -o 
38-S 


1-7 

47 
100 
16-5 

23-5 
30-0 

35-4 



When the wind blows upon a surface other than plane, the 
pressure on the projected area depends upon the form of the 
surface. The following table gives some idea of the relative 
wind- resistance of various surfaces, as found by various 
experiments : — 



Flat plate 

Parachute (concave surface), depth 



diameter 



Sphere ... 

Elongated projectile 

Cylinder 

Wedge (base to wind) 

,, (edge to wind), vertex angle 90° 

Cone (base to wind) 

,, (apex to wind), vertex angle 90° 
» . ,1 „ 60° 

Lattice girders 



o'36-0'4i 

0-5 

0-S4-0-S7 

0'8-o'97 

0'6-07 

0-9S 

054 
about o°8 



The pressure and velocity of the wind increase very much 
as the height above the ground increases (Stevenson's experi- 
ments). 

Feet above ground 

Velocity in miles per hour .. 



s 


1 


IS 


25 


52 


4 


6 


65 


7-5 


7 


t? 


18 


21 


23 


«3 


23 


25 


30 


32 


19 


28 


31 


35 


40 


Z6, 


32 


34 


37 


43 



' Deduced from Hutton's experiments by Unwin (see " Iron Roofs and 
Bridges "). 



S96 



Mechanics applied to Engineering, 



These figures appear to show that the pressure varies 
roughly as the square root of the height above the ground. 

The wind pressure as measured by small gauges is always 
higher than that found from gauges offering a large surface 
to the wind, probably because the highest pressures are only 
confined to very small areas, and are much greater than the 
mean taken on a larger surface. 

Forth Bridge Experiments. 



Date. 


Small 

revolving 

gauge. 


Small 
fixed 
gauge. 


Large fixed gauge, 15' X zo'. 


Mean. 


Centre. 


Corner. 


Mar. 31, 1886 
Jan. 25, 1890 


26 

27 


31 

24 


>9 
18 


28J 
23* 


22 

22 



In designing structures, it is usual to allow for a pressure 
of 40 lbs. per square foot. In very exposed positions this may 
not be excessive, but for inland structures, unless exceptionally 
exposed, 40 lbs. is unquestionably far too high an estimate. 

In sheltered positions in towns and cities the wind pressure 
rarely exceeds 10 lbs. per square foot. 

For further information on this question the reader should 
refer to special works on the subject, such as Walmisley's 
" Iron Roofs " and Charnock's " Graphic Statics " (Broadbent 
and Co., Huddersfield), Chatley's " The Force of the Wind " 
(Griffin), Husband and Harby's "Structural Engineering" 
(Longmans). 

Weight of Roof Coverings. — For preliminary estimates, 
the weights of various coverings may be taken as — 



Covering. 

Slates 

Tiles (flat) 
Corrugated iron 
Asphalted felt 
Lead ... 
Copper 
Snow 



Weight per sq. foot 
in pounds. 

8-9 
12-20 

ii-3i 
2-4 
S-8 

i-ii 

S 



Weight of Roof Structures. — For preliminary estimates, 
the following formulas will give a fair idea of the probable 
weight of the ironwork in a roof. 



Structures. 



597 



Let W = weight of ironwork per square foot of covered area 
{i.e. floor area) in pounds ; 

D = distance apart of principals in feet ; 

S = span of roof in feet. 
Then for trusses — 



w=e^3 



8 



and for arched roofs — 



W = ^- + 3 

Distribution of Load on a Roof. — It will often save 
trouble and errors if a sketch be made of the load distribution 
on a roof in this manner. 



wrniD 




Fig. 584. 



The height of the diagram shaded normal to the roof is 
the weight of the covering and ironwork (assumed uniformly 
distributed). The height of the diagonally shaded diagram 
represents the wind pressure on the one side. The lowest 
section of the diagram on each side is left unshaded, to indicate 
that if both ends of the structure are rigidly fixed to the 
supporting walls, that portion of the load may be neglected 
as far as the structure is concerned. But if A be on rollers, 
and B be fixed, then the wind-load only on the A side must be 
taken into account. 

When the slope of the roof varies, as in curved roofs, the 
height of the wind diagram must be altered accordingly. An 
instance of this will be given shortly. 

The whole of the covering and wind-load must be con- 
centrated at the joints, otherwise bending stresses will be set up 
in the bars. 

Method of Sections. — Sometimes it is convenient to 
check the force acting on a bar by a method known as the 



598 Mechanics applied to Engineering. 

method of sections — usually attributed to Ritter, but really 
due to Rankine — termed the method of sections, because the 
structure is supposed to be cut in two, and the forces required 
to keep it in equilibrium are calculated by taking moments. 

Suppose it be re- 
*" quired to find the force 

/ I. jc^. i acting along the bar/^. 

^^--;;/V<J^ . ... .^ i Take a section through 

X^^)/ / X'^J ^ \'"^i-': the Structure a^ ; then 

^-(^\ / p \/\N? three forces,/a, qe, pq, 

[ ^"'^^\/ — aZ — L v X/^^j y. must be applied to the 

I / \ -y^i *^"'- ^^"^ ^° ^^^P '^'^ 

J, \' Structure in equili- 

FiG. 585. brium. Take moments 

about the point O. 

The forces pa and qe pass through O, and therefore have no 

moment about it ; but pq has a moment pq X y about O. 

pqXy = Wi*i + Waaij + Wa«3 

- W,x, + W^, + W^, 
pq = 

By this method forces may often be arrived at which are 
difficult by other methods. 

Forces in Roof Structures. — We have already shown 
in Chapter IV. how to construct force or reciprocal diagrams 
for simple roof structures. Space will only allow of our now 
dealing with one or two cases in which difficulties may arise. 

In the truss shown (Fig. 586), a difficulty arises after the 
force in the bar tu has been found. Some writers, in order to 
get over the difficulty, assume that the force in the bar rs is 
the same as in ut. This may be the case when the structure is 
evenly loaded, but it certainly is not so when wind is acting on 
one side of the structure. We have taken the simplest case of 
loading possible, in order to show clearly the special method 
of dealing with such a case. 

The method of drawing the reciprocal diagram has already 
been described. We go ahead in the ordinary way till we 
reach the bar st (Fig. 586). In the place of sr and rq substitute 
a temporary bar xy, shown dotted in the side figure. With this 
alteration we can now get the force ey or eq; then qr, rs, etc., 
follow quite readily ; also the other half of the structure. 

There are other methods of solving this problem, but the 
one given is believed to be the best and simplest. The author 



Structures. 



599 



is indebted to Professor Barr, of Glasgow University, for this 
method. 

When the wind acts on a structure, having one side fixed and 
the other on rollers, the only difficulty is in finding the reactions. 
The method of doing this by means of a funicular polygon is 
shown in Fig. 587. .The funicular polygon has been fully 
described in Chapters IV. and XII, hence no further description 
is necessary. The direction of the reaction at the fixed support 




Fig. 586. 



is unknown, but as it must pass through the point where the roof 
is fixed, the funicular polygon should be started from this point. 
The direction of the reaction at the roller end is vertical, 
hence from/ in the vector polygon a perpendicular is dropped 
to meet the ray drawn parallel to the closing line of the 
funicular polygon. This gives us the point a ; then, joining ba, 
we get the direction of the fixed reaction. The reciprocal 



6oo 



Mechanics applied to Engineering. 



diagram is also constructed ; it presents no difficulties beyond 
that mentioned in thfe last paragraph. 

In the figure, the vertical forces represent the dead weight 
on the structure, and the inclined forces the wind. The two 
are combined by the parallelogram of forces. 

In designing a structure, a reciprocal diagram must be 
drawn for the structure, both when the wind is on the roller 




Vector polygon 

for finding 
the reactions. 



Fig. 587. 

and on the fixed side of the structure, and each member of 
the structure must be designed for the greatest load. 

The nature of the forces, whether compressive or tensional, 
must be obtained by the method described in Chapter IV. 

Island Station Boof. — This roof presents one or two 
interesting problems, especially the stresses in the main post, 
The determination of the resultant of the wind and dead load 
at each joint is a simple matter. The resultant of all the forces 
is given \yjpa on the vector polygon in magnitude and direction 
(Fig. 588). Its position on the structure must then be deter- 
mined. This has been done by constructing a funicular polygon 



Structures. 



60 1 



in the usual way, and producing the first and last links to meet 
in the point Q. Through Q a line is drawn parallel to pa in 




Vector 

polygon for 

finding the 

reaction. 



Fig. 588. 



the vector polygon. This resultant cuts the post in r, and may 
be resolved into its horizontal and vertical components, the 
horizontal component producing bending 
moments of different sign, thus giving the 
post a double curvature (Fig. 589). 

The bending moment on the post is 
obtained by the product j>a X Z, where Z 
is the perpendicular distance of Q from 
the apex of the post. 

When using reciprocal diagrams for 
determining the stresses in structures, we 
can only deal with direct tensions and 
compressions. But in the present instance, 
where there is bending in one of the members, we must 
intiroduce an imaginary external force to prevent this bending 
action. It will be convenient to assume that the structure is 
pivoted at the virtual joint r, and that an external horizontal 
force F is introduced at the apex Y to keep the structure in 
equilibrium. The value of F is readily found thus. Taking 
moments about r, we have — 




Fig. 589. 



6o2 Mechanics applied to Engineering. 

F X >-Y = ^ X 3 

z is the perpendicular distance from the point Y to the resultant. 

On drawing the reciprocal diagram, neglecting F, it will be 
found that it will not close. This force is shown dotted on the 
reciprocal diagram / /, and on measurement will be found to be 
equal to F. 

Dead and Live Loads on Bridges. — The dead loads 
consist of the weight of the main and cross girders, floor, ballast, 
etc., and, if a railway bridge, the permanent way ; and the live 
loads consist of the train or other traffic passing over the 
bridge, and the wind pressure. 

The determination of the amount of the dead loads and 
the resulting bending moment is generally quite a simple 




Fig. sgo. 

matter. In order to simplify matters, it is usual to assume 
(in small bridges) that the dead load is evenly distributed, and 
consequently that the bending-moment diagram is parabolic. 

In arriving at the bending moment on railway bridges, an 
equivalent evenly distributed load is often taken to represent 
the actual but somewhat unevenly distributed load due to a 
passing train. The maximum bending moment produced by a 
train which covers a bridge (treated as a standing load) can be 
arrived at thus (Fig. 590). Take a span greater than twice 
the actual span, so as to get every possible combination of loads 
that may come on the structure. Construct a bending-moment 
diagram in the ordinary way, then find by trial where the greatest 
bending moment occurs, by fitting in a line whose length is 
equal to the span. A parabola may then be drawn to enclose 



Structures. 



603 



this diagram as shown in the lower figure ; then, \i d = depth 

wP 
of this parabola to proper scale, we have -— = d, where a 

is the equivalent evenly distributed load due to the train. 
(The small diagram is not to scale in this case.) 

Let W, = total rolling load in tons distributed on each pair 
of rails ; 
S = span in feet. 

Then for English railways W, = i-6S + 20 

Maximum Shear due to a Rolling Load, Concentrated 
Rolling Load. — The shear 
at any section is equal to 
the algebraic sum of the 
forces acting to the right or 
left of any section, hence 
the shear between the load 
and either abutment is ^"'- s?'- 

equal to the reaction at the abutment. We have above — 

Ri/= Wa;, 




R, = W^ 



likewise Rj 



W^ 



When the load reaches the abutment, the shear becomes W, 

W 
and when in the middle of the span the shear is — . The 

shear diagram is shown shaded vertically. 

Uniform Rolling Load, whose Length exceeds the Span. — Let 



(VrrrrrCrC^ 




w = the uniform live load per foot run, and zv^ = the uniform 
dead load per foot run. 

Let M = ^ 



6o4 



Mechanics applied to Engineering. 



The total load on the structure = ■wn = Vf 
then Ri/= — = — 

2 2 

Ri = — «2 = Kw" 
2/ 

where K is a constant for any given case. But as the train 
passes from the right to the left abutment, the shear between 
the head of the train and the left abutment is equal to the left 
reaction Rj, but this varies as the square of the covered 
length of the bridge, hence the curve of shear is a parabola. 
When the train reaches the abutment, I = n; 

then Ri = — = — 
2 2 

The curves of shear for both dead and live loads are shown 
in Fig. 593. When a train passes from right to left over the 

point /, the shear is re- 
versed in sign, because 
the one shear is positive, 
and the other negative. 
The distance x between 
the two points /, / is 
known as the " focal 
distance " of the bridge. 
The focal distance x 
can be calculated thiis : The point / occurs where the shear 
due to the live load is equal to that due to the dead load. 




Fig. 



593- 



wn 
The shear due to the live load = — j 



dead ,, = -^ = w^ 
2 



m^ (I \ 



Solving, we get — 



n = l^U + M^ - /M 
and X = I — 2n 

= /(i - 2 VmTIP + 2M) 



Structures. 



605 



Determination of the Forces acting on the 
Members of a Girder.— In the case of the girder given 




\W'P^''''''"4'^ Changes 




Fig. 594. 

as an example, the dead load w^ = 075 ton per foot, or 
?-^^ ^^ = 5 tons per joint. 

The live load w = 175 ton per foot run, or 175 X 13*5 
= 23-6 tons on each bottom joint, thus giving 5 tons on 
each top joint, and 28'6 tons on each bottom joint. 

The forces acting on the booms can be obtained either by 



6o6 Mechanics applied to Engineering. 

constructing a reciprocal diagram for the structure when fully 
loaded as shown, or by constructing the bending-moment 
diagram for the same conditions. The depth of this diagram 
at any section measured on the moment scale, divided by the 
depth of the structure, gives the force acting on the boom at 
that section. The results should agree if the diagrams are 
carefully constructed. 

The forces in the bracing bars, however, cannot be obtained 
by these methods, unless separate reciprocal diagrams are con- 
structed for several (in this instance six) positions of the train, 
since the force in each bar varies with each position of the 
train. The nature of the stress on the bracing bars within 
the focal length changes as the train passes ; hence, instead of 
designing the focal members to withstand the reversal of stress, 
it is usual, for economic reasons, to counterbrace these panels 
with two tie-bars, and to assume that at any given instant only 
one of the bars is subjected to stress, viz. that bar which at the 
instant is in tension, since the other tie-bar is not of a suitable 
section to resist compression. 

All questions relating to moving loads on structures are 
much more readily solved by " Influence Lines " than by the 
tedious method of constructing reciprocal diagrams for each 
position of the moving load. By this means the greatest 
stresses which occur in the various members of the structure 
can be determined in a fraction of the time required by the 
older methods. 

Deflection of Braced Structures. — The deflection 
produced by any system of loading can be calculated either 
algebraically by equating the work done by the external forces 
to the internal elastic work done on the various members of 
the structure, or graphically by means of the Williott diagram. 

Readers should refer to Warren's " Engineering Construc- 
tion in Iron, Steel, and Timber," Fidler's " Practical Treatise 
on Bridge Construction," Husband and Harby's " Structural 
Engineering," Burr and Falk's " Influence Lines for Bridges 
and Roofs." 

Girder with a Double System of Triangulation. — 
Most girders with double triangulation are statically indeter- 
minate, and have to be treated by special methods. They can, 
however, generally be treated by reciprocal diagrams without 
any material error (Fig. 595). We will take one simple case 
to illustrate two methods of treatment. In the first each system 
is treated separately, and where the members overlap, the 
forces must be added : in the second the whole diagram will 



Structures. 



607 



be constructed in one operation. The same result will, of 
course, be obtained by both methods. 





Fig. 595. 



/ V^ 



6o8 Mechanics applied to Engineering. 

In dealing with the second method, the forces mn, hg acting 
on the two end verticals are simply the reactions of Fig. A. 
There is less liability to error if they are treated as two upward 
forces, as shown in Fig. C, than if they are left in as two vertical 
bars. It will be seen, from the reciprocal diagram, that the 
force in qs is the same as that in rt, which, of course, must be 
the case, as they are one and the same bar. 

Incomplete and Redundant Framed Structures. — 
If a jointed structure have not sufficient bars to make it retain 
its original shape under all conditions of loading, it is termed 
an "incomplete" structure. Such a structure may, however, 
be used in practice for one special distribution of loading 
which never varies, but if the distribution should ever be 
altered, the structure will change its shape. The determination 
of the forces acting on the various members can be found by 
the reciprocal diagram. 

But if a structure have more than sufficient bars to make it 
retain its original shape, it is termed a " redundant " structure. 
Then the stress on the bars depends entirely upon their relative 
yielding when loaded, and cannot be obtained from a reciprocal 
diagram. Such structures are termed " statically indeterminate 
structures." Even the most superficial treatment would occupy 
far too much space. If the reader wishes to follow up the 
subject, he cannot do better than consult an excellent little 
book on the subject, " Statically Indeterminate Structures," by 
Martin, published at Engineering Office. 

Pin Joints. — In all the above cases we have assumed 
that all the bars are jointed with frictionless pin joints, a 
condition which, of course, is never obtained in an actual 
structure. In American bridge practice pin joints are nearly 
always used, but in Europe the more rigid riveted joint finds 
favour. When a structure deflects under its load, its shape is 
slightly altered, and consequently bending stresses are set up 
in the bars when rigidly jointed. Generally speaking, such 
stresses are neglected by designers. 

Plate Girders. — It is always assumed that the flanges of 
a rectangular plate girder resist the whole of the bending 
stresses, and that the web resists the whole of the shear stresses. 
That such an assumption is not far from the truth is evident 
from the shear diagram given on p. 596. 

In the case of a parabolic plate girder, the flanges take some 
of the shear, the amount of which is easily determined. 

The determination of the bending moment by means of 
a diagram has already been fully explained. The bending 



Structures. 609 

moment at any point divided by the corresponding depth of 
the girder gives the total stress in the flanges, and this, divided 
by the intensity of the stress, gives the net area of one flange. 
In the rectangular girder the total flange stress will be greatest 
in the middle, and will diminish towards the abutments, 
consequently the section of the flanges should correspondingly 
diminish. This is usually accomplished by keeping the width 
of the flanges the same 
throughout, and reducing r ■ — > 



3_ 



T^ 



the thickness by reducing , 

the number of plates. The \ i i 

bending-moment diagram N^ 

lends itself very readily to 

the stepping of the plates. 

Thus suppose it were found Fig. 597. 

that four thicknesses of 

plate were required to resist the bending stresses in the flanges 

in the middle of the girder ; then, if the bending-moment 

diagram be divided into four strips of equal thickness, each 

strip will represent one plate. If these strips be projected up> 

on to the flange as shown, it gives the position where the plates 

may be stepped.^ 

The shear in the web may be conveniently obtained from 
the shear diagram (see Chapter XII.). 
Then if S = shear at any point in tons, 

f, = permissible shear stress, usually not exceeding 
3 tons per square inch of gross section of web, 
d^ = depth of web in inches, 
/ = thickness of plate in inches (rarely less than f 
inch), 

we have — 

The depth is usually decided upon when scheming the 
girder ; it is frequently made from -g- to -^ span. The thickness 
of web is then readily obtained. If on calculation the thickness 
comes out less than | inch, and it has been decided not to use 
a thinner web, the depth in some cases is decreased accordingly 
within reasonable limits. 



' It is usual to allow from 6 inches to 12 inches overlap of the plates 
beyond the points thus, obtained, 

2 R 



6io 



Mechanics applied to Engineering. 



The web is attached to the flanges or booms by means of 
angle irons arranged thus : 





Fig. 599. 



Fig. 598. 

The pitch / of the rivets must be such that the bearing 

and shearing stresses are 
within the prescribed 
Hmits. 

On p. 389 we showed 
that, in the case of any 
rectangular element subject 
to shear, the shear stress 
is equal in two directions 
at right angles, i.e. the 
shear stress along ef = 
shear stress along ed, 
which has to be taken by 
the rivets a, a. Fig. 598. 

The shep-r per (gross) inch run of web plate along ef 

The shearing resistance of each rivet is (in double shear) 

4 
Whence, to satisfy shearing conditions — 

This pitch is, however, often teo small to be convenient ; 
then two (zigzag) rows of rivets are used, and/ = twice the 
above value. 

The bearing resistance of each rivet is — 

dtf,=Up 



Structures. 



6ll 



The bearing pressure /, is usually taken at about 8 tons per 
square inch. We get, to satisfy bearing-pressure conditions — 

p = ^d (for a single row of rivets) (approximately) 
p = 6d (for a double row of rivets) „ 

The joint-bearing area of the two rivets b, b attaching the 
angles to the booms is about twice that of a single rivet (a) 
through the web ; hence, as far as bearing pressure is concerned, 
single rows are sufficient at b, b. A very common practice is 
to adopt a pitch of 4 inches, putting two rows in the web at a, a, 
and single rows at b, b. 

The pitch of the rivets in the vertical joints of the web 
(with double cover plates) is the same as in the angles. 

The shear diminishes from the abutments to the middle 
of the span, hence the thickness cff the web plates may be 
diminished accordingly. It often happens, however, that it is 
more convenient on the whole to keep the web plate of the 
same thickness throughout. The pitch / of the rivets may then 





-^^ 




V^"^ 


r^/ 


r^ 


\ \ '^ 


) // 


r 


w 


¥/ 




] 


d 


b 


/ 






( 


^ 







1^ ^ 


1 


1 




i~\ 


-^=H 


h , \\ ^ - , ; 


\ 


KJ 
















v_/ 


y^ ■ 


A 


















rs 





Fig. 6qo. 



be increased towards the middle. It should be remembered, 
however, that several changes in the pitch may in the end cost 
more in manufacture than keeping the pitch constant, and 
using more rivets. 

The rivets should always be arranged in such a manner 
that not more than two occur in any one section, in order to 
reduce the section of the angles as little as possible. 

Practical experience shows that if a deep plate girder be 
constructed with simply a web and two flanges, the girder will 
not possess sufficient lateral stiffness when loaded. In order to 
provide against failure from this cause, vertical tees, or angles, 
are riveted to the web and flanges, as shown in Fig. 600. That 



6i2 Meclianics applied to Engineering. 

such stiffeners are absolutely necessary in many cases none 
will deny, but up to the present no one appears to have arrived 
at a satisfactory theory as to the dimensions or pitch required. 
Rankine, considering that the web was liable to buckle 
diagonally, due to the compression component of the shear, 
treated a narrow diagonal strip of the web as a strut, and- 
proceeded to calculate the longest length permissible against 
buckling. Having arrived at this length, it becomes a simple 
matter to find the pitch of the stiffeners, but, unfortunately for 
this theory, there are a large number of plate girders that have 
been in constant use for many years which show no signs of 
weakness, although they ought to have buckled up under their 
ordinary working load if the Rankine theory were correct. 
The Rankine method of treatment is, however, so common 
that we must give our reasons for considering it to be wrong in 
principle. From the theory of shear, we know that a pure 
shear consists of two forces of equal magnitude and opposite 
in kind, acting at right angles to one another, each making an 
angle of 45° with the roadway ; hence, whenever one diagonal 
strip of a web is subjected to a compressive stress, the other 
diagonal is necessarily subjected to a tensile stress of equal 
intensity. Further, we know that if a long strut of length / be 
supported laterally (even very flimsily) in the middle, the 

effective length of that strut is thereby reduced to -, and in 

general the effective length of any strut is the length of its 
longest unstayed segment. But even designers who adhere to 
the Rankinian theory of plate webs act in accordance with this 
principle when designing latticed girders, in which they use 
thin flat bars for compression members, which are quite 
incapable of acting as long struts. But, as is well known, they 
do not buckle simply because the diagonal ties to which they 
are attached prevent lateral deflection, and the closer the 
lattice bracing the smaller is the liability to buckling; hence 
there is no tendency to buckle in the case in which the lattice 
bars become so numerous that they touch one another, or 
become one continuous plate, since the diagonal tension in 
the plate web effectually prevents buckling along the other 
diagonal, provided that the web is subjected to shear only. What, 
then, is the object of using stiffeners ? Much light has recently 
been thrown on this question by Mr. A. E. Guy (see "The 
Flexure of Beams : " Crosby Lockwood and Son), who has very 
thoroughly, both experimentally and analytically, treated the 
question of the twisting of deep narrow sections. It is well 



Structures. 613 

known that for a given amount of material the deeper and 
narrower we make a beam of rectangular section the stronger 
will it be if we can only prevent it from twisting sideways. 
Mr. Guy has investigated this point, and has made the most 
important discovery that the load at which such a beam will 
buckle sideways is that load which would buckle the same beam 
if it were placed vertically, and thereby converted into a strut. 

If readers will refer to the published accounts (" Menai and 
Conway Tubular Bridges," by Sir William Fairbairn) of the 
original experiments made on the large models of the Menai 
tubular bridge by Sir William Fairbairn, they will see that 
failure repeatedly occurred through the twisting of the girders ; 
and in the later experiments two diagonals were put in in order 
to prevent this side twisting, and finally in the bridge itself the 
ends of the girders were supported in such a way as to prevent 
this action, and in addition substantial gusset stays were riveted 
into the corners for the same purpose. Some tests by the 
author on a series of small plate girders of 15 feet span, showed 
in every case that failure occurred through their twisting. 

The primary function, then, of web stiffeners is believed to 
be that of giving torsional rigidity to the girder to prevent side 
twisting, but the author regrets that he does not see any way of 
calculating the pitch of plate or tee-stiffeners to secure the 
necessary -stiffness ; he trusts, however, that, having pointed out 
what he believes to be the true function of stiffeners, others may 
be persuaded to pursue the question further. 

This twisting' action appears to show itself most clearly 
when the girder is loaded along its tension flange, i.e. when the 
compression flange is free to buckle. Probably if the load were 
evenly distributed on flooring attached to the compression 
flange, there would be no need for any stiffeners, because the 
flooring itself would prevent side twisting ; in fact, in the United 
States one sees a great many plate girders used without any 
web stiffeners at all when they are loaded in this manner. 

But if the flooring is attached to the bottom of the girder, 
leaving the top compression flange without much lateral support, 
stiffeners will certainly be required to keep the top flange 
straight and parallel with the bottom flange. The top flange in 
such a case tends to pivot about a vertical axis passing through 
its centre. For this reason the ends should be more rigidly 
stiffened than the middle of the girder, which is, of course, the 
common practice, but it is usually assigned to another cause. 

There are, however, other reasons for using web stiffeners. 
Whenever a concentrated load is applied to either flange it 



6 14 Mechanics applied to Engineering. 

produces severe local stresses ; for example, when testing rolled 
sections and riveted girders which, owing to their shortness, do 
not fail by twisting, the web always locally buckles just under 
the point of application of the load. This local buckling is 
totally different from the supposed buckling propounded by 
Rankine. By riveting tee-stiffeners on both sides of the web, 
the local loading is more evenly distributed over it, and the 
buckling is thereby prevented. Again, when a concentrated 
load is locally applied to the lower flange, it tends to tear the 
flange and angles away from the web. Here again a tee or 
plate stiffener well riveted to the flange and web very effectually 
prevents this by distributing the load over the web. 

In deciding upon the necessary pitch of stiffeners /,, there 
should certainly be one at every cross girder, or other con- 
centrated load, and for the prevention of twisting, well-fitted 
plate stiffeners near the ends, pitched empirically about 2 feet 
6 inches to 3 feet apart ; then alternate plate and tee stiffeners, 
increasing in pitch to not more than 4 feet, appear to accord 
with the best modern practice. 

Weight of Plate Girders. — For preliminary estimates, 
the weight of a plate girder may be arrived at thus — 

Let w = weight of girder in tons per foot run ; 

W = total load on the girder, not including its own 
weight, in tons. 

W 
Then w = - rouehly 
500 ^ ■' 

Arched Structures.— We have already shown how to 
determine the forces acting on the various segments of a 
suspension-bridge chain. If such a chain were made of 
suitable form and material to resist compression, it would, 
when inverted, simply become an arch. The exact profile 
taken up by a suspension-bridge chain depends entirely upon 
the distribution of the load, but as the chain is in tension, and, 
moreover, in stable equilibrium, it immediately and automatically 
adjusts itself to any altered condition of loading ; but if such a 
chain Avere inverted and brought into compression, it would be 
in a state of unstable equilibrium, and the smallest disturbance 
of the load distribution would cause it to collapse immediately. 
Hence arched structures must be made of such a section that 
they will resist change of shape in profile ; in other words, they 
must be capable of resisting bending as well as direct stresses. 

Masonry Arches. — ^In a masonry arch the permissible 



Structures. 615 

bending stress is small in order to ensure that there may be 
no, or only a small amount of, tensile stress on the joints of the 
voussoirs, or arch stones. Assuming for the present that there 
may be no tension, then the resultant line of thrust must lie 
within the middle third of the voussoir (see p. 541). In 
order to secure this condition, the form of the arch must 
be such that under its normal system of loading the line 
of thrust must pass through or near the middle line of the 
voussoirs. Then, when under the most trying conditions of 
live loading, the line of thrust must not pass outside the middle 
third. This condition can be secured either by increasing the 
depth of the voussoirs, or by increasing the dead load on 
the arch in order to reduce liie ratio of the live to the dead 
load. Many writers still insist on the condition that there shall 
be no tension in the joints of a masonry structure, but every 
one who has had any experience of such structures is perfectly 
well aware that there are very few masonry structureis in which 
the joints do not tend to open, and yet show no signs of 
instability or unsafeness. There is a limit, of course, to the 
amount of permissible tension. If the line of thrust pass 
throtigh the middle third, the maximum intensity of compres- 
sive stress on the edge of the voussoir is twice the mean, and 
if there be no adhesion between the mortar and the stones, the 
intensity of compressive stress is found thus — 

Let T = the thrust on the voussoirs at any given joint per 

unit width ; 
d = the depth or thickness of the voussoirs ; 
X = the distance of the line of thrust from the middle 

of the joint; 
P = the maximum intensity of the compressive stress 

on the loaded edge of the joint. 
The distance of the line of thrust from the loaded edge is 

- — X, and since the stress varies directly as the distance from 
2 

this edge, the diagram of stress distribution will be a triangle 

with its centre of gravity in the line of thrust, and the area of 

the triangle represents the total stress ; hence — 



pxag-.) 



and P = 



= T 

2T 



<;-) 



6i6 Mechanics applied to Engineering. 

T 
The mean stress on the section = — 

d 

hence, when — 

d ^ . . r 

* = T, P = 2 X mean intensity of stress 

d „ . . 

X = -, P = 2^ X mean intensity 
4 

A masonry arch may fail in several ways; the most 
important are — 

1. By the crushing of the voussoirs. 

2. By the sliding of one voussoir over the other. 

3. By the tilting or rotation of the voussoirs. 

The first is avoided by making the voussoirs sufficiently 
deep, or of sufficient sectional area to keep the compressive 
stress within that considered safe for the material. This 
condition is fulfilled if — 

2'67T 

— -J- = or < the permissible compressive stress 

The depth of the arch stones or voussoirs d in feet at the 
keystone can be approximately found by the following 
expressions : — 

Let S = the clear span of the arch in feet j 
R, = the radius of the arch in feet. 

— a/S a/S 

Then d = ^— for brick to ^^ for masonry 

2-2 3 

or, according to Trautwine — 



d = 



where K = i for the best work ; 

I •12 for second-class work; 

I '33 for brickwork. 

The second-mentioned method of failure is avoided by so 

arranging the joints that the line of thrust never cuts the normal 

to th3 joint at an angle greater than the friction angle. Let // 

be tie line of thrust, then sliding will occur in 9ie manner 




Structures. 617 

shown by the dotted lines, if the joints be arranged as shown 
at M— that is, if the angle a exceeds the friction angle <^. But 
if the joint be as shown at aa, sliding cannot occur. 

The third-mentioned method of failure only occurs when 
the line of thrust passes right outside the section ; the voussoirs 
then tilt till the line of thrust passes through the pivoting points. 
An arch can never fail in this way if the line of thrust be kept 
inside the middle halt. 




Fig. 601. Fig. 602. 

The rise of the arch R (Fig. 505) will depend upon local 
conditions, and the lines of thrust for the various conditions of 
loading are constructed in precisely the same manner as the 
link-and-vector polygon. A line of thrust is first constructed for 
the distributed load to give the form of the arch, and if the line 
of thrust comes too high or too low to suit the desired rise, it is 
corrected by altering the polar distance. Thus, suppose the rise 
of the line of thrust were Rj, and it was required to bring it to Rj. 
If the original polar distance were OoH, the new polar distance 

required to bring the rise to Rj would be dH = OqH X ^• 

After the median line of the arch has been constructed, 
other link polygons, such as the bottom right-hand figure, are 
drawn in for the arch loaded on one side only, for one-third 
of the length, on the middle only, and any other ways which 
are likely to throw the line of thrust away from the median 
line. After these lines have been put in, envelope curves 
parallel to the median line are drawn in to enclose these lines 
of thrust at every point ; this gives us the middle half of the 
voussoirs. The outer lines are then drawn in equidistant from 
the middle half lines, making the total depth of the voussoirs 
equal to twice the depth of the envelope curves. 

An infinite number of lines of thrust may be drawn in for 
any given distribution of load. Which of these is the right 
one ? is a question by no means easily answered, and whatever 



6i8 



Mechanics applied to Engineering. 



answer may be given, it is to a large extent a matter of opinion. 
For a full discussion of the question, the reader should refei 
to I. O. Baker's " Treatise on Masonry " (Wiley and Co., New 
York); and a paper by H. M. Martin, I.C.E. Proceedings, 
vol. xciii. p. 462. 



7v 



y 



/ 



B' 



d 



OU 




Fig. 603. 



From an examination of several successful arches, the 
author considers that if, by altering the polar distance OH, a 
line of thrust can be flattened or bent so as to fall within the 
middle half, it may be concluded that such a line of thrust is 
admissible. One portion or point of it itiay touch the inner 
and one the outer middle half lines. As a matter of fact, an 



Structures. 



6ig 



exact solution of the masonry arch problem in which the 
voussoirs rest on plane surfaces is indeterminate, and we can 
only say that a certain assumption is admissible if we find that 
arches designed on this assumption are successful. 

Arched Ribs. — In the case of iron and steel arches, the 
line of thrust may pass right outside the section, for in a 
continuous rib capable of resisting tension as well as compres- 
sion the rib retains its shape by its resistance to bending. 
The bending moment varies as the distance of the line of 
thrust from the centre of gravity of the section of the rib. 
The determination of the position of the line of thrust is 
therefore important. 

Arched ribs are often hinged at three points — at the 




Fig. 604. 



springings and at the crown. It is evident in such a case that 
the line of thrust must pass through the hinges, hence there 
is no difficulty in finding its exact position. But when the arch 
is rigidly held at one or both springings, and not hinged at the 
CMOwn, Ihe position of the line of thrust may be found thus : ' 




Fio. 606. 
' See also a paper by Bell, I.C.E., vol. xxxiii. p. 68. 



620 Mechanics applied to Engineering. 

Consider a short element of the rib mn of length /. When 
the rib is unequally loaded, it is strained so that mn takes up 
the position nin'. Both the slope and the vertical position 
of the element are altered by the straining of the rib. First 
consider the efifect of the alteration in slope ; for this purpose 
that portion of the rib from B to A may be considered as being 
pivoted at B. Join BA ; then, when the rib is strained, BA 
becomes BAj. Thus the point A, has received a horizontal 
displacement AD, and a vertical displacement AjD. The two 
triangles BAE, AAjD are similar; hence — 

AD V . „ AAi . ,. , 

6 being expressed in circular measure. Likewise — 

AjD X , ^ AAi ... 

AA; = AB A,D=^.^ = e.* . . (11.) 

A similar relation holds for every other small portion of the 
rib, but as A does not actually move, it follows that some of the 
horizontal displacements of A are outwards, and some inwards. 
Hence the algebraic sum of all the horizontal displacements 
must be zero, or 'XQy — o (iii.) 

Now consider the vertical movement of the element. If 
the rib be pivoted at C, and free at A, when mn is moved to 
niti, the rib moves through an angle B„ and the point A 
receives a vertical displacement S . B^ We have previously seen 
that A has also a vertical displacement due to the bending of 
the rib ; but as the point A does not actually move vertically, 
some of the vertical displacements must be upwards, and some 
downwards. Hence the algebraic sum of all the vertical 
displacements is also zero, or — 

%6x^'&B, = o (iv.) 

By similar reasoning — 

Mx-, + Si9a = o 
or S(S - a;)0 + S^A = o (v.) 

If the arch be rigidly fixed at C — 

and "SSx = o (vL) 



Structures. 



621 



If it be fixed at A — 



Sa^o 



and S(S — x)6 = o from v. 
If it be fixed at both ends, we have by addition — 

S(S9 -xe + xff) = o 
2Se = o 

Then, since S is constant — 

26 = o (vii.) 

Whence for the three conditions of arches we have — 

Arch hinged at both ends, %y6 = o from iii. 
„ fixed „ „ "Zyd, and 39 = o from iii. and 

vii. 
„ „ „ one end only, ^yO, and 1,x6 = o from 

i. and vi. 

We must now find an expression for 6. Let the full line 
represent the portion of the unstrained 
rib, and the dotted line the same when 
strained. 

Let the radius of curvature before strain- 
ing be po) and after straining p^. 

Then, using the symbols of Chapter 
XIII., we have — 

M„ = 5^, and M^ = ^ 

Po Pi 

Then the bending moment on the rib due 
to the change of curvature when strained 

IS — ^ Fig. 607. 



-l-i 




M = Mi- M„= Ya(---\ 
\pi Pa' 



(viii.) 



But as / remains practically constant before and after the 
strain, we have — 

61 = -, and 5o = 



Pi Po 

and 0. - 5o = (9 = /- - -) 

\Pi. po' 



or /= . 



I 

Pi 



I 
P» 



(ix.) 



622 



Mechanics applied to Engineering. 



EI 



Then we have from viii. and ix. — 

M/ = Eie 
ande=- 

If the arched rib be of constant cross-section, 

constant ; but if it be not so, then the length / must be taken 

, I ■ 
so that y IS constant. 

The bending moment M on the rib is M = F . ab, where 
the lowest curved line through cb is 
the line of thrust, and the upper 
dark line the median line of the rib. 
Draw ac vertical at c; 
dc tangential at c; 
ab normal to dc ; 
ad horizontal. 
Let H be the horizontal thrust 
on the vector polygon. Then the 
triangles adc, dd d , also adb, cda. 
are similar ; hence — 



Fig. 608. 




F cd \ ab 


ac 
7d 


ab ad n 




ac cd F 





or — = _ = 



orY .ab = Yi.ac=y[. 
but H is constant for any given case. 
Let ac = z. 

Hence the expression S5_y = o 

/ . 

may be written Si\Ty = o, since ^y is constant 

or l^z .y = 
Then, substituting in a similar manner in the equations above, 
we have — 

Arch hinged at both ends, "S^z .y = o 
„ fixed „ „ Ss . J" = o, and S3 = o 

„ „ at one end only, Ss . 7 = o, and S.XZ = o 

Thus, after the median line of the arch has been drawn, 
a line of thrust for uneven loading is constructed; and the 



Structures. 



623 



median line is divided up into a number of parts of length /, 
and perpendiculars dropped from each. The horizontal dis- 
tances between them will, of course, not be equal; then all 
the values z Y. y must be found, some z's being negative, and 




Fig. 609. 



some positive, and the sum found. If the sum of the negative 
values are greater than the positive, the line of thrust must be 
raised by reducing the polar distance of the vector polygon, 
and vice versa if the positive are greater than the negative. 
The line of thrust always passes through the hinged ends. In 
the case of the arch with fixed ends, the sum of the a's must also 




Fig. 610. 

be zero ; this can be obtained by raising or lowering the line of 
thrust bodily. When one end only is fixed, the sum of all the 
quantities x . z must be zero as well as zy; this is obtained by 
shifting the line of thrust bodily sideways. 

Having fixed on the line of thrust, the stresses in the rib 
are obtained thus : — 

The compressive stress all over the rib at any section is — 

Jc ^ 

where T is the thrust obtained from the vector polygon, and A 
is the sectional area of the rib. 



The skin stress due to bending is — 



Or/= — ; 



M T . nb 



(see Fig. 608) 



624 Mechanics applied to Engineering. 

and the maximum stress in the material due to both — 

T , H.2 
°^ = A + ^ 

Except in the case of very large arches, it is never worth 
while to spend much time in getting the exact position of the 
worst line of thrust ; in many instances its correct position may 
be detected by eye within very small limits of error. 

Effect of Change of Length and Temperature on 
Arched Ribs. — Long girders are always arranged with ex- 
pansion rollers at one end to allow for changes in length as the 
temperature varies. Arched ribs, of course, cannot be so treated 
— hence, if their length varies due to any cause, the radius of 
curvature is changed, and bending stresses are thereby set up. 
The change of curvature and the stress due to it may be 

arrived at by the following 
approximation, assuming 
the rib to be an arc of a 
circle : — 

Let N, = n(^i - t\ 

N^ = R2 + ?1 
4 

4 
N" - Ni^' = R= - R,» 




Fio. 611. 



Substituting the value of Ni and reducing, we get — 

_£L+?£1=(R + R,)(R_R,) 

n^ n 
But R - Rj = 8 
and R + Ri = 2R (nearly) 

The fraction - is very small, viz. <^^ and is rarely more than 
n tj 



jJ^j ; hence the quantity involving its square, viz. 

is negligible. 

l"hen we get — 



N» 



9000000' 



Structures. 


62s 


*-«R 


. , . (X.) 


But (^y^N'^-Ra 




also(^y=p^-(p-R)» 





N^ - R2 = p-" - (p - R)» 

from which we get — 

N^ = 2pR 
Substituting in x., we have — 





8 


n 


and 


P 


- 2R 


also 


Pi 


-2R, 



The bending moment on the rib due to the change of 
curvature is — 

M = EI| — ) from viiL 

Vp Pi/ 

and the corresponding skin stress is— 

M My M^ 
^~ Z- I - 2I 

where d is the depth of the rib in inches if R and N are taken 
in inches. 

Then, substituting the value of M, we have — 



i'Eld / R RA 



Ni' may without sensible error (about i in 2000) be written 
N»; then— 

/=||(R-RO 

E^ 



/- jja 



2 s 



626 Meclianics applied to Engineering. 

The stress at the crown due to the change of curvature on 
account of the compression of the rib then becomes — 

-• _ 2E^p 

and 1=/' (see p. 374) 

where _/j is the compressive stress all over the section of the rib 
at the crown ; hence — 






J ^12 vj-S 



taking /„ as a preliminary estimate at about 4 tons per square 

inch ; then - =' — — (about). In the case of the change of 

curvature due to change of temperature, it is usual to take the 
expansion and contraction on either side of the mean tempera- 
ture of 60° Fahr. as \ inch per 100 feet for temperate climates 
such as England, and twice this amount for tropical climates. 
Hence for England — 

n 1200 4800 
putting E = 12,000 tons per square inch; 






•' va 



Thus in England the stress due to temperature change of 
curvature is about five-eighths as great as that due to the com- 
pression change of curvature. 

The value of p varies from I'zsN to 2'sN, and d from 
N N 
— to — . 
30 20 

Hence, due to the compression change of curvature — 

/= o'6 to o"8 ton per square inch 

and due to temperature in England^ 

/ = 0-4 to o"5 ton per square inch 

In the case of ribs fixed rigidly at each end, it can be 



Structures. 627 

shown by a process similar to that given in Chapter XIII., of 
the beam built in at both ends, that the change of curvature 
stresses at the abutments of a rigidly held arch is nearly twice 
as greatj and the stress at the crown about 50 per cent, greater 
than the stress in the hinged arch. 

An arched rib must, then, be designed to withstand the 
direct compression, the bending stresses due to the line of 
thrust passing outside the section, the bending stresses due to 
the change of curvature, and finally it must be checked to see 
that it is safe when regarded as a long strut ; to prevent side 
buckling, all the arched ribs in a structure are braced together. 

Effect of Sudden Loads on Structures. — If a bar be 
subjected to a gradually increasing stress, the strain increases in 
proportion, provided the elastic limit is not passed. The 
work done in producing the strain is given by the shaded area 
abc in Fig. 612. 

Let, X = the elastic strain produced ; 

/ = the unstrained length of the bar ; 
/ = the stress over any cross-section ; 
E = Young's Modulus of Elasticity ; 
A = sectional area of bar. 

Then .* = ir 
ill 

The work dorie in straining! /Aa: _/°A/ 

a bar of section A / ~ 2 ~ 2E , 

A/ P 
= T^E 

= \ vol. of bar X modulus of 
elastic resilience 

The work done in stretching a bar beyond the elastic Umit 
has been treated in Chapter X. 

If the stress be produced by a hanging weight (Fig. 613), 
and the whole load be suddenly applied instead of being 
gradually increased, then, taking A as i square inch to save the 
constant repetition of the symbol, we have — 




Fig. 6i2. 



The work stored in the weight w^ 



due to falling through a height jcf 

= fx 



But when the weight reaches c (Fig. 613), only a portion of 
the energy developed during the fall has been expended in 
stretching the bar, and the remainder is still stored in the 



628 



Mechanics applied to Engineering. 



falling weight ; the bar, therefore, continues to stretch until the 
kinetic energy of the weight is absorbed by the bar. 

* • fx 

The work done in stretching the bar by an amount « is — ; 

hence the kinetic energy of 
the weight, when it reaches c, 
is the difference between the 
total work done by the weight 
and the work done in stretch- 

fx fx 

ing the bar, ox fx = — . 

The bar will therefore con- 
tinue to stretch until the work 
taken up by it is equal to the 
total work done by gravity 
on the falling weight, or until 
the area ahd'xz equal to the 
area ageh. Then, of course, 
the area bed is equal to the 
area ach. 

When the weight reaches 
h, the strain, and therefore the 
stress, is doubled ; thus, when 
a load is suddenly applied to 
a bar or a structure, the stress 
produced is twice as great as if the load were gradually applied. 
When the weight reaches ^, the tension on the bar is greater 
than the weight; hence the bar contracts, and in doing so 
raises the weight back to nearly its original position at a ; it 
then drops again, oscillating up and down until it finally "comes 
to rest at c. See page 265. 

If the bar in question supported a dead weight W, and a 
further weight w^ were suddenly applied, the momentary load 
on the bar would, by the same process of reasoning, be 
W + 2ze/„. 

If the suddenly applied load acted upwards, tending to 
compress the bar, the momentary load would be W — 2Wo. 
Whether or not the bar ever came into compression would 
entirely depend upon the relative values of W and Wo. 

The special case when Wo = 2W is of interest ; the momen- 
tary load is then — 

W- 2(2 W) = -3W 

the negative sign simply indicating that the stress has been 




StrucHires. 



629 




(ii) 



(ill) 

Fig. 614. 



reversed. This is the case of a revolving loaded axle. Con- 
sider it as stationary for the moment, and overhanging as 
shown — 

The upper skin of the axle in 
case (i) is in tension, due to W. 
In order to relieve it of this stress, 
i.e. to straighten the axle, a force 
— W must be applied, as in (ii) ; 
and,- further, to bring the upper 
skin into compression of the same 
intensity, as in (i), a force — 2W 
must be applied, as shown in (iii). 
Now, when an axle revolves, every 
portion of the skin is alternately 
brought into tension and compres- 
sion ; 'hence we may regard a 
revolving axle as being under the 
same system of loading as in (iii), 
or that the momentary load is 
three times the steady load. 

The following is a convenient method of arriving at the 
momentary force produced by a suddenly applied load. 

The dead load W is shown by heavy lines with arrows 
pointing up for tension and down for compression, the suddenly 



-n 



■ZIV 



^ 



< 






^ 






l?^~ 



i 

CM 

^1 



» I 






\'V 






I 



Fig. 615. 



630 Mechanics applied to Engineering. 

applied load Wi is shown by full light lines, the dynamic 
increment by broken lines. The equivalent momentary load is 
denoted W„. In some instances in which the suddenly applied 
load is of opposite sign to that of the dead load, the dead 
load may be greater than the equivalent momentary load ; for 
example, cases 5, 6, 7, 8. In case 7 the momentary load is 
compressive ; when designing a member which is subjected 
to such a system of loading it should be carefully considered 
from the standpoint of a strut to withstand a load W^, and a 
tie to withstand a load W. In case 8 the member must be 
designed as a strut to support a load — W, and be checked as 
a tie for a load W„. 

In case 3, W may represent the dead load due to the weight 
of a bridge, and Wj the weight of a train when standing on the 
bridge. W^ is the momentary load when a train crosses the 
bridge. 

There are many other methods in use by designers for 
allowing for the effect of live loads, some of which give higher 
and some lower results than the above method. 

Falling Weight. — When the weight W„ falls through a 
height h before striking the collar on the bottom of the bar, 
Fig. 613, the momentary effect produced is considerably 
greater than 2W„. 

Let/, = the static stress due to the load z£/„ 
_ w„ 
~ A 
/ = the equivalent momentary stress produced 



I 2 2E 



^=E 

Work done in stretching the bar — ' 

or the resilience of the bar per 

square inch of section 
Work done by the falling weight, \ ^ W , s _ .,, , s 

per sq. inch of section j ^ ^^ + •»; -W + x) 

/../(. V. + f) 

If a helical buffer spring be placed on the collar, the 
momentary stress in the bar will be greatly reduced. Let 



Structures. 63 1 

the spring compress i inch for a load of P lbs., and let the 
maximum compression under the falling weight be S inches, 
and h the height the weight falls before it strikes the spring. 
The momentary load on the spring = P8 
The momentary load on the bar —f,h. 
PS =/.A 
The total work done in stretching the ) _ kf^l P82 

bar and compressing the spring j 2E 2 

The work done by the falling weight = (W„(/^ + S + a:) 

from which the value of/^ can be obtained. 

Experiments on the Repetition of Stress.— In 1870 
Wohler published the results of some extremely interesting 
experiments on the effect of repeated stresses on materials, 
often termed the fatigue of materials ; since then many others 
have done similar work, with the result that a large amount of 
experimental data is now available, which enables us to con- 
struct empirical formulas to approximately represent the general 
results obtained. In these experiments many materials have 
been subjected to tensile, compressive, torsional, and bending 
stresses which were wholly or partially removed, and in some 
cases reversed as regards the nature of the stress imposed. 
In all cases the experiments conclusively showed that when 
the intensity of the stresses imposed approached that of the 
static breaking strength of the material, the number of repe- 
titions before fracture occurred was small, but as the intensity 
was reduced the number of repetitions required to produce 
fracture increased, and finally it was found that if the stress 
imposed was kept below a certain limit, the bar might be 
loaded an infinite number of times without producing fracture. 
This limiting stress appears to depend (i) upon the ultimate 
strength of the material {not upon the elastic limit), and (ii) 
upon the amount of fluctuation of the stress, often termed 
the " range of stress " to which the material is subjected. 

In a general way the results obtained by the different 
experimenters are fairly concordant, but small irregularities 
in the material and in the apparatus used appear to produce 
marked effects, consequently it is necessary to take the average 
of large numbers of tests in order to arrive at reliable data. 

The following figures, selected mainly from Wohler's tests, 
will serve to show the sort of results that may be expected 
from repetition of stress experiments : — 



632 Mechanics applied to Engineering. 

Material. 

Krupp's Axle Steel. 

Tensile strength, varying from 42 to 49 tons per sq. inch. 



Tensile 


tress applied 




Nominal bending stress 




in tons 
from 


per square 
inch 

to 


repetitions before 
fracture. 


in tons per square 

inch 
from to 


repetitions before 
fracture. 





38-20 


18,741 





26-25 


1,762,000 





33'40 


46,286 





25-07 


1,031,200 





28-65 


170,170 





24-83 


1,477,400 





26-14 


123,770 





23-87 


5,234,200 





23-87 


473,766 





23-87 


40,600,000 





22-92 


13,600,000 
(unbrolcen) 






(unbroken) 



Nominal ben 
revoU 
from 


ding stress in a 
ing azle 

to 


Number of repetitions 
before fracture. 


20-1 


— 20-1 


5S.IOO 


17-2 


-17-2 


1Z7.77S 


16-3 


-16-3 


797,525 


15-3 


-'5-3 


642,675 


11 


M 


1.665,580 


I) 


l» 


3,114,160 


14-3 


-I4'3 


4.163.37s 


,( 


r* 


45,050,640 



Material. 

Krupp's Spring Steel. 

Tangle strength, 57-5 tons per sq. inch. 







Tensile stress applied 




in tons p 
in 
from 


sr square 
ch 

to 


repetitions before 
fracture. 


in tons p 
in 
from 


er square 
ch 

to 


repetitions before 
iractuiv. 


4775 


7-92 


62,000 


38-20 


4-77 


99.700 


>* 


15-92 


149,800 


»J 


9-55 


176,300 


w 


23-87 


400,050 


»» 


14-33 


619,600 


)i 


27-83 


376,700 


»» 




2.135.670 


I) 


31-52 


19,673,000 
(unbroken) 


11 


19-10 


35,800,000 
(unbroken) 


42-95 


9-SS 


81,200 


33-41 


4-77 


286,100 


If 


14-33 


1,562,000 


1. 


9-55 


701,800 




19-10 


225,300 


II 


11-94 


36,600,000 


II 


23-87 


1,238,900 






(unbroken) 




»» 


300,900 








If 


28-65 


33,600,000 
(unbroken) 









Strttcturis, 

Material. 

Phceiiix Iron for Axles. 

Tensile strength, 21 '3 tons per sq. inch. 



633 



Nominal 
in a re> 


bending stress 
•olving axle. 


Rounded shoulder. 


Square shoulder. 


Conical shoulder. 


From 


To 


Number of repetitions before fracture. 


iS'3 




-15-3 


56,430 






134 • 




-13-4 


183,145 






II-5 




-ri-S 


909,84.0 






%^ 




- 9-6 


4,917,992 






8-6 




- 8-6 


19,186,791 


2,063,760 


535,302 


7-6 




- 7-6 


132,250,000 

(not brolien) 


14,695,000 


1,386,072 



The above figures show very clearly the importance of 
having well rounded shoulders in revolving axles. 

Material (Tests by Author). 

Mild steel cut from an over-annealed crank shaft. 

Elastic limit, 8'28 tons sq. inch ; tensile strength, 27^98 tons sq. inch. 



Nominal bending stress in a revolving 
axle. 


Number of repetitions before fracture. 


From 


To 




lo-o 

95 
90 

8-5 
8-0 


-lO'O 

- 9-S 

— 9'0 

= 8:^ 


418,100 

842,228 

729,221 
5,000,000 not broken 
5,000,000 not broken 



The figures given in the above table were obtained from a 
series of tests made on bars cut from a broken crank-shaft ; 
the material was in a very abnormal condition owing to pro- 
longed annealing, which seriously lowered the elastic limit. 
Other specimens were tested after being subjected to heat 
by a steel specialist, which raised the elastic limit nearly 50 
per cent., but it did not materially affect the safe limit of 
stress under repeated loading, thus supporting the view, held 
by many, that the capacity of a given material to withstand 
repeated loading depends more upon its ultimate or breaking 
stress than upon its elastic limit. 



634 



Mechanics applied to Engineering. 



Various theories have been advanced to account for the 
results obtained by repeated loading tests, but all are more 
or less unsatisfactory ; hence in designing structures we are 
obliged to make use of empirical formulas, which only ap- 
proximately fit experimental data. 

Many of the empirical formulas in use are needlessly com- 
plicated, and are not always easy of application ; by far the 
simplest is the " dynamic theory " equation, in which it is 
assumed that the varying loads applied to test bars by Wohler 
and others produce the same effects as suddenly applied loads. 
In this theory it is assumed that a bar will not break under 
repeated loadings if the "momentary stress" (see Fig. 614) 
does not exceed the stress which would produce failure if 
statically applied. Whether the assumptions are justified or 
not is quite an open question, and the only excuse for adopt- 
ing such a theory is that it gives results fairly in accord with 
experimental values, and moreover it is easily remembered 
and applied. 

The diagram, Fig. 616, is a convenient method of showing 



Static braaking stress i 




Fig. 616. 



to what extent repetition of stress experiments give results in 
accordance with the " dynamic theory." In this diagram all 



Structures. 63 5 

stresses are expressed as fractions of the ultimate static stress, 
which will cause fracture of the material. The minimum stress 
due to the dead load on the material is plotted on the line aoh, 
and the corresponding maximum stress, which may be applied 
over four million (and therefore presumably an infinite number 
of) times is shown by the spots along the maximum stress line. 
If the " dynamic theory " held perfectly for repeated loading, 
or fatigue tests, the spots would all lie on the line AB^, since 
the stress due to the dead load, i.e. the vertical height between 
the zero stress line and the minimum stress line plus twice the 
live load stress, represented by the vertical distance between 
the minimum and maximum stress lines, together are equal to 
the static breaking stress. 

The results of tests of revolving axles are shown in group 
A; the dynamic theory demands that they should be repre- 
sented by a point situated o'33 from the zero stress axis. 

Likewise, when the stress varies from o to a maximum, the 
results are shown at B ; by the dynamic theory, they should be 
represented by a point situated o'S from the zero axis. For 
all other cases the upper points should lie on the maximum 
stress line. Whether they do lie reasonably near this line 
must be judged from the diagram. When one considers the 
many accidental occurrences that may upset such experiments 
as these, one can hardly wonder at the points not lying 
regularly on the mean line. 

For an application of this diagram to the most recent work 
on the effect of repeated loading, readers should refer to the 
Proceedings of the Institution of Mechanical Engineers., November, 
1911, p. gro. 

Assuming that the dynamic theory is applicable to mem- 
bers of structures when subjected to repeated loads, we proceed 
thus — let the dead or steady load be termed W,„i„ , and the live 
or fluctuating load (W,„„ — W„[„), then the equivalent static 
load is — 

W = W ■ Z 2('W - W • ) 

* * c ^ rain, -r ^ V max. * ^ inin./ 

or using the nomenclature of Fig. 614 

W„ = W + 2(W + K'l - W) 
W^ = W + 2Wi 

The plus sign is used when both the dead and the live 
loads act together, i.e. when both are tension or both com- 
pression, and the minus when the one is tension and the other 
compression. 



636 Mechanics applied to Engineering. 

For a fuller discussion of this question, readers are referred 
to the following sources : — Wohler's original tests, see En- 
gineering, vol. xi., 1871; British Association Report, 1887, 
p. 424; Unwin's " Testing of Materials " ; Fidler's "Practical 
Treatise on Bridge Construction " ; Morley's " Theory of 
Structures " ; Stanton and Bairstow, " On the Resistance of 
Iron and Steel to Reversals of Direct Stress," Froc. Inst. Civil 
Engineers, 1906, vol. clxvi. ; Eden, Rose, and Cunningham, 
"The Endurance of Metals," Froc. Inst. Mech. Efigineers, 
November, 1911. 



CHAPTER XVIII. 

HYDRAULICS. 

In Chapter X. we stated that a body which resists a 
change of form when under the action of a distorting stress 
is termed a solid body, and if the bddy returns to its original 
form after the removal of the stress, the body is said to be an 
elastic solid {t:.g. wrought iron, steel, etc., under small stresses) ; 
but if it retains the distorted form it assumed when under 
stress, it is said to be a plastic solid {e.g. putty, clay, etc.). If, 
on the other hand, the body does not resist a change of form 
when mider the action of a distorting stress, it is said to be a 
fluid body ; if the change of form takes place immediately it 
comes under the action of the distorting stress, the body is said 
to be a. perfect fluid {e.g. alcohol, ether, water, etc., are very 
nearly so) ; if, however, the change of form takes place gradually 
after it has come imder the action of the distorting stress, the 
body is said to be a viscous fluid (e.g. tar, treacle, etc.). The 
viscosity is measured by the rate of change of form under a 
given distorting stress. 

In nearly all that follows in this chapter, we shall assume 
that water is a perfect fluid ; in some instances, however, we 
shall have to carefully consider some points depending upon 
its viscosity. 

Weight of Water. — The weight of water for all practical 
purposes is taken at 62'5 lbs. per cubic foot, or o'o36 lb. per 
cubic inch. It varies slightly with the temperature, as shown 
in the table on the following page, which is for pure distilled 
water. 

The volume corresponding to any temperature can be found 
very closely by the following empirical formula : — 

Volume at absolute temperature T, taking ( T^" + 2 so 000 

the volume at 39' 2° Fahr. or 500° ■! = ^i, 

absolute as i \ 

PresBure due to a Given Head. — If a cube of water of 



638 



Mechanics applied to Engineering, 



I foot side be imagined to be composed of a series of vertical 

columns, each of i square inch section, and i foot high, each 

62'^ 
will weigh — 5= 0-434 lb. Hence a column of water i foot 
144 

high produces a pressure of o'434 lb. per square inch. 



Temp. Fahr. 


3»°. 


ig-J>. 


50°. 


100°. 


T50°. 


«o°. 




Ice. 


Water. 




Weight per 1 ' 
cubic foot in } 
lbs ) 


57'2 


62-417 


62-425 


62-409 


62-00 


61 20 


60-14 


59-84 


Volume of a \ 
given weight, 
taking water 
at 39'2° Fahr. 
as I ) 


I '091 


I •0001 


i-oooo 


1-0002 


1-007 


I 020 


1-038 


1-043 



The height of the column of water above the point in 
question is termed the head. 

Let h = the head of water in feet above any surface ; 

p = the pressure in pounds per square inch on that 

surface ; 
w = the weight of a column of water i foot high and 
I square inch section ; 
= 0-434 lb. 

Then p = wh, or 0-434^ 

or h = ~ = 2-305/, or say 2-31/ 

Thus a head of 2-31 feet of water produces a pressure of 
I lb. per square inch, 

Taking the pressure due to the atmosphere as 14-7 lbs. per 
square inch, we have the head of water corresponding to the 
pressure of the atmosphere — 

14-7 X 2-31 = 34 feet (nearly) 

This pressure is the same in all directions, and is entirely inde- 
pendent of the shape of the containing vessel. Thus in Fig'. 617 — 



Hydraulics. 



639 



The pressure over any unit area of surface at « = /„ = o-i,j,i,h„ 

b = P, = o-434'4. 

and so on. 

The horizontal width of the triangular diagram at the side 
shows the pressure per square inch at any depth below the surface. 
Thus, if the height of the triangle be 
made to a scale of i inch to the foot, 
and the width of the base 0*434^, the 
width of the triangle measured in 
inches will give the pressure in pounds 
per square inch at any point, at the 
same depth below the surface. 

Compressibility of Water. — 
The popular notion that water is 
incompressible is erroneous; the 
alteration of volume under such 
pressures as are usually used is, 

however, very small. Experiments show that the alteration 
in volume is proportional to the pressure, hence the relation 
between the change of volume when under pressure may be 
expressed in the same form as we used for Young's modulus 

on p. 374- 

Let V = the dimmution of volume under any given pressure 
p in pounds per square inch (corresponding to x 
on p. 374); 
V = the original volume (corresponding to / on p. 37 4) j 
K = the modulus of elasticity of volume of water ; 
p =■ the pressure in pounds per square inch. 




Fig. 617. 



Then \ = ^ 



_/V 



orK=^ 

V 



K = from 320,000 to 300,000 lbs. per square inch. 

Thus water is reduced in bulk or increased in density by 
I per cent, when under a pressure of 3000 lbs. per square inch. 
This is quite apart from the stretch of the containing vessel. 

Total Pressure on an Immersed Surface. — If, for 
any purpose, we require the total normal pressure acting' on an 
immersed surface, we must find the mean pressure acting on 
the surface, and multiply it by the area of the surface. We 
shall show that the mean pressure acting on a surface is the 
pressure due to the head of water above the centre of gravity 
of the surface. 



640 Mechanics applied to Engineering. 

Let Fig. 618 represent an immersed surface. Let it be 
divided up into a large number of horizontal strips of length 

^//j, etc., and of width* each at 

=-=^-= ^f=: ^^";3 p^-=--- a depth h-^, th, etc., respectively 
. t^. — .^^^^^ from the surface. Then the 

/ '' ] total pressure on each strip is 
I^^3^^' PA^t PA^, etc., where pi, p^ 
etc., are the pressures corre- 
FiG. 618. spending to A,, ^, etc. 

But/ = wh, and UJ> = Oi, Ij) = ^a, etc. 
The total pressure on each strip = wcL^h^, wciji^, etc. 
Total pressure on whole surface = P„ = w(^Ai + aji^ + etc.) 

But the sum of all the areas a^, a^, etc., make up the whole area 
of the surface A, and by the principle of the centre of gravity 
(p. 58) we have — 

ajii + aji^ +, etc., = AHo 

where Ho is the depth of the centre of gravity of the immersed 
surface from the surface of the water, or — 

P„ = wAH, 



— '"n.'-H 



Thus the total pressure in pounds on the immersed surface 
is the area of the surface in square units X the pressure in 
pounds per square unit due to the head of water above the 
centre of gravity of the surface. 

Centre of Pressure. — The centre of pressure of a plane 
immersed surface is the point in the surface through which 
the resultant of all the pressure on the surface acts. 
It can be found thus — 

Let H„ = the head of water above the centre of pressure ; 
Ho = the head of water above the centre of gravity of 
the surface ; 
Q = the angle the immersed surface makes with the 

surface of the water ; 
lo = the second moment, or moment of inertia of the 
surface about a line lying on the surface of 
the water and passing through o ; 
I = the second moment of the surface about a line 
parallel to the above-mentioned line, and 
passing through the centre of gravity of the 
surface ; 



Hydraulics. 641 

Ro = the perpendicular distance between the two 

axes; 
K* = the square of the radius of gyration of the 

surface about a horizontal axis passing through 

the c. of g. of the surface ; 
«„ Oj, etc. = small areas at depths h-^, h^, etc., respectively 

below the surface and at distances x^^x^, etc., 

from O. 
A = the area of the surface. 
Taking moments about O, we have — 

p^OiXx ■^■pia^i +, etc. = P*, 

XT 

wh^a^Xi + wh^a^^ +, etc. = wAHo^ ° 



"sin B 



a/ sin Q{a^oc^ + a^ +, etc.) = z«/AH„-t ° 



'sin 61 
sin^ Q{a^x^ + a^^ +, etc.) = AH„H, 




Fig. 619. 

On p. 76 we have shown that the quantity in brackets on 
the left-hand side of the equation is the second moment, or 
moment of inertia, of the surface about an axis on the surface 
of the water passing through O. Then we have — 

sin^ e lo = sin^ e(I + ARo") = AH„H„ 
or sin^ B{k.K^ + AR,") = AH„H, 

Substituting the value of R,, we get — 

„ sin^ e K* + Ho" 
^' = H^ 

The centre of pressure also lies in a vertical plane which 
passes through the c. of g. of the surface, and which is normal 
to the surface. 

The depth of the centre of pressure from the surface of the 
water is given for a few cases in the following table : — 

z T 



642 



Mechanics applied to Engineering. 



Vertical surface. 


«'. 


H„. 


Ht. 


Rectangle of depth d with upper edge at surface 1 
of water / 


12 


2 


i^ 


Circle of diameter d with circumference touching \ 
surface of water / 


16 


2 


1^ 


Triangle of height d with apex at surface ofl 
water and base horizontal / 


18 


2</ 

3 


3rf 
4 



The methods of finding k* and H,, have been fiilly described 
in Chapter III. 

Graphical Method for finding the Centre of 

Pressure. — In some cases of irregularly shaped surfaces the 

algebraic method given above is not 

^^^^^^^ g , —^ easy of application, but the following 

^^^-^^^^^^"^3= graphic method is extremely simple. 
/ \ \ In the figure shown draw a series 

of lines across, not necessarily equi- 
distant; project them on to a base- 
line drawn parallel to the surface of 
hb the water. In the figure shown only one 
line, CM, is projected on to the base- 
line in bb. Join bb to a point on the 
surface of the water and vertically over 
the centre of gravity of the immersed 
surface, which cuts off a line dd ; then 
we have, by similar triangles — 

aa bb h^ 
dd ~ dd ~ h^ 

or the width of the shaded figure dd at any depth h^ below 
the surface is proportional to the total pressure on a very 
narrow strip aa of the surface ; hence the shaded figure may 
be regarded as an equivalent surface on which the pressure is 
uniform ; hence the c. of g. of the shaded figure is the centre 
of pressure of the original figure. 

It will be seen that precisely the same idea is involved here 
as in the modulus figures of beams given in Chapter XI. 




Fig. 620. 



Hydraulics. 



643 




Fig. 621. 



The total normal pressure on the surface is the shaded area 
A, multiplied by the pressure due to the head at the base-line, 
or — 

Total normal pressure = wA^j 

Practical Application of the Centre of Pressure.— 
A good illustration of a practical applica- 
tion of the use of the centre of pressure is 
shown in Fig. 6a i, which represents a self- 
acting movable flood dam. The dam AB, 
-usually of timber, is pivoted to a back 
stay, CD, the point C being placed at a 
distance = f AB from the top j hence, when 
the level of the water is below A the centre 
of pressure falls below C, and the dam is 
stable; if, however, the water flows over 
A, the centre of pressure rises above C, and 
the dam tips over. Thus as soon as a flood occurs the dam 
automatically tips over and prevents the water rising much 
above its normal. Each section, of course, has to be replaced 
by those in attendance when the flood has abated. 

Velocity of Flow due to 
a Given Head. — Let the tank 
shown in the figure be provided 
with an orifice in the bottom as 
shown, through which water flows 
with a velocity V feet per second. 
Let the water in the tank be kept 
level by a supply-pipe as shown, 
and suppose the tank to be very 
large compared with the quantity 
passing the orifice per second, and 
that the water is sensibly at rest yw. 6sa. 

and free from eddies. 

Let A = area of orifice in square feet ; 
Aj = the contracted area of the jet ; 
Q = quantity of water passing through the orifice in 

cubic feet per second ; 
V = velocity of flow in feet per second ; 
W = weight of water passing in pounds per second ; 
h = head in feet above the orifice. 

Then Q = A„V 
Work done per second by W lbs. of \ _ ^tt , r m, 
water falling through h feet / ~ loot-iDs. 




644 Mechanics applied to Engineering. 

) = 

But these two quantities must be equal, or — 



Kinetic energy of ■ the water on \ _ WV* 
leaving the orifice ) ~ 2g 




WA = , and h = — 

and V = V '2'gh 

that is, the velocity of flow is equal to the velocity acquired by 
a body in falling through a height of h feet. 

Contraction and Friction of a Stream passing 
through an Orifice. — The actual velocity with which water 
flows through an orifice is less than that due to the head, 
mainly on account of the friction of the stream on the sides of 
the orifice ; and, moreover, the stream contracts after it leaves 

the orifice, the reason for which 
will be seen from the figure. 
If each side of the orifice be 
regarded as a ledge over which 
a stream of water is flowing, it 
is evident that the path taken 
by the water will be the result- 
FiG. 623. ant of its horizontal and vertical 

movements, and therefore it 
does not fall vertically as indicated by the dotted lines, which it 
would have to do if the area of the stream were equal to the 
area of the orifice. Both the friction and the contraction can 
be measured experimentally, but they are usually combined in 
one coefficient of discharge K^, which is found experimentally. 
Hydraulic Coefficients. — The coefficient of discharge Y^ 
may be split up into the coefficient of velocity K„ viz. — 

actual velocity of fl ow 

and the coefficient of contraction K„ viz. — 

actual area of the stream 
area of the orifice 

Then the coefficient of discharge — 
K^ = K„K. 



Hydraulics. 645 

The coefficient of resistance — 

jr _ actual kinetic energy of the jet of water leaving the orifice 
kinetic energy of the jet if there were no losses 

The coefficient of velocity for any orifice can be found 
experimentally by fitting the orifice into the vertical side of a 
tank and allowing a jet of water to issue from it horizontally. 
If the jet be allowed to pass through a ring distant h^ feet 
below the centre of the orifice and h^ feet horizontally from it, 
then any given particle of water falls A, feet vertically while 
travelling h.^ feet horizontally, 

where ^1 = \gfi 

and/ = A /?^ 
^ g 
also h^ = v^ 

where v^ = the horizontal velocity of the water on leaving the 
orifice. If there were no resistance in the orifice, it would have 
a greater velocity of efflux, viz. — 

V = ij 2gh 

where h is the head of water in the tank over the centre of the 
orifice. 

Then », = '^^ 
andK. = -»=- ^ 



This coefficient can also be found by means of a Pitot tube. 
i.e. a small sharp-edged tube which is inserted in the jet in 
such a manner that the water plays axially into its sharp-edged 
mouth ; the other end of the tube, which is usually bent for 
convenience in handling, is attached to a glass water-gauge. 
The water rises in this gauge to a height proportional to the 
velocity with which the water enters the sharp-edged mouth- 
piece. Let this height be h^^ then Vi = ij 2gh^, from which the 
coefficient of velocity can be obtained. This method is not 
so good as the last mentioned, because a Pitot tube, however 
well constructed, has a coefficient of resistance of its own, and 
therefore this method tends to give too low a value for K,. 



646 



Mechanics applied to Engineering. 



The area of the stream issuing from the orifice can be 
measured approximately by means of sharp-pointed micro- 
meter screws attached to brackets on the under side of the 
orifice plate. The screws are adjusted to just touch the issuing 
stream of water usually taken at a distance of about three 
diameters from the orifice. On stopping the flow the distance 
between the screw-points is measured, which is the diameter of 
the jet ; but it is very diflScult to thus get satisfactory results. 
A better way is to get it by working backwards from the 
coefficient of discharge. 

Plain Orifice. — ^The edges should be chamfered off as 
shown (Fig. 624); if not, the water dribbles down the sides 
and makes the coefficient variable. In this case K, = about 
0*97, and K, = about 0-64, giving K^ = o'62. Experiments 
show that Ki decreases with an increase in the head and the 
diameter of the orifice, also the sharper the edges the smaller 
is the coefficient, but it rarely gets below o'sga, and sometimes 
reaches o'64. As a mean value K^ = o'62. 

Q = o'62A.\ 2gh 





Fig. 624. 



Fig. 625. 



Rounded Orifice. — If the orifice be rounded to the same 
form as a contracted jet, the contraction can be entirely avoided, 
hence K, = i ; but the friction is rather greater than in the 
plain orifice, K, = o"96 to cgS, according to the curvature 
and the roughness of the surface. The head h and the diameter 
of the orifice must be measured at the bottom, i.e. at the place 
where the water leaves the orifice ; as a mean value we may 
take — 

Q= o-qTK'J 2gh 



Hydraulics. 



647 




Pipe Orifice. — The length of the pipe should be not 
less than three times the diameter. 
The jet contracts after leaving the 
square corner, as in the sharp-edged 
orifice ; it expands again lower 
down, and fills the tube. It is 
possible to get a clear jet right 
through, but a very slight disturb- 
ance will make it run as shown. 
In the case of the clear stream, the 
value of K is approximately the /k \' 
same as in the plain orifice. When \^)nhi 
the pipe runs full, there is a sudden 
change of velocity from the con- 
tracted to the full part of the jet, 
with a consequent loss of energy 
and velocity of discharge. 

Let the velocity at ^ = Vj ; and the head = h^ 
the velocity at o = V, ; „ „ = h^ 

Then the loss of head = ^— = -'- (see p. 673) 

V 2 ^V. — V )^ 
and A,= — + '^ -^ 

The velocities at the sections a and 6 will be inversely as 
the respective areas. If K„ be the coefficient of contraction at 

6, we have Vj = j^; inserting this value in the expression 
given above, we get — 



Fig. 626. 



v„ = 



V'Hi-') 



/^gK 



The fractional part of this expression is the coefficient of 
velocity K, for this particular form of orifice. The coefficient 
of discharge Kj, is from 0*94 to 0-95 of this, on account of 
friction in the pipe. Then, taking a mean value — 

Q = 0-945 K.AV2PI 

The pressure at a is atmospheric, but at b it is less (see 



648 Mechanics applied to Engineering. 

p. 666). If the stream ran clear of the sides of the pipe into 
the atmosphere, the discharge would be — 

but in this case it is — 



Qi = K„Av'2^^. 
Let h^ — nhf 



Then Qi = YLjykiJ 2gnhi 

or the discharge is —^ — times as great as before ; hence we 
may write — 



Q, = =^X lL^y.K^2gnh^ 



-^)h when running a full stream 
and only h^ when running clear. The difference is due to a 
partial vacuum at b amounting to hAn{=~-\ — i \. 

If the head h^ be kept constant and the length of the pipe 
be increased it will be found that the quantity passing diminishes. 
The maximum quantity passes when D = 4//4j where D is the 
diameter of the pipe in feet, and / is the friction coefficient, 
see p. 679. 

Readers familiar with the Calculus will have no difficulty 
in obtaining this result, the relation can also be proved by 
calculating the quantity of water passing for various values of 
the length ab. 

When the pipe is horizontal n = 1, and the vacuum head is 



Hydraulics. 649 

The following results were obtained by experiment : — 

The diameter of the pipe = 0*945 inches 
Length h^ = 2 "96 ,, 
Ka = o"6i2 



Head ^j (inches) ... 


l6-i 


131 


lO'I 


8-1 


6-1 
0-786 


4-1 
0-780 


3'« 


Kd 


0799 


0797 


0797 


0-792 


o'773 


Vacuum head by ex- 
periment in inches 


167s 


I4'37 


11-95 


10-50 


9-00 


7-45 


6-35 


Vacuum head by cal- 
culation 


1 6 '60 


14-24 


11-97 


10-45 


8-85 


7-37 


6-56 



Before making this experiment the pipe must be washed out 
with benzene or other spirit in order to remove all grease, and 
care must be taken that no water lodges in the flexible pipe 
which couples the water-gauge to the orifice nozzle. 

Re-entrant Orifice or Borda's Mouthpiece. — If a 
plain orifice in the bottom of a tank be closed by a cover or 
valve on the upper side, the total 
pressure on the bottom of the tank 
will be P, where P is the weight of 
water in the tank; but if the orifice be 
opened, the pressure P will be reduced 
by an amount P„, equal to (i.) the down- 
ward pressure on the valve, viz. whh. ; 
and (ii.) by a further amount P„ due to 
the flow of water over the surface of 
the tank all round the orifice. Then 
we have — 



P„ = w/iA + P, 




Fig. 627. 



In the case of Borda's mouthpiece, the orifice is so far 
removed from the side of the tank that the velocity of flow 
over the surface is practically zero ; hence no such reduction of 
pressure occurs, or P, is zero. 

Let the section of the jet be a, and the area of the 
orifice A. 



650 Mechanics applied to Engineering. 

Then the total pressure due to the column \ _ . . 
of water over the orifice ' 

the mass of water flowing per second = 

the momentum of the water flowing per second = 

The water before entering the mouthpiece was sensibly at 
rest, hence this expression gives us the change of momentum 
per second. 

Change of momentum) . , , 

per second 1 "^ »™P"lse per second, or pressure 

waSf^ _ K/AV 

hence a = o'sA 
or K„ = o'5 

If the pipe be short compared to its diameter, the value of 
P, will not be zero, hence the value of K can only have this 
low value when the pipe is long. The following experiments 
by the author show the effect of the length of pipe on the 
coefficient : — 



Length of projecting pipe 
expressed in diameters 





A 


1 


A 


i 


2 


Kd 


0'6i 


0-56 


o-SS 


0-S4 


0-S3 


052 



If the mouthpiece be caused to run full, which can be 
accomplished by stirring the water in the neighbourhood of 
the mouthpiece for an instant, the coefficient of velocity will 
be (see " Pipe Orifice ") — 



. = — ^ = 0'71 



Experiments give values from 0-69 to 073. 



Hydraulics. 



6si 



Plain Orifice in a small Approach Channel. — When 
the area a of the stream passing through the orifice is appre- 
ciable as compared with the area of the approach channel A„, 
the value of K„ varies with the proportions between the two. 
With a small approach channel there is an imperfect con- 
traction of the jet, and according to Rankine's empirical 
formula — 



Tr\/' 



2-6i8 - r6i8 



A^ 



where A is the area of the orifice, and A„ is the area of the 
approach channel. 

The author has, however, obtained a rational value for 
this coefficient (see Engineering, March ii, 1904), but the 
article is too long for reproduction here. The value obtained 



IS — 



■K _ o'5 ( ,«* - 2«2 -I- I \ 



where « = the ratio of the radius of the approach channel tc 
the radius of the orifice. 

The results obtained by the two formulas are — 



ff. 


Kc Rational. 


Kc Rankine. 


2 


0-679 


0-672 


3 


0-645 


0-640 


4 


0-634 


0-631 


S 


0-631 


0-626 


6 


0-629 


0-624 


8 


0-628 


0-622 


10 


0-627 


0620 


100 


0-625 


o-6i8 


1000 


0-625 


o-6i8 



Diverging Mouthpiece. — This form of mouthpiece is of 
great interest, in that the discharge of a pipe can be greatly 



652 



Mechanics applied to Engineering. 



increased by adding a nozzle of this form to the outlet end, 
because the velocity of flow in the throat a is greater than the 

velocity due to the head of 
water h above it. The pressure 
at b is atmospheric ; ^ hence the 
pressure at a is less than atmo- 
spheric (see p. 666); thus the 
water is discharging into a 
partial vacuum. If a water- 
gauge be attached at a, and the 
vacuum measured, the velocity 
of flow at a will be found to 
be due to the head of water 
above it pltis the vacuum head. 
We shall shortly show that the energy of any steadily 
flowing stream of water in a pipe in which the diameter varies 
gradually is constant at all sections, neglecting friction. 
By Bernouilli's theorem we have (see p. 666) — 




Fig. 623. 



W 2g W 2g W 2g 

where — is the atmospheric pressure acting on the free 
w 

surface of the water. The pressure at the mouth, viz. /„ is also 
atmospheric ; hence £-=^. 



w w 



VV 



The velocity V is zero, hence — is zero. 

^g 

Then, assuming no loss by friction, we have — 







H-4 

or h 
orV. 


^g 
= ^2gA 


and the discharge 


— 








Q = 


= K,V.A, = 


= KAVa^A 



' This reasoning will not hold if the mouthpiece discharges into 

vacuum. 



Hydraulics. 



6S3 



In the case above, the mouthpifece is horizontal, but if it be 
placed vertically with b below, the proof given above still holds j 
the h must then be measured from b, i.e. the bottom of the 
mouthpiece, provided the conditions mentioned below are 
fulfilled. 

Thus we see that the discharge depends upon the area at b, 
and is independent of the area at a ; there is, however, a limit 
to this, for if the pressure at a be below the boiUng point corre- 
sponding to the temperature, the stream will not be continuous. 

From the above, we have — 



w ' 



2g "^ W 



If — ^ becomes zero, the stream breaks up, or when — 



2g 



= ^=34 feet 



Buty^ = ^ = «, or V„ = «Vj 

hence ^-X» li_^.l^(«2 - i) = 34 

2^ 2.?- 

or A(n^ — i) = 34 



In order that the stream may be continuous, ^n' — 1) 
should be less than 34 feet, and the maximum discharge will 
occur when the term to the left is 
slightly less than 34 feet. 

The following experiments 
demonstrate the accuracy of the 
statement made above, that the 
discharge is due to the head of 
water + the vacuum head. The 
experiments were made by Mr, 
Brownlee, and are given in the 
Proceedings of the Shipbuilders of* fig. 629. 

Scotland for 1875-6. 

The experiments were arranged in such a manner that, in 
effect, the water flowed from a tank A through a diverging 
mouthpiece into a tank B, a vacuum gauge being attached at 
the throat t. 

The close agreement between the experimental and the 



' A ' 


^=-iJ 


_B__ 




^^r\ 


-^ 



6S4 



Mechanics applied to Engineering. 



calculated values as given in the last two columns, is a clear 
proof of the accuracy of the theory given above. 



Head of water 
in tank A. 

Feet. 


Head of water 

in tank B. 

Feet. 

Hj. 


Vacuum at throat 

in feet of water. 

H« 


Velocity of flow at throat. 
Feet per second. 


Ha. 


By experiment. 






VwCHs+H,). 


69-24 

69-24 

69-24 

12-50 

12-50 

12-50 

8-00 

2-00 

0-25 


58-85 

50-78 

None 

8-50 

5-00 

1-50 

None 
None 
None 


None 
33-S 
33-S 
11-3 
33-S 
33-S 
33-S 
8-2 
0-52 


6s -97 
80-97 
81-43 
37-90 
S3-98 
54-60 
51-67 
24-74 
6-66 


66-78 
81-34 

?3 

S4-43 
54-43 
51-70 
25-63 
7-04 



Jet Fniup or Hydraulic Injector. — If the height of 
the column of water in the vacuum gauge at / (Fig. 629) be 
less than that due to the vacuum produced, the water will be 
sucked in and carried on with the jet. Several inventors have 
endeavoured to utilize an arrangement of this kind for saving 
water in hydraulic machinery when working below their full 
power. The high-pressure water enters by the pipe A ; when 
passing through the nozzles on its way to the machine cylinder, 
it sucks in a supply of water from the exhaust sump viS B, and 
the greater volume of the combined stream at a lower pressure 
passes on to the cylinder. All the water thus sucked in is a 
direct source of gain, but the efficiency of the apparatus as 
usually constructed is very low, about 30 per cent. The author 
and Mr. R, H. Thorpe, of New York, made a long series 

of experiments on jet pumps, 
and succeeded in designing 
one which gave an efficiency 
of 72 per cent 

An ordinary jet pump is 

shown in Fig. 630. The main 

trouble that occurs with such a 

form of pump is that the watei 

^"'■^^°- chums round and round the 

suction spaces of the nozzles 

instead of going straight through. Each suction space between 

the nozzles should be in a separate chamber provided with a 




Hydraulics. 



655 



back-pressure valve, and the spaces should gradually increase 
in area as the high-pressure water proceeds — that is to say, the 
first suction space should be very small, and the next rather 
larger, and so on. 

Rectangular Notch. — An orifice in a vertical plane with 
an open top is termed a notch, or sometimes a weir. The 
only two forms of notches commonly used are the rectangular 
and the triangular. 



dh 



» 



■B- f — 

'i 




Fig. 631. 



From the figure, it will be observed that the head of water 
immediately over the crest is less than the head measured 
further back, which is, however, the true head H. 

In calculating the quantity of water Q that flows over such 
a notch, we proceed thus — 

The area of any elementary strip as shown = '&.dh 

quantity of water passing strip perl v t? /^a 
second, neglecting contraction J ~ » • ^ • »" 
where V = velocity of flow in feet per second, 



\ \ hhdh 



= ijzgh, or 

Hence the quantity of water passing stripl _ . — -ojiji. 

per second, neglecting contraction f ~ ^ *^ 
the whole quantity of water Q passing"^ 

over the notch in cubic feet per>=V*^B 

second, neglecting contraction J 

- h' 
Q = -/^BfWn 

Q = PH^2^H 
introducing a coefficient to) ^ T?-2T>tT / — s 
allow for contraction fQ= ^^^^ ^ *^" 

where B and H are both measured in feet; where K has 
values varying from 0*59 to o'64 depending largely on the 



656 Mechanics applied to Engineering. 

proportions of the section of the stream, i.e. the ratio of the 
depth to the width, and on the relative size of the notch and 
the section of the stream above it. In the absence of precise 
data it is usual to take K = o'62. The following empirical 
formula by Braschmann gives values of |K for various heads H 
allowing for the velocity of approach. Let B^ = the breadth 
of the approach channel in feet. 

|K = (0-3838 + o-o386l+°-:?gli) 

Triangular Notch. — In order to avoid the uncertainty 
of the value of K, Professor James Thompson proposed the ■ 
use of V notches ; the form of the section of the stream then 
always remains constant however the head may vary. Experi- 



Fin. 633. 

ments show that K for such a notch is very nearly constant. 
Hence, in the absence of precise data, it may be used with 
much greater confidence than the rectangular notch. The 
quantity of water that passes is arrived at thus : 

Area of elementary strip = b . dh 

^ b B.-h , ^ B(H - >4) 

area of elementary strip = ' ~ — '- ■ dh 

rl 

velocity of water passing strip = V = ij 2gh = \'^ h^ 

quantity of water passing] -g/jj _ >■, 

strip per second, neglect- [ = -i— = — -^2gh^ . dh 

ing contraction I "■ 

whole quantity of water Q j f^ = H 

\dh 



lole quantity of water Q ] T^ = H 

passing over the notch in I B (jj - h)h' 

cubic feet per second, ~"H'^"^Ja = o 



neglecting contraction 



Hydraulics. 



657 



B _P = ^ 

J h = a 



^ = \j^g 



3 
3 



h = \\ 



Q = |V^(fH? - |Hi) = |ViiAH^ 

Introducing a coefficient for the contraction of the steam 
and putting B = 2H for a right-angled notch. 

where C has the following values. See Engineering, April 
8th and 15th, igio. 



H (feet) ... 


0-05 


o-io 


o-is 


0-20 


0-25 


0-30 


040 


C 


o'289 


0-304 


0-306 


0-306 


0-305 


0-304 


0-303 



Rectangular Orifice in a Vertical Plane. — When the 
vertical height of the orifice is small compared with the depth 
of water above it, the discharge is commonly taken to be the 
same as that of an orifice in a horizontal plane, the head being 
H, i.e. the head to the centre of the orifice. When, however, 
the vertical height of the orifice is not small compared with the 





Fio. 633. 



Fig. 634. 



depth, the discharge is obtained by precisely the same reasoning 
as in the two last cases ; it is — 

Q = KfBV'2i(H.i - H,?) 

K, however, is a very uncertain quantity; it varies with the 
shape of the orifice and its depth below the surface. 

Drowned Orifice. — When there is a head of water on 

2 u 



658 Mechanics applied to Engineering. 

both sides of an orifice, the discharge is not free ; the calculation 
of the flow is, however, a very simple matter. The head 
producing flow at any section xy (Fig. 634) is Hj — Hj = H j 
likewise, if any other section be taken, the head producing flow 
is also H. Hence the velocity of flow V = V 2^H, and the 

quantity discharged — 

Q= KAVz^H 
K varies somewhat, but is usually taken o"62 as a mean value. 
Flov7 under a Constant Head. — It is often found 
necessary to keep a perfectly constant head in a tank when 
making careful measurements of the flow of 
liquids, but it is often very difficult to accom- 
plish by keeping the supply exactly equal to 
the delivery. It can, however, be easily 
managed with the device shown in the figure. 
It consists of a closed tank fitted with an 
orifice, also a gland and sliding pipe open 
to the atmosphere. The vessel is filled, or 
nearly so, with the fluid, and the sliding pipe 
adjusted to give the required flow. The 
flow is due to the head H, and the negative 
pressure / above the surface of the water, for 



a. 



F^E5 



H-h 



Fig. 635. as the water sinks a partial vacuum is formed 

in the upper part of the vessel, and air 
bubbles through. Hence the pressure p is always due to the 
head h, and the effective head producing flow through the 
orifice is H — ^, which is independent of the height of water 
in the vessel, and is constant provided the water does not sink 
below the bottom of the pipe. The quantity of water 
delivered is — 

Q = KAv'2^^H - h) 

where K has the values given above for different orifices. 

Velocity of Approach. — If the water approaching a 
notch or weir have a velocity V„, the quantity of water passing 
will be correspondingly greater, but the exact amount will 
depend upon whether the velocity of the stream is uniform at 
every part of the cross-section, or whether it varies from point 
to point as in the section over the crest of a weir or notch. 

Let the velocity be uniform, as when approaching an orifice 
of area a, the area of the approach channel being A. 

Let V = velocity due to the head /4, i.e. the head over the 
orifice ; 
V = velocity of water issuing from the orifice. 



Hydraulics. 



659 



Then V„ = ^V, and V = V, + p 

V = —V + V 
A 



V = . 



Ka 



Broad-Crested Weir. — The water flows in a parallel 
stream over the crest of the weir if the sill is of sufficient 




Fig. 636. 

breadth to allow the stream lines to take a horizontal direction. 
Neglecting the velocity of approach, the velocity of the stream 
passing over the crest is, V = V zgh where h is the depth of 
the surface below that of the water in the approach channel. 
Let B = the breadth (in feet) of the weir at right angles 
to the direction of flow. 
Then the quantity passing over the weir in cubic feet per 
second is — 

Q = B(H->%)V2p 

The value of h, however, is unknown. If h be large in 
proportion to H, the section of the stream will be small, and 
the velocity large; on the other hand, if h be small in pro- 
portion to H, the section of the stream will be large and the 
velocity small, hence there must be some value of h which 
gives a maximum flow. 

Let /4 = «H; _ 

Q = BV'2^(H - nYi)>JnK 
Q = BVii X hV - n") 
dQ , — 3,, _i „ I, 



66o Mechanics applied to Engineering; 

This is a maximum when — 

5«~- = ^ffi or when n = \ 

Inserting the value of n in the equation for Q, we have — 

Q = o-sSsBHV'i^ 

The actual flow in small smooth-topped weirs agrees well 
with this expression, but in rough masonry weirs the flow is 
less according to the degree of roughness. 

Time required to Lower the Water in a Tank 
through an Orifice. — The problem of finding the time T 
required to lower the water in a dock or tank through a sluice- 
gate, or through an orifice in the bottom, is one that often 
arises. 

(i.) 2'ank of uniform cross-section. 

Let the area of the surface of the water be A.; 

„ „ „ stream through the orifice be K^A ; 
„ greater head of water above the orifice be Hi ; 
„ lesser ,, „ ^ ^ „ „ xlg • 

„ head of water at any given instant be h. 

The quantity of water passing through) _-k- a / — r j, 
the orifice in the time dt ^ - ii-.* A. v 2^A . dt 

Let the level of the water in the tank be lowered by an 
amount dh in the interval of time dt. 

Then the quantity in the tank is reduced by an amount 
KJth, which is equal to' that which has passed through the 
orifice in the interval, or — 

Y^i^,jlgh.dt= kjh 

dt = ^ f° /i -^dh 



p = H, 



'^=K:f7p' ^~*^* 



2A,(VHi-_VH ,) 
K.A^2^ 

The time required to empty the tank is — 



Hydraulics. 



661 



It is impossible to get an exact expression for this, because the 
assumed conditions fail when the head becomes very small ; 
the expression may, however, be used for most practical 
purposes. 

(ii.) Tapered tank of uniform breadth B. 

In this case the quan- 
tity in the tank is reduced ♦ 
by the amount 'Q.l.dh 
in the given interval of 1 ^ 
time dt. H? | 

But/=y ^ 

ill 

hence '&.l.dh = -^r^h dh 
«i 




Fig. 637. 



dt = 



BL 



HiK^AVz^ 



Ji'dh 



Integrating, we get- 



2BL(Hii - H,J ) 
3HiK,AV2i- 



(iii.) Hemispherical tank. 
In this case — 

1^= 2-SJi-h'' 

The quantity in the tank is 
reduced by ir(2Ry4 — h'^)dh in the ^ 
interval dt. 



dt = 



K,AV2ir 
T = 



(2R/4 - hyrUh 




Fig. 638. 



K,AV2^ 



(2R/*^ - h^) dh 
J H, 



i 



rate. 



T = _^ 5 _3 5 / 

KiAV2^ 

Hv.) C«J<? ;■« which the surface of the water falls at a uniform 



662 



Mechanics applied to Engineering. 



In this case — is constant ; 
at 

hence Kik^~2gh = A„ X a constant 

But K^Av 2^ is constant in any given case, hence the area of 
the tank A„ at any height h above the orifice varies as ^/h 
. or the vertical section of the tank 
must be paraboUc as shown. 

(v.) Time required to change the 
level when water is flowing into a 
tank at constant rate and leaving 
by an orifice. 

Let Aa = area of the surface of 
the water in square 
feet, when the depth 
of water above the 
outlet is h feet. 




Fic. 638 A. 



In a tank of geometrical form we may write 

A„ = C/?" where C and n are constants. 

Q, = the quantity of water in cubic feet per second running 

into the tank. 
A = the area of the orifice in square feet. 
T, = the time required to raise the level of the water from 

Hi to H„ feet. 
T( = the time required to lower the level of the water from 

H„ to Hi feet. 
The quantity of water flowing into the tank) _/-,,. 



in the time dt 



V 



The portion of the water which is retained in j _ * 
the tank in the time dt \~ " 



dh 



The quantity of water flowing out of the tank) _ t^ a / — i jj 
through the outlet in the time dt \~ ^"^'^ "^^"^ " 

When the water is entering the tank at a constant rate and 
leaving more slowly at a rate dependent upon the head, we 
have — 

Y^iK'Jlgh dt=. Qi dt - A„ dh 
dt{Q_t - K^aV^) = A,dh = Ch" dh 

/•Hm in J I 

■^ Jh, Q,-K,A^/2^-4 

This expression can be integrated, but the final expression is 



Hydraulics. 663 

very long, and moreover in practice the form of the tank or 
reservoir does not always conform to a geometrical law, hence 
we use an approximate solution which can be made as accurate 
as we please by taking a large number of layers between 
measured contour areas. The above expression then becomes 



'H 



A 



1 Qi-K^aV 2^0-^1 

A, M A2 Ih A,U 

+ 'Z ,, ,==- + — ZTT^"^ • ' • 



2(q,-nVho q,-nVh2 q,-nVHs 



2(Q,-NVHji 

The first and last terms are divided by 2 because we start and 
end at the middle sections of the upper and lower layers, 
hence the thickness of these layers is only one half as great 
as that of the intermediate layers. 

If we require to find the time necessary to lower the 
surface, i.e. when the water leaves more rapidly than it enters, 
we have — 



,4 



Ai SA , A2 8/% , As U 

+ ^^ ,— — + ^^ , r— — + 



2{NVHi-Qe) NVH^-Q, NVHs-Q, 



2(NVH„-Q,)3 

Where Aj is the area of the surface at a height Hj above 
the middle of the culvert j and A2 is the area at a height 
Ha = Hi + 8;^, and A3 at a height H3 = Hg + 8/4, and so on, 
and N = K^a^/ 2g. 

Example. — Let Q, = 48 cubic feet per second 
Hi = 12 ft., hh = 0-5 ft, K^AVzg = 7 
Ai = 140 sq. ft., Ag = 151, A3 s= 162, A4=i7o, Ag = 190 

Then the time required to raise the level from 12 to 14 feet 
would be — 

T ^ 140 X o-5_ 151 X 0-5 r62 X o-5_ 

2(48—7^/12) 48 — 7Vi2-5 48 — 7V13 

170 X 0-5 , 190 X o'5 
H —-^— -\ -p= = 14-3 seconds. 

48-7Vi3-s 2(48- 7V 14) 
(vi.) Time of discharge through a submerged orifice. — In Fig. 
639 we have — 



664 



Mechanics applied to Engineering, 




and 8Ha + mj^ 



8A 



8Hi = 



Ih 



A^U 



!+■ 



Aa X Ab U 



ht = 



Fig. 639. 



T = 



T = 



Aa X Ab 



(Aa+Ab)k^aV2^/* 



(Aa + Ab)K^aV2^- 

zAa X Ab i _ . 

(Aa + Ab)k,aV2/ '' 



p 



h 

-L- 






Fig. 640. 



where H and Hi are the initial 

arid final differences of heads. 

lime of discharge when 

the two tanks are connected 

by a pipe of length L. — 

The loss of head h, due to 

LV 
friction in the pipe is =:=- 

(see page 681), and the head 

h, dissipated in eddies at the 

V 

outlet is — 

2P- 



hence V = 



y h + h, _ / h /h_ 



KD^2^ 



L 

KD "*" ig 



Hence from similar reasoning to that given in the last 
paragraph we have — 



T = 



AaAbVc 



V H, 



\dh 



(Aa + Ab^AJ Hi 

2AaAb Vc ■ , 

(Aa + Ab)a(H -Hx') 

Where A is the area of the pipe which is taken to be bell- 
mouthed at entry, if it be otherwise the loss at entry (see 
p. 673) must be added to the friction loss. 



Hydraulics. 665 

Time required to lower the Water in a Tank when 
it flpws over a Weir or Notch.^ — Rectangular Notch. — By 
the methods already given for orifices we have — 

jdi= ^^%- i \-^dh 
zK^Bv 2^J Ha 

Right-angled Vee Notch. — In this case we get by similar 
reasoning — 

T = ^SA^ r .-= dh = _25A„^( H,-^-- - H.-^-^ j 
8K^//2^-iHa 8K,V2^A -1-5 / 

-T ^ 5A. / I i_\ 

Flow-through Pipes of Variable Section. — For the 

present we shall only deal with pipes running full, in which the 
section varies very gradually from point to point. If the varia- 
tion be abrupt, an entirely different action takes place. This 
particular case we shall deal with later on. The main point 
that we have to concern ourselves with at present is to show 
that the energy of the water at any section of the pipe is 
constant — neglecting friction. 

If W lbs. of water be raised from a given datum to a 
receiver at a certain height h feet above, the work done in 
raising the water is W^ foot-lbs., or h foot-lbs. per pound of 
water. By lowering the water to the datum, WA foot-lbs. of 
work will be done. Hence, when the water is in the raised 
position its energy is termed its energy of position, or — 

The energy of position = WA foot-lbs. 

If the water were allowed to fall freely, i.e. doing no 
work in its descent, it would attain a velocity V feet per 

V^ 

second, where V = »/ 2gh, or h = — . Then, smce no energy 

2g 

WV 
is destroyed in the fall, we have VJh = foot-lbs. of 

energy stored in the falling water when it reaches the datum, 

or — foot-lbs. per pound of water. This energy, which is 

due to its velocity, is termed its kinetic energy, or energy of 
motion ; or — WV 
The energy of motion = 



666 



Mechanics applied to Engineering. 



If the water in the receiver descends by a pipe to the 
datum level — for convenience we will take the pipe as one square 
inch area — the pressure / at the foot of the pipe will be wh lbs. 
per square inch. This pressure is capable of overcoming a 
resistance through a distance / feet, and thereby doing pi foot- 
lbs, of work ; then, as no energy is destroyed in passing along 

the pipe, we have// = W^ = -^ foot-lbs. of work done by the 



water under pressure, or ^ foot-lbs. per pound of water. This 
w 

is known as its pressure-energy, or — 

The pressure-energy = —£- 

Thus the energy of a given quantity of water may exist 
exclusively in either of the above forms, or partially in one 
form and partially in another, or in any combination of the 
three. 



Total energy perl _ (energy of)' (energy ofl , (pressure-) 
pound of water) ~ \ position \ \ motion ft energy ) 



ig w 



This may, perhaps, be more clearly seen by referring to the 
figure. 




Fig. 64Z 



Then, as no energy of the water is destroyed on passing 
through the pipe, the total energy at each section must be the 
same, or — 

;i, + Yk + A =^ 4. Yl + A ^ constant 

ig W 2g W 



Hydraultcs. 



667 



The quantity of water passing any given section of the pipe 
in a given time is the same, or— 

or AjVi = AjV, 

Yi = ^ 

V, A, 

or the velocity of the water varies inversely as the sectional 
area — 




Fig. 642. 

Some interesting points in this connection were given by 
the late Mr. Froude at the British Association in 1875. 

Let vertical pipes be inserted in the main pipe as shown ; 
then the height H, to which the water will rise in each, will be 
proportional to the pressure, or — 

H, = ^, and Hj = -^ 

Wl w 

and the total heights of the water-columns above datum — 

w w 

and the differences of the heights — 



^-^ + A 



w 



•■■-h = 



•W 2g 2g 

V ' — V 

H, - H, = -1? 11 

2.? 



from the equation given above. 

Thus we see that, when water is steadily running through 



668 Mechanics applied to Engineering. 

a full pipe of variable section, the pressure is greatest at the 
greatest section, and least at the least section. 

In addition to many other experiments that can be made 
to prove that such is the case, one has been devised by Pro- 
fessor Osborne Reynolds that beautifully illustrates this point. 
Take a piece of glass tube, say \ inch bore drawn down to a 
fine waist in the middle of, say, -^ inch diameter ; then, when 
water is forced through it at a high velocity, the pressure is so 
reduced at the waist that the water boils and hisses loudly. 
The pressure is atmospheric at the outlet, but very much less at 
the waist. The hissing in water-injectors and partially opened 
valves is also due to this cause. 

Ventnri Water-meter. — An interesting application of 
this principle is the Venturi water-meter. The water is forced 
through a very easy waist in a pipe, and the pressure measured 
at the smallest and largest section ; then, if the difierence of 
the heads corresponding to the two pressures be Ho in feet of 
water (Fig. 643)— 

V^ — V 
' „ ^ = H„, or Vi - V,^ = 2^Ho 

Let Aj = nKi, ; then Vj = — 
n 

hence V/ - (^J = 2^Ho and V^ = / J^Si. 

• '/Ho'=cVh; 



Q = A,V, = Aj 

V 


/ 2^ 


/ 


where C = A, 


/ 2. 



The difference of head is usually measured by a mercury 
gauge (shown in broken lines in Fig. 643), and the tubes 
above the surface of the mercury (sp. gravity 13-6) should be 
kept full of water, the mercury head H„ is most conveniently 
measured in inches. 

Then we get, Q = C^il3ljlJ& ^ cVr^^Ei;: 

There is a small loss of head in the short cone due to 
friction which can be allowed for by the use of a coefficient 



Hydraulics. 



669 



of velocity K, which is very nearly constant over a very wide 
range, its value is from 0-97 to o'gS, or say, o'gys, then — 

Q = K^cVroS V'H^ = CVh^ very nearly. 

At very low velocities of flow, where errors are usually of 
no importance, the value of K„ varies in a very erratic fashion, 
the reason for which is unknown at present; but for such 
velocities of flow as are likely to be used in practice the meter 
gives extremely accurate results. When used for waterworks 
purposes the meter is always fitted with a recorder and 
integrator, particulars of which can be obtained of Mr. Kent, 
of High Holborn, London. 




Fig. 643. 



The loss of head on the whole meter often amounts to 

about — - For experimental data on the losses in divergent 

pipes, readers should refer to a paper by Gibson, "The 
Resistance to Flow of Water through Pipes or Passages having 
Divergent Boundaries," Transactions of the Royal Society of 
Edinburgh, vol. xlviii. 

Radiating Currents and Free Vortex Motion. — Let 
the figure represent the section of two circular plates at a small 
distance apart, and let water flow up the vertical pipe and 
escape round the circumference of the plates. Take any small 
portion of the plates as shown ; the strips represent portions of 
rings of water moving towards the outside. Let their areas be 
Oi, «2j then, since the flow is constant, we have — 



670 



Mechanics applied to Engineering. 



V2 <h n , ''1 

Wiffi = Villi, or — = — = — hence w^ = v-r- 

»i a.i r^ r^ 

or the velocity varies inversely as the radius. The plates being 
horizontal, the energy of position remains constant ; therefore — 

2g W 2g W 

Substituting the value of v^ found above, we have — ■ 



!!l J.-6 _ '"'''"'' 



M 



2g W 2g. Ti 

Then, substituting — + - = H 
2g w 



.P2 

w 



from above, and putting 






r 
; ^1"^ = ^, we have — 
'2 




H-/4,= 



A 



Then, starting 
with a value for hy, 
the h^ for other posi- 
tions is readily calcu- 
lated and set down 
from the line above. 
If a large number 
of radial segments 
were taken, they 
would form a com- 
plete cylinder of 
water, in which the 
water enters at the 
^""" ^^ centre and escapes 

radially outwards. The distribution of pressure will be the 
same as in the radial segments, and the form of the water 
will be a solid of revolution formed by spinning the dotted 
line of pressures, known as Barlow's curve, round the axis. 

The case in which this kind of vortex is most commonly 
met with is when water flows in radially to a central hole, and 
then escapes. 

Forced Vortex. — If water be forced to revolve in and 
with a revolving vessel, the form taken up by the surface is 
readily found thus : 



Hydraulics. 



671 



Let the vessel be rotating n times per second. 

Any particle of water is acted upon 
by the following forces : — 

(i.) The weight W acting vertically 
downwards. 

(ii.) The centrifugal force act- 
ing horizontally, where V is its velocity 
in feet per second, and r its radius in 
feet. 

(iii.) The fluid pressure, which is 
equal to the resultant of i. and ii. 

From the figure, we have — 



w ■ 




ae 

be 



Fig. 645. 



which may be written — 



'Wgr be 



2 TT 

But — IS constant, say C ; 
g 



Then Cfyt^ = "1 
be 

But ac = r 

therefore C«^ = — 
be 



And for any given number of revolutions per second «* does 
not vary ; therefore be, the 
subnormal, is constant, and 
the curve is therefore a para- 
bola. If an orifice were made 
in the bottom of the vessel 
at 0, the discharge would be 
due to the head h. 

Loss of Energy due to 
Abrupt Change of Direc- 
tion. — If a stream of water flow down an inclined surface AB 
with a velocity Vj feet per second, when it reaches B the 
direction of flow is suddenly changed from AB to BC, and the 




672 



Mechanics applied to Engineering. 



layers of water overtop one another, thus causing a breaking-up 
of the stream, and an eddying action which rapidly dissipates 
the energy of the stream by the frictional resistance of the 
particles of the water; this is sometimes termed the loss by 
shock. The velocity V3 with which the water flows after 
passing the corner is given by the diagram of velocities ABD, 
from which we see that the component Vj, normal to BC, is 
wasted in eddying, and the energy wasted per pound of water 

IS _1- =— i 

As the angle ABD increases the loss of energy increases, 
and when it becomes a right angle the whole of the energy is 
wasted by shock (Fig. 646). 

If the surface be a smooth curve (Fig. 647) in which there 
is no abrupt change of direction, there will be no loss due to 



^ \ 





Fig. 646. 



Fig. 647. 



shock ; hence the smooth easy curves that are adopted for the 
vanes of motors, etc. 

If the surface against which the water strikes (normally) is 
moving in the same direction as the jet with a velocity 
y 
— , then the striking velocity will be — 

V. - X.' = V, 

n 

and the loss of energy per pound of water will be— 



2£ 2g- 2g\ n' 



Hydraulics. 673 

When « = I, no striking lakes place, and consequently no 
loss of energy J when «= 00 , i.e. when the surface is stationary, 

the loss is -i, i.e. the whole energy of the jet is dissipated. 

Loss of Energy due to Abrupt Change of Section. 
—When water flows along a pipe in which there is an abrupt 
change of section, as shown, we may regard it as a jet of water 
moving with a velocity Vj striking against a surface (in this case 
a body of water) moving in the same 

Y 
direction, but with a velocity — j hence I ci9^ — ~~ 

the loss of energy per pound of water * ^nzj^ Cr-^A 
is precisely the same as in the last para- ^^s>;r:rr 

(y -ViV ^-^^^ — 

I ' > ~) Fig. 648 (see also No. 3 

graph, viz. ^ ^ . The energy lost ^c'lg p- 67+). 

in this case is in eddying in the corners of the large section, 

as shown. As the water in the large section is moving - 

n 
as fast as in the small section, the area of the large section 
is n times the area of the small section. Then the loss of 
energy per pound of water, or the loss of head when a pipe 
suddenly enlarges « times, is — 

v.'(--;)' 

Or if we refer to the velocity in the large section as Vj, we 
have the velocity in the small section «Vi, and the loss of 
head — 

When the water flows in the opposite direction, i.e. from 
the large to the small section, the loss 
of head is due to the abrupt change of 
velocity from the contracted to the 
full section of the small stream. The 
contracted section in pipes under pres- 
sure is, according to some experiments 
made in the author's laboratory, from fig.' 649 Cs« also No. 4 facing 
0*62 to o"66; hence, «* = from i'6i p-674). 

to i'5 ; then, the loss of head = 

2 X 




6/4 Mechanics applied to Engineering. 

Total Loss of Energy due to a Sudden Enlargement 
and Contraction. — Let the section before the enlargement 
be termed i, the enlarged section 2, and the section after the 
enlargement 3, with corresponding suffixes for velocities and 
pressures. Then for a horizontal pipe we have — 



W- 2g W 2g 2g\ nJ 

W Ig; 






Gibson finds that the loss of energy is slightly greater than 
this expression gives. The author finds that the actual loss in 
some cases is nearly twice as great as the calculated. 

Experiments on the Character of Fluid Motion. — 
Some very beautiful experiments, by Professor Hele-Shaw, 
F.R.S., on the flow of -fluids, enable us to study exactly the 
fn^nner ,ip which th,e ^ow takes place in channels of various 
fofms. He' takes two sheets of glass and' fits them into a 
suitable frame, whichholds them in position at about yj^ inch 
apart. , Through this narrow space liquid is caused to flow under 
■pressare, and in order to demonstrate the exact manner in 
which the flow takes place, bands of. coloured liquid are 
injected at the inlet end. In the narrow sections of the 
channel, where the velocity of flow is greatest, the bands 
themselves are narrowest, and they widen out in that portion 
of the channel where the velocity is least. The perfect 
manner in which the bands converge and diverge as the 
liquid passes through a neck or a pierced diaphragm, is in 
itself an elegant demonstration of the behaviour of a perfect 
fluid (see Diagrams i and 5). The form and behaviour of the 
bands, moreover, exactly correspond with mathematical demon- 
strations of the mode of flow of perfect fluids. The author 
is indebted to Professor Hele-Shaw, for the illustrations given, 
which are reproduced from his own photographs. 

In the majority of cases, however, that occur in practice, 
we are unfortunately unable to secure such perfect stream-line 
motions as we have just described. We usually have to deal 
with water flowing in sitiuous fashion with very complex eddy- 
ings, which is much more difficult to ocularly demonstrate than 
true stream-line motion. Professor Hele-Shaw's method of 
showing the tumultuous conditions under which the water is 
moving, is to inject fine bubbles of air into the water, which 
make the disturbances within quite evident. The diagrams 2, 






ITofacep. 674. 



FLOW OF WATER DIAGRAMS. 
Kindly supplied by Processor Helt-Shaw, F.R.S. 



Hydraulics. 675 

3, 4, and 6, also reproduced from his photographs, clearly 
demonstrate the breaking up of the water when it encounters 
sudden enlargements and contractions, as predicted by theory. 
A careful study of these figures, in conjunction with the 
theoretical treatment of the subject, is of the greatest value in 
getting a clear idea of the turbulent action of flowing water.. 

Readers should refer to the original communications by 
Professor Hele-Shaw in the Transactions of the Naval Architects, 
1897-98, also the engineering journals at that time. 

Surface Friction. — When a body immersed in water is 
caused to move, or when water flows over a body, a certain 
resistance to motion is experienced ; this resistance is termed 
the surface or fluid friction between the body and the water. 

At very low velocities, only a thin film of the water actually 
in contact with the body appears to be affected, a mere skim- 
ming action ; but as the velocity is increased, the moving body 
appears to carry more or less of the water with it, and to cause 
local eddying for some distance from the body. Experiments 
made by Professor Osborne Reynolds clearly demonstrate the 
difference between the two 
kinds of resistances — the sur- 
face resistance and the eddy- 
ing resistance. Water is caused 
to flow through the glass pipe 
AB at a given velocity ; a bent 
glass tube and funnel C is 
fixed in such a manner that a 
fine stream of deeply coloured 
dye is ejected. When the water fig. 650. 

flows through at a low velocity, 

the stream of dye runs right through like an unbroken thread ; 
but as soon as the velocity is increased beyond a certain 
point, the thread breaks up and passes through in sinuous 
fashion, thus demonstrating that the water is not flowing 
through as a steady stream, as it did at the lower velocities. 

Friction in Pipes. — Contimeotis Flow. — When the flow 
in a pipe is continuous, i.e. not of an eddying nature, the 
resistance to flow is entirely due to the viscosity of the fluid. 
On p. 314 we showed that the resistance to shearing a viscous 
fluid is — 

where A is the wetted surface in square feet, K is the 




6/6 Mechanics applied to Engineering. 

coefficient of viscosity, S the speed of shearing (usually denoted 
by V in hydraulics) in feet per second, / is the thickness of 
the sheared element in feet, in the case of a pipe of radius 
R, / = R. Then, without going fully into the question, for 
which treatises on Hydraulics should be consulted — 

Let Pi = the initial pressure in pounds per sq. foot. 
Pa = the final pressure in pounds per sq. foot. 
L = the length of the pipe in feet. 
Aj, = the area of the pipe in sq. feet. 
F, = (Pi - Pe)A, = W„(Hi - H,)A^ 
F. = W„/iA^ 

where h„ is the loss of head in feet due to the resistance on a 
length of pipe L. 

Then W„/4.A, = ^^^ 






R 

(27rRL)KV _ 2LKV ^ 8LKV 
W„.(xR^)R ~ W„R' ~ W^D'' 
LV 
CD" 



This expression only holds for stream line, or con- 
tinuous flow, the critical velocity V„ at which the flow changes 
from continuous to sinuous is always much higher than the 
velocity V„i at which the flow changes from sinuous to con- 
tinuous. The critical velocity also largely depends upon the 
temperature of the water owing to a change in the viscosity. 
Let Vc = the critical velocity, i.e. the velocity in feet per 
second at which the flow changes from con- 
tinuous to sinuous. 
V„i = ditto at which the flow changes from sinuous to 
continuous. 
n and «i = coefficients which depend upon the temperature 
of the water. 
D = the diameter of the pipe in feet, 

then V. = - 
and V,i = g 



Hydraulics. 



677 



Temperature 
Fahrenheit. 


«. 


"i- 


c. 


32 


0-25 


0-040 


52, SCO 


40 


021 


0-034 


61,000 


60 


0-15 


0-025 


83,500 


80 


0'12 


0-019 


io7,coo 


100 


O'lO 


0-016 


134,000 


120 


o'o8 


0-013 


164,000 


140 


0-065 


o-oii 


204,000 


160 


o'oss 


0-009 


240,000 


180 


0-047 


0-008 


278,000 


200 


0-040 


0-006 


328,000 


212 


0-037 


o"oo6 


350,000 



Mr. E. C. Thrupp has, however, shown that the values 



^ +5 



















r i 


B 










1 

1 
1 












1 
1 












J 




A 






> 




"••■ 






^ 

«' 


^'' 











-I +1 +3 +5 +7 +9 

Logarithms of hydraulic gradient. 
Fig. 651. 



given in the above table only hold for very small pipes; in 
the case of large pipes, channels, and rivers the velocity at 



678 



Mechanics applied to Engineering. 



which the water breaks up is very much greater than this 
expression gives. See a paper on " Hydraulics of the Re- 
sistance of Ships," read at the Engineering Congress in 
Glasgow, 1901 ; also Engineering, December 20, 1901, from 
which the curves in Fig. 651 have been taken. The Osborne 
Reynolds' law is represented by AC and BD, whereas Thrupp 
shows that experimental values lie somewhere between AA 
and BB. 

The change points from continuous to sinuous flow are 
shown in Fig. 652. At low velocities of flow the loss of 















/ 




i 










y 






2 










/ 






u 

s 
•a 
•0 








/ 

1 








.a 






r^ 


/ 








J 






1 










1 




E 


1/ 


B 








•c 




/ 












^ 




/ 














/ 















Logarithm of velocity. 
Fig. 6sa. 

head varies simply as the velocity, therefore the slope of the 
line AB is i to i. At B the flow suddenly changes to sinuous 
flow, and at higher velocities the loss of head varies approxi- 
mately as the square of the velocity, hence the slope of the 
line CD is 2 to i. When the velocity is decreased the loss 
of head continues to vary as the square until E is reached, 
and below that it returns to the state in which it varies simply 
as the velocity. The point B corresponds to V„, and the 
point E to V„i. 

For further details of Reynolds' investigations, the. reader 
is referred to the original papers in the Philosophical Trans- 
actions for 1884 and 1893; also to Gibson's "Hydraulics and 



Hydraulics. 679 

its Applications," and Turner and Brightmore's " Waterworks 
Engineering." 

Sinuous Flow.— Experiments by Mr. Froude at Torquay 
(see Brit. Ass. Proceedings, 1874), on the frictional resistance 
of long planks, towed end-on through the water at various 
velocities, showed that the following laws appear to hold 
within narrow limits : — 

(i.) The friction varies directly as the extent of the wetted 
surface. 

(ii.^ The friction varies directly as the roughness of the 
surface. 

(iii.) The friction varies directly as the square of the 
velocity. 

(iv.) The friction is independent of the pressure. 
For fluids other than water, we should have to add — 
(v.) The friction varies directly as the density and viscosity 
of the fluid. 

Hence, if S = the wetted surface in square feet ; 

/= a coefficient depending on the roughness of 
the surface ; i.e. the resistance per square 
foot at I foot per second in pounds ; 
V = velocity of flow relatively tp the surface in 

feet per second ; 
R = frictional resistance in pounds ; 

Then, R = S/V^ 

Some have endeavoured to prove from Mr. Froude's own 
figures that the first of the laws given above does not even 
approximately hold. The basis of their argument is that the 
frictional resistance of a plank, say 50 feet in length, is not 
ten times as great as the resistance of a plank 5 feet in length. 
This effect is, however, entirely due to the fact that the first 
portion of the plank meets with water at rest, and, therefore, if 
a plank be said to be moving at a speed of 10 feet a second, it 
simply means that this is the relative velocity of the plank and 
the still water. But the moving plank imparts a considerable 
velocity to the surrounding water by dragging it along with it, 
hence the relative velocity of the rear end of the plank and the 
water is less than 10 feet a second, and the friction is corre- 
spondingly reduced. In order to make this point clear the 
author has plotted the curves in Figs. 653 and 654, which are 
deduced from Mr. Froude's own figures. It is worthy of note 
that planks with rough surfaces drag the water along with them 
to a much greater extent than is the case with planks having 



68o 



Mechanics applied to Engineering. 



1-3 



B 



\ 



0-3^ 
0-2 - 



^!??5 



r-Af 



-i. 



'^^ 



Qa<£^ 



COAff.JE 



SAfID 



VW£^ 



^4A^ 



*0 2S 30 

DISTANCE nan CVTIMren 

Fig. 653. 







" 




































































































































































^ „ 










































K 










































^l 


^ 


-^ 


— 













r. 


m 


F 


01 


* 
















^ ti 


i 




.... 












1 


'n i 


• A< 


fr- 


f i 
















V , 




\\ 


































.._ 






*> r 






^ 


::::: 


— - 


-«_ 


_ 




/ 


/A 


'^r 


S 


m\ 

















" 

a 5 










Cl 


>/» 


pj 


J£ 


1 


•A 


NL 


■> 
















™ 


si •' 










































^ ' 




















































































y 




^ 

















































































« ZO 2S 30 3S 

DISTANCE FROM curwATen 

Fig. 6s4- 



H 



Hydraulics. 68 1 

smooth surfaces, a result quite in accordance with what one 
might expect. 

The value of/ deduced from these experiments is — 



Surface covered with coarse sand ... 


o'oi32 lb. 


„ „ fine „ 
„ „ varnish 

tinfoil 


... 0-0096 „ 
... 0-0043 .. 
... 0-0031 „ 



Professor Unwin and others have also experimented on the 
friction of discs revolving in water, and have obtained results 
very closely in accord with those obtained by Mr. Froude. 

Reducing the expression for the frictional resistance to a 
form suitable for application to pipes, we have, for any length 
of pipe L feet, the pressure Pj in pounds per square foot at one 
end greater than the pressure Pa at the other end, on account 
of the friction of the water. Then, if A be the area of the 
pipe in square feet, we have — 

R = (Pi - Pa)A 

Then, putting Pi = h^„ and Pj = -^jW^, we have — 
R = W„A(^i - hi) = W„AA 

where h is the loss of head due to friction on any length of 
pipe L; then — 

W„A/4 = S/V 

or -^ = LttD/V* 

hence/J = -.— =-.— 

/_ LV^ _ LV^ 
° W„' R ~4KR 

where R = hydraulic mean depth (see p. 683). 

The coefficient -^ has to be obtained by experiment ; 
according to D'Arcy — 

K 3200V 12D/ 
where D is the diameter of the pipe in feet. 



682 Mechanics applied to Engineering. 

D'Arcy's experiments were made on pipes varying in 
diameter from \ inch up to 20 inches ; for small pipes his 
coefficient appears to hold tolerably well, but it is certainly 
incorrect for large pipes. 

The author has recently looked into this question, and 
finds that the following expression better fits the most recent 
published experiments for pipes of over 8 inch diameter 
(see a paper by Lawford, Proceedings I.C.E., vol. cliii. p. 
297) :— 

— = ( I + -r~ ) for clean cast-iron pipes 

K 5ooo\ 2D/ 

— = ( I H — ;:r I for incrusted pipes 

K 25oo\ 2D/ 

But the above expression at the best is only a rough 
approximation, since the value of / varies very largely for 
different surfaces, and the resistance does not always vary as 
the square of the velocity, nor simply inversely as D. 

The energy of motion of i lb. of water moving with a 

velocity V feet per second is — ; hence the whole energy of 
motion of the water is dissipated in friction when — 

V^ LV 
2g~ KD 

Taking K = 2400 and putting in the numerical value for g, 
we get L = 37D. This value 37, of course, depends on the 
roughness of the pipe. We shall find this method of regarding 
frictional resistances exceedingly convenient when dealing with 
the resistances of T's, elbows, etc., in pipes. 

Still adhering to the rough formula given above, we can 
calculate the discharge of any pipe thus : 

The quantity discharged in) ^ _ a v _ '^^"^ 
cubic feet per second j^ - ti — AV — — — 

From the same formula, we have — 

''2400DA 



vV^ 



Hydraulics. 683 

Substituting this value, we have — 



Q=38-SD^\/J=38-SD^>/ 



L 



Thrupp's Formula for the Plow of Water. — All 
formulas for the flow of water are, or should be, constructed 
to fit experiments, and that which fits the widest range of ex- 
periments is of course the most reliable. Several investigators 
in recent years have collected together the results of published 
experiments, and have adjusted the older formulas or have 
constructed new ones to better accord with experiments. 
There is very little to choose between the best of recent 
formulas, but on the whole the author believes that this 
formula best fits the widest range of experiments ; others are 
equally as good for smaller ranges. It is a modification of 
Hagen's formula, and was published in a paper read before the 
Society of Engineers in 1887. 

Let V = velocity of flow in feet per second ; 

R = hydraulic mean radius in feet, i.e. the area of the 
stream divided by the wetted perimeter, and 

is? for circular and square pipes: 

L = length of pipe in feet ; 

h = loss of head due to friction in feet ; 

S = cosecant of angle of slope = — ; 

Q = quantity of water flowing in cubic feet per second. 

Then V = 



C^S 



where x, C, n are coeflScients depending on the nature of the 
surface of the pipe or channel. 

For small values of R, more accurate results will be ob- 
tained by substituting for the index x the value x + y^ 

In this formula the effect of a change of temperature is not 
taken into account. The friction varies, roughly, inversely as 
the absolute temperature of the water. 



684 



Mechanics applied to Engineering. 



Siufacc. 


n. 


c. 


X. 


y- 


- 


Wrought-iron pipes 


l-8o 


0-004787 


0-65 


0-018 


0*07 


Riveted sheet-iion pipes 


1-825 


0-005674 


0-677 


— 





New cast-iron pipes 


/I-85 

\2-00 


0-005347 

0-006752 


0-67 
0-63 





— 


Lead pipes 


I-7S 


0-005224 


0-62 


— 





Pure cement rendering 


/1 74 
1 1 '95 


0-004000 
0-006429 


0-67 

o-6i 


— 


— 


Brickwork (smooth) 


z-oo 


0-007746 


o'6i 








„ (rough) 


2 -co 


0-008845 


0-625 


0-01224 


0-50 


Unplaned plank 


2-00 


0-008451 


0-615 


°03349 


0-50 


Small gravel in cement 


2'00 


0-OII8I 


0-66 


0-03938 


0-60 


Large „ „ 


2-00 


0-OI4I5 


0-705 


0-07590 


I -00 


Hammer-dressed masonry ... 


2-0O 


0-OIII7 


0-66 


0-07825 


I -00 


Earth (no vegetation) 


2 -CO 


0-01536 


0-72 







Rough stony earth 


2-00 


0-02144 


0-78 





— 



If we take x as 0-62, and « = 2, C = 0-0067, wc get — 



Q = 



which reduces to — 



3-01 C VS 



Similarly, for new cast-iron pipes — 



h = . 



320oD''»* 
taking « = 1-85, and x = 0*67. 

These expressions should be compared with the rougher 
ones given on pp. 681, 682. 

Virtual Slope. — If two reservoirs at different levels be 
freely connected by a main through which water is flowing, the 
pressure in the main will diminish from a maximum at the 
upper reservoir to a minimum at the lower, and if glass pipes 
be inserted at intervals in the main, the height of the water in 
each will represent the pressure at the respective points, and 
the difference in height between any two points will represent 
the loss of head due to friction on tiiat section. If a straight 
line be drawn from the surface of the water in the one reservoir 
to that in the other, it will touch the surface of the water in all 
the glass tubes in the case of a main of uniform diameter and 
roughness. The slope of this line is known as the " virtual 



Hydraulics. 



685 



slope " of the main. If the lower end of the main be partially 
closed, it will reduce the virtual slope ; and if it be closed 
altogether, the virtual slope will be nil, or the line will be 
horizontal, and, of course, no water will flow. The velocity 
of flow is proportional to the virtual slope, the tangent of the 

angle of slope is the -^ in the expressions we use for the 
i-t 

friction in pipes. 

The above statement is only strictly true when there is no 

loss of head at entry into the main, and when the main is of 

uniform diameter and roughness throughout, and when there 

are no artificial resistances. When any such irregularities do 

exist, the construction of the virtual slope line offers, as a rule, 

no difficulties, but it is no longer straight. 




Fig. 655. 



In the case of the pipe shown in full lines the resistance is 
uniform throughout, but in the case of the pipe shown in 
broken line, there is a loss at entry a, due to the pipe project- 
ing into the top reservoir ; the virtual slope line is then 
parallel to the upper line until it reaches b, when it drops, due 
to a sudden contraction in the main ; from ^ to ^ its slope is 
steeper than from a to b, on account of the pipe being smaller 
in diameter ; at c there is a drop due to a sudden enlargement 
and contraction, the slope from ^ to 1/ is the same as from b to 
c, and at d there is a drop due to a sudden enlargement, then 
from dXo e the line is parallel to the upper line. The amount 
of the drop at each resistance can be calculated by the methods 
already explained. 

The pressure at every point in the main is proportional to 
the height of the virtual slope line above the main ; hence, if 
the main at any point rises above the virtual slope line, the 
pressure will be negative, i.e. less than atmospheric, or there 
will be a partial vacuum at such a point. If the main rises 
more than 34 feet above the virtual slope line, the water will 



686 Mechanics applied to Engineering. 

break up, and may cause very serious trouble. In waterworks 
mains great pains are taken to keep them below the virtual 
slope line, but if it is impracticable to do so, air-cocks are 
placed at such summits to prevent the pressure falling below 
that of the atmosphere ; the flow is then due to the virtual slope 
between the upper reservoir and this point. In certain cases 
it is better to put an artificial resistance in the shape of a 
pierced diaphragm or a valve on the outlet end of the pipe, in 
order to raise the virtual slope line sufficient to bring it above 
every point of the main, or the same result may be accom- 
plished by using smaller pipes for the lower reaches. 

Flow of Water down an Open Channel on a Steep 
Slope. — 

Let u = the initial velocity of the water in feet per second. 
V = the velocity of the water after running along a 

portion of the channel of length /(feet) measured 

on the slope. 
6 = the angle of the slope to the horizontal. 
H = the vertical fall of the channel in the length / then 

H = / sin e. 
h = the loss of head in feet due to friction while the 

water is flowing along the length /. 
R = the hydraulic mean depth of the channel. 
K = four times the constant in D'Arcy's formula for 

pipes (multiplied by 4 to make it applicable to 

channels, and using the hydraulic mean depth 

instead of the diameter). 

The velocity of a particle of water running down a slope 
is the same as that of a particle falling freely through the 
same vertical height, if there is no friction. When there is 
friction we have— 

V^ = u^ + 2^(H - A) 
The loss of head due to friction is — 



/ (?/' + 2^'-H)KR ^ / {t{' + 2glsine)K R 
V KR-l-2^/ V KR + 2^/ 



Hydraulics. 



687 



This expression is used by calculating in the first place 
the value for V, taking for H some small amount, say 10 feet. 
Then all the quantities under the root are constant, except u, 
hence we may write — 



v. = -v/!^ 



+ n sin 0)in 



m + n 
for the first 10 feet, then the u for the second 10 feet becomes 



-^ 



and for the third 10 feet- 

V. 



(Vi^ + n sin 



_ / (V/ + n sin e)m 



and so on for each succeeding 10 feet. 

The following table shows a comparison between the 
results obtained by Mr. Hill's formula ' and that given above — 

« = 15 feet per second, 

R=i-S3- 

K = 15130 (deduced from Mr. Hill's constant). 
e= 12° 40'. 
H = 10 feet. / = 45 "7 feet. 







Values of V. 


Fall reckoned from 














slope in feet. 




Hill. 


Author. 







15-00 


15-0 


10 




28-34 


27-9 


20 




36-22 


35-6 


30 




41-95 


413 


40 




46-42 


45-9 


5° 




50-03 


49-S 


60 




53'oi 


S2-6 


70 




55-50 


55-2 


80 




57-60 


57:4 


90 




59-39 


59-3 


100 




60-93 


60-9 



* Proceedings Institution of Civil Engineers, vol. clxi., p. 345. 



688 Mechanics applied to Engineering. 

If the slope varies from point to point the angle 5 must be 

taken to suit : similarly, if the hydraulic mean depth varies 

the proper values of R must be inserted in the expression. 

An interesting application of the above theory to the 

formation of ponds at the approach end of culverts will be 

found in a paper by the author, " The Flooding of the Approach 

End of a Culvert," Proc. Inst. Civil Engineers, vol. clxxxvi. 

Flow through a Pipe with a Restricted Outlet. — 

When a pipe is fitted with a valve or nozzle at the outlet end, 

the kinetic energy of the escaping water is usually quite a 

large fraction of the potential energy of the water in the upper 

reservoir ; but in the absence of such a restriction, the kinetic 

energy of the escaping water is quite a negligible quantity in 

the case of long pipes. 

The velocity of the water can be found thus : 

Let H = head of water above the valve in feet ; 

L = length of main in feet ; 

V = velocity of flow in the main in feet per second ; 

K = a constant depending on the roughness of the 

pipe (see p. 68 1) ; 

D = diameter of the pipe in feet j 

Vi = velocity of flow tlu'ough the valve opening ; 

n = the ratio of the valve opening to the area of the 

V 
pipe, or « = =^ . 
»i 

The total energy per pound of water is H foot-lbs. This 
is expended (i.) in overcoming friction, (ii.) in imparting kinetic 
energy to the water issuing from the valve ; or — 

LV» V,^ 

V 
Substitutmg the value of Vj = — , we have — 



H 



KD "*■ zgn^ 

On p. 717 we give some diagrams to show how the 
velocity of the water in the main varies as the valve is closed ; 
in all cases we have neglected the frictional resistance of the 
valve itself, which will vary with the type employed. - In the 
case of a long pipe it will be noticed that the velocity of flow 



Hydraulics. 689 

in the pipe, and consequently the quantity of water flowing, is 
but very slightly affected by a considerable closing of the valve, 
e.g. by closing a fully opened valve on a pipe 1000 feet long 
to o"3 of its full opening, the quantity of water has only been 
reduced to 0*9 of its full flow. But in the case of very short 
pipes the quantity passing varies very nearly in the same 
proportion as the opening of the valve. 

Resistance of Knees, Bends, etc. — ^We have already 
shown that if the direction of a stream of water be abruptly 
changed through a right angle, the whole 
of its energy of motion is destroyed ; a similar 
action occurs in a right-angled knee or elbow 
in a pipe, hence its resistance is at least 
equivalent to the friction in a length of pipe 
about 37 diameters long. In addition to this 
^^^^^ loss, the water overshoots the corner, as shown 
Fig. 656. in Fig- 656, and causes a sudden contraction 
and enlargement of section with a further loss 
of head. The losses in sockets, sudden enlargements, etc., 
can be readily calculated ; others have been obtained by experi- 
ment, and their values are given in the following table. When 
calculating the friction of systems of piping, the equivalent 
lengths as given should be added, and the friction calculated 
as though it were a length of straight pipe. 




2 Y 



690 



Mechanics applied to Engineering. 



Nature of resistance. 



Equivalent length of straight 

pipe expressed in diameters, 

on the basis of L =: 36D, 



15" 
30° 
45° 



ij inch check valve 
ij inch ball check valve 

Sluice and slide valves «= 



Unwin 



Right-angled knee or elbow (experiments) 

Right-angled bends, exclusive of resist- 
ance of socketsi at ends, radius of bend, 
= 4 diameters 

Ditto including sockets (experiments) 

Sockets (screvred) calculated from the! 
sudden enlargement and contraction ( 
(average sizes) 

Ditto by experiment 

Sudden enlargement to a square-ended) 

, large area 

pipe, where n = — =7^ 

' "^ small area 

Sudden contraction 

Mushroom valves 

{handle 
turned 
throue;h 



port area 



area of opening 

„. J J. V area of pipe 

Pierced diaphragm » = .f \ 

area of hole 
Water entering a re-entrant pipe, such as\ 

a Borda's mouthpiece ... ... ...J 

Water entering a square-ended pipe flush) 

with the side of the tank / 



/30-40 in plain pipe 

\ 50-90 with screwed elbow 

3-1 S 
22-30 
24 
16-20 

H-ff 

12 approx. 

120-400 

27 

200 

HOC 

700-1500 

2000-3000 

ioo(« — l)' 
36(1 -Sb - i)» 
18 
9-12 



Velocity of Water in Pipes. — Water is allowed to flow 

at about the velocities given below for the various purposes 
named : — 

Pressure pipes for hydraulic purposes for long mains 3 to 4 feet per sec. 

Ditto for short lengths ' Up to 25 „ 

Ditto through valve passages ' Up to 50 „ 

Pumping mains _ 3 to S «t 

Waterworks mains 2 to 3 „ 



' Such velocities are unfortunately common, but they should be avoided 
if possible. 



CHAPTER XIX. 

HYDRAULIC MOTORS AND MACHINES. 

The work done by raising water from a given datum to a 
receiver at a higher level is recoverable by utilizing it in one 
of three distinct types of motor. 

r. Gravity machines, in which the weight of the water is 
utilized. 

2. Pressure machines, in which the pressure of the water is 
utilized. 

3. Velocity machines, in which the velocity of the water is 
utilized. 

Gravity Machines. — In this type of machine the weight- 
energy of the water is utihzed by causing the water to flow 
into the receivers of the machine at the higher level, then to 
descend with the receivers in either a straight or curved path 
to the lower level at which it is discharged. If W lbs. of water 
have descended through a height H feet, the work done = 
WH foot-lbs. Only a part, however, of this will be utiUzed by 
the motor, for reasons which we will now consider. 





Fig. 637. Fig. 658. 

The illustrations. Figs. 657, 658, show various methods of 



692 



Mechanics applied to Engineering. 



utilizing the weight-energy of water. Those shown in Fig. 657 
are very rarely used, but they serve well to illustrate the 
principle involved. The ordinary overshot wheel shown in 
Fig. 658 will perhaps be the most instructive example to 
investigate as regards efficiency. 

Although we have termed all of these machines gravity 
machines, they are not purely such, for they all derive a small 
portion of their power from the water striking the buckets on 
entry. Later on we shall show that, for motors which utilize 
the velocity of the water, the maximum efficiency occurs when 
the velocity of the jet is twice the velocity of the buckets or 
vanes. 

In the case of an overshot water-wheel, it is necessary to 
keep down the linear velocity of the buckets, otherwise the 
centrifugal force acting on the water will cause much of it to 
be wasted by spilling over the buckets. If we decide that the 
inclination of the surface of the water in the buckets to the 
horizontal shall not exceed 1 in 8, we get the peripheral 
velocity of the wheel V„ = zVRj where R is the radius of the 
wheel in feet. 

Take, for example, a wheel required for a fall of 15 feet. 
The diameter of the wheel may be taken as a first approxima- 
tion as 12 feet. Then the velocity of the rim should not 
exceed 2 v' 6 = say 5 feet per second. Then the velocity of 
the water issuing from the sluice should be 10 feet per 
second ; the head h required to produce this velocity will be 

h = — , or, introducing a coefficient to 

allow for the friction in the sluice, we may 

write It ^= = 1-6 foot. One-half 

of this head, we shall show later, is lost 
by shock. The depth of the shroud is 
usually from 075 to i foot ; the distance 
from the middle of the stream to the c. 
of g. of the water in the bucket may be 
taken at about i foot, which is also a 
source of loss. 

The next source of waste is due to 
the water leaving the wheel before it reaches the bottom. 
The exact position at which it leaves varies with the form 
of buckets adopted, but for our present purpose it may be 
taken that the mean discharge occurs at an angle of 45° as 




FiQ. 659. 



Hydraulic Motors and Machines. 



693' 



shown. Then by measurement from the diagram, or by a 
simple calculation, we see that this loss is o'isD, A clearance 
of about o's foot is usually allowed between the wheel and 
the tail water. We can now find the diameter of the wheel, 
remembering that H = 15 feet, and taking the height from the 
surface of the water to the wheel as 2 feet. This together 
with the 0*5 foot clearance at the bottom gives us D = i2'5 
feet. 

Thus the losses with this wheel are — 

Half the sluice head = o'8 foot 
Drop from centre of stream to buckets = I'o „ 
Water leaving wheel too early,) _ ..„ 
o-is X 12-5 feet ]-^9 » 

Clearance at bottom = o's „ 



4' 2 feet 

15— 4"2 
Hydraulic efficiency of wheel = — -- — = 72 per cent. 

The mechanical efficiency of the axle and one toothed 
wheel will be about 90 per cent., thus giving a total efficiency 
of the wheel of 65 per cent. 

With greater falls this efficiency can be raised to 80 per cent. 

The above calculations do not profess to be a complete 
treatment of the overshot wheel, but they fairly indicate the 
sort of losses such wheels are liable to. 

The loss due to the water leaving too early can be largely 
avoided by arranging the wheel as shown in Fig. 660. 




Fig. 660. 



Fig. 661. 



Pressure Machines. — In these machines the water at the 
higher level descends by a pipe to the lower level, from whence 
it passes to a closed vessel or a cylinder, and acts on a movable 



694 



Mechanics applied to Engineering. 



piston in precisely the same manner as in a steam-engine. 
The work done is the same as before, viz. WH foot-lbs. for 



f 


ft .0 


^_____^ 


n — — 





- 


V 







n 




, \ 









^ 








Fig. 662, 



the pressure at the lower level is W„H lbs. per square foot j and 
the weight of water used per square foot of piston = W„L = W, 







I 



Fig. 663. 



where L is the distance moved through by the piston in feet. 
Then the work done by the pressure water = W„LH = WH 
foot-lbs. Several examples of pressure machines are shown 
in Figs. 661, 662, 663, a and b. Fig. 661 is an oscillating 
cylinder pressure motor used largely on the continent. Fig. 

662 is an ordinary hydraulic pressure riveter. Fig. 663 (a) is a 
passenger lift, with a wire-rope multiplying arrangement. Fig. 

663 {b) is an ordinary ram lift. For details the reader is referred 



Hydraulic Motors and Machines. 



69s 



to special books on hydraulic machines, such as Blaine ' or 
Robinson.'' 

The chief sources of loss in efficiency in these motors are — 

1. Friction of the water in the mains and passages. 

2. Losses by shock through abrupt changes in velocity of 
water. 

3. Friction of mechanism. 

4. Waste of water due to the same quantity being used when 
running under light loads as when running with the full load. 

The friction and shock losses may be reduced to a minimum 
by careful attention to the design of the ports and passages ; 
re-entrant angles, abrupt changes of section of ports and 
passages, high velocities of flow, and other sources of loss 
given in the chapter on hydraulics should be carefully avoided. 

By far the most serious loss in most motors of this type is 
that mentioned in No. 4 above. Many very ingenious devices 
have been tried with the object of overcoming this loss. 

Amongst the most promising of those tried are devices 
for automatically regulating the length of the stroke in pro- 
portion to the resistance overcome by the motor. Perhaps 
the best known of these devices is that of the Hastie engine, 
a full description of which will be found in Professor Un win's 
article on Hydromechanics in the " Encyclopaedia Britannica." 

In an experiment on this engine, the following results were 
obtained : — - 



Weight in pounds lifted \ 
22 feet / 


Jchain'l 
I only J 


427 


633 


745 


857 


969 


1081 


"93 


Water used in gallons at \ 
80 lbs. per square incH / 


7-5 


10 


14 


16 


17 


20 


21 


22 


Efficiency per cent, (actual) 


— 


,Si 


S4 


SO 


60 


S» 


61 


feS 


Efficiency per cent, if stroke \ 
were of fixed length ... / 


— 


23 


34 


40 


46 


53 


59 


65 



The efficiency in lines 3 and 4 has been deduced from 
the other figures by the author, on the assumption that the 
motor was working full stroke at the highest load given. 

The great increase in the efficiency at low loads due to 
the compensating gear is very clear. 

Cranes and elevators are often fitted with two cylinders of 
diflferent sizes, or one cylinder and a differential piston. When 
lightly loaded, the smaller cylinder is used, and the larger one 

' " Hydraulic Machinery " (Spon). 

' " Hydraulic Power and Machinery " (Griffin), 



696 



Mechanics applied to Engineeritig. 



only for full loads. The valves for changing over the con- 
ditions are usually worked by hand, but it is very often found 
that the man in charge does not take advantage of the smallei 
cylinder. In order to place it beyond his control, the ex- 
tremely ingenious device shown in Fig. 664 is sometimes used. 




FULL P/fESSUKE c 



The author is indebted to Mr. R. H. Thorp, of New York, 
the inventor, for the drawings and particulars from which the 
following account is taken. The working cylinder is shown at 
AB. 'Wlien working at full power, the valve D is in the 
position shown in full lines, which allows the water from B to 
escape freely by means of the exhaust pipes E and K ; then 
the quantity of water used is given by the volume A. But when 
working at half-power, the valve D is in the position shown in 
dotted lines ; the water in B then returns vid the pipe E, the 
valve D, and the pipe F to the A side of the piston. Under 
such conditions it will be seen that the quantity of high-pressure 
water used is the volume A minus the volume B, which is 
usually one-half of the former quantity. The position of the 
valve D, which determines the conditions of full or half power, 
is generally controlled by hand. The action of the automatic 
device shown depends upon the fact that the pressure of the 
water in the cylinder is proportional to the load lifted, for if the 
pressure were in excess of that required to steadily raise a light 



Hydraulic Motors and Machines. 



697 



load, the piston would be accelerated, and the pressure would 
be reduced, due to the high velocity in the ports. In general, 
the man in charge of the crane throttles the water at the inlet 
valve in order to prevent any such acceleration. In Mr. 
Thorp's arrangement, the valve D is worked automatically. 

In the position shown, the crane is working at full power ; 
but if the crane be only lightly loaded, the piston will be 
accelerated and the pressure of the water will be reduced by 
friction in passing through the pipe C, until the total pressure 
on the plunger H will be less than the total full water-pressure 
on the plunger G, with the result that the valve D will be forced 
over to the right, thus establishing communication between 
B and A, through the pipes E and F, and thereby putting the 
crane at half-power. As soon as the pressure is raised in A, 
the valve D returns to its full-power position, due to the area 
of H being greater than that of G, and to the pendulum 
weight W. 

It very rarely happens that a natural supply of high-pressure 
water can be obtained, conse- 
quently a power-driven pump has 
to be resorted to as a means of 
raising the water to a sufSciently 
high pressure. In certain simple 
operations the water may be 
used direct from the pump, but 
nearly always some method of 
storing the power is necessary. 
If a tank could be conveniently 
placed at a sufficient height, the 
pump might be arranged to 
deliver into it, from whence the 
hydraulic installation would draw 
its supply of high-pressure water. 
In the absence of such a con- 
venience, which, however, is 
seldom met with, a hydraulic 
accumulator (Fig. 665) is used. 
It consists essentially of a vertical 
cylinder, provided with a long- 
stroke plunger, which is weighted 
to give the required pressure, 
Fig. 66s.* usually from 700 to 1000 lbs. per 

square inch. With such a means 
of storing energy, a very large amount of power — ^far in excess 




698 Mechanics applied to Engineering. 

of that of the pump — may be obtained for short periods. In 
fact, this is one of the greatest points in favour of hydraulic 
methods of transmitting power. The levers shown at the side 
are for the purpose of automatically stopping and starting the 
pumps when the accumulator weights get to the top or bottom 
of the stroke. 

Energy stored in an Accumulator. — 

\l s = the stroke of the accumulator in feet j 
d = the diameter of the ram in inches ; 
fi = the pressure in pounds per square inch. 

Then the work stored in foot-lbs. = o-']2>^cPps 
Work stored per cubic foot of water in 1 , , . 

foot-lbs. [ = '44/ 

Work stored per gallon of water = -^r; — = 23'o4^ 

Number of gallons required per minute \ _ 33»°°p _ 143' 

at the pressure / per horse-power | ~ 23'o4/ "" p 

Number of cubic feet required per minute 1 _ 33>°°° _ 229-2 

at the pressure/ l ~ 144/ ~ / 



Effects of Inertia of Water in Pressure Systems. — 

In nearly all pressure motors and machines, the inertia of the 

water seriously modifies the pressures actually obtained in the 

cylinders and mains. For this reason such machines have to 

be run at comparatively low piston speeds, seldom exceeding 

100 feet per minute. In the case of free piston machines, such 

as hydraulic riveters, the pressure on the rivet due to this cause 

is frequently twice as great as would be given by the steady 

accumulator pressure. 

In the case of a water-pressure motor, the water in the 

mains moves along with the piston, and may be regarded as a 

part of the reciprocating parts. The pressure set up in the 

pipes, due to bringing it to rest, may be arrived at in the same 

manner as the " Inertia pressure," discussed in Chapter VI. 

Let w = weight of a column of water 1 square inch in 

section, whose length L in feet is that of the 

main along which the water is flowing to the 

motor = o"434L ; 

area of plunger or piston 

m = the ratio p -; ? — : -. — 

area of section of water main 



Hydraulic Motors and Machines. 699 

/ = the pressure in pounds per square inch set up in 
the pipe, due to bringing the water to rest at the 
end of the stroke (with no air-vessel) ; 
N = the number of revolutions per minute of the 

motor ; 
R = the radius of the crank in feet. 
Then, remembering that the pressure varies directly as the 
velocity of the moving masses, we have, from pp. 182, 187 — 

p = o"ooo34»2(o'434L)RN^ ( i ± - ] 

p = o"ooor5«LRN^( i ± - ) 

Relief valves are frequently placed on long lines of piping, 
in order to relieve any dangerous pressure that may be set 
up by this cause. 

Pressure due to Shock. — If water flows along a long 
pipe with a velocity V feet per second, and a valve at the 
outlet end is suddenly closed, the kinetic energy of the water 
will be expended in compressing the water and in stretching the 
walls of the pipe. If the water and the pipe were both 
materials of an unyielding character, the whole of the water 
would be instantly brought to rest, and the pressure set up 
would be infinitely great. Both the water and the pipe, how- 
ever, do yield considerably under pressure. Hence, even after 
the valve is closed, water continues to enter at the inlet end 
with undiminished velocity for a period of i seconds, until the 
whole of the water in the pipe is compressed, thus producing a 
momentary pressure greater than the static pressure of the 
water. The compressed water then expands, and the distended 
pipe contracts, thus setting up a return-wave, and thereby 
causing the water-pressure to fall below the static pressure. 
Let K = the modulus of elasticity of bulk of water 

= 300,000 lbs. per square inch (see p. 405) ; 
X = the amount the column of water is shortened, 

due to the compression of the water and to 

the distention of the pipe, in feet ; 
/ = the compressive stress or pressure in pounds per 

square inch due to shock ; 
w = the weight of a unit column of water, i.e. i sq. 

inch section, i foot long, = 0-434 lb. ; 
L = the length of the column of flowing water in 

feet: 



700 Mechanics applied to Engineering. 

d = the diameter of the pipe in inches j 
T = the thickness of the pipe in inches ; 
/, = the tensile stress in the pipe (considered thin) 

due to the increased internal pressure/; 
E = Young's modulus of elasticity for the pipe 

material. 

Then/, = ^ 

2 1 

The increase in diameter dae to thel _ fd^ 

increased pressure I 2TE 

tip ltd 
The increase in cross-section = =^=rp X — 

2TE 2 

The increase in volume of the pipe perl _ \jfd 
square inch of cross-section ) TE 

Let a portion of the pipe in question be represented by 
Fig. 666. Consider a plane section of the pipe, ab, distant L 
from the valve at the instant the valve is suddenly closed. On 

account of the yielding of 
the pipe and the compres- 
sion of the water, the 
plane ab still continues 
to move forward until the 
spring of the water and 
the pipe is a maximum, i.e. when the position dV is reached, 
let the distance between them bearj then, due to the elastic 
compression of the water, the plane ab moves forward by an 

amount x^= ~ (see p. 374), and a further amount due to the 
distention of the pipe of a;, = -,==-, hence— 

X ill 



a, a' 


VtOve 


^ 


■X' 




S^ L 

Fig. 666. 


—X 


' 



*=-^(^+te) 



and/ = 



X 



Ki+A) 



But since x is proportional to L in an elastic medium, the 
pressure/ is therefore independent of the length of the pipe. 

At the instant of closing the valve the pressure in the 
immediate neighbourhood rises above the static pressure by 
an amount / and a wave of pressure starting at the valve is 
transmitted along the pipe until it reaches the open end, the 
velocity of which V„ is constant. The time 4 taken by the 



Hydraulic Motors and Machines. 701 

wave in traversing a distance - is -^r; and the distance x^ 

n «V„ 

which the plane ab traverses in this time is - , but 

n 



*=Xe+^]4X=^^- 



Xn X niL _ \, 
Hence — = - = ■—- = mS[, 

Thus the velocity with which the plane ab travels is constant. 
Let the velocity of the water at the instant of closing the valve 
be V, and since the velocity of ab is constant the water con- 
tinues to enter the pipe at the velocity V for a period of t 
seconds after closing the valve, i.e. until the pressure wave 

reaches the open end of the pipe, hence V = - 

The change of mo-) _ mass of the change of velocity in 
mentum in the time t )~ water ^ the time t 

ft _ °'434L X 
Substituting the value of x, we have — 



L_ / 



\/«'(^+T^) 



and when the elasticity of the pipe is neglected— 



^V^ 



The quantity — is the velocity with which the compression 

wave traverses the pipe, pr the velocity of pulsation. Inserting 
numerical values for the symbols under the root, we get the 
velocity of pulsation 4720 feet per second, i.e. the velocity of 
sound in water when the elasticity of the pipe is neglected. 
The kinetic energy of the column of| _ o-434LV^ 
water per sq. inch of section 5 '^ 

The work done in compressing the water) _f{x„ + x^ 
and in the distention of the pipe j ~ " 2 



2 Vk ^ TE/ 



702 



Mechanics applied to Engineering. 

2g 2 VK TE/ 

and/= ■ 






^■^'V K ■ d 
When the elasticity of the pipe is neglected — 
/i = 63-5V 

A comparison between calculated and experimental results 
are given below. The experimental values are taken from 
Gibson's "Hydraulics and its Applications," p. 217. 



Sudden Closing of Valve. 



Velocity in feet per second 


0-6 


20 


3'o 


7-S 


Observed pressure lb. sq. inch 


43 


"3 


173 


426 


Calculated/ = 63-41; .... 


38 


127 


190 


476 


Calculated allowing for elasticityl 
of pipe / 


35 


116 


17s 


436 



When the valve is closed uniformly in a given time, the 
manner in which the pressure varies at each instant can be 
readily obtained by constructing (i.) a velocity-time curve; 
(ii.) a retardation or pressure curve, as explained on p. 140. 
But the pressure set up cannot exceed that due to a suddenly 
closed valve, although it may closely approach it. 

When the pressure wave reaches the open end of the pipe, 

the whole column of water is under compression to its full 

extent, it then expands, and when it reaches its unstrained 

volume the water at the open end is travelling outwards with 

a velocity V (very nearly, there is a small reduction due to 

molecular friction) and overshoots the mark, thus producing 

a negative pressure, i.e. below the static pressure in the pipe, 

to be followed by a pressure wave and so on. The time during 

which the initial pressure is maintained is therefore the time 

taken by a compression wave in traversing the pipe and 

2L 
returning, viz. — seconds. Hence, if the time occupied in 



Hydraulic Motors and Machines. 703 

closing the valve is not greater than this, the pressure set up 
at the instant of closing will be approximately that given by 
the above expression for f, but the length of time during 
which the pressure is maintained at its full amount will be 
correspondingly reduced. If the period of closing be greater 

than — seconds, the pressure set up at the instant of closing 

will be less than/, but a rigid solution of the problem then 
becomes somewhat complex. 

Maximum Power transmitted by a Water-Main. — 
We showed on p. 580 that the quantity of water that can be 
passed through a pipe with a given loss of head is — 



Q = 38-50^ \/^ 



Each cubic foot of water falling per second through a height of 
one foot gives — • 

62-5 X 60 

= 0*1135 horse-power 

33,ooo_ •'^ ^ 

henceH.P. = o-ii3sQH 

where H is the fall in feet, and Q the quantity of water in cubic 
feet per second. 

Then, if h be the loss of head due to friction, the horse- 
power delivered at the far end of the main L feet away is — 



H.P. = o-ii35Q(H--4) 
Substituting the value of Q from above, we have — 

ap; = 0II3S X 38-50^ (H - h)\/ J 

Let h = «H. Then, by substitution and reduction, we get 
the power delivered at the far end — 



, , /«H»I 

H.P. = 4-37 (i - «) V -17 

ig-inWB^i - «)' 



H.P.'' 
These equations give us the horse-power that can be 



704 Mechanics applied to Engineering. 

transmitted with any given fraction of the head lost in friction j 
also the permissible length of main for any given loss when 
transmitting a certain amount of power. 

The power that can be transmitted through a pipe depends 
on (i.) the quantity of water that can be passed ; (ii.) the effective 
head, i.e. the total head minus the friction head. 

Power transmitted P = Q(H — ^) X a constant 

or P = AV( H - j^ I X a constant 



(ALV \ 
AHV ^ 1 X a constant 
2400D/ 



2401 
Then— 



^=(^H-f^X a constant 



When the power is a maximum this becomes zero ; then — 

LV H 
240oD ~ 3 

or « = — 
3 

hence the maximum power is transmitted when \ of the head 
is wasted in friction. 

Those not familiar with the differential method can arrive 
at the same result by calculating out several values of V — V^, 
until a maximum is found. 

Whence the maximum horse-power that can be transmitted 
through any given pipe is — 



H.P. {max) = 1-67 isj 5!£' 



obtained by inserting n =\va. the equation above. 

N.B. — H, D, and L are all expressed in feet 

Velocity MacMnes.— In these machines, the water, 
having descended from the higher to the lower level by a pipe, 
is allowed to flow freely and to acquire velocity due to 
its head. The whole of its energy then exists as energy of 
motion. The energy is utilized by causing the water to 
impinge on moving vanes, which change its direction of flow, 
and more or less reduce its velocity. If it left the vanes with 



Hydraulic Motors and Machines. 



705 



110 velocity relative to the earth, the whole of the energy would 
be utilized, a condition of affairs which is never attained in 
practice. 

The velocity with which the water issues, apart from 
friction, is given by — 

V= ./JgR 

where H is the head of water above the outlet. When friction 
is taken into account — 



V = V2^(H - h) 

where h is the head lost in friction. 

Relative and Absolute Velocities of Streams. — We 
shall always use the term " absolute velocity," as the velocity 
relative to the earth. 

Let the tank shown in Fig. 667 be mounted on wheels, or 
otherwise arranged so that it can be moved along horizontally 




Ftr. 667. 



at a velocity V in the direction indicated by the arrow, and let 
water issue from the various nozzles as shown. In every case 
let the water issue from the tank with a velocity v at an angle 6 
with the direction of motion of the tank ; then we have — 



Nozzle. 


Velocity rel. 
to un1< V. 


Velocity rel. to ground Vo. 


». 


Cos*. 


A 
B 
C 


V 
V 
V 
V 


\ -V 



180° 
90° 

e 


1 

— I 




D 


^V« + w' + 2vV cos 9 


cos 



2 Z 



7o6 Mechanics applied to Engineering. 

The velocity Vo will be clear from the diagram. The expression 
for D is arrived at thus : 

y = ab cos 6 = v . cos 6 
and X = V .sm 6 

Vo = ^(V+^f + *= 

Substituting the values of x and y, and remembering that 
cos' + sin° 6 = 1, we get the expression given above. If the 
value of cos 6 for A, B, and C be inserted in the general 
expression D, the same results will be obtained as those given. 

Now, suppose a jet of water to be moving, as shown by the 
arrow, with a velocity V„ relative to the ground ; also the tank 
to be moving with a velocity V relative to the ground ; then it 
is obvious that the velocity of the water relatively to the tank 
is given by ab or v. We shall be constantly making use of this 
construction when considering turbines. 

Pressure on a Surface due to an Impinging Jet.— 
When a body of mass M, moving with a velocity V, receives 
an impulse due t-o a force P for a space of time /, the velocity 
will be increased to Vj, and the energy of motion of the body 
will also be increased ; but, as no other force has acted on the 
body during the interval, this increase of energy must be equal 
to the work expended on the body, or^ 

The work done on") . [distance through which 

the body I = "»P"'«^ X \ it is exerted 

= increase in kinetic energy 
The kinetic energy ~l MV 

before the impulse j ~ ^ 
The kinetic energy 1 MVi" 

after the impulse J ~ ~2 
Increase in kinetic 1 M 

energy J ~ T '^ ~ * ) 

The distance through which the impulse is exerted is — 

V, + V , 



2 



M V, 4- V 

hence ^(V,» - V) = P/ ' ^ 

or P^ = M(V, - V) 
or impulse in time / = change of momentum in time / 



Hydraulic Motors and Machines. 



707 




Let a jet of water moving with a velocity V feet per second 
impinge on a plate, as shown. After 
impinging, its velocity in its original direc- 
tion is zero, hence its change of velocity 
on striking is V, and therefore — 

Yt = MV 

orP=-V 

M 
But -T is the mass of water delivered per fig. 668. 

second. 

Let W = weight of water delivered per second. 

r^, W M 

Then — = -r 

S i 

, -^ WV 

and P = 

g 

For another method of arriving at the same result, see 

P- 593- 

It should be noticed that the pressure due to an impingmg 
jet is just twice as great as the pressure due to the head of 
water corresponding to the same velocity. This can be shown 
thus : 

. = ^^ 

p = wh = 

where w = the weight of a unit column of water. 

We have W = o/V. Substituting this value of wV — 



/ = 



WV 
2^ 



The impinging jet corresponds to a dynamic load, and a 
column of water to a steady load (see p. 627). 

In this connection it is interesting to note that, in the case 
of a sea-wave, the pressure due to a wave of oscillation is 
approximately equal to that of a head of water of the same 
height as the wave, and, in the case of a wave of translation, to 
twice that amount. 



7o8 



Mechanics applied to Engineering. 



Pressure on a Moving Surface due to an Imping- 
ing Jet. — Let the plate shown in the Fig. 669 be one of a 
series on which the jet impinges at very 
short intervals. The reason for making 
this stipulation will be seen shortly. 

Let the weight of water delivered per 
second be W lbs. as before ; then, if the 
plates succeed one another very rapidly as 
in many types of water-wheels, the quantitj- 
impinging on the plates will also be sensi- 
bly equal to W. The impinging velocity 




Fig. 669. 



is V 



— , or 



V f I — - J ; hence the pressure in pounds' 
weight on the plates is — 



WV 



P = 



And the work done per second on the plates in foot-lbs. — 

wv<i-J) 



n 



(i.) 



and the energy of the jet is — 



WV' 



(ii.) 



i. 2( \\ 

hence the efficiency of the jet= Tr = ^l^i ~«> 

The value of « for maximum efficiency can be obtained by 
plotting or by differentiation.* It will be found that n = 2. 
The efficiency is then 50 per cent., which is the highest that 
can be obtained with a jet impinging on flat vanes. A common 
example of a motor working in this manner is the ordinary 

' Efficiency = t) = — ( i j 

22 _, _ 

1) = — = 2n ' — 2»^ 



d-n 



= — 2tt~' + 4« ' = o, when t) is a maximum 



dn 
or 2H~' = 4«~' 

-i = 4, whence « = 2 



Hydraulic Motors and Machines. 



709 



undershot water-wheel ; but, due to leakage past the floats, axle 
friction, etc., the efficiency is rarely over 30 per cent. 

If the jet had been impinging on only one plate instead of 
a large number, the quantity of water that reached the plate 

per second would only have been W f i — - ) > then, sub- 
stituting this value for W in the equation above, it will be seen 

2 ( I \ 2 
that the efficiency of the jet = - I i I , and the maximum 

efficiency occurs when « = 3, and is equal to about 30 per cent. 

Pressure on an Oblique 
Surface due to an Impinging 
Jet. — The jet impinges obliquely 
at an angle Q to the plate, and splits 
up into two streams. The velocity V 
may be resolved into Vj normal and 
V„ parallel to the plate. After im- 

• ■ ..u » T. 1 V Fig. 670. 

pingmg, the water has no velocity 

normal to the plate, therefore the normal pressure — 

WVi WV sin Q 
~ g ~ g 
Pressure on a Smooth Curved Surface due to an 





Fig. 671. 



Impinging Jet. — We will first consider the case in which tjie 
surface is stationary and the water slides on it without shock ; 



•JIO 



Mechanics applied to Engineering. 



how to secure this latter condition we will consider shortly. 
We show three forms of surface (Fig. 671), to all of which the 
following reasoning applies. 

Draw ab to represent the initial velocity V of the jet in 
magnitude and direction ; then, neglecting friction, the final 
velocity of the water on leaving the surface will be V, and its 
direction will be tangential to the last tip of the surface. Draw 
ac parallel to the final direction and equal to ab, then be repre- 
sents the change of velocity Vj ; hence the resultant pressure 
on the surface in the direction of cb is — 



P = 



WVi 
g 



Then, reproducing the diagram of velocities above, we 
have — 

_j; = V sin e 
a; = V cos B 
y,^=(V -xf+f 

Then, substituting the values of x and _v 
and reducing, we have — 




Fig. 672. 



V, = Vv'2(i -cose 



The component parallel to the jet is V - .»: = V(i - cos 0). 
Thus in all the three cases given above we have the pressure 
parallel to the jet — 

_ WV(i - cos e) 

^0 — 



' The effect of friction is to reduce the length ac (Fig. 671), hence 
when 8 is less than 90°, the pressure is greater, and when fl is greater than 
90°, it is less than Po. The following results were obtained in the 
author's laboratory. The calculated value being taken as unity. 



Po by experiment 



Cone (45°). 


Hollowed 
cone (55°). 


Flat. 


1 
Approx. 
hemisphere 


I '4 


I -2 


i-o 


07 

t 



Pel ton 
bucket 

(162°). 



08 



Hydraulic Motors and Machines. 



711 



1 



■a 






£ 

i 



•Ss 




^'-' ^ 


<b 


w " 




c 


> 


iVl 


■K 


^ 




> 




1 


^ 




0--.S O 



>l« 



> 8 



1> 

o 



> « -i 









ilh 










Hydratilic Motors and Machines. 



713 



Pelton or Tangent Wheel Vanes. — The double vane 
shown in section is usually known as the Pelton Wheel 
Vane; but whether Pelton should have the credit of the in- 
vention or not is a disputed point. In this type of vane the 
angle B approaches 180°, then i — cos 6 = 2, and the resultant 
pressure on such a vane is twice as great as that on a flat vane, 
and the theoretical efficiency is 100 per cent, when n= 2 ; 
but for various reasons such an efficiency is never reached, 
although it sometimes exceeds 80 per cent., . including the 
friction of the axle. 

A general view of such a wheel is shown in Fig. 673. 




Fig. 673. 



It is very instructive to examine the action of the jet of 
water on the vanes in wheels of this type, and thereby to 
see why the theoretical efficiency is never reached; 

(1) There is always some loss of head in the nozzle itself; 
but this may be reduced to an exceedingly small amount by 
carefully proportioning the internal curves of the nozzle. 

(2) The vanes are usually designed to give the best effect 
when the jet plays fairly in the centre of the vanes; but in 
other positions the effect is often very poor, and, consequently, 
as each vane enters and leaves the jet, serious losses by shock 
very frequently occur. In order to avoid the loss at entry, 
Mr. Doble, of San Francisco, after a very careful study of the 
matter, has' shown that the shape of vane as usually used is 

' Reproduced by the kind permission of Messrs. Gilbert Gilkes and 
Co., Kendal. 



714 



Mechanics applied to Engineering. 



very faulty, since the water after striking the outer lip is 
abruptly changed in direction at the corners a and b, where 
much of its energy is dissipated in eddying ; then, further, on 




Fig 674. 



leaving the vane it strikes the back of the approaching vane, 
and thereby produces a back pressure on the wheel with a 









Fig. 675.— Doblo " Tangent Wheel " buckets. 

consequent loss in efficiency. This action is shown irf Fig. 
674. The outer lip is not only unnecessary, but is distinctly 
wrong in theory and practice. In the Doble vane (Fig, 675) 



Hydraulic Motors and Machines. 



715 



the outer lip is dispensed with, and only the central rib retained 
for parting the water sideways, with the result that the efficiency 
of the Doble wheel is materially higher than that obtained 
from wheels made in the usual form. 

(3) The angle B cannot practically be made so great as 
180°, because the water on leaving the sides of the vanes 
would strike the back edges of the vanes which immediately 
follow ; hence for clearance purposes this angle must be made 
somewhat less than 180°, with a corresponding loss in efficiency. 

(4) Some of the energy of the jet is wasted in overcoming 
the friction of the axle. 

In an actual wheel the maximum efficiency dQes not occur 




1-0 2-0 3-0 

Matio of Jet to ivheel velocity 

Fig. 676. 

when the velocity of the jet is twice that of the vanes, but 
when the ratio is about 2 ■2. 

The curve shown in Fig. 676 shows how the efficiency 
varies with a variation in speed ratio. The results were 
obtained from a small Pelton or tangent wheel in the author's 
laboratory; the available water pressure is about 30 lbs. per 
square inch. Probably much better results would be obtained 
with a higher water pressure. 

This form of wheel possesses so many great advantages 
over the ordinary type of impulse turbine that it is rapidly 
coming into very general use for driving electrical and other 
installations ; hence the question of accurately governing it is 
one of great importance. In cases in which a waste of water 
is immaterial, excellent results with small wheels can be 



7i6 



Mechanics applied to Engineering. 



obtained by the Cassel governor, in which the two halves of 
the vanes are mounted on separate wheels. When the wheel 
is working at its full power the two halves are kept together, 
and thus form an ordinary Pelton wheel ; when, however, the 
speed increases, the governor causes the two wheels to partially 
separate, and thus allows some of the water to escape between 
the central rib of the vanes. For much larger wheels Doble 
obtains the same result by affixing the jet nozzle to the end of 
a pivoted pipe in such a manner that the jet plays centrally on 
the vanes for full power, and when the speed increases, the 
governor deflects the nozzle to such an extent that the jet 
partially or fully misses the tips of the vanes, and so allows 
some of the water to escape without performing any work on 
the wheel. 

But by far the most elegant and satisfactory device for 
regulating motors of this type is the conical expanding nozzle, 
which effects the desired regulation without allowing any waste 
of water. The nozzle is fitted with an internal cone of special 
construction, which can be advanced or withdrawn, and thereby 
it reduces or enlarges the area of the annular stream of water. 
Many have attempted to use a similar device, but have failed 
to get the jet to perfectly coalesce after it leaves the point of 
the cone. The cone in the Doble ' arrangement is balanced 
as regards shifting along the axis of the nozzle ; therefore the 
governor only has to overcome a very small resistance in 
altering the area of the jet. Many other devices have been 
tried for varying the area of the jet in order to produce the 
desired regulation of speed, but not always with marked 
success. Another method in common use for governing and 
for regulating the power supplied to large wheels of this type 
is to employ several jets, any number of which can be brought 
to play on the vanes at will, but the arrangement is not 
altogether satisfactory, as the efficiency of the wheel decreases 
materially as the number of jets increases. In some tests 
made in California the following results were obtained : — 



Number of jets. 


Total horse-power. 


Horse-power per jet. 


2 

3 
4 


390 
480 


'55 

130 

los 

go 



' A similar device is used by Messrs. Gilbert Gilkes & Co., Kendal. 



Hydraulic Motms and Machines 



717 



The problem of governing water-wheels of this type, even 
when a perfect expanding nozzle can be produced, is one of 
considerable difficulty, and those who have experimented upon 
such motors have often obtained curious results which have 
greatly puzzled them. The theoretical treatment which follows 
is believed to throw much light on many hitherto unexplained 
phenojnena, such as (i.) It has frequently been noticed that 
the speed of a water motor decreases when the area of the jet 
is increased, the head of water, and the load on the motor, 




01 0-2 0-3 O* 0-5 0-6 0-7 0-8 0-9 10 
c/osed. Satio of Valv» opening to Aretv ofPipe=N. Fu/t op en. 

Fig. 677. 



remaining the same, and via versd, when the area of the jet is 
decreased the speed increases. If the area of the jet is regu- 
lated by means of a governor, the motor under such circum- 
stances will hunt in a most extraordinary manner, and the 
governor itself is blamed ; but, generally speaking, the fault is 
not in the governor at all, but in the proportions of the pipe 
and jet. (ii.) A governor which controls the speed admirably 
in the case of a given water motor when working under certain 
conditions, may entirely fail in the case of a similar water 
motor when tht; conditions are only slightly altered, such as ar? 
alteration in the length or diameter of the supply pipe. 



7i8 Mechanics applied to Engineering. 

On p. 688 we showed that the velocity (V) of flow at any 
instant in a pipe is given by the expression — 



V 



/IT 



L . I 
KD + ign'' 



In Fig. 677 we give a series of curves to show the manner 

in which V varies with the ratio of the area of the jet to the 

area of the cross-section of the pipe, viz. n. From these curves 

it will be seen that the velocity of flow falls off very slowly at 

first, as the area of the jet is diminished, and afterwards, as the 

" shut " position of the nozzle is approached, the velocity falls 

V 
very rapidly. The velocity of efflux V, = — of the jet itself is 

also shown by full lined curves. 

The quantity of water passing any cross-section of the main 
per second, or through the nozzle, is AV cubic feet per second, 
or 62-4AV lbs. per second. 

The kinetic energy of the stream issuing from the nozzle 
is — 

62-4AV 
2gn^ 



Inserting the value of V, and reducing, we get — 



^, , • • r , O76DV H \3 

The kinetic energy of the stream = 2 — / j I2 

VkD "*" 2P« V 



Which may be written — 



K. = ^i — :— r where B = g 



^ n 
C 1 



~ («5)i/ ^ A3 - (B«J -j- n-\)\ 
= C(B«J 4- n-l) -\ 
5" = C { - f(B«^ + n-\)-\^nl - %n-\)\ 



Hydraulic Motors and Machines. 

which may be written — 

C{-f«S(B«^ + i)-^ X |«-S(2B«^ - i)} 
and C{-(B«''+i)-^X(2B«^-i)} =o 

when it has its maximum value, but (B«^ + i) is greater 
unity, hence 2B«^ — i = o 

, 1 KD 
2B 4^L 



719 



than 



and n = sj"^ = 4-4\/t 



taking an average value for K. 



4^L 




01 02 03 O-l- 0-5 0-6 0-7 0-8 09 {o"'^'^' 
e/tse<f. Matio of Valve opening to Area, of Pipe^N, , Fulfopen, 

Fig. 678. 



720 Mechanics applied to Engineering. 

Curves showing how the kinetic energy of the stream varies 
with n are given in Fig. 678. Starting from a fully opened 
nozzle, the kinetic energy increases as the area of the jet is 
decreased up to a certain point, where it reaches its maximum 
value, and then it decreases as the area of the jet is further 
decreased. The increase in the kinetic energy, as the area of 
the jet decreases, will account for the curious action mentioned 
above, in which the speed of the motor was found to increase 
when the area of the jet was decreased, and vice versL The 
speed necessarily increases when the kinetic energy increases, 
if the load on the motor remains constant. If, however, the 
area of the jet be small compared with the area of the pipe, 
the kinetic energy varies directly as the area of the jet, or 
nearly so. Such a state is, of course, the only one consistent 
with good governing. 

We have shown above that — 

the kinetic energy is a maximum when n = 4"4'v/ — 

hence for good governing the area of the jet must be less 
than — 



4*4 (area of the cross-section of the pipe)/v/ — 

■45Vt; 



L 

/D"» 
or, 3-. 



L 

Poncelot Water-wheel Vanes. — In this wheel a thin 
stream of water having a velocity V feet per second glides up 

V 
curved vanes having a velocity - in the same direction as the 

stream. The water moves up the vane with the velocity V — — 

n 
relatively to it, and attains a height corresponding to this ve- 
locity, when at its uppermost point it is at rest relatively to the 

vane ; it then falls, attaining the velocity — ( V j, neglecting 

friction ; the negative sign is used because it is in the reverse 
direction to which it entered. Hence, as the water is moving 



Hydraulic Motors and Machines. 



721 



forward with the wheel with a velocity -, and backward 

n 

relatively to the wheel with a velocity - ( V J, the absolute 

velocity of the water on leaving the vanes is — 




Fig. 679. 



-Y_(v-^) = v(?-r) 



hence, when « = 2 the absolute velocity of discharge is zero, or 
all the energy of the stream is utilized. The efficiency may 
also be arrived at thus — 

Let a = the angle between the direction of the entering 
stream and a tangent to the wheel at the point of entry. Then 
the component of V along the tangent is V cos a. 

The pressure exerted in the ) _ 2W (^ y _ _ ^ ^ 
direction of the tangent ' ^f \ 'n ) ' 

The work done per second "1 aWV / _ i \ 

on the plate ]~ gn ^ n f 

And the hydraulic efficiency| _i{ ^q^ a — -^ 
of the wheel J ~ « \ " « / 

3^ 




722 Mechanics applied to Engineering. 

The efficiency of these wheels varies from 65 to 75 per 
cent., including the friction of the axle. 

Form of Vane to prevent Shock. — In order that the 
water may glide gently on to the vanes of any motor, the tangent 

to the entering tip of the vanes 

must be in the same direction 

as the path of the water relative 

to the tip of the wheel ; thus, 

in the figure, if ab represents 

the velocity of the entering 

stream, ad the velocity of the 

Fig. 680. vane, then db represents the 

relative path of the water, and 

the entering tip must be parallel to it. The stream then gently 

glides on the vane without shock. 

Turbines. — Turbines may be conveniently divided into 
two classes : (1) Those in which the whole energy of the water 
is converted into energy of motion in the form of free jets or 
streams which are delivered on to suitably shaped vanes in 
order to reduce the absolute velocity of the water on leaving 
to zero or nearly so. Such a turbine wheel receives its 
impulse from the direct action of impinging jets or streams ; 
and is known as an " impulse " turbine. When the admission 
only takes place over a small portion of the circumference, it is 
known as a "partial admission" turbine. The jets of water 
proceeding from the guide-blades are perfectly free, and after 
impinging on the wheel-vanes the water at once escapes into 
the air above the tail-race. 

(2) Those in which some of the energy is converted into 
pressure energy, and some into energy of motion. The water 
is therefore under pressure in both the guide-blades and in the 
wheel passages, consequently they must always be full, and 
there must always be a pressure in the clearance space between 
the wheel and the guides, which is not the case in impulse 
turbines. Such are known as "reaction" turbines, because 
the wheel derives its impulse from the reaction of the water as 
it leaves the wheel passages. There is often some little 
difficulty in realizing the pressure effects in reaction turbines. 
Probably the best way of making it clear is to refer for one 
moment to the simple reaction wheel shown in Fig. 681, in 
which water runs into the central chamber and is discharged at 
opposite sides by two curved horizontal pipes as shown ; the 
reaction of the jets on the horizontal pipes causes the whole to 
revolve. Now, instead of allowing the central chamber to 



Hydraulic Motors and Machines. 



723 



revolve with the horizontal pipes, we may fix the central chamber, 
as in Fig. 682, and allow the arms only to revolve ; we shall get a 
crude form of a reaction turbine. It will be clear that a water- 
tight joint must be made between the arms and the chamber, 
because there is pressure in the clearance space between. It will 
also be seen that the admission of water must take place over 





FjG. 68i. 



Frc. 6S2. 



the whole circumference, and, further, that the passages must 
always be full of water. A typical case of such a turbine is 
shown in Fig. 689. These turbines may either discharge into 
the air above the tail-water, or the revolving wheel may dis- 
charge into a casing which is fitted with a long suction pipe, 
and a partial vacuum is thereby formed into which the water 
discharges. 

In addition to the above distinctions, turbines are termed 
parallel flow, inward flow, outward flow, and mixed-flow 
turbines, according as the water passes through the wheel 
parallel to the axis, from the circumference inwards towards 
the axis, from the axis outwards towards the circumference, or 
both parallel to the axis and either inwards or outwards. 

We may tabulate the special features of the two forms of 
turbine thus : 



724 Mechanics applied to Engineering. 

Impulse. Reaction. 

All the energy of the water is Some of the energy of the water 

converted into kinetic energy before is converted into kinetic energy, and 

being utilized. some into pressure energy. 

The water impinges on curved The water is under pressure in 

wheel-vanes in free jets or streams, both the guide and wheel passages, 

consequently the wheel passages also in the clearance space ; hence 

must not be filled. the wheel passages are always full. 

The water is discharged freely As the wheel passages are always 

into the atmosphere above the tail- full, it will work equally well when 

water ; hence the turbine must be discharging into the atmosphere or 

at the foot of the fall. into water, i.e. above or below the 

tail-water, or into suction pipes. 
The turbine may be placed 30 feet 
above the foot of fhe fall. 

Water may be admitted on a Water must be admitted on the 

portion or on the whole circum- whole circumference of the wheel, 
ference of the wheel. 

Power easily regulated without Power difficult to regulate with- 

much loss. out loss. 

In any form of turbine, it is quite impossible to so arrange 
it that the water leaves with no velocity, otherwise the wheel 
would not clear itself. From 5 to 8 per cent, of the head 
is often required for this purpose, and is rejected in the tail- 
race. 

Form of Blades for Impulse Turbine. — The form 
of blades required for the guide passages and wheel of a 
turbine are most easily arrived at by a graphical method. The 
main points to be borne in mind are — the water must enter the 
guide and wheel passages without shock. To avoid losses 
through sudden changes of direction, the vanes must be smooth 
easy curves, and the changes of section of the passages gradual 
(for reaction turbines specially). The absolute velocity of the 
water on leaving must be as small as is consistent with making 
the wheel to clear itself. 

For simplicity we shall treat the wheel as being of infinite 
radius, and after designing the blades on that basis we shall, 
by a special construction, bend them round to the required 
radius. In all the diagrams given the water is represented 
as entering the guides in a direction normal to that in which 
the wheel is moving. Let the velocity of the water be 
reduced 6 per cent, by friction in passing over the guide- 
blades, and let 7 per cent, of the head be rejected in the 
tail-race. 

The water enters the guides vertically, hence the first tip 
of the guide-blade must be vertical as shown. In order to 



Hydraulic Motors and Machines. 



725 



find the direction of the final tip, we proceed thus : We have 
decided that the water shall enter the wheel with a velocity of 



.S1&. 




WH£SL 



Fig. 683. 




Fig. 684. 

94 per cent, of that due to the head, since 6 per cent, is lost 
in friction, whence — 

Vo = o-94V'2^H: 
also the velocity of rejection — 



V TOO 



We now set down ab to represent the vertical velocity with 
which the water passes through the turbine wheel, and from 
b we set off be to represent Vq ; then ac gives us the horizontal 
component of the velocity of the water, and = v 0^94^ — 0-27^ 
= o"9^ = Vi : cb gives us the direction in which the water leaves 
the guide-vanes ; hence a tangent drawn to the last tip of the 

' We omit n/2^H to save constant repetition. 



726 Mechanics applied to Engineering. 

guide-vanes must be parallel to cb. We are now able to con- 
struct the guide-vanes, having given the first and last tangents 
by joining them up with a smooth curve as shown. 

Let the velocity of the wheel be one-half the horizontal 

velocity of the entering stream, or V„ = — = 0*45 ; hence the 

horizontal velocity of the water relative to the wheel is also 0*45 . 
Set off dV as before = o'27, and dd horizontal and = 0-45 : we 
get dV representing the velocity of the water relative to the 
vane ; hence, in order that there may be no shock, a tangent 
drawn to the first tip of the wheel-vane must be parallel to 
dV \ but, as we want the water to leave the vanes with no 
absolute horizontal velocity, we must deflect it during its 
passage through the wheel, so that it has a backward velocity 
relative to the wheel of —0-45, and as it moves forward with 
it, the absolute horizontal velocity will be — o'45 -1- 0*45 = o. 
To accomplish this, set off de = 0-45. Then ffe gives us the 
final velocity of the water relative to the wheel; hence the 
tangent to the last tip of the wheel-vane must be parallel to lie. 
Then, joining up the two tangents with a smooth curve. We get 
the required form of vane. 

It will be seen that an infinite number of guides and vanes 
could be put in to satisfy the conditions of the initial and final 
tangents, such as the dotted ones shown. The guides are, foi 
frictionah reasons, usually made as short as is consistent with 
a smooth easy-connecting curve, in order to reduce the surface 
to a minimum. The wheel-vanes should be so arranged that 
the absolute path of the water through the wheel is a smooth 
curve without a sharp bend. 





9.-^^ 


rss 


""---. y 


J'yy Ji •: -K 


^23 


* z 3 - 





Fig. 6S5. 



The water would move along the absolute path K« and 
along the path YJi relative to the wheel if there were no vanes 



Hydraulic Motors and Machines. Jzj 

to deflect it, where hi is the distance moved by the vane while 
the water is traveUing from kto i; but the wheel-vanes deflect 
it through a horizontal distance hg, hence a particle of water 
at g has been deflected through the distance gj by the vanes, 
where gh = ij. The absolute paths of the water corresponding 
to the three vanes, i, 2, 3, are shown in the broken lines bear- 
ing the same numbers. In order to let the water get away 
very freely, and to prevent any possibility of them choking, the 
sides of the wheel-passages are usually provided with venti- 
lation holes, and the wheel is flared out. The efficiency of the 
turbine is readily found thus : 

The whole of the horizontal component of the velocity of 
the water has been imparted to the turbine wheel, hence — 

the work done per pound of) Vi" (o'gV)* 

water \~ 2g ~ 2g 

the energy per pound of the \ _ V^ 

water on entering '~ 2g 

the hydraulic efficiency = ^ = o'g" = 81 per cent. 

The losses assumed in this example are larger than is usual 
m well-designed turbines in which the velocity of rejection V, 
= o'i2e,)J 2gS.; hence a higher hydraulic efficiency than that 
found above may readily be obtained. The total or overall 
efficiency is necessarily lower than the hydraulic efficiency on 
account of the axle friction and other losses. Under the most 
favourable conditions an overall efficiency of 80 per cent, may 
be obtained; but statements as to much higher values than 
this must be regarded with suspicion. 

In some instances an analytical method for obtaining the 
blade angles is more convenient than the graphical. Take the 
case of an outward-flow turbine, and let, say, 5 per cent, of 
the head be wasted in friction when passing over the guide- 
blades, and the velocity of flow through the guides be 
o'i2^n/ 2gYi. Then the last tip of the guide-blades will make 
an angle 61 with a tangent to the outer guide-blade circle, 

where sin ft = = ©•i28, and ft = 7" 22'. The horizontal 

°"97S 
component of the velocity Vi = 0*125 cot 6 = o'gey. 

The circumferential velocity of the inner periphery of the 
wheel "V„ = o"483. The inlet tip of the wheel-vanes makes an 
angle ft with a tangent to the inner periphery of the wheel ; 



72t 



Mechanics applied to Engineering, 



or, to what is the same thing, the outer guide-blade circle, 

where tan e^ = -—J'^ = o'zS9. and 6^ = 14° 31'. 
0403 
Let the velocity of flow through the wheel be reduced to 

o'o8 V^^H at the outer periphery of the wheel, due to widening 

or flaring out the wheel-vanes, and let the outer diameter of 




-Os-- 



Fig. 686. 



the wheel be i'3 times the inner diameter; then the circum- 
ferential velocity of the outer periphery of the wheel will be 
0*483 X i'3 = 0-628, and the outlet tip of the wheel-vanes 
will make an angle B^ with a tangent to the outer periphery of 
o-o8 



the wheel, where tan B^ = 



= 0-127, 3-nd 6t — 7° 16'. 




0-628 

Pressure Turbine. — Before pro- 
ceeding to consider the vanes for a 
pressure turbine, we will briefly look at 
its forerunner, the simple reaction 
wheel. Let the speed of the orifices 
be V ; then, if the water were simply 
left behind as the wheel revolved, the 
velocity of the water relative to the 
orifices would be V, and the head 
required to produce this velocity — 



A = 



V2 



Let ^, 



\v-y 



Fig. 687. 



the height of the surface 
of the water or the head 
above the orifices. 



Then the total head! _ , , , _V2 , , 
producing flow, Hf -'*+^'-^"'"^' 



Hydraulic Motors and Machines. 7^9 

Let V = the relative velocity with which the water leaves 
the orifices. 

Then z'^ = 2^H = V^ + 2g/i^ 

The velocity of the water relative to the ground = » — V. 

If the jet impinged on a plate, the pressure would be 

per pound of water; but the reaction on the orifices is equal 
to this pressure, therefore the reaction — 



R = ^ 



and the work done per second ) ii v — ^(^ ~ ^) 
by the jets in foot-lbs. I ~ ^ 



(i.) 



energy wasted in discharge! (z/ — V)^ ■. 

water per pound in foot-lbs. ) ^ ' ■ • '"•) 

total enerev per 1 ■ , •• » — V* 



energy per 



1' 



1. + u. = 



pound of water f 2g 

i. 2V 

hydraulic efficiency — 



i. 4- ii. ~ w + V 

As the value of V approaches v, the efficiency approaches 
100 per cent., but for various reasons such a high efficiency 





Fig. 689. 

can never be reached. The hydraulic eflSciency may reach 65 
per cent., and the total 60 per cent. The loss is due to the 
water leaving with a velocity of whirl v — Y. In order to 
reduce this loss, Fourneyron, by means of guide-blades, gave 
the water an initial whirl in the opposite direction before it 
entered the wheel, and thus caused the water to leave with 



730 Mechanics applied to Engineering. 

little or no velocity of whirl, and a corresponding increase in 
efBciency. 

The method of arranging such guides is shown in Fig. 688 ; 
they are simply placed in the central chamber of such a wheel 
as that shown in Fig. 682. Sometimes, however, the guides 
are outside the wheel, and sometimes above, according to the 
type of turbine. 

Form of Blades for Pressure Turbine. — As in the 
impulse turbine, let, say, 7 per cent, of the head be rejected in 
the tail-race, and say 13 per cent, is wasted in friction. Then 
we get 20 per cent, wasted, and 80 per cent, utilized. 

Some of the head may be converted into pressure energy, 
and some into kinetic energy; the relation between them is 
optional as far as the efficiency is concerned, but it is con- 
venient to remember that the speed of the turbine increases as 

the ratio ==^ increases ; hence within the limit of the head of 

water at disposal any desired speed of the turbine wheel can 
be obtained, but for practical reasons it is not usual to make 
the above-mentioned ratio greater than i*6. Care must always 
be taken to ensure that the pressure in the clearance space 
between the guides and the wheel is not below that of the 
atmosphere, or air may leak in and interfere with the smooth 
working of the turbine. In this case say one-half is converted 
into pressure energy, and one-half into kinetic energy. If 80 
per cent, of the head be utilized, the corresponding velocity 
will be — 



V=\/ 



H X 80 „ ,__- 
'^ Tnn =o-89^/2^H 



Thus 89 per cent, of the velocity will be utilized. To find 
the corresponding vertical or pressure component V, and the 
horizontal or velocity component Vj, we draw the triangle of 
velocities as shown (Fig. 690), and find that each is 0-63 v'z^j'H. 
We will determine the velocity of the wheel by the principle 
of momentum. Water enters the wheel with a horizontal 
velocity V» = o'63, and leaves with no horizontal momentum. 

V 
Horizontal pressure per pound of water = — lbs. 



useful work done in foot-lbs. per second perl _ V^V, _ 



\- 



pound of water I g 

since we are going to utilize 80 per cent, of the head. 



= 0-8H 



Hydraulic Motors and Machines. 



731 



Hence V„ = 



o%H 



0-63 / 2^H 
V„= o•64^/2^H 



o-63>/2 



We now have all the necessary data to enable us to 
determine the form of the vanes. Set down ab to represent 
the velocity of flow through the wheel, and ac the horizontal 
velocity. Completing the triangle, we find cb = 0-69, which 
gives us the direction in which the water enters the wheel or 
leaves the guides. The guide-blade is then put m by the 
method explained for the impulse wheel. 



^ WHEEL 




Fig. 69s. 



To obtain the form of a wheel-vane. Set off ed to 
represent the horizontal velocity of the wheel, and ef parallel to 
cb to represent the velocity of the water on leaving the guides ; 
then df represents the velocity of the water relative to the 
wheel, which gives us the direction of the tangent to the first 
tip of the wheel-vanes. Then, in order that the water shall 
leave with no horizontal velocity, we must deflect it during its 
passage through the wheel so that it has a backward velocity 
relative to the wheel of — 0-64. Then^ setting down gh = o"2 7, 
and^'= — 0-64, we get hi as the final velocity of the water, which 
gives us the direction of the tangent to the last tip of the vane. 

In Figs. 693, 694, 695 we show the form taken by the 



732 Mechanics applied to Engineering. 

wheel-vanes for various proportions between the pressure and 
velocity energy. 




Fig. 693. 




Ficj. 694. 



099 , 




Fig. 595. 

In the case in which V„ = o, the whole of the energy is 
converted into kinetic energy ; then V^ = 0*89, and — 

0-8P-H 
V = ■ 

V„ = 0-45 Va^H 

Or the velocity of the wheel is one-half the- horizontal 
velocity of the water, as in the impulse wheel. The form of 
blades in this case is precisely the same as in Fig. 683, but 
they are arrived at in a slightly different manner. 

For the sake of clearness all these diagrams are drawn with 



Hydraulic Motors and Machines. 733 

assumed losses much higher than those usually found in 
practice. 

Centrifugal Head in Turbines. — It was pointed out 
some years ago by Professor James Thompson that the centri- 
fugal force acting on the water which is passing through an 
inward-flow turbine may be utilized in securing steady running, 
and in making it partially self-governing, whereas in an outward- 
flow turbine it has just the opposite effect. 

Let R„ = external radius of the turbine wheel ; 
Rj = internal „ „ „ 

V, = velocity of the outer periphery of the wheel ; 
Vj = „ „ inner „ „ 

CO .= angular velocity of the wheel ; 
w = weight of a unit column of water. 
Consider a ring of rotating water of radius /• and thickness 
dr, moving with a velocity v. 

The mass of a portion of the ring of area| _ ^^^ 
a measured normal to a radius > g 

The centrifugal force acting on the mass = dr 

■wan? 
= r . dr 



'he centrifugal force acting on all the') wao? {^' 

masses lying between the radius R(>-= I r,i 

and R. j "^ J R, 



or- 



The centrifugal head = f — ^ — — ^ j 



g 

wa 



The centrifugal head per square inch) V,^ — Vj^ 
per pound of water J ~ 2g 

This expression gives us the head which tends to produce 
outward radial flow of the water through the wheel due to 
centrifugal force — it increases as the velocity of rotation 
increases ; hence in the case of an inward-flow turbine, when 
an increase of speed occurs through a reduction in the external 
load, the centrifugal head also increases, which thereby reduces 
the effective head producing flow, and thus tends to reduce 



734 



Mechanics applied to Engineering 



the quantity of water flowing through the turbine, and thereby 
to keep the speed of the wheel within reasonable limits. On 
the other hand, the centrifugal head tends to increase the flow 
through the wheel in the case of an outward-flow turbine when 
the speed increases, and thus to still further augment the speed 
instead of checking it. 

The following results of experiments made in the author's 
laboratory will serve to show how the speed affects the quantity 
of water passing through the wheel of an inward- flow turbine : — 



GiLKEs' Vortex Turbine. 

External diameter of wheel 075 foot 

Internal „ „ 0-375 >. 

Static head (H,) of water above turbine 24 feet 

Guide blades Half open 





Quantity of water 

passing in cubic feet 

per second. 


Centrifugal liead He. 




Revolutions per 
■minute. 


«**-%/ ?KH, - H.) 





0-68 





068 


100 


0-68 • 


018 


0-68 


200 


0-67 


072 


067 


300 


0-66 


1-6 


066 


400 


0-64 


2-9 


0-64 


500 


0'62 


4-5 


o-6i 


600 


0-59 


6-5 


0-58 


700 


o-SS 


8-8 


0"S4 


800 


0-50 


"•s 


0-49 


900 


0-42 


H-S 


0-43 



The last column gives the quantity of water that will flow 
through the turbine due to the head H, — H„. The coefiicient 
of discharge K^ is obtained from the known quantity that 
passed through the turbine when the centrifugal head was 
zero, i.e. when the turbine was standing. 

IDimensions of Turbines. — The general leading dimen- 
sions of a turbine for a given power can be arrived at thus — 



Hydraulic Motors and Machines. 73 S 

Let B = breadth of the water-inlet passages in the turbine 
wheel ; 
R = mean radius of the inlet passage where the water 

enters the wheel ; 
Vj, = velocity of flow through the turbine wheel ; 
H = available head of water above the turbine ; 
Q = quantity of water passing through the turbine in 

cubic feet per second ; 
P = horse-power of the turbine ; 
17 = the efiBciency of the turbine ; 
N = revolutions of wheel per minute. 
Then, if B be made proportional to the mean radius of the 
water opening, say — 

B = aR ^ 

and V, = b\/2g&. \ where a, b, and c are constants 

V = cs/^im 

then Q = 2TrRBV, = 2abTr^^\/ 2gS., neglecting the thickness 
of the blades 

p = 62HQH, ^ „. ^^^R2 Hv/^ 
55° ' -^ ^ 

rV ^_= 

V„ = m\/2g'H.{c' — P), where ro = -^ = 1 for impulse 
turbines 

The velocity of the wheel V„ is to be measured at the mean 
radius of the water inlet. Then — 



27rRN = 6oV„ = 6oms/2gH.(c-' - P) 
^ _ 6oms/2gB.(c' - &') 
2TrR 

By substituting the value of R and reducing, we get- 
jj ^ l82wHVg3i;(<^ ^^) 



736 



Mechanics applied to Engineering. 







These equations enable us to find the necessary inlet area 
for the water, and the speed at which the turbine must run in 
order to develop the required power, having given the available 
head and quantity of water. The constants can all be 
determined by the methods already described. 

Projection of Turbine Blades. — In all the above cases 
we have constructed the vanes for a turbine of infinite radius, 

sometimes known as a 
" turbine rod." We shall 
now proceed to give a 
construction for bending 
the rod round to a tur- 
bine of small radius. 

The blades for the 
straight turbine being 
given, draw a series of 
lines across as shown; 
in this case only one is 
shown for sake of clear- 
ness, viz. ab, which cuts 
the blade in the point c. 
Project this point on 
to the base-line, viz. d. 
From the centre o de- 
scribe a circle dV touch- 
ing the line ab. Join od, cutting the circle dV in the point /. 
This point / on the circular turbine blade corresponds to the 
point c on the straight blade. Other points are found in the 
same manner, and a smooth curve is drawn through them. 

The blades for the straight turbine are shown dotted, and 
those for the circular turbine in full lines. 

EflSciency of Turbines. — The following figures are taken 
from some curves given by Professor Unwin in a lecture 
delivered at the Institution of Civil Engineers in the Hydro- 
mechanics course in 1884-5 ■ — 




1'"|G. 696. 



Type of turbine. 


Efficiency per cent, at various 
sluice-openings. 




Full. 


0-9. 


0-8. 


0-7. 


0-6. 


0-5. 


0-4. 


Impulse (Girard) 

Pressure (Jonval) (throttle- valve) 
Hercules 


80 
11 


80 
82 


80 

46 
80 


80 

35 
7S 


81 


81 
16 
63 


ss 



Hydraulic Motors and Machines. 737 

Losses in Turbines. — The various losses in turbines of 
course depend largely on the care with which^they are designed 
and manufactured, but the following values taken from the 
source mentioned above will give a good idea of the magnitude 
of the losses. 

Loss due to surface friction, eddying, etc,,! » , 

in the turbine . ... .'j ^ to 14 per cent. 
Loss due to energy rejected in tail-race . 3 to 7 „ 
„ shaft friction . . , 2 to 3 „ 



3 B 



CHAPTER XX. 



PUMPS. 



Nearly all water-motors, when suitably arranged, can be 

made reversible — that is to say, that if sufficient power be 

supplied to drive a water-motor backwards, it will raise water 

from the tail-race and deliver it into the 

^^— ^ head-race, or, in other words, it will act as a 

/L-sN pump- 

The only type of motor that cannot, for 
practical purposes, be reversed is an impulse 
motor, which derives its energy from a free 
jet or stream of water, such as a Pelton 
wheel. 

We shall consider one or two typical 
cases of reversed motors. 

Reversed Gravity Motors : Bucket 
Pumps, Chain Pump, Dredgers, Scoop 
Wheels, etc. — The two gravity motors 
shown in Figs. 657, 658, will act perfectly as 
pumps if reversed; an example of a chain 
pump is shown in Fig. 697. The floats are 
usually spaced about 10 feet apart, and the 
slip or the leakage past the floats is about 20 
per cent. They are suitable for lifts up to 
60 feet. The chain speed varies from 200 
to 300 feet per minute, and the efficiency is 
about 63 per cent. 

The ordinary dredger is also another 
pump of the same type. 

Reversed overshot water-wheels have 
been used as pumps, but they do not readily 
lend themselves to such work. 
A pump very similar to the reversed undershot or breast 
wheel is largely used for low lifts, and gives remarkably good 




Fig. 697. 



Pumps. 



739 



results ; such a pump is known as a " scoop wheel " (see Fig. 
698). 

The circumferential speed is from 6 to 10 feet per second. 
The slip varies from 5 per 
cent, in well-fitted wheels 
to 20 per cent, in badly 
fitted wheels. 

The diameter varies 
from 20 to 50 feet, and 
the width from i to 5 
feet; the paddles are 
pitched at about 18 inches. 
The total efficiency, in- 
cluding the engine, varies 
from 50 to 70 per cent. 

Reversed Pressure Motors, or Reciprocating Pumps. 
— If a pressure motor be driven from some external source the 
feed pipe becomes a suction pipe, and the exhaust a delivery 
pipe ; such a reversed motor is termed a plunger, bucket, or 




Fia 



I 

^ 



a 



<i 






£^ 



I' llllllllll 



^ I 



Fig. 699. Fig. 700. 

piston pump. They are termed single or double acting accord- 
ing as they deliver water at every or at alternate strokes of the 
piston or plunger. In Figs. 669, 700, 701, and 702 we show 
typical examples of various forms of reciprocating pumps. 



740 



Mechanics applied to Engineering. 



Fig. 669 is a bucket pump, single acting, and gives an 
intermittent discharge. It is only suitable for