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 shortcut 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 workedout 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 HeleShaw, 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 Steamengine .... 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 02'
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 = 0305 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 = 00929 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.
= 2992 inches of mercury.
= 33'9o feet of water.
= I4'7 lbs. per. sq. inch.
= I '033 kg. per sq. cm.
= 00703 kg. per sq. cm.
= 2307 feet of water.
= 2036 inches of mercury.
= 68970 dynes per sq. cm.
I lb. per sq. foot = 479 dynes per sq. cm.
I kilo, per sq. cm. = 14223 lbs. per sq. inch.
I cubic inch = 16387 c. cms.
I cubic foot = 00283 cubic metre.
I cubic yard = 07646 c. metre.
I c. cm. = 006103 c. inch.
I c. metre = 3531 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) = — — : ; fr — ~ = 
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 52 — 2*3 = 29 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 ,
'^ ( = 0447 metre per second
I kilometre " \ = °'^'l ^^ ,
( = 0278 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_ = 57296°
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
573
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, 0XViv^=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 vi 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^ 4ii
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 = 322 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 exteinal 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 onethird 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 footpounds, 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 footpoundal. Such a unit of work is,
however, never used by engineers ; the unit nearly always used
in England is the "footpound," 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 322 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. Footpound.
Kilogiam, Metre. Kilogrammetre.
Dyne. Centimetre. Erg.
I footpound = 32*2 footpoundals.
„ = 13,560,000 ergs.
Power. — Power is the rate of doing work. Unit power
is unit work done in unit time, or one footpound 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 horsepower,
which is 33,000 footpounds of work done per minute.
Horsepower
_ footpounds of work done in a given time
~ time (in minutes) occupied in doing the work X 33,000
I horsepower = 33jOoo footpounds per minute
= 7*46 X 10° ergs per second.
I French horsepower = 32,500 footpounds per minute
= 736 X 10^ ergs per second.
I horsepower = 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
' Spaceaverage.
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 contraclockwise, 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 contraclockwise.
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 contraclockwise 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 footlbs. 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 counterefficiency is the reciprocal of the efficiency,
and is always greater than unity.
_ ^ „ . work supplied to the machine
Counterefficiency = , — &£ ^^ ^. —
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
77694 (see Phil. Trans., vol. igo, pp. 301422) 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 footlbs. 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^
= (Mizi" + 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 footpounds 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 footpoundal = 4S3'6 X 3o"48' = 421,390 ergs
and I footpound = 322 footpoundals
= 322 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 loft. circle.
Fis. II.
Length of arc = —7
2irr0 rO
or =• ^
360 573
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 manysided 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
" Pocketbook," 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^ = 04.,
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){sb)(s—c)
FrG, ig.
Quadrilateral,
Area of figure =
bh
Fic. 9a
Trapezium,
A.
Area of figure = ( j li
Fig. «.
Irrtgular straightlined 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^ = 31416^'
or — = oi^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 rightangled
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 = —did^
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 + 24g
+ 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
= {KV'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
HY ^
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 = = mh
38
Mechanics applied to Engineering.
Hyperbola.
Area of figure = XY log.r
log, = 231 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 = "^ — = oyS^aTV
4 ^
Mensuration.
41
Suspend a knittingneedle 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+4A44AJ)
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 fulllined 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 flattopped 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.
_ nd.^ _ ltd?
6 6
= JW
d?)
Slice of sphere.
/T
"^
f
"^
..^
—
'■■^.,
,/■'
Volume of slice = {3R(Yj,''Yi'')Y,»lY,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
Kyf')dy
t
2R/ _/
y=Y,
y=Y,
_s_
±rJi_
= J zRC^iri^) _ Y,3Y.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 onehalf the volume of the sphere found by the other
method.
Paraboloid.
Volume of\_^„2„ ^a
paraboloid / 2^ "■"'^ 8° "
= iSyR'H, or o39D''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
• 032
Iron — cast
., 026 „
„ wrougtit
■ 0278 ,,
Steel
.. 0283
Lead
.. 0412 „
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 contraclockwise 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 poundsfeet 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.
<rljr>^ — ^ ■
*S «5 "i
Fig. 60.
T'
Fig. St.
Cloclcwise moments
about the point a.
Contraclockwise 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.
Wor
2 2
If w = dis
tributed load
per unit length,
w/= W
Wor —
2 2
W/
W = weight of
long arm of
lever
W, = weight of
Contraclockwise 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 contraclockwise 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
safetyvalve, 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 bellcrank lever be fitted with a knife
edge, and the weights travel along, as in some
types of testingmachines, the resultant passes
through the knifeedge, 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 ci, c,, c,, etc. //
The triangles def and Oba Ui;^ .
are similar ; .' /{
,^ , fd Oa / R M
thereforeV = ^, 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 twothirds 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'6i')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.
Exparabolic 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 = ^ = 5H
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
: whi, + 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 hi{K\ hi\ hiV, 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 straightedge, 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 parallelogiam.
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^"
II„_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
or77(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 (BB.)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
lOII
06
I 021
oS
I 039
04
I '065
03
no?
02
■ •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
alsoD,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
orD^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 baseline, 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 " Textbook 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^ .
(«Re)
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 ihiottgh
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 kycy
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^yffdy
= j(4Ry+y4R/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/1y = ^
. 2L 3 ^5 4 \y=o
IT r4R° R^ 4Rn
second moment of sphere = iV^R"
L3
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 righthand 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 fivesided 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 nonadjacent 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 98 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 cranepost is ad, or 79 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 cranepost at y may be very
much reduced by placing a balance weight Wj on the crane, as
shown. The forces acting on the balanceweight 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 cranepost The pressure on the lower
part of the cranepost 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 onehalf 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 cranepost. — 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 cranepost 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 oa, 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 oa\ 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
oa.\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 oad along OuD equal
to oa.\d, found by the preceding parallelogram, then the force
acting down the leg B is Oyj), and that down C is oaC
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^JV  ^
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 lefthand 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 arrowheads 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 arrowheads 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 arrowheads 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 arrowheads 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 arrowheads 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 connectingrod 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 connectingrod 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, connectingrods, crossheads, 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 pointpaths 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 pointpaths 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 pointpaths, 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
pointpath 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
fourbar 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 pointpath 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 crossbars 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 pointpaths 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 pointpaths. We will show
this by taking one or two simple
cases. In the fourbar 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 pointpaths 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 nonadjacent 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~ 2i6
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 129 X 03
= 036
V5 psPs 229 X 047
Likewise —
velocity 7 _ R? _ 2'3i _
velocity 8 Rs 220 ~
velocity 6 p^ x velocity 8 X R
hence
velocity 7 Ps X K.7 X velocity 8
= 038
Ps X R7 076 X 231
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 6 032
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 _ 23
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 —
47 ri.4
and ^a., X ii., = ?i,, X 4.7
Substituting this value above, we have —
^c 1.6 X ^as
'1.8
= M, or = iiP by similar triangles
X 47 47 54
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.
Steamengine Mechanism. — On p. 126 we showed
how a fourbar mechanism may be developed into the ordinary
steamengine mechanism, which is then often called the
" slidercrank 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 crankpin is known, and the velocity of the crosshead
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 connectingrod, 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 crosshead
But the velocity of the crankpin may be taken to be constant.
Let it be represented by the radius of the crankcircle OP;
then to the same scale Te represents the velocity of the cross
head. Set up fg = P<? at several positions of the crankpin,
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 crankpin.
When the connectingrod 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 steamengine 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
crankshaft in its main bearings ; (2) the crankpin in the big
end brasses of the connectingrod ; (3) the gudgeonpin in the
small end brasses.
Let the radius of the crankshaft journal be r„ that of the
crankpin be r^, and the gudgeonpin 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 gudgeonpin 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 connectingrod 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 crankarm and the connectingrod 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 connectingrod, i.e.
at the extreme " in " end of the stroke. Let the connectingrod
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 crankarm and the connectingrod 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 gudgeonpin 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 connectingrod, in reality the piston rod in this
case, and the link c oscillates about its centre, which was the
gudgeonpin in the ordinary steamengine, 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 quickreturn
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
quickreturn motion on shaping machines.
The cutting stroke takes place during the
slow swing of c, i.e. when the crankpin 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 quickreturn 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 steamengine, 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 steamengine
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 pistonrod.
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/
eS
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 timeinterval
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 subnormals 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 = uPC
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 connectingrod of the Musgrave engine can
be readily treated by this construction. A is the crankpin,
whose velocity and direction of motion are known. The
pistons are attached to the corners of the triangle by means of
short connectingrods, a suspension link DE serves to keep the
connectingrod 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^,,
R9
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 fourbar
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 linkmotion. Having given the speed of the engine
and the dimensions of the valvegear, we proceed to find the
velocity of the slidevalve.
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 valverod, 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 noncircular 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 wellknown 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 abovementioned 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 wellknown 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 semicircumference 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 guidebar 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 contraclockwise, 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 halftime 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 abovementioned 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 —
,. , II03WRN2
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,
II03WRN2
which should therefore be not less than 5 W,
when the follower simply rests on the cam, and
no!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
II03WRN2 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 simplelooking 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 159 XmX6)
Where 8/ is the difference in length of two adjacent radius
vectors, measured in feet. The numeral in the denominator
becomes igi 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 = 053 to
052P. 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.
doublepointed 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 pencilpoint 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
pencilpoint 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 fulllined curves are those
actually drawn by the pencils, the remainder of the teeth are
dotted in by copying the fulllined 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 crossband 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 doublepointed pencil, which is pressed
against two flanges on the discs ; then, when the discs turn, the
pencilpoints 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 straightedge 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. HeleShaw, 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
anticlockwise 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 drillingmachine (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 oo86
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 oldfashioned 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
motorcars 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 changespeed
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 contraclockwise fashion.
Mechanisms.
173
If a be fixed, the bar c must be moved in contraclockwise
fashion to cause c and x to approach, but b will still continue
to move in contraclockwise fashion. Let c be rotated through
one complete revolution in contraclockwise 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 contraclockwise 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 contraclockwise
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 ^ = 15.
V„+. = 2.
V„+. = =5.
lOO
100
ISO
lOO
5
 100 + 2 X 5= 90
I50+2SX 5 = i37"S
— lOO
5
100J2X s= no
150+25 X 5= i62s
100
s
100— 2X 5=IIO
 15025 X 5= 1625
lOO
20
 100+2 X 20=— 60
 150+25 X 20= — 100
lOO
50
 100+2 X 50=
 150+25 X 50= 25
100
60
 100+2 X 60= 20
 150+25 X 60=
lOO
70
 100+2 X 70= 40
 150+25 X 70= 25
lOO
100
100+2X100= 100
150+25x100= 100
— 100
100
IOO+2XIOO= 300
150+25x100= 400
100
— 100
100— 2X100= 300
— 150 — 25 X 100= —400
40
30
 40+2 X 30= 20
 60+25 X 30= 15
40
40
60
S
2X5 = 10
25x5=125
100
2( 100) =200
25(I00)=2SO
Humpage's Gear. — This compound epicyclic bevil train
is used by Messrs. Humpage, Jacques, and Pedersen, of Bristol,
as a variablespeed 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 =  ii'S X M + i
= 0541 + I = 0459
Whence for one revolution of E, C makes
483
°'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:
0541
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 _ — 0541 + I _ 1
Revs. C Ne 383 + I 1053
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 STEAMENGINE.
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 crankpin was
assumed to be constant. We there showed that, for an infinitely
long connectingrod or a slotted crosshead (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 crankpin. 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 highspeed engines, unless some means are adopted for
counteracting these disturbing effects.
We will first consider the case of an infinitely long con
nectingrod, 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 crankpin,
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 SteamEngine.
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 connectingrod 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 crankpin, 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 connectingrod be infinitely long, then it
always remains parallel
to the centre line of
the engine ; hence the
action is the same as
if the connectingrod
were rigidly attached • ,/
to the crosshead 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 baseline. We
show such a diagram in Fig. 2 04.
The height of any point in the
sloping line ab above the baseline
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 crankpin 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 pistonhead, pistonrod, tailrod (if any), crosshead,
small end of connectingrod and half the plain part of the rod.
A more accurate method of finding the portion of the connect
ingrod 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 airpumps or other connections are attached to
the crosshead, they may approximately be taken into account
in calculating the weight of the reciprocating parts j thus —
Fig. 205.
weight of \ piston ) piston and ailrods f both crossheads f small
area of
end of con. rod ( ' ' 1 air
2
piston
irpumpplungerf ^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 Connectingrods. — In Fig. 159
a velocity diagram is given for a short connectingrod ; repro
ducing a part of the figure in Fig. 206, we have —
Short rod —
crosshead velocity _ OX
crankpin velocity OCj
Infinite rod — ,
crosshead velocity _ OXj
crankpin velocity OCi
velocity of crosshead with short rod _ OX
velocity of crosshead 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 connectingrod 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 SteamEngine. 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 crosshead 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 connectingrod 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
shortrod velocity curve is the same as that of the longrod
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 shortrod 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
crosshead 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 shortrod 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 crankpin, assumed
constant and represented by R ;
V = the linear velocity of the crosshead.
' 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 crosshead /„ = — = —„• —
■" 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/ = oooo34ze/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 crankpin 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 lll 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 highpressure 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 SteamEngine.
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 jactual 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 Crankpin. — The diagram given in
/^
'~y<^
"^•',^
/ ^
''
N^^^^ ~~^
^
/
/ /■
1;*^— .
■~'\
J— \— v
/
1^
4
\\
.
w
Fig. 210.
/
Fig. 208 represents the pressure transmitted to the crankpin
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 rodends, 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 steamengines 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 highspeed 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
wellknown Willans engine an aircushion 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 = oooo34WRN2(^i f i^  W f i47A
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  i47Aa
Dynamics of the SteamEngine. 191
Or for adiabatic compression —
I47A„(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 steamengine, 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 crosshead or to a tailrod,
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 TwistingMoment Diagrams. ^ From the
diagrams of real pressures transmitted to the crankpin that
Fig. 211.
we have just constructed, we can readily determine the twisting
moment on the crankshaft 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 crankpin. 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 crankcircle 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 crankshaft 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 tripleexpansion 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 SteamEngine.
193
Diameter of cylinders —
High pressure
Intermediate
Low pressure
Stroke
9*01 inches,
I4"2S .1
2247 >.
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 pistonrod due to the working
fluid and to the inertia of the moving parts ;
Pi = the component of the eflfort acting along the
connectingrod ;
, . 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 twistingmoment diagram we have just
constructed shows very clearly that the turning effort on the
crankshaft 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 singlecylinder 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 crankshaft 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 twistingmoment diagram, plotted for convenience
Dynamics of the SteamEngine.
195
to a straight instead of a circular baseline, the length of the
base being equal to the semicircumference of the crankpin
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 crankpin 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 horsepower of engine x 33000 ... .
where N is the number of revolutions of the engine per minute
in a doubleacting 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 SteaniEngine.
197
K = o'oi to o'oa for ordinary electriclighting engines, but for
public lighting and traction stations it often gets as low
as o"ooi6 to 00025 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 24 to 45 foottons (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 connectingrod, 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 Doubleacting SteamEngines.'
Cutoff.
Single cylinder.
Two cylinders.
Cranks at right angles.
Tiiree cylinders.
Cranks at 120°
O'l
0'35
0088
0040
0'2
o'33
0082
0037
04
031
0078
0034
06
029
0072
0032
08
028
0070
0031
End of stroke
027
0068
0030
' The values of m vary much more in the case of two and three
cylinder engines than in singlecylinder 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 OiLENGINES.
Otto cycle.
Double acting.
Number of cylinders.
X
2
4
I
3
Exploding at every
cycle ....
Single
37 to
45
—
—
23 to
28
—
Twin or
tandem
—
iS to
I 8
03 to
04
—
03 to
04
When missing every
alternate charge.
Single
85 to
98
25 to
30
—
—
—
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 Doubleacting
SteamEngines.'
Cutoff.
Single cylinder.
Two cylinders.
Cranks at right angles.
Three cylinders.
Cranks at 120°.
oi
18
4'4
2'0
02
17
4" I
I "9
04
16
3'9
I "8
06
IS
36
'7
08
14
3S
16
End of stroke
«3
34
iS
GasEngines (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 inchtons per square
inch of metal sheared or punched through.
' See note at foot of p. 197.
200
Mechanics applied to Engineering.
GasEngine 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 onefourth of the explosionstroke effort, and
the excess energy is therefore approximately threefourths
of the whole explosionstroke 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 4stroke case is given in Fig. 216. It has
been constructed in precisely the same manner as Figs. 207,
208, and 215. When oil or gasengines 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 highspeed flywheel stores a much larger amount of
energy than the same weight of metal in the engine flywheel.
The peripheral speed of large castiron flywheels has to be
kept below a mile a minute (see p. 201) on account of their
Dynamics of the S teamEngine.
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 boilershell 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
/= 0096 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 SteamEngine. 203
is intermediate between that due to the abovementioned
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 poundinches 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 026 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
o26V„^A/'
'^ ■'" 12 X 322 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
o26V„^A/' ^ V^'A/'
■^^ ~ 22 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
lowworking 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 steamengine 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. Builtup
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 harddrawn steel wire
is wound under tension. Such wheels are of necessity more
costly than castiron 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 SteamEngine.
205
the cost of the builtup wheel may be even less than that of a
castiron 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
056
. ) » » » »
192
3.700
Jointed rim, ,, „
38"
14,500
065
Solid rim, 3 arms, „
363
13,200
038
.. .» ».
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 GasEngine." " 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 onefourth 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 onefourth 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 ,. ^ ..
mchpounds, approximately ; the bendmg stress is
Dynamics of the SteamEngine. 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 Couplingrods. —
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 poundinches ; 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 ) ^ ^ oooo34«/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 halfdepth 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^ '
ooooor2R„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 onehalf 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 SteamEngine.
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 +
96iEI
 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 Connectingrods. — In the case of
a couplingrod of uniform section, in which each particle
describes a circle of the same radius as the couplingcrank pin,
the centrifugal force produces an evenly distributed load ; but
in the case of a connectingrod the swing, and therefore the
centrifugal force at any sec
tion, varies from a maximum
at the crankpin to zero
at the gudgeonpin. The
centrifugal force acting on
any element distant x from
the gudgeonpin 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 connectingrod.
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 gudgeonpin, due to the centrifugal
force acting on the rod, be R,. Then, since the centrifugal
force varies directly as the distance x from the gudgeonpin,
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 gudgeonpin ) 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
/
^=V3
By substitution of the values of x and R, and by reduction,
we have —
_ _c^ _ c/"
■Minai. — / — ■
9V3 150
^ ^ ^ A . X RNV^
and the bendmg stress/ = —z ^
° "^ i64000K^
which is about onehalf as great as the stress in a couplingrod
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 straightedges, 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 —
000034WRN2 = oooo34WiRiN*
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 SteamEngine. 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^yi = c^y^i + hy
oooo34N^(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 contraclockwise 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 teamEngine.
215
/to
A
B
Tcrr
Fig. 224.
Provided the abovementioned 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 SteamEngines. — 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 contraclockwise mome.n\.,p^y„
due to the acceleration pressure p^ tend
ing to turn the whole engine round in a
contraclockwise 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
holdingdown 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 twocylinder 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
connectingrod 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
bedplate, 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
nectingrods about the gudgeonpins 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, crosshead, and connecting
rod of a singlecylinder engine, and let a " bobweight " be
substituted for the piston and crosshead of the cylinder A.
Then, provided the " bobweight " 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 " bobweights," 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 "bobweights" 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 threecylinder
engines,^the "bobweights" then become the weights of the
reciprocating parts of the two cylinders Aj and Aj.
Any threecylinder engine can be balanced in a similar man
ner by the addition of two extra cranks or eccentrics to drive
suitable "bobweights," 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 crossheads 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 threecylinder 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 crankshaft.
Dynamics of the Steam Engine.
217
A fourcylinder 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 crankshaft ; 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 connectingrod 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„ „ „ couplingcrank.
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 SteamEngine, 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
crankpin, and neglecting the obliquity of the connectingrod.
Then, for the present, assuming the balance weight to
rotate in the plane of the crankpin, in order that the recipro
cating parts may be balanced, we must have —
C = C
oooo34Wbp . Rb . N^ = oooo34Wp .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 connectingrod is neglected, may for every purpose be
regarded as though their weight were concentrated in a heavy
ring round the crankpin.
Now we come to a muchdiscussed point. We showed
above that with a short connectingrod 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 connectingrod which may be regarded
as rotating with the crankpin and the portion as reciprocating
with the crosshead may be most readily obtained by find
ing the centre of gravity of the whole rod — let its distance
from the crankpin 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— jJistobe 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 onehalf the plain part as
reciprocating, and the big end with onehalf the plain part as
rotating.
Insidecylinder Engine (uncoupled). In this case
we have —
Wp = weight of (piston  pistonrod f crosshead  small
end of connectingrod  \ plain part of rod) ;
W, = weight of (crankpin 1 crank webs ' f big end of
connectingrod P \ plain part of rod).
If we arrange balance weights so that their c. of g. rotates
in the same plane as the crankpins, their combined weight
would be Wg,  Wup, placed at the radius Rb, and if we only
counterbalance twothirds of the reciprocating parts, we get —
W
R(Wp + W,)
See p. 229,
_ JV^3»vp T T',y ... ,
»'B0 — t5 V"'J
Dynamics of the SteamEngine.
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 crankpin 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 = o76Wb,
Substituting from ii., we have —
o76R(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
35 ^ ^
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.
Outsideeylinder Engine (uncoupled). — Wp and W,
are the same as in the last paragraph. If the plane of rotation
of the crankpin 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 =
* =
cy
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
insidecylinder engines, but in this casejc = o'8f to o'<)c. On
substitution, we get Wb = i'I3Wb„ to iosW^o, and fl = 6° to 3°.
The same reasoning applies to the couplingrod balance
weights Wbo in the next paragraphs.
Insidecylinder Engine (coupled). — In this case we
have Wp the same as in the previous cases.
Wo = the weight of coupling crankweb 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 drivingwheel of the fourwheel coupled
engine, we have Wb arrived at in precisely the same manner
as in the case of the insidecylinder uncoupled engine, and
^^ Rb •
The portion of the coupling rod included in the Wo is, in
this case, onehalf the whole rod. The balance weight Wbo is
placed diametrically opposite the coupling crankpin. 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 crankpin.
Fig. 232.
In the sixwheel 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 crankpins ;
the balance weights then become very much greater. They are
treated in precisely the same way.
Some locomotivebuilders 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.
Outsidecylinder Engine (coupled).
W is the same as before ;
W, is the weight of crankweb ' and pin  coupling rod
from a\.ob ■\ big end of connectingrod + 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 sixwheel 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
crankpins and wheels nearly coincide, as already explained
when dealing with the uncoupled outsidecylinder engine.
On some narrowgauge 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 crankwebs 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 crankpin. Then, by
precisely similar reasoning to that given above —
where W.^'^^tyVp + W.fW.)
In some cases _)> is only osC, ; then —
Wb = is8Wbo, 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
oooo34W^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 SteamEngine.
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 castiron weight —
Wb = o52/(Ai + Aj, + A3)
For a wroughtiron or caststeel weight —
Wb = o56/(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,tA. (^^«PS8)
md WgRj = \ or
(056;
i?(Airi + AaZa + 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 crankpin.
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 SteamEngine.
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
1409
—
(542)
(I200)
—
60
979
430
70
719
260
80
S'Si
168
90
43S
ii6
100
352
083
no
291
o'6i
120
2 45
046
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 2inch 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—!°
o8i6/ ^ 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
1409
180
979
4 "3°
210
719
2 60
240
551
168
270
4'35
116
300
3"5z
083
330
2'9I
0'6i
360
245
046
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 ~ oooo34WR,N!'
,HXoooo34WN2 + 2P) = W + 2P/
„ W + 2PI . ,
"• = 000034WN' + 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 SteamEngine.
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
crossedarm 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 crossedarm
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„ = 000034WRNV/''  (R + xf
^=s/ ^^+^
000034 kV/''  (R + xf
X H„  H , „ RH„ o8i6 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*
oooo34RV/'(Ra:)^
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 crossedarm 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 wellknown 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. onehalf 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 SteamEngine.
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 15 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 SteamEngine.
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 knifeedges 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.
Crankshaft 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 crankshaft
governor, which alters the cutoff 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 sphericalended 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 cutoff is as late as the
slidevalve 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 leathercovered 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 SteamEngine.
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 crankshaft 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) = 859 feet per sec^ond.
X 60
SV8SQ
and the acceleration kt —7^ = 42*95 feet per sec. per sec,
and the force T = n_='' ^ — ? = 234 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
■ 02? 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 = oooo34WRN2H
= oooo34WN''(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 SteamEngine. 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 ballbearings or with
knifeedges (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 SteamEngine.
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 wellknown 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.
Crossedarm 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 fourball
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 oooo34WRN=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 000034 X 2'S X n^ X 450^* X I'sS = 260 poundinches
and for the centripetal moment —
The load on the spring =111 lbs. ; and R, = 278
centripetal moment = in x 278 = 310 poundinches
McLaren's Crankshaft 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 crankshaft.
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 2percent 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, crossedarm 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 : —
Footpounds of energy
Type of governor. stored per inch diametei
of cylinder.
For trip gears and where small resistances have to
be overcome ... ... ... .• ... o'5o*7S
For fairly well balanced throttlevalves 0751
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 slidevalves, about 100 footpounds of energy per inch
diameter of the highpressure cylinder is found to give good
results ; but in other cases, with unbalanced slidevalves, 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 nonreversible,
only about onehalf 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 slidevalve. Experiments show
that the latter amounts to about onefifth to onesixth of the
total pressure on the back of a slidevalve (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 slidevalve moves
while the springs move through their extreme range.
Generally speaking, it is better to so design the governor
that the valvegear cannot react upon it, then no amount of
pressure on the valvegear will after the height of the governor ;
that is to say, the reversed efficiency of the mechanism which
alters the cutoff 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 crosshead,, 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 crosshead is
T, WV^ * /■^
^^ = 7r^r •. ■ ■ • • «•
where W = weight of the crosshead in pounds.
V = velocity of the crankpin in feet per sec.
R = radius of the crank in feet.
X = displacement of crosshead 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 crosshead
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 crosshead 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 crosshead when it is displaced an
amount x from its middle or zeroforce 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 poundfdot units.
Mass of the body ( — ). where W is
in pounds.
The couple acting on the body
C = lA in poundfoot 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 = 2rsJ s— = 2'r\/ rr— = 2K\J —
^ ^ig ^ i2W^ V v^gGd"
t = o904>/ ^^ (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 onethird the weight of the spring. Thus
the W in the expression for the time of vibration should
include onethird the weight of the spring in addition to that
of the weight itself.
Expression (viii.) now becomes —
Example. — D = 2 inches d = 02 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 0267
V 12,000,000 X o'ooib
= 04668 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 133" or o'li foot for
every 50 lbs. Then
/.
'■ = 2W/y/
80 X oii
= 047 sec.
50 X 322
The weight of the spring = o69«^D
= 069 X 8 X 004 X 2
Wj = 044 lb.
Then allowing for the weight of the spring —
/ 8 X 8015 X 8 ^ ,
t = ooo4x/ 2 = o"4074 second
'V 12,000,000x00010
say 047 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
Springcontrolled
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
322x144X8 ^V"^322 X i44^V 12 / 43
= 18 poundsfeet units.
03 100 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 poundsfeet, and M, in poundsinches
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, onethird 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^'
Oziiird^ ,
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 = 0372 inch, L = 544 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 = ysi 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 =
265EI 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
+ Vr 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 = 0995 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 18945, 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
050
20,931
HS
033
28.095
168
025 to O'lO
31,000
176
Very small
31.590
178
Zero
32,864
iSi
TABLE IL
Value of r = i
h
0>
V5.
ioo too7S
32,864
18:
05 to 07s
36.884
192
o'33
41,026
203
025
43,289
208
0'20 to 014
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 selfcontained 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 contraclockwise 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 righthand 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 righthand 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 superelevation, and
when the dead load on each wheel is 10,000 pounds. Rail
centres 5 feet.
N = 2155 n = r75
Gyroscopic couple =
2000 X 2'8^ X 2is'5 X f75
2937
= 2015 poundsfeet.
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
superelevation.)
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 lefthand pivot x, or if a weight were suspended from the
righthand pivot x.
In the case of a motorcar, 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 crankshaft 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 shortbristled 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 050
020 to o'6o
0'I5 to o'30
Leather on wood
,, metal
Stone on stone
025 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 jnessuro
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=
24VV
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
0S4
069
034
074
0S9
0'46
043
026
063
052
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(.gfg)
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 = — rrr — ^ =  — 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 = ^ — ■ — 77— 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 cranepost used largely on
board ship. By raising and lowering the twisted cranepost,
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 squarethreaded
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 = 113 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 + 26+.^)
(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.
Vthread.
Sq. thread.
Vthread.
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
451
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'oo7o"oi5
O"o6o'io
OOS0'20
3S
0"020'022
0'04oo6
004
oogo'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 AB
Fig. 295.
I/aJf Cross Section
an C0rUreLvi£
is known as " endthrust," 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 endthrust 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
gunmetal 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 endthrust 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
00082
00076
0'0072
82
147
212
276
00053
00035
00029
00026
00024
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
0OOI2
None
4000
O'OOIO
■ —
6000
00008
7
76° F.
8000
O'OOIO
84
89° F.
10,000
oooii
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 "endthrust" 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 fourpoint radial bearing, the
outer cones screw into the casing, and thereby permit of adjust
ment as the bearing wears. B is a threepoint bearing capable
of similar adjustment. C is a threepoint 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 fourpoint 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 threepoint 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 twopoint 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 sheetiron 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 twopoint 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 onehalf the number can
be inserted unless some special device be resorted to. But
Friction.
305
since the loadcarrying 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
458
438
437
4' 36
Thus P„ =
437W
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
oneinch 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 welldesigned
bearing.
The following results, which may be regarded as typical,
lend support to the statements (i), (2), (3) : —
Load, in lbs.
Friction moment,
inchlbs.
1000
So
2000
60
3000
84
4000
128
Sooo
lyo
5ooo
20'4
7000
252
8030
304
9000
360
10,000
40'o
Speed, in revs, per
min. (approx.)...
Friction moment,
inchlbs.
S
20'0
5°
198
100
197
500
198
800
20'I
1000
20'0
Temperature, Fahrenheit .. .
Friction moment, inchlbs.
58°
376
65°
381
77°
390
39"o
98°
387
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
firstclass 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 welllubricated 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
089
090
0*64
087
082
084
_
087
o8o ! 086
090
087
087
057
oSS
0S9
—
—
127
I .35
124
II2
II4
125
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 crosssectional 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=KAK ^"'^ ^'^—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 oilborne 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 abovementioned
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 abovementioned 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
frictionspeed 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&rautLua conjtroU£d.a£ 120°F.
C Specimen A. Tempsralure ctSaiuettto vafi/.
+ Specimen fl. Temperature controUeoLat t20'*F.
4 Speoimen B. TemperaUire 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 0332,
and the coefficient at 120° F. would be 0005 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 023 per cent, of the whole.
Load in lbs. sq. inch
Coefficient of (A
friction tB
150
00143
00083
250
0OII2
00062
3S0
00091
00054
450
00082
o;ooso
750
00075
00045
950
00083
00047
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
0013
100
ooo8
250
0005
500
0004
750
00043
007
0I3S
014
015
0185
1000
0009
018
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
00084
00329
00252
00350
0192
O171
0147
0090
229
S2
S8
26
Friction. 319
6. Experiments by Tower and others show that a steel
shaft in a gunmetal 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
connectingrod 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 frictiontesting 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 oilbath
lubrication, on account of the difficulty of getting regular
lubrication by any other system. It was found that the
bearing was completely oilborne, 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 oilhole 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 oilholes 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
0035
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
abovementioned 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 oilborne thrust bearing. See a
paper by Newbigin in the Proceedings of the Institution of
Civil Engineers, Session 19131914; 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 oilhole
on the resultant line of pressure, to which a screwdown valve
was attached. When the oilhole 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 00004 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.
OZS O'J 07J tO
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 wasonly 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
forcepump is used to force the oil into the bearing ; it then
becomes equivalent to bath lubrication. Many highspeed
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
706
I '00
132
221
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 watercirculation 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 sphericalseated 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 frictiontesting 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
oilborne, as in Tower's oilbath
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 connectingrod
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, castiron, 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 6inch 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
47
«is
23 S
05075
In tolerably cool places, as in
marine and stationary en
gines
075I
0305
04OS
015025
In hot places and where heat is
not readily conducted away
04oS
0I03
02025
~
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.
Crankpins. — Locomotive ...
Marine and stationary
Shearing machines
Gudgeon pins. — Locomotive
Marine and stationary
Railway car axles
Ordinary pedestals, — Gunmetal
Good white metal
Collar and thrust bearings, — Gunmetal
Good white metal
Lignum vilse ...
Slide blocks. — Castiron or gunmetal
Good white metal
Chain and rope pulleys for cranes, — Gunmetal 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 footpounds j ~
horsepower absorbed =
12
WttDN/*
_ /itWDN
12 X 33)000 126,000
A convenient roughandready 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 footpounds / "~ 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 footpounds =
5 73
ftWDN
189,000
horsepower 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 = zttzV/ ■ 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
fulllined 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 = ViiJpr
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^ _ 2irr'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 = 2jr 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 waterpressure ;
d = diameter of plunger in inches ;
p = waterpressure in pounds per square inch.
Then F = o"o8/ I — j when in good condition
F = 008/ + ^ „ 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 liftingtackle, 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/^ _ WAW/^
''' 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 twosheaved 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 footlbs. 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 = 064
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 selfsustaining, i.e. it will not run back when left to
itself. The least frictional resistance necessary to ensure that it
shall be selfsustaining 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
nonreturn mechanism may be selfsupporting 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 selfsupporting property of a machine is, for many
purposes, highly convenient, especially in handlifting tackle,
such as screwjacks, 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.
9597
9396
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.
Twosheaved.
Threesheaved.
pounds.
Old lin.
New Jin.
Old iin.
New iin.
Old }in.
New Jin.
rope.
rope.
rope.
rope.
rope.
rope.
«4
94
90
_
_
28
94S
905
80
75
30
24
56
95
91
84
78S
5°
35
112
96
92
86
915
60
41
i6g
87S
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
singlechain 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 righthand chain will be
97 per cent, of that in the lefthand 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 righthand chain
can be transmitted to the lefthand 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  094R
= 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 Highefficiency Selfsustaining Pulley
Block. — In pulley blocks of the Weston type the efficiency
rarely exceeds 45 per cent., but in geared selfsustaining blocks
it may reach nearly 90 per cent.
The selfsustaining 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 handchain.
D = the diameter of the handchain 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 handchain 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 fullline 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 /
Selfsustaining geared\
blocks J
I + 0052/,
■Vr
1 + OWr
V,
I + oo09»,
Vr
2v,dVas.
3 lbs.
iS lbs.
W
W{i+oosz',.) + Kz',
W
W(l+0IZ/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 wellformed knifeedges, 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 machinecut toothed wheels, including the
friction on the axles —
t\ = o'g6 —
for rough unfinished teeth —
1/ = o'go —
25N
25N
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
0986
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 2inch diameter shaft running at
100 revolutions per minute requires about i horsepower 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 horsepower of the shafting.
Then F =
NLD«
80,000
The horsepower 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«
_ horsepower transmitted  friction horsepower
Hence the efficiency of line shafting on which there are
numerous pulleys is —
horsepower 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 onehalf 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 3inch 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,rR3Lmchlbs.
the maximum]
twisting mo , ^3
ment at the )='i inchlbs. (see p. 576)
motoi end I ^
of the shaft I
^^^ f ??"^"*^^ ) the effective twisting moment at the far end
of the trans \ = — r — rr—. 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 = fjiWnR^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 footlbs. per
square foot of belt passed over the pulleys.
Mechanical EfSciency of Steamengines. — The
work absorbed in overcoming the friction of a steamengine 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 2133
Ciankpin 57
Crosshead and gudgeonpin 45
Valve and rod ... 2°5 balanced, 22 unbalanced
Eccentric strap 4S
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. ...
275
O'O
27s
9 "25
563
362
1023
750
273
1114.
766
348
1234
^•09
325
1395
II '09
286
1429
1125
304
Experimental Engine, The University, Leeds.
Syphon and Pad Lubrication.
LH.P.
B.H.P.
Friction H.P.
248
00
248
Si6
235
281
683
394
289
830
561
269
1150
870
28o
1384
1082
302
1702
1389
3«3
2230
1909
321
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.
498
102 '7
147 I
1936
44'S
970
140*6
i860
53
S7
6'S
76
2175
2095
80
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 steamengine
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 doubleacting, and  — '^—
5° S°
= 88 per cent, if singleacting.
The mean effective pressure in a gasengine cylinder seldom
exceeds 75 lbs. square inch. Thus the mechanical efficiency
' is from 81 to 87 per cent.
The friction horsepower, 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
Oneton steam hoists or windlasses
Hydraulic windlass
„ jack
Cranes (steam)
Travelling overhead cranes
3040
4047
4045
7585
5070
6080
8090
6070
3050
Friction. 35 1
T .. draw bar H.P. ,^ „^
Locomotives o5~75
1. ri.r.
Twoton testingmachine, worm and wheel, screw and
nut, slide, two collars ... ... ... 23
Screw displacer— hydraulic pump and testingmachine,
two cup leathers, toothedgearing four contacts, three
shafts (bearing area, 48 sq. inches), area of flat slides,
18 sq. inches, two screws and nuts 23
(About 1000 H.P. engines,
spurgearing, and engine
friction 74
Rope drives 70
Belt , 71
Direct (400H.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 = 04343.
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 —
Horsepower 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 singleply 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. 07 2 X 8o7</V
and H.P.= — ^ =
33.000
600
for singleply belting;
andH.P«'V
.300
for doubleply 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 horsepower is —
vN
H.z= 50 square feet per minute for singleply, and
wV 25 square feet per mtaute for doubleply
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 —
o43V„'' ^ 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 singleply belting is about 022 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 horsepower 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 horsepower transmitted when V = 4700 feet per
minute, and the maximum horsepower per square inch of
rope = i7*i.
The curves in Fig. 337 show how the horsepower 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 — 314 —
fi = = 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 25 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 7222 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 26 x 03 = o'78.
Friction. 359
The value of ei^' now becomes lo'i when the rope embraces
half the pulley. The factor of safety on drivingropes 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 horsepower 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 „ „ oooo8 to 00012
Ihrust ■jOneflat,one vrace, 3 ,. „ mean 00018
"^^""Sslxwoyraces, 4 » .. .. o'o°S5
Gunmetal 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
Referencebooks 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. HeleShaw, 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 sawcuts 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 SS
§■£■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 stressstrain 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
tJuesT
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
245
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.
Loadstrain 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 stressstrain 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 testingmachine. a strcssstrain 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
stressstrain 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 testbar 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 onethousandth 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
141//
will be = 041.
Thus, when the modulus stress is reached in shear the
strain is 041 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 drawplate, 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 wiredrawing 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
drawplate 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 wiredrawing 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 harddrawn steel wires
had a lower modulus than the same wires after annealing.
The results were —
o*i6o
0*160
0*210
o'aio
0174
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,
96
3'S
on 3
ins.
19*0
19*5
5 n
34'S
54'o
468
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
00895
0126
6000
140
146
096
Ii7\
24
0085
0136
6870
180
185
097
85
30
0091
0195
S«40
235
246
096
127
V 30
0085
0187
SSSo
825
91
091
49
3"
0136
0522
7200
472
475
099
91
Steel
4?
0081
0220
5400
200
204
09S
90 wire
■^s
0023
0232
6330
309
3175
098
13^
ropes
108
0065
0358
6020
410
423
097
"5
222
0044
0337
5560
35 '4
463
076
105
222
0039
0264
6530
249
300
083
95'
19
0231
0796
,3770
695
827
0S4
87 ^l"
7
0231
0293
35yo
225
303
074
7'7 cable
Work done in fracturing a Bar. — Along one axis a
loadstrain 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 inchtons in fracturing the bar
and TTvj = work done in inchtons 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 inchtons per) _ ^ . 2 ,^ _ , .
square inch of section of bar )~ "t" 3^ /
= (L + 2M)
work done in inchtons per cubic inch = — 7(L + 2M)
^ X , . .
But Y X 100 = tf, the percentage of extension
hence the work done in inchtons per) £_,y , ^>
cubic inch ) ~ 3oo'' '
The work done in inchtons 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 inchtons
39'9 .. ..
Ordinary state
Turned shank
i:;
14 ins.
14 »
2' 1 5 inchtons
lb8 „ „
Ordinary state
Turned shanlc
Jin.
f..
14 ins.
14 .1
175 inchtons
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 stressstrain 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 crosssection 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 crosssection, 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 crosssection ; 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 071 = 05 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. 364366.
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 loadstrain 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 barrelshaped), 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 contraclockwise 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 doubleshear 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
Gunmetal
Copper
Aluminium
I4'6
io'9
79
22'0
266
480
135
150
88
135
I2"9
107
i8i
20'9
34'o
152
IIO
S7
092
ii8
136
•082
079
071
II3
073
065
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 . .
198
248
o8o
Mild steel . . .
222
284
078
Copper. . . .
104
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 textbooks 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^^ =/uih = 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
/»=fvf+^"
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 = o42E.
« = 4, G = E = o4oE.
« = 3. G = f E = 038E.
« = 2, G = E = 033E.
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 thincylinder 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 isj =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
36
Steel
11,000
36 to 4"6
Brass
6,400
31 to 33
Copper
10,500
29 to 30
Flint glass ...
2,400
3'9
Indiarubber ...
20
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 = fr (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 builtin beam of
breadth a, depth t, span b —
p.ab.~=fr
12
Plr
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 testingmachine, 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 <^ff +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 rivethole, 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 = 02 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 rivethole 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 Coverplate 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
lapriveted 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
rivetholes, 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 005 inch for e on the lefthand 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 = o83'^ + o'25 inch
Jt
w — d= 1*20 inches
Thus the space between the rivetholes (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 = igi"^ + 02S inch
Jt
Punched steel —
Drilled iron —
Drilled steel —
w — d = zt6 inches
w — d = ig2 inches
w — d = vg6 inches
w — d = V]2 inches
Double Coverplate 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
rivetholes 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 —
(wdcY:)tf, = ^(d + cj^
4 5
Fig. 394.
■wd= 3347 + 025
It
r \ i
w — d = 4*07 inches
k "*"■% \
Punched steel —
000
w — d= 3*59 inches
K)
Drilled iron —
1
w — d= 387 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 Coverplate Joint. — This joint
may fail by —
(i) Tearing through the outer row of rivetholes.
' (2) Tearing through the  \ ^m^
inner row of rivetholes 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>dc)tf, = ^{d + cf^^'
4 5
w — d = 2'7i inches
If we make (3) = (2), we get
{w2d 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=iW7
Punched.
Drilled.
Punched.
Drilled.
1'20 + d.
i+d.
109 + rf.
089 + d.
in.
in.
\
°67 GS)
187
167
176
156
A
1
5
078 (i)
198
178
I '87
167
'
%
087 (?)
207
187
196
176
\
09S HI)
2IS
195
204
184
2i6 + d.
196 + d.
1.92 + d.
lyi + d.
1
i
067 (JU
(a) 263
283
(a) Z43
263
(a) 221
2S9
(a) 201
■239
ji ■
\
o78(i)
294
274
(a) 251
270
(a) 231
250
B 1
\
i
087 (?)
303
2S3
(«) 276
279
(a) 256
259
;
09S «i)
3II
291
287
267
fttf
103(1)
319
299
295
27s
I
I 10(11)
326
306
302
282
407 + d.
3'87 + </.
359 + d.
339 + d.
1
\
078 (1)
468
4^
465
399
379
1
t
087 (?)
494
474
440
420
C 1
\
095 (ti)
502
482
454
434
II
i
I 03 (I)
Sio
490
462
442
I
1 10(11)
SI7
497
469
449
14
Ji7(ift)
524
504
476
456
Stress, Strain, and Elasticity.
415
Thick
ness of
plate, t.
Diameter
of rivet.
Steel plates and rivets
(drilled holes).
Inner row.
Outer row.
84 +rf.
1
1
in,
3
1
I
I
I
I
in.
09S (}■§)
103 (I)
iio(ij)
117(1^)
123 (ii)
130 (ift)
467
471
4;7S
481
485
93S
9"43
950
D i
.■:•.•)
1
...........
•
9"57
963
1
, 1
970
271 + d.
^ f
I
078 (3)
087 (?)
09S (H)
103(1)
I'IO(lJ)
ii7(ij)
I 75
179
i;83
187
igi
194
349
388
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 (wdc)f^ = nf^{d + c)t
and/s
_ \wd 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 rivetholes
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 Tiebar 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 crosssection 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 crosssections 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 iinch 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 onetenth 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 onehalf 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 thicksided 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?.rHr
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 castiron cylinders with
open ends. Wellfitting plugs were inserted in each end, and
the cylinder was filled with paraffin wax ; the plugs were then
forced in by the 100ton testing machine, a delicate extenso
meter, capable of reading to ^ ,'0 „ of an inch, was fitted on
/rttenor of
Cylinder
per ^
iMeril
pressure
'US
2607!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.
Builtup 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,
castiron 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
1215
1012
2124
1719
( 11,000
1 13,000
5,000)
6,000 j
1030
1540
Plates with grain
1315
—
—
20Z2
—
—
—
510
713
,, across ,,
H13
—
—
1820
—
.—
—
~
26
37
Best Yorkshire, )
with grain ... }
Best Yorkshire,!
across grain i
1314
lz14
—
2023
—
1719
12,000
—
1530
4050
1314
1920
IO3Z
13ao
Steel, oi % C. ...
1314
lOII
2122
—
1617
( 13.000
(14,000
5,0001
6,000 J
2730
4550
„ o=%C....
1718
1617
1314
3032
—
3426
„
2023
2732
„ os%C....
2021
—
1617
3435
—
2829
„
„
1417
1720
„ lo%C....
2829
2224
2123
5^55
■^
4247
14,000
„
45
78
Rivet steel
1517
—
1214
2628
—
2122
■ —
.—
3035
3050
Steel castings ...
TOII
(not
neal
an
ed)
3025
• 
(■12,000
to
} ~(
SI3
613
„ ,,
ISI7
(ann
ealed)
3040
/ —
^
^12,500
} ~ \
rozo
1535
Tool steel
3545
4050
—
4070
(unh
en
ard
ed)
/ 14,000
to
^ ~~ f
15
15
^
6g8o
—
—
6080
Chard
ened)
USjOoo
/  \
ni!
nil
Nickel steel Ni(
4% )
2530

—
3540




3035
505S
4565
—
__
9095
—
—
—
—
1012
2427
Manganese steel)
Mn I % ...;
15


3035




2833

., 2 %
IS
—
—
505S
—
57
—
Chrome steel (
Cr 5 %  i
3040


6075




1015
—
Chrome Vana }
6585
wat
cr
70go
(hard
ened)
—
—
1015
4555
ft „ •••
6075
oil
oil
6580
(hard
ened)
—
—
818
4555
Chrome nickel ...
4550
—
~" ,
5S6o
(ann
ealed)
—
—
1315
5055
.. ......{
100
i°5
}air
air
105
xzo
(hard
ened)
—
—
68
3035
ti »i ••• \
115
130
^oil
oil{
120
130
(hard
ened)


58
810
Cast iron
■jno
mar
lim
ked)
it S
7X1
3S6o
813
' 6,000
V 10,000
2.500 \
to
4,000 J
pra
cally
cti
nil
Copper
24
1013
anne
aled
1215
2025
1112
1 7.000
to
—
3540
5060
1012
bard
drawn
1620
—
. 7.500
—
35
4055
Gunmetal
34
S6
2 '55
gi6
3050
812
i S.ooo
l 5.5c"
2,000 (
2,500)
815
lOiS
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
24
710
56
4,000
2,000
1012
1215
Delta, bull metal.
etc.—
Cast
58
1214
—
1420
6070
—
( 5.50°
\ '°
1 —
^ 
816
1022
Rolled
IS2S
l622
2734
4560
—
( 6,000
1734
2750
Phosphor bronze
79
2426
—
6,000
2,500
Muntz metal
2025
(roll
ed)
810
—
=530
—
—
—
—
1020
3040
Aluminmm
27

710

56
J 3,000
( 5,000
} 1,700
4iS
3070
Duralumin
4S
—
—
2224
(
with
( 4,200
i 4,500
\
}
1012
3038
Oak
46
(
grain
0*2
0*07
500700
Soft woods
13
x3{
to
04
450500
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 knifeedge, 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 yL 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 onehalf the
distance of the outer blocks from the pivot has only increased
the strength of the beam by \, or, in other words, the fourblock
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. zAja" + zAjI'a' ^_ gtj._
Fig. 414.
= ■^2((zy= j a^yy^ + aij/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 halfsec
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 halfsection, 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 baseline as
shown, viz. db'. Join «'<?, l/o, cutting
By similar triangles we have —
^^■^ ^1=7=7. °^'^ = «^(7)
Beams. 437
Several fully workedout sections will be given later on.
By way of illustration, we will work out the strength of the
fourblock 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 castiron 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 castiron
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 baseline touching the
outside of the compression flange ; set off the tension baseline
parallel to the neutral axis, at the same distance from it as the
compression baseline, viz. _j'^; project the parts of the section
down to each baseline, 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 baseline is made to touch the outside of the tension
flange, and the compression baseline cuts the figure ; the parts
of the section above the compression baseline have been pro
jected down on to it, and the modulus figure beyond it passes
Beams. 441
outside the section ; at the baselines 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 baselines 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 baseline 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 Castiron 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 castiron 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 testingmachine 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 builtup sections, if there are
rivets in the tension flanges, the area of the rivetholes
should be deducted from the BA Thus, if there are «
rivets of diameter d in any one crosssection, 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.
0II8S'
The moment of inertia (see p. 88) = —
S
§1
12 VzS'
= oii8S»
BiH,'+B.,IV3A'
3H,
H, = 07H approx.
H, = 03H „
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,'*,VVi'
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.
Bullheaded
rail.
Examples of Modulus Figures.
Fig. 439.
Flatbottomed
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 ra449
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 baseline 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
JVmJtraZpl^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 tramrails 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 0405 of full size to suit the size
of page.
a in square inches.
«2
I,
in inches.
a'
b
Between
Tolal.
m and 2
038
038
014
2*20
007
2 » 3
o'6o
098
096
235
041
3 >. 4
o'6o
158
2' 50
3 "03
082
4 ., 5
o"64
222
493
225
2'19
5 „ 6
03S
257
66o
072
917
6 „ 7
0'12
269
724
042
172
O'lO
279
778
042
I8S
8 „ 9
003
282
795
042
189
9 „ 10
— 009
2 '73
745
042
177
10 „ II
—042
231
534
042
127
II „ 12
023
2o8
433
ioo
4*33
12 „ 13
082
126
159
530
030
13 .. P
— I 26
Depth of section 65 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
loaddeflection 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 crosssection, 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 socalled 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 = 0785 square inch.
— ^=(i65 ^)i2 X 0785
c = 58 inches. e = 107 inches.
472 Mechanics applied to Engineering.
Moment of resistance for the concrete —
For the stress in the rods —
^2 X 58
/Xo78s('
+ 107
j= loi,
400 inchlbs.
/= 8860 lbs. per square inch
The moment of inertia —
/6 X 5'8'V2 X 58 , \ . , 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^^ ^ = 265 X 12(225  c)
c=r^s'
^ = 315"
<?= i5'35
Beams 473
Moment of resistance of concrete —
\ ^(! X 7'i5 + 1534)  (^^^J3i5
X ^(1 X 3'iS + iS'35)4oo(i4oo)4oo = 560,000 in.lbs.
Stress on rods —
560,000 =/ X 265 X 20'I
/= 10,500 lbs. square inch.
Moment of inertia —
1400 X 7" 1 5 = 10,000 inch*units
or 265 X 1535 X 12 X 20I = 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.
WR,+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 (— ).
Bendingmoment Diagrams. — In order to show the
variation of the bending moments at various parts of a beam,
we frequently make use of bendingmoment 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
" bendingmoment 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, = WR,
In the cantilever in Fig. 469, let W = 800 lbs., be = 675
feet, bd = 45 feet.
The bending moment ai c = W .be
= 800 (lbs.) X 675 (feet)
= 5400 (Ibs.feet)
Let I inch on the bendingmoment diagram = 12,000 (lbs.
, /ii. r ,.\ ■ u 12000 Ibs.feet
feet), or a scale of 1 2,000 (Ibs.feet) per inch, or j. — jv — .
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 03 (inch) X "°°° O^sfeet) ^ ^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 bendingmoment 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).
„, 0225 (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,/^ = 22 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 95(feet)+5o(lbs.) X 7(feet)+4o(lbs.) X 3(feet )
115 (feet)
\ =755 0bsfeet)^
115 (feet) ^ ^
.R,=Wi + Wa + W3R,
R, =3o(lbs.)+So(lbs.)+4o(lbs.)6s6s (lbs.) = 543S (lbs.)
The bending moment at/, taking moments to the left of/,
= Ri«/Wie^W3^
= 6565 (lbs.) X 63 (feet)  30 (lbs.) X 4"3 (feet)  50 (lbs.)
X I 8 (feet) = 1946 (lbs.feet)
The bending moment at/, taking moments to the right oif,
= R,bf Wsef
= 54'3S (lbs.) X 52 (feet) — 40 (lbs.) X 22 (feet)
= 1946 (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 builtin 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 contraclockwise 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 baseline, and the
negative part below the baseline. 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
contraclockwise direction ; hence the shear diagram between
a and c will be above the baseline, 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= tWj „ 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'+MJMf"
iiiiiiiiiiiiwiMiiiiiiir yry_
Fig. 475.
IV
+ shear ; then bringing the + portions above the baseline 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
ww,
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 Bendingmoment 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.%27, lbs. /« = 5 feet
find (W. = 678 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 baseline is — o'4oi sq. inch, and the positive part above the
baseline 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)  1322 (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 bendingmoment 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 bendingmoment
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 bendingmoment 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
incmoment
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
shearingforce 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
righthand section ;ir is W . 4 ; likewise
with the lefthand 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 righthand 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 onehalf 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
bendingmoment 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
inffmoment
dUsnun
Ibs.inches.
in inches.
Scale of W
fff Ibfc. =11 tncn.
Scale of /,
i full siie.
n
Remarks.
The bendingmoment 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/ =483/,
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'
ingmoment
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 bendingmoment 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 bendingmoment diagram ^= j; or
R./X4
K/ = ^ =
= =R,/.
0/
if X /. W,/, X r.
= W/.
and the depth ofl
the bendingmo > = 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 bendingmoment 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 bendingmoment 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 bendingmoment 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 . 05)
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.
,^ = (p8)^+(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 bendingmoment
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 bendingmoment 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 bendingmoment 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 bendingmoment 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 bendingmoment 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 = = Elj5
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(lL')
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 bendingmoment 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 bendingmoment diagram
measured in inches ;
OH = the polar distance in inches ;
M =fthe bendingmoment in inchlbs.;
M = w.w.D.OH;
hence i inch depth on the bendingmoment diagram represents
M .
=r = m .n . OH mchlbs. j and i square inch of the bending
moment diagram represents tn .rfi . OH inchinchlbs. \ hence —
f area of bendingmoment 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
ingmoment 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 bendingmoment
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 bendingmoment diagram obtained by
taking the original bendingmoment diagram as a load diagram.
Hence, if we construct a second shear diagram, still treating
the original bendingmoment 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^' +
Jig / ixiJji — K2iA2
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 railwayplate 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 endingmoment diagram shown below
the beam by the method given on p. 498. Then the bending
moment at any section is M = »«.«. D . OH inchlbs., using
the same notation as on p. 499. Then i inch on the vertical
M
scale of the bendingmpment diagram = — = »«.«. OH
inchlbs., and i inch on the horizontal scale = n inches.
Hence one square inch on the diagram = m . «'0H inchlbs.
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 bendingmoment diagram may be found
readily by a linkandvector polygon, similar to that employed
for the bendingmoment diagram itself.
Treat the bendingmoment 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 baseline, 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 bendingmoment diagram
also I inch on the length of the bending1 _
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 bendingmoment diagram at the
point in inches, and OH is the polar distance, also expressed
in inches. Hence i inch depth of the bendingmoment diagram
represents Y=r= wz« . OH inchlbs., and i square inch of the
bendingmoment diagram represents ot«*OH inchinchlbs.
Hence the area xQ represents arDww^OH inchinchlbs. ; 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 bendingmoment 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
ii
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 = ^ = 06.
Referring to the table, ^ = 00107.
Then 8 =
00107 X 3 X 240
= 024 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"
0175
037S
059
076
00095
00177
00194
00139
00285
00885
003S8
00556
02114
Find the deflection under the 2 tons load,
tons per sq. inch. I = 630.
E = 12,000
8 =
02114 X 240'
I2000 X 630
= 039 inch.
By a graphical process 8 = 040 inch.
Deflection of Builtin 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
builtin 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.
■02S—*
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 bendingmoment diagram between those
sections, the net area of the bendingmoment diagram for a
builtin beam must also be zero.
A builtin beam may be regarded as a freeended 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 —
fegadfgcb^ o
But in order that this condition may be satisfied, the area
of the piermoment diagram adcb must be equal to the area of
the bendingmoment 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 heightline of the
bendingmoment diagram for the freeended beam.
Thus when the ends are built in, the maximum bending
moment is reduced to onehalf, 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 o289'L '' ^ wL»
2 2 24
Maximum bending moment on cantilever spans —
w\^ + ^' = «' X 0289L X o2iiL+"'^°'"''^'
2 2
_ «/L'
13
Deflection of central span —
g ^ 5K/(o578L)* ^ w\}
' 384EI 689EI
Deflection of cantilevers due to distributed load —
« _ w(o'2iiLy _ wh*
^ 8El 4038EI
Deflection of Beams.
Deflection due to halfload on central part —
5 _ zfLa X Li' _ w X o'289L x o2ii'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 bandingmoment
diagram due to the external loads between P and Q is equal
to the moment of the pier bendingmoment 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 builtin 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 builtin 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 onehalf 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./(^L9 = ^\3L84)
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«
^^ = —rv^ 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
_ 4s46wU _ WL'
^~ 384EI ~ 884EI
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" _ oo309PL»
884EI ~ EI
W
P = —  = 037W
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 bendingmoment 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 g6 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 15
3
6
.ffMOT, evenly distributed load 48
96
192
Cantilever, „ ,, 2
4
8
Beam, two symmetrically placed loads, as in} .,
Fig. 423 S"^^
93
l86
.ffMOT, irregular loading (approx.) 55
II
22
Cantilever, „ „ 175
35
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'iiX >
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
N
s
Q
8
N
5
_,^
, s
.~i
tM
ST'^
Q
1
Q
T,
5
1
M
a
+
a
all
'mm
Q
H
CO
P 6
— ,D
u
M
ll
,
J
tS
1^
1 S
a
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 Tiebar.— 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 tiebar 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:=RARA+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 = (yA){e!,eO = {yA)^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.}
^) <^^) ' 4a>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 = 1453 sq. ins. Aj = 1574 sq. ins. Ri = 5 ins.
h = 0384 in. M = 85 in.tons. yi = 2'46 ins,
y, = 281 ins. Then/, = 125 tons sq. in.
548
Mechanics applied to Engineering.
The Andrews Pearson theory gave i3"9 tons sq. in., the
common theory 88 tons sq. in., and by experiment i3'2 tons
sq. in.
By this theory/, = 6*2 tons sq. in., the AndrewsPearson
theory gave 4"6, and the common theory 76 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 punchingmachine with
a boxshaped 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.45) 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 oneeighth 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
crosssection).
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=XMx§X^EI = 7^7 = 7^
3 2
a„dW = 9^
' 96EI 96EI
The value obtained by Euler was —
_ g^EI _ 987EI _ 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 just 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 builtin beams.
The squareended 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 roundended 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
HL4LU
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 castiron 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 closegrained).
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,
crossgirders, roofcoverings, rooftrusses, and many other
details which a designer constantly has to deal with.
Straightline Strut Formula. — The more experience
S66
Mechanics applied to Engineering.
one gets in the testing of fullsized 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 straightline 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
HtffJU
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+ JU
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 onetenth ' 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 WfWi
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
onetenth or »ven ouetwenlieth.
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 bendingmoment 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 = 02 ^
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 0355 ' 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 tonsinches
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
pinholes in the other disc coupUng. Before W was applied
the pinholes 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, = oi96D{ii)
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 onehalf the depth of the keyway j thus,
the strength of a 2inch shaft having a sunk keyway 025 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, = 0I96D3
z =
019603(1
m^J
z.
irTid^
Z, = or96D<^
Z„ = 0208S3
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 reen
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
aGD 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 „ „
Horsepower 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 footlbs.  ^
where N = number of revolutions per minute.
27rPrN
The horsepower transmitted = = H P
'^ 12 X 33000 •"'
then —
_ 12 X 33000 X H.P. /D»
^'■~ 2irN ~ 51
_3 _ 12 X 33000 X H.P. X 51 _ 321400 H .P.
2tN/.  N^:
643 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,gif^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 castiron torsion
testpieces, 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 castiron 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
255
0°
0°
777
1925
4000
267
12°
11°
nyo
2240
475°
271
14°
14°
1228
22SS
4820
231
17°
15"
1308
2128
4628
240
19°
i6»
2606
137s
4320
208
,,0
31°
2644
766
3520
i62
38°
37°
3084
Zero
3084
i6'o
43°
45° Mean of a
Pure shear ...
130
0°
0° large number
„ tension...
...
115
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 inchtons, and a maximum twisting moment
of 450 inchtons. 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 inchtons.
D' 420 _ . ,
= ~ — D = Q'l; mches.
102 5 ^ =
The equivalent twisting moment (Guest) —
= Vsoo^ f 450^ = 540 inchtons.
— = ^^^ D = 07 mches.
513
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 righthand 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 _ 255WD ,. .
then 8 = ^SP'^ (from i. and ii.)
Substituting the value of / = mrD —
g^ 255D'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^
2S5D
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/, _ 314 X 70,000 _ 1^
D^ G 11,000,000 50
and so(M  1) = —
For a close coiled spring, p= d, then —
Vso(Mi)=?
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 255D X Qd ^^'""^ "• ^*i ">•)
Jl^nQ
(substituting the value of /) = , ^ (inchlbs.)
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 o6gnd^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) 556 224 503 895 1398
h also gives the number of footpounds 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 = ' „ (inchlbs.) steel
24,680,000 ^ '
weight = o88«S''D (steel)
Height a squaresection spring) , fl_ ,, .
will lift itself (steel) ] "■  260,600,000 ^ >
/^ (lbs. per square inch) ... 30,000 60,000 90,000 120,000 150,000
iCfeet) 345 138 311 553 86'3
It will be observed that in no respect is a squaresection
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 poundsinches ;
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 —
Opencoiled 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
0998
IO°
0992
IS"
0986
20°
0978
30°
0950
45°
0900
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 crosssection 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 _ oo8o7 X 1466'' X V . ^,,
r = ! = o'ooi;4V*
322 ^^
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 onehalf) 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
i8i
26'4
333
381
400
410
70
137
20'0
257
306
346
376
96
ii7o
228
255
245
20'0
i4"o
69
129
i7'3
197
197
173
129
17
62
132
214
292
34 o
38S
17
47
100
165
235
300
354
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'360'4i
05
0S40S7
0'8o'97
0'607
09S
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
75
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.
89
1220
ii3i
24
S8
iii
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 windload 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 windload 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 bendingmoment 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 bendingmoment
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, = i6S + 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
= 236 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 bendingmoment
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 tiebars, 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 tiebar 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
bendingmoment 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 bendingmoment
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 shepr 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 bearingpressure conditions —
p = ^d (for a single row of rivets) (approximately)
p = 6d (for a double row of rivets) „
The jointbearing 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 teestiffeners 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 teestiffeners 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, wellfitted
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
suspensionbridge 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 suspensionbridge 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
22 3
or, according to Trautwine —
d =
where K = i for the best work ;
I •12 for secondclass work;
I '33 for brickwork.
The secondmentioned 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 thirdmentioned 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
linkandvector 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 righthand figure, are
drawn in for the arch loaded on one side only, for onethird
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.
li
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 crosssection,
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^(pR)»
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—
/=(RRO
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 vjS
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 fiveeighths 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^
/ = 04 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 crosssection ;
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.
3820
18,741
2625
1,762,000
33'40
46,286
2507
1,031,200
2865
170,170
2483
1,477,400
2614
123,770
2387
5,234,200
2387
473,766
2387
40,600,000
2292
13,600,000
(unbrolcen)
(unbroken)
Nominal ben
revoU
from
ding stress in a
ing azle
to
Number of repetitions
before fracture.
201
— 201
5S.IOO
172
172
1Z7.77S
163
163
797,525
153
'53
642,675
11
M
1.665,580
I)
l»
3,114,160
143
I4'3
4.163.37s
,(
r*
45,050,640
Material.
Krupp's Spring Steel.
Tangle strength, 575 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
792
62,000
3820
477
99.700
>*
1592
149,800
»J
955
176,300
w
2387
400,050
»»
1433
619,600
)i
2783
376,700
»»
2.135.670
I)
3152
19,673,000
(unbroken)
11
1910
35,800,000
(unbroken)
4295
9SS
81,200
3341
477
286,100
If
1433
1,562,000
1.
955
701,800
1910
225,300
II
1194
36,600,000
II
2387
1,238,900
(unbroken)
»»
300,900
If
2865
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
153
56,430
134 •
134
183,145
II5
riS
909,84.0
%^
 96
4,917,992
86
 86
19,186,791
2,063,760
535,302
76
 76
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 overannealed 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
loo
95
90
85
80
lO'O
 9S
— 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 crankshaft ;
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= 0434 lb. Hence a column of water i foot
144
high produces a pressure of o'434 lb. per square inch.
Temp. Fahr.
3»°.
igJ>.
50°.
100°.
T50°.
«o°.
Ice.
Water.
Weight per 1 '
cubic foot in }
lbs )
57'2
62417
62425
62409
6200
61 20
6014
5984
Volume of a \
given weight,
taking water
at 39'2° Fahr.
as I )
I '091
I •0001
ioooo
10002
1007
I 020
1038
1043
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 ;
= 0434 lb.
Then p = wh, or 0434^
or h = ~ = 2305/, or say 231/
Thus a head of 231 feet of water produces a pressure of
I lb. per square inch,
Taking the pressure due to the atmosphere as 147 lbs. per
square inch, we have the head of water corresponding to the
pressure of the atmosphere —
147 X 231 = 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 « = /„ = oi,j,i,h„
b = P, = o434'4.
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 abovementioned 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 lefthand 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 baseline,
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 supplypipe 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 / ~ lootiDs.
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 sharpedged tube which is inserted in the jet in
such a manner that the water plays axially into its sharpedged
mouth ; the other end of the tube, which is usually bent for
convenience in handling, is attached to a glass watergauge.
The water rises in this gauge to a height proportional to the
velocity with which the water enters the sharpedged 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 sharppointed 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 screwpoints 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 064, 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= oqTK'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 sharpedged
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 095 of this, on account of
friction in the pipe. Then, taking a mean value —
Q = 0945 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) ...
l6i
131
lO'I
81
61
0786
41
0780
3'«
Kd
0799
0797
0797
0792
o'773
Vacuum head by ex
periment in inches
167s
I4'37
1195
1050
900
745
635
Vacuum head by cal
culation
1 6 '60
1424
1197
1045
885
737
656
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 watergauge to the orifice nozzle.
Reentrant 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
056
oSS
0S4
0S3
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 069 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\/'
26i8  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
0679
0672
3
0645
0640
4
0634
0631
S
0631
0626
6
0629
0624
8
0628
0622
10
0627
0620
100
0625
o6i8
1000
0625
o6i8
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 —
H4
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 18756.
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,).
6924
6924
6924
1250
1250
1250
800
200
025
5885
5078
None
850
500
150
None
None
None
None
33S
33S
113
33S
33S
33S
82
052
6s 97
8097
8143
3790
S398
5460
5167
2474
666
6678
8134
?3
S443
5443
5170
2563
704
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 highpressure 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
backpressure valve, and the spaces should gradually increase
in area as the highpressure 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 = (03838 + oo386l+°:?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 rightangled notch.
where C has the following values. See Engineering, April
8th and 15th, igio.
H (feet) ...
005
oio
ois
020
025
030
040
C
o'289
0304
0306
0306
0305
0304
0303
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
Hh
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 crosssection, 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
BroadCrested 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 = osSsBHV'i^
The actual flow in small smoothtopped 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 crosssection.
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^= 2SJih''
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 QiK^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{NVHiQe) NVH^Q, NVHsQ,
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 = 05 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 o5_ 151 X 05 r62 X o5_
2(48—7^/12) 48 — 7Vi25 48 — 7V13
170 X 05 , 190 X o'5
H —^— \ p= = 143 seconds.
487Vi3s 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
Rightangled Vee Notch. — In this case we get by similar
reasoning —
T = ^SA^ r .= dh = _25A„^( H,^  H.^^ j
8K^//2^iHa 8K,V2^A 15 /
T ^ 5A. / I i_\
Flowthrough 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^ footlbs., or h footlbs. per pound of
water. By lowering the water to the datum, WA footlbs. 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 footlbs.
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 = footlbs. of
energy stored in the falling water when it reaches the datum,
or — footlbs. 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^ = ^ footlbs. of work done by the
water under pressure, or ^ footlbs. per pound of water. This
w
is known as its pressureenergy, or —
The pressureenergy = —£
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 watercolumns 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 waterinjectors and partially opened
valves is also due to this cause.
Ventnri Watermeter. — An interesting application of
this principle is the Venturi watermeter. 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 136) 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 097 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^ = vr
»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 breakingup
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 p674).
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 HeleShaw,
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 HeleShaw, 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 streamline
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 streamline motion. Professor HeleShaw'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 HeltShaw, 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 HeleShaw in the Transactions of the Naval Architects,
189798, 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
025
0040
52, SCO
40
021
0034
61,000
60
015
0025
83,500
80
0'12
0019
io7,coo
100
O'lO
0016
134,000
120
o'o8
0013
164,000
140
0065
ooii
204,000
160
o'oss
0009
240,000
180
0047
0008
278,000
200
0040
0006
328,000
212
0037
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 endon 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.
13
B
\
03^
02 
^!??5
rAf
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
... 00096 „
... 00043 ..
... 00031 „
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 castiron 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=38SD^\/J=38SD^>/
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

Wroughtiron pipes
l8o
0004787
065
0018
0*07
Riveted sheetiion pipes
1825
0005674
0677
—
New castiron pipes
/I85
\200
0005347
0006752
067
063
—
Lead pipes
I7S
0005224
062
—
Pure cement rendering
/1 74
1 1 '95
0004000
0006429
067
o6i
—
—
Brickwork (smooth)
zoo
0007746
o'6i
„ (rough)
2 co
0008845
0625
001224
050
Unplaned plank
200
0008451
0615
°03349
050
Small gravel in cement
2'00
0OII8I
066
003938
060
Large „ „
200
0OI4I5
0705
007590
I 00
Hammerdressed masonry ...
20O
0OIII7
066
007825
I 00
Earth (no vegetation)
2 CO
001536
072
Rough stony earth
200
002144
078
—
If we take x as 062, and « = 2, C = 00067, wc get —
Q =
which reduces to —
301 C VS
Similarly, for new castiron pipes —
h = .
320oD''»*
taking « = 185, 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
it
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, aircocks 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 KRl2^/ 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=iS3
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.
1500
150
10
2834
279
20
3622
356
30
4195
413
40
4642
459
5°
5003
49S
60
53'oi
S26
70
5550
552
80
5760
57:4
90
5939
593
100
6093
609
* 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 footlbs. 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 rightangled 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
Rightangled knee or elbow (experiments)
Rightangled 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 squareended)
, 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 reentrant pipe, such as\
a Borda's mouthpiece ... ... ...J
Water entering a squareended pipe flush)
with the side of the tank /
/3040 in plain pipe
\ 5090 with screwed elbow
31 S
2230
24
1620
Hff
12 approx.
120400
27
200
HOC
7001500
20003000
ioo(« — l)'
36(1 Sb  i)»
18
912
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 footlbs. 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 weightenergy 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 waterwheel, 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 ^= = 16 foot. Onehalf
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,) _ ..„
ois X 125 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 steamengine.
The work done is the same as before, viz. WH footlbs. 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
footlbs. 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 wirerope 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 ;
reentrant 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 /
75
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 halfpower, 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 highpressure
water used is the volume A minus the volume B, which is
usually onehalf 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 waterpressure
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 halfpower. As soon as the pressure is raised in A,
the valve D returns to its fullpower 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 highpressure
water can be obtained, conse
quently a powerdriven 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 highpressure 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 footlbs. = o']2>^cPps
Work stored per cubic foot of water in 1 , , .
footlbs. [ = '44/
Work stored per gallon of water = ^r; — = 23'o4^
Number of gallons required per minute \ _ 33»°°p _ 143'
at the pressure / per horsepower  ~ 23'o4/ "" p
Number of cubic feet required per minute 1 _ 33>°°° _ 2292
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 waterpressure 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 airvessel) ;
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 returnwave, and thereby
causing the waterpressure 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, = 0434 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 crosssection = =^=rp X —
2TE 2
The increase in volume of the pipe perl _ \jfd
square inch of crosssection ) 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 _ o434LV^
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 = 635V
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
06
20
3'o
7S
Observed pressure lb. sq. inch
43
"3
173
426
Calculated/ = 6341; ....
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 velocitytime 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 WaterMain. —
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 = 3850^ \/^
Each cubic foot of water falling per second through a height of
one foot gives — •
625 X 60
= 0*1135 horsepower
33,ooo_ •'^ ^
henceH.P. = oii3sQH
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. = oii35Q(H4)
Substituting the value of Q from above, we have —
ap; = 0II3S X 3850^ (H  h)\/ J
Let h = «H. Then, by substitution and reduction, we get
the power delivered at the far end —
, , /«H»I
H.P. = 437 (i  «) V 17
iginWB^i  «)'
H.P.''
These equations give us the horsepower 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—
^=(^Hf^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 horsepower that can be transmitted
through any given pipe is —
H.P. {max) = 167 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 to 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 seawave, 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 waterwheels, 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 footlbs. —
wv<iJ)
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»^
dn
= — 2tt~' + 4« ' = o, when t) is a maximum
dn
or 2H~' = 4«~'
i = 4, whence « = 2
Hydraulic Motors and Machines.
709
undershot waterwheel ; 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
io
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
10 20 30
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 horsepower.
Horsepower 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 waterwheels 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 02 03 O* 05 06 07 08 09 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 crosssection 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 crosssection of the main
per second, or through the nozzle, is AV cubic feet per second,
or 624AV lbs. per second.
The kinetic energy of the stream issuing from the nozzle
is —
624AV
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 nl) \
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"^ = 44\/t
taking an average value for K.
4^L
01 02 03 Ol 05 06 07 08 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 crosssection of the pipe)/v/ —
■45Vt;
L
/D"»
or, 3.
L
Poncelot Waterwheel 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 guideblades are perfectly free, and after
impinging on the wheelvanes the water at once escapes into
the air above the tailrace.
(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 guideblades 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 tailwater, 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 mixedflow
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
wheelvanes 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
tailwater, 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
tailrace.
The water enters the guides vertically, hence the first tip
of the guideblade 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 = o94V'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^ — 027^
= o"9^ = Vi : cb gives us the direction in which the water leaves
the guidevanes ; hence a tangent drawn to the last tip of the
' We omit n/2^H to save constant repetition.
726 Mechanics applied to Engineering.
guidevanes must be parallel to cb. We are now able to con
struct the guidevanes, having given the first and last tangents
by joining them up with a smooth curve as shown.
Let the velocity of the wheel be onehalf 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 = 045 : 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 wheelvane 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 —045, 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 = 045. Then ffe gives us the
final velocity of the water relative to the wheel; hence the
tangent to the last tip of the wheelvane 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 easyconnecting curve, in order to reduce the surface
to a minimum. The wheelvanes 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 wheelvanes 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 wheelpassages 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 welldesigned 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 outwardflow 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 guideblades will make
an angle 61 with a tangent to the outer guideblade 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 wheelvanes 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 guideblade 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 wheelvanes, 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 = 0628, and the outlet tip of the wheelvanes
will make an angle B^ with a tangent to the outer periphery of
oo8
the wheel, where tan B^ =
= 0127, 3nd 6t — 7° 16'.
0628
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 ^,
\vy
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 footlbs. I ~ ^
(i.)
energy wasted in discharge! (z/ — V)^ ■.
water per pound in footlbs. ) ^ ' ■ • '"•)
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 guideblades, 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 tailrace, 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 abovementioned 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 onehalf is converted
into pressure energy, and onehalf into kinetic energy. If 80
per cent, of the head be utilized, the corresponding velocity
will be —
V=\/
H X 80 „ ,__
'^ Tnn =o89^/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 063 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 footlbs. per second perl _ V^V, _
\
pound of water I g
since we are going to utilize 80 per cent, of the head.
= 08H
Hydraulic Motors and Machines.
731
Hence V„ =
o%H
063 / 2^H
V„= o•64^/2^H
o63>/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 = 069, which
gives us the direction in which the water enters the wheel or
leaves the guides. The guideblade is then put m by the
method explained for the impulse wheel.
^ WHEEL
Fig. 69s.
To obtain the form of a wheelvane. 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 wheelvanes. 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 — 064. Then^ setting down gh = o"2 7,
and^'= — 064, 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.
wheelvanes 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 —
08PH
V = ■
V„ = 045 Va^H
Or the velocity of the wheel is onehalf 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
inwardflow turbine may be utilized in securing steady running,
and in making it partially selfgoverning, 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 inwardflow 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 outwardflow 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 „ „ 0375 >.
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.)
068
068
100
068 •
018
068
200
067
072
067
300
066
16
066
400
064
29
064
500
0'62
45
o6i
600
059
65
058
700
oSS
88
0"S4
800
050
"•s
049
900
042
HS
043
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 waterinlet 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 = horsepower 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 baseline, 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 18845 ■ —
1'"G. 696.
Type of turbine.
Efficiency per cent, at various
sluiceopenings.
Full.
09.
08.
07.
06.
05.
04.
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 t