LIBRARY
UNIVERSITY OF CALIFORNIA.
Clas.
STRENGTH OF MATERIALS
STRENGTH OF
MATERIALS
A. MANUAL FOR STUDENTS
OF ENGINEERING
BY WILLIAM CHARLES POPPLEWELL
M.SC. (VICT.), ASSOC. MEM. INST. C.E.
Lecturer on Strength of Materials, Theory of Structures, and Hydraulics, at
the Manchester Municipal School of Technology ; formerly Assistant
to the Professor of Engineering in the University of Edin-
burgh ; and Senior Assistant Lecturer in Engineering
at the Yorkshire College
OFTH&
cRSlTY
OF
l\vb*
OLIVER AND BOYD
EDINBURGH: TWEEDDALE COURT
LONDON: 10 PATERNOSTER ROW, E.G.
1907
TKQIBB
PREFACE
•
THE greater part of the matter contained in the following pages
is based on the notes of lectures given to the day and evening
students at the Manchester Municipal School of Technology
during the last five sessions.
The book is intended for the use of those students of engineer-
ing who are desirous of obtaining a working knowledge of the
fundamental principles involved in problems of machine and
structural design. It should be found useful to candidates for
the Third and Honours stages of the Examinations of the Board
of Education, the examination for the admission of Associate
Members to the Institution of Civil Engineers, as well as the
examinations in the Engineering Schools of the Universities.
It will be seen that special attention has been paid to the
unequal distribution of stress, and to the limits of elasticity in
iron and steel. Many of the examples quoted are taken from
experimental results obtained by the writer or his students. It
is also to be noted that the majority of the proofs given are
similar to those used in most of the text-books.
The author desires to acknowledge his indebtedness to the
many writers of books and scientific papers to which he has
referred in collecting these notes. W. C. P.
MANCHESTER, 1907.
175497
CONTENTS
PAGE
INTRODUCTORY ........ xi
CHAPTEE I
STRESS, STRAIN, AND ELASTICITY
Stress; Load; Intensity of Stress; Strain. Different Kinds of
Stress and Strain : Simple Stresses. Distribution of Stress —
Uniform Stress : Elasticity— Plasticity ; Hooke's Law ; Young's
Modulus for Direct Elasticity ; Lateral Contraction and Dilata-
tion 1
CHAPTEE II
DIRECT, TANGENTIAL, AND OBLIQUE STRESSES
Stress on an Area not at Right Angles to the Axis of a Body under
Simple Tension or Compression. Effect of two Direct Stresses
at Eight Angles : The Nature of Shear Stress ; Shear Strain-
Modulus of Eigidity ; Elasticity of Volume or Cubic Elasticity ;
Relation between E, G, K . . . . . .15
CHAPTEE III
STRESSES IN BEAMS — BENDING AND SHEARING ACTIONS
Distribution of Direct Stress in a Beam Section : Moment of Eesist-
ance ; Moments of Inertia ...... 25
CHAPTEE IV
GRAPHICAL METHOD FOR DETERMINING THE MOMENT OF INERTIA
Moment of Inertia ....... 39
vii
PAGE
viii CONTENTS
CHAPTEE V
DEFLECTION OF BEAMS
First Method ; Second Method ...... 47
CHAPTER VI
SHEAR STRESS IN LOADED BEAMS
Distribution of Shear Stress in Loaded Beams. Deflection Due to
Shear. Continuous Beams ...... 69
CHAPTEE YII
RELATION BETWEEN LOAD AND STRESS IN A PRISMATIC BAR
The General Problem. Particular Cases. Instances of Unequal
Stresses due to Eccentric Loading. Method of Measuring the
Strains ........ 78
CHAPTEE VIII
PILLARS, STRUTS, OR COLUMNS
Strength of Pillars, Struts, or Columns . . . . .92
CHAPTEE IX
TORSION AND SPRINGS
Elastic Circular Shaft ; Angle of Twist of a Shaft ; Horse-power
Transmitted. Torsional Strength of Shafts in the Plastic State.
Loads and Deformations of Springs : Helical Springs of Round
Steel ......... 99
CHAPTEE X
TORSION COMBINED WITH BENDING
The Effect of Torsion Combined with Bending . . . .110
CONTENTS ix
CHAPTER XI
STEENGTH OF CYLINDERS
PAGE
Thin Cylinders ; Thick Cylinders . . . . .115
CHAPTER XII
RIVETED JOINTS
Strength and Efficiency of Riveted Joints . . . .125
CHAPTER XIII
STRENGTH OF MATERIALS AS FOUND FROM THE
RESULTS OF TESTS
Testing; Testing Machines; Appliances for Measuring Elastic
Deformations . . . . . . .131
CHAPTER XIV
THE LIMITS OF ELASTICITY
Limit of Proportionality ; Yield Point ; Illustrations of the Three
Limits ; Unsymmetrical Loading ; Changes of Limit by Previous
Loading ........ 138
CHAPTER XV
THE MATERIALS USED IN CONSTRUCTION
Iron and Steel : The Low-Carbon Steels ; Medium Steels ; High-
Carbon or Hard Steels; Steel Castings; Cast Iron;
Malleable Cast Iron. Copper : Alloys of Copper. Vitreous
Materials : Stone ; Bricks ; Cement, Mortar, and Concrete ;
Portland Cement Concrete. Repeated and Reversed Stresses :
Safe Stresses Allowable in Practice. The Microstructure of
Metals .... 149
x CONTENTS
APPENDIX —
PAQB
General Table of Strengths and Weights . . . .167
Elastic or Young's Modulus . . . . .168
Weights of Materials . . . . . .169
ADDITIONAL EXAMINATION QUESTIONS . . . .171
INDEX 179
INTRODUCTORY
ALL material used by the engineer, whether it forms part of some
piece of mechanism or of a fixed structure, has to withstand the
application of force. The structural part must be so proportioned
by the engineer who initiates its design as to be able to carry the
loads which come upon it without injury to itself, and at the same
time without the employment of more material than is necessary.
It is for the carrying out of this safe and economical design that
a knowledge of the " Strength of Materials " is required.
The causes producing the loads which come upon the various
parts of a structure, and their magnitudes as depending on outside
effects and upon the form of the structure as a whole, will not be
discussed in any detail in what follows. That part of the subject
is generally dealt with under the title of " Theory of Structures."
Here the reader is more particularly concerned with a knowledge
of the relations between the ascertained loads and the dimensions
and forms as affecting the stresses in structural members.
The subject of " Strength of Materials " naturally divides itself
into two parts. The former of these deals with the nature and
intensity of the forces which come upon a part, as depending on
its form and the loads which it has to carry. The second portion
of the subject relates more especially to the effect which these
forces have upon the internal structure of the material itself.
One is analytical and depends upon mathematical proofs, the other
is descriptive and experimental.
This order will be observed in the following chapters, the
general laws bearing upon the relations of stress and strain, which
apply equally to all materials, being taken first ; to be followed by
a more detailed discussion of the effect of stress upon particular
materials.
STRENGTH OF MATERIALS
CHAPTEE I
STRESS, STRAIN, AND ELASTICITY
Stress. — Stress is the force exerted by a portion of material
upon that part adjacent to it.
A stress may act normally to a surface, when it will be either
a compressive or a tensile stress. In the former of these the
tendency is for the two portions of material on opposite sides of
the section to be pressed against one another ; in the latter case
the tendency is towards a separation of the parts. Or, the stress
may be tangential to the surface in question, with a tendency to
cause the portion on one side of the section to slide upon the
other: this is called a shear stress. Again, the stress may be
partly normal and partly tangential.
Load. — This term is applied to the total force which acts upon
a structural part. A body acted upon by a load is said to be in a
state of stress. Thus, in the case of a metal rod which is with-
standing a pair of loads or forces pulling away from one another
at the opposite ends, the whole of the material between the two
ends is said to be in a state of tensile stress, and it is only the
mutual adhesion between the individual particles which prevents
the metal from being torn asunder at any point.
Intensity of Stress. — Stress is generally defined as being so
many units of force acting upon a unit of the area referred to.
The load, on the other hand, is defined as so many units of force,
irrespective of the extent of the surface upon which it acts. Thus
i A
2 STRENGTH OF MATERIALS
in the case of the above tension bar, the load might be given as
so many tons, and if the extent of the area of the cross-section
were so many square inches, the stress would be given as so many
tons acting on each square inch, or as so many tons per square
inch.
The principal units employed for the measurement of stress
are as follows : —
In Great Britain. — For the metals and timber the stresses are
usually given in tons per square inch, and, less frequently, in
pounds per square inch.
For brickwork, masonry, and concrete, stresses are given in
tons per square foot, and sometimes in pounds per square foot.
In the United States of America the units employed are
respectively pounds per square inch and pounds per square foot.
In the countries using the metric system, kilogrammes or
grammes per square centimetre or millimetre are the units.
Strain. — Strain is the deformation brought about by stress.
Stress never occurs without changing the shape of the piece of
material on which it acts.
For example, in the case of the steel bar before referred to, a
tension load or pull is put upon the bar, causing tensile stress
throughout its length, with the result that the bar is stretched :
this stretch is the strain.
The strain may be temporary or permanent, or part of it may
be temporary and part permanent. Under the ordinary working
loads which are put upon engineering materials the strains
accompanying the stresses are relatively of very small magni-
tude, and are almost always temporary, disappearing on the
removal of the loads. When, however, the loading is carried
beyond the working limits the strains become larger, and the
material only partially recovers itself on the removal of the load,
leaving the remainder of the deformation as permanent strain.
Different Kinds of Stress and Strain.
Simple Stresses. — There are three kinds of simple stress,
namely : Tension or pulling ; compression or thrusting ; and
simple shear.
In Tensile Stress the loads act outwards along the axis of the
piece of material, giving rise to a tensile stress on any section
STRESS, STRAIN, AND ELASTICITY 3
normal to the axis. The strain in this case consists of the amount
the bar is stretched ; this is shown by the dotted portion on Fig.
I (a).
In Compressive Stress the line of action of the loads is the
same, but they tend towards instead of away from one another,
(*)
\
O>-
(I)
FIG. 1.
giving rise to a stress which compels the particles into closer
union and at the same time causes a shortening, Fig. 1 (&).
The above are called direct stresses.
In Simple Shear the loads act parallel to one another in
4 STRENGTH OF MATERIALS
opposite directions, tending to cause the two portions of the
material acted upon to slide one upon the other.
The strain caused by shear stress is one of distortion. This is
shown by the dotted portion, Fig. 1 (c).
Torsion is the particular case of shear stress which occurs
when a shaft is twisted, Fig. 1 (d). The effect is to cause any
two adjacent normal sections of the shaft to revolve relatively to
one another. The strain in this case is measured by the angle of
rotation of one end of the shaft with respect to the other.
Bending or Cross-breaking is the kind of stress met with where
the load is applied in such a way that it causes a bar to bend, as
shown on Fig. 1 (/). Here the bending action of the load results
in a curving of the beam, the material on the convex surface being
lengthened and put in tension, while that on the concave side is in
compression. Thus there are two kinds of simple stress occurring
at the same time. The strain in this case is the amount of deflec-
tion of the centre from its original position, measured in a direction
at right angles to the axis.
Distribution of Stress — Uniform Stress.
It has been said that stress is expressed as the amount of the
force (compression, tension, or shear) acting upon each unit of area
exposed to it. When the load is caused to act in such a way that
the intensity of the stress is the same at all points of the sectional
area considered, it is said to be uniform. In such a case the
stress is
where P is the total force or load and A the area of the section
considered. This applies to simple tension, compression, and
shear.
But it is possible to apply the load in such a way that the
intensity of the stress is not the same at all points of the area.
For instance, if the pull in a tie bar is applied along a line which
lies outside the geometrical axis of the bar, the intensity of the
stress on a section at right angles to the axis is not the same at all
points. In such a case the above equation only serves to give the
average stress.
STRESS, STRAIN, AND ELASTICITY 5
Also, a stress may or may not vary uniformly. The principal
cases of stress are summarised in the following table : —
Nature of Stress.
Kind of Strain.
How Strain is Measured.
Tension \
Compression/
Change in length
(Change in length)
(Original length)
Shear
Distortion
(Angle of distortion)
(Unit angle)
Torsion
Twisting
(Angle of twist)
Bending
Deflection
(Amount of deflection)
(Span)
Elasticity — Plasticity. — It has been said that the strain in a
piece of material under stress is sometimes temporary and some-
times permanent, or partly temporary and partly permanent. If,
after the removal of the load, the strain wholly disappears by
reason of the material recovering its original form and dimensions,
the strain is said to have been temporary and the state of the
material to be elastic.
When, however, the material fails to recover its original
dimensions and some of the strain remains after the removal of
the stress, the material is said to have been strained beyond its
elastic limit, and to have acquired permanent set.
Almost all the materials of engineering exhibit this property
of taking permanent set after a certain stress has been reached.
In some, permanent set is found after the application of very
small stresses.
Materials like wrought iron and mild steel are found to arrive
at a point in the loading when the strain goes on increasing with
little or no increase in load. When this point has been reached
the material is said to have become plastic.
In materials where this plastic state is reached, the period
6 STRENGTH OF MATERIALS
extending from the end of the elastic to the beginning of the
plastic stage is spoken of as the semi-plastic stage. The part of
this subject which relates to the stresses and strains in the plastic
and semi-plastic stages will be left for later consideration.
For the present, all the material dealt with will be assumed to
be perfectly elastic, homogeneous and isotropic — that is, to have
the same properties at all points and in every direction.
Hooke's Law. — The first law connecting stress and strain in
elastic bodies is that enunciated by Hooke, and which says,
"Stress is proportional to Strain." Thus, if a bar is strained
a given amount under a certain stress, it will be strained
double this amount under twice the stress, three times for thrice
the stress, and so on. The same is true for every kind of
strain, whether it be change in length, distortion, deflection or
twist.
In Fig. 2 is shown a prismatic bar to which is applied a tensile
load, P, acting along its axis. If the area of the cross-section of
P — - '
FIG. 2.
the bar taken at right-angles to its axis be called A, then the
tensile stress upon the bar will be :
Previous to the application of the load two points are marked
upon the surface of the bar, at a distance apart, I. When the
load is applied this distance is increased by an amount x, so that
the distance between the points is now (l+x) instead of I.
rv*
The strain is then -j-s and as, according to Hooke's Law. stress
is proportional to strain,
Young's Modulus for Direct Elasticity. — The amount of
elastic strain in tension or compression corresponding to a given
STRESS, STRAIN, AND ELASTICITY 7
stress varies for different materials, and is defined by what is
called Young's Modulus, or the Elastic Modulus or Coefficient.
If the bar is extended an amount x, by the application of a
stress /, then, according to Hooke's Law, in order to extend the
bar an amount I, the stress must be proportionately greater. This
stress is the elastic modulus and is constant for a given material.
It may be defined otherwise as the stress which would be
required to extend (or compress) a body through a distance equal
to its own original length, on the assumption that the material
remained perfectly elastic.
This constant for direct elasticity is usually denoted by the
symbol E. Thus:
* /
I E
or, E-£
stress sL.
or again, E = — = x
strain -j-
Variation in length x
where Strain = — . . — ^- — = -r
Original length /
T* *
If — , the strain, is made equal to unity, the stress / then
h
becomes E, the elastic modulus, which from this point of view may
be defined as the stress required to produce unit strain.
For example, if a bar of steel 1 sq. in. in section is found
to stretch ^^th of an incn ^n 10 ins.' of its length when a load
of 10,000 Ibs. is gradually applied, the value of the Young's
Modulus will be
10,000x10
= ~i~
300
= 30,000,000 Ibs. per sq. in.
This is somewhere near the usual modulus for steel and
wrought iron, which is fairly constant, and is found to vary
somewhere between 27,000,000 and 31,000,000 Ibs. per sq. in.
For cast iron it is about 15,000,000, and for copper 16,000,000 Ibs.
per sq. in.
In the above case the bar will have been stretched -^V^h
a
STRENGTH OF MATERIALS
of its own length under a stress of 10,000 Ibs. per sq. in.
The working stress on structural steel does not often exceed
15,000 Ibs. on the sq. in., or one-and-a-half times the above
stress. This means that the strain will be increased in the
same ratio, or the greatest stretch to be expected in steel under
the working conditions will not exceed about -ornyirth Part of
its length.
A knowledge of the modulus of any given material is im-
portant in enabling calculations to be made as to the strains
likely to occur in structural parts when subjected to working
loads, and also calculations of the probable strains in complex
structures built of materials having widely differing moduli.
What has been said about elastic strain in tension applies
equally to compression.
The uses and meanings of the terms which have been denned
will be made clearer by following the details of the two examples
given below. The figures given were found in making experi-
ments upon bars of the materials mentioned.
Example 1. — Experiment on the elastic extension of a round
wrought iron bar, 1 in. in diameter. The bar was placed in a
testing machine, and loads applied in uniform increments.
Extensions corresponding to the loads were measured by means
of a Ewing's Extensometer, the units of whose scale correspond
to sloths of an inch.
The following are the loads and the corresponding extensions
as given by the extensometer readings : —
Loads — Tons
0
i
1
1J
2
2i
3
3i
4
44
5
Readings
o-oo
O'll
0-31
0-51
0-71
0-92
1-12
1-33
1-53
174
1-95
Loads — Tons
5i
6
64
H
/
71
8
8|
9
»4
10
Readings
2-15
2-35
2-56
2-76
2-97
3-18
3-40
3-62
3-83
4-05
These readings are shown plotted on the accompanying
diagram (Fig. 3).
STRESS, STRAIN, AND ELASTICITY
9
To find the elastic modulus, take pairs of readings having
differences in loads of 5 tons :
4-05 3-83 3-62 3'40 3'18 2'97 2'76
1-95 174 1-53 1-33 1-12 '92 '71
'51
2'35 2'15
'31 '11
2-10 2-09 2-09 2-07 2 "06 2 '05 2 "05 2 "05 2_04 2 '04
The total for ten readings being 20*64, the average extension
2-064 .
for a difference in load of 5 tons is r._ ms.
500
TONS
15
10
O 002
0 004
0-006
0008
0-010
FIG. 3. — Load-extension diagram for a bar of wrought iron.
Loads vertical. Extensions horizontal.
Since within the limits of the experiment strain is proportional
to stress,
let / = stress in the material.
I = original length.
x = extension produced by/.
E = elastic modulus (Young's Modulus).
10
STRENGTH OF MATERIALS
If the bar is extended an amount x for an increase of stress /,
it will be stretched an amount I by a stress E, or,
E = • — , as before.
x
. 5 tons x 2240 ,,
f 07854 " lbs' Per «!• m'
I = 8 ins. (the measured length on the Ewing instru-
ment).
2-064 .
X =
E =
5 x 2240 x 8 x 500
0-7854x2-064
= 27,600,000 lbs. per sq. in.
Example 2. — A compression experiment on a circular cast
iron bar.
Diameter of bar 1/244 ins. Total length, 2 "5 ins.
Eeadings were taken on 1'25 ins.
A Ewing's Extensometer was used to measure the compres-
sions, the units of whose scale are -g-^nr^h8 °f an inch.
Table of Loads and Compressions. (For diagram, see Fig. 4.)
Load increasing :
Load in tons
i
2
4
6
8
10
12 14
16
18
20
22
Readings, \
TOTO in- •/
0
0-6
1-4
2-22
3-03
3-84
4-7 5'6
6-57
7-73
9-18
11-3
Loads decreasing :
Loads in \
tons . . /
22
20
18
16
14
12
10
8
6
4
2
*
Readings, ^
v-fa-t in. J
11-3
10-72
10-05
9-37
8-68
7-95
7-2
6-42
5-63
4-76
3-91
2-95
The bar was loaded up to 22 tons in steps of 2 tons, and the
compressions noted (see first table). Then the loads were taken
off and the extensions noted (see second table). To find the
STRESS, STRAIN, AND ELASTICITY
11
average compression, the bar was taken as being elastic up to 12
tons, and the following differences were taken :
4-7
2-24
3-84
1-4
2-44
3-03
0-6
2-43
The average compression or differ-
ence in length for 6 tons load
= 2-44 units,
2-44 .
Wbich = 2500 mS"
= x
TONS
25
to
0-0008" 0-0016" 0-0024" 0-003Z" 0-0040" 00048
FIG. 4. — Load-compression diagram for a cast iron bar.
Loads vertical. Compressions horizontal.
From which —
6x2240x1-25x2500
(1-244)2 x 0-7854x2-44
= 14,160,000 Ibs. per sq. in.
12
STRENGTH OF MATERIALS
Example 3. — Find the amount of stretch of a copper trolley
wire under a working stress of \ ton per sq. in. in a length of
400 yds.
Diameter of wire = \ in.
E = 16,000,000.
Working stress = \ ton per sq. in. = 1120 Ibs. per sq. in.
Let x — amount of stretch in inches.
1120x 400x3x12
= 1-008 ins.
Answer: Amount of stretch = T008 ins.
The following two examples will serve to show how calcula-
tions are made which involve the use of E for
two materials which are under stress at the
same time :
i
FIG. 5.
Example 4. — (See Fig. 5.) — A steel bolt
2 ins. in diameter passes through a cir-
cular casting 4 ins. in diameter and 10
ins. long. Find how far along the screw the
nut will have to travel, after it just touches
the casting, in order to put a tension of 5
tons per sq. in. on the bolt.
E for steel = 30,000,000 per sq. in.
„ cast iron = 15,000,000 „
Let the bolt be stretched an amount x^
„ the casting be shortened an amount #2
„ area of section of bolt be a^ = 314 sq. ins.
„ area of section of casting be a2 = 9 '42 sq. ins.
The total tensile load on the bolt must equal the total compressive
load on the casting. Let this
=W=/,x «,=./>,
where ft = the tensile stress on the steel,
fc = the compressive stress on the cast iron.
STRESS, STRAIN, AND ELASTICITY
ftl 5 x 2240 x 10
13
30,000,000
= 0-00373 in.
and
a, I
5x3-14x2240x10
9-42 x 15,000,000
= 0-00248 in.
The total movement of the nut down the
screw is
^+#2 = 0*00621 in.
Example 5. — (See Fig. 6.) — A pillar of
ferro-concrete is 3 ft. long, 10 ins. by 12 ins.
cross-section. There are four bars of steel
If ins. diameter running from end to end.
The total load on the pillar is 50 tons. Find the
load carried by the steel and the load carried by
the concrete.
E for steel = 29,000,000 Ibs. per sq. in.
E for concrete = 1,860,000
Let Ws, Wc be the loads on the steel and con-
crete respectively.
xg, xc be the compressions on the steel and
concrete.
W
i
FIG. 6.
as = Area of steel = 07854 x (175)2x4
= 9 '6 2 sq. ins.
ac = Area of concrete = 120-9 '6 2
= 110-38 sq. ins.
For the steel,
For the concrete,
W x36
S
9-62 x 29,000,000
x =
Wc x 36
110-38x1,860,000
but x must = a?..
s c
Wgx36
Wcx36
9-62 x 29,000,000 1 10-38 x 1,860,000
j^.
OF THE \
/ERSITY )
OF /
14 STRENGTH OF MATERIALS
or, W, 9-62x2900 1-355
w; ~~'' 110-38~x~186 = 1
50 x 1-355
Wg = — «-^r«- - 28-77 tons.
Lateral Contraction and Dilatation. — When an elastic body
is extended under a tensile stress, it is found that a lateral
shrinkage takes place. If the longitudinal strain is called
0, and the lateral strain — that is, the ratio of the diminution in
thickness to the original thickness — is called 0, then it is found
FIG. 7. — Poisson's Ratio.
that for a given material there is a constant relation existing
between these two. Thus, if 0 = T and d> = — -=— , then u =
I cL (p
where — is known as "Poisson's Katio." For indiarubber a is
M
found by experiment to have a value, for small strains, of 2.
For hard solids it varies from 3 to 4, the latter value being
that of the metals. It will be seen at a later stage that Poisson's
Eatio becomes of especial importance when theoretical relations
have to be established between Young's Modulus for direct stresses
and the corresponding constants for shear and volumetric varia-
tion. The meaning of the above is illustrated on Fig. 7.
CHAPTEE II
DIRECT, TANGENTIAL, AND OBLIQUE STRESSES
THE stress referred to so far has been either simple direct stress,
as in tension and compression, or simple shear stress. If an
imaginary plane surface in a body in a state of stress be taken
anywhere in the material, the stress upon it may be either wholly
normal, as in Fig. 8 (a) ; wholly tangential, as shown at (6) ; or
it may be inclined to a normal to the surface at an angle a (c).
The first two cases are quite simple. In the last case, the oblique
stress / may be resolved into its two components, one acting
iUJlii
(I)
FIG. 8.
normally to the surface, and the other tangentially. The normal
component may be compressive or tensile. These normal and
tangential components will have the following values :
The oblique stress being called fr,
The normal component stress fn = fr cos a, and
The tangential component stress fs = fr since.
Stress on an Area not at Right Angles to the Axis of a
Body under Simple Tension or Compression.
In Fig. 9, a prismatic body is subjected to a direct load P
along its axis, which gives rise to a simple direct stress on an
15
16
STRENGTH OF MATERIALS
area XY normal to the axis. What will be the character and
magnitude of the stress on any other area AB inclined to the
first at an angle a ?
In the first place, the uniform direct
stress on the normal surface XY is
1 /
1 /
A
vJ/'
1 ^x
1
1
B
where P is the whole direct load (compressive
or tensile), and a is the area of the normal
section XY.
The oblique stress on the surface AB will
be
P cos a
Jr ~ a ~ 7t
cos a
t
P
FIG. 9.
This oblique stress can be resolved into its
normal and tangential component stresses.
Thus, the normal component
/ = f cos a
/« J r
2
P cos a cos a
while the tangential component stress is
j^ = f^ sin a =
= fsin a cos a
This may be written, --— ~— — .
P cos a sin a
Sin 2a has a maximum value
when 2 a = 90°, or a = 45°. So that the shear stress fs has "its
greatest value when the angle between the two normals is 45°,
and when
Effect of two Direct Stresses at Right Angles.
Consider a cube, BODE (Fig. 10), the area of whose face is (a).
Let two pairs of normal forces, P' and P'', act on pairs of opposite
DIRECT, TANGENTIAL, AND OBLIQUE STRESSES 17
faces, and let compressive forces be plus and tensile force minus.
The intensity of stress on BC and ED is
PX
I M
f.
J
and on BE and CD,
Consider the effect of these in
producing stress on any other
section LM taken through the
centre and normal to the plane of
the paper.
The normal stress on LM due to P7 is
B
FIG. 10.
where a is the angle between the direction of P' and the normal
to LM. Similarly,
/"„ =/" cos s(|--«)
/// • o
sm2a
Therefore, the total normal stress on LM is
/. =/'„+/*„
= f ' cos 2 a + f" sin 2 a
Similarly, the tangential stress on LM due to P'
= f'g = f sin a cos a
tending to cause the part MBEL to slide in the direction M to L.
Also the tangential stress due to P" is
/•// /•// . /37T \ /3?r \
J $ ---- J sin ( -^- + a) cos (— + a)
= - f" sin a cos a
Therefore the total tangential stress on LM due to P' and P'' is
f - - f j_ f"
J$ - J s +J s
= f sin a cos a - f" sin a cos a
- (/' - /") sin a cos a
The value of this tangential stress attains a maximum when
sin a cos a is a maximum : that is to say, when a = 45°.
B
18
STRENGTH OF MATERIALS
When the tangential stress is a maximum, and a = 45°, the
total normal stress,
fn -- f cos 2 45° + /" sin 2 45°
- Kf +/")
Nature of Shear Stress. — In Fig. 11 (a) is shown a
piece of material rigidly held in sockets near its ends. These
sockets are being forcibly moved, the upper one to the right
and the lower one towards the left. The effect is to give rise to
(*)
A -_v B
B
C D
FIG. 11.
shearing stresses on planes at right angles to the axis of the bar,
shown in the figure by AB and CD. Consider a small cube
of the material between these two planes (Fig. 11 (&)). On the
upper and lower faces of this cube there will be a pair of equal
and opposite shearing stresses /s, one acting in the direction AB
and the other from C towards D. The tendency of this pair of
stresses is to rotate the cube in the direction ABCD, there being
a couple of a magnitude equal to the stress fs, multiplied by the
DIRECT, TANGENTIAL, AND OBLIQUE STRESSES 19
vertical distance between AB and CD. This rotation does not
take place, so that there must be a couple of equal magnitude
acting in the opposite direction, to keep the cube in the given
position. This couple consists of a second pair of stresses acting
on the two faces, AD and CB, of the cube, and as the arm
of this couple is equal to that of the first, it follows that the
intensity of the stresses on AD and CB must be the same as
those on AB and CD.
Next consider one pair of these four stresses acting on adjacent
faces of the cube, namely, those on AB and CB respectively.
These are forces of equal magnitude, acting in directions at right
angles to one another. The direction of their resultant will bisect
the angle between them, arid, by the parallelogram of forces, will
have a magnitude equal to
Pt - VT/.AB
Thus the pair of stresses will result in a single force, acting in a
direction from D to B, and tending to pull half of the cube, ABC,
away from the other half, ADC. In other words, this force, Pp in
tending to cause separation across the plane, AC, is exerting a
tensile stress on this surface, which is resisted by the internal
stress in the material, preventing separation of the surfaces from,
taking place.
The intensity of this tensile stress will be
(area on which the stress acts)
x/2AB
That is to say, there will be a tensile stress of equal intensity on
a section of the cube taken through its diagonal, or being at an
angle of 45° with the direction of the shearing stress.
By precisely similar reasoning it can be shown that the same
shear stresses give rise to a compression stress, also of equal
intensity to that of the shear stress, on a section taken through
the other diagonal. This is shown on the figure.
The fact that these tangential and direct stresses do exist
at the same time is very clearly demonstrated in the testing of
20 STRENGTH OF MATERIALS
several kinds of material. For example, in the torsion test to
destruction of a cast iron shaft, failure takes place, not by shear-
ing, which is the original stress induced by the twisting, but by
the material being torn asunder across a surface forming a spiral
at an angle of 45° with the axis of the shaft. Here the tangential
stress is accompanied by a tensile stress, and as the material is
more ready to fail by pulling than by sliding, failure takes place
in this way.
Another instance is to be found in the failure of most brittle
materials under a crushing stress. The compressive stress induces
shear stresses on planes inclined to the direction of the load, and
the material usually fails by slipping along these planes. This is
found to occur in crushing tests of such materials as cast iron,
stone, cement, and so forth.
In mild steel tension tests the fractures most often take
place on one plane forming an angle
with the axis, or a pair of such planes
forming a truncated pyramid and a
corresponding recess. In the case of a
round bar the fracture takes the form
of an incomplete cone and cup.
Shear Strain — Modulus of Rigidity.
— Let a cube of unit length of side, one of
whose faces is the square BODE (Fig. 12),
FIG 12 be subjected to a uniform shear stress
fs. It has been shown that on the pairs
of opposite faces there will be stress fs along CB, fs along ED,
fs along EB, and fs along CD. Also there will be a tensile
stress ft acting normally to the diagonal plane EC, and a com-
pressive stress fc on the diagonal plane BD. Moreover, it has
been shown that
/.=/«- /,
The effect of the shear stress is to distort the figure from the
square to a new shape, Bc^E, the lengths of the sides remaining
unchanged. CBc is called the angle of distortion. Call this
angle /3. In elastic materials the angle of distortion is propor-
tional to the shear stress, or
£«•/,
In order to produce distortion through an angle whose value in
DIRECT, TANGENTIAL, AND OBLIQUE STRESSES 21
circular measure is unity, it is necessary to apply a shear stress
G where
G_ J^
J s r*
Js shear stress
ft angular strain
T
G is called the Modulus of Rigidity, or 'shear modulus.
It is clear that the angle
DBrf = A
Next, to see how E and G are related
It is obvious that when BODE is distorted into the shape
Bft£E, the diagonal BD is lengthened so as to become l&d. This
lengthening may be considered to have been brought about partly
by the tensile stress in the direction BD, and partly by the
squeezing caused by the compressive stress in a direction at right
angles to BD.
The elongation in direction BD due to /, is
/,BD
E
and in direction BD due to fe is
/CBD
where is Poisson's ratio for the material in question.
BD = \/TBC = \f~2. 1. So that the total elongation,
x/2
* = ^E
E
Next, consider the lengthening of BD as caused by the distor-
tion. This elongation is
x = Ed - BD
= 2BC cos (^ -•£•)-
|sin|- + cos|cos|
22 STRENGTH OF MATERIALS
But for very small angles,
-4-1
and here
BC = 1
so that
X
and also
jj_f(t±*\
E' s \ ,, I
\ /* /
so that, finally,
fs x/T _ /u + 1
G
_
E 20* + 1)
In the case of the metals, where p = 4 (nearly),
Cr _2_
E : 5
That this is true may be shown by experiments in torsion and
tension upon samples of the same material.
Elasticity of Volume or Cubic Elasticity (Fig. 13). — Let all
six faces of the cube of unit length of side be pressed upon by a
uniform stress p, which in this case is a pressure. The effect of
this pressure will be to diminish the linear dimensions of the cube
in all three directions, and in this way reduce its volume. If this
decrease in volume be called v and the original volume V, then
the volumetric strain will be -^, and =r = ^-, where K is a
coefficient called the Modulus of Cubic Elasticity. This may be
written
pressure _JL
volumetric strain —
or
DIRECT, TANGENTIAL, AND OBLIQUE STRESSES 23
The following will show how K and E are related : —
Call the linear strain in any direction parallel to one edge of
the cube x, due to the pair of pressures in the same direction.
There will be a retardation or a dilatation in this direction, due to
the other two pairs of pressures in directions at right angles to the
first = —
B
FIG. 13.
X
Thus the total strain in direction AB = x — 2 —
AC = x-2-
u.
A T) _ r O X
» )) }) xxj_/ — i/6 — — —
M
For volumetric strain the total diminution is
and
but V = 1, so that here
K =
P .
(•-*-}
\ p/
24 STRENGTH OF MATERIALS
But x = •£- for linear strain, so that
P
K =
.E
3/A-6
Relation bet-ween E, G, K. — It has previously been shown
that
G P
E "
and it is now seen that
K
E " 3/x-6
From these two equations the relation is obtained that
6K 2G
^ ~~ 3K-E == E-2G
or F 9KG
" 3K + G
CHAPTEE III
(V
STRESSES IN BEAMS— BENDING AND SHEARING ACTIONS
THE effect produced by a force acting normally to the axis of a
bar or prism is shown in Fig. 14.
Here is a bar of elastic material fixed or built into some rigid
substance, and in this way forming a cantilever. The bar is
supposed to have no weight. It is acted upon by a force, P,
pushing upwards at its outer end and at right angles to its axis.
Consider any normal sec-
tion of the cantilever at AB. •/////// //\ A
The effect of P on the material
in the plane AB will be two-
fold. In the first place, the
tendency of P is to cause that
portion of the cantilever to the
right of AB to slide upwards,
as shown at b, leaving the
portion to the left stationary.
The force, P, with which the
part is pushed upwards is
called the shearing force on
AB. This causes shearing
stress on the plane AB and
this is opposed by the resistance offered by the material to sliding.
There will thus be shear stresses on the vertical section, and,
from what has previously been said, it is clear that there must
also be shearing stresses of equal magnitude on horizontal planes
taken through AB.
The precise distribution of this shear stress over the section
of the beam will be considered more in detail in a later paragraph.
(O
FIG. 14.
26 STRENGTH OF MATERIALS
For the present, it will be sufficient to investigate the second part
of the effect of P.
If the beam were to be cut through at AB the shear effect of
P would be to cause actual sliding to take place. When this
movement is prevented by the application of a downward force
Fs = P acting at AB, it is evident that the further tendency of
P will be to cause a separation between the planes at the lower
part of the section, and, if there should be an opening between
these at the top of the section, to close it. That is to say, there
will be a tendency to sever the lower fibres by tension and to
crush the upper fibres.
If x is the perpendicular distance from the line of action of P
to the section AB, then the bending moment on the section is
M = P x x
In the present simple case the bending moment is due to a
i
FIG. 15.
single force P. The bending moment on a section of a beam is in
most cases due to a number of such forces, to a uniformly distri-
buted load, or to both. It will be sufficient at present to consider
the effect of a bending moment M, without any reference to the
manner in which it is produced.
The following example will help to make clear what is the
nature of the stress caused by bending in one of the simplest
cases, namely, one in which the bending moment is resisted by
stresses in the flanges of a girder.
Example. — A bridge, 40 ft. span and 12 ft. wide, is carried
by two plate girders (Fig. 15). The total load per sq. ft.
of platform is 7 cwts. Stress allowed in flange, 6J tons per
STRESSES IN BEAMS 27
sq. in., depth of girder 4 ft. 6 ins., and width of flange 12 ins.
Find the thickness of flange, t.
Total load on bridge (40 x 12) x 7
~~20~~
= 168 tons.
Total load on one girder
= W= *|8 » 84 tons.
The bending moment is greatest at the centre, and is
W l W l
w/
This is equal to the moment of resistance
= (total force on flange) x (depth of girder)
= (£.12. 6J) x (54), inch-tons, where t is the required
thickness.
Therefore
t 12.6J.54 =
or _ 84.40.12.4
= 8.12.25.54
= 1 14 ins. = 1 J ins. say.
In the example just considered, it is assumed — which is
approximately the case — that the tensile stress in the lower
flange and the compressive stress in the top flange are both
uniform over the flange section.
Where the section of the beam is solid, or where the thickness
of the flange is great relatively to the depth of the beam, the fact
that the stress varies from point to point must be taken into con-
sideration.
Distribution of Direct Stress in a Beam Section.
When a beam is curved by bending, as seen in Fig. 16, it is
evident that the layers of the material on the inside or concave
surface are compressed and those on the convex surface are in a
state of tension. Further, it is not difficult to see that these
28
STRENGTH OF MATERIALS
stresses are greatest at the surfaces, and diminish as the centre of
the section is approached.
If the portion to the right of the vertical section, CD, were to
be hinged so as to be free to turn about a point, F, it would tend
to revolve in the direction indicated by the arrow. This tendency
is resisted by the compressive stresses exerted by the left-hand
portion on that part of the section OF above F and by the tensile
stress on FD, and the summation of the moments of these repre-
sents the moment of resistance, which is equal in magnitude but
opposite in sense to the bending moment of the external forces.
Those stresses above' F act to the right, and those below to
FIG. 16.
the left. As the external forces are wholly vertical, the algebraic
sum of these horizontal stresses must be zero.
Now refer to Fig. 17.
At (a) two adjacent and parallel sections, AB and CD, are
shown as they would appear on the unloaded beam. At (b) the
same portion of the beam is shown when loaded.
The lines AB and CD, representing the sections, now appear
to be no longer parallel, but inclined to one another and meeting
in some point Q.
This convergence of the lines is brought about by the shorten-
ing of AC under the compressive stress, and the lengthening of
BD under tension. C'D' is drawn parallel to AB, at a distance
away from AB equal to EF. CD and CT)' intersect at F.
Consider the portion below EF. The fibres of the material
at BD will have been extended an amount DT), and in a fibre
LM, at a distance x from EF, the extension is less. The material
of the beam is supposed to be elastic, so that, as the stretch of
the fibres diminishes in amount towards the centre, the stress also
STRESSES IN BEAMS
29
diminishes, until some point F is reached where there is no longer
any elongation and the stress is zero.
Beyond F the stress changes sign and becomes compressive,
increasing in amount as the distance from F towards C increases.
The surface GH, which includes all points where there is no
(*)
FIG. 17.
stress, is called the Neutral Surface, and its intersection with the
plane section AB is a line NET, called the Neutral Axis.
In the theory relating to the bending stresses in beams the
following assumptions are made : —
1. The material remains elastic under all stresses imposed.
2. Its elastic modulus is the same for tension as for compression.
30 STRENGTH OF MATERIALS
For iron and steel experiment shows this to be
practically true. In the case of such materials as con-
crete the difference is not great.
3. A plane section of the learn, normal to the Neutral Surface,
suffers no distortion when the learn is loaded but remains
a plane.
This means that the straight lines AB and CD in the figure
remain straight when the beam is loaded. The extension (or
compression) of any fibre LM, at a distance FM or x from the
neutral axis F, is MM', which forms the base of the triangle
FMM'. This is one of a series of similar triangles having a
common apex at F.
The lengths of their bases are proportional to their altitudes, x,
and, as the stress is proportional to the extension (or compression),
it follows that, if the above assumptions be sound, the stress at any
point must le proportional to the distance of that point from the
neutral axis.
In Fig. 17 (c) let KS be a strip of the section of the beam at
AB, drawn parallel to the neutral axis NIST, at a distance x from it,
and let its length be a and its width Sx.
If / is the maximum tensile stress at the edge of the section
farthest from NN, and at a distance y from it, and q is the stress
on the strip, then
_q_ x_
/ y
X ,
or q = — / = ex
yj
where c is a constant.
The total force on the strip is
q a &r = c x a 8x
and the total force on the section is the integral
c I a x dx
Below NN this is tension, and above NN it is compression ; and
as the whole of the stress on the section acts parallel to the
neutral surface, and the external forces are wholly normal to this
STRESSES IN BEAMS 31
direction, it follows that the algebraic sum of the direct stresses
on the section must be zero, or
ilaxdx = 0
From this it follows that the line from which x is measured,
that is to say, the neutral axis, must pass through the centre of
gravity of the section.
Moment of Resistance. — The force on one strip being
the moment of this force about the neutral axis is
= q a 8x x = c a .r2 8x
This is the resisting moment of one strip. The moment of
resistance of the whole section is the integral
taken between the limits x = — y and a; = -f yv y being the distance
between the neutral axis and the point where the maximum stress
/ occurs, and yv the distance from the neutral axis to the extreme
edge of the section opposite to y.
(Bending Moment) = (Moment of Eesistance), or
+2/1
M = c\ ax^dx
but c = — , and the integral
y
l
is the Moment of Inertia of the section about its neutral axis.
Writing this as I, the above equation becomes
M = ^-l
* y
M J
or _.'..,£-
i y
The fraction " is often called the Modulus of the Section, and
c/
is usually denoted by the symbol Z. The above equation then
becomes M = /Z
32
STRENGTH OF MATERIALS
Moments of Inertia. — The moment of inertia of an element
of area SCD about any axis is the product of the area So) and the
square of the perpendicular distance of its
^^fo centroid from the axis. Thus in the case
represented in the accompanying figure (Fig.
x 18) the moment of inertia
i
x __<L__ Y I = &o*2
In the same way, the moment of inertia of
an area which can be split up into a number of such elements
is the summation of the moments of inertia of the several
elements
I = I xz do), or = I a xz dx
where adx is the element of area.
Where the area in question is a regular geometrical figure the
integration can generally be effected,
and the value of I found mathemati-
cally. Thus in the case of the rect-
angle on Fig. 19, its I about its own
neutral axis, or one drawn through
its centre of gravity, is
•+y
a x2 dx
r
= 1
X
when a = b = constant, and y — -
±y
FIG. 19.
Integrating,
and the modulus
-*K]
2
bd?
This case occurs very frequently and the result should be
remembered.
STRESSES IN BEAMS
33
The moment of inertia of any section may be written in the
form, I = AF
or k2 —
where A = the area of the section in question,
and k = the radius of gyration.
The radius k is such that if the area A could be concentrated
at the end of this radius, the effect would be the same.
The following rules are extremely useful when moments of
inertia have to be found of areas whose figures are either rect-
angular or elliptical, or are complex areas made up of these
figures.
1. The moment of inertia of an area about an axis through
its centroid. and coinciding with or perpendicular to one of its
principal axes, is
[(area of the figure) x (the square of the rectangular semi-axis)]
~~3 or 4
The divisor (3) is used where the figure is rectangular, and
the (4) where it is elliptical.
Examples : —
(a) Moment of inertia of a
circle about its diameter (d ^
= 2r), Fig. 20.
Here the semi-axes are the
two radii, one in the neutral
axis and one perpendicular to
it. So that, FIG. 20.
I =
irr-
(b) Similarly, for a rectangle of width (b) and deptii (d) about
a neutral axis parallel to (6), Fig. 19.
id\(i
bd3
STRENGTH OF MATERIALS
2. The moment of inertia of an area about an axis other than
its own, but parallel to it, is the sum of (the moment of inertia
about its neutral axis) -f (the product of its area into the square of
the perpendicular distance between the two axes).
Thus (Fig. 21), if I is the moment of inertia of the given area
about its neutral axis, A its area,
h the perpendicular distance be-
tween the two axes, and Ia the
moment of inertia about the new
axis XY, then
I, = I + AA'
/\ — ~ — —
FIG. 21.
3. The moment of inertia of a complex area about any axis
A D
"T
X<M I '
' ~~ 1 1
'I
1
1
*-2~»
1
1
E H
L O V«
* 4 i •
1
<M 1
' 1— ' 'H
r
-|. Y
Y ' '
F G
M N
1
1
1
c
1
1
1
1
U 12* *
FIG. 22.
is the sum of the moments of inertia of the several parts about
this same axis.
If Ij, I2, I3, etc., are the respective moments of inertia about
the given axis of the several parts which make up the entire area,
and I is the moment of inertia of the whole, then
I = Ij + I2 + I8 + etc.
STRESSES IN BEAMS
35
The following worked-out examples should help the reader to
a more complete appreciation of the rules which have just been
laid down : —
(12)3
= 288
Example 1. — To find the moment of inertia of the section
(Fig. 22) about XY, its neutral axis.
Moment of inertia of rectangle ABCD about
the axis XY ....
Moment of inertia of rectangle LMNO about
the axis XY ....
Moment of inertia of rectangle EFGH about
the axis XY is same as that of rectangle
LMNO
Moment of inertia of entire figure about the
axis XY
1 n
3
10
T
T
Modulus of section =
= 294-e (
its moment of inertia
= 49-1
E L
A..
_ 10 -
M P
X ; * —
N
O
is -
Fig. 23.
Example 2. — To find the moment of inertia of the given section
about its neutral axis XY (Fig. 23).
36
STRENGTH OF MATERIALS
Moment of inertia of rectangle AD
about the axis XT .
Moment of inertia of rectangle EL
about the axis XT . .
Moment of inertia of rectangle MNOP
about the axis XY' . .
Moment of inertia of entire figure
about the axis XY
Modulus of figure =
2Jx(12)3 5x14
:4
:4
12 2
12 2
10x(2)3 20
12 3
5x144 5x144
20
2 2
3
0 2180 7 ft
/inch-\
Units/
1 M
its moment of inertia
726-6
6
121-1
Example 3. — To find moment of inertia of same section as last,
but about neutral axis at right angles (Fig. 24).
T'
*£ ^
i
i
>
F ! D
i
i
H
i
i
i
Xvl
^ N»
1 Y
1
1
i
I
1
s
1
i L
s
T ; 0
A_
.
idi
M
„
N
Fi
G. 24.
i
Moment of inertia of rectangle
ABCD about the axis XY
12 x (2
125 9375
« + 8
9500
STRESSES IN BEAMS
37
Moment of inertia of rectangle
STF about the axis XY
Moment of inertia of rectangle
LMNO about XY is the same
as that of rectangle ABCD .
Moment of inertia of entire figure
about axis XY
2x(10)3 1000
9500
8
9500
9500
1000
6
Modulus of section
9500 1000
4 6
28500 + 2000
12
30500
12
= 2581*6 /inch-\
\ units/
moment of inertia
= 338-8
Example 4 (Fig. 25). — Section of brickwork pier. To find its
moment of inertia.
1
A
— c
?---
--- f •
— ~i —
" f"
1
1
.-J Y
i
s
1
1
X _ _
*
f-
i
i
. i.
c
B
I '
FIG. 25.
First, to find the centre of gravity or centroid in order to fix
the position of the neutral axis.
Taking moments about BC,
(Entire area) x g = (Area AJ x 3£ + (Area A2) x 1
38 STRENGTH OF MATERIALS
18x31 + 6x1 69
24 = 24
= 2-875 feet
The moment of inertia of the entire area about XY is
I = (Moment of I of Ax) + (Moment of I of A2)
f- 18 x 0-39) + (~~ + 6 x 3-52
- (13-50 + 7-02) + (2 + 21-1)
= 43-60 feet-units.
CHAPTER IV
GRAPHICAL METHOD FOR DETERMINING THE
MOMENT OF INERTIA
WHERE the shape of the area whose moment of inertia is required
is that of one of the simple geometrical figures, such as a square,
rectangle, triangle, circle, or the combination of a number of these,
there is little difficulty in determining the required value by the
methods which have just been described. But in many cases which
occur in practice the area is of such irregular shape that the
problem cannot be attacked mathematically, and a graphical
method, depending upon the following principles, must be em-
ployed.
It has previously been shown (see Fig. 17) that the moment of
resistance is equal to the integral
- — a j?2 dx
Here, it will be remembered, / is the maximum stress at a
distance y from the neutral axis, and a is the length of the strip
which forms the element of area. The above integral may be
written
The actual stress on the element in question is — /, or / reduced
u
in the proper ratio, — .
y
Now, without altering its value, the same integral may be
written as
/[f -][/*
89
40
STRENGTH OF MATERIALS
The meaning suggested by this way of expressing it, is that instead
x
of reducing the stress in the ratio of — , the length of the strip is
u
reduced in the same ratio, and the stress is taken as having the
constant value/. In other words, the force due to a reduced stress
acting on the actual element of area (a 8x) is replaced by a force
consisting of the original maximum stress acting on a reduced area.
The ratio of reduction is the same in both cases, so that the total
force on the strip must be the same.
If this reduction is made in each element of area of which the
original section is composed, these new elements added together
will make up a new figure, called the modulus figure. The
FIG. 26.
original maximum stress / acting on this area will give a moment
of resistance equal to the one represented by the above integral.
The resultant of the uniform stress on the modulus figure must
pass through the C.G. of this figure.
The moment of resistance on one side of the neutral axis will
then be
= / x (area of modulus figure) x (distance of its C.G. from
the neutral axis)
= /xZ
The actual way of finding the modulus graphically is as
follows : —
Consider any figure as shown in the accompanying sketch,
Fig. 26.
METHOD FOR DETERMINING MOMENT OF INERTIA 41
(1) Find the centre of gravity. This is most conveniently
done by cutting out the figure in thin cardboard and balancing it
above a knife edge in two positions, roughly at right angles to
one another. The intersection of the two lines about which the
figure balances will be the required C.G.
(2j Draw the neutral axis XY through the C.G. at right angles
to the direction in which the load is applied.
(3) Draw a base-line BL parallel to the neutral axis and
touching the figure at the point distant farthest from XY, this
distance being y, Also draw a second base-line BL on the other
side of XY and the same distance, y, from it.
(4) Draw a number of lines, as CD, parallel to XY and cutting
the boundaries of the figure at C and D. Drop perpendiculars
CE and DF on to BL.
Join E and F to G. The lines thus drawn will cut CD in H
andK.
C and D are points on the boundary line of the original
figure ; H and K will be the corresponding points on the
modulus figure. The remaining points are to be found
in the same way. By following out this construction
throughout, the two shaded modulus figures will be
obtained. These are such that a "constant stress upon
them equal to the maximum stress f will have the same
effect as the varying stress on the original figure," by
reason of the proof already given.
(5) As the nature of the stress on one of these areas is tensile
and on the other compressive, and as the algebraic sum of the
forces acting at any point of the beam parallel to its axis is zero,
it follows that the area of the modulus figure above XY must be
equal to the portion below XY.
(6) Measure the two areas. This is most conveniently done by
using a planirneter.
Call the area above XY o^, and the one below XY o>2.
When the two areas have been measured, a>x should be
equal to w2. The difference ought not to be more than
1 per cent.
This equality of the two areas provides an excellent check on
the accuracy of the previous work. When their difference
is found to be too great, the finding of the C.G. of the
42
STRENGTH OF MATERIALS
original figure and the construction used for obtaining the
modulus figures must be revised or repeated.
(7) Find the centres of gravity of the two modulus figures.
It will be found most convenient to cut out the figures in card-
board and balance, as in the case of the original figure. Call the
distances of these C.G-.'s from XY respectively kl and h2.
Then the moment of resistance about XY will be
FIG. 27.
the resultant of the total force on each
modulus figure acting at its C.G., and
the arms of the moments being Ax
and hy
But (01 = co2 = w, so that the moment
of resistance is
Here the modulus Z, is represented by
wXJ.
It is rarely that o^ comes out quite
equal to eo2. When they are sufficiently
alike, co must be taken as their mean value.
To find the moment of inertia when the modulus has been
determined, it is only necessary
to multiply Z by the distance
from the neutral axis to the base-
line, or
I = Zy
Where the figure is symmetri-
cal about the neutral axis, y = yl
and the base-lines touch the figure
at both the top and the bottom.
Several examples of the mod-
ulus figures for symmetrical areas
are given below. Of these,
Fig. 27 shows the modulus
figure for the rectangle shown on Fig. 19.
Fig. 28 shows the modulus figure for the circle shown on
Fig. 20.
i
FIG. 28.
METHOD FOR DETERMINING MOMENT OF INERTIA 43
Fig. 29 shows the modulus figure for the section shown on
Fig. 22.
Hf J
FIG. 29.
FIG. 30.
Fig. 30 shows the modulus figure for the section shown on
Fig. 24.
FIG. 31.
Fig. 31 shows the modulus figure for the section shown on
Fig. 23.
In cases where the original figure is not symmetrical about the
neutral axis, as in Fig. 26, the modulus figures must be constructed
from base-lines at a distance from the neutral axis corresponding
to the point where the maximum stress of greatest importance
occurs.
For instance, the area on Fig. 32 represents the section of a
44
STRENGTH OF MATERIALS
cast iron frame. Cast iron is known to be much weaker in tension
than in compression, and it is therefore necessary to design the
section on the basis of a maximum safe tensile stress. This occurs
at the edge of the large flange, and the base-line must be taken at
the distance of this edge from the neutral axis.
The modulus will have a value differing according to which
base-line is used.
i
y,
\
\
-i-
\
FIG. 32.
It is to be noted, however, that the value of the moment
of inertia will be the same whichever base-line is used.
Thus, in an unsymmetrical section, let
y = distance of one base-line from neutral axis ;
Z = the modulus as found from this ;
2/x = distance of other base-lines from neutral axis ; and
Zx = the corresponding modulus.
Then, the moment of inertia
From which Zl y
z = y~i
or Zx = Z —
In this way Zl can be found from Z by multiplying it by the
inverse ratio of the /s.
The two pairs of modulus figures giving Z and Zx are shown
for the section in Fig. 32.
The examples given here of the finding of the modulus in the
cases of areas made up of regular geometrical figures are chiefly
given for the purpose of showing how the method is carried out.
METHOD FOR DETERMINING MOMENT OF INERTIA 45
As a rule, for figures of this type it is much easier, and, at the
same time, more accurate, to find the modulus by calculation in
the manner already explained.
On Fig. 33 is shown a good typical example of the kind of
FIG. 33.
area in which it is not possible to find Z by calculation, namely,
that of a tram-rail section.
The modulus has in this case been found as described, working
from the tension base-line at the lower edge of the flange.
It is sometimes found convenient, though not necessary, to
46 STRENGTH OF MATERIALS
" mass together " a separated section. In the above case it is
split by the groove in which the flange of the wheel runs.
This " massing " is done by drawing a number of horizontal
lines across the part to be dealt with, and transferring the out-
standing widths to the solid part, as indicated in the drawing.
The new part thus formed is shown by the dotted line. This
process does not alter the value of the modulus about a neutral
axis parallel to these horizontal lines.
The actual measurements taken from Fig. 33 were as follow :— ~
Area of modulus figure.
Three measurements were^
made on each side theV
neutral axis
Top.
3-89
3-87
3;89
3-88
Bottom.
3-89
3-90
391
3-90
means
The average of the two means so obtained is 3 '89 sq. ins.
J was measured as 5 '61 ins., and y as 3 '425 ins. From these
the modulus
Z - 3-89x5-61
= 21-82 inch-units,
and the moment of inertia,
I = Zx.y
= 21-82x3-42
= 74-74
FIG. 34.
The modulus figure shown on Fig. 34 is for the same section as
that previously given on Fig. 25.
CHAPTEE V
DEFLECTION OF BEAMS
IN this connection the first thing to be done is to find an expres-
sion for the radius of curvature E at any point of the deflected
beam, this curvature generally varying from point to point.
In Fig. 35, which is similar to Fig. 17, the two plane sections
AB and CD, originally parallel, are taken sufficiently near to one
another to make the radius of curvature sensibly constant from
E to F. The effect of the bending is to cause AB and CD to
become inclined to one another and to meet in Q, which is the
centre of curvature for the short length considered. The bottom
fibres are lengthened an amount D'D.
Call BD' = EF = AC' = L, and D'D = s.
s f
Then -T- = ^r, according to Hooke's Law.
Li &
Now C'D' has been drawn parallel to AB, so that the triangles
QEF and FDT) are similar.
T?T? TVT)
It therefore follows that -^-^ = ^^,
t^Jcj
Here,
EF = L,
QE = E (at the neutral surface),
DD'. = s, and
so that the above may be written,
JL _^_
"tt : %
1 / r. f M-.
= -^r— but ^- = —
L y L y I J
47
48
STRENGTH OF MATERIALS
This equation is fundamental in relation to the deflection
of beams. Its meaning is that the reciprocal of the radius of
curvature of the neutral surface at any point in a loaded beam is
equal to the bending moment at this point divided by the product
of the elastic modulus of the material and the moment of inertia
of the section.
FIG. 35.
It is necessary here to take note of the following definitions : —
1. The shearing force S at any section of a beam is the algebraic
sum of all the forces normal to its axis acting on one side of
the section.
For the sake of uniformity, those shearing forces which act
as at (a) (Fig. 35) are taken as minus, and. those as at (5) &splus.
DEFLECTION OF BEAMS
49
2. The bending moment M at any section of a learn is the algebraic
sum of all the moments acting on the beam on one side of
that section.
Those as at (a) are minus, and as at (b) are plus.
Next, to find the deflection. There are two ways of doing
this.
First Method. — CASE I. Deflection at the end of a cantilever
carrying one concentrated load.
FIG. 36.
It is only necessary in what follows to consider the curvature
of the neutral surface (shown by strong lines in the figures).
On Fig. 36 is shown the cantilever in question, having a length
I and carrying a single load W at its outer end. Take any two
adjacent points A and C at a horizontal distance x from the outer
end, and at a distance apart Sx.
D
OF THE
•-.
| UNIVERSITY |
50 STRENGTH OF MATERIALS
Radii are drawn from A and C to meet in Q, and corre-
sponding tangents from these points cut off from the vertical
line through the outer end of the cantilever a short length, e.
This will be the deflection at the outer end due to the curvature
of the element of length Sx. The total deflection A will be the
summation of all these increments of deflection, from a point
where x = 0 to x — I.
Now, by similar triangles,
8x_ j_
R" : x
1 e
or ~rr — s
R x 6.r
but it has already been shown that
JL M.
TT : El
rn, , M x Sx
Therefore e = -^-r-
& i
At the point in question, the bending moment,
M = ~Wx
so that
W a*
El
The total deflection is the integral of this, or,
W [i
W/3 The minus sign indicates a
3 E I downward deflection.
CASE II. Cantilever loaded uniformly. (Fig. 37.)
In this the load is applied uniformly and is equal to w units
of weight per unit of length. As before,
M x 8x
"BIT
But the bending moment at the point AC is
M = (load on portion, x) X (distance of the C.G-. of this load
from the section),
or
DEFLECTION OF BEAMS
51
so that
w .r3
2 El
and the whole deflection is
2El
tr3 dx
8EI
W/3
8E I
but wl = W where W is
the total load.
FIG. 37.
The same method can be used for supported beams, or the
above two results may be utilised as follows : —
CASE III. Beam freely resting upon two supports whose distance
apart is I, and carrying a concentrated load W at the centre
(Fig. 38).
W
FIG. 38.
The upward reaction at- each support is half the load, or,
-^-, and the neutral axis is horizontal at the centre The beam
2i
can therefore be regarded as made up of two similar inverted
cantilevers, of length --, and loaded at their outer ends with up-
52
STRENGTH OF MATERIALS
W
ward forces = -~-. The vertical distance between the centre and
the points which touch the supports will, from the former of the
above equations, be
3E1
W/3
CASE IV. Beam resting upon two supports whose distance apart
is I, and carrying a uniformly distributed load w per foot run
(Fig. 39).
FIG. 39.
The total load W = w .1
The beam is symmetrical about the centre, where the maximum
deflection will occur. The reactions at the support = —=-—. The
2t 2i
beam can be taken as a pair of cantilevers, each starting from the
W
centre and having an upward force -~- at the outer end or support,
W
deflecting it upwards, and a uniform load — tending to deflect it
downwards.
The required deflection A will be the difference between the
upward and the downward deflections of the cantilever, whose
length is --, that is,
3EI
8EI
5
384
W/3
ET
DEFLECTION OF BEAMS
53
CASE V. Beam fixed horizontally at the supports and carrying
a single concentrated load "W at the centre. As before, the span is I.
(Fig. 40.)
It is necessary that the neutral axis be held perfectly hori-
zontal at the supports, and that these be on the same level.
It will be seen that, starting from A, the beam is first deflected
with the concave part facing downwards, up to some point C.
After this, as far as the centre, the concave part is upwards. So
that at C there is a change in the direction of curvature. The two
points in the beam where this occurs, C and D, are called points
of contrary flexure.
_
-I —
FIG. 40.
As there is no curvature at C and D, there can be no bending
moment. So that the conditions would be fulfilled if there were
to be flexible joints at C and D, or if the middle portion CD were
to be suspended by links, as shown in the lower figure.
The beam can then be regarded as made up of four cantilevers :
W
AC, EC, BD, and ED. The same force -~- pulls down on the end
, *•
of AC as pulls upward at the end of EC.
As the change of curvature is gradual at C and D, the angle of
slope of the end C of AC must be the same as that of C on EC. 4
The four cantilevers are then in all respects similar, and their
deflections A! and A2 must be equal.
54 STRENGTH OF MATERIALS
The required deflection of the beam at its centre will be made
up of these two added together, or,
A = Aj+ A2=2A1
= 9
W
2 V4
3EI
W/3
192 EJ
Second Method. — The second way of treating beam deflections
is more general than the first, and is based upon the assumption
that for a curve whose abscissae and ordinates are respectively x
and y, the second differential coefficient, -—^ =^ where E is the
ctx Jtc
radius of curvature at the point xy. This is a very slightly modi-
fied form of a well-known mathematical result.
It has already been shown that -rr- = =^, so that the above
K hii
may be written
d^ M d*y
-ft = or E I -J + = M
dx* h, I ax2
CASE I. Cantilever of length I, carrying a single load W at its
ou'er end (Fig. 41).
The bending moment at any point,
M = - W (/ - x)
First, EIg = M=_w(/_ir)
Now integrate E j | = _ w /, + ^f + c (a constant) .
/j/)j
~ is the tangent of the angle of slope of the curve at X. When
dii
x — 0, ~ = 0, as the line is horizontal, and therefore C must = 0
ax
dii W I2
also ; when x = I, at the outer end, El-/ = 7:—
dx
DEFLECTION OF BEAMS
55
Again integrating, ^ _ W W + Wr» + fi
^i D
2^ is the vertical distance of any point below the horizontal
through the support. When x = 0, y = 0, and therefore B = 0, so
that
y~ ~
6 El
FIG. 41.
The deflection at the outer end when x = I is
_
=
W/3
3EI
which is the same as was found by the first method.
In this and the following cases the diagrams showing the
variation of the shearing force S, and the bending moment M,
are drawn below the principal figure.
56
STRENGTH OF MATERIALS
CASE II. Cantilever of length I, carrying a uniformly dis-
tributed load w per foot (Fig. 42).
Here the total load is W = wl, and the bending moment at any
point X is
M== _w(i_v\(L^L\- wl2 wix _^
First, -p T d2y
FIG. 42.
Integrating, this gives
,dy ID I2 x w I x2 iv x*
dx^ -- 2~ +~~2 --- 6~ + ^ constant) ;
when a; = 0, ^ = 0, so that 0 = 0.
ax
Integrating again,
w
wlx* 10
~-y - , g 24 + B (a constant) ;
when x = 0, y = 0, so that B = 0.
DEFLECTION OF BEAMS
57
At the extreme outer end of the cantilever, when x = l, the
w 1B
slope is — „ -p -.., and the deflection is
wl±
or, in terms of the total load,
wl*
8~I7
W/3
S f
M
FIG. 43.
CASE III. Beam supported at ends of span I, and carrying a
concentrated load W in the centre (Fig. 43).
Measuring x from the left-hand support A, the bending
W Wl
moment at any point x is M= +-^\ and at the centre is --
Nbw,
58 STRENGTH OF MATERIALS
Integrating,
When
E j = + C (a constant).
dx 4
_
= ~ ~
therefore, Q = W P ^
or W/2
16
Then dy _Wrf W/2
dx 4 16
Integrating again,
_ ,
E I y = -TT^ --- T^i — h B (a constant).
<y 12 16
When a? = 0, y = 0, and therefore B = 0.
Then the deflection at any point x is
_W,r3 _ W/ 2r
^ ~ 12ET T6ET
The deflection at the centre, when x = -~-, is
z
W/3 W/3
A = WWl ~ 32 El
W/3
48 El
the minus sign again indicating that the deflection is measured
downwards.
The slope -~- is for that portion of the beam between the left-
hand support and the centre. Beyond the centre the value of M
is different owing to the presence of W.
At the left-hand support x = 0, and the slope
.'_ dy W a-2 W/2
1 ~ dx ~ 4E1 16 El
W/2
16 El
From the symmetry of loading, the numerical value of i at the
other support is the same, but the sign is changed — that is,
W/2
* = Tern
DEFLECTION OF BEAMS
59
CASE IV. Beam supported at ends of span I, and uniformly
loaded with w per foot run (Fig. 44).
Total load on beam = W = wl
ID I
Eeactions at supports, Ex = E2 = — —
Zi
Bending moment at any point X is
M =
As before,
2 J*~( wx j-9-
wl
Integrating, £
when
therefore,
FIG. 44.
x-L *_o
2> dx~ ()}
constant),
or
60 STRENGTH OF MATERIALS
SO that -n, T dy wlx2 mr3 wl*
dx ~~ ~^~ "~6 24"
Integrating again,
wlxB wx* wl3x ,
E ly = ~T2 ---- 24 ---- 24~ + (a constant)
when a? = 0, y = 0, and therefore B = 0, so that the deflection at X is
IV I X* WO?
At the centre, where x = y, the deflection is
w;/4 ?r/4
38E~48El
384 El
5 W/3
"384 IT
The slope at any point X is
fa '' 4 El ~TEl " 24^T
at the left-hand support this is
~24EI
CASE V. Seam fixed at ends of span I, and carrying a concen-
trated load W in the centre (Fig. 45).
It is assumed in this case that (a) the two ends of the free
part A and B are in the same horizontal line, and (&) that the ends
are so restrained that the neutral surface is horizontal at these
points, and consequently the tangents of the angles of slope,
First, suppose the beam to rest freely on the supports, as in
W/2
Case III. The slope at the supports is then = _ ... .
A bending moment M: will now have to be applied at each
support, in order to bend the overhanging ends downwards, as
dij
indicated by the arrows, until ~ = 0.
DEFLECTION OF BEAMS
61
Neglecting this moment, the reactions at A and B caused by
W
W are =-«-• The bending moment at any point of the beam X is
M = —ft— — M! when Mj is the moment necessary to keep the
neutral surface horizontal at the supports. Every part of the
beam is subjected to this same moment.
To find
Integrating,
VVI
M fe
FIG. 45.
At the supports, where #="0, :r = Q> •'• C = 0; when x = -^
again = 0, and 0 = -~ - Mj -^
or M ^W/
M! - -g-
62 STRENGTH OF MATERIALS
Then „ d W** Vflx
-T<r -T6 +B
When x=0, y = 0, and B = 0
When
6 64 /El
W/3
192 El
The deflection, of any other point at x is
y El \ 12 " 16 J
The points F F are points of contrary flexure, where the curva-
ture changes and there is no bending moment. To find these, put
M =- 0
W x Wl
•2 "T
from which x _/_
The slope at any part is
lx
CASE VI. Beam of span I fixed at the supports and uniformly
loaded with w per foot run (Fig. 46).
The conditions of supporting the beam are precisely similar to
those in the last case. The total load is wl = W, and the re-
W wl
actions at the supports clue to this alone &?&=—- = ——
Zi Zi
The bending moment at any point X is
2 2
d*i/ _ wlx '^2_M
it A 4 o r
DEFLECTION OF BEAMS
63
di/
At the support, where x = Q, -^ = 0 ; .'. 0 = 0, and, at the centre,
-¥ = 0 ; and x=-^-t that is
from which
wl* wl* I
1 -—" Ml
12
7_ O
M
Then
FIG. 46.
,., , dy ic I x2 w .r3 wl^
dx 4 ~~6 T2
24
when a? = 0, y = 0 ; . '. B = 0, and when a; = —
A
384 El
W/8
384 E I
64
STRENGTH OF MATERIALS
The slope at any point of this beam is
dy 1 iw I x2 w x^ w /2 x
There are two points of contrary flexure, where the bending
moment is zero, one on each side the centre. To find the positions
of these, put
w I x w x2 wl2
from which
which are the distances of the points FF from A.
— / ,|
R
R*
M
FIG. 47.
e reactions at the supports in the case of a learn
uniformly loaded and resting upon three supports, equally spaced and
in the same horizontal plane (Fig. 47).
This problem is best solved by making use of the fact that
" the deflection of a beam acted upon by several loads is the sum
of the deflections due to the several loads."
DEFLECTION OF BEAMS 65
First, suppose the centre support to be taken away and the
beam to rest freely on the remaining supports, whose distance
apart is /. Then, if the total load is W, the deflection in the
centre is
5 W/3
A ~ 384 E I
Now, raise the central support until the centre is restored to the
level of the outer supports. Call the force required to effect this
P. The case now resolves itself into an upward load P applied in
the centre of a freely supported beam of I span, and producing a
central deflection = A-
Considered from this point of view,
P/3
A~ 48^1
Also,
__
__
384 E I
so that P/3 5 W[3
48 El " 384 El
and _ 240
" 384
Case of a cantilever uniformly loaded, with the outer end resting
on a support at the same level as the fixed end (Fig. 48).
This may be treated in a manner similar to that employed in
the last case. By equating the (downward deflection caused by
the uniform load acting on the free cantilever) to the (upward
deflection caused by a load at the outer end), this isolated load,
which is the reaction on the support, is found to be 3/8 wl, or
3/8 W.
The point of contrary flexure is 3/4 I from the outer end, or
1/4 I from the fixed end.
Example 1. — In a rolled steel beam (symmetrical about the
neutral axis), the moment of inertia of the section is 72 inch-units.
E
66
STRENGTH OF MATERIALS
The beam is 8 ins. deep, is laid across an opening of 10 ft., and
carries a distributed load of 9 tons. Find the maximum fibre
stress, also the central deflection, taking E at 13,000 tons per
sq. in.
M
FIG. 48.
Let A = deflection in inches ;
M = bending moment in inch-tons (maximum bendin
moment) ;
/ = extreme fibre stress in tons per sq. in. ;
W = total distributed load in tons ;
/ = width of opening in inches ;
y = distance from neutral axis to extreme edge
section in inches ;
I = moment of inertia of section in inch-units.
The maximum bending moment occurs at the centre, and is
W/ 9 x 10 x 12
~~ Q W
O O
= 135 inch-tons.
of
DEFLECTION OF BEAMS 67
The maximum fibre stress,
My_ 135x4
T"= ~72~~
= 7'5 tons per sq. in.
The central deflection,
5 W/3
84 ET
5x9x120x120x120
A 384 El
384x13000x72
= 0-216 ins.
Example 2 (I.C.E.). — Suppose that three beams or planks, A,
B, and C, of the same material, are laid side by side across a span
/=100 ins., and a load W = 600 Ibs. is laid across them at the
centre of the span, so that they must all bend together. The
beams are all 6 ins. wide, but while A and C have a depth of
3 ins., the depth of the middle beam B is twice as great. How
much of the weight W will be carried by each of the three beams,
and what will be the extreme fibre stress in each ?
Led I = length of beams in inches ;
E = modulus ;
Ij = moment of inertia of 6" x 6" beam in inch-units ;
I2 = moment of inertia of two 6" X 3" beams in inch-units ;
A = deflection of beams in inches ;
Mj_ = maximum bending moment of 6" x 6" beam in inch-lbs. ;
M9 = maximum bending moment of two 6" X 3" beams in inch-
lbs. ;
y — distance from neutral axis to extreme edge of section ;
/ = extreme fibre stress in Ibs. per sq. in.
Suppose the two 6" x 3" beams to be taken as one 12" x 3"
beam.
The load being placed across all three beams, and all bending
together, the deflection will be the same in each case.
Let x — the load carried by the 6" x 6" beam.
Then W — x — the load carried by the two 6" x 3" beams.
Deflection = A = ^ ^ , = 7o~pr~j
x (w - x)
or -r- = *-rs — '-
68 STRENGTH OF MATERIALS
But fi = * = 6 = 108 inch-units
and I
2
x
108
108(600-.r)
'T 27
a- = 480 Ibs.
600 -x = 120 Ibs.
and J x (600 -ar) = 60 Ibs.
Load carried by 6" X 6" beam is 480 Ibs.
Load carried by each 6" x 3" beam is 60 Ibs.
Bending moment for 6" X 6" beam
WL = 480x100 =; 12000inch.lbs.
4 4
Bending moment for 6" X 3" beam (each)
WL 60 x 100
^
Maximum fibre stress for 6" x 6" beam
12000x3
and f =
Ia 108
= 333'S Ibs. per sq. in.
I, 27
CHAPTEE VI
SHEAR STRESS IN LOADED BEAMS
Distribution of Shear Stress in Loaded Beams
IT is not difficult to see that there must be shear stress produced
on vertical sections of horizontal loaded beams by the shearing
forces which tend to cause sliding. The distribution of this stress
over the section will now be considered.
If there is shear stress at any point on a vertical section, it
has been shown that there must also be a shear stress of equal
intensity on a horizontal plane going through the" same point. It
is this horizontal stress that is first determined, and the shear
stress on a vertical section taken as being equal.
FIG. 49.
The simplest illustration of the existence of a horizontal shear
stress is to be found in the case of a rectangular timber beam
which has been sawn into planks, as shown in Fig. 49. On these
being supported at the ends, it is seen that they are deflected
downwards by their own weight to a very much greater extent
than would be the case with the original solid beam. The reason
is that the horizontal layers are free to slide upon one another,
and become so many individual beams. In the original beam the
tendency to slide in this way is resisted by the horizontal shear
stress in the material.
To find the actual intensity of the shear stress fs at any point
(59
70
STRENGTH OF MATERIALS
whose distance from the neutral axis of the vertical section is r,
the following method may be used :
Eeferring to Fig. 50, consider two vertical sections AA and
BB, enclosing a slice of the material of the beam in question, and
whose distance apart is Sz.
Next, consider the equili-
brium of that portion of the
slice which lies above any
horizontal plane LLP distant
r from the neutral axis. This
portion is acted upon by hori-
zontal forces only, and is in
equilibrium. Three forces act
upon it, namely, the horizontal
direct stress on AL = P; the
contrary horizontal stress on
BLL — Pj_ ; and the shear force
on LLj = F. Let the former
S of these direct forces be the
g greater.
Using the same notation
as before,
•c.
p = -
Also P, = P/' — adx
J r L y
where /^ and/g are the maxi-
mum direct stresses at A and
B, and Y=fJ8z, fs being the
required shear stress, j the
length, and Sz the width of
the shear surface. But, if A
is the vertical area above
LL, and X the distance of
its centre of gravity from the neutral axis,
AX = \ x a d .
Therefore
f
P = — AX
y
SHEAR STRESS IN LOADED BEAMS
71
and
Also, F = P — Pr and, in the limit
y
AX
If M is the bending moment at AA, and M — dM the bending
moment at BB,
M = /'--, and M-dM = /' — ,
A y 1> y
from which
Therefore
c?M
l/s,/
FIG. 51.
But -^ is the shearing force S at the section, from which
it follows that M
f _ SAX
I 7
which is the shear stress
required.
Example. — To find the
distribution of shear stress
on the vertical section of a
rectangular beam of width
b and depth d = 2h, when subjected to a total shearing force S
(Fig. 51).
Using the value found above,
, SAX
' ~ lj
A is the area of the section above LL (Fig. 50), = &(A — ?•)
X is the distance of its C.G. from the neutral axis, = —
I is the moment of inertia of the whole section, =
j is constant and = &.
12
72
STRENGTH OF MATERIALS
Patting these values in the above equation, the required stress,
If the values of fs are found for a number of different values
of r, and the results plotted from a vertical base-line, the variation
of fs will be shown by the parabolic curve MOQ.
FIG. 52.
On Fig. 52 are shown curves of shear stress for a beam of I
section and for a tram-rail.
Deflection Due to Shear.
The way in which the deflection caused by the bending
moment on a beam is found was discussed in the last chapter.
In addition to the deflection so found, there is an additional
SHEAR STRESS IN LOADED BEAMS 73
amount to be added, caused by the effect of the shearing forces
in producing shear distortion.
In beams where the ratio y— r is great, the deflection due to
depth
shear is so small as to be negligible ; but in short, deep beams it
becomes of more importance.
Professor Perry has shown that for a rectangular cantilever
loaded at the end only, the total deflection at the end is
W73 _6_ Wl
AI" "3EI + T Gbd
and of the two parts which go to make up this total, that on the
left is the one already found ; that on the right gives the deflection
due to shear. Here G is the shear modulus.
Continuous Beams. Tj
Let a continuous beam rest freely on supports which are on
the same level at A, B, C, D (Fig. 53). At the first and last
1
h* — X ~ ~*1
i
1
i
— /- -i- — r
i
i
i j'"
\
\
-*""<!
i
i
[
\
\ I
4 .
A
B
^
C
FIG. 53.
D
Z
support, A and Z, there will be no bending moment ; at each of
the others there will be a bending moment due to the difference in
inclination of the tangents at A and the other supports.
Let M^, E^, ZA) and MB, E£, ZB, etc., be the respective bending
moments, reactions, and angles of slope at A, B, etc.
First, let the beam carry uniform loads per ft. run, wt w't w",
etc., in the respective spans /, I', I".
The bending moment at a point distant x from A is
and
from which R M^ ~ MB + wl_ .„
74 STRENGTH OF MATERIALS
«
Therefore wx2 , ,. x wlx
-El*
— -L*l j o
Whence, on integrating,
M . — M,, r2
when a; = 0, - = ^. so that C = El z A ; when a; = /, ~ = zs
doc ctx
Therefore
^,.73 / ?y./3
El Sy) = MAl + -6- - (M^ - M£) T - -j- • + El ZA
Again, integrating (2),
when aj = 0, ,y = 0, and when x = l, y = Q, so that the constant
that is
or
In a similar manner, taking B as origin and considering the
span BC = /', the following result is obtained :—
/ V/'/2
-T2
/ wl*- I I w/3
M +M -.-
^6 B 3 24
Therefore
SHEAR STRESS IN LOADED BEAMS 75
From which
3 (i)
This last is the equation of 3 moments, and is similar for each
pair of spans. That is,
M/' + 2MI)(I" + 1'") + M/" = (iii)
and so on.
There are five unknown quantities in the above three equations
(i), (ii), and (iii). It generally happens that the beam is allowed
to rest freely on the supports at the two extreme ends, so that
M^ = 0 and Mi( = 0, and there will be three unknowns to be found
from three equations, so that these values are determinate.
V/A &
ABC
FIG. 54.
Example 1.— Case of uniform load on a continuous beam rest-
ing on three supports, separated by two equal spans (Fig. 54).
Here, using equation (i), M^ /+ 2MB (/+/') + MC/' =
so that
1] ~" ~8
and
wl2 ?/;/2
= -£- wl=R by reason of symmetry.
Total load = 2wl, so that
76 STRENGTH OF MATERIALS
calling the total load
3
and
This is the same result as was previously obtained in another
way. '
Example 2. — Case of uniform load in a continuous beam rest-
ing on four supports, separated by three equal spans (Fig. 55).
^ A A
A B C D
FIG. 55.
Using equations (i), (ii), and (iii) —
M^ and M^ = 0, w = w' = w", and also M£ = Mc, l = l'= I",
so that it is only necessary to write (i) as
which becomes
or
and
R _ M A ~ MB wl
__icl_ wl _ 2
and from symmetry
SHEAR STRESS IN LOADED BEAMS
The total load on the beam is 3wl, so that
or if the total load 3wl is called W,
and
CHAPTER VII
RELATION BETWEEN LOAD AND STRESS IN A PRISMATIC BAR
REFERRING to the accompanying figure (Fig. 56), the upper view
represents the elevation of a prismatic bar or piece of material,
/, m, o, p, whose cross-section on a plane at right angles to the axis
is shown in the lower view. The geo-
metric axis of the bar will pass through
the centre of gravity of this section. Let
a load P act either along or parallel to
this axis, as shown by the dotted line.
This may be either a compressive or a
tensile load. Suppose for the present
that it is a load in compression. If it
acts along the axis of the piece, it will
result in a uniform compressive stress
which is the same at all points of
the section. Its magnitude will be
p
fd = --, where A is the area in question.
A.
In most cases of loading of this kind,
uniformity of stress is aimed at but is
rarely attained. If the line of the load
P is shifted ever so little away from the
geometric axis of the prisrn, the stress
at once becomes a variable one, with a
maximum value greater and a mini-
FIG 56 mum less than fd. The farther the
load is moved away from the axis, the
greater becomes the difference between the mean and the extreme
stresses. The precise relation which exists between the load and
the extreme and mean stresses for any given section, when the line
78
LOAD AND STRESS IN A PRISMATIC BAR 79
of loading is so moved away from the geometric axis, is given in
what immediately follows.
Again referring to Fig. 56, the load P is supposed no longer to
act at the centre of gravity of the area O, but its line of action,
still parallel to the axis, has been moved away from this point, and
now passes through some new point, Q, at a distance a from 0
measured along FG, one of the principal axes of the figure. The
point Q may or may not lie in this line, but for the present it is to
be supposed that it does.
It has been shown that if M is the bending moment on a
section of a prism, M = , where I is the moment of inertia of
u
the area about the neutral axis CD. In the present case the load
P is one of compression, and, in acting as it does at some distance
from the centroid of the area, gives rise to the bending moment M.
This tends to bend the piece in the direction indicated by the
arrows, causing compression of the material at F and tension at G.
Both stresses have a maximum value at these points, and diminish
uniformly as they approach the neutral axis. If the section is
symmetrical about the neutral axis and the elastic properties are
the same in compression as in tension, it has been shown that the
TV *"
value of the maximum stress, /6 = — ^—, will be the same at F as
at G. The precise value of the bending moment will in the
present case be M = Pa, and therefore fb = - — ^—. This gives the
value of the maximum tensile and compressive stresses due to the
bending action caused by the load in acting away from the centroid
of the area,
It has just been shown that the uniform stress is the load
p
divided by the area of the section, or/d=-r-.
.A.
The actual combined stress at either edge, F or G, will be the
algebraic sum of the uniform and bending stresses.
Thus, at F, which is on the same side of the neutral axis as
the point where the load acts, the two stresses will have the same
sign. In the present instance this is one of compression. If the
uniform stress fd be given the plus sign, then the stress due to
bending, f My
h = ± —
80
STRENGTH OF MATERIALS
according as it is on the same or the opposite side of the neutral
axis as the load P. At any point in the section there will be a com-
bined stress, consisting on the one part of the uniform stress and
on the other of the stress due to the bending.
At F, the total or effective stress will be
,. f f
«/€ = Jd^Jb = ~A"
and at G,
Je
My
1
My
As the moment of inertia I = A . kz where, as before, A is the
area and k is the radius of gyration about CD, the above expres-
sion may for convenience be written
f - J!. + J?L
' • ~ ' ±
d - 4- —
A
The above expression gives a relation between the dimensions
of the section of the piece of material under stress, the magnitude
of the load, the distance of the line
I of action of the load from the centre
!
of gravity, and the maximum and
minimum stresses resulting. It is
applicable to sections of any form,
so long as the moments of inertia can
be found. In the case of the more
usual sections, which consist wholly
of, or are made up of regular figures,
such as rectangles and circles, the
values of the radius of gyration can
readily be found by the usual mathe-
matical methods ; where the section has a more complicated form
the modulus must be found by the graphical method, and from it
the radius of gyration determined.
Particular Cases. — One or two particular cases will now
be considered.
Circle. — In Fig. 57 is shown a circular section of radius r and
diameter d. In this case the distance from the neutral axis or
FIG. 57.
LOAD AND STRESS IN A PRISMATIC BAR
81
T' °r 16
centre of gravity to the edge of the section is the radius of the
circle, and the expression giving the maximum stress is
P
and as in the circle
the above becomes
By again subtracting the uniform stress resulting from the
load acting down the geometrical axis of the bar from the
maximum stress obtained as above, the stress due to bending
alone, which may be called the excess caused by the eccentric
loading, is obtained. That is,
T?
Excess
s f , „
=/.-/,- '- ' =.f,
$
which reduces to T— — ; or, when put in the form of a percentage
-A. T
by which the maximum exceeds the uniform or mean, it becomes
4^100 SalOO ,
- or -- - — , where d is the diameter of the circle.
T a
In the following table are given values of this percentage for
different degrees of eccentricity, varying from 0 to Q'SQd ; that is
to say, at various points from the centre to the circumference
of the circle.
TABLE OF PERCENTAGE EXCESS OF MAXIMUM OVER UNIFORM OR MEAN
STRESS IN A CIRCULAR SECTION.
a
t
Remarks.
0
0
At centre
0-05d
40 per cent.
O'Wd
80
0-20d
160
0-30d
240
0-±0d
320 . „
0-50rf
400
At circumference
82
STRENGTH OF MATERIALS
Thus, it will be seen that by moving the line of action of the
load no more than one-tenth of the diameter of the circle away
from its centre, the stress at the circumference is about
!*- -h - -
b -
i
FIG. 58.
doubled, and by taking it to the circumference the stress is
magnified fivefold.
Rectangle. — In the case of a rectangular section (Fig. 58)
similar results are found. Here, as before,
100
6a
100
where a has the same meaning as before, and h is the depth
of the section measured at right angles to the neutral axis.
TABLE OF PERCENTAGE EXCESS OF MAXIMUM OVER UNIFORM OR MEAN
STRESS IN A RECTANGULAR SECTION.
a
e
Remarks.
0
0
At centre
0-05A
30 per cent.
0-10A
60 „
0-20/*
120 „
0-30A
180
040A
240
0-50A
300
At edge
LOAD AND STRESS IN A PRISMATIC BAR
83
By comparing the figures in these two tables it will at once be
seen that the increase is more marked in the case of the circle
than in the rectangle.
d = JO
_L
FIG. 59.
Example of G-irder Section (Fig. 59) : —
— * -{"•«:!:'
i = 4J
13 J sq. ins.
y = 5 ins.
6x(10y 5ix(8i)3
12 12
= 16-5
e = per cent, excess = -M- x 100
k2
A nj 1 x5 x 100
when « = 0-ld, e = = 30 per cent.
1 0*0
2x5 x 100
„ a = 0-2d, e = -- = 60
3x5 x 100
„ a = 0-3d, e = -- - 90
A . , 4x 5 x 100
„ a =0-4d, ^ = - =120
5x 5 x 100
e = -- =150
84
STRENGTH OF MATERIALS
TABLE OF PERCENTAGE EXCESS OF MAXIMUM OVER UNIFORM OR MEAN
STRESS IN THE I SECTION SHOWN ON FIG. 59.
a
e
Remarks.
0
0
At centre
0-lOd
30 per cent.
0-20d
60 „
0-3Qd
90 „ .
0-40d
120
0-50d
150
At edge
It will be seen that the more the material is spread away from
the centre the smaller is the effect produced by moving the load-
line any given distance from the neutral axis.
Position of the load P which will give a zero stress at one edye of
the section.
The minimum stress is
Put this equal to zero
from which a = — , which means that if P acts along a line which
k*
passes through the section at a distance — from its centre of
gravity measured at right angles to the neutral axis, the stress
due to bending will just neutralise the direct stress at the edge of
the section on the side farthest away from the load.
For a circle, where &2 = - and y = -
That is to say, P must act within the middle quarter of the
diameter in order that the stress at any part of the section may
not change signs, or, in other words, the load-line in cutting any
LOAD AND STRESS IN A PRISMATIC BAR
85
section must not be outside a circle which is concentric with that
forming the outer boundary and whose diameter is = -^- •
In the case of a rectangle of height h and width I,
and
so that
&2 =
y =
a =
T
//
This means that the load-line must be kept within the middle
third of the depth, in order that the stress may be of the same
kind at all parts of
~T
i-
A
_L
FIG. 60.
the section.
The last result be-
comes of great im-
portance in such cases
as masonry arches and
walls, where it is never
permissible to put any
part of the material
under tensile stress.
If the load-line gets
outside the middle third of the section, the intended compressive
stress becomes a tension stress at one edge.
In the I section above,
k2 = 16'5 ins.
y = 5 ins.
and #> 16-5' v.
a = = - • = 3'3 ins.
y °
The depth is 10 ins., so that a can be written as a= e^ X 3'3,
or the load-line must be kept within the middle two-thirds of the
depth in the direction of the centre line of the web.
The dotted lines in Figs. 57, 58, 59, indicate the boundaries
within which the load-line must act in order to ensure that the
stress is all tension or all compression.
The diagram on Fig. 60 shows how the direct stress and the
bending stress combine to produce the resultant stress. This is
for the case shown on Fig. 64.
86
STRENGTH OF MATERIALS
Instances of Unequal Stresses due to Eccentric Loading.
The only way to determine the stress occurring in any
portion of material under stress is to measure the compression
or extension of a given length on the surface of the material.
It is well known that so long as the stresses in the material are
not carried beyond the elastic limit, they are proportional to the
strains or deformations ; so that if, when the material is in this
condition, the extensions or compressions can be measured on
the surface of the material at given points, the deformations so
found will be measures of the stresses at these points.
M
X
FIG. 61.
Method of Measuring the Strains (Extensions or
Compressions).
For the purpose of ascertaining the amount of compression or
extension of a structural part or specially prepared specimen,
the writer has made use of the Martens mirror apparatus, the
principle of whose working is indicated in the sketch on Fig. 61.
Here the material forms part of a bar in tension, XY, and it is
desired to measure the extension of the material on the right-
hand and left-hand surfaces respectively. Take the right-hand
side. Here a light metal bar B is held against the surface of the
LOAD AND STRESS IN A PRISMATIC BAR 87
material. The lower end of this bar has a sharp edge or point,
which penetrates the surface to a very small extent and prevents
the measuring bar B from slipping. At the upper end is a groove
in which rests one side of a rocking prism, whose other edge
rests against the surface of XY. Before loading the prism lies
in a horizontal position, but when a tension load is applied the
edge of the prism resting against the surface of the material
is raised as the bar is extended, while the edge against B remains
stationary relatively to the bar; in this way the prism is tilted
slightly.
The prism carries a mirror, M, as indicated in the sketch. The
observer, on looking through a telescope at A, sees reflected in
the mirror a view of a graduated scale Bp and a hair-line in
the diaphragm of the telescope appears to lie across the scale.
The more the material of the bar stretches, the further the prism
is tilted and a different part of the scale brought into view.
The distances and graduations of the scale are so proportioned
that the apparent movement of the hair-line across the scale pro-
vides a definite measure of the amount of stretch of the material.
It will be seen that in the sketch there are two arrangements
of this kind, one on each face of the bar. The apparatus can be
used equally well for compression as for tension.
Example 1. — A masonry pier, having the section already
given on Eig. 25, carries a load whose line of action passes
through a point 1 ft. 6 ins. from the wide edge. To find the maxi-
mum load to give a compressive stress which does not exceed 12
tons per sq. ft. :
The moment of inertia, I, has already been found, and = 43*6
foot-units.
Let W = load in tons,
Area of pier = (18 + 6) = 24 sq. ft.
W
Uniform compressive stress = ^ tons per sq. ft. = 0'0417 W.
Compressive stress at X due to bending moment
g = 2-875', and a = 3-500-2-875
= 0-625
M = = (W x 0-625) Ib.-ft.
y = (1-5 + 0-625) = 2-125 ft.
88
Therefore
=
STRENGTH OF MATERIALS
W x 0-625 x 2-125
-. 12 = 0-0305 W + 0-0417 W
12
W =
0-0722
- 166 tons.
Example 2. — On Fig. 62 is shown a sketch which represents
a mild steel test -piece when being
P subjected to a compressive load be-
tween the two platens T T of a testing
machine.
The thick lines M M represent the
bars B B of the Martens mirror appar-
atus on Fig. 61. On Fig. 63 are shown
curves of loads and compressions for two
different conditions of loading.
Those marked (a) give the loads
and the corresponding compressions
when the flat ends of the bar rest
against the platens of the machine on
the assumption that the line of load
will coincide with the axis of the bar
and consequently the stress will be the
same at all points. That this assump-
tion is not borne out by the observations
taken during the experiment is shown
by the curves. The eccentricity of the
load-line, which gives rise to this in-
equality of stress, is probably due to the
platens of the machine not being quite
parallel.
The other pair of curves represents the result of applying
the load, not through the flat ends of the bar, but through a pair
of steel balls placed in the centres of the ends, and seem to indi-
cate that the load-line is now very near the axis. The two
curves almost coincide, showing that the stress is practically
uniform.
The figures from which the above curves were plotted were
obtained from actual experiment.
FIG. 62.
LOAD AND STRESS IN A PRISMATIC BAR
89
Example 3 (Fig. 64). — This is an experiment not unlike the
last. Here are two cubical blocks of concrete, one being bedded
upon the other through a joint of cement mortar. The Martens
mirrors were used as before.
First the load was placed at P P in the centre, and the curves
(a) (a) (Fig. 65) obtained. These almost coincide, indicating
uniformity of stress.
The load was then allowed to act at Px P1; just on the edge
of the middle third of the section. It has been shown that with
COMPRESSIONS.
tf?
J
FIG. 63.
this arrangement of loading the compressive stress is a maximum
at the right-hand edge and zero on the left-hand edge. This
is practically the case in the present experiment, as is shown by
the curves I and &. That the result is not quite what it should
be is caused by the difficulty in applying the load precisely at the
edge of the middle third.
Example 4. — The two curves on Fig. 66 were obtained from an
experiment similar to the last. The sample of material was in
this case a brickwork pier 18 ins. square and 36 ins. high (see
Fig. 97). The load was applied through a very carefully prepared bed
90
STRENGTH OF MATERIALS
P Pf
I"
P
FIG. 64.
COMPRESSIONS.
FIG. 65.
LOAD AND STRESS IN A PRISMATIC BAR
91
of neat cement mortar by first setting the pillar on the lower platen,
then putting a layer of soft cement on the top, bringing down
the top platen with a little pressure, and leaving it so until the
mortar had set. Between the mortar and the platen was a thin
COMPRESSIONS.
FIG.
layer of cardboard. After this process it would naturally be
thought that the stress would be perfectly uniform. The curves
indicate that this is far from being the case, and shows how
difficult it is to obtain uniformity under these conditions.
CHAPTEE VIII
PILLARS, STRUTS, OR COLUMNS
IN the case of a short prism (Fig. 67a), whose length is /, smallest
diameter d, and area of cross-section A, the crushing load which
causes failure is
P-/A or * _/
where / is the crushing stress for the material in question. For
this to hold good, the ratio -y must not be more than about 3 to 1.
Cb
Where the ratio becomes much larger than this, P is less than /A,
p
or -r- is less than /.
In this case the structural part is spoken of as a pillar, strut,
or long column.
In a short column failure takes place by simple crushing, and
in a long column partly by crushing and partly by bending. The
greater the ratio —r the more is the strength dependent on the
ct
effect of the length, until a point is reached, where -=- is greater
K
than about 150, at which the failure is practically due to bending
alone and simple crushing is negligible, k = radius of gyration.
p
For the first of the three cases — = /.
.A.
For the third case the following formula, due to Euler, is to be
92
PILLARS, STRUTS, OR COLUMNS
93
used. Euler's Formula for very long columns (Fig. 67c), whose
ends are perfectly free, is
P 21? F
A :: **V
where P, /, and A have the same meanings as before,
E is the direct elastic modulus,
and k is the least radius of gyration of the cross-section.
For the second case (Fig. 676), which is the one most commonly
FIG. 67.
met with in practice, various formulas have been devised. It is
apparent that the longer the column relatively to its diameter, the
more easily will it collapse under a given load, so that a formula
P P
must be used where, instead of having -A-=/, -r- must be less
A. A
than /, the diminution becoming greater as I increases. The
94
STRENGTH OF MATERIALS
formulas in use, of which the three best known are given below,
are all more or less empirical.
Gordon's Formula was based by Professor Gordon on the
results of Hodgkinson's long series of experiments. It is
L /
A /2
Here it will be seen that / is diminished by increasing its denomi-
nator, which is made to depend on the square of the ratio -j- and
upon an empirical constant c. This constant varies with both the
form of the section and the material, with the result that a large
number of constants must be known to fit all cases that may arise.
The constants originally given for use in this formula hardly apply
to the materials and forms of section now in use, and it is not
necessary to give them here.
Rankine's Modification of Gordon's Formula is similar in general
form, being
P /
A" /a
In this case the empirical constant a varies only with the
material, the difference in- shape being accounted for by the
change in the radius of gyration Jc. This is more satisfactory
than Gordon's, as fewer constants are required
The following constants are given by several authorities for
use in Kankine's formula : —
Material.
/Ibs. per sq. in.
a
Wrought iron
36,000
inrVo
Mild steel .
48,000
TACTS'
Hard steel .
70,000
i
53TR7
Cast iron .
80,000
TFW
In this formula, / is somewhat higher than the Yield Point
Stress in compression for the first three, which are ductile
PILLARS, STRUTS, OR COLUMNS
95
materials, and is the Crushing Stress for cast iron. As the load
on a column increases, there is a small deflection to one side or the
other; causing a maximum compressive stress on the concave side.
When this stress passes the yield stress of the material, the deflec-
tion increases rapidly, and buckling follows.
The constants a, given here, all refer to pillars in which the
ends are free to turn, either by being hinged or having rounded
ends. The length I is the distance between the centres about
which the ends can turn.
Where the column is rigidly fixed at both ends, the proper
constant is ~ ; where it is fixed at one end and free at the other,
4
the constant is --a.
FIG.
Fig. 68 shows the effect of fixing one or both ends. Whatever
method of fixing is employed, the effect must be to keep the axis
of the unloaded column at the fixed end in the line of action of
the load.
A more recent and more workable formula is that of Professor
Johnson of Wisconsin, who utilised the results of a large number
of destruction tests of columns made by Considere and Tetmajer.
96
STRENGTH OF MATERIALS
On Fig. 69 are plotted three curves showing the relation
p
between the buckling stress -r- for columns having different
.A.
I P
values of -=-. One shows the values of —r- as given by Euler's
formula, which for columns of ordinary lengths is seen to be quite
inapplicable. The second is that of Kankine, which gives much
more reasonable results. Johnson plotted the results of Consi-
dere's and Tetmajer's tests in a similar manner, and found that the
parts of the curves between the line of zero -j- and the point of
rC
contact with the Euler curve are approximately parabolas whose
equations are of the form
-i- = f'~ B — where B = — L
A <y k* 4?r2E
Here / is diminished as I increases, not by an increasing divisor,
but by taking away a larger quantity.
Jbs per
60,000
40.000
2O 4O 60 80 100 120 I&O 160 180 ZOO
20.000
The symbol / denotes the compression yield point for the
ductile materials and the ultimate crushing stress for brittle
materials.
PILLARS, STRUTS, OR COLUMNS
97
— then denotes the stress on the cross-section of the column
A
which will cause it to fail by buckling or fracturing, according
p
to the material. To get the safe stress, -r- must be divided bv
A.
a factor of safety.
It must also be noted that k is the least radius of gyration
of the section.
The values given by Johnson for the constants B and /
are: —
P /2
TABLE OF VALUES OF B AND f IN FORMULA — = f — B -^
Material.
Condition of Ends.
-r- not greater
than
fibs.
per sq. in.
B.
Wrought Ironj
Pin (hinged)
Flat
170
210
34,000
34,000
0-67
0-43
Mild Steel . |
Pin (hinged)
Flat
150
190
42,000
42,000
0-97
0-62
Cast Iron . \
Rounded
Flat
70
120
60,000
60,000
6-25
2-25
It is necessary to point out that the values of / which are
given above are those chosen by Johnson, as average values arising
from his own experience. Where accuracy is desired, it will be
necessary to find the value of / directly from compression tests of
samples of the material in question, and calculate B accordingly.
The following example will serve to illustrate the use of
Johnson's formula: —
Steel pillar of the section shown on Fig. 70, loaded through
rollers whose centres are 49J inches apart.
Depth = 6 ins.
Width of flange = 3 ins.
Thickness of flange = 0*45 in.
of web = 0'39 in.
98 STRENGTH OF MATERIALS
From the dimensions it is found that
A = 4'66 sq. ins.
I - 2-05.
&2 = 4 = 0-440.
A.
I = 49-25 ins.
f = 40,000 Ibs. per sq. in. by experiment.
Then,
and
B "
FIG. 70.
:-33J
P = 4-66(40,000 - 1-33^
= 68 tons nearly.
The actual collapsing load of this was found by experiment to
be 69 tons.
OF THE
UNIVERSITY
Of
/
CHAPTEE IX
TORSION AND SPRINGS
BY torsion is meant the effect produced by the action of a pair of
equal and opposite couples at the ends of a prism or shaft acting
in planes at right angles to its axis. The kind of stress caused by
torsion most commonly occurs in rotating shafts used for trans-
mitting power.
Elastic Circular Shaft. — In Fig. 71 is shown a view of a
short length, I, of a solid circular shaft, whose ends consist of two
FIG. 71.
circles having centres 0 and Q at the ends of the axis OQ. These
circular planes are perpendicular to the axis.
This forms a short length taken out of a shaft such as that
shown on Fig. 72.
Suppose a line AB parallel to the axis to be drawn on the
surface of the shaft before any force is applied to it.
The pair of equal and opposite couples or twisting moments,
T T, are now allowed to act on the shaft. The effect will be
to rotate the ends relatively to one another, in the directions
100 STRENGTH OF MATERIALS
indicated by the arrows. If A be taken as stationary, the effect
of the twisting will be to move the point B to C, and the line AB
into a new position AC, making an angle /3 with AB. This line
AC will form part of a spiral.
The movement of B to C will mean that the radius QB = r
takes up a new position QC after turning through an angle
BQC = 0. For the purposes of the following proof the material is
assumed to be elastic. This being the case, it is reasonable to
assume that the radius QB, which is a straight line, remains
straight when twisting takes place. If E is any point in this
radius intermediate between B and Q, where QE = #, E will move
to F when B moves to C, and
EF x_
BC = r
From this it follows that a line DE, which is parallel to the
axis before the twist is applied, will assume a new position DF,
the angle EDF being called /3r
Now, BC EF EF
_
or
For stresses within the elastic limit, the angles of distortion
are relatively small, and the above may be written
A
V T
This angular distortion is due to the shear stress produced by
the twisting moment and acting circumferentially about Q. It
TORSION AND SPRINGS
101
has previously been shown that the angle of distortion is pro-
portional to the shear stress producing it. Or, if fs is the shear
stress at BC, on the surface of the shaft, and s is the shear stress
at any other surface EF, then
or, in other words, the shear stress in a circular shaft subjected to
torsion has a maximum value at the
surface of the shaft, and at any other
radius x is proportional to that radius.
Next, to find the torsional resisting
moment. On Fig. 73 the circle repre-
sents the section of the shaft having a
diameter d and radius r. Let this
section be divided up into a number of
elemental rings of • radius = x, and
width = Sx. The area of one such ring
= 27rxSx. The circumferential force
on the ring = s (2irxSx)
f,
as
U j — J
Also, the moment of this force about the centre is
-*,
27T
-
This is the torsional resisting moment of one of the elemental
rings, and the total resisting moment will be
[r TT
xzdx = — f r3
Jo 2 Js
or
This is equal and opposite to the external twisting moment T,
> where d =
102 STRENGTH OF MATERIALS
For a hollow shaft (Fig. 74) having inner and outer diameters
respectively = d2 and dv and radii r2 and rp
T T = ^
l' = |-/s
I
FIG. 74. 16^s
The expressions
7T 7T ,. 7T / A - 7T
— r4. — a4, — 1 r,4 — r0 I, and —
2 ' 32 2 \ l 2 / 32
are the polar moments of inertia, I , where
Angle of twist of a Shaft. — To find the angle 0 through
which the circular end at B is rotated relatively to that at A,
T>/~1
= <f> (radians), or BC = <f>r
Also for small angles,
?rr = ^r = ft (radians), or BC = /?/
Ar> /
So that BC = <#»• = /?/ or <#> = /?-£-
Again, for the angular distortion /3 = ^f , where G- is the shear
modulus ; so that the above may be written
/* / 2T / 32T/
* ~ G T VG^
2T
because / has been shown to = — »-
TTf
For hollow shafts
217
TORSION AND SPRINGS 103
Expressed in degrees, these angles are respectively
_2T j_ 180 360 T/
TrG r4 TT = 7T2Gr4
For hollow shafts
,0 360 T/
Using diameters instead of radii
5760T/
and for hollow shafts,
5760T/
It is the usual practice not to allow 0° to exceed 1° in a length
l=20d.
Horse-power transmitted. — If T is expressed in Ib. -inches,
T
then -= will be the twisting moment expressed in Ib.-feet. And
j.Z
the work done by T in one minute will be — ^ - where N is the
LA
number of revolutions made by the shaft in one minute. The
horse-power transmitted will therefore be
TN27T
=
12x33,000 16 x 12x33,000
from which approximately,
H P
~ 322,000
From this the diameter d of a shaft to transmit a given H.P.
at speed of revolution, N, can be found, as
d _ 3/322,000 H.P. = 67.g /H.P.
t
The usual values given to fs in practice are :
For wrought iron shafts /s = 5000 Ibs. per sq. in.
„ mild steel shafts/s = 6000 Ibs. per sq. in.
„ medium-carbon steel shafts fs = 8000 Ibs. per sq. in.
104
STRENGTH OF MATERIALS
Torsional Strength of Shafts in the Plastic State.
"When the twisting moment on a shaft is increased to such a
point that the material is being stressed some way beyond the
limit of elasticity, experiment seems to show that the shear stress
s at any radius x is no longer proportional to x, but has become
much more nearly constant and equal to the stress fs at the
surface. The author has found that
in ductile materials like soft steel
and the best wrought iron s is
practically = fs when the point of
fracture is reached. Therefore, tak-
ing s as being equal to/s, the former
equations become :
For solid shafts,
And, for hollow shafts,
For shafts under working con-
ditions the elastic limit is never
reached, and the former set of
formulas are to be used for provid-
ing a relation between the work-
ing torsional moment, the safe
shear stress, and the dimensions
of the shaft.
Where it is desired to find
the twisting moment necessary to fracture the shaft, the latter
set of formulas are to be used, where fs is the ultimate shearing
strength of the material.
Loads and Deformations of Springs.
Helical springs of round steel. — In Fig. 75 is shown a
spring of this type intended to be loaded in tension. The
following applies equally well to a compression spring.
TORSION AND SPRINGS 105
Here, let
P = the load on the spring acting along its axis.
D = the mean diameter of the coil.
d = the diameter of the wire.
n = number of complete coils.
G = shear modulus of the material.
Then the length of wire in the spring,
The effect of the load is to put upon the wire, throughout its
length, a torsional moment whose value is
PD
T = -JT- ; = PK where E is the radius of the coil,
&
and this results in a twist of the wire, as shown in Fig. 76.
Consider one coil. Throughout the length I of the coil there
is the twisting moment T, due to a force P acting at the end of an
arm — . The twisting of the wire will allow the point A to move
relatively to B an amount S where
*-*
2
<j> being the angle of twist for the length in question, or.
But it has previously been shown that
32T/
so that 8PD3
for one coil. The total extension for a spring of n coils will
then be
A =
The corresponding formula for springs of square metal is :
106
STRENGTH OF MATERIALS
It is to be noted that the above are only true so long as the
strains are relatively small, so that S is sensibly = ^-, and not
a
D
- tan
*, 8
__£ j
1
FIG. 76.
For spring steel the value of G- varies from 12,000,000 Ibs. per
sq. in. to 14,000,000 Ibs. per sq. in.
TORSION AND SPRINGS 107
Example 1. — A solid steel shaft is to transmit 600 H.P. at a
speed of 58 revolutions per minute. Find the diameter of the
shaft. The stress in the metal must not exceed 8000 Ibs. per
sq. in.
HP =
12x33,000
and 3 / H7iVxT37000 x~l 2
-v
= 7-46 ins.
3 /600 x 33,000 jxJ2 x 16
2 xV2~x" 58^8000
Example 2. — A shaft 3 ins. in diameter, running at 250
revolutions per minute, transmits 50 H.P. Find the maximum
stress.
Using the same symbols as above,
12x33,000
H.P. x 12x33,000
and fs =
50x12x33000x16
2 x 7T2 x 250 x (3)3
/. The maximum stress = 2378 Ibs. per sq. in.
^Example 3. — A spring of steel of circular section containing
five coils is subjected to compressive loads in a testing machine.
Loads are applied and readings taken of the lengths. A piece
of the steel is then subjected to tension loads. Loads are
applied and readings taken on a length of 15 cm. The outside
diameter of the spring is 4*8 ins. and the diameter of the steel
0'96 in. Find the shear modulus and the modulus of direct
elasticity.
108
STRENGTH OF MATERIALS
Let G- = shear modulus in Ibs. per sq. in.
E = modulus of elasticity in Ibs. per sq. in.
P = load on spring (compressive) in Ibs.
D = mean diameter of spring in inches.
d = diameter of the steel in inches.
A = the compression on the spring in inches.
n — the number of coils in the spring.
Mean diameter of spring,
D = 4-8-0-96
= 3-84 ins.
TABLE OF READINGS TAKEN WHEN THE SPRING WAS COMPRESSED.
Loads, tons .
*
ft
1
1
u
1ft
IS
Length of 5 coils, ins.
7-39
7'24
7-09
6-95
6-81
6-66
6-52
Compressions
—
0-15
0-15
0-14
0-14
0-15
0-14
Loads, tons .
2
2|
2i
2|
3
3|
Length of 5 coils, ins.
6-38
6-25
6-11
5-98
5-84
5-67
Compressions
0-14
0-13
0-14
0-13
0-14
0-17
The average compression (from \ to 3 tons) for \ ton
1-55
11
= 01409 in.
The shear modulus
8 x 560 x 5 x (3-84)3
0-1409 x (0-96)4
- 10,590,000 Ibs. per sq. in.
In the tension test on a piece of the same steel, the readings
were taken on 15 cm. = 5*905 ins.
TORSION AND SPRINGS
109
Let E = the elastic modulus in Ibs. per sq. in.
/ = the length on which readings were taken in inches.
/ = the stress in the metal in Ibs. per sq. in.
x = the elongation in inches.
W = the load applied to produce x, in Ibs.
A = the cross-sectional area of the metal in sq. ins.
TABLE OF EXTENSIONS.
Load in Tons . . • k
4
I
1
U
1*
If
2
Extensions, TTnnfth in. . 0
0-14
0-29
0-45
0-605
0-76
0-915
1-065
I
Differences
0-14
0-15
0-16
0-155
0-155
0-155
0-15
Load in tons
2*
2*
2|
3
3i
3*
3|
4
Extensions, T^th in. .
1-22
1-38
1-525
1-68
1-825
1-985
2-14
2-295
Differences .
0.155
0-16
0-145
0-155
0-145
0-16
0-155
0-155
The average extension (from J to 4 tons) for J ton
2-295
= = 0-15,3 YoVo^hs or an in.,
lo
i.e., a; = 0-000153 in.
The stress per sq. in. of sectional area of the metal
E =
The modulus of elasticity
n
x
Wl
Ax
560 x 5-905
0-7854 x (0-96)* x 0-000153
= 29,860,000 Ibs. per sq. in.
The ratio of
The modulus of elasticity : The shear modulus
is as :— 29,860,000 : 10,590,000
i.e., 2-819 : 1
CHAPTEE X
TORSION COMBINED WITH BENDING
IT often occurs in practice that a shaft has not only to withstand
a torsional moment T, but is so supported and loaded that there
is a bending moment acting at the same time. The most usual
instances of this kind occur in shafts which run in bearings some
distance apart and carry heavy wheels or pulleys between the
bearings, and in the crank-shafts of engines.
On Fig. 77 is shown a typical case. A force acts at right
angles to the centre line of a crank of radius B, giving rise to a
twisting moment
T - PB
In plan the line of action of P is distant H from the centre of the
no
TORSION COMBINED WITH BENDING
111
nearer bearing, so that on the section of the shaft at XY there is,
in addition to T, a bending moment
M = PH
On Fig. 78 is shown an enlarged view of the portion of the
shaft near 0, the force
P being supposed normal
to the paper. In the
centre of the shaft, at
the point marked by the
circle and on the section
made by XY, there will be
a maximum direct stress
due to M. If P acts to-
wards the paper, this will
be a tensile stress ; if from
the paper, the material will
be in compression at this
point. At the same point
there will also be a shear
stress caused by the twist-
ing moment T. Call the
direct stress / and the
shear stress ff There
will be a shear stress
along YX, and it has previously been shown that there must be
a shear stress of equal intensity on planes at right angles to YX.
Next, consider an elemental triangular solid ABC, of uniform
thickness, at the surface of the shaft, and which is kept in
equilibrium by the stresses acting on its faces, and which is of
such proportions that the resultant stress in its diagonal AC is
wholly normal. Call the angle CAB = 0.
On AB is the shear stress fs, on BC is the direct stress / and
also the shear stress fs, and on AC is the resulting normal
stress fn.
The total normal stress on AC = /nAC. Eesolving this hori-
zontally and vertically :
(Horizontal component) =/wAC sin 0 =
(Vertical component) = /WAC cos 0 = fsBG
FIG. 78.
112 STRENGTH OF MATERIALS
From which
/AB
/.-/- ^=fs™^
and
Subtracting (b) from (a),
-/=/.(cot 0-tan
- 2/8 cot 2#
tan 20 = -- (1)
This gives two values of 20 differing by 180°, and two values
of 0 differing by 90°.
Now multiply the above two equations (OL) and (&) together :
This last equation will be made use of for combining the torsion
and bending moments.
Here r =. Mr 4M
and f 2T
./<- ..q
Now /„'-//„ = /.'
This is a quadratic whose solution gives
/» = T ± VT
2M 4M2
"
" ^ = M ±
?rr3
Putting -^- /n = TE where T^ is a twisting moment which,
acting alone on the shaft in question, would produce a stress of
magnitude fn in the material, and is called the equivalent twisting
TORSION COMBINED WITH BENDING
113
moment, the above may be written, taking the plus value as
being .the greater,
or, if M^ is the equivalent bending moment,
that is,
ME= JM +
+ T2
The stress fn, brought about by the torsion and bending
moments acting simultaneously, may be taken as being a tensile
stress on AC, a compressive stress on HG at right angles to AC,
or a shear stress on planes making angles of 45° with these. See
Fig. 79.
FIG. 79.
' From equation (1), knowing fs and /, it is possible to calculate
the angle which the plane, whose stress is wholly normal, makes
with the axis.
Where there is no bending, and/=o,
and
Tan 20 = (infinity)
20 = 90°
0 = 45°
114
STRENGTH OF MATERIALS
That this is so is very forcibly illustrated in the case of a cast
iron shaft broken under pure torsion, when it is seen that the
surface of failure takes a spiral
form, the helix making an angle of
45° with the axis. Such a fracture
is shown on Fig. 80 (a).
Where there is bending as well
as torsion, the angle 6 is greater
than as indicated in Fig. 80 (b).
The formula
FIG. 80.
is the one given by Eankine, but,
according to the more complete theory of elasticity, when Poisson's
Eatio is taken into account, this is not strictly true. Grashof
gives the more correct value as
Example. — A steel shaft is supported in two bearings 40 ins. from
centre to centre, and midway between these carries a pulley weigh-
ing 10 tons. 410 H.P. are transmitted through the shaft at 98
revolutions per minute. Find the diameter when the stress in the
material does not exceed 3J tons per sq. in.
M = ™^ = 100 in.-tons,
2i
(H. P.) 12 33,000
T = L 2.-N2240 — H8 m.-tons.
Here
and
Then TE
= 100 + V(100)2+ (118)2
= 255 in.-tons.
And, finally, the diameter
j __ \\ -&'
^^r
255 x 16
- 7'2 ins.
CHAPTEE XI
STRENGTH OF CYLINDERS
Thin Cylinders. — In a cylinder of length I (see Fig. 81), whose
thickness, t, is small relatively to its internal diameter, d, and
which is subjected to an internal pressure of intensity, p, the
total force tending to
lift the upper half from
the lower half is pdl, or
p2rl where r = -~- — the
2i
internal radius.
To this separation is
opposed the resisting
stress in the material.
The area on which this
stress / acts is tl on
each side, or, altogether,
2tl, so that the total
opposing pdl is
Therefore,
pdl = 2ftl, or pd = 2ft, or pr = ft.
If the ends of the cylinder are closed, as in the case of a
boiler, there is a stress fe induced on the annular section made by
a plane AB at right angles to the axis of the cylinder. Now the
total pressure on the end is p -^- d2.
This is resisted by fe acting on a ring whose area is very
nearly irdt. So that,
force
f2tl.
FlG 81<
pd =
or
pd _ pr
47 ~ 2t
115
116
STRENGTH OF MATERIALS
But, from the last equation, the stress on a longitudinal section
'
is •£-, and, therefore,
t
Stress on a transverse section
Stress on a longitudinal section
PL
1L
PL
Thick Cylinders. — In what has just been said about thin
cylinders, where the ratio — is large, the stress is assumed to be
t
sensibly uniform throughout the thickness.
Where — is small and has a value of from about 8 to 3, the
t
cylinder is called thick, and must be treated in a different manner.
For a cylinder of this kind, subjected to an internal pressure,
the greatest tensile
stress is on the inside
surface. The pressure
upon the thin cylinder
lying next the inside
is balanced partly by
the resistance to stress
of its own material and
partly by the support
it receives from the
material which lies im-
mediately outside it.
FIG 82 So that the internal
pressure upon the
second layer is less than that upon the first, and so on, each
successive layer from inside to outside being subjected to a smaller
tensile stress. The precise nature of the stress variation in the
walls of a thick cylinder is given in the following. The section
of such a cylinder is shown in Fig. 82.
In this proof it is assumed that a plane section -of the cylinder
before pressure is applied remains plane when under stress; that
is, there is no distortion of the plane section. The material is
supposed to be elastic, and it must also be supposed that the
stresses in the material only vary with the distance from the
centre — i.e., at the same radius there is the same stress. Take a
STRENGTH OF CYLINDERS 117
slice of the cylinder of thickness EF. Consider the elemental
ring of thickness Sr. Since the stress is practically uniform
throughout this ring, it is only necessary to consider the small
portion of it ABCD. This portion is shown more in detail on
Fig. 83.
Three stresses act on this piece, namely,
p, radially,
/, circumferentially,
and /p longitudinally.
These are taken as having plus signs when acting away from the
centre of the elemental cube ABCD. The strains, radially,
circumferentially, and longitudinally, caused by the three stresses,
are:
Due to stress p. Due to stress/. Due to stress fr
«=^£ a = ^£ a- -A.
w/E wE E
~P_ f -/i
2 /xE E a2~ ^E
-»= -i- - if *- 4
lli //JCi //Jjj
- being Poisson's Eatio. Hence total strains are :
A JL (P+/)
1 ' E yuE
A _/- (P+/i)
E y^E
A P C/+/l)
^Vq i~^ - ^^,
E pR
Now, as the plane normal section is always plane, the longi-
tudinal strain Al will be a constant, but -— is a constant, and
often /j = 0. " Hence, p + / — constant. This may be put :
P+/-2X (1)
For radius r, by theory of thin cylinders, 2pr = force tending to
fracture ring outwardly.
118
STRENGTH OF MATERIALS
Also, 2(p + Sp)(r+8r) = force tending to crush ring inwards
and 2fSr = circumferential force tending to push ring outwards.
These forces balance, hence
2/Sr, or (p + 8p) (r + 8r) - pr =/Sr
Hence, in limit, when Sp and Sr are indefinitely diminished,
d(pr)=fdr
Hence
+ rdp = f
This gives
or,
(2)
- 2Xr
Hence, integrating,^2 = Xr2 +Y (Y= constant). Therefore,
and
/-«-•£
STRENGTH OF CYLINDERS 119
These are the fundamental equations, and may be applied to
the following cases : —
Case of Hydraulic Cylinder.
Let P = internal pressure and Prt = external pressure, from (3),
. Y
therefore,
/T> "p N "p 2 T? 2
11 — Jr ) -*-^i ^o — V.1 Jlt'l — "*" JLI'')
Y = p /_ R 2 and X = - R 2_p 2
and
Neglecting external pressure, Pa = 0, the maximum stress
(at KT) is given by
F = _j_^|_rqj (5)
of external pressure being great compared with internal
pressure.
In this case P = 0. Equation (4) then gives :
PT?2i"PR2 OT>P2
Jtu/ + Jr Ko ^Jr.lt0
This is evidently a compressive stress.
Example 1. — A steel boiler 8 ft. diameter, with working pressure
150 Ibs. per sq. in. and safe stress = 6 tons per sq. in. To find t :
R = 4 ft. = 48 ins. t/'= 6 tons per sq. in. p = 150 Ibs. per sq. in.
48 x 150
••' ' - 61T2240 ' °'53 ln'
120 STRENGTH OF MATERIALS
Example 2. — Hydraulic cylinder, 6 ins. inside radius, internal
pressure,^? = 1100 Ibs. per sq. in. Safe F=l ton per sq.. in. Find
thickness : —
i ^
V_ R.,2 - V R,2
F : 8,1+11,1 =' V
h R
1100 + - 2240 -224R°*62
Rr K2
(1100R22+ 1100x36) = (2240R22 - 2240 x 36)
1140R22 = 3340x36 = 120100
R22 = 105-3
R2 = 10-2
Thickness of material = 10'2 — 6 = 4'2 ins.
Variation of circumferential and radial stresses in thick
cylinders subjected to internal pressure P alone.
This is the most usual case in practice. Here Pa = 0 and
/= circumferential stress at any radius r
2
and p = radial stress at any distance r from the centre,
Taking various values of r, the corresponding values of f and
p are plotted on Fig. 84. The working is given in Example 5
below. It will be seen that the circumferential stress /, which is
the most important to be considered in this connection, varies
from a maximum value on the inside surface to a minimum on the
outside.
This means that the material on the inside is doing more than
its share in withstanding tension; and, in order to equalise the
tension stress under the maximum pressure in such cylinders as
guns, it is usual to impose an initial compressive stress in the
material. This stress is greatest on the inside surface, and
STRENGTH OF CYLINDERS
121
diminishes towards the outside, the result being that when the
internal pressure is applied, the algebraic sum of the initial and
FIG. 84.
induced stresses is more nearly uniform than would otherwise be
the case.
In guns this initial
compressive stress is at-
tained by winding succes-
sive layers of wire round
the inner tube, the wire
being kept under a uniform
tension, or one which has
a predetermined variation, f
This effect is shown in
Fig. 85; here the initial
compression stresses are
plotted upwards and ten-
sile stresses downwards.
The stresses indicated by FlG 85
the curves are there due
to the winding alone, before there is any internal pressure.
122
STRENGTH OF MATERIALS
On Fig. 86 is shown how the initial stress due to cooling of
the metal in cast cylinders varies from point to point. The
metal on the inside and on the outside cools first, and as the
intermediate part cools it shrinks, putting itself in circumferential
tension, while the inner and outer parts are in compression.
It has been shown that
P_ R22 - RT2
F = R^ + R!*
and from this it follows that if F be the ultimate tensile stress of
a brittle material like cast iron, or the yield stress of a more
ductile material, then when P is
made = F, the E2 must be infinitely
great. In other words, it is im-
possible in a thick cylinder with
no initial stress to use a pressure
which is greater than F, however
thick the walls may be.
If this were attempted, failure
would take place by cracks be-
ginning on the inside and extend-
ing outwards.
Example 3. — A thick cylinder
is built up so that the initial
tensile stress of the outer and the compressive stress of the inner
skin are both 3 tons per sq. in. Calculate the resultant stress of
both skins when under internal fluid pressure of 4J tons per
sq. in. Diameter, 20 ins. external and 10 ins. internal.
Considering cylinder without initial stresses,
f^ = stress on inner skin =
95 15
= — — = — = 7'5 tons per sq. in.
PR ^
fQ = stress on outer skin = *
= 7'5 x J = 1J tons per sq. in.
Hence resultant stress on outer skin
= If + 3 = 4J tons per sq. in.
Eesultant stress on inner skin
= 7 J _ 3 = 41 tons per sq. in.
FIG. 86.
STRENGTH OF CYLINDERS 123
Example 4. — Find maximum and minimum stresses in the
walls of a thick cylinder ; internal diameter 8 ins. and external
diameter 14 ins. Internal fluid pressure, 2000 Ibs. per sq. in.
T? 2 I "D 2
Maximum stress is at inner radius, and = PW^ — ^r?
= 2000 x— ' | = 2000 || = 3940 Ibs. per sq. in.
Minimum stress is at outer radius, and = ^ * f =^_ — _L J
130,000 16
= — ^3—49 = 1286 Ibs. per sq. m.
Example 5. — In a cylinder where Ex = l, E2 = 4, find the
values of / at radii increasing by half -inches. P = 1000 Ibs., and
plot the curve (Fig. 84).
At inside radius
P(R22 + RT2) 1000(16 + 1) _ 1000x17
j • > o T) o i /* I T~eE — OO 1 OS*
At any other radius, r,
f 1000x1 16
At radius I'D ins. j = — -^ x 1
15 2-25
= 66-6 x 8-1 = 539-5 Ibs.
2 „ / = 66-6 x 5 - 339 „
2-5 „ / = 66-6 x 3-56 = 237/ „
3 „ /= 66-6x2-7 = 180 / „
3-5 „ / = 66-6 x 2-3 = 153 J „
4 „ / = 66-6 x 2 =133 „
2PR 2
or at outside radius/ = ^ — ^
"
2 x 1000 x 1 2000
15 = 15" = 133 lbs<
124 STRENGTH OF MATERIALS
2
PR 2 R 2 T^
Also, using formula,^-^1 pg2-^-^ the values of the radial
r2
stresses are found as follow : —
At radius 1 in. p = 1000 Ibs. per sq. in.
„ 1-5 ins. p = 408 „ „
2 „ p - 200 „
2-5 „ p = 104 „
3 „ p = 52 „
„ 3'5 „ p = 21 „ „
0
„
Strength of Thin and Thick Spherical Shells.— Using the same
notations as before, for thin shells
p 2t_ 4*
7 r rf
And for thick shells
jP 2R28-2R18
F "
CHAPTEE XII
RIVETED JOINTS
THE full discussion of the subject of riveted joints in all its
detail really comes under the head of " Machine Design," but it
may not be out of place here to notice some of the chief problems
so far as they relate to the strength properties of the materials
used.
Of the many forms of joint employed for connecting together
the edges of iron and steel plates, those illustrated on the following
sketches may be taken as among the more usual.
I
I
I
)
1
. * —
JO O
0
O
O
o
FIG. 87.
FIG. 88.
The single riveted lap-joint shown on Fig. 87 is the simplest.
The edges of the two plates overlap, and there is one row of
rivets which pass through the two plates.
Fig. 88 represents a single riveted butt-joint, where the main
plates come edge to edge, and each is connected to an auxiliary
plate, called a cover plate, by a single row of rivets.
125
126
STRENGTH OF MATERIALS
The double riveted lap-joint on Fig. 89 has two rows of rivets
arranged diagonally or zigzag.
On Fig. 90 is shown a similar joint to that on Fig. 88, but
with double riveting and two cover plates.
FIG. 89.
The usual theory upon which the proportions of riveted joints
are made to depend is somewhat as follows : —
It is assumed that a joint may fail in one of five different
•>VsK\\S>
O
1
O
0 !
O
o
1
1
O
O
O
o
1
1
1
O
— _^^
n i
— — '
g
FIG. 90.
ways. Considering the simplest case of a strip of the plate
joined by a single rivet, these are (Fig. 91) : —
(a) By the shearing of the rivet.
(&) By the tearing of the plate through the rivet hole.
(c) By the rivet breaking through the plate in front of it.
RIVETED JOINTS
127
(d) By the rivet crushing the material of the plate in front
of it.
(e) By the rivet shearing out that part of the plate which
is in front of it.
1- —
FIG. 91.
Now, calling
p = the pitch; that is, the distance from centre to centre
measured along the line of the rivets ; this will be
the same as the width of the strip ;
t = the thickness of the plate ;
d = the diameter of the rivet ;
I = the lap of the plate; that is, the distance from the
edge of the rivet hole to the edge. of the plate;
ft = the tensile resistance of the plate ;
fg = the shearing resistance of the rivet ;
fc = the crushing strength of the plate ;
then the resistance of the joint, in Fig. 91, to failure in each
of the five ways enumerated above, will be :
7T
(a) Eesistance to shearing of rivet = fs — d2
(&) Eesistance to tearing of plate =ftt(p — d)
tl2
(c) Eesistance to breaking through plate = C-r
128 STRENGTH OF MATERIALS
This is a case of a short beam, fixed at the ends, and of
span d, depth I, and breadth t, C being a constant depending on
the material.
The value of C is given by some authorities as about 45 tons.
(d) Kesistance of plate to crushing =fctd
(e) Eesistance of plate to shearing = 3/X~o~
\ Zi
An economically designed joint should be as ready to fail in
one way as another ; that is —
Generally speaking, (d) and (e) are not considered, as it is
found that if the joint designed according to the usual rules is
strong enough to resist (a), (b), and (c), it will be sufficiently strong
for the other two.
Assuming that t is given, the diameter of the rivets must first
be found. To do this, it is found that the following empirical
rule gives the most satisfactory results for plates of the more
usual sizes: —
d = 1'25\/ t for boiley work
or d = I'l \/ t for bridge work
Now, taking the former of these values, and equating (&)
and (a),
But d12 = (1-25)2*
and fg = 0'8ft, very nearly
p-d -- 0-8x^-x(l-25)2
or p == d + 0-98 in.
This is from purely theoretical considerations, but it is found
from experiment that the pitch must be somewhat greater,
especially in the case of punched holes, where the metal of the
plate is damaged by the punching for a small distance round
the hole.
The following values, based on the experiments on riveted
RIVETED JOINTS 129
joints carried out by the Institution of Mechanical Engineers,
may be used : —
For iron plates and ri vets (Punched ' ' /» = «*+! '50 ins.
I Drilled . . p = d+ 1-40 ins.
For steel plates and rivets{Punched ' ' P = d+l'l3 ins.
iDrilled . . p = d+lin.
To find the overlap, /, equate (c) to (&).
= 0-8 *J~d
Practical conditions, however, would seem to make it neces-
sary to increase this to about l'l\/d. This means that, instead
of (c) being equal in strength to (b), it is so far stronger that there
is not likely to be any possibility of failure taking place in this
direction.
It is important to remember that the pitch is partly dependent
on the resistance of the rivet material to shear. In most cases the
rivets are made from a material which is softer and more ductile
than that used for the plates.
For a given ductile material the shearing strength may be
taken as approximately equal to four-fifths of the tensile strength.
Efficiency of Riveted Joints. — The tearing resistance of the
plate on a section taken through the centres of the holes isftt (p — d),
and the tensile resistance of the same length of the original plate
isfttp. The former of these is obviously less than the latter, on
account of the metal being cut away in making the rivet holes.
For any given joint the ratio of one of these to the other is called
the efficiency of the joint. In other words, the efficiency is the
fraction of the strength of the plate which the joint provides.
Theoretically, the efficiency is -—*-. -- or *• - , but it is actually
found that the efficiency is somewhat less for the reasons already
mentioned as to the tensile strength per square inch of the drilled
or punched plate being less than that of the uncut plate. If some
I
130
STRENGTH OF MATERIALS
of these efficiencies are calculated it will be found that they are
greater for the smaller sizes of plates, and that, for a given size of
plate, the efficiency is greater where there are two rows of rivets
than where there is only one, and still greater for treble riveted
joints.
The following figures will serve to give an approximate idea of
the values to be expected : —
EFFICIENCIES OF DIFFERENT RIVETED JOINTS.
Punched Holes.
Drilled Holes.
Size of plates ....
Jin.
fin.
lin.
4 in.
fin.
lin.
Single Riveting —
Iron plates — Iron rivets
Steel plates — Steel rivets
56
51
51
47
48
44
59
54
54
49
50
46
Double Riveting —
Iron plates — Iron rivets
Steel plates— Steel rivets .
72
68
68
64
65
61
74
70
70
66
66
63
Treble Riveting —
Steel plates — Steel rivets
78
74
71
The above efficiencies are all given as percentages.
CHAPTEK XIII
STRENGTH OF MATERIALS AS FOUND FROM THE
RESULTS OF TESTS
IN designing any kind of engineering structure the form and
dimensions are made such that the loads which are likely to
come upon it will not give rise to more than a certain pre-
determined stress in any part. In order to decide upon the
amount of stress which may be allowed upon any given material
when loaded in some particular manner, the strength of the
material must be determined by experiment. An ideal kind of
experiment which might be made upon a structure would be to
test the structure as a whole with loads which are as nearly the
actual loads as possible, and afterwards to carry them beyond
their working values and see what happened. But it is not often
that full-sized pieces can be tested, partly because the making of
an actual structural piece which is to be immediately destroyed
is costly, and, what is more important, there are few testing
machines which will accommodate full-sized pieces, except very
small ones. Testing of this kind is, in spite of the difficulties in
the way, sometimes carried out, and always yields valuable results.
Testing. — By testing is generally meant the loading in a testing
machine of samples of the material, either specially prepared or
cut from the material which is being used. When in the test-
ing machine observations are taken for the purpose of finding
out what happens to the material when under the load.
These tests may be carried out, first, for the purpose of
ascertaining new facts about a given material, as to its behaviour
under certain given conditions, which is really a form of research ;
and, in the second place, with the idea of ascertaining whether
material supplied by a maker possesses all the qualities which
have been specified by the user. These last are commercial tests.
131
132 STRENGTH OF MATERIALS
Specimens of the metals may be tested in tension, compression,
shear, bending, and torsion ; but, except in the case of cast iron, it is
generally sufficient to make the first of these tests for the purpose
of ascertaining the quality of the material. The reason is probably
that the tensile test is the simplest and most easy to carry out,
and because it leads to some five or six different quantities, each of
which is capable of providing information regarding some different
quality in the material.
In a tension test a sample of the material, generally in the
form of a bar, is taken hold of at each end and pulled with an
increasing amount of force. The function of the testing machine
is to apply this pull at the will of the experimenter and at the
same time to indicate its magnitude.
The observations usually made during the application of load
to a test bar are, in addition to the magnitude of the load itself,
the following : —
1. A series of measurements of the deformations accompany-
ing the loads may be taken by using an exten-
someter, so as to make it possible for the modulus of
elasticity to be calculated. This is rarely done in com-
mercial work, though there may be occasions when an
exact determination is required to satisfy a specification.
2. Such observations are taken as are necessary to fix the
yield point load. Something has already been said
about the elastic limit, proportional limit, and the yield
point, and this will be again discussed more fully. For
the present it is sufficient to say that the load on the
specimen is increased until the yield point has been
reached, and is then noted. The yield point may
be found by using dividers, and seeing when a certain
marked length of the bar begins to rapidly increase with-
out any increase in load ; or by noting the point when
the weigh beam of the testing machine is seen to drop
quickly on to its lower stop owing to the rapid stretch of
the material.
3. After the yield point is passed the load is still further
increased until fracture takes place, and the load which
is required to bring about rupture is noted. This
is called the maximum load, or, in some cases, the
AS FOUND FROM THE RESULTS OF TESTS 133
ultimate load. The yield point load and the maximum
load are divided by the original cross-sectional area of
the bar to obtain the yield point stress and the
maximum stress.
This last result is generally the one to which most import-
ance is attached.
4. The permanent elongation of the material, as measured
upon a length marked on the bar before the test, is also
measured and noted. This may be upon a length of 10
ins., 8 ins., or any other length down to 2 ins., according to
the size of the bar. The result is given as a percentage,
5. The amount the cross-section of the bar has been reduced is
measured at the point of fracture, and is called the
reduction or contraction of area, and is expressed as a
percentage of the original area.
Nos. 4 and 5 yield information as to the ductility of the
material.
The results usually required in specifications and by inspectors
of material are :—
The yield point stress ;
The maximum stress ; and
The ductility, as indicated by the permanent elongation
after fracture.
The two last are always insisted upon in tests of ductile
material such as wrought iron, mild steel, and bronze. The yield
point stress and the reduction in area are less frequently required.
The principal forms of test bars used for ductile materials are
indicated by the sketches on Fig. 92.
Here, A is a round bar whose middle portion has been turned
to a smaller diameter.
B is a plain round or square bar.
C is a short turned specimen with screwed ends, such as
might be cut out of a large forging.
D is a plate or flat bar specimen with the middle part
milled out to a smaller width.
E is a similar bar untouched.
134 STRENGTH OF MATERIALS
F is the kind of specimen used for cast iron, the centre
part being turned and the ends formed to fit spherical
dies.
B
FIG. 92.
Testing Machines. — The essential points of any testing
machine for carrying out the tests which have been and which
will be mentioned, are : —
(a) Some kind of device for holding the ends of the specimen
in such a manner that they cannot be pulled away by the
load which is applied. This may be in the form of steel
wedges with serrated surfaces, or the ends of the bar
may be screwed so as to fit into corresponding sockets
in the jaws of the machine.
(6) There must be some way of applying any desired load to
the bar.
(c) The last, and what is the most important part of the
machine, is the arrangement for measuring the magnitude
of the applied load.
AS FOUND FROM THE RESULTS OF TESTS
135
(d) Ib is also necessary that the grips at one end of the
specimen should be capable of being moved at the will
of the experimenter, so as to enable him to take up any
stretch which may occur in the bar during the test.
A skeleton drawing of a simple form of testing machine is
given on Fig. 93. This represents the vertical type made by
Messrs Buckton of Leeds.
Here the specimen is at A and is held in a vertical position, its
lower end being attached to a screw by means of which it is pulled
downwards as desired and a tension put upon the bar. The upper
end of the specimen is connected to the short arm oi a lever or
FIG. 93.
steelyard at the point D ; this lever turns about a fulcrum C, its
other end being pulled downwards by a moving weight E. The
amount of the tension on the specimen depends upon the magni-
tude of the weight E, and, at the same time, upon the ratio of EC
to DC. Most machines in use in this country are made to depend
upon this plan of having a screw or hydraulic device by which the
load is applied and a weighing lever for the purpose of measuring
this load. Where it is desired to have the specimen in a hori-
zontal instead of a vertical position, a knee-lever is interposed
between the weighing lever and the specimen.
For compression the links suspended from D are fixed to a
136 STRENGTH OF MATERIALS
plate, and the links from the screw are attached to another plate,
this being pulled downwards by the screw towards the first plate,
which is placed below it.
In another type of testing machine which is coming into
use in this country, the load is applied by the pressure
of some fluid such as oil acting on a plunger or ram, and the
magnitude of the load is measured by the pressure of the fluid.
So long as there is no appreciable friction in the ram, this fluid
pressure will be proportional to a measure of the load. For a
more detailed description of the several forms of testing machines
the reader is referred to one of the books which deal exclusively
with the subject.*
Appliances for Measuring Elastic Deformations. — Where
measurements for the purpose of calculating Young's Modulus
have to be made, some form of extensometer must be applied to the
specimen under observation. The mirror instrument devised by
Professor Martens of Charlottenberg has already been described in
Chap. VII.
Besides the sketch on Fig. 61, which explains the principle of
action of the Martens mirror apparatus, views are given on Figs.
94 and 95.f These are taken from photographs ^of the apparatus,
when actually applied to a vertical bar. The bar shown here is
one used by the writer for the purpose of explaining the action of
the apparatus to students, and for enabling them to obtain some
little practice with it before using it on a specimen under stress.
It is made telescopic, so that its total length can be varied at will
by simply turning the milled head of a differential screw. The
appliance can be put upon any table or convenient stand, and,
besides being applicable for demonstration purposes, it is found
extremely useful when an extensometer is to be calibrated or
tested, or when two extensometers are to be compared. The
mirrors are shown more in detail on Fig. 96.
There are two measuring bars, one on the surface of the bar
next the telescope, and the other on the surface farther from it.
By adopting this arrangement the observer can see how the stress
varies at the different parts of the section. This is not possible
* See the Author's book on Testing of Materials, published by the
Scientific Publishing Co., Manchester.
t The photographs on Figs. 94 arid 95 were kindly taken in the photo-
graphic department of the Manchester School of Technology.
[Between pp. 13G and 137.
AS FOUND FROM THE RESULTS OF TESTS 137
with most extensometers, where only the average strain is
measured and the observer cannot tell how nearly his load is
being applied along the axis of the specimen. In the Martens
mirror apparatus a number of measuring bars are usually sup-
plied of different lengths, some for flat and some for rounded
specimens.
This instrument is very useful in that it can be just as readily
applied to actual parts of machines or structures under stress as
to test specimens. On Tig. 97 it is shown when being used for
measuring the compressions of a brickwork pier.
Its fixing and adjustment is sometimes a little tedious at first,
but when once set it is found to give extremely good results.
Among the other extensometers in use in this country is that
of Ewing, in which a micrometer-microscope is combined with a
lever; and the simple lever instruments of the Kennedy type,
where the change in length of the specimen moves the short arm
of a lever with a high velocity ratio, the outer end of the long arm
carrying a pointer which moves over a graduated scale.
Where material is tested under a compressive stress, it is
generally necessary to bed the specimen in some substance such
as Portland cement, plaster of Paris, lead, or millboard, in order
to distribute the stress uniformly over the surface. This is found
to be necessary in the case of such substances as stone, brick,
cement, and concrete ; where the specimen is of metal, it is usually
possible to machine the surface so as to get a flat surface to rest
against the compression plates of the machine.
Besides tensile and compressive tests, bending or cross-breaking
tests are frequently resorted to. These are used in the case of
timber beams, iron and steel girders, rails, and so forth.
The torsional strength of small shafts can be found experi-
mentally by applying a torsional moment in a special machine ;
and direct shearing tests are also carried out for some purposes.
There is little difficulty in determining and interpreting the
maximum stress of materials tested to destruction and the elonga-
tion and reduction after fracture, but it is found that, as regards
the limits of elasticity, there is sometimes considerable ambiguity ;
this point will be discussed in the following chapter.
CHAPTEE XIV
THE LIMITS OF ELASTICITY
IT has already been pointed out that when a bar is loaded it is
altered in length, and that if when the load is taken off the bar
returns to its original dimensions, it is said to be perfectly elastic.
It has further been shown that as the loading is increased, there
comes a point beyond which the change in length is partly
temporary and partly permanent, and that the stress at which
this begins to take place is the elastic limit of the material. This
is the simplest, and in many ways the most scientific description
of the elastic limit ; but, as will be seen, there are great difficulties
in the way of determining this point, and its very existence
depends upon the precision of the instruments used in its deter-
mination.
Limit of Proportionality. — It is found from experiment that
the proportionality between load and deformation or stress and
strain exists until a stress is reached beyond which the strains
begin to increase for equal increments of stress. Take the case of
a bar under tension in a testing machine : the loads are applied so
much at a time, the increase being the same at every step. Means
are taken for measuring the small extensions accompanying the
tension stresses, and it is found that for every equal increase of
load a correspondingly equal increase of length takes place. This
goes on until a certain load is reached, beyond which the incre-
ments of stretch are no longer equal, but continually increase.
This point has been given the name of limit of proportionality, or
P-limit. This term clearly defines its own meaning, and could
not be improved upon, but, possibly with some reason, it has often
been called the elastic limit.
So far, it is seen that there are two ways in which the elastic limit
stress may be denned — viz., as the point beyond which permanent
138
THE LIMITS OF ELASTICITY 139
set is given to the bar, and otherwise as the stress beyond which
the strain is no longer proportional to the stress. If these two
points were found to coincide, there would be no difficulty about
the elastic limit ; but, unfortunately, it is usually found that the
first permanent set takes place at a stress below that at which
proportionality appears to cease. It is therefore necessary to
clearly distinguish between these two ; they are so often confused,
and hastily assumed to be one and the same point. If the original
term, elastic limit, is retained for the first — and that would seem to
be the most reasonable course to adopt — then the second can still
be called P-limit.
It has been said that, using the most precise and accurate
measuring instruments at present constructed, the elastic limit is
found to lie below the P-limit ; but the opinion is strongly held
by some authorities — notably M. Fremont — that the elastic limit
and the P-limit are coincident, and that it is only want of
accuracy and precision in our measuring instruments which makes
it appear that they occupy distinctly different positions.
In his investigations into the position of the elastic limit
of metals, M. Fremont has adopted the method of using a micro-
scope to examine the polished surface of the metal under stress
and has taken as the elastic limit a point at which a slight dulling
of the surface is shown by the microscope ; and he maintains that
this is the only true elastic limit. In the course of his examina-
tion he found that this limit only appeared at first in one or more
isolated spots, until the change gradually covered the whole of
the surface ; and in order to overcome the difficulty of noting the
first appearance he used bars of gradually varying section, in
which there was one small part where the stress was always
greatest, and at which the appearance would change as soon as
the stress at the smallest area reached the elastic limit stress.
As a result of these investigations of M. Fremont, a third and
new definition of the elastic limit is created, which only tends to
add to the already existing confusion as to what really fixes
this point.
Besides these, there have at different times been other attempts
to solve the difficulty. Of these, Styffe wanted to fix the limit
as a point which was made to depend on the increase of per-
manent strain as depending on the rate of increase of stress.
By this method a point is obtained for the limit which lies above
140 STRENGTH OF MATERIALS
the P-limit. In a similar manner Wertheim attempted to fix
the limit as a stress where the permanent strain was a definite
fraction of the original length of the specimen, namely o0o00.
There are many objections to this plan, chiefly on account of the
delicacy of the operation and the great care required in carrying
it out, and also because only part of this apparently permanent
strain is really permanent, the remaining portion disappearing
after the bar has rested for a time.
Yield Point. — The five points which have been mentioned as
elastic limits are all more or less entitled to some claim on the
term, but this does not apply to the point so often mentioned
in commercial tests as the elastic limit, but which is really the
yield point. This is the stress, readily found, where the strain
rapidly increases in amount as the load is slowly increased or
is stationary, and is clearly discernible by means of compass
measurements and by the appearance of the surface of the bar.
The various ways of fixing this point include the dropping of
the testing-machine beam on to its bottom stop, advocated by
Kennedy; the change in the appearance of the surface of the
bar recommended by Styffe ; the perceptible thickening of a line
inscribed on the bar by a pair of compasses, which method is
used by many experimenters in this country; and the employ-
ment of an autographic diagram, also much used and recom-
mended by Unwin. A way of finding the yield point which the
author has often employed is to use an extensometer and note
the load at which the pointer or cross -wire begins to creep
quickly along the scale without any increase in load. This,
though probably the most reliable method, is somewhat elaborate,
and requires more time than can be spared in ordinary commercial
testing operations.
Illustrations of the Three Limits. — It will be well here to
inspect one or two diagrams which serve to illustrate what has
been said regarding the several points which have been men-
tioned as elastic limits.
The curves on Fig. 98 have been plotted from the results of a
tension test on a turned bar of wrought iron, the measurements
of extension being taken with a Martens mirror extensometer
capable of reading directly to To-o-oo-th of an inch, and by
estimation to 50^00th of an inch. The extensions were
measured on a 10 cm. length of the bar.
THE LIMITS OF ELASTICITY
141
The wrought iron bar in question was put into the testing
machine, and loads applied by increments of 2 tons. Before
each new load was applied the load was removed, and the
readings on the extensometer scales, if any, taken. In this way
it was possible to determine the point at which the reading line
did not return to zero after loading — the reading at zero load
A
BACK
FRONT.
O O-OOOI O-OOO2 O-OOO3 O-OOO4
PERMANENT SET-INCHES
FIG. 1. — BAR OF WROUGHT IRON IN TENSION.
IJIAM., 1'62 IN. MEASURED LENGTH, 4 INS.
FIG. 98.
giving the permanent set produced by the last loading. By doing
this, it was possible to determine, with a fair degree of accuracy,
at what load the bar first began to take permanent set.
The curves on Fig. 98 represent the permanent set for each
increment of 2 tons up to 22 tons load. Here loads are plotted
vertically and extensions horizontally. It will be se*en that until
about 12 tons have been reached there is no permanent set, the
reading line of the extensometer coming back to zero after each
142
STRENGTH OF MATERIALS
loading. At this load a permanent set is first noticed, and
increases for each successive load. The two curves represent the
readings on the two sides of the bar, one being marked on the
diagram as " back," the other as " front." It will be seen that the
permanent set, besides commencing at a definite point, increases
by regular amounts, and also that this increase is more rapid
on the front of the bar than on the back. From this it is
gathered that the bar is not held evenly in the grips, so that the
line of pull lies between the axis of the bar and its front side,
causing the stress on this side to be greater than at the back.
25
20
O'OOOl O OOO2 O-OOO.3 INCHES.
15
o
"PT
INCHES
O-OOI
0-002
0-003
Fig. 99.
The point where the lines first leave the vertical zero line is
marked on the diagram E.L., being the load at which the elastic
limit, as above defined, has been reached.
Now look at Fig. 99. This refers to the same test as the
last. Two curves have been plotted, the upper marked "P.S.
mean," and is an average of the two plotted on Fig. 98, again
showing the elastic limit. On the other curve have been plotted
the total average extensions. It will be seen that up to a certain
load the points of the curve fall on a perfectly straight line, thus
indicating true proportionality between load and extension, or
THE LIMITS OF ELASTICITY 143
between stress and strain. This point where the curve leaves
the straight and begins to turn slightly towards the right, is
marked P.L., indicating the P-limit, or limit of proportionality.
This point occurs at a load higher than the elastic limit just
found.
This illustrates what has previously been said about the
relative positions of the elastic limit and the P-limit, namely, that
the latter is found to be higher than the former ; but it is quite
conceivable that, as M. Fremont considers likely, these two points
really coincide, and would be found to coincide if the measurements
of strains could be made with greater precision than is now the
case. The way in which the proportionality limit is usually found,
and this is probably the most reliable way, is to measure the
strains corresponding to known stresses, and plot one against the
other to a fairly large scale on true squared paper. A straight-
edge is then laid along the series of points so obtained. It will
be found that the straight-edge can be made to coincide with the
earlier part of the series of points, and it is clearly seen where the
line begins to leave the straight. It would seem that this point
of departure ought to occur precisely at the same load as that at
which it was first found that there was any permanent set, and
that the more nearly they would so coincide, the more precise were
the strain measurements and the larger and more accurate the
plotting. As a matter of fact, it is found to be difficult, with
specimens of the ordinary sizes, even when estimating to fifty-
thousandths of an inch, to make the increments of strain quite
uniform. It is also difficult to obtain perfect uniformity in the
increments of the load, and this leads to small unevennesses
in the strains, apart from the intrinsic want of precision in their
own readings. Taking all these considerations into account, the
writer believes that the elastic limit ought to coincide with the
P-limit. Whether greater precision in measuring would fix the
elastic limit more accurately is open to doubt when the form of
the set curve is examined : the curve appears to run into the line
of zero strains at a definite angle, and not to approach it as to a
tangent.
Next, referring to Fig. 100, it will be found that a stress strain
diagram has been plotted which shows the P-limit and the yield
point for the same bar. In this case the extensions were taken
with a Ewing extensometer on 1J ins. of a tension specimen of
144
STRENGTH OF MATERIALS
high-carbon steel f of an inch in diameter. The diagram forms a
straight line up to the proportionality limit, which is shown at
4 tons, corresponding to 13'06 tons per sq. in. Beyond this point
the line diverges from the straight, curves more and more rapidly
towards the right, and eventually becomes horizontal, in this way
showing a rapid increase of length without any increase in load.
This is the yield point, and is marked at 6 tons, or 19 '74 tons per
sq. in. on the diagram. It may be added that the maximum
strength of this specimen was found to be about 4*7 tons per sq. in.
7*
.._/L..
S.
RL
NCHE.S. 0-001
0-002
FIG. 100.
To anyone who is observing the extensometer readings, this yield
point is determined with great certainty by noting when the
reading line begins to creep rapidly along the scale without any
increase in the load. It is not necessary to take a series of
observations, but simply to watch the scale as the load is being
gradually increased, and it is at once seen when the creeping
begins. As already pointed out, for commercial purposes the use
of an extensometer is generally found to occupy too much time,
and the yield point is more usually found by using a pair of
dividers or watching the drop of the beam, though the latter is
THE LIMITS OF ELASTICITY
145
apt to be misleading in some cases. Another plan which has
been mentioned is to let the bar tell its own tale by drawing
an autographic diagram. Three such diagrams are shown on
Fig. 101, the yield point being indicated in each case. In the
curve taken from a mild steel bar (a), the position of the point
is unmistakable, and is shown by the distinct horizontal jump in
the curve where yielding takes place. It is to be noticed that
the early part of this step is similar to the curved part of the
diagram on Fig. 100, but drawn to a much smaller horizontal scale ;
the portion referred to is lettered (I) (m) in both diagrams.
E X T E N 5 I ONS
FIG. 101.
The second of these curves, (b), has a much less distinctly
marked yield point, and the third one, (c), for a bar of wrought
iron, is even worse. In cases like the last, the autographic diagram
is apt to fail as a means of locating the yield point, unless great
care is used in manipulating the apparatus and it is in the hands
of a very experienced observer. In any case, the actual quanti-
tative value of the yield point load should not be scaled from the
diagram, but read off on the scale of the machine as soon as it is
seen that the pencil is leaving the elastic line.
The writer has tried having three observers watching for the
146
STRENGTH OF MATERIALS
yield point on the same test. One used an extensometer, the second
tried the measured length of the bar with a pair of dividers, while
the third watched for the drop of the beam. It was found that
the extensometer man noted the point first, the dropping of the
beam came second, and the dividers were last.
Unsymmetrical Loading. — It has already been pointed out
that in very many cases the bar under test is loaded unsymmetri-
cally, with the result that there is a greater stress on one side
than on the other. The curves of extension for the two sides of
30
20
10
INCHES o 001 o«oo2 o-oo3 o-oo4
FIG. 102.
such an unequally loaded bar are shown in Fig. 102; this refers
to a cast steel round bar, rather more than 2 sq. ins. in area. It will
be seen that the two limits are not very far apart, that on the right-
hand diagram occurring slightly earlier than in the case of the
other. As the extensions, and consequently the stresses, of the
right-hand side are greater than those for the left, as shown by
the greater slope of the curve, the proportionality limit will occur
first here. The effect of this slight yielding appears to throw the
THE LIMITS OF ELASTICITY 147
load nearer the other side and bring the stress to a more nearly
uniform condition, so that the proportionality limit is reached on
the left side almost as soon as on the right. This approach to
uniformity is shown on the diagram by the fact that the two
dotted lines are nearly parallel. This would appear to show that
where the loading is slightly eccentric, the general P-limit will
only be very little in excess of the same point for the more greatly
stressed portion of the bar.
The above are only isolated examples taken from a large
number of similar experiments.
Changes of Limit by Previous Loading. — In what has been
said so far, it has been assumed that the limit, whether elastic
limit, P-limit, or yield point, is a fixed point for a given material ;
but this only so long as the material is always in precisely the
same condition.
It is well known that if the P-limit has been found for a bar
under a tension load, and the load is raised to a point somewhat
in excess of the limit just determined, then on reloading the bar
with the intention of finding the limit a second time, this will now
be found to be at a higher stress, not far from the maximum stress
applied in the first loading. In other words, the limit will have
been raised by this previous loading, and not only so, but it may
be raised time after time with successive loadings. Moreover,
these artificially raised limits can be made to again fall to some-
where near the original point in several ways, such as resting for
a time, by the application of heat, or by hammering. If, however,
the bar has been previously loaded beyond the limit in tension,
then the loading is reversed and the stress is carried beyond the
limit in compression ; it will be found that on now reloading in
tension, the limit, instead of having been raised by the previous
loading, will have been depressed. This appears to lead to the
conclusion that the effect of previous loading beyond the limit is
to raise the limit for loading in the same direction, and to depress
it by loading with the same stress in the opposite direction, all
these being artificially produced limits. Further, it is found that
while an artificially raised limit is lowered by a reversed loading,
an artificially depressed limit is raised by a reverse loading. The
result is that after a series of reversed loadings, carried in each
case just beyond the limit, the limit settles down to a fixed value
which is the same for compression as for tension ; this point has
148 STRENGTH OF MATERIALS
been styled the natural limit for the material. The limit found
on the first loading of the bar is generally higher than the natural
limit, and has probably been artificially formed in the process of
manufacture or by after-treatment. It has been found that bars
return to their primitive limit after resting for a few weeks or
months, or more quickly by the application of a temperature such
as that of boiling water, or by hammering.
Looking back at what has just been said, it will be seen that
there are really two well-defined limits found during the loading
of a bar of iron or steel in tension. These may be called the limit
of proportionality (or P-limit) and the yield point. The former of
these is less easy to determine than the second, and its exact
location partly depends on the precision of the measuring instru-
ments used in its determination, and in all probability it coincides
with the true elastic limit. It may safely be said that whenever
the elastic limit is spoken of in connection with a commercial test,
what is really meant is the yield point. Of course, if it is
generally agreed to call this point the elastic limit, well and good,
but this should be clearly stated, and the existing ambiguity
removed once for all. The objections to this use of the term will
be gathered from what has been said above; and, in addition, it
is well known that many of the high-carbon steels show no
yield point, and a limit can only be determined by using instru-
ments of precision.
Besides defining what is meant by the limit given in com-
mercial tests, it would be well if the manner of finding it could
be clearly defined, and one standard method adopted. This would
lead to uniformity, and enable useful comparisons to be made.
And lastly, where the limit given is to be in any sense used as
a criterion of the quality of the metal, not only should the nature
of the limit and its manner of determination be clearly understood,
but the history of the test specimen should be known as regards
its treatment between leaving the rolls and being tested.
The greater part of the matter in the above chapter* is taken by
permission from the author's article in the Engineering Review, April,
1904.
CHAPTEE XV
THE MATERIALS USED IN CONSTRUCTION
THE materials of engineering naturally divide themselves into
three chief classes, namely :
1. Metals, of which the most important are. the several alloys of
iron, copper, and steel.
2. Vitreous materials, including stones of various kinds, brick
and terra-cotta, lirne, cement, mortar, and concrete.
3. Materials which do not come under the above, heads, such as
timber, ropes, belting, and others of minor importance.
The above will be taken in the order given, and their chief
strength properties, so far as they affect the engineer, will be
briefly discussed.
Iron and Steel.
The metal Iron, alloyed with greater or lesser quantities of
other elementary substances and containing various impurities,
goes to form what are now among the most useful and important
materials employed by the engineer.
There are a great number of these alloys of iron, ranging from
the softest kind of wrought iron to the hardest steel, but they
may be roughly classed under the following heads : —
Wrought Iron,
Mild or Low-carbon Steels,
Medium Steels,
High-carbon Steels,
Cast Iron.
The above are placed in the order of their percentage of
carbon, wrought iron containing the lowest and cast iron the
149
150 STRENGTH OF MATERIALS
highest. In preparing any of these varieties of iron, the
material operated upon is pig-iron, the relatively crude form of
iron which is obtained from >the smelting of the ore. As the
strength properties of these materials depend to a very large
extent on the processes of manufacture, as well as upon the
ingredients which are alloyed or mixed with the iron, the methods
used in the principal operations will be very briefly indicated.
In addition to other impurities, the pig-iron of commerce
contains carbon (2J to 5 per cent.), silicon, and manganese.
Wrought iron, on the other hand, contains little more than traces
of these, so that the process in which wrought iron is prepared
from pig consists essentially in burning the alloyed metals out
of the iron, and leaving what is approximately pure iron. In
the puddling process, by which much of the wrought iron
is made, the pig-iron is brought to a molten condition and
exposed to the action of a decarburising or oxidising flame in
a reverberatory furnace. After a time the carbon, silicon, and
manganese are burnt out of the mass, which has now become
thick and sticky. This is worked up into a ball and placed
between squeezers to get rid of the iron oxide or scale, and the
billet so obtained is afterwards rolled into plates and bars. The
rolling does not take place in one process, but is repeated in a
series of operations. That is to say, in rolling bar iron, a number
of the bars rolled from the original billet are bundled together,
heated, and again rolled out into a single bar. This treatment
results in the fibrous and laminated appearance seen in the
fracture of a test specimen of wrought iron.
The result of the process of manufacture of wrought iron from
pig is to get rid of almost all the carbon, as well as the greater
part of all the other ingredients, leaving little more than traces
of carbon, silicon, manganese, sulphur, phosphorus, and copper.
The three last exist only as undesirable impurities.
The material having this composition, as produced by the
puddling or the crucible process, and afterwards treated by
hammering and rolling, is known as wrought iron, and has the
following distinguishing qualities : — It is soft, tough, ductile, and
fibrous in structure ; it is equally strong in tension and in com-
pression ; it is capable of being easily pressed and hammered into
various shapes when it has been softened by being brought to
a red heat, and less readily when cold ; two or more pieces can
THE MATERIALS USED IN CONSTRUCTION 151
be welded into one by hammering at a white heat ; it is only at
very high temperatures that wrought iron is reduced to a fluid
state, thus rendering it unsuitable for castings ; and, lastly, unlike
the hard steels, it cannot be hardened by heating and quenching.
Of the ingredients included in wrought iron, carbon is
generally less than 010 per cent., often as low as 0'03 per cent,
and occasionally as high as 019 per cent. ; of manganese there
is little more than a trace; the silicon varies from a trace to
013 per cent. ; the sulphur varies from nothing to 0'025 per
cent. ; and there is of phosphorus 0'02 to 0'20 per cent.
The effect of a deleterious quantity of phosphorus is to cause
metal to be cold short, or to show a want of ductility when
worked cold, and in a like manner sulphur and copper produce
the same effect when the iron is being worked at a red heat, or
makes it red short.
When tested under a tensile load it is found that the yield
point of wrought iron is from 12 to 17 tons per sq. in., and the
maximum stress is from 19 to 25 tons per sq. in. A good average
value for the maximum stress is 23 tons per sq. in. The elonga-
tion after fracture varies from 10 to 30 per cent, and the reduction
in area from 10 to 50 per cent. The appearance of the fracture
should be fibrous, without showing too much lamination. A test
bar broken under a steady load should not show a crystalline
fracture.
In compression there is no maximum, as squeezing of the
material can go on indefinitely. The definite point to be n^ted
in the loading is the yield point, and this occurs at about the
same stress as for the same material in tension.
The shearing strength is found to be from 75 to 85 per cent,
of the tensile. A common value for this ratio used in design is
80 per cent., or four-fifths.
The elastic modulus for wrought iron is in the neighbourhood
of 27,000,000 to 29,000,000 Ibs. per sq. in.
It is found that wrought iron attains its maximum strength at
a temperature of from 500° to 600° F. ; after this temperature has
been reached the strength falls off rapidly.
A typical stress strain diagram for wrought iron is shown on
Fig. 103.
Steel. — Steel is iron containing from 0*05 to T50 per cent, of
alloyed carbon, with the addition of small quantities of manganese,
152
STRENGTH OF MATERIALS
silicon, sulphur, and phosphorus. The composition of low-
carbon steel, used for such purposes as structural work and
boilers, is somewhat as follows: —
Carbon . . 0170 to 0'275 per cent.
Silicon V . 0-020 „ 0'090
Manganese . 0'050 „ 0700
Sulphur . - . 0-010 „ 0-040
Phosphorus . O'OIO „ 0'050
40
10
HS
MNS
10
20
30
PERCENTAGE ELONGATION.
FIG. 103. — Here CI refers to cast iron ; WI, wrought iron ; MS, mild steel;
MMS, medium steel ; HS, high-carbon steel.
The Low-Carbon Steels are prepared from pig-iron, either by
burning out the carbon and silicon and putting back sufficient
carbon to give it the requisite qualities, as in the Bessemer pro-
cess, or by melting pig-iron containing too large a proportion of
carbon with a second ingredient which contains less carbon, such
as scrap wrought iron or haematite iron ore, in such propor-
THE MATERIALS USED IN CONSTRUCTION 153
tions that the resulting material contains just the right percentage
of carbon : this is what is done in the Siemens and Siemens-
Martin open hearth processes. In all these methods the metal so
produced is run into moulds so as to obtain ingots or slabs of the
steel, and these are afterwards hammered and rolled into the
desired bars or plates or section bars.
The general properties of the low-carbon or mild steels so
produced are similar to those of wrought iron, with one or two
modifications. The strength properties are higher, both as regards
the yield point and maximum strength, and also the ductility as
given by the elongation and reduction; the appearance of the
fracture, instead of being fibrous, should have a fine silky or
velvety appearance.
The yield point when under tensile stress varies from 13 to 22
tons per sq. in., and the maximum stress from 25 to 32 tons per
sq. in., while the elongation is from 20 to 30 per cent., and the
reduction in area from 40 to 60 per cent.
Like wrought iron, the compressive yield point stress of mild
steel is the same as that in tension ; and the shear strength is
about four-fifths of the tensile.
Young's modulus generally falls between 28,000,000 and
30,000,000 Ibs. per sq. in.
The maximum tensile strength occurs at 500° or 600° F.
A typical stress diagram is given on Fig. 103.
Medium Steels. — In what are known as the medium steels
the percentage of carbon is higher, the strengths are higher, and
the ductility somewhat less. These medium steels cover a con-
siderable range as regards their strength properties. They are
made by both the open hearth and Bessemer processes, and have
the following compositions : —
Composition of Medium Steels,
Carbon - /. ,. 0*300 to 0450 per cent.
Silicon . . 0-050 „ O'OGO
Manganese . 0450 „ O'GOO
Sulphur . - : i . 0-020 „ 0'040
Phosphorus fVl. 0'030 „ 0'070
These medium steels are used for such purposes as engine
forgings, piston rods, wheel tyres, rails, and any similar purposes
154 STRENGTH OF MATERIALS
where a steel is required which is not very hard, but which is
harder and stronger than ordinary low-carbon structural steel.
The yield point stress in tension of the medium steels varies
from 14 to 22 tons per sq. in., and the maximum stress from 32
to 44 tons per sq. in. ; while the elongation is from 12 to 20 per
cent. The compressive and shearing strengths follow the same
law as in rnild steel. While the elastic modulus is very little
higher, rarely being very far from 30,000,000 Ibs. per sq. in.
It will be seen that the general properties of the medium
steels are similar to those of the low- carbon steels, but in advance
of them in the same way as mild steel is in advance of wrought
iron.
High-Carbon or Hard Steels. — The percentage of carbon
present in these steels varies from 0*45 per cent, to 1*50 per cent.,
and for some special purposes is still higher. Under this heading
are included many varieties of steel produced in a number of
ways and used for purposes where either great strength or great
hardness is required. They include hard steel for forgings, steel
for guns, springs, and every kind of cutting tool. In many of the
purposes for which hard steel is used, it is necessary not only to
have strength and hardness, but to be able to control the degree
of this hardness. This is the one characteristic property possessed
by the high-carbon steels which does not belong to wrought
iron and the lower carbon steels. It is well known that this
hardening can be brought about by heating the steel to red-
ness, and cooling it quickly by plunging in water or oil. The
degree of hardness depends upon the temperature of the steel when
cooling takes place. It is usual with engineering tools to first
make the steel very hard by quenching in cold water, and after-
wards to let it down to the required " temper " by gradually rais-
ing the temperature until the required point is reached, and then
again quenching in water. In the process of this reheating, the
brightened surface of the metal changes colour, each tint corre-
sponding to a certain degree of hardness.
The tensile strength of the hard steels may vary from 34 tons
per sq. in. up to considerably over 100 tons per sq. in. Their
ductility is not great in those steels with very high tensile
strengths, being as low as 3 or 4 per cent. The condition of hard
steel often approaches that of brittleness. The fractures usually
show a very fine crystalline appearance.
THE MATERIALS USED IN CONSTRUCTION 155
Besides the more usual kinds of steel having the analyses
which have been indicated, there are special steel alloys
which have strength properties peculiar to themselves. Among
these may be mentioned nickel steel, in which a quantity of nickel
is alloyed with the other constituents, manganese steel, and chrome
steel.
The elastic modulus for high-carbon steel is not very much
higher than for the low-carbon and medium steels, but occasionally
it is found to be as high as 32,000,000 Ibs. per sq. in.
Stress strain diagrams for medium and hard steels are shown
on Fig. 103.
Steel Castings. — The tensile strength of steel which is run
into moulds so as to form castings is not high, varying from 17 to
28 tons per sq. in. Its ductility is low and it often contains blow-
holes, which militate against its strength and make it unsuitable
for castings which are to have accurately machined surfaces.
Steel castings are, however, much stronger than similar castings
made from cast iron, and considerable advances have been made
of late years by which sounder castings are being produced.
Cast Iron. — Cast iron differs greatly from the other varieties
of iron which have been mentioned. It contains a larger proportion
of carbon, which is partly combined and partly in a state of
mechanical mixture. In the harder kinds the carbon is almost
all combined, while in the softer kinds, such as are used for parts
which have to be machined, it exists mostly in the form of mixed
graphite. The total quantity of carbon present varies from rather
less than 2 per cent, to something over 4 per cent. In addition
to the carbon there is also silicon, generally from 1 to 3 per cent. ;
manganese from 0'2 to 2;7 per cent. ; between 0'2 and 1/5 per
cent, of phosphorus ; and from a trace to 2 '5 per cent, of sulphur.
The strength of cast iron is much greater in compression than
in tension. Its tensile strength may be from 8 to 14 tons per
sq. in., while the compressive strength varies from 25 to 60 tons
per sq. in.
Tests of direct shear of cast iron are difficult to make and of
little use, but the coefficient of torsional strength, that is, the
stress as calculated from the usual torsional formula corresponding
to the twisting moment which it takes to fracture a shaft, works
out at from 14 to 18 tons per sq. in. in the more ordinary kinds.
156 STRENGTH OF MATERIALS
There is practically no permanent set in cast iron under stress,
beyond what can be measured with an extensometer. But though
the permanent set is small, it takes place at very low loads, and
under no stress can cast iron be called truly elastic. An approxi-
mate elastic modulus can be calculated at low stresses, and this is
found to vary from 9,000,000 to 16,000,000 Ibs. per sq. in.; an
average value is about 15,000,000 Ibs. per sq. in., or about half
that of wrought iron and steel.
The most satisfactory test for cast iron is that made upon a
beam of the metal loaded in the centre. The most common
dimensions of such a test beam are 1 in. wide, 2 ins. deep, and
loaded in the centre of a 36-in. span. A beam of this kind should
withstand about 3000 Ibs. with a deflection at the centre of
TFths of an inch.
As cast iron is never perfectly elastic, the beam formula does
not hold good up to the breaking load, but a value corresponding to
the maximum stress from the beam formula may be calculated
and called the coefficient of bending strength. It is found to
be about one-and-a-half times the tensile strength.
A typical load strain diagram for cast iron is shown on
.Fig. 103.
Malleable Cast Iron.— If ordinary castings are treated by
subjecting them at a high temperature to the oxidising influence
of haematite ore or manganese dioxide, the carbon in the iron is
reduced and the metal becomes more "steely." The resulting
material is called malleable cast iron. It does not possess the
characteristic brittleness of cast iron, and may be bent and
hammered without fracture. Its strength is increased by the
process, and is found to be from 16 to 22 tons per sq. in. in
tension.
Copper.
For certain purposes the use of copper is almost indispensable.
It is very ductile, an excellent conductor of heat, and has a high
electrical conductivity. The strength of copper varies according
to the state of the metal. The tensile strength varies from about
7 tons per sq. in. in the poorer sorts of castings to 24 tons per
sq. in. in the hard drawn wire used for overhead conductors.
Kolled copper, whether in plates or bars, has a maximum strength,
THE MATERIALS USED IN CONSTRUCTION
157
in the unannealed state, of about 14 tons per sq. in. When hard
drawn wire or hard rolled plate or bar is softened by heating to
redness and quenching in water, the tensile strength is greatly
diminished, being brought down to two-thirds its former value,
while its ductility is greatly increased. In the hard state the
elongation is from 1 to 6 per cent., while the same metal
annealed may have an elongation after fracture of 40 to 50
per cent.
Hard drawn or rolled copper should have a proportional limit
JO
201
B
g
O 10
10
20
30
50
PERCENTAGE ELONGATION.
FJG. 104. — Here A refers to copper trolley wire annealed; C, the same in
original state; B, gun-metal ; D, aluminium bronze.
of 7 or 8 tons per sq. in. ; in softened copper there is practically no
limit.
The elastic modulus for copper is from 16,000,000 to 18,000,000
Ibs. per sq. in.
Alloys of Copper. — The principal alloys of copper are the
various bronzes containing copper and tin, along with small per-
centages of other metals such as zinc, aluminium, phosphorus, and
manganese. The strengths of these alloys vary enormously
according to the proportions of the different ingredients, and may
158 STRENGTH OF MATERIALS
be from 9 tons per sq. in. in the case of the poorest quality of cast
yellow brass to 40 tons per sq. in. in aluminium or manganese
bronze. A good quality of gun-metal casting, such as is used
for engine parts, and containing copper, say, 32 parts, tin 4, and
zinc 1, should have a maximum tensile strength of 14 tons per
sq. in.
Typical stress strain diagrams for copper and some of its
alloys are given on Fig. 104.
Vitreous Materials.
Stone. — The strength of stone is generally specified as the
number of tons per square foot of area which are required to
crush it. In order to find this for any given sample, a cubical
piece, from 3 ins. to 12 ins. length of side, is taken and
placed between the platens of a compression testing machine.
The load should be uniformly distributed over the surface, either
by inserting thin layers of millboard between the platens and
the specimen, or, better, by setting it in a bed of plaster of
Paris.
The actual crushing strength varies from 150 tons per sq. ft.
in the soft Ancaster stone, 300 for the softer kinds of sandstone,
500 for Portland limestone, 600 or 700 for the harder sandstones,
1000 for Welsh basalt, and 1500 for granite. There may be
considerable variety in different samples from the same quarry.
Bricks. — Bricks should be tested for strength in the same
manner as stone, great care being taken to level any inequalities
exposed to pressure by previously filling up with cement or plaster
of Paris.
Bricks tested in this manner yield results of which the fol-
lowing are a few typical examples : —
Kind of Brick. Crushing Stress,
tons per sq. ft.
London stock ..... 140
Common wire cut, Manchester . .260
Leicester wire cut . . . .290
Accrington plastic . . . .250
Blue Staffordshire . . . .360
Blue Brindle, Staffordshire . . .480
THE MATERIALS USED IN CONSTRUCTION 159
The writer has found that the strength of bricks is from 2J
to 5 times the strength of the brickwork in which they are used.
The strength of the bricks themselves is only a very rough
indication of the strength of the resulting brickwork. In order
to obtain good brickwork from good bricks, it is most important
that the cementing material be of the very best quality.
It is useless to employ good bricks when the mortar is
bad. The results of all the experiments which have been
made on this point go to show that the ultimate strength of
a given sample of brickwork depends very largely upon the
mortar.
Cement, Mortar, and Concrete. — The most important cement
used for engineering work is what is known as Portland cement.
This is an artificial product, made by first mixing an earth which
is composed principally of lime with one in which clay pre-
dominates. These two are ground together, wet in the right
proportions, made up into lumps, dried, and then burnt like
bricks. A hard material results from the calcination, and this,
when ground into a fine powder, is the required cement. It
consists of compound silicates and aluminates of lime. When
this cement is mixed into a paste with water it sets hard, so
as to form a kind of artificial stone. The time of setting varies
from half an hour to several hours. In order to determine the
quality of a given sample of cement, a number of tests are
applied; but it is only necessary to refer here to the tests for
tensile and compressive strength. The former is found by
applying a tensile load to a sample made up into a suit-
able form for holding, and called a briquette. The tensile
strength varies from 200 to 800 Ibs. on the sq. in. A
good cement should have a strength of 400 Ibs. per sq. in.
when tested at the end of a week, and should improve with
age.
Crushing tests of neat cement (that is, cement alone) and
mixtures of sand and cement yield results which have a more
direct bearing on the suitability of a cement for engineering
work, where most of the stress is compressive. They are,
however, more costly to make. The compressive strength of
neat cement is from 150 to 250 tons per sq. ft., while mixtures
of sand and cement, such as are used for mortar, have been
160
STRENGTH OF MATERIALS
shown by tests made by the writer to be somewhat as follow,
for a fairly good cement: —
Ratio.
Crushing Strength, tons per sq. ft.
Sand : Cement.
7 Weeks Old.
20 Weeks Old.
1 : 1
80
100
2 : 1
30
50
3 : 1
25
30
4 : 1
20
25
5 : 1
15
20
Portland Cement Concrete. — This is one of the most valuable
materials which engineers, whether civil or mechanical, 'have at
their disposal. By concrete is meant a compact mass, composed of
small pieces of broken stone, gravel, broken brick, or other similar
substance cemented together by being set in a binding matrix of
cement and sand. This broken stone and sand, which is mixed with
the cement, is called the aggregate or ballast ; the proportion of
aggregate to cement varies, a common ratio being 5 or 6 to 1.
The cement and ballast are mixed dry, then sufficient water is
added to produce a thin paste with the sand and cement, and the
resulting mixture is tipped or shovelled into the position it has
to occupy, and allowed to set.
The ultimate strength of the concrete so formed depends on
the character of the aggregate employed, on the quality of the
cement, on the manner and thoroughness of the mixing, on the
age, and on the ratio of cement to aggregate. The writer has
found that concrete made with gravel and good cement in the
proportion of 5 to 1 has a crushing stress at the end of thirteen
weeks of 210 tons per sq. ft., and after thirty-four weeks of 225
tons per sq. ft. Mr Deacon gives the results of tests of some
samples of the concrete used on the Vyrnwy works as 107 tons
per sq. ft. after one month, and 185 tons after thirty-six months.
These are good results, but in some cases the strength may be as
low as 50 tons per sq. ft.
The vitreous materials, including stone, brick, brickwork,
cement, and concrete are all fairly elastic, and take little per-
manent set up to considerably above the working stresses.
THE MATERIALS USED IN CONSTRUCTION
161
On Fig. 105 is shown a stress strain diagram obtained during
the loading of a brickwork pier. This is typical of almost all
I50r
100
O-l 0-2 03
COMPRESSION— Inches per foot of height.
FIG. 105.
vitreous materials. The modulus of elasticity varies somewhat.
The following are a few results obtained by the writer :—
Material.
Hard bricks
Softer bricks
Good brickwork .
Poorer brickwork
Portland cement concrete
Elastic Modulus — Ibs. per sq. in.
5,000,000 to 6,000,000
1,500,000 „ 3,000,000
1,200,000
500,000
1,500,000
2,000,000
1,000,000
2,500,000
L
162 STRENGTH OF MATERIALS
Repeated and Reversed Stresses.
The effect of the repetition or reversal of the applied stresses
has a most important bearing on the life of machine parts. The
evidence which is available on this point is by no means com-
plete, but sufficient has been obtained to help the designer in the
proportioning of his parts. The earliest experiments on this
subject were those of Wohler and Bauschinger, and these have
been supplemented more recently by Eeynolds and Smith, and by
Stanton at the National Physical Laboratory.
It is impossible to give anything like a complete account here
of the results obtained, but a few of the more general laws arrived
at may be mentioned.
Most of these experiments have been made on wrought iron
and steel. The stresses were applied in different ways, and ab
different speeds of repetition and reversal.
Three chief methods have been adopted, namely: either the
same stress ranging from zero to a certain fixed value, either in
tension or compression, was applied many times in succession
until rupture took place, or the successive application was from
a certain stress in compression to either one of the same or a
different magnitude in tension ; or, again, the range was from a
maximum compressive stress to a tensile stress of the same
magnitude, obtained by rotating a shaft under a constant bending
moment. The speed of repetition in Wohler's tests was approxi-
mately 60 per minute. In the National Physical Laboratory
experiments, the speed was intended to be more nearly that most
commonly met with in modern practice, and was in the neigh-
bourhood of 800 reversals per minute.
The main fact revealed by these experiments appears to be
that the ultimate failure of a specimen depends, not so much on
the magnitude of the stress imposed, but upon the range of stress.
Speaking roughly, the same result would be produced by applying
a stress extending from 10 tons per sq. in. in compression to 10
tons in tension, or a through range of 20 tons, as if the bar were
subjected to a range of stress which reached from zero to 20 tons
per sq. in. in tension.
Besides the range of stress being a factor in the ultimate failure
THE MATERIALS USED IN CONSTRUCTION 163
of the material, this was found to depend also on the number of
repetitions or reversals.
With a certain range of stress it was found that the number
of repetitions or reversals might be unlimited. When this range
was exceeded, the number of repetitions which was needed to
produce failure was smaller the greater the stress.
This limiting range has been found to nearly coincide with
what has been called the natural elastic limit of the material,
and which has already been defined. It has therefore been
suggested by Unwin that it might be possible to determine the
natural limit, and to take this as the safe range of stress
within which any number of repetitions or reversals would not
cause failure.
Unwin quotes the following results, taken from Wohler's
figures, as showing how the limiting number of reversals is
affected by the range. They refer to samples of Krupp
steel which were subjected to reversals of equal stress in
tension and compression by rotation under a fixed bending
moment.
Kange of Approximate Number of Repeti-
Stress. tions to cause Fracture.
Tons per sq. in. Thousands.
40-2 55
344 128
32-6 798
30-6 1,666
28-6 4,163
28-6 45,050
The following are three typical sets of results from the National
Physical Laboratory experiments of Stanton.* The number of
reversals per minute was in all the cases quoted very nearly
795 per minute. The range of stress was from a compressive
stress to a tensile stress, the ratio of tension to compression being
1-4 to 1.
* Min. Proc. Inst.C.E., vol. iii.
164 STRENGTH OF MATERIALS
Wrought Iron, having an elastic limit of 14*28 tons per sq. in.
and a maximum of 2 3 '76 tons per sq. in.
Range of Approximate Number of Repeti-
Stress. tions to cause Fracture.
Tons per sq. in. Thousands.
27-86 101
26-86 200
25-80 253
24-60 217
23-43 373
21-25 1,000^|
20-20 1,116 \ Not broken.
19-05 1,028J
Mild Steel. — 0'33 per cent, carbon, with 14'30 elastic limit
and 28'30 maximum stress, in tons per sq. in.
Range of Approximate Number of Repeti-
Stress. tions to cause Fracture.
Tons per sq. in. Thousands.
3017 68
28-06 97
26-4 236
25-78 1,914
25-71 1,330
23-29 2,000. Not broken.
Piston-rod Steel. — 0'446 per cent, carbon, with 19'62 elastic
limit and 43 '8 5 maximum stress, in tons per sq. in.
Range of Approximate Number of Repeti-
Stress. tions to cause Fracture.
Tons per sq. in. Thousands.
30-16 155
29-93 120
28-85 308
28-80 752
27-88 1,032 '
27-88 3,049 broken'
The few figures here quoted appear to show that for every
material there is a certain range of stress within which the loading
may be indefinitely repeated without any perceptible weakening ;
but if it is extended beyond this range, fracture will eventually
occur after a certain number of repetitions.
THE MATERIALS USED IN CONSTRUCTION
165
Safe Stresses Allowable in Practice.
In designing structures and machine parts, it is customary to
find the stress which may be applied to any material by dividing
the maximum stress as found from an ordinary tensile test by
a factor of safety. A much more reasonable and scientific plan
is to say that the material may be subjected to a certain
range of stress as fixed by the results of some such experiments
as those mentioned, or one which has a direct relation to the
natural elastic limit.
In the light of our existing knowledge of the strength
properties of the materials of construction and the effect of
repetition and reversal of stress, the following may be taken as
allowable working stresses.
Material.
Simple
Compression.
Simple
Tension.
Shear Stress
in Torsion.
Tons per sq. in.
Tons per sq. in.
Tons per sq. in.
Average Wrought Iron .
5f
5f
3f
Do. across grain .
5|
5|
Mild Steel .
6
6
3f
Medium Steel
7£
n
5|
High-Carbon Steel
9|
9|
9*
Steel Castings
5
5
4
Cast Iron
5
1*
1J
These are for steady loads in one direction only. Where
the load is repeated so as to alternate many times between no
stress and stress of one kind, the above must be multiplied by f ;
and where the stress alternates many times between tension and
compression of the same amount, they must be multiplied by \.
The Micro-structure of Metals.
In recent years much has been do.ne to add to our knowledge
of the internal structure of metals, and of the effects of stress
upon this structure. A few of the main facts may be mentioned.
When a polished surface of iron or steel is placed under a
microscope of sufficient power, it is seen that the surface appears
to be divided up into irregular areas like a map. These areas
166 STRENGTH OF MATERIALS
are the sections of the crystals which combine to form the
material. It is further seen that these crystal surfaces have not
all the same appearance, some being bright and others dull.
The size of the grains depends upon the thermal treatment,
rapid cooling leading to small crystals and slow cooling to
large ones.
When the surface is examined after the application of a stress
somewhat beyond the elastic limit, it is found that the grains are
elongated in the direction of stress when the load is tensile.
After the lapse of a considerable time at the ordinary tempera-
ture, or the application of a moderate temperature for a short
time, the crystals resume their former shapes. This corresponds
to what takes place when the elastic limit has been raised by
stress, when it is found that the limit is again lowered to its
original value after some time has elapsed or heat has been
applied.
It will be remembered that fracture of materials in tension
or compression generally resolves itself into failure by shear, and
that this is most noticeable in the compression of brittle materials
like cement and cast iron and the tensile fracture of some kinds
of steel.
Professor Ewing has observed and pointed out that a some-
what similar effect is produced in the constituent grains of iron
and steel when strained some little way past the proportional
limit. Under a microscope of high power it is seen that each
grain of material so strained is covered with rows of parallel
lines, there being generally three or four systems of these
inclined to one another. These lines are really minute steps
representing the section of " slip planes " with the surface of the
material. It thus appears that the permanent set takes place
by an internal sliding, not of one grain upon another, but along
parallel surfaces in the individual grains. It is by means of
this slip that the change of shape of the crystals can take place.
APPENDIX
GENERAL TABLE OF STRENGTHS AND WEIGHTS
Material.
Ultimate Tensile
Strength.
Crushing
Strength.
IRON AND STEEL.
Tons per sq. in.
Tons per sq. in.
Wrought iron —
Merchant bar ....
19 to 22
11 to 14
Ship plates .....
21 „ 23
Best Yorkshire ....
23 „ 25
12 to 16
Rivet iron .....
24 „ 26
13 „ 17
Low-carbon steel, 0'10% C. .
23
13
Bridge steel, 0-20% C. .
26 to 28
14 to 17
Boiler steel, 0 '25% to 0 -30% C. .
28 ,, 32
14 „ 18
Tyre steel, 0-28% C. .
36 „ 48
Medium steels, 0'30% to 0'45% C.—
Forging steel ....
33 „ 35
15 to 18
Hard forging steel
35 „ 40
16 „ 20
Piston-rod steel, 0'45% C. .
44
Nickel steel
24 to 55
12* to 38
24 „ 30
11 „ 15
High-carbon steel, 0 '45% to 1 '50% C. —
Manganese steel ....
33 „ 42
...
Spring steel, 0 -50% C. .
60 „ 70
Tungsten steel ....
70
...
Hadfield's forged cast steel .
46 to 125
...
Manganese steel wire .
100
Pianoforte wire ....
100 to 150
...
Ordinary steel castings .
Hadfield's unhammered castings
Cast iron, very soft for machining .
17 „ 28
32 „ 64
8 „ 10
25 to 35
hard ....
14
60
Malleable iron castings .
16 to 22
21
COPPER.
14
Ordinary cast ....
6 to 10
Plate, hard rolled ....
14 „ 15
...
Hard drawn trolley wire .
24
...
Hard drawn wire, small diameter .
28 to 30
...
BRONZE, ETC.
Ordinary gunmetal castings .
Good gunmetal castings .
Rolled bronze, hard ....
10
14
27
...
Aluminium bronze . .
27 to 40
...
Yellow brass, good ....
Aluminium, rolled bar
12
9 to 12
Zinc
1 „ 3
Tin
2 »»
...
Lead .
107
168
APPENDIX
TABLE OF STRENGTHS AND WEIGHTS -continued.
Material.
Ultimate Tensile
Strength.
Crushing
Strength.
VITREOUS MATERIALS.
Tons per sq. in.
Tons per sq. ft.
Stone—
Softer kinds of sandstone
150 to 300
Portland limestone
...
500
Hard sandstones ....
600 to 700
Welsh basalt ....
...
1000
Granite .....
1500
Brick—
London stock ....
140
Common wire cut ....
260
Leicester wire cut
...
290
Accrington plastic
Blue Staffordshire
...
250
360 to 480
Brickwork —
Common bricks and mortar .
50
„ better mortar
...
80
Good class bricks and best mortar
...
100 to 200
Portland cement, good average
300 to 500
150
Mortar, 3 sand, 1 cement
...
30
„ 2 „ 1 .
...
50
1 „ 1 •
. 100
Portland cement concrete, 5 to 1
...
100 to 220
Breaking Strength of a
TIMBER.
1'xV'xl" beam.
Central load, Ibs.
White pine .....
450
360
Memel pine .....
500
500
Ash
700
580
Beech
600
550
Oak
550
530
Teak
800
750
Greenheart .....
850
780
Pitch pine . .
550
570
ELASTIC OR YOUNG'S MODULUS.
Material.
Elasticity.
Million Ibs. per sq. in.
Wrought iron and steel ....
27 to 32
Cast iron ......
9 16
Copper ......
16 18
Gunmetal ......
13 16
Aluminium ......
10 12
Hard bricks .....
5 6
Softer bricks .
1 3
Good brickwork .....
1 2
Portland cement concrete . .
1-5 2'5
Timber, average along the grain .
1-5
APPENDIX
WEIGHTS OF MATERIALS.
169
Material.
Lbs. in 1 cub. ft.
Wrought iron .
485
Steel .
499
Cast iron
450
Copper
552
Gunmetal
528
Brass .
525
Tin
455
Zinc .
437
Lead, sheet .
711
Aluminium
166
,, cast
160
Sandstone
150
Portland stone
151
Welsh basalt .
172
Granite
165
Bricks, soft .
110
,, hard .
134
Portland cement concrete
138
Lime mortar .
105
Earth, average
100
White pine
30
Memel pine
35
Pitch pine
40
Beech .
43
Ash .
45
Spruce
32
Teak .
55
Greenheart
71
Oak .
53
Lignum vitae .
83
L2
ADDITIONAL EXAMINATION QUESTIONS
CHAPTER I. — ELASTICITY.
1 . Explain the meaning of the following terms : — Stress, strain, elastic
limit, yield point, and modulus of elasticity.
2. Sketch typical load strain diagrams (a) for a tensile test of wrought iron
up to the yield point, and (6) for a compression test of cast iron.
Point out the differences in behaviour of these materials.
3. Explain the meaning of the expression " Elastic Modulus." How would
you set about finding its value for a given sample of cast iron ? Sketch
the form of load strain diagram you would expect to obtain.
4. In a tensile test of a steel bar 1 in. in diameter, it was found that the elastic
extension for each ton of load was 0'000985 in. on a length of 10 ins.
Find the modulus of elasticity. [29,000,000 Ibs. per sq. in.
5. A copper trolley wire 50 ft. long sustains a pull of 250 Ibs., and under
this load is found to be stretched 0*07 in. Find the elastic modulus,
the diameter of the wire being 0'40 in. [17,050,000 Ibs. per sq. in.
CHAPTER III.— BEAMS.
6. If M is the bending moment at a point in a beam, I the moment of
inertia of the section at this point, / the maximum stress in the
material at the section, and y the distance from the neutral axis of
the section to the part where the maximum stress occurs, prove
that M. _ J_
I '" y
7. A girder resting on supports 70 ft. apart carries two rolling loads of 10
tons and 15 tons respectively. The loads roll across at a constant
distance apart of 8 ft. Find the maximum bending moment on the
girder.
8. What will be the stress caused in the flanges of a plate girder, 80 ft. span,
172 STRENGTH OF MATERIALS
depth of 7 1 ft., and flanges 16 ins. wide, and If ins. thick, when
carrying a uniform load of 1 '60 tons per sq. ft. of length ?
[6'08 tons per sq. in.
9. The slide bars of a horizontal engine are 40 ins. long, and have a
rectangular section l£ ins. wide. Find the depth of the bars if the
greatest vertical force in the centre of a bar is 500 Ibs. and the
maximum stress allowable in the material is 2 tons per sq. in.
[2'31 ins.
10. A cast iron beam 1'92 ins. deep, 1'03 ins. wide, when tested on a span
of 36 ins., is found to break with a central load of 2760 Ibs. Find
what would be the central breaking load of a similar beam 3^ ins.
deep, 1§ ins. wide, resting on supports 45 ins. apart. [9870 Ibs.
11. A plate girder has a span of 70 ft. and a depth of 8 ft. The flanges are
16 ins. wide and 1^ ins. thick. Find what load per foot run the
girder must carry in order that the stress in the metal of the flanges
may not exceed 6 tons per sq. in. [1 '88 tons per ft.
12. What is the greatest bending moment which each point of a beam
supported at the ends is liable to under a uniform load of 1 ton per
ft. and a moving load of 10 tons, the span being 15 ft.1? Find an
expression for this, and also sketch the bending moment diagram.
13. If a beam of pitch-pine timber 1 in. wide, 1 in. deep, and 1 ft. span
breaks with a central load of 800 Ibs., find the safe distributed load
that may be put upon a beam of the same material 15 ft. span, 4 ins.
wide, and 8 ins. deep, using a factor of safety of 6. [4551 Ibs.
14. Estimate the safe distributed load that can be carried by a pitch-pine
beam 5 ins. wide, 9 ins. deep, and resting upon supports 18 ft. apart.
[6000 Ibs.
15. Find the weight of a steel girder 20 ft. long, 6 ins. deep, and having
flanges 5 ins. wide. The thickness of the metal is f in. in the flanges
and f in. in the web. [767 Ibs.
16. A railway bridge 130 ft. span is covered for half its length by a train
weighing 2 tons per ft. of length, one end being at the centre of the
span. Find the bending moment and the shearing force at a point
in the middle of the covered portion of the bridge. Also, sketch the
diagrams of bending moment and shearing force.
[2113 tons-feet], [32'5 tons.
17. Find the depth at the centre of a cross-girder 18 ft. span, flanges 7 ins.
wide and f in. thick, when the distributed load is 2| tons per foot
run. The stress in the metal is not to exceed 6 tons per sq. in.
[2 ft. 9 ins.
18. In a beam 40 ft. span, resting freely at its ends, two loads of 5 tons are
placed, each 10 ft. from the centre. Find the bending moment
ADDITIONAL EXAMINATION QUESTIONS 173
and shearing force at each support, at the centre, and under each
load. Also, draw the bending moment and shearing force diagrams.
[At supports, M 0, S 5 ; at centre, M 50, SO; at loads, M 50, S 5.
CHAPTER IV.— GRAPHICAL MOMENTS OF INERTIA.
19. Solve Example 2, Chapter IV., by calculation.
20. Prove the rule for finding graphically the modulus of the section of
a rectangular beam, and point out how it is connected with the
ordinary beam formula as obtained by purely mathematical reasoning.
21. The upper flange of a cast iron girder is 5 ins. wide and 1 in. thick, the
lower flange is 8 ins. wide and 2 ins. thick, the total depth is 10 ins.,
and the web 1 in. thick. Find graphically the moment of inertia of
the section. [341 inch-units.
CHAPTER V. — DEFLECTION OF BEAMS.
22. Prove that at any point in a loaded beam the relation between the
radius of curvature, the bending moment, the elastic modulus of the
material, and the moment of inertia of the section is -^^^
23. Find the central deflection of a tram rail on a clear span of 10 ft. when
the elastic modulus is 29,000,000 Ibs. per sq. in. The maximum
stress in the material is 6 tons per sq. in. The centre of gravity of
the section is 3'57 ins. from the bottom edge.
[Central load = 4'l tons ; central deflection = 0' 198 in.
24. A plate girder has a span of 80 ft. and a depth of 7 ft. 6 ins. Its flanges
are 16 ins. wide and 1*75 ins. thick. The stress in the metal of the
flanges must not exceed 6 tons per sq. in. Find the uniformly dis-
tributed load per foot that may be put upon the girder, and the deflec-
tion in the centre under this load. The elastic modulus may be
taken as 28,000,000 Ibs. per sq. in. [w = 1 '57 ; A = 1 '025 ins.
25. Find the load per foot run that is carried by a steel girder 30 ft. span and
18 ins. deep, the flanges being 8 ins. wide and 1 £ ins. thick, when the
maximum stress in the material is 7 tons per sq. in. Also, find the
deflection at the centre under this load. E = 29,000,000.
[w = 0*809 ton ; A =0'81 in.
26. A road bridge over a river is carried across three spans, of which the
central one is 150 ft. and the two shore spans 90 ft. each. The
girders are in three portions, made up of two shore parts, each 110 ft.
long, and a central part, also of 110 ft. in length. These shore por-
174 STRENGTH OF MATERIALS
tions form cantilevers which project beyond the river piers, and from
the ends of these the centre girder is freely hung. With a uniform
load of 2 tons per ft. throughout the length of the bridge, sketch
the diagrams of shearing force and bending moment.
27. A steel joist whose moment of inertia is 55*9 is tested with a central
load on a span of 5 ft., and it is found that for every 8 tons incre-
ment of load there is a central deflection of 0'0506 in. Find the
modulus of elasticity.
28. Prove the rule for finding the load carried by each pier of a continuous
girder supported on three points equidistant and on the same level.
Sketch the bending moment diagram for this case.
CHAPTER VI.— SHEAR STRESS IN BEAMS.
29. Knowing the shearing force at any section of a loaded beam and the
dimensions of its section, deduce an expression which will give the
intensity of the shear stress at any point in the section.
30. In Question 24 the web of the girder is % in. thick. Find the mean and
maximum shearing stresses in the web close to one abutment.
31 . Sketch the curves showing the variation of shearing stress in the cases
of — (1) A beam of rectangular section; (2) A beam of I section;
(3) A tram rail.
CHAPTER VII. — ECCENTRIC LOADING.
32. A solid steel bar 3 ins. in diameter carries a tensile load of 40 tons.
What will be the stress in the metal when the load is applied along
the axis of the bar; and also, what will be the minimum and
maximum stresses when the direction of the load is shifted ^ in.
from the axis 1
[Tensile = 13'2 tons per sq. in. ; compressive = l*8 tons per sq. in.
33. A load of 20 tons is applied axially to a pier of brickwork 1^ ft. square
in section, at a point 6 ins. from the centre of the section measured
in a direction parallel to one side. Find the maximum and minimum
stresses in the material in tons per sq. ft.
[Maximum stress, 26'6 ; minimum stress, 8'8.
34. A rectangular bar of steel 4 ins. wide and 2 ins. thick is bent so as to
form a semicircular plate, the narrow edges being curved and the
wide sides remaining flat. The radius of curvature of the centre
line is 5 ins. Find the tensile loads that must be applied diametri-
ADDITIONAL EXAMINATION QUESTIONS 175
cally at the two ends which will cause a maximum stress in the
material of 5 tons per sq. in. ; also, find the minimum stress. Prove
the formula.
[W = 5'33 tons ; minimum stress = -4'33 tons per sq. in.
CHAPTER VIII. — STRUTS.
35. Estimate what will be the probable collapsing load of mild steel strut
which is freely hinged at the ends, 1 5 ft. long between the hinges,
and has a solid circular section 3 ins. in diameter.
CHAPTER IX. — TORSION.
36. If a length of shafting 3 ins. in diameter and running at 100 revolutions
per minute will transmit 50 horse-power, what horse-power can be
transmitted by a 5 -inch shaft of the same material when running at
180 revolutions per minute 1 [416*6 H.P.
37. If a wrought iron shaft 3f ins. in diameter will safely transmit 90 horse-
power at 130 revolutions per minute, find the size of steel shaft to
transmit 160 horse-power at 72 revolutions, the ratio of the torsional
strength of steel to that of wrought iron being as 7 to 5. [4*8 ins.
38. A solid steel shaft is to transmit 600 horse-power at a speed of 58
revolutions per minute, and the stress in the material is not to exceed
8000 Ibs. per sq. in. Find the diameter. [7'46 ins.
39. What horse-power can be transmitted by a hollow shaft, whose diameters
are 10 ins. and 5 ins., when running at 150 revolutions per minute,
the stress in the material being limited to 8000 Ibs. per sq. in.
[3500 H.P.
40. A wrought iron shaft, 4 ins. in diameter, and 30 ft. long, is used to
transmit power. Find the maximum twisting moment allowable in
order that the shear stress in the metal may not exceed 4 tons per
sq. in., and find through how many degrees the shaft will be twisted
under this moment. G = 1 2,000,000 Ibs. per sq. in.
[T = l 12,600 inch-lbs.; H.P. = 50'2; 0 = 77°.
41. A helical spring, made from round steel 0'96 in. in diameter, has an out-
side diameter of 4'8 ins. and consists of 10 complete coils. When
unloaded, the coils nearly touch one another. Find the amount this
spring will be extended under a load of 1} tons, when the shear
modulus of the material is 1 1,000,000 Ibs. per sq. in. [ A = 1*36 ins.
42. A hollow shaft, whose outer is twice its inner diameter, is to be used
to transmit 8000 horse-power while making 76 revolutions per
176 STRENGTH OF MATERIALS
minute, and the stress in the material must not exceed 7500 Ibs. per
sq. in. Find the outer diameter ; also, find the angle of twist on a
length of 50 ft. [D2 = 16'9 ins. ; 0 = 2T.
CHAPTEE X.— TORSION AND BENDING.
43. In the middle of a free span of shafting, 10 ft. long between the bear-
ings, there is a pull of 1200 Ibs., at right angles to its centre line, due
to a belt. The shaft is transmitting 30 horse-power at a speed of
200 revolutions per minute. The shear stress in the material must
not exceed 8000 Ibs. per sq. in. Find the equivalent twisting moment
and the diameter. [T^ - 73,200 ; M = 36,000 ; d=3'6 ins. ; T = 9450.
44. Prove the formula you make use of in the last question.
45. Find the shear stress in the metal of a shaft 3 ins. diameter which is
transmitting 25 horse-power at a speed of 250 revolutions per minute.
If there were to be a bending moment of 10,000 inch-lbs. acting on
the shaft in addition to the twisting moment, what would the
diameter then have to be ? [1 188 Ibs. per sq. in. ; 4'53 ins.
46. The distance from the centre of the crank pin to the centre of the near
bearing, in the case of an overhung engine crank 9 ins. radius, is 10
ins. Find the diameter of the journal when the maximum pressure
on the crank pin is 12,000 Ibs., and the stress in the material does
not exceed 7500 Ibs. per sq. in. [d=5*4 ins.
CHAPTER XI. — CYLINDERS.
47. Find the thickness of a copper pipe to carry steam at a pressure of 250
Ibs. per sq. in. The diameter is 7 ins., and the stress on the metal is
not to exceed 2000 Ibs. per sq. in. [£=0'43 in.
48. Find the diameter of a boiler steam drum in mild steel, whose diameter
is to be 3 ft., and the working pressure 160 Ibs. per sq. in. The
stress in the material must not exceed 10,000 Ibs. per sq. in. [3% of
an inch.
CHAPTER XII. — RIVETED JOINTS.
49. Two steel plates, 1 in. thick and 7 ins. wide, are allowed to overlap at
their ends and are connected by three 1^-in. steel rivets. The
overlap is sufficiently great to prevent any possibility of the rivets
tearing through the edges. If the rivets are made from the same
steel as the plates, would you expect the joint to fail by the tearing
ADDITIONAL EXAMINATION QUESTIONS 177
of the plates between the rivet holes or by the shearing of the rivets 1
At what load do you suppose this would take place ?
[By tearing at 91 tons.
50. Design a longitudinal double-riveted lap-joint for a boiler steam drum
in steel, 3 ft. diameter, 1 60 Ibs. working pressure.
CHAPTEES XIII., XIV., AND XV.
51. Make a sketch of a typical load strain diagram as obtained during the
tension test of a bar of cast iron. Define and quote values for
(a) the maximum stress in tension and compression, and (6) the
elastic modulus for cast iron.
52. State briefly the chief strength properties of wrought iron, steel, and
cast iron, and say for what purposes each is best suited.
53. What is a load strain diagram ? Sketch a typical diagram for a tensile
test of wrought iron, and explain how the diagram shows the elastic
limit and yield point. In what respects does cast iron differ from
wrought iron as regards its elastic properties 1
54. Describe the effect of repeatedly applying a tensile load which is slightly
above the P-limit to a bar of steel ; and how does this result differ
from that obtained when the load, instead of being always tensile, is
alternately tensile and compressive 1 State what you know as to
how the results of Wohler's experiments, and others of a similar
nature, affect questions of design.
55. Sketch typical load strain diagrams (a) for a tensile test of wrought
iron up to the yield point, and (6) for a compression test of cast iron.
Point out the differences in behaviour of these materials.
56. What is the usual bending test for cast iron 1 State how it is made,
and what kind of results you would expect to obtain from average
material.
57. Describe the test you would employ for the purpose of ascertaining the
principal tensile strength properties of a strip of mild steel plate.
Make a diagrammatic sketch of a testing machine suitable for this
purpose with which you are acquainted. Give the results you would
expect from a good sample of structural steel, and sketch the load
strain diagram.
58. A round bar of mild steel is tested in tension, with the following
results : —
Original dimensions : 1*002 ins. diameter, 10 ins. long.
Final dimensions : 0'659 in. diameter, 13'06 ins. long.
Load at yield point, 13' 15 tons ; load, maximum, 21 '90 tons.
178 STRENGTH OF MATERIALS
From these data, find stress at yield point, maximum stress, per-
centage extension on ]0 ins., and percentage reduction in area.
[1670 tons per sq. ip. ; 27*80 tons per sq. in. ;
30'6 per cent. ; and 57 per cent.
59. Sketch the typical load strain diagram for a mild steel tension test
specimen, indicating the chief points on the curve.
60. State clearly how you would set about finding out whether a given
sample of Portland cement was good or bad. How would you
expect the properties of two samples of the same cement to differ,
one of which was coarsely ground and the other ground extremely
fine?
61. In two samples of steel one is found to contain 0*2 per cent, of carbon
and the other 0'5 per cent. Apart from the effects of any other
constituents, what differences would you expect to find in the
strength properties of these two ?
62. State what you know as to the percentage of carbon in steel, and its
bearing on the strength properties of different steels.
63. A cast iron beam, 1/90 ins. deep, 0'98 in. wide, when tested on a span
of 36 ins., is found to break under a central load of 2500 Ibs. If the
corresponding dimensions had been 2 ins. and 1 in., what would you
expect to have been the breaking load 1 State whether you consider
this to be a good result, and how deflection would have been shown
before fracture. Do you consider this to be a good test as compared
with a tensile test, and, if so, why ? [2830.
INDEX
ALLOYS of copper, 157
Angle of twist, 102
BEAMS, deflection of, 47
Beams, shearing in, 69
Beams, stresses in, 25
Bending and torsion, 110
Bricks, strength of, 158
Brickwork, strength of, 159
CAST iron, 155
Cement, Portland, 159
Columns, 92
Concrete, 160
Continuous beams, 73
Contraction and dilatation, 14
Copper, 156
Cubic elasticity, 22
Cylinders, thin, 115
Cylinders, thick, 116
DEFLECTION of beams, 47
Deflection due to shear, 72
Deformation, measurement of, 136
Distribution of stress, 4
ECCENTRIC loading, 78, 146
Efficiency of riveted joints, 129
Elasticity, direct, 6
Elasticity of volume, 22
Extensometer, Martens, 136
GRAPHICAL determination of I, 39
Gunmetal, 157
HARD steel, 154
Helical springs, 104
High-carbon steel, 154
179
Hooke's Law, 6
Horse-power transmitted by shafts,
103
INERTIA, moment of, 32
Intensity of stress, 1
Iron, cast, 155
Iron, malleable cast, 156
Iron, wrought, 150
LATERAL contraction, 14
Limit of proportionality, 138
Limits of elasticity, 138
Load, 1
Loading, unsymmetrical, 78, 146
Low-carbon steel, 152
MALLEABLE cast iron, 156
Martens mirror extensometer,
136
Medium steel, 153
Microstructure of metals, 165
Mild steel, 152
Modulus of elasticity, 6
Modulus of rigidity, 21
Moment of inertia, 32
Moment of inertia found graphi-
cally, 39
Moment of resistance, 27
Mortar, strength of, 159
OBLIQUE stresses, 15
PILLARS, 92
Plasticity, 5
Portland cement, 159
Piston-rod steel, 164
Poisson's Ratio, 14
180
Previous loading, effect of, 147
Proportionality, limit of, 138
REPEATED stresses, 162
Reversed stresses, 162
Riveted joints, 125
SAFE stresses, 165
Shafts, 99
Shear stress in beams, 69
Springs, 104
Steel, mild, 152
Steel, medium, 153
Steel, high-carbon, 154
Stone, 158
Struts, 92
INDEX
TESTING, 131
Thick cylinders, 116
Torsion, 99
Torsion, combined with bending,
110
UNSYMMETRICAL loading, 78, 146
YIELD point, 140
Young's Modulus, 6
VITREOUS materials, 1 58
WROUGHT iron, 1 50
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