Simpl

1-1 INTRODUCTION

Strength of Materials extends the study of forces that was begun in

Engineering Mechanics, but there is a sharp distinction between the two
subjects. Fundamentally, the field of mechanics covers the relations
between forces acting on rigid bodies; in statics, the bodies are in
equilibrium, whereas in dynamics, they are accelerated but can be put in
equilibrium by applying correctly placed inertia forces.
In contrast to mechanics, strength of materials deals with the
relations between externally applied loads and their internal effects on
bodies. Moreover, the bodies are no longer assumed to be ideally rigid;
the deformations, however small, are of major interest. The properties of
the material of which a structure or machine is made affect both its
choice and the dimensions that will satisfy the requirements of strength
and rigidity.
The difference between mechanics and strength of materials can
be further emphasized by the following example. It is a simple problem
in statics to determine the force required at the end of a crowbar to pry
up a given load (Fig. 1-1). A moment summation about the fulcrum
determines P. This statics solution assumes the crowbar to be both rigid
enough and strong enough to permit the desired action. In strength of
 a

Simple Stress

Figure 1-1. Crowbar must neither break nor bend excessively.

materials, however, the solution must extend further. We must investi-

gate the bar itself to be sure that it will neither break nor be so flexible
that bends without lifting the load.
it

Throughout this book we shall study the principles that govern

these two fundamental concepts of strength and rigidity. In this first
chapter we start with simple axial loadings; later we consider twisting
loads and bending loads; and finally we discuss simultaneous combina-
tions of these three basic types of loadings.

1-2 ANALYSIS OF INTERNAL FORCES

Consider a body of any shape acted upon by the forces shown in Fig.
1-2. In engineering mechanics, we would startby determining the
resultant of the applied forces to determine whether or not the body
remains at rest. If the resultant is zero, we have static equilibrium —
condition generally prevailing in structures. If the resultant is not zero,
we may apply inertia forces to bring about dynamic equilibrium. Such
cases are discussed later under dynamic loading. For the present we
shall consider only cases involving static equilibrium.
In strength of materials, we make an additional investigation of
the internal distribution of the forces. This is done by passing an

exploratory section a -a through the body and exposing the internal

Figure 1-2. Exploratory section a-a through loaded member.

1 -2 Analysis of Internal Forces

Figure 1-3. Components of internal effects on exploratory section a-a.

forces acting on the exploratory section that are necessary to maintain

equilibrium of a free-body diagram of either segment. In general, the
internal forces reduce to a force and a couple which, for convenience,
are resolved into components normal and tangent to the section, as
shown in Fig. 1-3.
The origin of the reference axes is always at the centroid which is
the key reference point of the section. Although we are not yet ready to
show why this is so, we shall prove it as we progress; in particular, we
shall prove it for normal forces in the next article. If the X axis is
normal to the section, the section is known as the X surface or, more
briefly, the X face. The orientation of the Y and Z axes in the plane of
the section is usually chosen to coincide with the principal axes of
inertia.

The notation used 1-3 identifies both the exploratory

in Fig.
section and the direction moment component. The first
of the force or
subscript denotes the face on which the component acts; the second
subscript indicates the direction of the particular component. Thus P
is the force on the X face acting in the Y direction.
Each component reflects a different effect of the applied loads on
the member and is given a special name, as follows:

Pxx Axial force. This component measures the pulling (or

pushing) action over the section. A pull represents a
tensile force which tends to elongate the member,
whereas a push is a compressive force which tends to
shorten it. It is often denoted by P.

Pxv>
xyi Pxz Shear force. These are components of the total resis-
tance to sliding the portion to one side of the explora-
tory section past the other. The resultant shear force is
usually designated by V, and its components by Vy and
Vz to identify their directions.
 1 Simple Stress

A/. Torque. This component measures the resistance to

twisting the member and is commonly given the
symbol T.

M^M X
Bending moments. These components measure the resis-

tance to bending the member about the Y ox Z axes

and are often denoted merely by My or M z
.

From the preceding discussion, it is evident that the internal effect

of a given loading depends upon the selection and orientation of the
exploratory section. In particular, if the loading acts in one plane, say
the XY plane as is frequently the case, the six components in Fig. 1-3
reduce to only three, viz., the axial force Pxx (or P), the shear force P
(or K), and the bending moment M
xz (or M
). Then, as shown in Fig.

l-4a, these components are equivalent to the single resultant force R. A

little reflection show that if the exploratory section had been
will
oriented differently, like b-b in Fig. l-4b where it is perpendicular to
/?, the shearing effect on the section would reduce to zero and the

tensile effect would be at a maximum.

The purpose of studying strength of materials is to ensure that the
structures used will be safe against the maximum internal effects that
may be produced by any combination of loading. We shall learn as our
study proceeds that it is not always possible or convenient to select an
exploratory section that is perpendicular to the resultant load; instead,
we may have to start by analyzing the effects acting on a section like
a-a in Figs. 1-2 and l-4a, and then learn how these effects combine to

^— Normal
L component

component

(a) Normal and shear (b) When exploratory section b-b is

components on arbitrary perpendicular to resultant R
section a-a. of applied loads, only normal
forces are produced.
Figure 1-4.
1 -3 Simple Stress 5

produce maximum internal effects like those on section b-b in Fig.

l-4b. This procedure we Chapter 9, which deals with
shall study later in
combined stresses. For the present, we restrict our study to conditions
of loading in which the section of maximum internal effect is evident by
inspection.

1-3 SIMPLE STRESS

One of the basic problems of the engineer is to select the proper

material and correctly use and proportion it so as to enable a structure
or machine to do most efficiently what it is designed to do. For this
purpose, it is essential to determine the strength, stiffness, and other
properties of materials. A tabulation of the average properties of com-
mon metals is given in Appendix B, Table B-l, on page 634.
Let us consider two bars of equal length but different materials,
suspended from a common support as shown in Fig. 1-5. If we knew
nothing about the bars except that they could support the indicated
maximum loads [500 N (Newtons) for bar 1 and 5000 N for bar 2], we
could not tell which material is stronger. Of course, bar 2 supports a
greater load, but we cannot compare strengths without having a com-
mon basis of comparison. In this instance, the cross-sectional areas are
needed. So let us further specify that bar 1 has a cross-sectional area of
10 mm and bar 2 has an area of 1000
2 2
mm
Now it is simple to .

compare their strengths by reducing the data to load capacity per unit
area. Here we note that the unit strength of bar 1 is

= SOON SOON
a, =50xl()6N/m2
10 mm2 10 X 10" 6 m 2

and bar 2 has a unit strength

= 5000 N 5000 N = 5 X 10
6
N/m 2
a,
1000 mm lOOOxKT'm2
2

Thus the material of bar 1 is ten times as strong as the material of bar 2.

Barl Bar 2

500 N 5000 N
Figure 1-5. Bars supporting maximum loads.
6 1 Simple Stress

The unit strength of a material is usually defined as the stress* in

the material. Stress is expressed symbolically as

-1 »-»
where o (Greek lowercase letter sigma) is the stress or force per unit
area, P is the applied load, and A is the cross-sectional area. Observe
that maximum stress in tension or compression occurs over a section
normal to the load, as indicated in Fig. l-4b. Shearing stress is dis-
cussed in the next section.
From can be seen that the units for stress are the units
Eq. (1-1) it

of force divided by the units of area. In SI (which is the official

abbreviation for the international system of units, Le Systeme Interna-
tional d'Unites), force is measured in newtons (N) and area is measured
in square meters (m 2 ). Thus the units for stress are newtons per square
meter (N/m 2 ). Frequently, one newton per square meter is referred to
as one pascal (Pa). Since the prefix M (read as "mega") refers to
6
multiples of 10 in SI, in the above example, the stress in bar 1 may be
expressed as 50 MN/m 2
(or 50 MPa) and that in bar 2 as 5 MN/m2 (or
5 MPa).
Even as simple an expression as Eq. (1-1) requires careful discus-
sion. Dividing load by area does not give the stress at all points in the
cross-sectional area; it merely determines the average stress. A more
exact definition of stress is obtained by dividing the differential load dP
by the differential area over which it acts:

°= § d- la>
Next let us see under what conditions a = P/A will accurately
define the stress at all points of the cross section. The condition under
which the stress is constant or uniform is known as simple stress. We
shall show now that a uniform stress distribution can exist only if the
resultant of the applied loads passes through the centroid of the cross
section/
Suppose that a cutting plane isolates the lower half of one of the
bars in Fig. 1-5. Then, as shown in Fig. 1-6, the resisting forces over
the cut section must balance the applied load P. A typical resisting force
is dP. Applying the conditions of equilibrium, we obtain

[2Z =0] P = fdP= fodA

[SA/V = 0] Pb = fx dP = fx(o dA)
•Some engineers use the terms stress or total stress as synonymous with
load or force, and unit stress or stress intensity when referring to the intensity of
load per unit area. In this book, stress will always denote force per unit area.
There are certain exceptions to this rule; they are caused by stress con-
centration (see p. 512) and by abrupt changes in cross section, and at points in
the vicinity of the applied loads (see p. 7).
1-3 Simple Stress

Figure 1-6. For uniform stress, P must pass through the centroid C.

If we specify that the stress distribution is to be constant over the

cut section, o may be written outside the integrals in the above equa-
tions to obtain

P= a f dA = oA
and therefore,

Pb = (oA)b = ofxdA
Then, canceling the common factor a, we obtain
, fxdA _
b = -A~= X
from which the coordinate b of the point C
is recognized as being the x

coordinate of the centroid of the section. By taking a moment summa-

tion about the X axis, we could similarly show that y defines the v
coordinate of C We conclude that a uniform stress distribution is

obtained only when the resultant of the applied loads passes through the
centroid of that surface.
It does not follow, however, that positioning the load through the
centroid of the section always results in a uniform stress distribution.
For example, in Fig. 1-7 is shown the profile of a flat bar of constant
thickness. The load P is applied at the center line of the bar. At sections
b-b and /-/, the stress distribution is uniform and illustrates the

principle discussed earlier; but at the other indicated sections the

stresses are not uniform.
At section e-e, the stress distribution is not uniform because the
line of action of P obviously does not pass through the centroid of the
Nor are the stresses uniformly distributed all across sections c-c
section.
and d-d because, although the action line of P does pass through the
 1 Simple Stress

Figure 1-7. Exceptions to uniform stress distribution occur at sections a-a, c-c,
d-d, and e-e.

centroids of these sections, here there are abrupt changes in section. At

such sections, the stresses are usually highly localized and can be
determined only by the mathematical theory of elasticity or some
experimental method such as photoelasticity. Also, the stress is not
uniform across section a-a because here the section is too close to the
point where the load is applied. Unless a section is located at a distance
from the end of the rod at least equal to the minimum width of the rod,
we will not obtain a uniform stress distribution.*
In order to visualize why sections c-c, d-d, and a-a do not have
uniform stress, imagine that the applied force P produces stress lines
which radiate out from the load and distribute themselves throughout
the body as indicated by the dashed lines in the figure. Although this
concept is not actually correct, it does indicate the existence of stress
concentration wherever the shape of the body interferes with the "free
flow" of the stress lines. The bunching of these lines about the hole in
section c-c, and around the sharp corner of section d-d, which indi-
cates stress concentration, contrasts with the relatively smooth flow of
stress around the radius between sections e-e and /-/.

•See S. Timoshenko and J. N. Goodier, Theory of Elasticity, 2nd ed.,

McGraw-Hill. New York, 1951, p. 33.
1 -3 Simple Stress

ILLUSTRATIVE PROBLEMS

101. An aluminum tube is rigidly fastened between a bronze rod

and a steel rod as shown in Fig. l-8a. Axial loads are applied at the
positions indicated. Determine the stress in each material.

Aluminum
Steel
Bronze = 1000
A mm2
A = 700 mm 2 A = 800 mm 2
20 kN 10 kN
15 kN 15 kN

500 mm-4" — 600 mm- 700 mm

(a)

20 kN

20 kN 15 kN ^

20 kN 15 kN 15 kN f-
(b)

Figure 1-8.

Solution: To calculate the stress in each section, we must first de-

termine the axial load in each section. The appropriate free-body
diagrams are shown in Fig. l-8b, from which we determine the axial
load in each section to be Pb — 20 kN (compression), Pa = 5 kN
(compression), and P5 — 10 kN (tension). The stresses in each section
are

a — Ol = 20 kN 20 x 10
3
N
A 700 mm 2
700 X 10" 6 m 2

= 28.6 X 10
6
N/m 2 = 28.6 MPa Ans.

o„ = 5 kN 5 X 10
3
N
1000 mm 2
1000 X 10~ 6 m 2

=5 X 10
6
N/m = 5 MPa 2
Ans.

= 10 kN 10 X 10*N
a,
800 mm 2
800 X 10~ 6 m 2

= 12.5 x 10
6
N/m = 2
12.5 MPa Ans.
10 1 Simple Stress

The stresses in the bronze and aluminum sections are compressive,

whereas the stress in the steel section is tensile.
Note that neither the lengths of the sections nor the materials from
which the sections are made affect the calculation of the stresses.
As you can see from this example, the first step in calculating the
stress in a member is to determine the internal force carried by the
member. This determination is accomplished by the analysis of correctly
drawn free-body diagrams. Note that in this example, it would have
been easier to determine the load in the steel section by taking the
section lying to the right of the exploratory section in the steel.

102. For the truss shown in Fig. l-9a, determine the stress in
members AC and BD. The cross-sectional area of each member is 900
mm 2
.

Solution: The three assumptions used in the elementary analysis of

trusses are as follows:

1. Weights of the members are neglected.

2. All connections are smooth pins.

3. All external loads are applied directly to the pin joints.

Using the above three assumptions, the members of the truss may be
analyzed as two-force members —
the internal force system carried by
any member reduces to simply a single force (tension or compression)
acting along the line of the member.
The free-body diagram of the entire truss is shown in Fig. l-9a.
An equilibrium analysis of this free-body diagram results in the follow-
ing values for the external reactions: Ay = 40 kN, Hy = 60 kN, and
Hx =0.
To determine the force in member AC, we pass an imaginary
cutting plane which isolates joint A (section ©, Fig. l-9a). The free-
body diagram of joint A is shown in Fig. l-9b. Here, AB and AC
represent the forces in members AB and AC, respectively. Note that
both members have been assumed to be in tension. Analyzing the
free-body diagram in Fig. l-9b,

[2 Y - 0] QA y
+\AB =
AB= -\Ay = -
-f(40) = -66.7 kN
[2X - 0] © AC +\AB =
AC = -\AB = -|(-66.7) = 53.4 kN

The minus sign indicates that the 66.7 kN force in member AB is

1-3 Simple Stress 11

® D

4 panels at 4 m = 16 m

(a)

£D

±E
*~AC

(b)

Figure 1-9.

compressive. The member AC is 53.4 kN, tension.

force in
To determine the force in member BD, we pass an imaginary
cutting plane which exposes the force in member BD (section ©, Fig.
l-9a). The free-body diagram of the portion of the truss to the left of
section © is shown in Fig. 1 -9c. (The portion of the truss to the right of

section © could also have been used.) The forces in members BD, BE,
and CE are assumed to be tensile. To calculate the force BD, we
eliminate the forces BE and CE by taking a moment summation about
their point of intersection, E, and write

[2ME = 0] -4,(8) + 30(4) -*Z)(4)-0

4BD m -SAy + 120 8(40) + 120
= -200
=
BD -50 kN

Therefore, the force in member BD is 50 kN, compression.

12 1 Simple Stress

The stresses in members AC and BD, are

— ~ 53.4 kN 53.4 X 10
3
N
a °ac
900 mm 2
900 X 10" 6 m2
- 59.3 x 10
6
N/m = 2
59.3 MPa Arts.
(tension)

" 50 kN 50 x 10
3
N
°bd
900 mm 2
900 X 10" 6
m2
= 55.6 X 10
6
N/m 2 = 55.6 MPa Ans.
(compression)

In truss analysis, the method of analyzing a single joint, as shown

in Fig. 1 —9b, is method of joints. The analysis of a
referred to as the
section of the truss composed of two or more joints, as shown in Fig.
l-9c, is called the method of sections. It must be reemphasized that the
force internal to a member of a truss lies along the line of the member
only because sufficient assumptions are made which reduce all members
to two-force members. As discussed in Art. 1-2, the internal forces for
an arbitrarily loaded member are considerably more complicated than
simply an axial force.

PROBLEMS
103. Determine the largest weight which can be supported by W
the two wires shown in Fig. P-103. The stresses in wires AB and AC are
not to exceed 100 MPa and 150 MPa, respectively. The cross-sectional
areas of the two wires are 400 mm 2
for wire AB and 200 mm2 for wire
AC.
Ans. W = 33.5 kN

Figure P-103.

104. For the truss shown in Fig. P-104, calculate the stresses in
members DF % C£, and BD. The cross-sectional area of each member is
1200 mm 2
. Indicate tension (T) or compression (C).
Ans. DF = 188 MPa (C); CE = 1 13 MPa (T);
BD = 80.1 MPa(C)
1-3 Simple Stress 13

/7V7

100 kN 200 kN
Figure P-104.

105. For the truss shown in Fig. P-105, determine the cross-sec-
tional areas of bars BE, BF, and CF so that the stresses will not exceed
100 MN/m 2
in tension or 80 MN/m 2
in compression. A reduced stress
in compression is specified to avoid the danger of buckling.
Arts. A BE = 625 mm2 ; A BF = 427 mm 2
; A CF = 656 mm 2
14 1 Simple Stress

106. The bars of the pin-connected frame in Fig. P-106 are each
30 mm by 60 mm in section. Determine the maximum load P that can
be applied without exceeding the allowable stresses specified in Problem
105.

Figure P-106.

107. A cast-iron column supports an axial compressive load of

250 kN. Determine the inside diameter of the column if its outside
diameter is 200 mm
and the limiting compressive stress is 50 MPa.
108. Determine the outside diameter of a hollow steel tube that
will carry a tensile load of 500 kN at a stress of 140 MN/m 2
. Assume
the wall thickness to be one-tenth of the outside diameter.
Arts. D — 107 mm
109. Part of the landing gear for a light plane is shown in Fig.
P-109. Determine the compressive stress in the strut AB caused by a
landing reaction R = 20 kN. Strut AB is inclined at 53.1° with BC.
Neglect weights of the members. Ans. a = 65.7 MN/m 2

Hollow strut
OD - 40 mm
ID 30 mm

1 200 mm-*
Flgure P-109.
450 mm-

110. A steel tube is rigidly attached between an aluminum rod

and a bronze rod as shown in Fig. P-l 10. Axial loads are applied at the
positions indicated. Find the maximum value of P that will not exceed a
1 -3 Simple Stress 15

stress in aluminum of 80 MPa, in steel of 150 MPa, or in bronze of 100

MPa.

Steel
Bronze
Aluminum A = 500 mm 2
A = 400 mm 2
A = 200 mm 2

k lm 2m
Figure P-110.
2.5 m

A
homogeneous 150-kgj bar AB carries a 2-kN force as
111.
shown P— 111. The bar is supported by a pin at B and a
in Fig.
10-mm-diameter cable CD. Determine the stress in the cable.

4 m

2kN
Figure P-111.

112. Determine the weight of the heaviest cylinder which can be

placed in the position shown in Fig. P— 112 without exceeding a stress of
50 MN/m2 in the cable BC. Neglect the weight of bar AB. The
cross-sectional area of cable BC is 100 mm2 .

6 m

Figure P-112.
16 1 Simple Stress

113. A 1000-kg homogeneous bar AB is suspended from two

cables AC and BD y each with cross-sectional area 400 mm 2
, as shown in
Fig. P— 1 Determine the magnitude P and location x of the largest
13.

additional force which can be applied to the bar. The stresses in the
cables AC and BD are limited to 100 MPa and 50 MPa, respectively.
Arts. P - 50.2 kN; x = 0.602 m

1.8 m 1.8 m

Figure P-113.

1-4 SHEARING STRESS

Shearing stress differs from both tensile and compressive stress in that it

is caused by forces acting along or parallel to the area resisting the

forces, whereas tensile and compressive stresses are caused by forces
perpendicular to the areas on which they act. For this reason, tensile
and compressive stresses are frequently called normal stresses, whereas a
shearing stress may be called a tangential stress.
A shearing stress is produced whenever the applied loads cause
one section of a body to tend to slide past its adjacent section. Several
examples are shown in Fig. 1-10. In (a) the rivet resists shear across its
cross-sectional area, whereas in the clevis at (b) the bolt resists shear
across two cross-sectional areas; case (a) may be called single shear and
case (b) double shear. In (c) a circular slug about to be punched out of
is

a plate; the resisting area is similar to the milled edge of a coin. In each
case, the shear occurs over an area parallel to the applied load. This
may be called direct shear in contrast to the induced shear that may
occur over sections inclined with the resultant load, as was illustrated in
Fig. l-4a.
1 -4 Shearing Stress 17

-f>-

^-
T7
(a) (b) h (c)

Figure 1-10. Examples of shear.

The discussion concerning uniform normal stresses in the preced-

ing article might lead us to conclude also that a uniform shearing stress
will exist when the resultant shearing force V passes through the
centroid of the cross section being sheared. If this were true, the
shearing stress t (Greek lowercase letter tau) could be found from

t = (1-2)

Actually, the shearing stress across a section is practically never uni-

formly distributed (e.g., see Art. 5-7), so Eq. (1-2) must be interpreted
as giving merely the average shearing stress. This does not limit the
usefulness of Eq. (1-2) provided we use an average shearing stress that
takes into account the actual nonuniform distribution. Moreover, the
shearing stress distribution does approach uniformity when both
the distance between the applied shearing loads and the depth of the
shearing area are small. These are the conditions that prevail in Fig.
1-10 and in the following problems.

PROBLEMS

114. As to be punched out of a plate

in Fig. l-10c, a hole is

having an ultimate shearing 300 MPa. (a) If the compressive

stress of
stress in the punch is limited to 400 MPa, determine the maximum
thickness of plate from which a hole 100 mm
in diameter can be
punched, (b) If the plate is 10 mm
thick, compute the smallest diameter
hole which can be punched. Arts, (a) / = 33.3 mm; (b) d — 30.0 mm
18 1 Simple Stress

115. The end chord of a timber truss is framed into the bottom
chord as shown in Fig. P— 1 15. Neglecting friction, (a) compute dimen-
sion b if the allowable shearing stress is 900 kPa; and (b) determine
dimension c so that the bearing stress does not exceed 7 MPa.
Ans. (a) b — 321 mm; (b) c = 41.2 mm

P=50kN

y^
4fe
i\ r°\
i.f-
1 ^ i

F gure P- 115.

116. In the landing gear described in Problem 109, the bolts at A

and B are in single shear and the one at C double shear. Compute
is in
the required diameter of these bolts if the allowable shearing stress is 50
MN/m 2
.

117. A 750-mm pulley, loaded as shown in Fig. P-117, is keyed

to a shaft of 50-mm diameter. Determine the width b of the 75-mm-long
key if the allowable shearing stress is 70 MPa. Ans. b = 1 1.4 mm

10 kN

\
10 mm ,y^Jb mm
r b -<-

6kN
Figure P-117.
1-4 Shearing Stress 19

118. The bell crank shown in Fig. P— 118 is in equilibrium, (a)

Determine the required diameter of the connecting rod A B if its axial
stress is limited to 100 MN/m
2
(b) Determine the shearing stress in the
.

pin at D if its diameter is 20 mm.

P-*

,j*.
/////////?/.

30 kN
Figure P-118.

119. The mass of the homogeneous bar AB shown in Fig. P-119

is2000 kg. The bar is supported by a pin at B and a smooth vertical
surface at A Determine the diameter of the smallest pin which can be
.

used at B if its shear stress is limited to 60 MPa. The detail of the pin
support at B is identical to that of the pin support at D shown in Fig.
P-118. Ans. d= 14.9 mm

S777

Figure P-119.

120. Two blocks of wood, 50 mm

wide and 20 mm
thick, are
glued together as shown in Fig. P-120. (a) Using the free-body diagram
concept illustrated in Fig. l-4a, determine the shear load and from it
20 1 Simple Stress

the shearing stress on the glued joint if P = 6000 N. (b) Generalize the
procedure of part (a) to show that the shearing stress on a plane inclined
at an angle to a transverse section of area A is r = P sin 20/2/1.

Figure P-120.

121. A rectangular piece of wood, 50 mm by 100 mm in cross

section, is used as a compression block as shown in Fig. P— 121.
Determine the maximum axial force Pwhich can be safely applied to
the block if the compressive stress in the wood is limited to 20 MN/m 2

and the shearing stress parallel to the grain is limited to 5 2

MN/m The .

grain makes an angle of 20° with the horizontal, as shown. (Hint: Use
the results of Problem 120.) Arts. P - 77.8 kN

100 mm
Figure P-121.

1-5 BEARING STRESS

Bearing stress differs from compressive stress in that the latter is the
internal stress caused by a compressive force whereas the former is a
contact pressure between separate bodies. Some examples of bearing
stress are the soil pressure beneath piers and the forces on bearing
plates. We now consider the contact pressures between an axle and its
bearing, or between a rivet or bolt and the contact surface of the plate
against which it pushes.
1 -5 Bearing Stress 21

pb
c©

Projected area of
rivet hole

Figure 1-11. Exaggerated bearing deformation of upper plate.

Pb'A b a b =(td)ab

In Fig. 1-11, the result of an excessive bearing stress is to cause

yielding of the plate or of the rivet, or both. The intensity with which
the rivet bears against the rivet hole is not constant, but actually varies
from zero at the edges of the hole to a maximum
back of the
directly in
rivet. The is avoided
difficulty inherent in a variable stress distribution
by the common practice of assuming the bearing stress ob to be
uniformly distributed over a reduced area which is the projected area of
the rivet hole. Then the bearing load is expressed by
Pb = * b °b = {«*)<** (1-3)

This result is analogous to that for a cylinder subjected to a

uniform internal pressure (see the next article, especially Fig. 1-14).
There, as we shall see, the net force is equal to the uniform pressure
multiplied by the projected area.

ILLUSTRATIVE PROBLEM

122. Figure 1-12 shows a W460 X 97 beam riveted to a W610 X

125 girder by two 100 X 90 X 10 -mm angles with 19-mm-diameter
Z3J

*££
^ , &***
li
V7777
2222-W610X 125 girder
web = 11.9 mm
W460 X 97 beam, web =11.4 mm
Figure 1-12. Strength of beam and girder connection.
22 1 Simple Stress

rivets. (Refer to Appendix B for the properties of structural sections.)

For the shop-driven rivets that attach the angles to the beam, assume
t = 80 MPa and ob = 170 MPa. For the field-driven rivets (riveted on
the job), assume r = 70 MPa and ob = 140 MPa. The web of the girder
is 11.9 mm thick, and the web of the beam is 1 1 .4 thick. Determine mm
the allowable end reaction.

Solution: At the girder, the shearing resistance is that of 8 field-driven

rivets in single shear; hence we have
2
[P -Ar] P- 8(-^)(19 x 10" 3
) (70 x 10
6
) = 159 kN
The bearing resistance at the girder depends on the minimum
thickness of the connection, which in this case is the 10-mm thickness of
the clip angle. We obtain for 8 field-driven rivets in bearing:

[P = Aob ] P= 8(19 X 10" 3 )(10 X 10" 3 )(140 X 10

6
)

= 213 kN
At the beam, there are 4 shop-driven rivets in double shear, giving
a total of 8 single-shear areas. With an allowable shearing stress of 80
MPa, this makes the shear resistance greater here than at the girder.
The bearing resistance at the beam depends on the web thickness
of the beam. Since this is smaller than the combined thickness of the
two clip angles, for the 4 rivets in bearing, we obtain
[P = Aob ] P- 4(19 X 10- 3 )(11.4 X 10" 3 )(170 X 10
6
)

« 147 kN
The safe beam reaction is the smallest of the above values, that is,
147 kN; it is limited by the bearing of the shop-driven rivets against the
W460 x 97 beam.

PROBLEMS

123. In Fig. 1-11, assume that a 20-mm-diameter rivet joins the

plates which are each 100 mm wide, (a) If the allowable stresses are 140
MN/m 2
for bearing in the plate material and 80 MN/m 2
for shearing
of the rivet, determine the minimum thickness of each plate, (b) Under
the conditions specified in part (a), what is the largest average tensile
stress in the plates. Ans. (a) 8.98 mm; (b) 35.0 MN/m 2

124. The shown in Fig. P-124 is fastened by three

lap joint
20-mm-diameter Assuming that P = 50 kN, determine (a) the
rivets.
shearing stress in each rivet, (b) the bearing stress in each plate, and (c)
the maximum average tensile stress in each plate. Assume that the
applied load P is distributed equally among the three rivets.
 —
1 -5 Bearing Stress 23

130 mm
o- <> -e

25 mm
_L_ 25 mm
C^ C^ <cx

—^y—o—C7 L^
Figures P-124 and P-125.

125.' For the lap joint in Problem 124, determine the maximum
safe load P which may be applied if the shearing stress in the rivets is
limited to 60 MPa, the bearing stress in the plates to 110 MPa, and the
average tensile stress in the plate to 140 MPa. Arts. 56.5 kN
126. In the clevis shown in Fig. 1 — 10b, on page 17, determine
the minimum bolt diameter and the minimum thickness of each yoke
that will support a load P= 55 kN without exceeding a shearing stress
of 70 MPa and a bearing stress of 140 MPa.
127. A
22.2-mm-diameter bolt having a diameter at the root of
the threads of 18.6 mm
is used to fasten two timbers as shown in Fig.

P-127. The nut is tightened to cause a tensile load in the bolt of 34 kN.
Determine (a) the shearing stress in the head of the bolt, (b) the shearing
stress in the threads, and (c) the outside diameter of the washers if their
inside diameter is 28 mm
and the bearing stress is limited to 6 MPa.

Figure P-127.
24 1 Simple Stress

128. Figure P-128 shows a roof truss and the detail of the
riveted connection at joint B. Using allowable stresses of t = 70 MPa
and ob = 140 MPa, how many 19-mm-diameter rivets are required to
fasten member BC to the gusset plate? Member BE1 What is the largest
average tensile or compressive stress in BC and BE1
Ans. For BC, 1 rivets; for BE, 5 rivets

129. Repeat Problem 128 if the rivet diameter is 22 mm and all

other data remain unchanged.

D
14 mm
gusset
B ^f F
6 m ^ plate

a/m
/\ 4
c
4 m
r
4 m
G
4
\ m /\
H

96 kN 200 kN 96 kN
75 X 75 X6 mm
(a)

Figures P-128 and P-129.

1-6 THIN-WALLED CYLINDERS

A under a pressure of p N/m 2 is

cylindrical tank carrying a gas or fluid
subjected to tensile forces which resist the bursting forces developed
across longitudinal and transverse sections. Consider first a typical
longitudinal section A- A through the pressure-loaded cylinder in Fig.
I — 13a. A free-body diagram of the half-cylinder isolated by the cutting

plane A -A is shown in Fig. 1 — 13b.

dF^pdA=pL(D2)dO
(a) (b)

Figure 1-13. Analytical determination of bursting force F.

1-6 Thin-Walled Cylinders 25

The elementary force acting normal to an element of the cylinder

located at an angle 9 from the horizontal diameter is

dF = pdA = pL^-dB

A similar force (not shown) acts on the symmetrically placed element on

the other side of the vertical center line. Since the horizontal compo-
nents of such pairs of forces cancel out, the bursting force F is the
summation of the vertical components of these elementary forces:

V
F = f{pL^ <#) sin = />Ly [ -cos B]
Q

which reduces to
F = pDL
It is apparent that the total bursting force F, acting normal to the
cutting plane A -A, is resisted by the equal forces P acting on each cut
surface of the cylinder wall. Applying a vertical summation of forces,
we obtain

[2 V= 0] F = pDL = IP — F^i^e- (1-4)

A simpler method of determining the bursting force F is indicated

in Fig. 1-14. Here the lower half of the cylinder is occupied by a fluid.
Since a fluid transmits pressure equally in all directions, the pressure
on the cylinder is the same as that in Fig. 1-13. From the
distribution
accompanying free-body diagram, it is apparent that the bursting force
Fy acting over the flat surface of the fluid, equals the pressure intensity/?
multiplied by the area DL over which it acts, or F = pDL as before.
The stress in the longitudinal section that resists the bursting force
F is obtained by dividing it by the area of the two cut surfaces. This
gives

°=A
F _pDL = pD
2tL It
_12^T^ „ ,.

(a) (b)

Figure 1-14. Direct evaluation of bursting force F.

26 1 Simple Stress

This stress is usually called the tangential stress because it acts tangent
to the surface of the cylinder; other common names are circumferential
stress, hoop stress, and girth stress. The stress computed by Eq. (1-5) is

the average stress; for cylinders having a wall thickness equal to 1/10 or
less of the inner radius, it is practically equal to the maximum stress at
the inside surface. (See Art. 13-11 for the stress distribution in thick-
walled cylinders.)
If we consider next a free-body diagram of a transverse section
(Fig. 1-15), we see that the bursting force acting over the end of the
cylinder is resisted by the resultant P of the tearing forces acting over
the transverse section. The area of a transverse section is the wall
thickness multiplied by the mean circumference, or tt(D + t)t; if / is

small compared to D, it is closely approximated by irDt. Thus we obtain

2
7TD
[r-r\ irDtOj = p

or

a, - S±L _ 3T*^> £^ fc^^ (1_6)

where a, denotes what is called the longitudinal stress because it acts

parallel to the longitudinal axis of the cylinder.

F - nD 2
P

"P^(nDt)(7 l

Figure 1-15. Bursting force on a transverse section.

Comparing Eqs. (1-5) and (1-6) shows that the longitudinal stress
is one-half the value of the tangential stress. In effect, this is equivalent
to stating that, if the pressure in a cylinder is raised to the bursting
point, failure will occur along a longitudinal section or longitudinal
seam of the cylinder. When a cylindrical tank is composed of two sheets
riveted together, as in Fig. 1-16, the strength of the longitudinal joint
should be twice the strength of the girth joint. In other words if, as is

often the case, the longitudinal joint is not twice as strong as the girth
1-6 Thin-Walled Cylinders 27

Longitudinal joint

Girth joint

Figure 1-16.

joint, the permissible internal pressure will depend on the strength of the
longitudinal joint.
Equations (1-5) and (1-6) have been developed primarily to
determine the relation stated in the above paragraph, not as equations to
be memorized. It is generally best to compute the stresses by determining
the resisting load P from a free-body diagram and then computing the
stress by using a — P/A.
For this purpose, Fig. 1-14 is replaced by the
equivalent skeleton diagram in Fig. 1-17, which also establishes the
relation IP = pDL.
F= P DL

Figure 1-17.

When the ends of the cylinder are not squared off as in Fig. 1-15,
but are rounded or dished as in Fig. 1-18; the bursting force on a
transverse section may still be computed as the product of the internal
pressure multiplied by the projected area of the transverse section. Thus,
using the concept discussed in connection with Fig. 1-14, we may
imagine the volume between the transverse section A -A and the

Dished or
concave end

Rounded or
convex end
Figure 1-18.
28 1 Simple Stress

rounded end in Fig. 1-18 to be full of a fluid. The resultant longitudinal

force will equal the product of the pressure intensity multiplied by the
shaded area of the transverse section.
As another application of the concept of a fluid to transmit
pressure, consider a pump chamber cast in several parts, with projecting
flanges that are bolted together as shown in Fig. 1-19. The bursting
force to be resisted by the bolts in section A -A is proportional to the
cross-sectional area at A -A and is expressed by F, = p(ttD ]
2
/4); simi-
larly, the bursting force resisted by the bolts in section B-B is F2 =
2
P (ttD 2 /4).

Figure 1-19.

The principles just discussed for determining the tangential stress

in thin-walled cylinders may also be applied to computing the contact
pressure exerted by hoops shrunk upon cylinders or the tensile stress
developed in a thin rotating ring. In the latter case, for example, the
bursting force is generated by the centrifugal force developed in one-
may be obtained (Fig. 1-20) by assuming the
half of the ring. Its value
mass of the half-ring concentrated at its center of gravity, whence we
have

F— mroi
1
(a)

in which <o is the angular velocity in radians per second and m is the

» F=m\

Figure 1-20. Free-body diagram of one-half of rotating ring.

1-6 Thin-Walled Cylinders 29

mass of one-half of the ring. For a thin ring, m is given by

m = pV = pnrArc

where p (Greek lowercase letter rho) is the mass per unit volume of the
ring, A is the cross-sectional area of the ring, and rc is the radius of the
mean circumference. From mechanics, the value of f for a semicircular
ring is r = 2rc / m. Substituting these values reduces Eq. (a) to
'

F= (pAwre )l ^L 2
= 2pAv 2 (b)

where v = rc co is the peripheral velocity of the ring.

From equilibrium of the free-body diagram in Fig. 1-20 we have

IP = F
Hence the stress is

P pAv 2 2 , v

Thus mass density and the square of

the stress varies directly with the
must be taken to use
the peripheral velocity. In applying Eq. (c), care
consistent units. With p in kilograms per cubic meter and v in meters
per second, o will be determined in newtons per square meter.

ILLUSTRATIVE PROBLEM

130. A large pipe, called a penstock in hydraulic work, is 1.5 m

in diameter. Here it is composed of wooden staves bound together by
steel hoops, each 300 in cross-sectional mm 2
is used to area, and
conduct water from a reservoir to a powerhouse. If the maximum tensile
stress permitted in the hoops is 130 MPa, what is the maximum spacing
between hoops under a head of water of 30 m? (The mass density of
water is 1000 kg/m 3 .)

Solution: The pressure corresponding to a head of water of 30 m is

given by

[p = pgh] p = (1000 kg/m )(9.81 m/s )(30 m)

3 2

= 294 x 103 N/m2 - 294 kPa

If the maximum spacing between hoops is denoted by L, then, as
shown in Fig. 1-21, each hoop must resist the bursting force on the
length L. Since the tensile force in a hoop is given by P= Ao, we
obtain from the free-body diagram

[pDL = IP]
(294 X 3
10 )(1.5)L = 2(300 X 10" 6 )(130 X 10
6
)

whence
L 0.177 m= 177 mm Arts.
30 1 Simple Stress

r P = A(T

P = A<T
<

k*,-+ +-L
Figure 1-21. Spacing of hoops in a penstock.

PROBLEMS
131. Show that the stress in a thin-walled spherical shell of
diameter D
and wall thickness / subjected to internal pressure p is given
by a « pD/At.
132. A cylindrical pressure vessel is fabricated from steel plates
which have a thickness of 20 mm. The diameter of the pressure vessel is
500 mm and its length is 3 m. Determine the maximum internal pressure
which can be applied if the stress in the steel is limited to 140 MPa. If
the internal pressure were increased until the vessel burst, sketch the
type of fracture which would occur. Arts. 1 1.2 MPA

133. Find the limiting peripheral velocity of a rotating steel ring

if the allowable stress is 140 MN/m 2
and the mass density of steel is
will the stress reach 200 MN/m
2
7850 kg/m 3 At what . angular velocity
if the mean radius is 250 mm? Ans. 134 m/s; 640 rad/s

134. A water tank is 8 m in diameter and 12 m high. If the tank

is to be completely filled, determine the minimum thickness of the tank
plating if the stress is limited to 40 MPa. Ans. 1 1.8 mm
135. The strength per meter of the longitudinal joint in Fig. 1-16
is 480 kN, whereas for the girth joint it is 200 kN. Determine the
maximum diameter of the cylindrical tank if the internal pressure is 1.5

MN/m 2
.

136. A pipe carrying steam at 3.5 MPa has an outside diameter

of 450 mm
and a wall thickness of 10 mm. A gasket is inserted between
the flange at one end of the pipe and a flat plate used to cap the end.
How many 40-mm diameter bolts must be used to hold the cap on if the
allowable stress in the bolts is 80 MPa, of which 55 MPa is the initial
stress? What circumferential stress is developed in the pipe? Why is it
necessary to tighten the bolts initially, and what will happen if the steam
1-6 Thin-Walled Cylinders 31

pressure should cause the stress in the bolts to be twice the value of the
initial stress? Ans. 17 bolts; 75.3 MPa
137. A spiral-riveted penstock 1.5 m in diameter is made of steel
plate 10 mm thick. The pitch of the spiral or helix m. The
is 3 spiral
seam is a single-riveted lap joint consisting of 20-mm-diameter rivets.

Using r = 70 MPa and ob = 140 MPa, determine the spacing of the

rivets along the seam for a water pressure of 1.25 MPa. Neglect end
thrust. What is the circumferential stress?
Ans. 43.7 mm; 93.8 MPa
Repeat Problem 137, using a 2-m-diameter penstock
138.
fastened with 30-mm-diameter rivets, with all other data remaining
unchanged.
139. The tank shown in Fig. P-139 is fabricated from 10-mm
steel plate. Determine the maximum longitudinal and circumferential
stresses caused by an internal pressure of 1.2 MPa.
Ans. 17.9 MPa; 60 MPa

140. The tank shown in Fig. P-140 is fabricated from steel plate.
Determine the minimum thickness of plate which may be used if the
stress is limited to 40
2
MN/m
and the internal pressure is 1.5 MN/m2 .

400 mm

Figures P-139 and P-140.

SUMMARY
Axial loads result in a uniform stress distribution that may be
determined from

a= AP (1-D

Shearing stresses and bearing stresses are also computed by divid-

ing the load by the resisting area, but the results represent average
values. In particular, the bearing area of a rivet against a plate is given
by the projected area of the rivet hole.
32 1 Simple Stress

The stresses in thin-walled cylinders subjected to internal pressure

are most readily obtained by applying the conditions of equilibrium to a

free-body diagram of either a longitudinal or a transverse section,
depending on whether circumferential or longitudinal stress is involved.
Ihe resisting forces thereby exposed are assumed to be uniformly
distributed over the corresponding cut surfaces.