The bolts used for the connections of this steel framework ...

wafflecanadianMechanics

Jul 18, 2012 (2 years and 3 months ago)

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The bolts used for the connections of this steel framework are subjected to stress. In
this chapter we will discuss how engineers design these connections and their
fasteners.
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Stress
CHAPTER OBJECTIVES
In this chapter we will review some of the important principles of statics and
show how they are used to determine the internal resultant loadings in a body.
Afterwards the concepts of normal and shear stress will be introduced, and
specific applications of the analysis and design of members subjected to an
axial load or direct shear will be discussed.
1.1 Introduction
Mechanics of materials is a branch of mechanics that studies the
relationships between the external loads applied to a deformable body
and the intensity of internal forces acting within the body. This subject
also involves computing the deformations of the body, and it provides a
study of the body’s stability when the body is subjected to external forces.
In the design of any structure or machine, it is first necessary to use the
principles of statics to determine the forces acting both on and within its
various members. The size of the members, their deflection, and their
stability depend not only on the internal loadings, but also on the type of
material from which the members are made. Consequently, an accurate
determination and fundamental understanding of material behavior will
be of vital importance for developing the necessary equations used in
mechanics of materials. Realize that many formulas and rules for design,
as defined in engineering codes and used in practice, are based on the
fundamentals of mechanics of materials, and for this reason an
understanding of the principles of this subject is very important.
3
1
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Historical Development.
The origin of mechanics of materials
dates back to the beginning of the seventeenth century, when Galileo
performed experiments to study the effects of loads on rods and beams
made of various materials. For a proper understanding, however, it was
necessary to establish accurate experimental descriptions of a material’s
mechanical properties.Methods for doing this were remarkably improved
at the beginning of the eighteenth century.At that time both experimental
and theoretical studies in this subject were undertaken primarily in
France by such notables as Saint-Venant, Poisson, Lamé, and Navier.
Becausetheireffortswerebasedonmaterial-bodyapplicationsof mechanics,
theycalledthis study“strengthof materials.”Currently,however,it is usually
referred to as “mechanics of deformable bodies” or simply “mechanics of
materials.”
Over the years,after many of the fundamental problems of mechanics
of materials had been solved,it became necessary to use advanced
mathematical and computer techniques to solve more complex problems.
As a result,this subject expandedintoother subjects of advancedmechan-
ics such as the theory of elasticity and the theory of plasticity.Research in
these fields is ongoing,not only to meet the demands for solving advanced
engineering problems,but to justify further use and the limitations upon
which the fundamental theory of mechanics of materials is based.
1.2 Equilibrium of a Deformable Body
Since statics plays an important role in both the development and
application of mechanics of materials, it is very important to have a good
grasp of its fundamentals. For this reason we will review some of the
main principles of statics that will be used throughout the text.
External Loads.
A body can be subjected to several different types
of external loads; however, any one of these can be classified as either a
surface force or a body force, Fig.1–1.
Surface Forces.
As the name implies,surface forces are caused by the
direct contact of one body with the surface of another.In all cases these
forces are distributed over the area of contact between the bodies.If this
area is small in comparison with the total surface area of the body,then
the surface force can be idealized as a single concentrated force,which is
applied to a point on the body.For example,the force of the ground on
the wheels of a bicycle can be considered as a concentrated force when
studying the loading on the bicycle.If the surface loading is applied along
a narrow area,the loading can be idealized as a linear distributed load,
Here the loading is measured as having an intensity of
along the area and is represented graphically by a series of arrows along
the line s.The resultant force of w(s) is equivalent to the area under
the distributed loading curve,and this resultant acts through the centroid
Cor geometric center of this area.The loading along the length of a beam
is a typical example of where this idealization is often applied.
F
R
force>lengthw1s2.
w(s)
Concentrated force
idealization
Linear distributed
load idealization
Surface
force
Body
force
s
C
G
F
R
W
Fig.1–1
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F
F
Type of connection Reaction
Cable
Roller
One unknown: F
One unknown: F
F
Smooth support One unknown: F
External pin
Internal pin
F
x
F
y
F
x
F
y
Two unknowns: F
x
, F
y
F
x
F
y
M
Fixed support Three unknowns: F
x
, F
y
, M
Two unknowns: F
x
, F
y
Type of connection Reaction
u
u
u
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Body Force.
A body force is developed when one body exerts a force
on another body without direct physical contact between the bodies.
Examples include the effects caused by the earth’s gravitation or its
electromagnetic field. Although body forces affect each of the particles
composing the body, these forces are normally represented by a single
concentrated force acting on the body. In the case of gravitation, this
force is called the weight of the body and acts through the body’s center
of gravity.
Support Reactions.
The surface forces that developat the supports
or points of contact between bodies are called reactions.For two-
dimensional problems,i.e.,bodies subjected to coplanar force systems,
the supports most commonly encountered are shown in Table 1–1.Note
carefully the symbol used to represent each support and the type of
reactions it exerts on its contacting member.In general,one can always
determine the type of support reaction by imagining the attached member
as being translated or rotated in a particular direction.If the support
prevents translationinagivendirection,thenaforce must be developedon
the member in that direction.Likewise,if rotation is prevented,a couple
moment must be exerted on the member.For example,a roller support can
only prevent translation in the contact direction,perpendicular or normal
to the surface.Hence,the roller exerts a normal force F on the member at
the point of contact.Since the member can freely rotate about the roller,a
couple moment cannot be developed on the member.
TABLE 1–1
Many machine elements are pin connected
in order to enable free rotation at their
connections. These supports exert a force
on a member, but no moment.
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Equations of Equilibrium.
Equilibrium of a body requires both
a balance of forces, to prevent the body from translating or having
accelerated motion along a straight or curved path, and a balance of
moments, to prevent the body from rotating. These conditions can be
expressed mathematically by the two vector equations
(1–1)
Here, represents the sum of all the forces acting on the body, and
is the sum of the moments of all the forces about any point O
either on or off the body. If an x,y,z coordinate system is established
with the origin at point O, the force and moment vectors can be resolved
into components along the coordinate axes and the above two equations
can be written in scalar form as six equations, namely,
(1–2)
Often in engineering practice the loading on a body can be
represented as a system of coplanar forces. If this is the case, and the
forces lie in the x–y plane, then the conditions for equilibrium of the
body can be specified by only three scalar equilibrium equations; that is,
(1–3)
In this case, if point O is the origin of coordinates, then moments will
always be directed along the z axis, which is perpendicular to the plane
that contains the forces.
Successful application of the equations of equilibrium requires
complete specification of all the known and unknown forces that act
on the body. The best way to account for these forces is to draw the
body’s free-body diagram. Obviously, if the free-body diagram is drawn
correctly, the effects of all the applied forces and couple moments can be
accounted for when the equations of equilibrium are written.
©M
O
= 0
©F
y
= 0
©F
x
= 0
©M
x
= 0 ©M
y
= 0 ©M
z
= 0
©F
x
= 0 ©F
y
= 0 ©F
z
= 0
©M
O
©F
©M
O
= 0
©F = 0
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section
F
4
F
2
(
a
)
F
1
F
3
*The body’s weight is not shown, since it is assumed to be quite small, and therefore
negligible compared with the other loads.
Internal Resultant Loadings.
One of the most important
applications of statics in the analysis of mechanics of materials problems
is to be able to determine the resultant force and moment acting within a
body, which are necessary to hold the body together when the body is
subjected to external loads. For example, consider the body shown in
Fig.1–2a, which is held in equilibrium by the four external forces.*
In order to obtain the internal loadings acting on a specific region within
the body, it is necessary to use the method of sections. This requires that
an imaginary section or “cut” be made through the region where the
internal loadings are to be determined. The two parts of the body are
then separated, and a free-body diagram of one of the parts is drawn,
Fig.1–2b. Here it can be seen that there is actually a distribution of
internal force acting on the “exposed” area of the section. These forces
represent the effects of the material of the top part of the body acting on
the adjacent material of the bottom part.
Although the exact distribution of internal loading may be unknown,
we can use the equations of equilibrium to relate the external forces on
the body to the distribution’s resultant force and moment, and
at any specific point O on the sectioned area, Fig.1–2c. When doing so,
note that acts through point O, although its computed value will not
depend on the location of this point. On the other hand, does
depend on this location, since the moment arms must extend from O to
the line of action of each external force on the free-body diagram. It will
be shown in later portions of the text that point Ois most often chosen at
the centroid of the sectioned area, and so we will always choose this
location for O, unless otherwise stated. Also, if a member is long and
slender, as in the case of a rod or beam, the section to be considered is
generally taken perpendicular to the longitudinal axis of the member.
This section is referred to as the cross section.
M
R
O
F
R
M
R
O
,F
R
F
1
F
2
(
b
)
Fig.1–2
F
R
F
1
F
2
O
M
R
O
(c)
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Three Dimensions.
Later in this text we will show how to relate
the resultant loadings, and to the distribution of force on the
sectioned area, and thereby develop equations that can be used for
analysis and design. To do this, however, the components of and
acting both normal or perpendicular to the sectioned area and within the
plane of the area, must be considered, Fig.1–2d. Four different types of
resultant loadings can then be defined as follows:
Normal force, N.
This force acts perpendicular to the area. It is
developed whenever the external loads tend to push or pull on the two
segments of the body.
Shear force, V.
The shear force lies in the plane of the area and is
developed when the external loads tend to cause the two segments of the
body to slide over one another.
Torsional moment or torque,T.
This effect is developed when the
external loads tend to twist one segment of the body with respect to
the other.
Bending moment,M.
The bending moment is causedby the external
loads that tend to bend the body about an axis lying within the plane of
the area.
In this text, note that graphical representation of a moment or torque
is shown in three dimensions as a vector with an associated curl. By the
right-hand rule, the thumb gives the arrowhead sense of the vector and
the fingers or curl indicate the tendency for rotation (twist or bending).
Using an x,y,z coordinate system, each of the above loadings can be
determined directly from the six equations of equilibrium applied to
either segment of the body.
M
R
O
,F
R
M
R
O
,F
R
(d)
O
F
1
F
2
N
T
M
V
Torsional
Moment
Bending
Moment
Shear
Force
M
R
O
F
R
Normal
Force
O
(c)
M
R
O
F
1
F
2
F
R
Fig.1–2 (cont.)
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Coplanar Loadings.
If the body is subjected to a coplanar system
of forces,Fig.1–3a,then only normal-force,shear-force,and bending-
moment components will exist at the section,Fig.1–3b.If we use the
x,y,z coordinate axes, with origin established at point O, as shown on the
left segment, then a direct solution for N can be obtained by applying
and V can be obtained directly from Finally, the
bending moment can be determined directly by summing moments
about point O(the z axis), in order to eliminate the moments
caused by the unknowns Nand V.
©M
O
= 0,
M
O
©F
y
= 0.©F
x
= 0,
section
F
4
F
3
F
2
F
1
(a)
Fig.1–3
O
V
M
O
N
x
y
Bending
Moment
Shear
Force
Normal
Force
(b)
F
2
F
1
In order to design the members of this
building frame, it is first necessary to find
the internal loadings at various points along
their length.
Important Points

Mechanics of materials is a study of the relationship between the
external loads on a body and the intensity of the internal loads
within the body.

External forces can be applied to a body as distributed or
concentrated surface loadings, or as body forces which act through-
out the volume of the body.

Linear distributed loadings produce a resultant force having a
magnitude equal to the area under the load diagram, and having a
location that passes through the centroid of this area.

A support produces a force in a particular direction on its
attached member if it prevents translation of the member in that
direction, and it produces a couple moment on the member if it
prevents rotation.

The equations of equilibrium and must be
satisfied in order to prevent a body from translating with accel-
erated motion and from rotating.

When applying the equations of equilibrium, it is important to
first draw the body’s free-body diagram in order to account for all
the terms in the equations.

The method of sections is used to determine the internal resultant
loadings acting on the surface of the sectioned body. In general,
these resultants consist of a normal force, shear force, torsional
moment, and bending moment.
©M = 0©F = 0
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The following examples illustrate this procedure numerically and also
provide a review of some of the important principles of statics.
Procedure for Analysis
The method of sections is used to determine the resultant internal
loadings at a point located on the section of a body. To obtain these
resultants, application of the method of sections requires the
following steps.
Support Reactions

First decide which segment of the body is to be considered. If
the segment has a support or connection to another body, then
before the body is sectioned, it will be necessary to determine the
reactions acting on the chosen segment of the body. Draw the free-
body diagram for the entire body and then apply the necessary
equations of equilibrium to obtain these reactions.
Free-Body Diagram

Keep all external distributed loadings, couple moments, torques,
and forces acting on the body in their exact locations, then pass an
imaginary section through the body at the point where the
resultant internal loadings are to be determined.

If the body represents a member of a structure or mechanical
device, the section is often taken perpendicular to the longitudinal
axis of the member.

Draw a free-body diagram of one of the “cut” segments and
indicate the unknown resultants N, V, M, and T at the section.
These resultants are normally placed at the point representing the
geometric center or centroid of the sectioned area.

If the member is subjected to a coplanar system of forces, only N,
V, and Mact at the centroid.

Establish the x,y,z coordinate axes with origin at the centroid
and show the resultant components acting along the axes.
Equations of Equilibrium

Moments should be summed at the section, about each of the
coordinate axes where the resultants act. Doing this eliminates
the unknown forces N and V and allows a direct solution for M
(and T).

If the solution of the equilibrium equations yields a negative
value for a resultant, the assumed directional sense of the
resultant is opposite to that shown on the free-body diagram.
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EXAMPLE 1.1
Determine the resultant internal loadings acting on the cross section
at C of the beam shown in Fig.1–4a.
(
a
)
A
B
C
3 m 6 m
270 N/m
Fig.1–4
180 N/m
540 N
2 m
4 m
V
C
M
C
N
C
(
b
)
BC
1.5 m
0.5 m
1 m
180 N/m
90 N/m
540 N
135 N
V
C
M
C
N
C
1215 N
3645 Nm
CA
(c)
SOLUTION
Support Reactions.This problem can be solved in the most direct
manner by considering segment CB of the beam, since then the
support reactions at Ado not have to be computed.
Free-Body Diagram.Passing an imaginary section perpendicular to
the
longitudinal axis of the beam yields the free-body diagram of
segment CBshown in Fig.1–4b.It is important to keep the distributed
loading exactly where it is on the segment until after the section is
made.Only then should this loading be replaced by a single resultant
force.Notice that the intensity of the distributed loading at C is
found by proportion,i.e.,from Fig. 1–4a,
The magnitude of the resultant of the distributed load is
equal to the area under the loading curve (triangle) and acts through
the centroid of this area.Thus,which
acts fromCas shown in Fig.1–4b.
Equations of Equilibrium.Applying the equations of equilibrium
we have
Ans.
Ans
.
Ans.
NOTE:The negative sign indicates that acts in the opposite
direction to that shown on the free-body diagram. T
ry solving this
problem using segment AC, by first obtaining the support reactions at
A, which are given in Fig.1–4c.
M
C
M
C
= -1080 N
#
m
-M
C
- 540 N 12 m2 = 0d+©M
C
= 0;
V
C
= 540 N
V
C
- 540 N = 0+
c
©F
y
= 0;
N
C
= 0
-N
C
= 0:
+
©F
x
= 0;
1>316 m2 = 2 m
F =
1
2
1180 N>m216 m2 = 540 N,
w = 180 N>m.
w>6 m = 1270 N>m2>9 m,
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EXAMPLE 1.2
Determine the resultant internal loadings acting on the cross section
at Cof the machine shaft shown in Fig.1–5a.The shaft is supported by
bearings at Aand B, which exert only vertical forces on the shaft.
SOLUTION
We will solve this problem using segment AC of the shaft.
Support Reactions.A free-body diagram of the entire shaft is
shown in Fig.
1–5b. Since segment AC is to be considered, only the
reaction at Ahas to be determined. Why?
The negative sign for indicates that acts in the opposite sense to
that shown on the free-body diagram.
Free-Body Diagram.Passing an imaginary section perpendicular
to
the axis of the shaft through C yields the free-body diagram of
segment AC shown in Fig.1–5c.
Equations of Equilibrium.
Ans
.
Ans.
Ans.
NOTE:The negative signs for and indicate they act in the
opposite directions on the free-body diagram. As an exercise, calculate
the reaction at B and try to obtain the same results using segment
CBDof the shaft.
M
C
V
C
M
C
= -5.69 N
#
m
M
C
+ 40 N10.025 m2 + 18.75 N10.250 m2 = 0d+©M
C
= 0;
V
C
= -58.8 N
-18.75 N - 40 N - V
C
= 0+
c
©F
y
= 0;
N
C
= 0:
+
©F
x
= 0;
A
y
A
y
A
y
= -18.75 N
-A
y
10.400 m2 + 120 N10.125 m2 - 225 N10.100 m2 = 0d+©M
B
= 0;
225 N
C
D
200 mm
100 mm 100 mm
50 mm50 mm
800 N/m
B
(
a
)
A
Fig.1–5
0.275 m
0.125 m
(800 N
/
m)(0.150 m) = 120 N
0.100 m
225 N
A
y
B
y
(b)
B
(c)
40 N
18.75 N
0.250 m
0.025 m
M
C
V
C
C
A
N
C
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EXAMPLE 1.3
The hoist in Fig.1–6a consists of the beam AB and attached pulleys,
the cable, and the motor. Determine the resultant internal loadings
acting on the cross section at Cif the motor is lifting the 500-lb load W
with constant velocity. Neglect the weight of the pulleys and beam.
SOLUTION
The most direct way to solve this problem is to section both the cable
and the beam at C and then consider the entire left segment.
Free-Body Diagram.See Fig.1–6b.
Equations of Equilibrium.
Ans
.
Ans.
Ans.
NOTE:As an exercise, try obtaining these same results by
considering just the beam segment AC, i.e., remove the pulley at A
from the beam and show the 500-lb force components of the pulley
acting on the beam segment AC.
M
C
= -2000 lb
#
ft
500 lb 14.5 ft2 - 500 lb 10.5 ft2 + M
C
= 0d+©M
C
= 0;
-500 lb - V
C
= 0

V
C
= -500 lb+
c
©F
y
= 0;
500 lb + N
C
= 0


N
C
= -500 lb:
+
©F
x
= 0;
4 ft
2 ft6 ft
0.5 ft
0.5 ft
A
C
D
B
W
(
a
)
(b)
4.5 ft
C
0.5 ft
500 lb
500 lb
V
C
N
C
M
C
A
Fig.1–6
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EXAMPLE 1.4
Determine the resultant internal loadings acting on the cross section
at G of the wooden beam shown in Fig.1–7a. Assume the joints at
A,B,C,D, and E are pin connected.
SOLUTION
Support Reactions.Here we will consider segment AG for the
analysis. A free-body diagram of the entire structure is shown in
Fig.
1–7b.Verify the computed reactions at Eand C.In particular,note
that BC is a two-force member since only two forces act on it. For this
reason the reaction at C must be horizontal as shown.
Since BA and BD are also two-force members, the free-body
diagram of joint B is shown in Fig.1–7c. Again, verify the magnitudes
of the computed forces and
Free-Body Diagram.Using the result for the left section AG
of the beam is shown in Fig.
1–7d.
Equations of Equilibrium.Applying the equations of equilibriumto
segment AG, we have
Ans.
Ans
.
Ans.M
G
= 6300 lb
#
ft
M
G
- 17750 lb2
A
3
5
B
12 ft2 + 1500 lb12 ft2 = 0d+©M
G
= 0;
V
G
= 3150 lb
-1500 lb + 7750 lb
A
3
5
B
- V
G
= 0+
c
©F
y
= 0;
7750 lb
A
4
5
B
+ N
G
= 0 N
G
= -6200 lb:
+
©F
x
= 0;
F
BA
,
F
BD
.F
BA
(
a
)
300 lb/ft
2 ft 2 ft 6 ft
1500 lb
A
B
G
D
C
3 ft
E
3 ft
6 ft
(6 ft)  4 ft
(6 ft)(300 lb/ft)  900 lb
1500 lb
E
y

 2400 lb
E
x
 6200 lb
F
BC
 6200 lb
(
b
)
2
3
1
2
6200 lb
3
4
5
(c)
B
F
BA
 7750 lb
F
BD
 4650 lb
(d)
N
G
M
G
V
G
2 ft
3
4
5
7750 lb
1500 lb
A
G
Fig.1–7
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EXAMPLE 1.5
Determine the resultant internal loadings acting on the cross section
at Bof the pipe shown in Fig.1–8a. The pipe has a mass of and
is subjected to both a vertical force of 50 N and a couple moment of
at its end A. It is fixed to the wall at C.
SOLUTION
The problem can be solved by considering segment AB, which does
not involve the support reactions at C.
Free-Body Diagram.The x,y,z axes are established at B and the
free-body diagram of segment ABis shown in Fig.
1–8b. The resultant
force and moment components at the section are assumed to act in
the positive coordinate directions and to pass through the centroid of
the cross-sectional area at B. The weight of each segment of pipe is
calculated as follows:
These forces act through the center of gravity of each segment.
Equations of Equilibrium.Applying the six scalar equations of
equilibrium, we have*
Ans.
Ans
.
Ans.
Ans.
Ans.
Ans.
NOTE:What do the negative signs for and indicate?
Note that the normal force
whereas the shear
force is Also, the torsional
moment is and the bending moment is
M
B
= 2130.32
2
+ 102
= 30.3 N
#
m.
T
B
= 1M
B
2
y
= 77.8 N
#
m
V
B
= 2102
2
+ 184.32
2
= 84.3 N.
N
B
= 1F
B
2
y
= 0,
1M
B
2
y
1M
B
2
x
1M
B
2
z
= 0©1M
B
2
z
= 0;
1M
B
2
y
= -77.8 N
#
m
1M
B
2
y
+ 24.525 N 10.625 m2 + 50 N 11.25 m2 = 0©1M
B
2
y
= 0;
1M
B
2
x
= -30.3 N
#
m
- 9.81 N 10.25 m2 = 0
1M
B
2
x
+ 70 N
#
m - 50 N 10.5 m2 - 24.525 N 10.5 m2©1M
B
2
x
= 0;
1F
B
2
z
= 84.3 N
1F
B
2
z
- 9.81 N - 24.525 N - 50 N = 0©F
z
= 0;
1F
B
2
y
= 0©F
y
= 0;
1F
B
2
x
= 0©F
x
= 0;
W
AD
= 12 kg>m211.25 m219.81 N>kg2 = 24.525 N

W
BD
= 12 kg>m210.5 m219.81 N>kg2 = 9.81 N
70 N
#
m
2 kg>m
0.75 m
50 N
1.25 m
B
A
0.5 m
C
D
70 Nm
(a)
0.625 m
70 N∙m
(
b
)
y0.625 m
A
50 N
0.25 m
0.25 m
x
z
9.81 N
24.525 N
B
(F
B
)
z
(M
B
)
z
(M
B
)
y
(M
B
)
x
(F
B
)
y
(F
B
)
x
Fig.1–8
*The magnitude of each moment about an axis is equal to the magnitude of each
force times the perpendicular distance from the axis to the line of action of the force.
The direction of each moment is determined using the right-hand rule, with positive
moments (thumb) directed along the positive coordinate axes.
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 15
16
C
HAPTER
1 S
TRESS
PROBLEMS
1–1.Determine the resultant internal normal force
acting on the cross section through point A in each column.
In (a), segment BCweighs 180 lb￿ft and segment CDweighs
250 lb￿ft. In (b), the column has a mass of 200 kg￿m.
1–2.Determine the resultant internal torque acting on the
cross sections through points Cand Dof the shaft.The shaft
is fixed at B.
1–3.Determine the resultant internal torque acting on the
cross sections through points B and C.
*1–4.A force of 80 N is supported by the bracket as
shown. Determine the resultant internal loadings acting on
the section through point A.
3 ft
2 ft
2 ft
1 ft
B
A
C
500 lbf
t
350 lbft
600 lbft
Prob.1–3
3 kip
3 kip
5 kip
10 ft
4 ft
4 ft
8 in.8 in.
A
C
D
(a)
B
8 kN
3 m
1 m
6 kN6 kN
4.5 kN4.5 kN
200 mm200 mm
A
(b)
200 mm200 mm
Prob.1–1
200 mm
50 mm 50 mm
C
A
D
B
70 N m
150 mm300 mm
Prob.1–5
C
250 Nm
D
400 Nm
300 Nm
0.4 m
0.1 m
0.3 m
0.1 m
0.2 m
A
B
Prob.1–2
0.1 m
0.3 m
30
80 N
A
45
Prob.1–4
1–5.
Determine the resultant internal loadings acting on
the cross section through point Dof member AB.
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 16
P
ROBLEMS
17
1–6.The beam AB is pin supported at A and supported
by a cable BC. Determine the resultant internal loadings
acting on the cross section at point D.
1–7.Solve Prob. 1–6 for the resultant internal loadings
acting at point E.
1–9.The force acts on the gear tooth.
Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point Aof section a–a.
F = 80 lb
1–10.The beam supports the distributed load shown.
Determine the resultant internal loadings on the cross
section through point C. Assume the reactions at the
supports Aand B are vertical.
1–11.The beam supports the distributed load shown.
Determine the resultant internal loadings on the cross
sections through points D and E. Assume the reactions at
the supports Aand B are vertical.
4 ft
B
C
6 ft
8 ft
3 ft
A
D
1200 lb
E
3 ft
Probs.1–6/7
*1–8.The boom DF of the jib crane and the column DE
have a uniform weight of 50 lb￿ft. If the hoist and load
weigh 300 lb, determine the resultant internal loadings in
the crane on cross sections through points A, B, and C.
5 ft
7 ft
D F
C
B A
300 lb
2 ft 3 ft
E
8 ft
Prob.1–8
a
30
a
F  80 lb
0.23 in.
45
A
0.16 in.
Prob.1–9
6 ft
8 ft
C
A
B
300 lb/ft
4.5 ft
400 lb/ft
6 ft
4.5 ft
ED
Probs.1–10/11
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 17
18
C
HAPTER
1 S
TRESS
1–17.Determine the resultant internal loadings acting on
the cross section at point B.
M
4 ft
3 ft
4 ft
C B
1.5 ft
A
0.25 ft
4 ft
3 ft
D
Probs.1–15/16
*1–12.Determine the resultant internal loadings acting
on (a) section a–a and (b) section b–b. Each section is
located through the centroid, point C.
1–15.The 800-lb load is being hoisted at a constant speed
using the motor M, which has a weight of 90 lb. Determine
the resultant internal loadings acting on the cross section
through point B in the beam. The beam has a weight of
40 lb￿ft and is fixed to the wall at A.
*1–16.Determine the resultant internal loadings acting
on the cross section through points C and Dof the beam in
Prob. 1–15.
45


8 ft
4 ft
45


A
C
B
b
a
a b
600 lb/ft
Prob.1–12
1–13.Determine the resultant internal normal and shear
forces in the member at (a) section a–a and (b) section b–b,
each of which passes through point A. Take The
650-Nloadis appliedalongthecentroidal axis of themember.
1–14.Determine the resultant internal normal and shear
forces in the member at section b–b, each as a function of
Plot these results for The 650-N load is
applied along the centroidal axis of the member.
0° … u … 90°.
u.
u = 60°.
A
b
a
u
b
a
650 N
650 N
Probs.1–13/14
A
C
12 ft3 ft
60 lb/

ft
B
Prob.1–17
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 18
P
ROBLEMS
19
1–18.The beam supports the distributed load shown.
Determine the resultant internal loadings acting on the
cross section through point C. Assume the reactions at the
supports Aand B are vertical.
1–19.Determine the resultant internal loadings acting on
the cross section through point Din Prob. 1–18.
3 m
3 m
DC
A B
0.5 kN/m
1.5 kN/m
3 m
Probs.1–18/19
*1–20.The wishbone construction of the power pole
supports the three lines, each exerting a force of 800 lb on
the bracing struts. If the struts are pin connected at A, B,
and C, determine the resultant internal loadings at cross
sections through points D, E, and F.
4 ft
D
E
A
B
F
C
4 ft
6 ft
6 ft
800 lb
800 lb
800 lb
Prob.1–20
1–21.The drum lifter suspends the 500-lb drum. The linkage
is pin connected to the plate at A and B. The gripping action
on the drum chime is such that only horizontal and vertical
forces are exerted on the drum at G and H. Determine the
resultant internal loadings on the cross section through point I.
1–22.Determine the resultant internal loadings on the
cross sections through points K and J on the drum lifter in
Prob. 1–21.
8 in.
5 in.
3 in.
C
A B
E
D
F
I
K
60
60
2 in.
5 in.
J
HG
5 in.
500 lb
1–23.The pipe has a mass of 12 kg￿m. If it is fixed to the
wall at A, determine the resultant internal loadings acting on
the cross section at B. Neglect the weight of the wrench CD.
300 mm
200 mm
150 mm
60 N
60 N
400 mm
150 mm
B
A
x
y
z
C
D
Prob.1–23
Probs.1–21/22
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 19
20
C
HAPTER
1 S
TRESS
1–25.Determine the resultant internal loadings acting on
the cross section through point B of the signpost. The post
is fixed to the ground and a uniform pressure of acts
perpendicular to the face of the sign.
7 lb>ft
2
1–26.The shaft is supported at its ends by two bearings A
and B and is subjected to the forces applied to the pulleys
fixed to the shaft. Determine the resultant internal loadings
acting on the cross section through point D. The 400-N
forces act in the direction and the 200-N and 80-N forces
act in the direction. The journal bearings at A and B
exert only y and z components of force on the shaft.
+y
-z
*1–24.The main beam AB supports the load on the wing
of the airplane.The loads consist of the wheel reaction of
35,000 lb at C,the 1200-lb weight of fuel in the tank of the
wing,having a center of gravity at D,and the 400-lb weight of
the wing,having a center of gravity at E.If it is fixed to the
fuselage at A,determine the resultant internal loadings on
the beamat this point.Assume that the wing does not transfer
any of the loads to the fuselage,except through the beam.
6 ft
4 ft
2 ft
1.5 ft
D E
1 ft
A
B
C
35,000 lb
z
x
y
Prob.1–24
4 ft
z
y
6 ft
x
B
A
3 ft
2 ft
3 ft
7 lb/ft
2
Prob.1–25
B
C
200 mm
200 mm
300 mm
A
200 N
200 N
400 N
400 N
150 mm
400 mm
80 N
80 N
z
x
y
150 mm
D
Prob.1–27
B
C
200 mm
200 mm
300 mm
A
200 N
200 N
400 N
400 N
150 mm
400 mm
80 N
80 N
z
x
y
150 mm
D
Prob.1–26
1–27.
The shaft is supported at its ends by two bearings A
and B and is subjected to the forces applied to the pulleys
fixed to the shaft. Determine the resultant internal loadings
acting on the cross section through point C. The 400-N
forces act in the direction and the 200-N and 80-N forces
act in the direction. The journal bearings at A and B
exert only y and z components of force on the shaft.
+y
-z
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 20
A B
C
90
6 in.
P
ROBLEMS
21
1–29.The bolt shank is subjected to a tension of 80 lb.
Determine the resultant internal loadings acting on the
cross section at point C.
1–30.The pipe has a mass of . If it is fixed to the
wall at A, determine the resultant internal loadings acting
on the cross section through B.
12 kg>m
*1–28.Determine the resultant internal loadings acting
on the cross section of the frame at points F and G. The
contact at E is smooth.
1–31.The curved rod has a radius r and is fixed to the wall
at B. Determine the resultant internal loadings acting on
the cross section through A which is located at an angle
from the horizontal.
u
4 ft
1.5 ft
1.5 ft
3 ft
A
B
C
E
D
G
80 lb
5 ft
2 ft
2 ft
30
F
Prob.1–28
Prob.1–29
M
V
N
d
u
M  dM
T  dT
N  dN
V  dV
T
Prob.1–33
1 m
2 m
2 m
B
A
y
z
x
C
800 Nm
3
4
5
750 N
Prob.1–30
r
A
P
u
B
A
B
C
45
90
D
O
r
22.5
Prob.1–32
*1–32.
The curved rod AD of radius r has a weight per
length of If it lies in the horizontal plane, determine the
resultant internal loadings acting on the cross section
through point B. Hint: The distance from the centroid C of
segment AB to point Ois CO = 0.9745r.
w.
1–33.A differential element taken from a curved bar is
shown in the figure. Show that
and dT>du = M.dM>du = -T,
dV>du = -N,dN>du = V,
Prob.1–31
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 21
22
C
HAPTER
1 S
TRESS
1.3 Stress
It was stated in Section 1.2 that the force and moment acting at a
specified point on the sectioned area of a body, Fig.1–9, represents the
resultant effects of the actual distribution of force acting over the
sectioned area, Fig. 1–10a. Obtaining this distribution of internal loading
is of primary importance in mechanics of materials. To solve this problem
it is necessary to establish the concept of stress.
Consider the sectioned area to be subdivided into small areas, such as
shown dark shaded in Fig.1–10a. As we reduce to a smaller and
smaller size, we must make two assumptions regarding the properties of
the material. We will consider the material to be continuous, that is, to
consist of a continuumor uniform distribution of matter having no voids,
rather than being composed of a finite number of distinct atoms or
molecules. Furthermore, the material must be cohesive, meaning that all
portions of it are connected together, rather than having breaks, cracks,
or separations. A typical finite yet very small force acting on its
associated area is shown in Fig. 1–10a. This force, like all the others,
will have a unique direction, but for further discussion we will replace it
by its three components, namely, and which are taken
tangent, and normal to the area, respectively. As the area approaches
zero, so do the force and its components; however, the quotient of the
force and area will, in general, approach a finite limit. This quotient is
called stress, and as noted, it describes the intensity of the internal force on
a specific plane (area) passing through a point.
¢F
¢A
¢F
z
,¢F
y
,¢F
x
,
¢A,
¢F,
¢A¢A
F
1
F
2
F
1
F
A
F
F
z
z
y
x
F
x
F
y
z
(c)
x
y
(b)
zz
x
y
(a)
x
y
t
yz
s
y
t
yx
t
xz
s
x
t
xy
Fig.1–10
F
1
F
2
O
M
R
O
F
R
Fig.1–9
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 22
S
ECTI ON
1.3 S
TRESS
23
Normal Stress.
The intensity of force, or force per unit area, acting
normal to is defined as the normal stress, (sigma). Since is
normal to the area then
(1–4)
If the normal force or stress “pulls” on the area element as shown in
Fig. 1–10a, it is referred to as tensile stress, whereas if it “pushes” on
it is called compressive stress.
Shear Stress.
The intensity of force, or force per unit area, acting
tangent to is called the shear stress, (tau). Here we have shear
stress components,
Note that the subscript notation z in is used to reference the
direction of the outward normal line, which specifies the orientation of
the area Fig.1–11. Two subscripts are used for the shear-stress
components, and The z axis specifies the orientation of the area,
and x and y refer to the direction lines for the shear stresses.
General State of Stress.
If the body is further sectioned by
planes parallel to the x–z plane, Fig. 1–10b, and the y–z plane, Fig. 1–10c,
we can then “cut out” a cubic volume element of material that
represents the state of stress acting around the chosen point in the body,
Fig.1–12. This state of stress is then characterized by three components
acting on each face of the element. These stress components describe
the state of stress at the point only for the element orientated along the
x,y,z axes. Had the body been sectioned into a cube having some other
orientation, then the state of stress would be defined using a different set
of stress components.
Units.
In the International Standard or SI system, the magnitudes of
both normal and shear stress are specified in the basic units of newtons
per square meter This unit, called a pascal is
rather small, and in engineering work prefixes such as kilo-
symbolized by k, mega- symbolized by M, or giga-
symbolized by G, are used to represent larger, more realistic values of
stress.* Likewise, in the U.S. Customary or Foot-Pound-Second system
of units, engineers usually express stress in pounds per square inch (psi)
or kilopounds per square inch (ksi), where 1 kilopound 1kip2 = 1000 lb.
110
9
2,110
6
2,
110
3
2,
11 Pa = 1 N>m
2
21N>m
2
2.
t
zy
.t
zx
¢A,
s
z
t
zy
= lim
¢A:0
¢F
y
¢A
t
zx
= lim
¢A:0
¢F
x
¢A
t¢A
¢A
¢A
s
z
= lim
¢A:0

¢F
z
¢A
¢F
z
s¢A
*Sometimes stress is expressed in units of where However, in
the SI system, prefixes are not allowed in the denominator of a fraction and therefore it is
better to use the equivalent 1 N>mm
2
= 1 MN>m
2
= 1 MPa.
1 mm = 10
-3
m.N>mm
2
,
x
y
z
T
zx
T
zy
s
z
Fig.1–11
x
y
z
z
zx
zy
yz
yx
xz
x
xy
y
s
s s
t
t
t
t
t
t
Fig.1–12
(1–5)
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 23
24
C
HAPTER
1 S
TRESS
1.4 Average Normal Stress
in an Axially Loaded Bar
Frequently structural or mechanical members are made long and
slender. Also, they are subjected to axial loads that are usually applied to
the ends of the member. Truss members, hangers, and bolts are typical
examples. In this section we will determine the average stress
distribution acting on the cross section of an axially loaded bar, such as
the one having the general form shown in Fig.1–13a. This section
defines the cross-sectional area of the bar, and since all such cross
sections are the same, the bar is referred to as being prismatic. If we
neglect the weight of the bar and section it as indicated, then, for
equilibrium of the bottom segment, Fig.1–13b, the internal resultant
force acting on the cross-sectional area must be equal in magnitude,
opposite in direction, and collinear to the external force acting at the
bottom of the bar.
Assumptions.
Before we determine the average stress distribution
acting over the bar’s cross-sectional area, it is necessary to make two
simplifying assumptions concerning the material description and the
specific application of the load.
1.It is necessary that the bar remains straight both before and after
the load is applied, and also, the cross section should remain flat or
plane during the deformation, that is, during the time the bar
changes its volume and shape. If this occurs, then horizontal and
vertical grid lines inscribed on the bar will deform uniformly when
the bar is subjected to the load, Fig.1–13c. Here we will not
consider regions of the bar near its ends, where application of the
external loads can cause localized distortions. Instead we will focus
only on the stress distribution within the bar’s midsection.
2.In order for the bar to undergo uniformdeformation,it is necessary
that P be applied along the centroidal axis of the cross section,and
the material must be homogeneous and isotropic.Homogeneous
material has the same physical and mechanical properties
throughout its volume,and isotropic material has these same
properties in all directions.Many engineering materials may be
approximated as being both homogeneous and isotropic as
assumed here.Steel,for example,contains thousands of randomly
oriented crystals in each cubic millimeter of its volume,and since
most problems involving this material have a physical size that is
much larger than a single crystal,the above assumption regarding
its material composition is quite realistic.It should be mentioned,
however,that steel can be made anisotropic by cold-rolling,(i.e.,
rolling or forging it at subcritical temperatures).Anisotropic
materials have different properties in different directions,and
although this is the case,if the anisotropy is oriented along the bar’s
P
P
(
a
)
P
P
External force
Cross-sectional
area
Internal force
(b)
(c)
P
P
Region of
uniform
deformation
of bar
Fig.1–13
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 24
S
ECTI ON
1.4 A
VERAGE
N
ORMAL
S
TRESS I N AN
A
XI ALLY
L
OADED
B
AR
25
axis,then the bar will also deform uniformly when subjected to an
axial load.For example,timber,due to its grains or fibers of wood,
is an engineering material that is homogeneous and anisotropic,
and due to the customary orientation of its fibers it is suited for the
following analysis.
Average Normal Stress Distribution.
Provided the bar is
subjected to a constant uniform deformation as noted, then this
deformation is the result of a constant normal stress Fig.1–13d. As a
result, each area on the cross section is subjected to a force
and the sum of these forces acting over the entire cross-
sectional area must be equivalent to the internal resultant force P at the
section. If we let and therefore then, recognizing
is constant, we have
(1–6)
Here
normal stress at any point on the cross-sectional area
resultant normal force, which is applied through the
centroid of the cross-sectional area. P is determined using the
method of sections and the equations of equilibrium
area of the bar
The internal load Pmust pass through the centroid of the cross-section
since the uniform stress distribution will produce zero moments about
any x and y axes passing through this point, Fig. 1–13d. When this occurs,
These equations are indeed satisfied, since by definition of the centroid,
and (See Appendix A.)
Equilibrium.
It should be apparent that only a normal stress exists
on any volume element of material located at each point on the cross
section of an axially loaded bar. If we consider vertical equilibrium of the
element, Fig.1–14, then applying the equation of force equilibrium,
s = s¿
s1¢A2 - s¿1¢A2 = 0©F
z
= 0;
1
x dA = 0.
1
y dA = 0
0 = -
L
A

x dF = -
L
A

xs dA = -s
L
A

x dA1M
R
2
y
= ©M
y
;
0 =
L
A

y dF =
L
A

ys dA = s
L
A

y dA1M
R
2
x
= ©M
x
;
A = cross-sectional
P = internal
s = average
s =
P
A
P = sA

L
dF =
L
A

s dA+
c
F
Rz
= ©F
z
;
s¢F:dF,¢A:dA
¢F = s ¢A,
¢A
s,
(d)
P
F  sA
P
y
x
x
z
y
A
s
Fig.1–13 (cont.)
¿
A
s
s
Fig.1–14
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 25
26
C
HAPTER
1 S
TRESS
In other words, the two normal stress components on the element must
be equal in magnitude but opposite in direction. This is referred to as
uniaxial stress.
The previous analysis applies to members subjected to either tension
or compression, as shown in Fig.1–15. As a graphical interpretation, the
magnitude of the internal resultant force P is equivalent to the volume
under the stress diagram; that is,
Furthermore, as a consequence of the balance of moments, this resultant
passes through the centroid of this volume.
Although we have developed this analysis for prismatic bars, this
assumption can be relaxed somewhat to include bars that have a slight
taper. For example, it can be shown, using the more exact analysis of the
theory of elasticity, that for a tapered bar of rectangular cross section, for
which the angle between two adjacent sides is 15°, the average normal
stress, as calculated by is only 2.2% less than its value found
from the theory of elasticity.
Maximum Average Normal Stress.
In our analysis both the
internal force P and the cross-sectional area A were constant along the
longitudinal axis of the bar, and as a result the normal stress is
also constant throughout the bar’s length. Occasionally, however, the bar
may be subjected to several external loads along its axis, or a change in
its cross-sectional area may occur. As a result, the normal stress within
the bar could be different from one section to the next, and, if the
maximum average normal stress is to be determined, then it becomes
important to find the location where the ratio P￿Ais a maximum. To do
this it is necessary to determine the internal force P at various sections
along the bar. Here it may be helpful to show this variation by drawing
an axial or normal force diagram. Specifically, this diagram is a plot of
the normal force P versus its position x along the bar’s length. As a sign
convention, P will be positive if it causes tension in the member, and
negative if it causes compression. Once the internal loading throughout
the bar is known, the maximum ratio of P￿Acan then be identified.
s = P>A
s = P>A,
P = sA 1volume = height * base2.
This steel tie rod is used to suspend a portion
of a staircase, and as a result it is subjected to
tensile stress.

P
P
P
P
Tension Compression
s
s
s
s
P

A
Fig.1–15
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Important Points

When a body that is subjected to an external load is sectioned,
there is a distribution of force acting over the sectioned area which
holds each segment of the body in equilibrium. The intensity of
this internal force at a point in the body is referred to as stress.

Stress is the limiting value of force per unit area, as the area
approaches zero. For this definition, the material at the point is
considered to be continuous and cohesive.

The magnitude of the stress components depends upon the type of
loading acting on the body, and the orientation of the element at
the point.

When a prismatic bar is made from homogeneous and isotropic
material, and is subjected to axial force acting through the
centroid of the cross-sectioned area, then the material within the
bar is subjected only to normal stress. This stress is assumed to be
uniform or averaged over the cross-sectional area.
Procedure for Analysis
The equation gives the average normal stress on the cross-
sectional area of a member when the section is subjected to an
internal resultant normal force P. For axially loaded members,
application of this equation requires the following steps.
Internal Loading

Section the member perpendicular to its longitudinal axis at the
point where the normal stress is to be determined and use the
necessary free-body diagram and equation of force equilibrium to
obtain the internal axial force P at the section.
Average Normal Stress

Determine the member’s cross-sectional area at the section and
compute the average normal stress

It is suggested that be shown acting on a small volume element
of the material located at a point on the section where stress is
calculated. To do this, first draw on the face of the element
coincident with the sectioned area A. Here acts in the same
direction as the internal force P since all the normal stresses on
the cross section act in this direction to develop this resultant. The
normal stress acting on the opposite face of the element can be
drawn in its appropriate direction.
s
s
s
s
s = P>A.
s = P>A
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EXAMPLE 1.6
The bar in Fig.1–16a has a constant width of 35 mm and a thickness of
10 mm. Determine the maximum average normal stress in the bar
when it is subjected to the loading shown.
(b)
9 kN
9 kN
12 kN
12 kN
P
AB
 12 kN
P
BC
 30 kN
P
CD
 22 kN 22 kN
P(kN)
x
12
22
30
(c)
12 kN
22 k
N
9 kN
9 kN
4 kN
4 kN
35 mm
A D
B C
(a)
(
d
)
30 k
N
85.7 MPa
35 mm
10 mm
Fig.1–16
SOLUTION
Internal Loading.By inspection, the internal axial forces in regions
AB,BC, and CDare all constant yet have different magnitudes. Using
the method of sections, these loadings are determined in Fig. 1–16b;
and the normal force diagram which represents these results
graphically is shown in Fig. 1–16c. By inspection, the largest loading is
in region BC, where
Since the cross-sectional area of
the bar is constant, the largest average normal stress also occurs
within this region of the bar.
Average Normal Stress.Applying Eq. 1–6, we have
Ans.
NOTE:The stress distribution acting on an arbitrary cross section of
the
bar withinregionBCis showninFig.1–16d.Graphically the volume
(or “block”) represented by this distribution of stress is equivalent to
the load of 30 kN;that is,30 kN = 185.7 MPa2135 mm2110 mm2.
s
BC
=
P
BC
A
=
30110
3
2N
10.035 m210.010 m2
= 85.7 MPa
P
BC
= 30 kN.
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EXAMPLE 1.7
The 80-kg lamp is supported by two rods AB and BC as shown in
Fig.1–17a. If AB has a diameter of 10 mm and BC has a diameter of
8 mm, determine the average normal stress in each rod.
SOLUTION
Inter
nal Loading.We must first determine the axial force in each
rod. A free-body diagram of the lamp is shown in Fig.
1–17b. Applying
the equations of force equilibrium yields
By Newton’s third law of action, equal but opposite reaction, these
forces subject the rods to tension throughout their length.
Average Normal Stress.Applying Eq. 1–6, we have
Ans.
Ans
.
NOTE:The average normal stress distribution acting over a cross
section of rod AB is shown in Fig.
1–17c, and at a point on this cross
section, an element of material is stressed as shown in Fig. 1–17d.
s
BA
=
F
BA
A
BA
=
632.4 N
p10.005 m2
2
= 8.05 MPa

s
BC
=
F
BC
A
BC
=
395.2 N
p10.004 m2
2
= 7.86 MPa
F
BA
= 632.4 NF
BC
= 395.2 N,
F
BC
A
3
5
B
+ F
BA
sin 60° - 784.8 N = 0+
c
©F
y
= 0;
F
BC
A
4
5
B
- F
BA
cos 60° = 0:
+
©F
x
= 0;
A
60
B
C
3
4
5
(a)
Fig.1–17
(
b
)
60
F
BA
F
BC
y
x
80(9.81)  784.8 N
B
3
4
5
632.4 N
8.05 MPa
8.05 MPa
(c)(d)
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EXAMPLE 1.8
The casting shown in Fig.1–18a is made of steel having a specific
weight of Determine the average compressive stress
acting at points Aand B.
g
st
= 490 lb>ft
3
.
SOLUTION
Internal Loading.A free-body diagram of the top segment of the
casting where the section passes through points A and B is shown
in
Fig.1–18b. The weight of this segment is determined from
Thus the internal axial force P at the section is
Average Compressive Stress.The cross-sectional area at the
section is
and so the average compressive stress
becomes
Ans.
NOTE:The stress shown on the volume element of material in
Fig.
1–18c is representative of the conditions at either point A or B.
Notice that this stress acts upward on the bottomor shaded face of the
element since this face forms part of the bottomsurface area of the cut
section,and on this surface,the resultant internal force P is pushing
upward.
= 9.36 psi
= 1347.5 lb>ft
2
= 1347.5 lb>ft
2
11 ft
2
>144 in
2
2
s =
P
A
=
2381 lb
p10.75 ft2
2
A = p10.75 ft2
2
,
P = 2381 lb
P - 1490 lb>ft
3
212.75 ft2 p10.75 ft2
2
= 0
P - W
st
= 0+
c
©F
z
= 0;
W
st
= g
st

V
st
.
0.75 ft
0.75 ft
2.75 ft
y
z
x
(a)
A
B0.75 ft
0.4 ft
Fig.1–18
2.75 ft
(b)
A
P
(c)
9.36 ps
i
B
W
st
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EXAMPLE 1.9
Member ACshown in Fig.1–19a is subjected to a vertical force of 3 kN.
Determine the position x of this force so that the average compressive
stress at the smooth support C is equal to the average tensile stress in
the tie rod AB.The rod has a cross-sectional area of and the
contact area at Cis 650 mm
2
.
400 mm
2
SOLUTION
Internal Loading.The forces at A and C can be related by
considering the free-body diagram for member AC, Fig.
1–19b. There
are three unknowns, namely, and x. To solve this problem
we will work in units of newtons and millimeters.
(1)
(2)
Average Normal Stress.A necessary third equation can be written
that requires the tensile stress in the bar AB and the compressive
stress at C to be equivalent, i.e.,
Substituting this into Eq. 1, solving for
then solving for we
obtain
The position of the applied load is determined from Eq. 2,
Ans.
NOTE:as required.0 6 x 6 200 mm,
x = 124 mm
F
C
= 1857 N

F
AB
= 1143 N
F
C
,F
AB
,
F
C
= 1.625F
AB
s =
F
AB
400 mm
2
=
F
C
650 mm
2
-3000 N1x2 + F
C
1200 mm2 = 0d+©M
A
= 0;

F
AB
+ F
C
- 3000 N = 0+
c
©F
y
= 0;
F
C
,F
AB
,
x
A
B
C
200 mm
(a)
3 kN
Fig.1–19
(b)
x
3 kN
A
200 mm
F
AB
F
C
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1.5 Average Shear Stress
Shear stress has been defined in Section 1.3 as the stress component that
acts in the plane of the sectioned area. In order to show how this stress
can develop, we will consider the effect of applying a force F to the bar in
Fig.1–20a. If the supports are considered rigid, and F is large enough, it
will cause the material of the bar to deform and fail along the planes
identified by AB and CD. A free-body diagram of the unsupported
center segment of the bar, Fig. 1–20b, indicates that the shear force
must be applied at each section to hold the segment in
equilibrium.Theaverage shear stress distributedover eachsectionedarea
that develops this shear force is defined by
(1–7)
Here
average shear stress at the section, which is assumed to be the
same at each point located on the section
internal resultant shear force at the section determined from
the equations of equilibrium
area at the section
The distribution of average shear stress is shown acting over the
sections in Fig. 1–20c. Notice that is in the same direction as V, since
the shear stress must create associated forces all of which contribute to
the internal resultant force Vat the section.
The loading case discussed in Fig. 1–20 is an example of simple or
direct shear, since the shear is caused by the direct action of the applied
load F. This type of shear often occurs in various types of simple
connections that use bolts, pins, welding material, etc. In all these cases,
however, application of Eq. 1–7 is only approximate. A more precise
investigation of the shear-stress distribution over the critical section
often reveals that much larger shear stresses occur in the material than
those predicted by this equation. Although this may be the case,
application of Eq. 1–7 is generally acceptable for many problems in
engineering design and analysis. For example, engineering codes allow
its use when considering design sizes for fasteners such as bolts and for
obtaining the bonding strength of joints subjected to shear loadings. In
this regard, two types of shear frequently occur in practice, which
deserve separate treatment.
t
avg
A =
V =
t
avg
=
t
avg
=
V
A
V = F>2
F
(a)
B
D
A
C
(b)
F
V
V
(c)
F
t
avg
Fig.1–20
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Single Shear.
The steel and wood joints shown in Figs.1–21a and
1–21c, respectively, are examples of single-shear connections and are
often referred to as lap joints. Here we will assume that the members are
thin and that the nut in Fig. 1–21a is not tightened to any great extent so
friction between the members can be neglected. Passing a section
between the members yields the free-body diagrams shown in
Figs.1–21b and 1–21d. Since the members are thin, we can neglect the
moment created by the force F. Hence for equilibrium, the cross-
sectional area of the bolt in Fig. 1–21b and the bonding surface between
the members in Fig. 1–21d are subjected only to a single shear force
This force is used in Eq. 1–7 to determine the average shear
stress acting on the colored section of Fig. 1–21d.
Double Shear.
When the joint is constructed as shown in Fig.1–22a
or 1–22c, two shear surfaces must be considered. These types of
connections are often called double lap joints. If we pass a section
between each of the members, the free-body diagrams of the center
member are shown in Figs. 1–22b and 1–22d. Here we have a condition
of double shear. Consequently, acts on each sectioned area and
this shear must be considered when applying t
avg
= V>A.
V = F>2
V = F.
F
F
(
a
)
F
(
b
)
V  F
V  F
F
(
c
)
F
(
d
)
F
Fig.1–21
The pin on this tractor is subjected to double
shear.
F
(
d
)
F
(
c
)
F
(
b
)
V
V
F
(
a
)
F
2
F
2
F
2
F
2
V
V
F
2
F
2
F
2
F
2
Fig.1–22
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Equilibrium.
Consider a volume element of material taken at a
point located on the surface of any sectioned area on which the average
shear stress acts, Fig. 1–23a. If we consider force equilibrium in the y
direction, then
force
stress area
And in a similar manner, force equilibrium in the z direction yields
Finally, taking moments about the x axis,
moment
force arm
stress area
so that
In other words, force and moment equilibrium requires the shear stress
acting on the top face of the element, to be accompanied by shear stress
acting on three other faces, Fig. 1–23b. Here all four shear stresses must
have equal magnitude and be directed either toward or away from each
other at opposite edges of the element. This is referred to as the
complementary property of shear, and under the conditions shown in
Fig.1–23, the material is subjected to pure shear.
Although we have considered here a case of simple shear as caused by
the direct action of a load, in later chapters we will show that shear stress
can also arise indirectly due to the action of other types of loading.
t
zy
= t
œ
zy
= t
yz
= t
œ
yz
= t
t
zy
= t
yz
-t
zy
1¢x ¢y2 ¢z + t
yz
1¢x ¢z2 ¢y = 0©M
x
= 0;
t
yz
= t
œ
yz
.
t
zy
= t
œ
zy
t
zy
1¢x ¢y2 - t
œ
zy
¢x ¢y = 0©F
y
= 0;
Pure shear
(a) (b)
￿
Section plane
x
y
z
y
z
x
t¿
zy
t
zy
t¿
yz
t
yz
t
t
t
t
Fig.1–23
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Procedure for Analysis
The equation is used to compute only the average shear
stress in the material. Application requires the following steps.
Internal Shear

Section the member at the point where the average shear stress is
to be determined.

Draw the necessary free-body diagram, and calculate the internal
shear force V acting at the section that is necessary to hold the
part in equilibrium.
Average Shear Stress

Determine the sectioned area A, and compute the average shear
stress

It is suggested that be shown on a small volume element of
material located at a point on the section where it is determined.
To do this, first draw on the face of the element, coincident
with the sectioned area A. This shear stress acts in the same
direction as V. The shear stresses acting on the three adjacent
planes can then be drawn in their appropriate directions
following the scheme shown in Fig. 1–23.
t
avg
t
avg
t
avg
= V>A.
t
avg
= V>A
Important Points

If two parts which are thin or small are joined together, the
applied loads can cause shearing of the material with negligible
bending. If this is the case, it is generally suitable for engineering
analysis to assume that an average shear stress acts over the cross-
sectional area.

Oftentimes fasteners, such as nails and bolts, are subjected to
shear loads. The magnitude of a shear force on the fastener is
greatest along a plane which passes through the surfaces being
joined. A carefully drawn free-body diagram of a segment of the
fastener will enable one to obtain the magnitude and direction of
this force.
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EXAMPLE 1.10
The bar shown in Fig.1–24a has a square cross section for which the
depth and thickness are 40 mm. If an axial force of 800 N is applied
along the centroidal axis of the bar’s cross-sectional area, determine
the average normal stress and average shear stress acting on the
material along (a) section plane a–a and (b) section plane b–b.
SOLUTION
Part (a)
Inter
nal Loading.The bar is sectioned, Fig.1–24b, and the internal
resultant loading consists only of an axial force for which
Average Stress.The average normal stress is determined from
Eq. 1–6.
Ans.
No shear stress exists on the section, since the shear force at the
section is zero.
Ans.
NOTE:The distribution of average normal stress over the cross
section is shown in Fig. 1–24c.
t
avg
= 0
s =
P
A
=
800 N
10.04 m210.04 m2
= 500 kPa
P = 800 N.
a
a
b
b
800 N
20 mm
60
(a)
20 mm
(b)
800 N
P  800 N
(c)
500 kPa
500 kPa
Fig.1–24
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Part (b)
Inter
nal Loading.If the bar is sectioned along b–b, the free-body
diagram of the left segment is shown in Fig.
1–24d. Here both a
normal force (N) and shear force (V) act on the sectioned area. Using
x, y axes, we require
or, more directly, using axes,
Solving either set of equations,
Average Stresses.In this case the sectioned area has a thickness
and depth of 40 mm and
respectively,
Fig. 1–24a. Thus the average normal stress is
Ans.
and the average shear stress is
Ans.
NOTE:The stress distribution is shown in Fig.1–24e.
t
avg
=
V
A
=
400 N
10.04 m210.04619 m2
= 217 kPa
s =
N
A
=
692.8 N
10.04 m210.04619 m2
= 375 kPa
40 mm>sin 60° = 46.19 mm,
V = 400 N
N = 692.8 N
V - 800 N sin 30° = 0+Q©F
y¿
= 0;
N - 800 N cos 30° = 0+R©F
x¿
= 0;
y¿x¿,
V sin 60° - N cos 60° = 0+
c
©F
y
= 0;
-800 N + N sin 60° + V cos 60° = 0:
+
©F
x
= 0;
V
800 N
60
(
d
)
30
y
y¿
x¿
x
30
60
N
800
N
(e)
375 kPa
217 kPa
375 kPa
Fig.1–24 (cont.)
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EXAMPLE 1.11
The wooden strut shown in Fig.1–25a is suspended from a 10-mm-
diameter steel rod, which is fastened to the wall. If the strut supports
a vertical load of 5 kN, compute the average shear stress in the rod at
the wall and along the two shaded planes of the strut, one of which is
indicated as abcd.
SOLUTION
Inter
nal Shear.As shown on the free-body diagram in Fig.1–25b,
the rod resists a shear force of 5 kN where it is fastened to the wall. A
free-body diagram of the sectioned segment of the strut that is in
contact with the rod is shown in Fig.
1–25c. Here the shear force
acting along each shaded plane is 2.5 kN.
Average Shear Stress.For the rod,
Ans.
For the strut,
Ans.
NOTE:The average-shear-stress distribution on the sectioned rod
and strut segment is shown in Figs.
1–25d and 1–25e, respectively.
Also shown with these figures is a typical volume element of the
material taken at a point located on the surface of each section. Note
carefully how the shear stress must act on each shaded face of these
elements and then on the adjacent faces of the elements.
t
avg
=
V
A
=
2500 N
10.04 m210.02 m2
= 3.12 MPa
t
avg
=
V
A
=
5000 N
p10.005 m2
2
= 63.7 MPa
c
5 kN
(a)
20 mm
40 mm
b
a
d
(
b
)
5 kN
V  5 k
N
force of
strut on rod
a
d
c
5 kN
V  2.5 kN
V  2.5 kN
force of
rod on strut
(c)
b
(d)
5 kN
63.7 MPa
Fig.1–25
(
e
)
5 kN
3.12 MPa
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The inclined member in Fig.1–26a is subjected to a compressive force
of 600 lb. Determine the average compressive stress along the smooth
areas of contact defined by AB and BC, and the average shear stress
along the horizontal plane defined by EDB.
(b)
3
4
5
600 lb
F
AB
F
BC
(
c
)
V
360 l
b
(d)
3
45
600 lb
160 psi
240 psi
(e)
360 lb
80 psi
(a)
1 in.
3
4
5
600 lb
1.5 in.
3 in.
2 in.
A
C
B
D
E
Fig.1–26
SOLUTION
Internal Loadings.The free-body diagram of the inclined member
is shown in Fig.
1–26b. The compressive forces acting on the areas of
contact are
Also, from the free-body diagram of the top segment of the bottom
member, Fig.1–26c, the shear force acting on the sectioned horizontal
plane EDB is
Average Stress.The average compressive stresses along the
horizontal and vertical planes of the inclined member are
Ans.
Ans
.
These stress distributions are shown in Fig.1–26d.
The average shear stress acting on the horizontal plane defined by
EDB is
Ans.
This stress is shown distributed over the sectioned area in Fig.1–26e.
t
avg
=
360 lb
13 in.211.5 in.2
= 80 psi
s
BC
=
480 lb
12 in.211.5 in.2
= 160 psi

s
AB
=
360 lb
11 in.211.5 in.2
= 240 psi
V = 360 lb:
+
©F
x
= 0;
F
BC
- 600 lb
A
4
5
B
= 0

F
BC
= 480 lb+
c
©F
y
= 0;
F
AB
- 600 lb
A
3
5
B
= 0

F
AB
= 360 lb:
+
©F
x
= 0;
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1–35.The anchor shackle supports a cable force of 600 lb.
If the pin has a diameter of 0.25 in., determine the average
shear stress in the pin.
1–37.The thrust bearing is subjected to the loads shown.
Determine the average normal stress developed on cross
sections through points B, C, and D. Sketch the results on a
differential volume element located at each section.
PROBLEMS
1–34.The column is subjected to an axial force of 8 kN,
which is applied through the centroid of the cross-sectional
area. Determine the average normal stress acting at section
a–a. Show this distribution of stress acting over the area’s
cross section.
*1–36.While running the foot of a 150-lb man is
momentarily subjected to a force which is 5 times his
weight. Determine the average normal stress developed in
the tibia T of his leg at the mid section a–a. The cross
section can be assumed circular, having an outer diameter
of 1.75 in. and an inner diameter of 1 in. Assume the fibula
F does not support a load.
0.25 in.
600 lb
Prob.1–35
750 lb
a
T
F
a
Prob.1–36
8 kN
a
a
75 mm
10 mm
10 mm
10 mm
75 mm
70 mm
70 mm
Prob.1–34
500 N
200 N
65 mm
140 mm
100 mm
B
D
C
150 N
150 N
Prob.1–37
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41
1–38.The small block has a thickness of 5 mm. If the
stress distribution at the support developed by the load
varies as shown, determine the force F applied to the block,
and the distance d to where it is applied.
1–42.The 50-lb lamp is supported by three steel rods
connected by a ring at A.Determine which rod is subjected
to the greater average normal stress and compute its value.
Take The diameter of eachrodis giveninthe figure.
1–43.Solve Prob. 1–42 for
*1–44.The 50-lb lamp is supported by three steel rods
connected by a ring at A. Determine the angle of orientation
of AC such that the average normal stress in rod AC is
twice the average normal stress in rod AD. What is the
magnitude of stress in each rod? The diameter of each rod is
given in the figure.
u
u = 45°.
u = 30°.
1–39.The lever is held to the fixed shaft using a tapered
pin AB, which has a mean diameter of 6 mm. If a couple is
applied to the lever, determine the average shear stress in
the pin between the pin and lever.
1–41.The cinder block has the dimensions shown. If
it is subjected to a centrally applied force of
determine the average normal stress in the material. Show
the result acting on a differential volume element of the
material.
P = 800 lb,
60 mm
120 mm
40 MPa
60 MPa
F
d
180 mm
Prob.1–38
20 N 20 N
250 mm
250 mm
12 mm
A
B
Prob.1–39
0.25 in.
A
D C
B
0.35 in.
0.3 in.
u
45
Probs.1–42/43/44
1 in.
1 in.
4 in.
2 in.
2 in.
1 in.
1 in.
2 in.
3 in.
3 in.
P
Probs.1–40/41
*1–40.
The cinder block has the dimensions shown. If the
material fails when the average normal stress reaches 120 psi,
determine the largest centrally applied vertical load P it can
support.
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 41
42
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HAPTER
1 S
TRESS
1–45.The shaft is subjected to the axial force of 30 kN. If
the shaft passes through the 53-mm diameter hole in the
fixed support A, determine the bearing stress acting on the
collar C. Also, what is the average shear stress acting along
the inside surface of the collar where it is fixed connected to
the 52-mm diameter shaft?
1–49.The open square butt joint is used to transmit a
force of 50 kip from one plate to the other. Determine the
average normal and average shear stress components that
this loading creates on the face of the weld, section AB.
1–46.The two steel members are joined together using a
60° scarf weld. Determine the average normal and average
shear stress resisted in the plane of the weld.
*1–48.The board is subjected to a tensile force of 85 lb.
Determine the average normal and average shear stress
developed in the wood fibers that are oriented along
section a–a at 15° with the axis of the board.
30 mm
25 mm
60
8 kN
8 kN
Prob.1–46
15
3 in.
a
1 in.
85 lb85 lb
a
Prob.1–48
52
0.5 in.
Prob.1–50
10 mm
30 kN
52 mm
60 mm
40 mm
53 mm
A
C
Prob.1–45
C
A B
20
30 mm
40 mm
775 N
Prob.1–47
30
30
50 kip
50 kip
2 in.
6 in.
A
B
Prob.1–49
1–47.
The J hanger is used to support the pipe such that
the force on the vertical bolt is 775 N. Determine the
average normal stress developed in the bolt BC if the bolt
has a diameter of 8 mm. Assume Ais a pin.
1–50.The specimen failed in a tension test at an angle of
52° when the axial load was 19.80 kip. If the diameter of the
specimen is 0.5 in., determine the average normal and
average shear stress acting on the area of the inclined
failure plane. Also, what is the average normal stress acting
on the cross section when failure occurs?
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 42
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ROBLEMS
43
1–51.A tension specimen having a cross-sectional area A
is subjected to an axial force P. Determine the maximum
average shear stress in the specimen and indicate the
orientation of a section on which it occurs.u
1–55.The row of staples AB contained in the stapler is
glued together so that the maximum shear stress the glue
can withstand is Determine the minimum
force F that must be placed on the plunger in order to shear
off a staple from its row and allow it to exit undeformed
through the groove at C. The outer dimensions of the staple
are shown in the figure. It has a thickness of 0.05 in. Assume
all the other parts are rigid and neglect friction.
t
max
= 12 psi.
*1–56.Rods ABandBChave diameters of 4 mmand6 mm,
respectively.If the load of 8 kN is applied to the ring at B,
determine the average normal stress in each rod if
1–57.Rods ABand BChave diameters of 4 mm and 6 mm,
respectively. If the vertical load of 8 kN is applied to the ring
at B, determine the angle of rod BC so that the average
normal stress in each rod is equivalent. What is this stress?
u
u = 60°.
*1–52.The joint is subjected to the axial member force of
5 kN. Determine the average normal stress acting on
sections AB and BC. Assume the member is smooth and is
50-mm thick.
1–54.The two members used in the construction of an
aircraft fuselage are joined together using a 30° fish-mouth
weld.Determine the average normal andaverage shear stress
on the plane of each weld.Assume each inclined plane
supports a horizontal force of 400 lb.
P
u
P
A
Prob.1–51
5 kN
40 mm
A
C
B
50 mm
60
45
Prob.1–52
800 lb 800 lb
30
1 in.
1 in.
1.5 in.
30
Prob.1–54
1–53.
The yoke is subjected to the force and couple
moment. Determine the average shear stress in the bolt
acting on the cross sections through Aand B. The bolt has a
diameter of 0.25 in. Hint:The couple moment is resisted by
a set of couple forces developed in the shank of the bolt.
500 lb
80 lbft
2 in.
2.5 in.
A
B
60
Prob.1–53
F
0.5 in.
0.3 in.
C
A B
Prob.1–55
8 kN
u
C
BA
Probs.1–56/57
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 43
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1–58.The bars of the truss each have a cross-sectional
area of Determine the average normal stress in
each member due to the loading State whether
the stress is tensile or compressive.
1–59.The bars of the truss each have a cross-sectional
area of If the maximumaverage normal stress inany
bar is not toexceed20ksi,determinethemaximummagnitude
P of the loads that can be applied to the truss.
1.25 in
2
.
P = 8 kip.
1.25 in
2
.
1–63.The railcar docklight is supported by the
diameter pin at A. If the lamp weighs 4 lb, and the extension
arm AB has a weight of , determine the average
shear stress in the pin needed to support the lamp. Hint:
The shear force in the pin is caused by the couple moment
required for equilibrium at A.
0.5 lb>ft
1
8
-in.-
*1–64.The two-member frame is subjected to the
distributed loading shown. Determine the average normal
stress and average shear stress acting at sections a–a and
b–b. Member CB has a square cross section of 35 mm on
each side. Take w = 8 kN>m.
*1–60.The plug is used to close the end of the cylindrical
tube that is subjected to an internal pressure of
Determine the average shear stress which the glue exerts on
the sides of the tube needed to hold the cap in place.
p = 650 Pa.
1–62.Solve Prob. 1–61 for pin B. The pin is subjected to
double shear and has a diameter of 0.2 in.
3 ft
4 ft 4 ft
P
0.75 P
E D
A
B C
Probs.1–58/59
3 ft
1.25 in.
B
A
Prob.1–63
P
40 m
m
35 mm
25 mm
Prob.1–60
A
20 lb
20 lb
5 in.
1.5 in.2 in.1 in.
E
C
B
D
Probs.1–61/62
4 m
B
A
C
3 m
b
b
a
a
w
Prob.1–64
1–61.
The crimping tool is used to crimp the end of the
wire E. If a force of 20 lb is applied to the handles,
determine the average shear stress in the pin at A. The pin
is subjected to double shear and has a diameter of 0.2 in.
Only a vertical force is exerted on the wire.
HEBBMC01_0132209918.QXD 7/9/07 2:51 PM Page 44
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ROBLEMS
45
1–67.The beam is supported by a pin at Aand a short link
BC. If determine the average shear stress
developed in the pins at A, B, and C. All pins are in double
shear as shown, and each has a diameter of 18 mm.
*1–68.The beam is supported by a pin at A and a short
link BC. Determine the maximum magnitude P of the loads
the beam will support if the average shear stress in each pin
is not to exceed 80 MPa. All pins are in double shear as
shown, and each has a diameter of 18 mm.
P = 15 kN,
P
2
P
1
P
n
A
m
A
2
A
1
d
1
d
2
d
n
L
1
L
2
L
m
x
Prob.1–66
30
C
1 m
P 4P 4P 2P
1.5 m 1.5 m
AB
0.5m 0.5m
Probs.1–67/68
1–65.
Member A of the timber step joint for a truss is
subjected to a compressive force of 5 kN. Determine the
average normal stress acting in the hanger rod C which has
a diameter of 10 mm and in member B which has a
thickness of 30 mm.
C
10
mm
F
C
m
m
4
0
m
m
30

6
0

5
kN
F
B
E
A
A
A
B
D
F
Prob.1–65
A
C
B
1.5 ft2 ft
0.5 ft
200 lb
u
Prob.1–69
￿
1–66.Consider the general problem of a bar made from
msegments, each having a constant cross-sectional area
and length If there are n loads on the bar as shown,
write a computer program that can be used to determine
the average normal stress at any specified location x. Show
an application of the program using the values
A
2
= 1 in
2
.P
2
= -300 lb,
d
2
= 6 ft,L
2
= 2 ft,A
1
= 3 in
2
,P
1
= 400 lb,d
1
= 2 ft,
L
1
= 4 ft,
L
m
.
A
m
1–69.The frame is subjected to the load of 200 lb.
Determine the average shear stress in the bolt at A as a
function of the bar angle Plot this function,
and indicate the values of for which this stress is a
minimum. The bolt has a diameter of 0.25 in. and is
subjected to single shear.
u