Solution manual for mechanics of materials 10th edition hibbeler sample

731
Solution
sallow =
p r
2 t
; 12(106
) =
300(103
)(1.5)
2 t
t = 0.0188 m = 18.8 mm Ans.
8–1.
A spherical gas tank has an inner radius of r = 1.5 m. If it is
subjected to an internal pressure of p = 300 kPa, determine
its required thickness if the maximum normal stress is not to
exceed 12 MPa.
Ans:
t = 18.8 mm
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732
.
Solution
sallow =
p r
2 t
; 15(103
) =
200 ri
2(0.5)
ri = 75 in.
ro = 75 in. + 0.5 in. = 75.5 in. Ans.
8–2.
A pressurized spherical tank is made of 0.5-in.-thick steel.
If it is subjected to an internal pressure of p = 200 psi,
determine its outer radius if the maximum normal stress
is not to exceed 15 ksi.
Ans:
ro = 75.5 in.

733
8–3.
The thin-walled cylinder can be supported in one of two
ways as shown. Determine the state of stress in the wall of
the cylinder for both cases if the piston P causes the internal
pressure to be 65 psi.The wall has a thickness of 0.25 in. and
the inner diameter of the cylinder is 8 in.
Solution
Case (a):
s1 =
pr
t
; s1 =
65(4)
0.25
= 1.04 ksi Ans.
s2 = 0 Ans.
Case (b):
s1 =
pr
t
; s1 =
65(4)
0.25
= 1.04 ksi Ans.
s2 =
pr
2t
; s2 =
65(4)
2(0.25)
= 520 psi Ans.
P
(a) (b)
P
8 in. 8 in.
Ans:
(a) s1 = 1.04 ksi, s2 = 0,
(b) s1 = 1.04 ksi, s2 = 520 psi

734
.
*8–4.
The tank of the air compressor is subjected to an internal
pressure of 90 psi. If the inner diameter of the tank is 22 in.,
and the wall thickness is 0.25 in., determine the stress
components acting at point A. Draw a volume element of
the material at this point, and show the results on the
element. A
Solution
Hoop Stress for Cylindrical Vessels: Since
r
t
=
11
0.25
= 44 7 10, then thin wall
analysis can be used.Applying Eq. 8–1
s1 =
pr
t
=
90(11)
0.25
= 3960 psi = 3.96 ksi Ans.
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
s2 =
pr
2t
=
90(11)
2(0.25)
= 1980 psi = 1.98 ksi Ans.
Ans:
s1 = 3.96 ksi,
s2 = 1.98 ksi

735
8–5.
Air pressure in the cylinder is increased by exerting forces
P = 2 kN on the two pistons, each having a radius of 45 mm.
If the cylinder has a wall thickness of 2 mm, determine the
state of stress in the wall of the cylinder.
47 mm
P
P
Solution
p =
P
A
=
2(103
)
p(0.0452
)
= 314 380.13 Pa
s1 =
p r
t
=
314 380.13(0.045)
0.002
= 7.07 MPa Ans.
s2 = 0 Ans.
The pressure P is supported by the surface of the pistons in the longitudinal
direction.
Ans:
s1 = 7.07 MPa, s2 = 0

736
.
8–6.
Determine the maximum force P that can be exerted on
each of the two pistons so that the circumferential stress in
the cylinder does not exceed 3 MPa. Each piston
has a radius of 45 mm and the cylinder has a wall thickness
of 2 mm.
47 mm
P
P
Solution
s =
p r
t
; 3(106
) =
p(0.045)
0.002
p = 133.3 kPa
P = pA = 133.31103
2 (p)(0.045)2
= 848 N Ans.
Ans:
P = 848 N

737
Solution
a) s1 =
pr
t
=
1.35(106
)(0.75)
0.008
= 126.56(106
) = 127 MPa Ans.
b) 126.56 (106
)(0.05)(0.008) = s1′(2)(0.04)(0.008)
s1′ = 79.1MPa Ans.
c) From FBD(a)
+ cΣFy = 0; Fb - 79.1(106
)[(0.008)(0.04)] = 0
Fb = 25.3 kN
(tavg)b =
Fb
A
-
25312.5
p
4(0.01)2
= 322 MPa Ans.
8–7.
A boiler is constructed of 8-mm-thick steel plates that
are fastened together at their ends using a butt joint
consisting of two 8-mm cover plates and rivets having a
diameter of 10 mm and spaced 50 mm apart as shown. If
the steam pressure in the boiler is 1.35 MPa, determine
(a) the circumferential stress in the boiler’s plate away
from the seam, (b) the circumferential stress in the outer
cover plate along the rivet line a–a, and (c) the shear
stress in the rivets.
a
8 mm
50 mm a
0.75 m
Ans:
(a) s1 = 127 MPa,
(b) s1′ = 79.1MPa,
(c) (tavg)b = 322 MPa

738
.
Solution
Normal Stress: Since the pipe has two open ends,
slong = s2 = 0 Ans.
Since
r
t
=
6
0.25
= 24 7 10, thin-wall analysis can be used.
sh = s1 =
pr
t
=
250(6)
0.25
= 6000 psi = 6 ksi Ans.
*8–8.
The steel water pipe has an inner diameter of 12 in. and a wall
thickness of 0.25 in. If the valve A is opened and the flowing
water has a pressure of 250 psi as it passes point B, determine
the longitudinal and hoop stress developed in the wall of the
pipe at point B.
A
B
Ans:
slong = 0,
sh = 6 ksi

739
Solution
Normal Stress: Since
r
t
=
6
0.25
= 24 7 10, thin-wall analysis can be used.
shoop = s1 =
pr
t
=
300(6)
0.25
= 7200 psi = 7.20 ksi Ans.
slong = s2 =
pr
2t
=
300(6)
2(0.25)
= 3600 psi = 3.60 ksi Ans.
The state of stress on an element in the pipe wall is shown in Fig. a.
8–9.
The steel water pipe has an inner diameter of 12 in.and a wall
thickness of 0.25 in. If the valve A is closed and the water
pressure is 300 psi, determine the longitudinal and hoop
stress developed in the wall of the pipe at point B. Draw the
state of stress on a volume element located on the wall.
A
B
Ans:
shoop = 7.20 ksi, slong = 3.60 ksi

740
.
Solution
s1 =
400
2(18)(1)
= 1600 psi Ans.
s1 =
pr
t
; 1600 =
p(8)
(18)
p = 25 psi Ans.
P1 =
s1
E
=
1600
29(106
)
= 55.1724(10-6
)
d = P1L = 55.1724(10-6
)(p)a8 +
1
16
b = 0.00140 in. Ans.
8–10.
The A-36-steel band is 2 in. wide and is secured around the
smooth rigid cylinder. If the bolts are tightened so that the
tension in them is 400 lb, determine the normal stress in
the band, the pressure exerted on the cylinder, and the
distance half the band stretches.
8 in.
in.1—
8
Ans:
s1 = 1.60 ksi, p = 25 psi, d = 0.00140 in.

741
Solution
Require
dF = dT; dF =
PL
AE
=
sL
E
, dT = a∆TL
s2(20)(12)
29(106
)
= (6.60)(10-6
)(60)(20)(12)
s2 = 11.5 ksi Ans.
s1 =
pr
t
=
600(10)
0.25
= 24 ksi Ans.
8–11.
The gas pipe line is supported every 20 ft by concrete piers
and also lays on the ground. If there are rigid retainers at
the piers that hold the pipe fixed,determine the longitudinal
and hoop stress in the pipe if the temperature rises 60° F
from the temperature at which it was installed. The gas
within the pipe is at a pressure of 600 lbin2.The pipe has an
inner diameter of 20 in. and thickness of 0.25 in. The
material is A-36 steel.
20 ft
Ans:
s2 = 11.5 ksi,
s1 = 24 ksi

742
.
Solution
+cΣFy = 0; p(0.225)2
450(103
) - tavg (2p)(0.225)(0.01) = 0;
tavg = 5.06 MPa Ans.
s1 =
p r
t
=
450(103
)(0.225)
0.02
= 5.06 MPa Ans.
s2 =
p r
2 t
=
450(103
)(0.225)
2(0.02)
= 2.53 MPa Ans.
*8–12.
A pressure-vessel head is fabricated by welding the circular
plate to the end of the vessel as shown. If the vessel sustains
an internal pressure of 450 kPa, determine the average
shear stress in the weld and the state of stress in the wall of
the vessel.
450 mm
10 mm
20 mm
Ans:
tavg = 5.06 MPa,
s1 = 5.06 MPa,
s2 = 2.53 MPa

743
Solution
dT = a ∆ TL
p(24) - p(23.99) = 6.60(10-6
)(T1 - 65)(p)(23.99)
T1 = 128.16°F = 128°F Ans.
Cool down:
dF = dT

FL
AE
= a ∆ TL

F(p)(24)
(1)(0.25)(29)(106
)
= 6.60(10-6
)(128.16 - 65)(p)(24)
F = 3022.21 lb
s1 =
F
A
; s1 =
3022.21
(1)(0.25)
= 12 088 psi = 12.1 ksi Ans.
s1 =
pr
t
; 12 088 =
p(12)
(0.25)
P = 252 psi Ans.
8–13.
An A-36-steel hoop has an inner diameter of 23.99 in.,
thickness of 0.25 in., and width of 1 in. If it and the
24-in.-diameter rigid cylinder have a temperature of 65° F,
determine the temperature to which the hoop should be
heated in order for it to just slip over the cylinder.What is the
pressure the hoop exerts on the cylinder, and the tensile stress
in the ring when it cools back down to 65° F?
24 in.
Ans:
T1 = 128°F, s1 = 12.1 ksi, p = 252 psi

744
.
Solution
Equilibrium for the Ring: From the FBD
+S ΣFx = 0; 2P - 2pri w = 0 P = pri w
Hoop Stress and Strain for the Ring:
s1 =
P
A
=
priw
(ro - ri)w
=
pri
ro - ri
Using Hooke’s Law
P1 =
s1
E
=
pri
E(ro - ri)
(1)
However, P1 =
2p(ri)1 - 2pri
2pri
=
(ri)1 - ri
ri
=
dri
ri
.
Then, from Eq. (1)

dri
ri
=
pri
E(ro - ri)
dri =
pri
2
E(ro - ri)
Ans.
8–14.
The ring, having the dimensions shown, is placed over a
flexible membrane which is pumped up with a pressure p.
Determine the change in the inner radius of the ring after
this pressure is applied. The modulus of elasticity for the
ring is E.
p
ro
w
ri
Ans:
dri =
pri
2
E(ro - ri)

745
Solution
Equilibrium for the Ring: From the FBD
+S ΣFx = 0;2P - 2pri w = 0 P = pri w
Hoop Stress and Strain for the Ring:
s1 =
P
A
=
priw
(ro - ri)w
=
pri
ro - ri
Using Hooke’s Law
P1 =
s1
E
=
pri
E(ro - ri)
(1)
However, P1 =
2p(ri)1 - 2pri
2pri
=
(ri)1 - ri
ri
=
dri
ri
.
Then, from Eq. (1)

dri
ri
=
pri
E(ro - ri)
dri =
pri
2
E(ro - ri)
Compatibility: The pressure between the rings requires
dr2 + dr3 = r2 - r3 (2)
From the result obtained above
dr2 =
pr2
2
E(r2 - r1)
dr3 =
pr3
2
E(r4 - r3)
Substitute into Eq. (2)

pr2
2
E(r2 - r1)
+
pr3
2
E(r4 - r3)
= r2 - r3
p =
E(r2 - r3)
r2
2
r2 - r1
+
r3
2
r4 - r3
Ans.
8–15.
The inner ring A has an inner radius r1 and outer radius r2.
The outer ring B has an inner radius r3 and an outer radius
r4, and r2 7 r3. If the outer ring is heated and then fitted over
the inner ring, determine the pressure between the two rings
when ring B reaches the temperature of the inner ring. The
material has a modulus of elasticity of E and a coefficient of
thermal expansion of a.
r1
r2
r3
A
B
r4
Ans:
p =
E(r2 - r3)
r2
2
r2 - r1
+
r3
2
r4 - r3

746
.
Solution
Normal Pressure: Vertical force equilibrium for FBD(a).
+ cΣFy = 0; 103p(242
)4 - N = 0 N = 5760p lb
The Friction Force: Applying friction formula
Ff = msN = 0.5(5760p) = 2880p lb
a) The Required Torque: In order to initiate rotation of the two hemispheres relative
to each other, the torque must overcome the moment produced by the friction force
about the center of the sphere.
T = Ff r = 2880p(2 + 0.12512) = 18190 lb # ft = 18.2 kip # ft Ans.
b) The Required Vertical Force: In order to just pull the two hemispheres apart, the
vertical force P must overcome the normal force.
P = N = 5760p = 18096 lb = 18.1 kip Ans.
c) The Required Horizontal Force: In order to just cause the two hemispheres to
slide relative to each other, the horizontal force F must overcome the friction force.
F = Ff = 2880p = 9048 lb = 9.05 kip Ans.
*8–16.
Two hemispheres having an inner radius of 2 ft and wall
thickness of 0.25 in. are fitted together, and the inside
pressure is reduced to -10 psi. If the coefficient
of static friction is ms = 0.5 between the hemispheres,
determine (a) the torque T needed to initiate the rotation
of the top hemisphere relative to the bottom one, (b) the
vertical force needed to pull the top hemisphere off the
bottom one, and (c) the horizontal force needed to slide
the top hemisphere off the bottom one.
Ans:
(a) T = 18.2 kip # ft,
(b) P = 18.1 kip,
(c) F = 9.05 kip

747
.
Solution
Normal Stress in the Wall and Filament Before the Internal Pressure is Applied:
The entire length L of wall is subjected to pretension filament force T. Hence, from
equilibrium, the normal stress in the wall at this state is
2T - (s′)w (2Lt) = 0 (s′)w =
T
Lt
and for the filament the normal stress is
(s′)fil =
T
wt′
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: In
order to use s1 = prt, developed for a vessel of uniform thickness, we redistribute
the filament’s cross-section as if it were thinner and wider, to cover the vessel with
no gaps. The modified filament has width L and thickness twL, still with cross-
sectional area wt subjected to tension T.Then the stress in the filament becomes
sfil = s + (s′)fil =
pr
(t + t′wL)
+
T
wt′
Ans.
And for the wall,
sw = s - (s′)w =
pr
(t + t′wL)
-
T
Lt
Ans.
Check: 2wt′sfil + 2Ltsw = 2rLp OK
8–17.
In order to increase the strength of the pressure vessel,
filament winding of the same material is wrapped around
the circumference of the vessel as shown. If the pretension
in the filament is T and the vessel is subjected to an internal
pressure p, determine the hoop stresses in the filament and
in the wall of the vessel. Use the free-body diagram shown,
and assume the filament winding has a thickness t′ and
width w for a corresponding length L of the vessel.
T
p
w
t¿
L
t
T
s1
s1
Ans:
sfil =
pr
t + t′wL
+
T
wt′
,
sw =
pr
t + t′wL
-
T
Lt

748
.
8–18.
Determine the shortest distance d to the edge of the plate at
which the force P can be applied so that it produces no
compressive stresses in the plate at section a–a. The plate has
a thickness of 10 mm and P acts along the centerline of this
thickness.
Solution
sA = 0 = sa - sb
0 =
P
A
-
M c
I
0 =
P
(0.2)(0.01)
-
P(0.1 - d)(0.1)
1
12 (0.01)(0.23
)
P(-1000 + 15000 d) = 0
d = 0.0667 m = 66.7 mm Ans.
a
500 mm
P
a
300 mm
200 mm
d
Ans:
d = 66.7 mm

749
.
8–19.
Determine the maximum distance d to the edge of the plate
at which the force P can be applied so that it produces no
compressive stresses on the plate at section a–a. The plate
has a thickness of 20 mm and P acts along the centerline of
this thickness.
Solution
Internal Loadings: Consider the equilibrium of the left segment of the plate
sectioned through section a–a, Fig. a,
S+ ΣFx = 0; N - P = 0 N = P
a+ΣMo = 0; P(d - 0.1) - M = 0 M = P(d - 0.1)
Section Properties: For the rectangular cross section,
A = 0.2(0.02) = 0.004 m2
I =
1
12
(0.02)(0.23
) = 13.3333(10-6
) m4
Normal Stress: It is required that sA = 0. For the combined loadings,
sA =
N
A
-
Mc
I
0 =
P
0.004
-
P(d - 0.1)(0.1)
13.3333(10-6
)
d = 0.1333 m = 133 mm Ans.
a
a
P
d
200 mm
Ans:
d = 133 mm

750
.
*8–20.
The plate has a thickness of 20 mm and the force
P = 3 kN acts along the centerline of this thickness such that
d = 150 mm. Plot the distribution of normal stress acting
along section a–a.
Solution
Internal Loadings: Consider the equilibrium of the left segment of the plate
sectioned through section a–a, Fig. a,
+S ΣFx = 0; N - 3 = 0 N = 3.00 kN
a+ΣMo = 0; 3(0.05) - M = 0 M = 0.150 kN # m
A = 0.2(0.02) = 0.004 m2
I =
1
12
(0.02)(0.23
) = 13.3333(10-6
) m4
Normal Stress: For the combined loadings, the normal stress at points A and B can
be determined from
s =
N
A
{
Mc
I
=
3.00(103
)
0.004
{
0.150(103
)(0.1)
13.3333(10-6
)
sA = 750(103
) - 1.125(106
) = -0.375(106
) Pa = 0.375 MPa (C) Ans.
sB = 750(103
) + 1.125(106
) = 1.875(106
) Pa = 1.875 MPa (T) Ans.
Using similar triangles, the location of the neutral axis can be determined
y
1.875
=
0.2 - y
0.375
; y = 0.1667 m
a
a
P
d
200 mm
Ans:
sA = 0.375 MPa (C),
sB = 1.875 MPa (T)

751
.
8–21.
If the load has a weight of 600 lb, determine the maximum
normal stress on the cross section of the supporting member
at section a–a.Also, plot the normal-stress distribution over
the cross section.
Solution
Internal Loadings: Consider the equilibrium of the free-body diagram of the bottom
cut segment shown in Fig. a.
a+ cΣFy = 0; N - 600 = 0 N = 600 lb
a+ΣMC = 0; 600(1.5) - M = 0 M = 900 lb # ft
Section Properties: The cross-sectional area and the moment of inertia about the
centroidal axis of the member are
A = p(12
) = p in2
I =
p
4
(14
) =
p
4
in4
Normal Stress: The normal stress is the combination of axial and bending stress.
Thus,
s =
N
A
{
Mc
I
By observation, the maximum normal stress occurs at point B, Fig. b.Thus,
smax = sB =
600
p
+
900(12)(1)
p4
= 13.9 ksi (T) Ans.
For Point A,
sA =
600
p
+ -
900(12)(1)
p4
= -13.6 ksi = 13.6 ksi (C) Ans.
Using these results, the normal stress distribution over the cross section is shown in
Fig. b.The location of the neutral axis can be determined from
p
13.9
=
2 - x
13.6
; x = 1.01 in.
Ans:
smax = sB = 13.9 ksi (T), sA = 13.6 ksi (C)
1.5 ft
1 in.a a
Section a – a

752
.
8–22.
The steel bracket is used to connect the ends of two cables.
If the allowable normal stress for the steel is
sallow = 30 ksi, determine the largest tensile force P that can
be applied to the cables.Assume the bracket is a rod having
a diameter of 1.5 in.
Solution
Internal Loading: Consider the equilibrium of the left segment of the bracket
sectioned through an arbitrary cross section, Fig. a.
+S ΣFx = 0; N - P = 0 N = P
a+ΣMo = 0; M - P(4.75) = 0 M = 4.75P
Section Properties: For the circular cross section,
A = pc2
= p(0.752
) = 0.5625p in2
I =
p
4
c4
=
p
4
(0.754
) = 0.24850 in4
Allowable Normal Stress: The maximum normal stress occurs at the bottom of the
circular cross section. For the combined loadings,
smax = sallow =
N
A
+
Mc
I
30(103
) =
P
0.5625p
+
4.75P(0.75)
0.24850
P = 2.01320(103
) lb
= 2.01 kip Ans.
4 in.
P P
Ans:
Pmax = 2.01 kip

753
.
8–23.
The steel bracket is used to connect the ends of two cables.
If the applied force P = 1.50 kip, determine the maximum
normal stress in the bracket. Assume the bracket is a rod
having a diameter of 1.5 in.
Solution
Internal Loading: Consider the equilibrium of the left segment of the bracket
sectioned through an arbitrary cross section, Fig. a.
+S ΣFx = 0; N - 1.50 = 0 N = 1.50 kip
a+ΣMo = 0; M - 1.50(4.75) = 0 M = 7.125 kip # in.
A = pc2
= p(0.752
) = 0.5625p in2
I =
p
4
c4
=
p
4
(0.754
) = 0.24850 in4
Maximum Normal Stress: The maximum normal stress occurs at the bottom of the
circular cross section. For the combined loadings,
smax =
N
A
+
Mc
I
=
1.50
0.5625p
+
7.125(0.75)
0.24850
= 22.35 ksi = 22.4 ksi (T) Ans.
4 in.
P P
Ans:
smax = 22.4 ksi (T)

754
.
*8–24.
The column is built up by gluing the two boards together.
Determine the maximum normal stress on the cross section
when the eccentric force of P = 50 kN is applied.
Solution
Section Properties: The location of the centroid of the cross section, Fig. a, is
y =
ΣyA
ΣA
=
0.075(0.15)(0.3) + 0.3(0.3)(0.15)
0.15(0.3) + 0.3(0.15)
= 0.1875 m
The cross-sectional area and the moment of inertia about the z axis of the cross
section are
A = 0.15(0.3) + 0.3(0.15) = 0.09 m2
Iz =
1
12
(0.3)(0.153
) + 0.3(0.15)(0.1875 - 0.075)2
+
1
12
(0.15)(0.33
) + 0.15(0.3)(0.3 - 0.1875)2
= 1.5609(10-3
) m4
Equivalent Force System: Referring to Fig. b,
+ cΣFx = (FR)x; -50 = -F F = 50 kN
ΣMz = (MR)z; -50(0.2125) = -M M = 10.625 kN # m
Normal Stress: The normal stress is a combination of axial and bending stress.Thus,
s =
N
A
+
My
I
By inspection, the maximum normal stress occurs at points along the edge where
y = 0.45 - 0.1875 = 0.2625 m such as point A.Thus,
smax =
-50(103
)
0.09
-
10.625(103
)(0.2625)
1.5609(10-3
)
= -2.342 MPa = 2.34 MPa (C) Ans.
150 mm
150 mm
250 mm
75 mm
300 mm
50 mm
P
Ans:
smax = 2.34 MPa (C)

755
.
8–25.
The column is built up by gluing the two boards together. If
the wood has an allowable normal stress of sallow = 6 MPa,
determine the maximum allowable eccentric force P that
can be applied to the column.
Solution
Section Properties: The location of the centroid c of the cross section, Fig. a, is
y =
ΣyA
ΣA
=
0.075(0.15)(0.3) + 0.3(0.3)(0.15)
0.15(0.3) + 0.3(0.15)
= 0.1875 m
The cross-sectional area and the moment of inertia about the z axis of the cross
section are
A = 0.15(0.3) + 0.3(0.15) = 0.09 m2
Iz =
1
12
(0.3)(0.153
) + 0.3(0.15)(0.1875 - 0.075)2
+
1
12
(0.15)(0.33
) + 0.15(0.3)(0.3 - 0.1875)2
= 1.5609(10-3
) m4
Equivalent Force System: Referring to Fig. b,
+cΣFx = (FR)x; -P = -F F = P
ΣMz = (MR)z; -P(0.2125) = -M M = 0.2125P
F =
N
A
+
My
I
By inspection, the maximum normal stress, which is compression, occurs at points
along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A.Thus,
-6(106
) =
-P
0.09
-
0.2125P(0.2625)
1.5609(10-3
)
P = 128 076.92 N = 128 kN Ans.
150 mm
150 mm
250 mm
75 mm
300 mm
50 mm
P
Ans:
Pmax = 128 kN

756
.
Solution
A = 0.75(0.5) = 0.375 in2
I =
1
12
(0.5)(0.753
) = 0.017578 in4
smax =
P
A
+
Mc
I
=
500
0.375
+
2000(0.375)
0.017578
= 44.0 ksi (T) Ans.
8–26.
The screw of the clamp exerts a compressive force of 500 lb
on the wood blocks. Determine the maximum normal stress
along section a–a. The cross section is rectangular, 0.75 in.
by 0.50 in.
4 in.
0.75 in.
a
a
Ans:
smax = 44.0 ksi (T)

757
.
Solution
A = 0.75(0.5) = 0.375 in2
I =
1
12
(0.5)(0.753
) = 0.017578 in4
smax =
P
A
+
Mc
I
=
500
0.375
+
2000(0.375)
0.017578
= 44.0 ksi (T)
smin =
P
A
-
Mc
I
=
500
0.375
-
2000(0.375)
0.017578
= -41.3 ksi (C)
y
41.33
=
(0.75 - y)
44.0
y = 0.363 in.
8–27.
The screw of the clamp exerts a compressive force of 500 lb
on the wood blocks. Sketch the stress distribution along
section a–a of the clamp. The cross section is rectangular,
0.75 in. by 0.50 in.
4 in.
0.75 in.
a
a
Ans:
smax = 44.0 ksi (T),
smin = -41.3 ksi (C)

758
.
*8–28.
The joint is subjected to the force system shown. Sketch the
normal-stress distribution acting over section a–a if the
member has a rectangular cross section of width 0.5 in. and
thickness 1 in.
a
A
B
a
1 in.
2 in.
500 lb
3
45
1.5 in.
250 lb
A
B
Solution
Internal Loadings: Consider the equilibrium of the lower segment of the joint
sectioned through a–a, Fig. a.
+S ΣFx = 0; V - 250 - 500a
3
5
b = 0 V = 550 lb
+ cΣFy = 0; N - 500a
4
5
b = 0 N = 400 lb
a+ ΣMo = 0; M + 500a
4
5
b(1.5) - 500a
3
5
b(2) - 250(2) = 0
M = 500 lb # in.
A = 0.5(1) = 0.5 in2
I =
1
12
(0.5)(13
) = 0.04167 in4
Normal Stress: For the combined loadings, the normal stress at points B and C,
Fig. b, can be determined from
s =
N
A
{
Mc
I
=
400
0.5
{
500(0.5)
0.04167
sC = 800 + 6000 = 6800 psi = 6.80 ksi (T) Ans.
sB = 800 - 6000 = -5200 psi = 5.20 ksi (C) Ans.
Using similar triangles,
1 - y
5.20
=
y
6.80
; y = 0.5667 in.
Using these results, the normal stress distribution over section a–a, shown in Fig. b,
can be sketched.
Ans:
sC = 6.80 ksi (T),
sB = 5.20 ksi (C)

759
.
8–29.
The joint is subjected to the force system shown. Determine
the state of stress at points A and B, and sketch the results
on differential elements located at these points.The member
has a rectangular cross-sectional area of width 0.5 in. and
thickness 1 in.
a
A
B
a
1 in.
2 in.
500 lb
3
45
1.5 in.
250 lb
A
B
Solution
Internal Loadings: Consider the equilibrium of the lower segment of the joint
sectioned through a–a, Fig. a,
+S ΣFx = 0; V - 250 - 500a
3
5
b = 0 V = 550 lb
+ cΣFy = 0; N - 500a
4
5
b = 0 N = 400 lb
a+ ΣMo = 0; M + 500a
4
5
b(1.5) - 500a
3
5
b(2) - 250(2) = 0
M = 500 lb # in.
A = 0.5(1) = 0.5 in2
I =
1
12
(0.5)(13
) = 0.04167 in4
For points A and B shown in Fig. b,
QB = 0 QA = y′A′ = 0.25[0.5(0.5)] = 0.0625 in3
Normal Stress: For the combined loadings, the normal stress at points A and B,
Fig. b, can be determined from
sA =
N
A
+
MyA
I
=
400
0.5
+
500(0)
0.04167
= 800 psi = 0.800 ksi (T) Ans.
sB =
N
A
-
Mc
I
=
400
0.5
-
500(0.5)
0.04167
= - 5200 psi = 5.20 ksi (C) Ans.
Shear Stress: Applying the shear formula,
tA =
VQA
It
=
550(0.0625)
0.04167(0.5)
= 1650 psi = 1.65 ksi Ans.
tB =
VQB
It
= 0 Ans.
Using these results, the state of stress acting on the differential elements at points
A and B shown in Fig. c can be sketched.
Ans:
sA = 0.800 ksi (T),
sB = 5.20 ksi (C),
tA = 1.65 ksi,
tB = 0

760
.
Solution
Support Reactions: Referring to the free-body diagram of the handle shown in
Fig. a,
a+ΣMD = 0; 100(0.25) - FC (0.05) = 0 FC = 500 N
Internal Loadings: Consider the equilibrium of the free-body diagram of the
segment shown in Fig. b.
ΣFy′ = 0; 500 - V = 0 V = 500 N
a+ΣMC = 0; M - 500(0.025) = 0 M = 12.5 N # m
Section Properties: The moment of inertia of the cross section about the centroidal
axis is
I =
1
12
(0.0075)(0.023
) = 5(10-9
) m4
Referring to Fig. c, QA and QB are
QA = 0
QB = y′A′ = 0.005(0.01)(0.0075) = 0.375(10-6
) m3
Normal Stress: The normal stress is contributed by bending stress only. Thus
s =
My
I
For point A, y = 0.01 m. Then
sA = -
12.5(0.01)
5(10-9
)
= -25 MPa = 25 MPa (C) Ans.
For point B, y = 0. Then
sB = 0 Ans.
8–30.
The rib-joint pliers are used to grip the smooth pipe C. If the
force of 100 N is applied to the handles, determine the state
of stress at points A and B on the cross section of the jaw at
section a–a. Indicate the results on an element at each point.
250 mm
100 N
100 N
a
Section a – a
a
A
B
C
25 mm
25 mm
10 mm
20 mm 7.5 mm
45°

761
.
8–30. Continued
Ans:
sA = 25 MPa (C), sB = 0,
tA = 0, tB = 5 MPa
Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,
tA =
VQA
It
= 0 Ans.
tB =
VQB
It
=
500[0.375(10-6
)]
5(10-9
)(0.0075)
= 5 MPa Ans.
The state of stress of points A and B are represented by the elements shown in
Figs. d and e respectively.

762
.
Solution
+S ΣFx = 0; NA - 150 cos 30° = 0
NA = 129.9038 lb
+ cΣFy = 0; VA - 150 sin 30° = 0
VA = 75 lb
a+ΣMA = 0; 150 cos 30°(1.5) + 150 sin 30°(2) - MA = 0
MA = 344.8557 lb # in.
sA =
P
A
+
Mc
I
=
129.9038
p(1
4)2
+
344.8557(1
4)
p
4 (1
4)4
= 28.8 ksi Ans.
tA = 0 (Since QA = 0) Ans.
8–31.
The 1
2-in.-diameter bolt hook is subjected to the load of
F = 150 lb. Determine the stress components at point A on
the shank. Show the result on a volume element located at
this point.
2 in.
1.5 in.30Њ
F = 150 lb
A
B2 in.
Ans:
sA = 28.8 ksi,
tA = 0

763
.
Solution
+S ΣFx = 0; NB - 150 cos 30° = 0; NB = 129.9038
+ cΣFy = 0; VB - 150 sin 30° = 0; VB = 75 lb
a+ΣMB = 0; 150 cos 30°(1.5) + 150 sin 30°(4) - MB = 0
MB = 494.8557 lb # in.
sB =
P
A
-
Mc
I
=
129.9038
p(1
4)2
-
494.8557(1
4)
p
4 (1
4)4
= -39.7 ksi Ans.
*8–32.
The 1
2-in.-diameter bolt hook is subjected to the load of
F = 150 lb. Determine the stress components at point B on
the shank. Show the result on a volume element located at
this point.
2 in.
1.5 in.30Њ
F = 150 lb
A
B2 in.
Ans:
sB = -39.7 ksi

764
.
8–33.
The block is subjected to the eccentric load shown.
Determine the normal stress developed at points A and B.
Neglect the weight of the block.
A B
C
a
a
100 mm150 kN
150 mm
Solution
Internal Loadings: Consider the equilibrium of the upper segment of the block
ΣFx = 0; N + 150 = 0 N = -150 kN
ΣMy = 0; My + 150(0.05) = 0 My = -7.50 kN # m
ΣMz = 0; Mz - 150(0.075) = 0 Mz = 11.25 kN # m
A = 0.1(0.15) = 0.015 m2
Iy =
1
12
(0.15)(0.13
) = 12.5(10-6
) m4
Iz =
1
12
(0.1)(0.153
) = 28.125(10-6
) m4
Normal Stresses: For the combined loadings, the normal stress can be determined
from
s =
N
A
-
Mzy
Iz
+
Myz
Iy
For point A, yA = 0.075 m and zA = 0.05 m
sA =
-150(103
)
0.015
-
11.25(103
)(0.075)
28.125(10-6
)
+
3 -7.50(103
)4(0.05)
12.5(10-6
)
= -70.0(106
) Pa = 70.0 MPa (C) Ans.
For point B, yB = 0.075 m and zB = -0.05 m
sB =
-150(103
)
0.015
-
11.25(103
)(0.075)
28.125(10-6
)
+
3 -7.50(103
)4(-0.05)
12.5(10-6
)
= -10.0(106
) Pa = 10.0 MPa (C) Ans.
Ans:
sA = 70.0 MPa (C),
sB = 10.0 MPa (C)

765
.
8–34.
The block is subjected to the eccentric load shown. Sketch
the normal-stress distribution acting over the cross section
at section a–a. Neglect the weight of the block.
A B
C
a
a
100 mm150 kN
150 mm
Solution
Internal Loadings: Consider the equilibrium of the upper segment of the block
ΣFx = 0; N + 150 = 0 N = -150 kN
ΣMy = 0; My + 150(0.05) = 0 My = -7.50 kN # m
ΣMz = 0; Mz - 150(0.075) = 0 Mz = 11.25 kN # m
A = 0.1(0.15) = 0.015 m2
Iy =
1
12
(0.15)(0.13
) = 12.5(10-6
) m4
Iz =
1
12
(0.1)(0.153
) = 28.125(10-6
) m4
Normal Stress: For the combined loadings, the normal stress can be determined
from
s =
N
A
-
Mzy
Iz
+
Myz
Iy
For point A, yA = 0.075 m and zA = 0.05 m
sA =
-150(103
)
0.015
-
11.25(103
)(0.075)
28.125(10-6
)
+
(-7.50)(103
)(0.05)
12.5(10-6
)
= -70.0(106
) Pa = 70.0 MPa (C) Ans.
For point B, yB = 0.075 m and zB = -0.05 m
sB =
-150(103
)
0.015
-
11.25(103
)(0.075)
28.125(10-6
)
+
(-7.50)(103
)(-0.05)
12.5(10-6
)
= -10.0(106
) Pa = 10.0 MPa (C) Ans.
For point C, yC = -0.075 m and zC = -0.05 m
sC =
-150(103
)
0.015
-
11.25(103
)(-0.075)
28.125(10-6
)
+
(-7.50)(103
)(-0.05)
12.5(10-6
)
= 50.0(106
) Pa = 50.0 MPa (T) Ans.
For point D, yD = -0.075 m and zD = 0.05 m
sD =
-150(103
)
0.015
-
11.25(103
)(-0.075)
28.125(10-6
)
+
(-7.50)(103
)(0.05)
12.5(10-6
)
= -10.0(106
) Pa = 10.0 MPa (C) Ans.

766
.
8–34. Continued
The location of neutral axis can be found using similar triangles.
y
10.0
=
0.15 - y
50.0
; y = 0.025 m
z
10.0
=
0.1 - z
50.0
; z = 0.01667 m
Using these result, the normal stress distribution over the cross section shown in
Fig. b can be sketched.
Ans:
sA = 70.0 MPa (C),
sB = 10.0 MPa (C),
sC = 50.0 MPa (T),
sD = 10.0 MPa (C)

767
.
Ans:
sA = 27.3 ksi (T),
sB = 0.289 ksi (T),
tA = 0,
tB = 0.750 ksi
Solution
Support Reactions:
+ cΣFy = 0; 2000 - 2F cos 30° = 0 F = 1154.70 lb
Internal Forces and Moment:
+ ΣFx = 0; 1154.70 sin 30° - N = 0 N = 577.35 lb
+ cΣFy = 0; V - 1154.70 cos 30° = 0 V = 1000 lb
a+ΣMB = 0; M - 1154.70 cos 30°(1.5) = 0
M = 1500 lb # ft
Section Properties:
A = 1(2) = 2.00 in2
I =
1
12
(1)(23
) = 0.6666 in4
QB = y′A′ = 0.5(1)(1) = 0.500 in3
QA = 0
Normal Stress:
s =
N
A
{
My
I
sA =
577.35
2.00
+
1500(12)(1)
0.6666
= 27 300 psi = 27.3 ksi (T) Ans.
sB =
577.35
2.00
+
1500(12)(0)
0.6666
= 289 psi = 0.289 ksi (T) Ans.
t =
VQ
It
tA = 0 Ans.
tB =
1000(0.500)
0.6666(1)
= 750 psi = 0.750 ksi Ans.
8–35.
The spreader bar is used to lift the 2000-lb tank. Determine
the state of stress at points A and B, and indicate the results
on a differential volume element.
1.5 ft0.5 ft
30Њ30Њ
A
B
1 in.
1 in.
A
B

768
.
Solution
Internal Loadings: Consider the equilibrium of the free-body diagram
of the drill’s right cut segment, Fig. a.
ΣFx = 0; N - 150a
4
5
b = 0 N = 120 N
ΣFy = 0; 150a
3
5
b - Vy = 0 Vy = 90 N
ΣMx = 0; 20 - T = 0 T = 20 N # m
ΣMz = 0; -150a
3
5
b(0.4) + 150a
4
5
b(0.125) + Mz = 0
Mz = 21N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis,
and the polar moment of inertia of the drill’s cross section are
A = p10.0052
2 = 25p110-6
2 m2
Iz =
p
4
10.0054
2 = 0.15625p110-9
2 m4
J =
p
2
10.0054
2 = 0.3125p110-9
2 m4
Referring to Fig. b, QA is
QA = 0
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
s =
N
A
-
Mzy
Iz
For point A, y = 0.005 m. Then
sA =
-120
25p110-6
2
-
21(0.005)
0.15625p110-9
2
= -215.43 MPa = 215 MPa (C) Ans.
*8–36.
The drill is jammed in the wall and is subjected to the torque
and force shown. Determine the state of stress at point A on
the cross section of the drill bit at section a–a.
150 N
3
4
5
125 mm
20 N·m
400 mm
a
a
5 mm
B
A
Section a – a
z
x
y
y

769
.
*8–36. Continued
Shear Stress: The transverse shear stress developed at point A is
c 1txy2V d
A
=
VyQA
Izt
= 0
The torsional shear stress developed at point A is
3(txz)T4A =
Tc
J
=
20(0.005)
0.3125p110-9
2
= 101.86 MPa
Thus,
1txy2A = 0 Ans.
1txz2A = c 1txz2T d
A
= 102 MPa Ans.
The state of stress at point A is represented on the element shown in Fig. c.
Ans:
sA = 215 MPa (C),
1txy2A = 0,
1txz2A = 102 MPa

770
.
Solution
Internal Loadings: Consider the equilibrium of the free-body diagram
of the drill’s right cut segment, Fig. a.
ΣFx = 0; N - 150a
4
5
b = 0 N = 120 N
ΣFy = 0; 150a
3
5
b - Vy = 0 Vy = 90 N
ΣMx = 0; 20 - T = 0 T = 20 N # m
ΣMz = 0; -150a
3
5
b(0.4) + 150a
4
5
b(0.125) + Mz = 0
Mz = 21 N # m
Section Properties: The cross-sectional area, the moment of inertia about the z axis,
and the polar moment of inertia of the drill’s cross section are
A = p10.0052
2 = 25p110-6
2 m2
Iz =
p
2
10.0054
2 = 0.15625p110-9
2 m4
J =
p
2
10.0054
2 = 0.3125p110-9
2 m4
Referring to Fig. b, QB is
QB = y′A′ =
4(0.005)
3p
c
p
2
10.0052
2 d = 83.333110-9
2 m3
Normal Stress: The normal stress is a combination of axial and bending stress. Thus,
s =
N
A
-
Mzy
Iz
For point B, y = 0. Then
sB =
-120
25p110-6
2
- 0 = -1.528 MPa = 1.53 MPa (C) Ans.
8–37.
The drill is jammed in the wall and is subjected to the torque
and force shown. Determine the state of stress at point B on
the cross section of the drill bit at section a–a.
150 N
3
4
5
125 mm
20 N·m
400 mm
a
a
5 mm
B
A
Section a – a
z
x
y
y

771
.
Shear Stress: The transverse shear stress developed at point B is
c 1txy2V d
B
=
VyQB
Izt
=
90c 83.333110-9
2 d
0.15625p110-9
2(0.01)
= 1.528 MPa
The torsional shear stress developed at point B is
c 1txy2T d
B
=
Tc
J
=
20(0.005)
0.3125p110-9
2
= 101.86 MPa
Thus,
1txz2B = 0 Ans.
1txy2B = c 1txy2T d
B
- c 1txy2V d
B
= 101.86 - 1.528 = 100.33 MPa = 100 MPa Ans.
The state of stress at point B is represented on the element shown in Fig. d.
8–37. Continued
Ans:
sB = 1.53 MPa (C), 1txz2B = 100 MPa

772
.
Solution
sD = -
P
A
-
My
I
= -
8(103
)
(0.1)(0.05)
-
12(103
)(0.03)
1
12(0.05)(0.1)3

sD = -88.0 MPa Ans.
tD = 0 Ans.
8–38.
The frame supports the distributed load shown. Determine
the state of stress acting at point D. Show the results on a
differential element at this point.
4 kN/m
D
B
A
C
E
1.5 m 1.5 m
20 mm
50 mm
20 mm
60 mm
3 m
3 m
5 m
D
E
Ans:
sD = -88.0 MPa, tD = 0

773
.
Solution
sE = -
P
A
-
My
I
=
8(103
)
(0.1)(0.05)
+
8.25(103
)(0.03)
1
12(0.05)(0.1)3
= 57.8 MPa Ans.
tE =
VQ
It
=
4.5(103
)(0.04)(0.02)(0.05)
1
12 (0.05)(0.1)3
(0.05)
= 864 kPa Ans.
8–39.
The frame supports the distributed load shown. Determine
the state of stress acting at point E. Show the results on a
differential element at this point.
4 kN/m
D
B
A
C
E
1.5 m 1.5 m
20 mm
50 mm
20 mm
60 mm
3 m
3 m
5 m
D
E
Ans:
sE = 57.8 MPa, tE = 864 kPa

774
.
*8–40.
The rod has a diameter of 40 mm. If it is subjected to the
force system shown, determine the stress components that
act at point A, and show the results on a volume element
located at this point.
Solution
Internal Loadings: Consider the equilibrium of the left segment of
the rod being sectioned, Fig. a.
ΣFx = 0; Nx - 1500 = 0 Nx = 1500 N
ΣFy = 0; Vy - 600 = 0 Vy = 600 N
ΣFz = 0; Vz + 800 = 0 Vz = -800 N
ΣMx = 0; Tx - 100 = 0 Tx = 100 N # m
ΣMy = 0; My + 800(0.3) = 0 My = -240 N # m
ΣMz = 0; Mz + 600(0.3) = 0 Mz = -180 N # m
Section Properties: For the circular cross section, Fig. b,
A = pc2
= p(0.022
) = 0.400(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.024
) = 40.0(10-9
)p m4
J =
p
2
c4
=
p
2
(0.024
) = 80.0(10-9
)p m4
(QA)z = 0 (QA)y = y′A′ =
4(0.02)
3p
c
p
2
(0.022
) d = 5.3333(10-6
) m3
Normal Stress: For the combined loadings, the normal stress at point A can be
determined from
sx = sA =
Nx
A
-
MzyA
Iz
+
MyzA
Iy
=
1500
0.400(10-3
)p
-
(-180)(0)
40.0(10-9
)p
+
(-240)(0.02)
40.0(10-9
)p
= -37.00(106
) Pa = 37.0 MPa (C) Ans.
Shear Stress: The transverse shear stress in z and y directions and the torsional shear
stress can be obtained using the shear formula tV =
VQ
It
and the torsion formula
tT =
Tr
J
respectively.
(txy)A = (tV)y - tT
=
60035.3333(10-6
)4
40.0(10-9
)p(0.04)
-
100(0.02)
80.0(10-9
)p
= -7.321(106
) Pa = -7.32 MPa Ans.
(txz)A = (tV)z = 0 Ans.
Using these results,the state of stress at point A can be represented by the differential
volume element show in Fig. c.
100 mm
300 mm
y
x
B
A
1500 N
800 N
600 N
100 Nиm
z
Ans:
sA = 37.0 MPa (C),
(txy)A = -7.32 MPa,
(txz)A = 0

775
.
8–41.
Solve Prob. 8–40 for point B.
Solution
Internal Loadings: Consider the equilibrium of the left segment of
the rod being sectioned, Fig. a.
ΣFx = 0; Nx - 1500 = 0 Nx = 1500 N
ΣFy = 0; Vy - 600 = 0 Vy = 600 N
ΣFz = 0; Vz + 800 = 0 Vz = -800 N
ΣMx = 0; Tx - 100 = 0 Tx = 100 N # m
ΣMy = 0; My + 800(0.3) = 0 My = -240 N # m
ΣMz = 0; Mz + 600(0.3) = 0 Mz = -180 N # m
A = pc2
= p(0.022
) = 0.400(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.024
) = 40.0(10-9
)p m4
J =
p
2
c4
=
p
2
(0.024
) = 80.0(10-9
)p m4
(QB)z = z′A′ =
4(0.02)
3p
c
p
2
(0.022
) d = 5.3333(10-6
) m3
(QB)y = 0
Normal Stress: For the combined loadings, the normal stress at point B can be
determined from
sx = sB =
Nx
A
-
MzyB
Iz
+
MyzB
Iy
=
1500
0.400(10-3
)p
-
(-180)(-0.02)
40.0(10-9
)p
+
(-240)(0)
40.0(10-9
)p
= -27.45(106
) = 27.5 MPa (C) Ans.
VQ
It
tT =
Tr
J
, respectively.
(txz)B = (tV)z - tT
=
-80035.3333(10-6
)4
40.0(10-9
)p(0.04)
-
100(0.02)
80.0(10-9
)p
= -8.807(106
) Pa = -8.81 MPa Ans.
(txy)B = (tV)y = 0 Ans.
Using these results, the state of stress at point B, can be represented by the
differential volume element shown in Fig. c.
100 mm
300 mm
y
x
B
A
1500 N
800 N
600 N
100 Nиm
z
Ans:
sB = 27.5 MPa (C),
(txz)B = -8.81 MPa,
(txy)B = 0

776
.
Solution
ΣFx = 0; Vx - 125 = 0; Vx = 125 lb
ΣFy = 0; 75 - Ny = 0; Ny = 75 lb
ΣFz = 0; Vz - 200 = 0; Vz = 200 lb
ΣMx = 0; 200(8) - Mx = 0; Mx = 1600 lb # in.
ΣMy = 0; 200(3) - Ty = 0; Ty = 600 lb # in.
ΣMz = 0; Mz + 75(3) - 125(8) = 0; Mz = 775 lb # in.
A = p(0.52
) = 0.7854 in2
J =
p
2
(0.54
) = 0.098175 in4
I =
p
2
(0.54
) = 0.049087 in4
(QA)x = 0
(QA)z =
4(0.5)
3p
a
1
2
b(p)(0.52
) = 0.08333 in3
(sA)y = -
Ny
A
+
Mxc
I
= -
75
0.7854
+
1600(0.5)
0.049087
= 16202 psi = 16.2 ksi (T) Ans.
(tA)yx = (tA)V - (tA)twist
=
Vx(QA)Z
I t
-
Tyc
J
=
125(0.08333)
0.049087(1)
-
600(0.5)
0.098175
= -2843 psi = -2.84 ksi Ans.
(tA)yz =
Vz(QA)x
It
= 0 Ans.
8–42.
The beveled gear is subjected to the loads shown.Determine
the stress components acting on the shaft at point A, and
show the results on a volume element located at this point.
The shaft has a diameter of 1 in. and is fixed to the wall at C. C
B
x
z
y
A
200 lb
125 lb75 lb
8 in.
3 in.
Ans:
(sA)y = 16.2 ksi (T),
(tA)yx = -2.84 ksi,
(tA)yz = 0

777
.
Solution
ΣFx = 0; Vx - 125 = 0; Vx = 125 lb
ΣFy = 0; 75 - Ny = 0; Ny = 75 lb
ΣFz = 0; Vz - 200 = 0; Vz = 200 lb
ΣMx = 0; 200(8) - Mx = 0; Mx = 1600 lb # in.
ΣMy = 0; 200(3) - Ty = 0; Ty = 600 lb # in.
ΣMz = 0; Mz + 75(3) - 125(8) = 0; Mz = 775 lb # in.
A = p(0.52
) = 0.7854 in2
J =
p
2
(0.54
) = 0.098175 in4
I =
p
2
(0.54
) = 0.049087 in4
(QB)z = 0
(QB)x =
4(0.5)
3p
a
1
2
b(p)(0.52
) = 0.08333 in3
(sB)y = -
Py
A
+
Mzc
I
= -
75
0.7854
+
775(0.5)
0.049087
= 7.80 ksi (T) Ans.
(tB)yz = (tB)V + (tB)twist
=
Vz(QB)x
It
-
Tyc
J
=
200(0.08333)
0.049087 (1)
+
600(0.5)
0.098175
= 3395 psi = 3.40 ksi Ans.
(tB)yx =
Vx(QB)z
It
= 0 Ans.
8–43.
The beveled gear is subjected to the loads shown.Determine
the stress components acting on the shaft at point B, and
show the results on a volume element located at this point.
The shaft has a diameter of 1 in. and is fixed to the wall at C. C
B
x
z
y
A
200 lb
125 lb75 lb
8 in.
3 in.
Ans:
(sB)y = 7.80 ksi (T),
(tB)yz = 3.40 ksi,
(tB)yx = 0

778
.
Solution
Referring to Fig. a,
ΣFx = (FR)x; -6 - 12 = F F = -18.0 kip
ΣMy = (MR)y; 6(1.5) - 12(1.5) = My My = -9.00 kip # in
ΣMz = (MR)z; 12(3) - 6(3) = Mz Mz = 18.0 kip # in
The cross-sectional area and moments of inertia about the y and z axes of the cross
section are
A = 6(3) = 18 in2
Iy =
1
12
(6)(3)3
= 13.5 in4
Iz =
1
12
(3)(63
) = 54.0 in4
The normal stress developed is the combination of axial and bending stress. Thus,
s =
F
A
-
Mz y
Iz
+
My z
Iy
For point A, y = 3 in. and z = -1.5 in.
sA =
-18.0
18.0
-
18.0(3)
54.0
+
-9.00(-1.5)
13.5
= -1.00 ksi = 1.00 ksi (C) Ans.
For point B, y = 3 in and z = 1.5 in.
sB =
-18.0
18.0
-
18.0(3)
54
+
-9.00(1.5)
13.5
= -3.00 ksi = 3.00 ksi (C) Ans.
*8–44.
Determine the normal stress developed at points A and B.
Neglect the weight of the block.
a
a
6 in.
6 kip
12 kip3 in.
A
D
B
Ans:
sA = 1.00 ksi (C),
sB = 3.00 ksi (C)

779
.
Solution
Referring to Fig. a,
ΣFx = (FR)x; -6 - 12 = F F = -18.0 kip
ΣMy = (MR)y; 6(1.5) - 12(1.5) = My My = -9.00 kip # in
ΣMz = (MR)z; 12(3) - 6(3) = Mz Mz = 18.0 kip # in
The cross-sectional area and the moment of inertia about the y and z axes of the
cross section are
A = 3 (6) = 18.0 in2
Iy =
1
12
(6)(33
) = 13.5 in4
Iz =
1
12
(3)(63
) = 54.0 in4
The normal stress developed is the combination of axial and bending stress. Thus,
s =
F
A
-
Mzy
Iz
+
Myz
Iy
For point A, y = 3 in. and z = -1.5 in.
sA =
-18.0
18.0
-
18.0(3)
54.0
+
-9.00(-1.5)
13.5
= -1.00 ksi = 1.00 ksi (C) Ans.
For point B, y = 3 in. and z = 1.5 in.
sB =
-18.0
18.0
-
18.0(3)
54.0
+
-9.00(1.5)
13.5
= -3.00 ksi = 3.00 ksi (C) Ans.
8–45.
Sketch the normal-stress distribution acting over the cross
section at section a–a. Neglect the weight of the block.
a
a
6 in.
6 kip
12 kip3 in.
A
D
B

780
.
For point C, y = -3 in. and z = 1.5 in.
sC =
-18.0
18.0
-
18.0(-3)
54.0
+
-9.00(1.5)
13.5
= -1.00 ksi = 1.00 ksi (C) Ans.
For point D, y = -3 in. and z = -1.5 in.
sD =
-18.0
18.0
-
18.0(-3)
54.0
+
-9.00(-1.5)
13.5
= 1.00 ksi (T) Ans.
The normal stress distribution over the cross section is shown in Fig. b
8–45. Continued
Ans:
sA = 1.00 ksi (C), sB = 3.00 ksi (C),
sC = 1.00 ksi (C), sD = 1.00 ksi (T)

781
.
Solution
Equilibrium: For the man
a+ΣMB = 0; 981(1) - 2FA(1.35) = 0 FA = 363.33 N
Section Properties:
r-
= 0.15 +
0.025
2
= 0.1625 m
LA
dA
r
= 2p(r - 2r2
- c2
)
= 2p(0.1625 - 20.16252
- 0.01252
)
= 3.02524(10-3
) m
A = p(0.01252
) = 0.490874(10-3
) m2
R =
A
1A
dA
r
=
0.490874(10-3
)
3.02524(10-3
)
= 0.162259 m
r-
- R = 0.1625 - 0.162259 = 0.240741(10-3
) m
Internal Force and Moment: The internal moment must be computed about the
neutral axis.
+ cΣFy = 0; -363.33 - N = 0 N = -363.33 N
a+ΣMo = 0; -M - 363.33(0.462259) = 0 M = -167.95 N # m
Normal Stress: Apply the curved-beam formula.
For tensile stress
(st)max =
N
A
+
M(R - r2)
Ar2(r-
- R)
=
-363.33
0.490874(10-3
)
+
-167.95(0.162259 - 0.175)
0.490874(10-3
)(0.175)0.240741(10-3
)
= 103 MPa (T) Ans.
For compressive stress,
(sc)max =
N
A
+
M(R - r1)
Ar1(r-
- R)
=
-363.33
0.490874(10-3
)
+
-167.95(0.162259 - 0.15)
0.490874(10-3
)(0.15)0.240741(10-3
)
= -117 MPa = 117 MPa (C) Ans.
8–46.
The man has a mass of 100 kg and center of mass at G.
If he holds himself in the position shown, determine the
maximum tensile and compressive stress developed in the
curved bar at section a–a. He is supported uniformly by two
bars, each having a diameter of 25 mm. Assume the floor is
smooth. Use the curved-beam formula to calculate the
bending stress.
G
aa
300 mm
150 mm
0.35 m 1 m
300 mm
Ans:
(st)max = 103 MPa (T),
(sc)max = 117 MPa (C)

782
.
8–47.
The solid rod is subjected to the loading shown. Determine
the state of stress at point A, and show the results on a
differential volume element located at this point.
Solution
Internal Loadings: Consider the equilibrium of the left segment of the rod being
sectioned, Fig. a.
ΣFx = 0; Nx - 100 = 0 Nx = 100 kN
ΣFy = 0; Vy - 10 = 0 Vy = 10 kN
ΣFz = 0; Vz + 20 = 0 Vz = -20 kN
ΣMx = 0; Tx + 20(0.03) + 10(0.03) = 0 Tx = -0.9 kN # m
ΣMy = 0; My + 20(0.2) + 100(0.03) = 0 My = -7.00 kN # m
ΣMz = 0; Mz + 10(0.4) = 0 Mz = -4.00 kN # m
A = pc2
= p(0.032
) = 0.9(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.034
) = 0.2025(10-6
)p m4
J =
p
2
c4
=
p
2
(0.034
) = 0.405(10-6
)p m4
(QA)z = z′A′ =
4(0.03)
3p
c
p
2
(0.032
) d = 18.0(10-6
) m3
(QA)y = 0
determined from
sx = sA =
Nx
A
-
MzyA
Iz
+
MyzA
Iy
=
100(103
)
0.9(10-3
)p
-
3 -4.00(103
)4(0.03)
0.2025(10-6
)p
+
3 -7.00(103
)4(0)
0.2025(10-6
)p
= 224.00(106
) Pa = 224 MPa (T) Ans.
VQ
It
tT =
Tr
J
, respectively.
(txz)A = (tV)z + tT
=
-20(103
)318.0(10-6
)4
0.2025(10-6
)p(0.06)
+
-0.9(103
)(0.03)
0.405(10-6
)p
= -30.65(106
) Pa = -30.7 MPa Ans.
(txy)A = (tV)y = 0 Ans.
volume element shown in Fig. c.
Ans:
sA = 224 MPa (T),
(txz)A = -30.7 MPa,
(txy)A = 0
30 mm
A
x
y
z
B
C
100 kN
10 kN
20 kN
200 mm
200 mm

783
.
*8–48.
the state of stress at point B, and show the results on a
differential volume element at this point.
Solution
sectioned, Fig. a.
ΣFx = 0; Nx - 100 = 0 Nx = 100 kN
ΣFy = 0; Vy - 10 = 0 Vy = 10 kN
ΣFz = 0; Vz + 20 = 0 Vz = -20 kN
ΣMx = 0; Tx + 20(0.03) + 10(0.03) = 0 Tx = -0.9 kN # m
ΣMy = 0; My + 20(0.2) + 100(0.03) = 0 My = -7.00 kN # m
ΣMz = 0; Mz + 10(0.4) = 0 Mz = -4.00 kN # m
A = pc2
= p(0.032
) = 0.9(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.034
) = 0.2025(10-6
)p m4
J =
p
2
c4
=
p
2
(0.034
) = 0.405(10-6
)p m4
(QB)z = z′A′ =
4(0.03)
3p
c
p
2
(0.032
) d = 18.0(10-6
) m3
(QB)y = 0
determined from
sx = sB =
Nx
A
-
MzyB
Iz
+
MyzB
Iy
=
100(103
)
0.9(10-3
)p
-
3 -4.00(103
)4(-0.03)
0.2025(10-6
)p
+
3 -7.00(103
)4(0)
0.2025(10-6
)p
= -153.26(106
) Pa = 153 MPa (C) Ans.
stress can be determined using the shear formula tV =
VQ
It
tT =
Tr
J
, respectively.
(txz)B = -tT + (tV)z
=
0.9(103
)(0.03)
0.405(10-6
)p
+
-20(103
)318.0(10-6
)4
0.2025(10-6
)p(0.06)
= 11.79(106
) Pa = 11.8 MPa Ans.
(txy)B = (tV)y = 0 Ans.
Using these results,the state of stress at point B can be represented by the differential
Ans:
sB = 153 MPa (C),
(txz)B = 11.8 MPa,
(txy)B = 0
30 mm
A
x
y
z
B
C
100 kN
10 kN
20 kN
200 mm
200 mm

784
.
8–49.
the state of stress at point C, and show the results on a
differential volume element at this point.
Solution
sectioned, Fig. a.
ΣFx = 0; Nx - 100 = 0 Nx = 100 kN
ΣFy = 0; Vy - 10 = 0 Vy = 10 kN
ΣFz = 0; Vz + 20 = 0 Vz = -20 kN
ΣMx = 0; Tx + 20(0.03) + 10(0.03) = 0 Tx = -0.9 kN # m
ΣMy = 0; My + 20(0.2) + 100(0.03) = 0 My = -7.00 kN # m
ΣMz = 0; Mz + 10(0.4) = 0 Mz = -4.00 kN # m
A = pc2
= p(0.032
) = 0.9(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.034
) = 0.2025(10-6
)p m4
J =
p
2
c4
=
p
2
(0.034
) = 0.405(10-6
)p m4
(QC)y = y′A′ = c
4(0.03)
3p
d c
p
2
(0.032
) d = 18.0(10-6
) m3
(QC)z = 0
Normal Stress: For the combine loadings, the normal stress at point C can be
determined from
sx = sC =
Nx
A
-
MzyC
Iz
+
MyzC
Iy
=
100(103
)
0.9(10-3
)p
-
3 -4.00(103
)4(0)
0.2025(10-6
)p
+
3 -7.00(103
)4(0.03)
0.2025(10-6
)p
= -294.73(106
) Pa = 295 MPa (C) Ans.
stress can be determined using the shear formula tV =
VQ
It
tT =
Tr
J
, respectively.
(txy)C = (tV)y - tT
=
10(103
)318.0(10-6
)4
0.2025(10-6
)p(0.06)
-
-0.9(103
)(0.03)
0.405(10-6
)p
= 25.94(106
) Pa = 25.9 MPa Ans.
(txz)C = (tV)z = 0 Ans.
Using these results,the state of stress at point C can be represented by the differential
Ans:
sC = 295 MPa (C),
(txy)C = 25.9 MPa,
(txz)C = 0
30 mm
A
x
y
z
B
C
100 kN
10 kN
20 kN
200 mm
200 mm

785
.
Solution
Require sA = 0
sA = 0 =
P
A
+
Mc
I
; 0 =
-P
pc2
+
(Pe)c
p
4
c4
e =
c
4
Ans.
8–50.
The post has a circular cross section of radius c. Determine
the maximum radius e at which the load P can be applied
so that no part of the post experiences a tensile stress.
Neglect the weight of the post.
P
c
e
Ans:
e =
c
4

786
.
Solution
Equivalent Force System: As shown on FBD.
Section Properties:
A = 2a(2a) + 2J
1
2
(2a)aR = 6a2
Iz =
1
12
(2a)(2a)3
+ 2J
1
36
(2a)a3
+
1
2
(2a)aaa +
a
3
b
2
R
= 5a4
Iy =
1
12
(2a)(2a)3
+ 2J
1
36
(2a)a3
+
1
2
(2a)aa
a
3
b
2
R
=
5
3
a4
Normal Stress:
s =
N
A
-
Mzy
Iz
+
My z
Iy
=
-P
6a2
-
Peyy
5a4
-
Pezz
5
3 a4
=
P
30a4
1 -5a2
- 6eyy - 18ez z2
At point B where y = -a and z = -a, we require sB 6 0.
0 7
P
30a4
3 -5a2
- 6(-a) ey - 18(-a)ez4
0 7 -5a + 6ey + 18ez
6ey + 18ez 6 5a Ans.
When ez = 0, ey 6
5
6
a
When ey = 0, ez 6
5
18
a
Repeat the same procedures for point A, C and D. The region where P can be
applied without creating tensile stress at points A, B, C, and D is shown shaded in
the diagram.
8–51.
The post having the dimensions shown is subjected to the
load P. Specify the region to which this load can be applied
without causing tensile stress at points A, B, C, and D.
x
y
z
A
a a a a
a
a
D
ez
ey
B
C
P
Ans:
6ey + 18ez 6 5a

787
.
Ans:
P =
dmaxp(r0
4
- ri
4
)
r0
2
+ ri
2
+ 4er0
Solution
smax =
P
A
+
Pero
p
4
(r0
4
- r4
i )
smax = Pc
1
p(r2
0 - r2
i )
+
4er0
p(r0
4
- r4
i )
d
smax =
P
p(r0
2
- ri
2
)
c 1 +
4er0
(r0
2
+ ri
2
)
d
smax =
P(r0
2
+ ri
2
+ 4er0)
p(r0
2
- ri
2
)(r0
2
+ ri
2
)
smax =
P(r0
2
+ ri
2
+ 4er0)
p(r0
4
- ri
4
)
P =
dmaxp(r0
4
- ri
4
)
r0
2
+ ri
2
+ 4er0
Ans.
*8–52.
The vertebra of the spinal column can support a maximum
compressive stress of smax,before undergoing a compression
fracture. Determine the smallest force P that can be applied
to a vertebra,if we assume this load is applied at an eccentric
distance e from the centerline of the bone, and the bone
remains elastic. Model the vertebra as a hollow cylinder
with an inner radius ri and outer radius ro.
‫؍‬ e
P
ro
ri

788
.
Solution
Tc
J
= max on perimeter =
PRr
J
tmax =
V
A
+
Tc
J
=
P
A
+
PRr
J
QED
8–53.
The coiled spring is subjected to a force P. If we assume the
shear stress caused by the shear force at any vertical section of
the coil wire to be uniform,show that the maximum shear stress
in the coil is tmax = PA + PRrJ, where J is the polar moment
of inertia of the coil wire and A is its cross-sectional area.
P
P
R
r2
Ans:
N/A

789
.
Solution
Member CD:
a+ ΣMC = 0;
3
5
FDE(16) - 28.8(8) = 0;
FDE = 24.0 kip
Segment:
+d ΣFx = 0; N -
4
5
(24.0) = 0; N = 19.2 kip
+ cΣFy = 0; V +
3
5
(24.0) - 19.8 = 0; V = 5.40 kip
a+ ΣMo = 0; -M - 19.8(5.5) +
3
5
(24.0)(11) = 0;
M = 49.5 kip # ft
A = 7(1.5) + 6(1) = 16.5 in2
y-
=
Σy-
A
ΣA
=
0.75(1.5)(7) + 4.5(6)(1)
16.5
= 2.1136 in.
I =
1
12
(7)(1.53
) + 7(1.5)(2.1136 - 0.75)2
+
1
12
(1)(63
) + 1(6)(4.5 - 2.1136)2
= 73.662 in4
QA = QB = 0
Normal Stress:
s =
N
A
{
My
I
sA =
19.2
16.5
-
49.5(12)(7.5 - 2.1136)
73.662
= -15.9 ksi = 15.9 ksi (C) Ans.
sB =
19.2
16.5
+
49.5(12)(5.3864)
73.662
= 44.6 ksi (T) Ans.
Shear Stress: Since QA = QB = 0,
tA = tB = 0 Ans.
8–54.
The frame supports a centrally applied distributed load of
1.8 kipft. Determine the state of stress at points A and B
on member CD and indicate the results on a volume
element located at each of these points. The pins at C
and D are at the same location as the neutral axis for the
cross section.
12 ft
5 ft
16 ft
A
B
D
C
1.8 kip/ft
1.5 in.
6 in.
1 in.
3 in. 3 in.
E
A
B
Ans:
sA = 15.9 ksi (C),
sB = 44.6 ksi (T),
tA = tB = 0

790
.
Solution
Internal Loadings: Consider the equilibrium of the right segment of the rod being
sectioned, Fig. a.
ΣFx = 0; Nx - 1200 = 0 Nx = 1200 lb
ΣFy = 0; Vy + 200 = 0 Vy = -200 lb
ΣFz = 0; Vz + 300 = 0 Vz = -300 lb
ΣMx = 0; Tx + 300(3) = 0 Tx = -900 lb # in.
ΣMy = 0; My + 300(9) = 0 My = -2700 lb # in.
ΣMz = 0; Mz + 1200(3) - 200(9) = 0 Mz = -1800 lb # in.
A = pc2
= p(0.52
) = 0.25p in.2
Iy = Iz =
p
4
(0.54
) = 0.015625p in.4
J =
p
2
(0.54
) = 0.03125p in4
(QA)z = z
9=
A=
=
4(0.5)
3p
c
p
2
(0.52
) d = 0.08333 in3
(QA)y = 0
determined from
sx = sA =
Nx
A
-
MzyA
Iz
+
MyzA
Iy
=
1200
0.25p
-
(-1800)(-0.5)
0.015625p
+
(-2700)(0)
0.015625p
= -16.81(103
) psi = 16.8 ksi (C) Ans.
Shear Stress: The transverse shear stress in z and y directions and the torsional
shear stress can be obtained using the shear formula tV =
VQ
It
and torsion formula
tT =
Tr
J
, respectively.
(txz)A = -tT + (tV)z
= -
-900(0.5)
0.03125p
+
-300(0.08333)
0.015625p (1)
= 4.074(103
) psi = 4.07 ksi Ans.
8–55.
The 1-in.-diameter rod is subjected to the loads shown.
Determine the state of stress at point A, and show the
results on a differential volume element located at this point.
B
A
y
z
x 200 lb9 in.
3 in.
300 lb
1200 lb
Ans:
sA = 16.8 ksi (C),
(txz)A = 4.07 ksi,
(txy)A = 0

791
.
Solution
Internal Loadings: Consider the equilibrium of the right segment of the rod being
sectioned, Fig. a.
ΣFx = 0; Nx - 1200 = 0 Nx = 1200 lb
ΣFy = 0; Vy + 200 = 0 Vy = -200 lb
ΣFz = 0; Vz + 300 = 0 Vz = -300 lb
ΣMx = 0; Tx + 300(3) = 0 Tx = -900 lb # in.
ΣMy = 0; My + 300(9) = 0 My = -2700 lb # in.
ΣMz = 0; Mz + 1200(3) - 200(9) = 0 Mz = -1800 lb # in.
A = pc2
= p(0.52
) = 0.25p in2
Iy = Iz =
p
4
(0.54
) = 0.015625p in4
J =
p
2
(0.54
) = 0.03125p in4
(QB)y = y9=
A=
=
4(0.5)
3p
c
p
2
(0.52
) d = 0.08333 in3
(QB)z = 0
determined from
sx = sB =
Nx
A
-
MzyB
Iz
+
MyzB
Iy
=
1200
0.25p
-
(-1800)(0)
0.015625p
+
(-2700)(0.5)
0.015625p
= -25.97(103
) psi = 26.0 ksi (C) Ans.
Shear Stress: The transverse shear stress in z and y directions and the torsional
shear stress can be obtained using the shear formula tV =
VQ
It
and torsion formula
tT =
Tr
J
, respectively.
(txy)B = -tT + (tV)y
=
900(0.5)
0.03125p
+
-200(0.08333)
0.015625p (1)
= 4.244(103
(txz)B = (tV)z = 0 Ans.
*8–56.
The 1-in.-diameter rod is subjected to the loads shown.
Determine the state of stress at point B, and show the
results on a differential volume element located at this point.
B
A
y
z
x 200 lb9 in.
3 in.
300 lb
1200 lb
Ans:
sB = 26.0 ksi (C),
(txy)B = 4.24 ksi,
(txz)B = 0

792
.
Solution
Point A:
sA =
Mc
I
=
10.5(103
)(0.05)
p
4(0.05)4
= 107 MPa (T) Ans.
tA =
Tc
J
=
3(103
)(0.05)
p
4(0.05)4
= 15.279(106
) = 15.3 MPa Ans.
Point B:
sB = 0 Ans.
tB =
Tc
J
-
VQ
It
= 15.279(106
) -
3000(4(0.05)/3p))(1
2)(p)(0.05)2
p
4(0.05)4
(0.1)
tB = 14.8 MPa Ans.
8–57.
The sign is subjected to the uniform wind loading.
Determine the stress components at points A and B on the
100-mm-diameter supporting post. Show the results on a
volume element located at each of these points.
D
y
x
CB
A
1 m
1.5 kPa
3 m
2 m
2 m
z
Ans:
sA = 107 MPa (T), tA = 15.3 MPa,
sB = 0, tB = 14.8 MPa

793
.
Solution
Point C:
sC =
Mc
I
=
10.5(103
)(0.05)
p
4(0.05)4
= 107 MPa (C) Ans.
tC =
TC
J
=
3(103
)(0.05)
p
2(0.05)4
= 15.279(106
) = 15.3 MPa Ans.
Point D:
sD = 0 Ans.
tD =
Tc
J
+
VQ
It
= 15.279(106
) +
3(103
)(4(0.05)/3p)(1
2)(p)(0.05)2
p
4(0.05)4
(0.1)
= 15.8 MPa
Ans.
8–58.
The sign is subjected to the uniform wind loading.
Determine the stress components at points C and D on the
100-mm-diameter supporting post. Show the results on a
volume element located at each of these points.
D
y
x
CB
A
1 m
1.5 kPa
3 m
2 m
2 m
z
Ans:
sC = 107 MPa (C), tC = 15.3 MPa,
sD = 0, tD = 15.8 MPa

794
.
8–59.
The bearing pin supports the load of 900 lb. Determine the
stress components in the support member at point A.
Represent the state of stress at point A with a differential
element.
30Њ
3 in.
A
A
B
B
4 in.
1 in.
900 lb
0.25 in.
0.5 in.
0.25 in.
Solution
Internal Loadings: Consider the equilibrium of the right segment of the supports
being sectioned, Fig. a.
a+ ΣFx = 0; N - 900 cos 30°
= 0 N = 779.42 lb
+Q ΣFy = 0; 900 sin 30°
- V = 0 V = 450 lb
a+ΣMo = 0; 900(3 sin 30°
- 1) - M = 0 M = 450 lb # in
Section Properties: For the square cross section,
A = 0.5(0.5) = 0.25 in2
I =
1
12
(0.5)(0.53
) = 0.0052083 in4
QA = 0
Normal Stress: For the combined loadings,the normal stress can be determined from
sA =
N
A
+
Mc
I
=
779.42
0.25
+
450(0.25)
0.0052083
= 24.72(103
) psi = 24.7 ksi (T) Ans.
tA =
VQA
It
= 0 Ans.
element in Fig. b.
Ans:
sA = 24.7 ksi (T),
tA = 0

795
.
*8–60.
The bearing pin supports the load of 900 lb. Determine the
stress components in the support member at point B.
Represent the state of stress at point B with a differential
element.
30Њ
3 in.
A
A
B
B
4 in.
1 in.
900 lb
0.25 in.
0.5 in.
0.25 in.
Solution
Internal Loadings: Consider the equilibrium of the right segment of the support
being sectioned, Fig. a.
a+ΣFx = 0; N - 900 cos 30°
= 0 N = 779.42 lb
+Q ΣFy = 0; 900 sin 30°
- V = 0 V = 450 lb
a+ΣMo = 0; 900(3 sin 30°
- 1) - M = 0 M = 450 lb # in
Section Properties: For the square cross section,
A = 0.5(0.5) = 0.25 in2
I =
1
12
(0.5)(0.53
) = 0.0052083 in4
For the area shown shaded in Fig. b,
QB = y-=
A=
= 0.12530.25(0.5)4 = 0.015625 in3
Normal Stress: For the combined loadings,the normal stress can be determined from
sB =
N
A
+
My
I
=
779.42
0.25
+ 0 = 3.1177(103
tB =
VQB
It
=
450(0.015625)
0.0052083(0.5)
= 2.700(103
element in Fig. c.
Ans:
sB = 3.12 ksi (T),
tB = 2.70 ksi

796
.
Solution
x =
Σx-
A
ΣA
=
(0.005)(0.04)(0.01) + 0.04(0.06)(0.01)
0.04(0.01) + 0.06(0.01)
= 0.026 m
A = 0.04(0.01) + 0.06(0.01) = 0.001 m2
I =
1
12
(0.04)(0.013
) + (0.04)(0.01)(0.026 - 0.005)2
+
1
12
(0.01)(0.063
) + 0.01(0.06)(0.040 - 0.026)2
= 0.4773(10-6
) m4
(smax)t =
P
A
+
Mx
I
=
8(103
)
0.001
+
1.808(103
)0.026
0.4773(10-6
)
= 106.48 MPa = 106 MPa Ans.
(smax)c =
P
A
-
Mc
I
=
8(103
)
0.001
-
1.808(103
)(0.070 - 0.026)
0.4773(10-6
)
= -158.66 MPa = -159 MPa Ans.
x
158.66
=
70 - x
106.48
; x = 41.9 mm
8–61.
The C-frame is used in a riveting machine. If the force at the
ram on the clamp at D is P = 8 kN, sketch the stress
distribution acting over the section a–a.
D
a
40 mm
10 mm
60 mm 10 mm
200 mm
a
P
Ans:
(smax)t = 106 MPa, (smax)c = -159 MPa

797
.
Solution
x =
ΣxA
ΣA
=
(0.005)(0.04)(0.01) + 0.04(0.06)(0.01)
0.04(0.01) + 0.06(0.01)
= 0.026 m
A = 0.04(0.01) + 0.06(0.01) = 0.001 m2
I =
1
12
(0.04)(0.013
) + (0.04)(0.01)(0.026 - 0.005)2
+
1
12
(0.01)(0.063
) + 0.01(0.06)(0.040 - 0.026)2
= 0.4773(10-6
) m4
s =
P
A
{
Mx
I
Assume tension failure,
180(106
) =
P
0.001
+
0.226P(0.026)
0.4773(10-6
)
P = 13524 N = 13.5 kN
Assume compression failure,
-180(106
) =
P
0.001
-
0.226P(0.070 - 0.026)
0.4773(10-6
)
P = 9076 N = 9.08 kN (controls) Ans.
8–62.
Determine the maximum ram force P that can be applied to
the clamp at D if the allowable normal stress for the
material is sallow = 180 MPa.
D
a
40 mm
10 mm
60 mm 10 mm
200 mm
a
P
Ans:
Pmax = 9.08 kN

798
.
Solution
Internal Forces and Moments: As shown on FBD.
ΣFx = 0; 1.50 + Nx = 0 Nx = -15.0 kip
ΣFy = 0; Vy - 10.8 = 0 Vy = 10.8 kip
ΣFz = 0; Vz = 0
ΣMx = 0; Tx - 10.8(6) = 0 Tx = 64.8 kip # ft
ΣMy = 0; My - 1.50(6) = 0 My = 9.00 kip # ft
ΣMz = 0; 10.8(6) + Mz = 0 Mz = -64.8 kip # ft
Section Properties:
A = p132
- 2.752
2 = 1.4375p in2
Iy = Iz =
p
4
134
- 2.754
2 = 18.6992 in4
(QC)z = (QD)y = 0
(QC)y = (QD)z =
4(3)
3p
c
1
2
(p)132
2 d -
4(2.75)
3p
c
1
2
(p)12.752
2 d
= 4.13542 in3
J =
p
2
134
- 2.754
2 = 37.3984 in4
Normal Stress:
s =
N
A
-
Mz y
Iz
+
My z
Iy
sC =
-1.50
1.4375p
-
(-64.8)(12)(0)
18.6992
+
9.00(12)(2.75)
18.6992
= 15.6 ksi (T) Ans.
sD =
-1.50
1.4375p
-
(-64.8)(12)(3)
18.6992
+
9.00(12)(0)
18.6992
= 124 ksi (T) Ans.
8–63.
The uniform sign has a weight of 1500 lb and is supported by
the pipe AB, which has an inner radius of 2.75 in. and an outer
radius of 3.00 in.If the face of the sign is subjected to a uniform
wind pressure of p = 150 lbft2
, determine the state of stress
at points C and D. Show the results on a differential volume
element located at each of these points. Neglect the thickness
of the sign, and assume that it is supported along the outside
edge of the pipe.
3 ft
6 ft
12 ft
B
A
y
x
z
C
D
FE
150 lb/ft2

799
.
8–63. Continued
Shear Stress: The tranverse shear stress in the z and y directions and the torsional
shear stress can be obtained using the shear formula and the torsion formula.
tV =
VQ
It
and ttwist =
Tr
J
, respectively.
(txz)D = ttwist =
64.8(12)(3)
37.3984
= 62.4 ksi Ans.
(txy)D = tVy
= 0 Ans.
(txy)C = tVy
- ttwist
=
10.8(4.13542)
18.6992(2)(0.25)
-
64.8(12)(2.75)
37.3984
= -52.4 ksi Ans.
(txz)C = tVz
= 0 Ans.
Ans:
sC = 15.6 ksi (T), sD = 124 ksi (T),
(txz)D = 62.4 ksi, (txy)D = 0,
(txy)C = -52.4 ksi, (txz)C = 0

800
.
Solution
Internal Forces and Moments: As shown on FBD.
ΣFx = 0; 1.50 + Nx = 0 Nx = -1.50 kip
ΣFy = 0; Vy - 10.8 = 0 Vy = 10.8 kip
ΣFz = 0; Vz = 0
ΣMx = 0; Tx - 10.8(6) = 0 Tx = 64.8 kip # ft
ΣMy = 0; My - 1.50(6) = 0 My = 9.00 kip # ft
ΣMz = 0; 10.8(6) + Mz = 0 Mz = -64.8 kip # ft
Section Properties:
A = p132
- 2.752
2 = 1.4375p in2
Iy = Iz =
p
4
134
- 2.754
2 = 18.6992 in4
(QF)z = (QE)y = 0
(QF)y = (QE)z =
4(3)
3p
c
1
2
(p)132
2 d -
4(2.75)
3p
c
1
2
(p)12.752
2 d
= 4.13542 in3
J =
p
2
134
- 2.754
2 = 37.3984 in4
Normal Stress:
s =
N
A
-
Mzy
Iz
+
Myz
Iy
sF =
-1.50
1.4375p
-
(-64.8)(12)(0)
18.6992
+
9.00(12)(-3)
18.6992
= -17.7 ksi = 17.7 ksi (C) Ans.
sE =
-1.50
1.4375p
-
(-64.8)(12)(-3)
18.6992
+
9.00(12)(0)
18.6992
= -125 ksi = 125 ksi (C) Ans.
*8–64.
Solve Prob. 8–63 for points E and F.
3 ft
6 ft
12 ft
B
A
y
x
z
C
D
FE
150 lb/ft2

801
.
Shear Stress: The tranverse shear stress in the z and y directions and the torsional
shear stress can be obtained using the shear formula and the torsion formula,
tV =
VQ
It
and ttwist =
Tr
J
, respectively.
(txz)E = -ttwist = -
64.8(12)(3)
37.3984
= -62.4 ksi Ans.
(txy)E = tVy
= 0 Ans.
(txy)F = tVy
+ ttwist
=
10.8(4.13542)
18.6992(2)(0.25)
+
64.8(12)(3)
37.3984
= 67.2 ksi Ans.
(txz)F = tVy
= 0 Ans.
*8–64. Continued
Ans:
sF = 17.7 ksi (C),
sE = 125 ksi (C),
(txz)E = -62.4 ksi,
(txy)E = 0,
(txy)F = 67.2 ksi,
(txz)F = 0

802
.
Solution
I =
1
4
(p)(0.014
) = 7.85398(10-9
) m4
QB = y-
A=
=
4(0.01)
3p
a
1
2
b(p)(0.012
) = 0.66667(10-6
) m3
QA = 0
sA =
Mc
I
=
12(0.01)
7.85398(10-9
)
= 15.3 MPa Ans.
tA = 0 Ans.
sB = 0 Ans.
tB =
VQB
It
=
150(0.6667)(10-6
)
7.85398(10-9
)(0.02)
= 0.637 MPa Ans.
8–65.
The pin support is made from a steel rod and has a diameter
of 20 mm. Determine the stress components at points A and
B and represent the results on a volume element located at
each of these points.
B
C
DA
80 mm
150 N
Ans:
sA = 15.3 MPa, tA = 0,
sB = 0, tB = 0.637 MPa

803
.
Solution
I =
1
4
(p)(0.014
) = 7.85398(10-9
) m4
QD = y′A=
=
4(0.01)
3p
a
1
2
b(p)(0.012
) = 0.66667(10-6
) m3
QC = 0
sC =
Mc
I
=
12(0.01)
7.85398(10-9
)
= 15.3 MPa Ans.
tC = 0 Ans.
sD = 0 Ans.
tD =
VQD
It
=
150(0.6667)(10-6
)
7.8539(10-9
)(0.02)
= 0.637 MPa Ans.
8–66.
Solve Prob. 8–65 for points C and D.
B
C
DA
80 mm
150 N
Ans:
sC = 15.3 MPa, tC = 0,
sD = 0, tD = 0.637 MPa

804
.
Solution
Normal stress due to axial force:
A = 2[0.5(3)] + 5(0.5) = 5.5 in2
sA =
P
A
=
60
5.5
= 10.9090 psi (T)
Normal stress due to bending:
r-
= 15 in. rA = 12 in. rB = 18 in.
Σ
L
dA
r
= Σb ln
r2
r1
= 3 ln
12.5
12
+ 0.5 ln
17.5
12.5
+ 3 ln
18
17.5
= 0.3752 in.
R =
A
1
dA
r
=
5.5
0.3752
= 14.6583 in.
r-
- R = 0.3417 in.
(sA)b =
M(R - rA)
ArA(r-
- R)
=
59.0(12)(14.6583 - 12)
5.5(12)(0.3417)
= 83.4468 psi (T)
(sB)b =
M(R - rB)
ArB(r-
- R)
=
59.0(12)(14.6583 - 18)
5.5(18)(0.3417)
= -69.9342 psi = 69.9342 psi (C)
sA = 83.4468 + 10.9090 = 94.4 psi (T) Ans.
sB = 69.9342 - 10.9090 = 59.0 psi (C) Ans.
8–67.
The handle of the press is subjected to a force of 20 lb. Due to
internal gearing, this causes the block to be subjected to a
compressive force of 80 lb.Determine the normal-stress acting
in the frame at points along the outside flanges A and B. Use
the curved-beam formula to calculate the bending stress.
3 in.
0.8 ft
1.50 ft
1 ft
A B
5 in.0.5 in. 0.5 in.
0.5 in.
80 lb
20 lb
Ans:
sA = 94.4 psi (T),
sB = 59.0 psi (C)

805
.
*8–68.
The bar has a diameter of 40 mm. Determine the state of
stress at point A and show the results on a differential
volume element located at this point.
Solution
sectioned, Fig. a.
ΣFx = 0; Nx - 1200 = 0 Nx = 1200 N
ΣFy = 0; Vy - 800a
3
5
b = 0 Vy = 480 N
ΣFz = 0; Vz + 800a
4
5
b = 0 Vz = -640 N
ΣMx = 0; Tx = 0
ΣMy = 0; My + 800a
4
5
b(0.2) = 0 My = -128 N # m
ΣMz = 0; Mz + 800a
3
5
b(0.2) = 0 Mz = -96 N # m
A = pc2
= p(0.022
) = 0.4(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.024
) = 40.0(10-9
)p m4
J =
p
2
c4
=
p
2
(0.024
) = 80.0(10-9
)p m4
(QA)z = z-=
A=
=
4(0.02)
3p
c
p
2
(0.022
) d = 5.3333(10-6
) m3
(QA)y = 0
Normal Stress: For the combined loading, the normal stress at point A can be
determined from
sx = sA =
Nx
A
-
MzyA
Iz
+
MyzA
Iy
=
1200
0.4(10-3
)p
-
(-96)(0.02)
40.0(10-9
)p
+
(-128)(0)
40.0(10-9
)p
= 16.23(106
) Pa = 16.2 MPa (T) Ans.
Shear Stress: Since Tx = 0, the shear stress in the z and y directions is contributed
by transverse shear stress only which can be obtained using the shear formula,
tV =
VQ
It
(txz)A = (tV)z =
-64035.3333(10-6
)4
40.0(10-9
)p(0.04)
= -0.6791(106
) Pa = -0.679 MPa Ans.
Using these results, the state of stress at point A can be represented by the volume
element shown in Fig. c.
200 mm
200 mm
y
z
x
B
A
800 N
3
4
5
1200 N
Ans:
sA = 16.2 MPa (T),
(txz)A = -0.679 MPa,
(txy)A = 0

806
.
8–69.
Solve Prob. 8–68 for point B.
200 mm
200 mm
y
z
x
B
A
800 N
3
4
5
1200 N
Solution
sectioned, Fig. a.
ΣFx = 0; Nx - 1200 = 0 Nx = 1200 N
ΣFy = 0; Vy - 800a
3
5
b = 0 Vy = 480 N
ΣFz = 0; Vz + 800a
4
5
b = 0 Vz = -640 N
ΣMx = 0; Tx = 0
ΣMy = 0; My + 800a
4
5
b(0.2) = 0 My = -128 N # m
ΣMz = 0; Mz + 800a
3
5
b(0.2) = 0 Mz = -96 N # m
A = pc2
= p(0.022
) = 0.4(10-3
)p m2
Iy = Iz =
p
4
c4
=
p
4
(0.024
) = 40.0(10-9
)p m4
J =
p
2
c4
=
p
2
(0.024
) = 80.0(10-9
)p m4
(QB)y = y-=
A=
=
4(0.02)
3p
c
p
2
(0.022
) d = 5.3333(10-6
) m3
(QB)z = 0
Normal Stress: For the combined loading, the normal stress at point B can be
determined from
sx = sB =
Nx
A
-
MzyB
Iz
+
MyzB
Iy
=
1200
0.4(10-3
)p
-
(-96)(0)
40.0(10-9
)p
+
(-128)(0.02)
40.0(10-9
)p
= -19.42(106
) Pa = 19.4 MPa (C) Ans.
Shear Stress: Since Tx = 0, the shear stress in z and y directions is contributed by
transverse shear stress only,which can be obtained using the shear formula,tV =
VQ
It
.
(txy)B = (tV)y =
48035.3333(10-6
)4
40.0(10-9
)p(0.04)
= 0.5093(106
) Pa = 0.509 MPa Ans.
(txz)B = (tV)z = 0 Ans.
Using these results, the state of stress at point B can be represented by the volume
element shown in Fig. c.
Ans:
sB = 19.4 MPa (C),
(txy)B = 0.509 MPa,
(txz)B = 0

807
.
Solution
A =
p
4
(0.752
) = 0.44179 in2
I =
p
4
(0.3754
) = 0.015531 in4
QA = 0
tA = 0 Ans.
sA =
My c
I
=
-1250(0.375)
0.015531
= -30.2 ksi = 30.2 ksi (C) Ans.
8–70.
The 3
4-in.-diameter shaft is subjected to the loading shown.
Determine the stress components at point A. Sketch the
results on a volume element located at this point. The
journal bearing at C can exert only force components Cy
and Cz on the shaft, and the thrust bearing at D can exert
force components Dx, Dy, and Dz on the shaft.
x
C
A
B
125 lb
2 in.
8 in.
8 in.
125 lb
D
20 in.
20 in.
10 in.
z
y
2 in.
Ans:
tA = 0, sA = 30.2 ksi (C)

808
.
Solution
A =
p
4
(0.752
) = 0.44179 in2
I =
p
4
(0.3754
) = 0.015531 in4
QB = y′A′ =
4(0.375)
3p
a
1
2
b(p)(0.3752
) = 0.035156 in3
sB = 0 Ans.
tB =
VzQB
It
=
125(0.035156)
0.015531(0.75)
= 0.377 ksi Ans.
8–71.
Solve Prob. 8–70 for the stress components at point B.
x
C
A
B
125 lb
2 in.
8 in.
8 in.
125 lb
D
20 in.
20 in.
10 in.
z
y
2 in.
Ans:
sB = 0, tB = 0.377 ksi

809
.
Solution
The location of the neutral surface from the center of curvature of the hook, Fig. a,
can be determined from
R =
A
Σ
LA
dA
r
where A = p(0.252
) = 0.0625p in2
Σ
LA
dA
r
= 2p1r - 2r2
- c2
2 = 2p11.75 - 21.752
- 0.252
2 = 0.11278 in.
Thus,
R =
0.0625p
0.11278
= 1.74103 in.
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in.
Referring to Fig. b, I and QA are
I =
p
4
(0.254
) = 0.9765625(10-3
)p in4
QA = 0
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c.
d+ ΣFx = 0; N - 80cos 45° = 0 N = 56.57 lb
+ cΣFy = 0; 80sin 45° - V = 0 V = 56.57 lb
a+ΣMo = 0; M - 80cos 45°(1.74103) = 0 M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress.Thus,
s =
N
A
+
M(R - r)
Ae r
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point A,
r = 1.5 in.Then
s =
56.57
0.0625p
+
(98.49)(1.74103 - 1.5)
0.0625p(0.0089746)(1.5)
= 9.269(103
The shear stress is contributed by the transverse shear stress only.Thus
t =
VQA
It
= 0 Ans.
The state of stress of point A can be represented by the element shown in Fig. d.
*8–72.
The hook is subjected to the force of 80 lb. Determine the
state of stress at point A at section a–a. The cross section is
circular and has a diameter of 0.5 in. Use the curved-beam
formula to calculate the bending stress.
a
a
80 lb
1.5 in.
A A
B
B
45Њ
Ans:
s = 9.27 ksi (T),
t = 0

810
.
Solution
The location of the neutral surface from the center of curvature of the hook, Fig. a,
can be determined from
R =
A
Σ
LA
dA
r
where A = p(0.252
) = 0.0625p in2
Σ
LA
dA
r
= 2p 1r - 2r2
- c2
2 = 2p 11.75 - 21.752
- 0.252
2 = 0.11278 in.
Thus,
R =
0.0625p
0.11278
= 1.74103 in
Then
e = r - R = 1.75 - 1.74103 = 0.0089746 in
Referring to Fig. b, I and QB are computed as
I =
p
4
(0.254
) = 0.9765625(10-3
)p in4
QB = y′A′ =
4(0.25)
3p
c
p
2
(0.252
) d = 0.0104167 in3
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c.
d+ ΣFx = 0; N - 80cos45° = 0 N = 56.57 lb
+ cΣFy = 0; 80sin45° - V = 0 V = 56.57 lb
a+ΣMo = 0; M - 80cos45° (1.74103) = 0 M = 98.49 lb # in
The normal stress developed is the combination of axial and bending stress.Thus,
s =
N
A
+
M(R - r)
Aer
Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point B,
r = 1.75 in.Then
s =
56.57
0.0625p
+
(98.49)(1.74103 - 1.75)
0.0625p (0.0089746)(1.75)
= 1.48 psi (T) Ans.
The shear stress is contributed by the transverse shear stress only.Thus,
t =
VQB
It
=
56.57 (0.0104167)
0.9765625(10-3
)p (0.5)
= 384 psi Ans.
The state of stress of point B can be represented by the element shown in Fig. d.
8–73.
The hook is subjected to the force of 80 lb. Determine the
state of stress at point B at section a–a.The cross section has
a diameter of 0.5 in. Use the curved-beam formula to
calculate the bending stress.
a
a
80 lb
1.5 in.
A A
B
B
45Њ
Ans:
s = 1.48 psi (T), t = 384 psi

811
.
Solution
L
dA
r = 2pa3.125 - 2(3.125)2
- (0.625)2
b = 0.395707
R =
A
L
dA
r
=
p(0.625)2
0.396707
= 3.09343 in.
M = 800(3.09343) = 2.475(103
)
s =
M(R - r)
Ar(r - R)
+
P
A
(st)max =
2.475(103
)(3.09343 - 2.5)
p(0.625)2
(2.5)(3.125 - 3.09343)
+
800
p(0.625)2
= 15.8 ksi Ans.
(sc)max =
2.475(103
)(3.09343 - 3.75)
p(0.625)2
(3.75)(3.125 - 3.09343)
+
800
p(0.625)2
= -10.5 ksi Ans.
R8–1.
The eye hook has the dimensions shown. If it supports a
cable loading of 800 lb, determine the maximum normal
stress at section a–a and sketch the stress distribution acting
over the cross section. Use the curved-beam formula to
calculate the bending stress.
3.75 in. 2.5 in.
a
800 lb
1.25 in.
a
800 lb
Ans:
(st)max = 15.8 ksi, (sc)max = -10.5 ksi

812
.
Solution
Support Reactions: Referring to the free-body diagram of member BC shown in
Fig. a,
a+ΣMB = 0; F sin 45°(1) - 20(9.81)(2) = 0 F = 554.94 N
S+ ΣFx = 0; 554.94 cos 45° - Bx = 0 Bx = 392.4 N
+ cΣFy = 0; 554.94 sin 45° - 20(9.81) - By = 0 By = 196.2 N
Internal Loadings: Consider the equilibrium of the free-body diagram of the right
segment shown in Fig. b.
S+ ΣFx = 0; N - 392.4 = 0 N = 392.4 N
+ cΣFy = 0; V - 196.2 = 0 V = 196.2 N
a+ΣMC = 0; 196.2(0.5) - M = 0 M = 98.1 N # m
Section Properties: The cross-sectional area and the moment of inertia of the cross
section are
A = 0.05(0.075) = 3.75110-3
2 m2
I =
1
12
(0.05)10.0753
2 = 1.7578110-6
2 m4
Referring to Fig. c, QE is
QE = y′A′ = 0.025(0.025)(0.05) = 31.25110-6
2 m3
Thus,
s =
N
A
{
My
I
For point E, y = 0.0375 - 0.025 = 0.0125m.Then
sE =
392.4
3.75110-3
2
+
98.1(0.0125)
1.7578110-6
2
= 802 kPa Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,
tE =
VQA
It
=
196.2331.25110-6
2 4
1.7578110-6
2(0.05)
= 69.8 kPa Ans.
The state of stress at point E is represented on the element shown in Fig. d.
R8–2.
The 20-kg drum is suspended from the hook mounted on
the wooden frame. Determine the state of stress at point E
on the cross section of the frame at section a–a. Indicate the
results on an element.
1 m
1 m
1 m
b
a
a
b
C
B
A
30Њ
1 m
0.5 m0.5 m
50 mm
75 mm
25 mm
Section a – a
E
75 mm
75 mm
25 mm
Section b – b
FD
Ans:
sE = 802 kPa, tE = 69.8 kPa

813
.
R8–3.
The 20-kg drum is suspended from the hook mounted on the
wooden frame. Determine the state of stress at point F on
the cross section of the frame at section b–b. Indicate the
results on an element.
1 m
1 m
1 m
b
a
a
b
C
B
A
30Њ
1 m
0.5 m0.5 m
50 mm
75 mm
25 mm
Section a – a
E
75 mm
75 mm
25 mm
Section b – b
FD
Solution
Support Reactions: Referring to the free-body diagram of the entire frame shown
in Fig. a,
a+ΣMA = 0; FBD sin 30°(3) - 20(9.81)(2) = 0 FBD = 261.6 N
+ cΣFy = 0; Ay - 261.6 cos 30° - 20(9.81) = 0 Ay = 422.75 N
S+ ΣFx = 0; Ax - 261.6 sin 30° = 0 Ax = 130.8 N
Internal Loadings: Consider the equilibrium of the free-body diagram of the lower
cut segment, Fig. b.
S+ ΣFx = 0; 130.8 - V = 0 V = 130.8 N
+ cΣFy = 0; 422.75 - N = 0 N = 422.75 N
a+ΣMC = 0; 130.8(1) - M = 0 M = 130.8 N # m
centroidal axis of the cross section are
A = 0.075(0.075) = 5.625110-3
2 m2
I =
1
12
(0.075)10.0753
2 = 2.6367110-6
2 m4
Referring to Fig. c, QE is
QF = y′A′ = 0.025(0.025)(0.075) = 46.875110-6
2 m3
Thus,
s =
N
A
{
My
I
For point F, y = 0.0375 - 0.025 = 0.0125 m.Then
sF =
-422.75
5.625110-3
2
-
130.8(0.0125)
2.6367110-6
2
= -695.24 kPa = 695 kPa (C) Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,
tA =
VQA
It
=
130.8c 46.875110-6
2 d
2.6367110-6
2(0.075)
= 31.0 kPa Ans.
The state of stress at point A is represented on the element shown in Fig. d.
Ans:
sF = 695 kPa (C),
tA = 31.0 kPa

814
.
Solution
Segment AB:
(smax)AB =
PAB
A
=
1500
(1.5)(1.5)
= 667 psi Ans.
Segment CD:
sa =
PCD
A
=
1500
(1.5)(1.5)
= 666.67 psi
sb =
Mc
I
=
1875(12)(0.75)
1
12 (1.5)(1.53
)
= 40 000 psi
(smax)CD = sa + sb = 666.67 + 40 000
= 40 666.67 psi = 40.7 ksi Ans.
*R8–4.
The gondola and passengers have a weight of 1500 lb and
center of gravity at G. The suspender arm AE has a square
cross-sectional area of 1.5 in.by 1.5 in.,and is pin connected at
its ends A and E. Determine the largest tensile stress
developed in regions AB and DC of the arm.
Ans:
(smax)AB = 667 psi, (smax)CD = 40.7 ksi
4 ft
5.5 ft
GG
1.25 ft
1.5 in.
1.5 in.
A
B C
DE

815
.
Solution
Internal Loadings: Considering the equilibrium for the free-body diagram of the
femur’s upper segment, Fig. a,
+ cΣFy = 0; N - 75 = 0 N = 75lb
a+ΣMO = 0; M - 75(2) = 0 M = 150 lb # in
centroidal axis of the femur’s cross section are
A = p112
- 0.52
2 = 0.75p in2
I =
p
4
114
- 0.54
2 = 0.234375p in4
s =
N
A
+
My
I
By inspection, the maximum normal stress is in compression.
smax =
-75
0.75p
-
150(1)
0.234375p
= -236 psi = 236 psi (C) Ans.
R8–5.
If the cross section of the femur at section a–a can be
approximated as a circular tube as shown, determine the
maximum normal stress developed on the cross section at
section a–a due to the load of 75 lb.
Ans:
smax = 236 psi (C)
a a
2 in.
75 lb
M
F
1 in.
0.5 in.
Section a – a

816
.
Solution
A = 0.03(0.03) = 0.9(10-3
) m2
I =
1
12
(0.03)(0.033
) = 67.5(10-9
) m4
Require sA = 0
sA = 0 =
P
A
+
Mc
I
0 =
-98.1 cos u
0.9(10-3
)
+
98.1 sin u(0.015)
67.5(10-9
)
0 = -1111.11 cos u + 222222.22 sin u
tan u = 0.005; u = 0.286° Ans.
R8–6.
A bar having a square cross section of 30 mm by
30 mm is 2 m long and is held upward. If it has a mass of
5 kg/m, determine the largest angle u, measured from the
vertical, at which it can be supported before it is subjected
to a tensile stress along its axis near the grip.
Ans:
u = 0.286°
2 m
u

817
.
Solution
a+ΣMB = 0; 12(3) + 10(8) - FA(10) = 0
FA = 11.60 kip
I = 2c
1
12
(0.25)(2)3
d = 0.333 in4
A = 2(0.25)(2) = 1 in2
At point C,
sC =
P
A
=
2(5.80)
1
= 11.6 ksi Ans.
tC = 0 Ans.
At point D,
sD =
P
A
-
Mc
I
=
2(5.80)
1
-
[2(5.80)](1)
0.333
= -23.2 ksi Ans.
tD = 0 Ans.
R8–7.
The wall hanger has a thickness of 0.25 in. and is used to
support the vertical reactions of the beam that is loaded as
shown.If the load is transferred uniformly to each strap of the
hanger, determine the state of stress at points C and D on the
strap at A. Assume the vertical reaction F at this end acts in
the center and on the edge of the bracket as shown.
Ans:
sC = 11.6 ksi, tC = 0,
sD = -23.2 ksi, tD = 0
10 kip
A B
2 kip/ft
2 ft 2 ft 6 ft
2 in.
3.75 in.
2.75 in.
3 in.
1 in.
2 in.
2 in.
F
C
D
1 in.

.
Solution
a+ΣMA = 0; FB(10) - 10(2) - 12(7) = 0; FB = 10.40 kip
I = 2c
1
12
(0.25)(2)3
d = 0.333 in4
; A = 2(0.25)(2) = 1 in2
At point C:
sC =
P
A
=
2(5.20)
1
= 10.4 ksi Ans.
tC = 0 Ans.
At point D:
sD =
P
A
-
Mc
I
=
2(5.20)
1
-
[2(5.20)](1)
0.333
= -20.8 ksi Ans.
tD = 0 Ans.
*R8–8.
The wall hanger has a thickness of 0.25 in. and is used to
support the vertical reactions of the beam that is loaded as
shown.If the load is transferred uniformly to each strap of the
hanger, determine the state of stress at points C and D on the
strap at B. Assume the vertical reaction F at this end acts in
the center and on the edge of the bracket as shown.
Ans:
sC = 10.4 ksi, tC = 0,
sD = -20.8 ksi, tD = 0
10 kip
A B
2 kip/ft
2 ft 2 ft 6 ft
2 in.
3.75 in.
2.75 in.
3 in.
1 in.
2 in.
2 in.
F
C
D
1 in.

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Solution manual for mechanics of materials 10th edition hibbeler sample