PROBLEM 14.21
A 400–lb vertical force is applied at D to a gear attached to the solid l–in.
diameter shaft AB. Determine the principal stresses and the maximum
shearing stress at point H located as shown on top of the shaft.
SOLUTION
PROBLEM 14.22
A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that
the mechanic applies a vertical 24–lb force at A, determine the principal
stresses and the maximum shearing stress at point H located as shown
3
consent of McGraw–Hill Education.
PROBLEM 14.23
The steel pipe AB has a 102–mm outer diameter and a 6–mm wall
thickness. Knowing that arm CD is rigidly attached to the pipe,
determine the principal stresses and the maximum shearing stress
consent of McGraw–Hill Education.
PROBLEM 14.23 (Continued)
PROBLEM 14.24
For the state of plane stress shown, determine the largest value of
y
σ
for which
the maximum in–plane shearing stress is equal to or less than 75 MPa.
PROBLEM 14.25
Solve Probs. 14.5 and 14.9, using Mohr’s circle.
PROBLEM 14.5 through 14.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 14.9 through 14.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in–plane shearing stress, (b) the maximum
in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
ave
60 MPa,
40 MPa,
35 MPa
50 MPa
2
x
y
xy
xy
σ
σ
τ
σσ
σ
= −
= −
=
+
= = −
Plotted points for Mohr’s circle:
ave
:( , ) ( 60 MPa, 35 MPa)
:( , ) ( 40 MPa, 35 MPa)
:( , 0) ( 50 MPa, 0)
x xy
y xy
X
Y
C
στ
στ
σ
−=− −
= −
= −
(a)
35
tan 3.500
10
GX
CG
β
= = =
74.05
137.03
2
b
β
θβ
= °
=−=− °
37.0
b
θ
=−°
180 105.95
152.97
2
a
aβ
θa
= °− = °
= = °
53.0
a
θ
= °
22 22
10 35 36.4 MPaR CG GX= + = +=
(b)
min ave
50 36.4R
σσ
= −=−−
min 86.4 MPa
σ
= −
max ave
50 36.4R
σσ
= +=−+
max
13.60 MPa
σ
= −
(a′)
45 7.97
dB
θθ
= + °= °
8.0
d
θ
= °
45 97.97
eA
θθ
= + °= °
98.0
e
θ
= °
(b′)
max 36.4 MPaR
τ
= =
max 36.4 MPa
τ
=
(c′)
ave
50 MPa
σσ
′= = −
PROBLEM 14.26
Solve Probs. 14.16 and 14.10, using Mohr’s circle.
PROBLEM 14.5 through 14.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 14.9 through 14.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in–plane shearing stress, (b) the maximum
in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
150 MPa
σ
=
x
30 MPa
σ
=
y
80 MPa
τ
= −
xy
ave
90 MPa
2
σσ
σ
+
= =
xy
Plotted points for Mohr’s circle:
:( , ) (150 MPa, 80 MPa)
στ
−=
x xy
X
:( , ) (30 MPa, 80 MPa)
y xy
Y= −
στ
ave
: ( , 0) (90 MPa, 0)
σ
=C
(150 30) 60
22
σσ
−−
= =
xy
22
(60) (80) 100= +=R
(a)
80
tan 2 60
θ
=
p
2 53.130
θ
= °
p
26.6 and 63.4
θ
=−° °
p
(b)
max ave
90 100
σσ
= += +R
max
190.0 MPa
σ
=
min ave 90 100
σσ
= −= −R
min
10.00 MPa
σ
= −
(a′)
45
θθ
= +°
sp
18.4 and 108.4
θ
=°°
s
(b′)
max
τ
=R
max
100.0 MPa
τ
=
(c′)
ave
σσ
′=
90.0 MPa
′=
σ
consent of McGraw–Hill Education.
PROBLEM 14.27
Solve Prob. 14.11, using Mohr’s circle.
PROBLEM 14.9 through 14.12 For the given state of stress, determine
(a) the orientation of the planes of maximum in–plane shearing stress, (b) the
maximum in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
18 ksi
s
=
x
12 ksi
s
= −
y
8 ksi
xy
=
τ
ave
3 ksi
2
ss
s
+
= =
xy
Plotted points for Mohr’s circle:
: ( , ) (18 ksi, 8 ksi)
x xy
X−= −
sτ
: ( , ) ( 12 ksi, 8 ksi)
y xy
Y= −
sτ
ave
: ( , 0) (3 ksi, 0)
s
=C
8
tan 0.5333
15
a
= = =
FX
CF
28.07
a
= °
114.04
2
θa
= = °
A
(a)
45 59.0
θθ
= + °= °
DA
59.0
θ
= °
D
45 30.1
EA
= − °=− °
θθ
30.1
θ
=−°
E
22 22
15 8 17 ksi= + = +=R CF FX
(b)
max
17.00 ksi
τ
= =R
max
17.00 ksi
τ
=
(c)
ave 3.00 ksi
ss
′= =
3.00 ksi
′=
s
consent of McGraw–Hill Education.
PROBLEM 14.28
Solve Prob. 14.12, using Mohr’s circle.
PROBLEM 14.9 through 14.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in–plane shearing stress, (b) the maximum
in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
2 ksi
s
=
x
10 ksi
s
=
y
3 ksi
τ
= −
xy
ave
2 10 6 ksi
22
ss
s
++
= = =
y
x
Plotted points for Mohr’s circle:
: ( , ) (2 ksi, 3 ksi)
x xy
X−=
sτ
: ( , ) (10 ksi, 3 ksi)
y xy
Y= −
sτ
ave
: ( , 0) (6 ksi, 0)
s
=C
3
tan 0.75
4
a
= = =
FX
FC
36.87
a
= °
118.43
2
θa
= = °
B
(a)
45 26.6
θθ
= − °=− °
DB
26.6
θ
=−°
D
45 63.4
EB
= + °= °
θθ
63.4
θ
= °
E
2222
4 3 5 ksi= + = +=R CF FX
(b)
max
5.00 ksi
τ
= =R
max 5.00 ksi
τ
=
(c)
ave 6.00 ksi
ss
′= =
6.00 ksi
′=
s
consent of McGraw–Hill Education.
PROBLEM 14.29
Solve Prob. 14.13, using Mohr’s circle.
PROBLEM 14.13 through 14.16 For the given state of stress, determine the normal
and shearing stresses after the element shown has been rotated through (a) 25°
clockwise, (b) 10° counterclockwise.
SOLUTION
ave
0,
8 ksi,
5 ksi
4 ksi
2
x
y
xy
xy
s
s
τ
ss
s
=
=
=
+
= =
Plotted points for Mohr’s circle:
: (0, 5 ksi)
:(8 ksi, 5 ksi)
:(4 ksi, 0)
X
Y
C
−
2222
5
tan 2 1.25
4
2 51.34
4 5 6.4031 ksi
θ
θ
= = =
= °
= + = +=
p
p
FX
FC
R FC FX
(a)
25
θ
= °
.
2 50
θ
= °
51.34 50 1.34
ϕ
= °− °= °
ave
cos
x
R
ss ϕ
′
= −
2.40 ksi
x
s
′
= −
sin
xy R
τϕ
′′=
0.1497 ksi
τ
′′
=
xy
ave cos
yR
ss ϕ
′= +
10.40 ksi
y
s
′
=
(b)
10
θ
= °
.
2 20
θ
= °
51.34 20 71.34
ϕ
= °+ °= °
ave
cos
x
R
ss ϕ
′
= −
1.951 ksi
s
′
=
x
sin
xy R
τϕ
′′=
6.07 ksi
xy
τ
′′=
ave cos
yR
ss ϕ
′= +
6.05 ksi
y
s
′=
PROBLEM 14.22
A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that
the mechanic applies a vertical 24–lb force at A, determine the principal
stresses and the maximum shearing stress at point H located as shown
3
consent of McGraw–Hill Education.
PROBLEM 14.23
The steel pipe AB has a 102–mm outer diameter and a 6–mm wall
thickness. Knowing that arm CD is rigidly attached to the pipe,
determine the principal stresses and the maximum shearing stress
consent of McGraw–Hill Education.
PROBLEM 14.23 (Continued)
PROBLEM 14.24
For the state of plane stress shown, determine the largest value of
y
σ
for which
the maximum in–plane shearing stress is equal to or less than 75 MPa.
PROBLEM 14.25
Solve Probs. 14.5 and 14.9, using Mohr’s circle.
PROBLEM 14.5 through 14.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 14.9 through 14.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in–plane shearing stress, (b) the maximum
in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
ave
60 MPa,
40 MPa,
35 MPa
50 MPa
2
x
y
xy
xy
σ
σ
τ
σσ
σ
= −
= −
=
+
= = −
Plotted points for Mohr’s circle:
ave
:( , ) ( 60 MPa, 35 MPa)
:( , ) ( 40 MPa, 35 MPa)
:( , 0) ( 50 MPa, 0)
x xy
y xy
X
Y
C
στ
στ
σ
−=− −
= −
= −
(a)
35
tan 3.500
10
GX
CG
β
= = =
74.05
137.03
2
b
β
θβ
= °
=−=− °
37.0
b
θ
=−°
180 105.95
152.97
2
a
aβ
θa
= °− = °
= = °
53.0
a
θ
= °
22 22
10 35 36.4 MPaR CG GX= + = +=
(b)
min ave
50 36.4R
σσ
= −=−−
min 86.4 MPa
σ
= −
max ave
50 36.4R
σσ
= +=−+
max
13.60 MPa
σ
= −
(a′)
45 7.97
dB
θθ
= + °= °
8.0
d
θ
= °
45 97.97
eA
θθ
= + °= °
98.0
e
θ
= °
(b′)
max 36.4 MPaR
τ
= =
max 36.4 MPa
τ
=
(c′)
ave
50 MPa
σσ
′= = −
PROBLEM 14.26
Solve Probs. 14.16 and 14.10, using Mohr’s circle.
PROBLEM 14.5 through 14.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 14.9 through 14.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in–plane shearing stress, (b) the maximum
in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
150 MPa
σ
=
x
30 MPa
σ
=
y
80 MPa
τ
= −
xy
ave
90 MPa
2
σσ
σ
+
= =
xy
Plotted points for Mohr’s circle:
:( , ) (150 MPa, 80 MPa)
στ
−=
x xy
X
:( , ) (30 MPa, 80 MPa)
y xy
Y= −
στ
ave
: ( , 0) (90 MPa, 0)
σ
=C
(150 30) 60
22
σσ
−−
= =
xy
22
(60) (80) 100= +=R
(a)
80
tan 2 60
θ
=
p
2 53.130
θ
= °
p
26.6 and 63.4
θ
=−° °
p
(b)
max ave
90 100
σσ
= += +R
max
190.0 MPa
σ
=
min ave 90 100
σσ
= −= −R
min
10.00 MPa
σ
= −
(a′)
45
θθ
= +°
sp
18.4 and 108.4
θ
=°°
s
(b′)
max
τ
=R
max
100.0 MPa
τ
=
(c′)
ave
σσ
′=
90.0 MPa
′=
σ
consent of McGraw–Hill Education.
PROBLEM 14.27
Solve Prob. 14.11, using Mohr’s circle.
PROBLEM 14.9 through 14.12 For the given state of stress, determine
(a) the orientation of the planes of maximum in–plane shearing stress, (b) the
maximum in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
18 ksi
s
=
x
12 ksi
s
= −
y
8 ksi
xy
=
τ
ave
3 ksi
2
ss
s
+
= =
xy
Plotted points for Mohr’s circle:
: ( , ) (18 ksi, 8 ksi)
x xy
X−= −
sτ
: ( , ) ( 12 ksi, 8 ksi)
y xy
Y= −
sτ
ave
: ( , 0) (3 ksi, 0)
s
=C
8
tan 0.5333
15
a
= = =
FX
CF
28.07
a
= °
114.04
2
θa
= = °
A
(a)
45 59.0
θθ
= + °= °
DA
59.0
θ
= °
D
45 30.1
EA
= − °=− °
θθ
30.1
θ
=−°
E
22 22
15 8 17 ksi= + = +=R CF FX
(b)
max
17.00 ksi
τ
= =R
max
17.00 ksi
τ
=
(c)
ave 3.00 ksi
ss
′= =
3.00 ksi
′=
s
consent of McGraw–Hill Education.
PROBLEM 14.28
Solve Prob. 14.12, using Mohr’s circle.
PROBLEM 14.9 through 14.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in–plane shearing stress, (b) the maximum
in–plane shearing stress, (c) the corresponding normal stress.
SOLUTION
2 ksi
s
=
x
10 ksi
s
=
y
3 ksi
τ
= −
xy
ave
2 10 6 ksi
22
ss
s
++
= = =
y
x
Plotted points for Mohr’s circle:
: ( , ) (2 ksi, 3 ksi)
x xy
X−=
sτ
: ( , ) (10 ksi, 3 ksi)
y xy
Y= −
sτ
ave
: ( , 0) (6 ksi, 0)
s
=C
3
tan 0.75
4
a
= = =
FX
FC
36.87
a
= °
118.43
2
θa
= = °
B
(a)
45 26.6
θθ
= − °=− °
DB
26.6
θ
=−°
D
45 63.4
EB
= + °= °
θθ
63.4
θ
= °
E
2222
4 3 5 ksi= + = +=R CF FX
(b)
max
5.00 ksi
τ
= =R
max 5.00 ksi
τ
=
(c)
ave 6.00 ksi
ss
′= =
6.00 ksi
′=
s
consent of McGraw–Hill Education.
PROBLEM 14.29
Solve Prob. 14.13, using Mohr’s circle.
PROBLEM 14.13 through 14.16 For the given state of stress, determine the normal
and shearing stresses after the element shown has been rotated through (a) 25°
clockwise, (b) 10° counterclockwise.
SOLUTION
ave
0,
8 ksi,
5 ksi
4 ksi
2
x
y
xy
xy
s
s
τ
ss
s
=
=
=
+
= =
Plotted points for Mohr’s circle:
: (0, 5 ksi)
:(8 ksi, 5 ksi)
:(4 ksi, 0)
X
Y
C
−
2222
5
tan 2 1.25
4
2 51.34
4 5 6.4031 ksi
θ
θ
= = =
= °
= + = +=
p
p
FX
FC
R FC FX
(a)
25
θ
= °
.
2 50
θ
= °
51.34 50 1.34
ϕ
= °− °= °
ave
cos
x
R
ss ϕ
′
= −
2.40 ksi
x
s
′
= −
sin
xy R
τϕ
′′=
0.1497 ksi
τ
′′
=
xy
ave cos
yR
ss ϕ
′= +
10.40 ksi
y
s
′
=
(b)
10
θ
= °
.
2 20
θ
= °
51.34 20 71.34
ϕ
= °+ °= °
ave
cos
x
R
ss ϕ
′
= −
1.951 ksi
s
′
=
x
sin
xy R
τϕ
′′=
6.07 ksi
xy
τ
′′=
ave cos
yR
ss ϕ
′= +
6.05 ksi
y
s
′=