PROBLEM 14.31
Solve Prob. 14.15, 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,
80 MPa,
50 MPa
40 MPa
2
x
y
xy
xy
σ
σ
τ
σσ
σ
=
= −
= −
+
= = −
Plotted points for Mohr’s circle:
: (0, 50 MPa)
: ( 80 MPa, 50 MPa)
: ( 40 MPa, 0)
X
Y
C
−−
−
22 22
50
tan 2 1.25
40
2 51.34
40 50
64.031 MPa
p
p
FX
CF
R CF FX
= = =
= °
= += +
=
θ
θ
(a)
25
θ
= °
.
2 50
θ
= °
51.34 50 1.34
ϕ
= °− °= °
ave
cos
x
R
σσ ϕ
′
= +
24.0 MPa
x
σ
′=
sin
xy R
τϕ
′′= −
1.497 MPa
τ
′′= −
xy
ave cos
yR
σσ ϕ
′= −
104.0 MPa
y
σ
′= −
(b)
10
θ
= °
.
2 20
θ
= °
51.34 20 71.34
ϕ
= °+ °= °
ave
cos
x
R
σσ ϕ
′
= +
19.51 MPa
x′= −
σ
sin
xy R
τϕ
′′= −
60.7 MPa
xy
τ
′′= −
ave cos
yR
σσ ϕ
′= −
60.5 MPa
y
σ
′
= −
PROBLEM 14.32
Solve Prob. 14.16, 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
60 MPa,
90 MPa,
30 MPa
15 MPa
2
x
y
xy
xy
σ
σ
τ
σσ
σ
= −
=
=
+
= =
Plotted points for Mohr’s circle:
: ( 60 MPa, 30 MPa)
: (90 MPa, 30 MPa)
: (15 MPa, 0)
X
Y
C
−−
30
tan 2 0.4
75
θ
= = =
pFX
FC
2 21.80 10.90
θθ
= °= °
pP
22 22
75 30 80.78 MPaR FC FX= + = +=
(a)
25
θ
= °
.
2 50
θ
= °
2 2 50 21.80 28.20
P
ϕθθ
= − = °− °= °
ave
cos
σσ ϕ
′
= −
x
R
56.2 MPa
σ
′= −
x
sin
τϕ
′′= −
xy R
38.2 MPa
τ
′′
= −
xy
ave cos
σσ ϕ
′= +
yR
86.2 MPa
y
σ
′
=
PROBLEM 14.32 (Continued)
(b)
10
θ
= °
.
2 20
θ
= °
2 2 21.80 20 41.80
ϕθ θ
= + = °+ °= °
p
ave cos
σσ ϕ
′= −
xR
45.2 MPa
σ
′
= −
x
53.8 MPa
τ
′′=
xy
ave cos
σσ ϕ
′= +
yR
75.2 MPa
σ
′=
y
consent of McGraw–Hill Education.
PROBLEM 14.33
Solve Prob. 14.17, using Mohr’s circle.
PROBLEM 14.17 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in–plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
SOLUTION
ave
4 MPa 1.6 MPa 0
2.8 MPa
2
x y xy
xy
σσ τ
σσ
σ
=−=− =
+
= = −
Plotted points for Mohr’s circle:
ave
: ( , ) ( 4 MPa, 0)
: ( , ) ( 1.6 MPa, 0)
:( , 0) ( 2.8 MPa, 0)
x xy
y xy
X
Y
C
στ
στ
σ
−=−
= −
= −
15 . 2 30
1.2MPa 1.2MPaCX R
θθ
=−° =−°
= =
(a)
sin30 sin30 1.2sin30
xy CX R
τ
′′ ′
=− °=− °=− °
0.600 MPa
xy
τ
′′
= −
(b)
ave cos30 2.8 1.2cos30
xCX
σσ
′′
= − °=− − °
3.84 MPa
x
σ
′
= −
consent of McGraw–Hill Education.
PROBLEM 14.34
Solve Prob. 14.18, using Mohr’s circle.
PROBLEM 14.18 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in–plane shearing stress
consent of McGraw–Hill Education.
PROBLEM 14.35
Solve Prob. 14.19, using Mohr’s circle.
PROBLEM 14.19 Two steel plates of uniform cross section
10 80 mm×
are welded together as shown. Knowing that centric 100-
kN forces are applied to the welded plates and that
25
β
= °
, determine
(a) the in–plane shearing stress parallel to the weld, (b) the normal stress
perpendicular to the weld.
SOLUTION
36
3) 3)
100 10 125 10 Pa 125 MPa
(10 10 (80 10
xP
A
σ
−−
×
== =×=
××
00
y xy
στ
= =
From Mohr’s circle:
(a)
62.5sin50
w
τ
= °
47.9 MPa
w
τ
=
(b)
62.5 62.5cos50
w
σ
=+°
102.7 MPa
w
σ
=
consent of McGraw–Hill Education.
PROBLEM 14.36
Solve Prob. 14.20, using Mohr’s circle.
PROBLEM 14.20 The centric force P is applied to a short post as shown. Knowing
that the stresses on plane a-a are
15 ksi
s
= −
and
5 ksi,
τ
=
determine (a) the angle
β
post.
SOLUTION
0
0
x
xy
y
P
A
s
τ
s
=
=
= −
(a) From the Mohr’s circle,
5
tan 0.3333
15
β
= =
18.4
β
= °
(b)
cos2
22
PP
AA
sβ
−= +
2( ) (2)(15)
1 cos2 1 cos2
P
A
s
ββ
−
= =
++
16.67 ksi=
16.67 ksi
PROBLEM 14.37
Solve Prob. 14.21, using Mohr’s circle.
PROBLEM 14.21 A 400–lb vertical force is applied at D to a gear
attached to the solid 1–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 14.38
Solve Prob. 14.22, using Mohr’s circle.
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 on top of the
3
4–in. diameter shaft.
consent of McGraw–Hill Education.
PROBLEM 14.31
Solve Prob. 14.15, 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,
80 MPa,
50 MPa
40 MPa
2
x
y
xy
xy
σ
σ
τ
σσ
σ
=
= −
= −
+
= = −
Plotted points for Mohr’s circle:
: (0, 50 MPa)
: ( 80 MPa, 50 MPa)
: ( 40 MPa, 0)
X
Y
C
−−
−
22 22
50
tan 2 1.25
40
2 51.34
40 50
64.031 MPa
p
p
FX
CF
R CF FX
= = =
= °
= += +
=
θ
θ
(a)
25
θ
= °
.
2 50
θ
= °
51.34 50 1.34
ϕ
= °− °= °
ave
cos
x
R
σσ ϕ
′
= +
24.0 MPa
x
σ
′=
sin
xy R
τϕ
′′= −
1.497 MPa
τ
′′= −
xy
ave cos
yR
σσ ϕ
′= −
104.0 MPa
y
σ
′= −
(b)
10
θ
= °
.
2 20
θ
= °
51.34 20 71.34
ϕ
= °+ °= °
ave
cos
x
R
σσ ϕ
′
= +
19.51 MPa
x′= −
σ
sin
xy R
τϕ
′′= −
60.7 MPa
xy
τ
′′= −
ave cos
yR
σσ ϕ
′= −
60.5 MPa
y
σ
′
= −
PROBLEM 14.32
Solve Prob. 14.16, 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
60 MPa,
90 MPa,
30 MPa
15 MPa
2
x
y
xy
xy
σ
σ
τ
σσ
σ
= −
=
=
+
= =
Plotted points for Mohr’s circle:
: ( 60 MPa, 30 MPa)
: (90 MPa, 30 MPa)
: (15 MPa, 0)
X
Y
C
−−
30
tan 2 0.4
75
θ
= = =
pFX
FC
2 21.80 10.90
θθ
= °= °
pP
22 22
75 30 80.78 MPaR FC FX= + = +=
(a)
25
θ
= °
.
2 50
θ
= °
2 2 50 21.80 28.20
P
ϕθθ
= − = °− °= °
ave
cos
σσ ϕ
′
= −
x
R
56.2 MPa
σ
′= −
x
sin
τϕ
′′= −
xy R
38.2 MPa
τ
′′
= −
xy
ave cos
σσ ϕ
′= +
yR
86.2 MPa
y
σ
′
=
PROBLEM 14.32 (Continued)
(b)
10
θ
= °
.
2 20
θ
= °
2 2 21.80 20 41.80
ϕθ θ
= + = °+ °= °
p
ave cos
σσ ϕ
′= −
xR
45.2 MPa
σ
′
= −
x
53.8 MPa
τ
′′=
xy
ave cos
σσ ϕ
′= +
yR
75.2 MPa
σ
′=
y
consent of McGraw–Hill Education.
PROBLEM 14.33
Solve Prob. 14.17, using Mohr’s circle.
PROBLEM 14.17 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in–plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
SOLUTION
ave
4 MPa 1.6 MPa 0
2.8 MPa
2
x y xy
xy
σσ τ
σσ
σ
=−=− =
+
= = −
Plotted points for Mohr’s circle:
ave
: ( , ) ( 4 MPa, 0)
: ( , ) ( 1.6 MPa, 0)
:( , 0) ( 2.8 MPa, 0)
x xy
y xy
X
Y
C
στ
στ
σ
−=−
= −
= −
15 . 2 30
1.2MPa 1.2MPaCX R
θθ
=−° =−°
= =
(a)
sin30 sin30 1.2sin30
xy CX R
τ
′′ ′
=− °=− °=− °
0.600 MPa
xy
τ
′′
= −
(b)
ave cos30 2.8 1.2cos30
xCX
σσ
′′
= − °=− − °
3.84 MPa
x
σ
′
= −
consent of McGraw–Hill Education.
PROBLEM 14.34
Solve Prob. 14.18, using Mohr’s circle.
PROBLEM 14.18 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in–plane shearing stress
consent of McGraw–Hill Education.
PROBLEM 14.35
Solve Prob. 14.19, using Mohr’s circle.
PROBLEM 14.19 Two steel plates of uniform cross section
10 80 mm×
are welded together as shown. Knowing that centric 100-
kN forces are applied to the welded plates and that
25
β
= °
, determine
(a) the in–plane shearing stress parallel to the weld, (b) the normal stress
perpendicular to the weld.
SOLUTION
36
3) 3)
100 10 125 10 Pa 125 MPa
(10 10 (80 10
xP
A
σ
−−
×
== =×=
××
00
y xy
στ
= =
From Mohr’s circle:
(a)
62.5sin50
w
τ
= °
47.9 MPa
w
τ
=
(b)
62.5 62.5cos50
w
σ
=+°
102.7 MPa
w
σ
=
consent of McGraw–Hill Education.
PROBLEM 14.36
Solve Prob. 14.20, using Mohr’s circle.
PROBLEM 14.20 The centric force P is applied to a short post as shown. Knowing
that the stresses on plane a-a are
15 ksi
s
= −
and
5 ksi,
τ
=
determine (a) the angle
β
post.
SOLUTION
0
0
x
xy
y
P
A
s
τ
s
=
=
= −
(a) From the Mohr’s circle,
5
tan 0.3333
15
β
= =
18.4
β
= °
(b)
cos2
22
PP
AA
sβ
−= +
2( ) (2)(15)
1 cos2 1 cos2
P
A
s
ββ
−
= =
++
16.67 ksi=
16.67 ksi
PROBLEM 14.37
Solve Prob. 14.21, using Mohr’s circle.
PROBLEM 14.21 A 400–lb vertical force is applied at D to a gear
attached to the solid 1–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 14.38
Solve Prob. 14.22, using Mohr’s circle.
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 on top of the
3
4–in. diameter shaft.
consent of McGraw–Hill Education.