SOLUTION CAP 9

Prévia do material em texto

Problem 9-1
Prove that the sum of the normal stresses x + y = x' + y' is constant. See Figs. 9-2a and 9-2b.
Solution:
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.
x'
x y
2
x y
2
cos 2 xy sin 2= (1)
y'
x y
2
x y
2
cos 2 xy sin 2= (2)
(1) + (2) :
LHS x' y'=
RHS
x y
2
x y
2
= RHS x y=
Hence,
x' y' x y= (Q.E.D.)
Problem 9-2
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: x 3MPa y 5MPa xy 8MPa 40deg
Solution: Set A m
2 180deg
Force Equilibrium: For the sectioned element,
Ax A cos Ay A sin
Fxx x Ax Fxy xy Ax
Fyy y Ay Fyx xy Ay
Given
+ Fx'=0; Fx' Fxy sin Fxx cos Fyx cos Fyy sin 0=
+ Fy'=0; Fy' Fxy cos Fxx sin Fyx sin Fyy cos 0=
Guess Fx' 1kN Fy' 1kN
Fx'
Fy'
Find Fx' Fy'
Fx'
Fy'
4052.11
404.38
kN
Normal and Shear Stress:
0A
F
A
lim=
x'
Fx'
A
x' 4.052 MPa Ans
x'y'
Fy'
A
x'y' 0.404 MPa Ans
The negative signs indicate that the sense of x' and x'y' are opposite to that shown in FBD.
Problem 9-3
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: x 0.200MPa y 0.350MPa
50deg xy 0MPa
Solution: Set A m
2 180deg
Force Equilibrium: For the sectioned element,
Ax A cos Ay A sin
Fxx x Ax Fxy xy Ax Fxy 0.00
Fyy y Ay Fyx xy Ay Fyx 0.00
Given
+ Fx'=0; Fx' Fxy sin Fxx cos Fyx cos Fyy sin 0=
+ Fy'=0; Fy' Fxy cos Fxx sin Fyx sin Fyy cos 0=
Guess Fx' 1kN Fy' 1kN
Fx'
Fy'
Find Fx' Fy'
Fx'
Fy'
122.75
270.82
kN
Normal and Shear Stress:
0A
F
A
lim=
x'
Fx'
A
x' 0.123 MPa Ans
x'y'
Fy'
A
x'y' 0.271 MPa Ans
Problem 9-4
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: x 0.650MPa y 0.400MPa
60deg xy 0MPa
Solution: Set A m
2 90deg
Force Equilibrium: For the sectioned element,
Ax A sin Ay A cos
Fxx x Ax Fxy xy Ax Fxy 0.00
Fyy y Ay Fyx xy Ay Fyx 0.00
Given
 + Fx'=0; Fx' Fxy sin Fxx cos Fyx cos Fyy sin 0=
+ Fy'=0; Fy' Fxy cos Fxx sin Fyx sin Fyy cos 0=
Guess Fx' 1kN Fy' 1kN
Fx'
Fy'
Find Fx' Fy'
Fx'
Fy'
387.50
454.66
kN
Normal and Shear Stress:
0A
F
A
lim=
x'
Fx'
A
x' 0.387 MPa Ans
x'y'
Fy'
A
x'y' 0.455 MPa Ans
The negative signs indicate that the sense of x' is opposite to that shown in FBD.
Problem 9-5
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: x 60MPa y 50MPa
30deg xy 28MPa
Solution: Set A m
2 180deg
Force Equilibrium: For the sectioned element,
Ax A cos Ay A sin
Fxx x Ax Fxy xy Ax Fxy 24248.71 kN
Fyy y Ay Fyx xy Ay Fyx 14000.00 kN
Given
+ Fx'=0; Fx' Fxy sin Fxx cos Fyx cos Fyy sin 0=
+ Fy'=0; Fy' Fxy cos Fxx sin Fyx sin Fyy cos 0=
Guess Fx' 1kN Fy' 1kN
Fx'
Fy'
Find Fx' Fy'
Fx'
Fy'
33251.29
18330.13
kN
Normal and Shear Stress:
0A
F
A
lim=
x'
Fx'
A
x' 33.251 MPa Ans
x'y'
Fy'
A
x'y' 18.330 MPa Ans
The negative signs indicate that the sense of x' is opposite to that shown in FBD.
Problem 9-6
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: x 90MPa y 50MPa
xy 35MPa 60deg
Solution: Set A m
2 90deg
Force Equilibrium: For the sectioned element,
Ax A sin Ay A cos
Fxx x Ax Fxy xy Ax
Fyy y Ay Fyx xy Ay
Given
+ Fx'=0; Fx' Fxy sin Fxx cos Fyx cos Fyy sin 0=
 + Fy'=0; Fy' Fxy cos Fxx sin Fyx sin Fyy cos 0=
Guess Fx' 1kN Fy' 1kN
Fx'
Fy'
Find Fx' Fy'
Fx'
Fy'
49689.11
34820.51
kN
Normal and Shear Stress:
0A
F
A
lim=
x'
Fx'
A
x' 49.69 MPa Ans
x'y'
Fy'
A
x'y' 34.82 MPa Ans
The negative signs indicate that the sense of x'y' isopposite to that shown in FBD.
Problem 9-7
Solve Prob. 9-2 using the stress-transformation equations developed in Sec. 9.2.
Given: x 5MPa y 3MPa xy 8MPa 40deg
Solution: 90deg
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 4.05 MPa Ans
The negative signs indicate that the sense of x' is a compressive stress.
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 0.404 MPa Ans
The negative signs indicate that the sense of x'y' is in the -y' direction.
Problem 9-8
Solve Prob. 9-4 using the stress-transformation equations developed in Sec. 9.2.
Given: x 0.650MPa y 0.400MPa
60deg xy 0MPa
Solution: 90deg
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 0.387 MPa Ans
The negative signs indicate that the sense of x' is a compressive stress.
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 0.455 MPa Ans
Problem 9-9
Solve Prob. 9-6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a
sketch.
Given: x 90MPa y 50MPa
xy 35MPa 60deg
Solution: 90deg
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 49.69 MPa Ans
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 34.82 MPa Ans
The negative signs indicate that the sense of x'y' is in the -y' direction.
Problem 9-10
Determine the equivalent state of stress on an element if the element is oriented 30° counterclockwise
from the element shown. Use the stress-transformation equations.
Unit Used: kPa 1000Pa
Given: x 0kPa y 300kPa
30deg xy 950kPa
Solution:
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 747.7 kPa Ans
y'
x y
2
x y
2
cos 2 xy sin 2
y' 1047.7 kPa Ans
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 345.1 kPa Ans
Problem 9-11
Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the
element shown.
Given: x 0.300MPa y 0MPa
60deg xy 0.120MPa
Solution:
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 0.0289 MPa Ans
y'
x y
2
x y
2
cos 2 xy sin 2
y' 0.329 MPa Ans
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 0.0699 MPa Ans
Problem 9-12
Solve Prob. 9-6 using the stress-transformation equations.
Given: x 90MPa y 50MPa
60deg xy 35MPa
Solution: 90deg
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 49.69 MPa Ans
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 34.82 MPa Ans
Problem 9-13
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: x 45MPa y 60MPa xy 30MPa
Solution:
(a) Principal Stress:
1
x y
2
x y
2
2
xy
2
1 52.97 MPa
Ans
2
x y
2
x y
2
2
xy
2
2 67.97 MPa
Ans
Orientation of Principal Stress:
tan 2 p
2 xy
x y
= p
1
2
atan
2 xy
x y
'p p 90deg
p 14.87 deg 'p 75.13 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'
x y
2
x y
2
cos 2 p xy sin 2 p
x' 52.97 MPa
Therefore, p1 p p1 14.87 deg Ans
p2 'p p2 75.13 deg Ans
(b)
max
x y
2
2
xy
2
max 60.47 MPa Ans
avg
x y
2 avg
7.50 MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2 s
x y
2 xy
= s
1
2
atan
x y
2 xy
's s 90deg
s 30.13 deg 's 59.87 deg
By observation,in order to preserve equilibrium along AB, max has to act in the
direction shown in the figure.
Problem 9-14
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: x 180MPa y 0MPa xy 150MPa
Solution:
(a) Principal Stress:
1
x y
2
x y
2
2
xy
2
1 264.93 MPa Ans
2
x y
2
x y
2
2
xy
2
2 84.93 MPa Ans
Orientation of Principal Stress:
tan 2 p
2 xy
x y
= p
1
2
atan
2 xy
x y
'p p 90deg
p 29.52 deg 'p 60.48 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'
x y
2
x y
2
cos 2 p xy sin 2 p
x' 264.93 MPa
Therefore, p1 p p1 29.52 deg Ans
p2 'p p2 60.48 deg Ans
(b)
max
x y
2
2
xy
2
max 174.93 MPa Ans
avg
x y
2 avg
90.00 MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2 s
x y
2 xy
= s
1
2
atan
x y
2 xy
's s 90deg
s 15.48 deg 's 74.52 deg
By observation, in order to preserve equilibrium along AB, max has to act in the
direction shown in the figure.
Problem 9-15
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: x 30MPa y 0MPa xy 12MPa
Solution:
(a) Principal Stress:
1
x y
2
x y
2
2
xy
2
1 4.21 MPa Ans
2
x y
2
x y
2
2
xy
2
2 34.21 MPa Ans
Orientation of Principal Stress:
tan 2 p
2 xy
x y
= p
1
2
atan
2 xy
x y
'p p 90deg
p 19.33 deg 'p 70.67 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'
x y
2
x y
2
cos 2 p xy sin 2 p
x' 34.21 MPa
Therefore, p1 'p p1 70.67 deg Ans
p2 p p2 19.33 deg Ans
(b)
max
x y
2
2
xy
2
max 19.21 MPa Ans
avg
x y
2 avg
15.00 MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2 s
x y
2 xy
= s
1
2
atan
x y
2 xy
's s 90deg
s 25.67 deg 's 64.33 deg
By observation, in order to preserve equilibrium along AB, max has to act in the
direction shown in the figure.
Problem 9-16
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: x 200MPa y 250MPa xy 175MPa
Solution:
(a) Principal Stress:
1
x y
2
x y
2
2
xy
2
1 310.04 MPa
Ans
2
x y
2
x y
2
2
xy
2
2 260.04 MPa
Ans
Orientation of Principal Stress:
tan 2 p
2 xy
x y
= p
1
2
atan
2 xy
x y
'p p 90deg
p 18.94 deg 'p 71.06 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'
x y
2
x y
2
cos 2 p xy sin 2 p
x' 260.04 MPa
Therefore, p1 'p p1 71.06 deg Ans
p2 p p2 18.94 deg Ans
(b)
max
x y
2
2
xy
2
max 285.04 MPa Ans
avg
x y
2 avg
25.00 MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2 s
x y
2 xy
= s
1
2
atan
x y
2 xy
's s 90deg
s 26.06 deg 's 63.94 deg
By observation, in order to preserve equilibrium along AB, max has to act in the
direction shown in the figure.
Problem 9-17
A point on a thin plate is subjected to the two successive states of stress shown. Determine the
resultant state of stress represented on the element oriented as shown on the right.
Given:
(a) x'_a 200MPa x'y'_a 0MPa
y'_a 350MPa a 30deg
(a) x'_b 0 x'y'_b 58MPa
y'_b 0 b 25deg
Solution:
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 of the text.
For element (a):
x_a
x'_a y'_a
2
x'_a y'_a
2
cos 2 a x'y'_a sin 2 a x_a 237.50 MPa
y_a
x'_a y'_a
2
x'_a y'_a
2
cos 2 a x'y'_a sin 2 a y_a 312.50 MPa
xy_a
x'_a y'_a
2
sin 2 a x'y'_a cos 2 a xy_a 64.95 MPa
For element (b):
x_b
x'_b y'_b
2
x'_b y'_b
2
cos 2 b x'y'_b sin 2 b x_b 44.43 MPa
y_b
x'_b y'_b
2
x'_b y'_b
2
cos 2 b x'y'_b sin 2 b y_b 44.43 MPa
xy_b
x'_b y'_b
2
sin 2 b x'y'_b cos 2 b xy_b 37.28 MPa
Combining the stress componenets of two elements yields
x x_a x_b x 193.1 MPa Ans
y y_a y_b y 356.9 MPa Ans
xy xy_a xy_b xy 102.2 MPa Ans
Problem 9-18
The steel bar has a thickness of 12 mm and is subjected to the edge loading shown. Determine the
principal stresses developed in the bar.
Given: d 50mm t 12mm
q 4
kN
m
L 500mm
Solution:
Normal and Shear Stress:
In accordance with the established sign convention.
x 0MPa y 0MPa
xy
q
t xy
0.333 MPa
In-plane Principal Stress: Apply Eq. 9-5.
1
x y
2
x y
2
2
xy
2
1 0.333 MPa Ans
2
x y
2
x y
2
2
xy
2
2 0.333 MPa Ans
Problem 9-19
The steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the
maximum in-plane shear stress and the average normal stress developed in the steel.
Given: ax 300mm ay 100mm t 10mm
qx 30
kN
m
qy 40
kN
m
Solution:
Normal and Shear Stress:
In accordance with the established sign convention.
x
qx
t x
3.00 MPa
y
qy
t y
4.00 MPa
xy 0
Maximum In-plane Shear Stress: Apply Eq. 9-7.
max
x y
2
2
xy
2
max 0.500 MPa Ans
Average Normal Stress: Apply Eq. 9-8.
avg
x y
2 avg
3.50 MPa Ans
Problem 9-20
The stress acting on two planes at a point is indicated. Determine the shear stress on plane a-a and the
principal stresses at the point.
Given: a 80MPa b 60MPa
45deg 60deg
Solution:
x b sin
xy b cos
Given
a
x y
2
x y
2
cos 2 xy sin 2=
a
x y
2
sin 2 xy cos 2=
Guess y 1MPa a 1MPa
y
a
Find y a
y
a
48.04
1.96
MPa Ans
Principal Stress:
1
x y
2
x y
2
2
xy
2
1 80.06 MPa Ans
2
x y
2
x y
2
2
xy
2
2 19.94 MPa Ans
Problem 9-21
The stress acting on two planes at a point is indicated. Determine the normal stress b and the principal
stresses at the point.
Given: a 4MPa x'y' 2MPa
bb 45deg 60deg
Solution:
Stress Transformation Equations :
Applying Eqs. 9-3 and 9-1 with bb 90deg
y a sin xy a cos x' b=
Given
x'
x y
2
x y
2
cos 2 xy sin 2=
x'y'
x y
2
sin 2 xy cos 2=
Guess x 1MPa x' 1MPa
x
x'
Find x x'
x
x'
7.464
7.464
MPa
b x'
b 7.464 MPa Ans
In-plane Principal Stress: Applying Eq. 9-5,
1
x y
2
x y
2
2
xy
2
1 8.29 MPa Ans
2
x y
2
x y
2
2
xy
2
2 2.64 MPa Ans
Problem 9-22
The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt
is 40 kN, determine the principal stresses at points A and B and show the results on elements located at
each of these points. The cross-sectional area at A and B is shown in the adjacent figure.
Given: h 50mm t 30mm P 40kN
a 100mm b 200mm
Solution: L 3a b
Support Reactions :
+ O=0; E' L P a b( ) 0=
E'
P a b( )
L
E' 24 kN
Internal Force and Moment : At Section A-B:
+ Fx=0; E V 0= V E'
+ O=0; M E a( ) 0= M E' a
Section Property :
A h t A 1500 mm2
I
1
12
t h3 I 312500 mm4
QB 0.5 h t( ) 0.25 h( ) QB 9375 mm
3
QA 0 (since A' = 0)
Normal Stress: M c
I
=
cA 0.5h A
M cA
I A
192.00 MPa
cB 0 B
M cB
I B
0 MPa
Shear Stress : V Q
I t
=
A
V QA
I t A
0.00 MPa
B
V QB
I t B
24.00 MPa
In-plane Principal Stress:
At A: xA A yA 0 xy A
Since no shear stress acts upon the element,
A1 yA A1 0 MPa Ans
A2 xA A2 192 MPa Ans
At B: xB B yB 0 xy B
2
2
2.1 22 xy
yxyx
B1 xy B1 24 MPa Ans
B2 xy B2 24 MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2 p
2B
xB yB
= p
1
2
90deg( ) 'p p 90deg
p 45 deg 'p 45 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'_B
xB yB
2
xB yB
2
cos 2 p B sin 2 p
x'_B 24.00 MPa
Therefore, p1 'p p1 45.00 deg Ans
p2 p p2 45.00 deg Ans
Problem 9-23
Solve Prob. 9-22 for points C and D.
Given: h 50mm t 30mm P 40kN
a 100mm b 200mm hD 10mm
Solution: L 3a b
Support Reactions :
+ E=0; P 2a( ) R L 0=
R
2P a
L
R 16 kN
Internal Force and Moment : At Section C-D:
+ Fx=0; R V 0= V R
+ O=0; M R b( ) 0= M R b( )
Section Property :
A h t A 1500 mm2
I
1
12
t h3 I 312500 mm4
QD hD t 0.5 h 0.5 hD QD 6000 mm
3
QC 0 (since A' = 0)
Normal Stress: M c
I
=
cC 0.5h C
M cC
I C
256.00 MPa
cD 0.5 h hD D
M cD
I D
153.6 MPa
Shear Stress : V Q
I t
=
C
V QC
I t C
0.00 MPa
D
V QD
I t D
10.24 MPa
In-plane Principal Stress:
At C: xC 0 yC C xy C
Since no shear stress acts upon the element,
C1 yC C1 256 MPa Ans
C2 xC C2 0 MPa Ans
At D: xD 0 yD D xy D
D1
xD yD
2
xD yD
2
2
D
2
D1 0.680 MPa Ans
D2
xD yD
2
xD yD
2
2
D
2
D2 154.28 MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point D,
tan 2 p
2 D
xD yD
= p
1
2
atan
2 D
xD yD
'p p 90deg
p 3.797 deg 'p 86.203 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'_D
xD yD
2
xD yD
2
cos 2 p D sin 2 p
x'_D 0.680 MPa
Therefore, p1 p p1 3.80 deg Ans
p2 'p p2 86.20 deg Ans
Problem 9-24
The grains of wood in the board make an angle of 20° with the horizontal as shown. Determine the
normal and shear stress that act perpendicular to the grains if the board is subjected to an axial load of
250 N.
Unit Used: kPa 1000Pa
Given: P 250N 20deg
h 60mm t 25mm
Solution:
x
P
h t x
0.1667 MPa
y 0
xy 0
90deg
x'
x y
2
x y
2
cos 2 xy sin 2 x' 19.50 kPa Ans
x'y'
x y
2
sin 2 xy cos 2 x'y' 53.57 kPa Ans
Problem 9-25
The wooden block will fail if the shear stress acting along the grain is 3.85 MPa. If the normal stress
x = 2.8 MPa, determine the necessary compressive stress y that will cause failure.
Given: x 2.8MPa xy 0MPa
grain 58deg x'y' 3.85MPa
Solution: grain 90deg
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2=
x'y'
x y
2
sin 2=
y
2 x'y'
sin 2 x
y 5.767 MPa Ans
Problem 9-26
The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the
principal stresses at points A and B and show the results on elements located at each of these points.
Given: bf 150mm tf 20mm
dw 150mm tw 20mm
a 2m b 1m
hB 50mm w 12
kN
m
Solution:
Internal Force and Moment : At Section A-B:
+ Fy=0; V w a 0= V w a
+ A=0; M w a 0.5a b( ) 0= M w a 0.5a b( )
Section Property : D dw tf
yc
0.5tf bf tf 0.5dw tf dw tw
bf tf dw tw
yc 52.50 mm
I1
1
12
bf tf
3 bf tf 0.5tf yc
2
I2
1
12
tw dw
3 dw tw 0.5dw tf yc
2
I I1 I2 I 16562500.00 mm
4
QA 0 (since A' = 0)
QB hB tw D yc 0.5 hB QB 92500 mm
3
Normal Stress: M c
I
=
cA yc A
M cA
I A
152.15 MPa
cB D yc hB B
M cB
I B
195.62 MPa
Shear Stress : V Q
I t
= A
V QA
I bf
A 0.00 MPa
B
V QB
I tw
B 6.702 MPa
In-plane Principal Stress:
At A: xA A yA 0 xy A
Since no shear stress acts upon the element,
A1 xA A1 152.15 MPa Ans
A2 yA A2 0 MPa Ans
At B: xB B yB 0 xy B
B1
xB yB
2
xB yB
2
2
B
2
B1 0.229 MPa Ans
B2
xB yB
2
xB yB
2
2
B
2
B2 195.852 MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2 p
2 B
xB yB
= p
1
2
atan
2 B
xB yB
'p p 90deg
p 1.96 deg 'p 88.04 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'_B
xB yB
2
xB yB
2
cos 2 p B sin 2 p
x'_B 194.94 MPa
Therefore, p1 'p p1 88.04 deg Ans
p2 p p2 1.96 deg Ans
Problem 9-27
The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal
stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the
results on properly oriented elements located at these points.
Given: do 15mm a 50mm P 0.6kN
Solution:
Internal Force and Moment : At Section A-B:
+ Fx=0; N P 0= N P
+ O=0; M P a( ) 0= M P a
Section Property :
A
do
2
4
A 176.71 mm2
I
do
4
64
I 2485.05 mm4
I
Mc
A
N
2.1Normal Stress:
cA 0.5do A
N
A
M cA
I A
87.15 MPa
cB 0.5do B
N
A
M cB
I B
93.94 MPa
In-plane Principal Stress:
At A: xA A yA 0 xy 0
Since no shear stress acts upon the element,
A1 yA A1 0 MPa Ans
A2 xA A2 87.15 MPa Ans
At B: xB B yB 0 xy 0
Since no shear stress acts upon the element,
B1 xB B1 93.94 MPa Ans
B2 yB B2 0 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7
A.max
xA yA
2
2
xy
2
A.max 43.6 MPa Ans
B.max
xB yB
2
2
xy
2
B.max 47.0 MPa Ans
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
)2tan( S_A s_A
45deg Anstan 2 s_A
xA yA
2 xy
=
Ans
's_A s_A 90deg 's_A 45 deg Ans
)2tan( S_Btan 2 s_B
xB yB
2 xy
= s_B 45deg Ans
Ans
's_B s_B 90deg 's_B 45 deg Ans
By observation, in order to preserve equilibrium along AB, max has to act in the
direction shown in the figure.
Average Normal Stress: Applying Eq. 9-8
avg_A
xA yA
2 avg_A
43.57 MPa
avg_B
xB yB
2 avg_B
46.97 MPa
Problem 9-28
The simply supported beam is subjected to the traction stress 0 on its top surface. Determine the
principal stresses at points A and B.
Problem 9-29
The beam has a rectangular cross section and is subjected to the loadings shown. Determine the
principal stresses and the maximum in-plane shear stress that are developed at point A and point B.
These points are just to the left of the 10-kN load. Show the results on properly oriented elements
located at these points.
Given: b 150mm d 375mm
F 10kN P 5kN L 1.2m
Solution:
Support Reactions : By symmetry, R1=R ; R2= R
+ Fy=0; 2R F 0= R 0.5F
+ Fx=0; H1 P 0= H1 P
Internal Force and Moment : At Section A-B:
+ Fx=0; H1 N 0= N H1
+ Fy=0; R V 0= V R
+ O=0; M R 0.5L( ) 0= M 0.5R L
Section Property :
A b d I
1
12
b d3
QA 0 QB 0 (since A' = 0)
Normal Stress: N
A
M c
I
=
cA 0.5d A
N
A
M cA
I A
0.942 MPa
cB 0.5d B
N
A
M cB
I B
0.764 MPa
Shear Stress : Since QA = QB = 0, A 0 B 0
In-plane Principal Stress:
At A: xA A yA 0 xy 0
Since no shear stress acts upon the element,
A1 yA A1 0 MPa Ans
A2 xA A2 0.942 MPa Ans
At B: xB B yB 0 xy 0
Since no shear stress acts upon the element,
B1 xB B1 0.764 MPa Ans
B2 yB B2 0 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max_A
xA yA
2
2
xy
2
max_A 0.471 MPa Ans
max_B
xB yB
2
2
xy
2
max_B 0.382 MPa Ans
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
)2tan( S_A s_A
45deg Anstan 2 s_A
xA yA
2 xy
=
Ans
's_A s_A 90deg 's_A 45 deg Ans
)2tan( S_Btan 2 s_B
xB yB
2 xy
= s_B 45deg Ans
Ans
's_B s_B 90deg 's_B 45 deg Ans
By observation, in order to preserve equilibrium along AB, max has to act in the
direction shown in the figure.
Average Normal Stress: Applying Eq. 9-8,
avg_A
xA yA
2 avg_A
0.471 MPa Ans
avg_B
xB yB
2 avg_B
0.382 MPa Ans
Problem 9-30
The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at
point A and at point B. These points are located at the top and bottom of the web, respectively.
Although it is not very accurate, use the shear formula to compute the shear stress.
Given: bf 200mm tf 10mm
tw 10mm dw 200mm
P 25kN 30deg
a 3m w 8
kN
m
Solution:
Internal Force and Moment : At Section A-B:+ Fx=0; P cos N 0= N P cos
+ Fy=0; V P sin w a 0=
V P sin w a
+ O=0; M P sin a( ) w a( ) 0.5a( ) 0=
M P a sin 0.5w a2
Section Property : D dw 2tf
A bf D bf tw dw A 6000 mm
2
I
1
12
bf D
3 bf tw dw
3 I 50.80 10 6 m4
QA bf tf
D
2
tf
2
QA 210000 mm
3 QB QA
Normal Stress: N
A
M c
I
=
cA 0.5dw A
N
A
M cA
I A
148.293 MPa
cB 0.5dw B
N
A
M cB
I B
141.077 MPa
Shear Stress : V Q
I t
=
A
V QA
I tw
A 15.09 MPa
B
V QB
I tw
B 15.09 MPa
In-plane Principal Stress:
At A: xA A yA 0 xy A
A1
xA yA
2
xA yA
2
2
A
2
A1 149.8 MPa Ans
A2
xA yA
2
xA yA
2
2
A
2
A2 1.52 MPa Ans
At B: xB B yB 0 xy B
B1
xB yB
2
xB yB
2
2
B
2
B1 1.60 MPa Ans
B2
xB yB
2
xB yB
2
2
B
2
B2 142.7 MPa Ans
Problem 9-31
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and
the maximum in-plane shear stress that is developed anywhere on the surface of the shaft.
Problem 9-32
A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is
subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30
mm.
Unit used: kPa 1000Pa
Given: do 30mm t 1mm
30deg P 10N
Solution:
Section Property : di do 2t
A
4
do
2 di
2 A 91.11 mm2
Normal Stress:
x
P
A x
109.76 kPa
y 0
xy 0
Shear stress along the seam:
x'y'
x y
2
sin 2 xy cos 2
x'y' 47.53 kPa Ans
Problem 9-33
Solve Prob. 9-32 for the normal stress acting perpendicular to the seam.
Unit used: kPa 1000Pa
Given: do 30mm t 1mm
30deg P 10N
Solution:
Section Property : di do 2t
A
4
do
2 di
2 A 91.11 mm2
Normal Stress:
x
P
A x
109.76 kPa
y 0
xy 0
Normal stress perpendicular to the seam:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 82.32 kPa Ans
Problem 9-34
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and
the maximum in-plane shear stress that is developed at point A. The bearings only support vertical
reactions.
Problem 9-35
The drill pipe has an outer diameter of 75 mm, a wall thickness of 6 mm, and a weight of 0.8 kN/m. If
it is subjected to a torque and axial load as shown, determine (a) the principal stresses and (b) the
maximum in-plane shear stress at a point on its surface at section a.
Given: do 75mm t 6mm L 6m
P 7.5kN Mx 1.2kN m w 0.8
kN
m
Solution:
Internal Force and Moment : At section a:
Fx=0; N P w L 0= N P w L
x=0; T Mx 0= T Mx
Section Property : di do 2t
A
4
do
2 di
2 J
32
do
4 di
4
Normal Stress: N
A
9.457 MPa
Shear Stress :
c 0.5do
T c
J
28.850 MPa
a) In-plane Principal Stresses:
x 0 y xy
for any point on the shaft's surface. Applying Eq. 9-5,
1
x y
2
x y
2
2
xy
2
1 24.51 MPa Ans
2
x y
2
x y
2
2
xy
2
2 33.96 MPa Ans
b) Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
x y
2
2
xy
2
max 29.24 MPa Ans
Problem 9-36
The internal loadings at a section of the beam are shown. Determine the principal stresses at point A.
Also compute the maximum in-plane shear stress at this point.
Given: bf 200mm tf 50mm
tw 50mm dw 200mm
Px 500kN My 30kN m
Py 800kN Mz 40kN m
Solution:
Section Property : D dw 2tf
A bf D bf tw dw A 30000 mm
2
Iz
1
12
bf D
3 bf tw dw
3 Iz 350.00 10
6 m4
Iy
1
12
2tf bf
3 dw tw
3 Iy 68.75 10
6 m4
QA 0 (since A' = 0)
Normal Stress: yA 0.5D zA 0.5bf
A
Px
A
Mz yA
Iz
My zA
Iy
A 77.446 MPa
Shear Stress : Since QA = 0, A 0
In-plane Principal Stress:
x A y 0 xy 0
Since no shear stress acts upon the element,
1 y 1 0 MPa Ans
2 x 2 77.45 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
x y
2
2
xy
2
max 38.72 MPa Ans
Problem 9-37
Solve Prob. 9-36 for point B.
Given: bf 200mm tf 50mm
tw 50mm dw 200mm
Px 500kN My 30kN m
Py 800kN Mz 40kN m
Solution:
Section Property : D dw 2tf
A bf D bf tw dw A 30000 mm
2
Iz
1
12
bf D
3 bf tw dw
3 Iz 350.00 10
6 m4
Iy
1
12
2tf bf
3 dw tw
3 Iy 68.75 10
6 m4
QB 0 (since A' = 0)
Normal Stress: yB 0.5D zB 0.5bf
B
Px
A
Mz yB
Iz
My zB
Iy
B 44.113 MPa
Shear Stress : Since QB = 0, B 0
In-plane Principal Stress:
x B y 0 xy 0
Since no shear stress acts upon the element,
1 x 1 44.113 MPa Ans
2 y 2 0.00 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
x y
2
2
xy
2
max 22.06 MPa Ans
Problem 9-38
Solve Prob. 9-36 for point C, located in the center on the bottom of the web.
Given: bf 200mm tf 50mm
tw 50mm dw 200mm
Px 500kN My 30kN m
Py 800kN Mz 40kN m
Solution:
Section Property : D dw 2tf
A bf D bf tw dw A 30000 mm
2
Iz
1
12
bf D
3 bf tw dw
3 Iz 350.00 10
6 m4
Iy
1
12
2tf bf
3 dw tw
3 Iy 68.75 10
6 m4
QC bf tf
D
2
tf
2
QC 1250000 mm
3
Normal Stress: yC 0.5dw zC 0
C
Px
A
Mz yC
Iz
My zC
Iy
C 5.238 MPa
Shear Stress : V Q
I t
=
C
Py QC
Iz tw
C 57.14 MPa
In-plane Principal Stress:
x C y 0 xy C
1
x y
2
x y
2
2
xy
2
1 54.58 MPa Ans
2
x y
2
x y
2
2
xy
2
2 59.82 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
x y
2
2
xy
2
max 57.20 MPa Ans
Problem 9-39
The wide-flange beam is subjected to the 50-kN force. Determine the principal stresses in the beam at
point A located on the web at the bottom of the upper flange. Although it is not very accurate, use the
shear formula to calculate the shear stress.
Given: bf 200mm tf 12mm
tw 10mm dw 250mm
P 50kN a 3m
Solution:
Internal Force and Moment : At Section A-B:
+ Fy=0; V P 0= V P
+ O=0; M P a( ) 0= M P a
Section Property : D dw 2tf
A bf D bf tw dw A 7300 mm
2
I
1
12
bf D
3 bf tw dw
3 I 95.45 10 6 m4
QA bf tf
D
2
tf
2
QA 314400 mm
3
Normal Stress: M c
I
=
cA 0.5dw A
M cA
I A
196.435 MPa
Shear Stress : V Q
I t
=
A
V QA
I tw
A 16.47 MPa
In-plane Principal Stress:
x A y 0 xy A
1
x y
2
x y
2
2
xy
2
1 197.81 MPa Ans
2
x y
2
x y
2
2
xy
2
2 1.37 MPa Ans
Problem 9-40
Solve Prob. 9-39 for point B located on the web at the top of the bottom flange.
Given: bf 200mm tf 12mm
tw 10mm dw 250mm
P 50kN a 3m
Solution:
Internal Force and Moment : At Section A-B:
+ Fy=0; V P 0= V P
+ O=0; M P a( ) 0= M P a
Section Property : D dw 2tf
A bf D bf tw dw A 7300 mm
2
I
1
12
bf D
3 bf tw dw
3 I 95.45 10 6 m4
QB bf tf
D
2
tf
2
QB 314400 mm
3
Normal Stress: M c
I
=
cB 0.5dw B
M cB
I B
196.435 MPa
Shear Stress : V Q
I t
=
B
V QB
I tw
B 16.47 MPa
In-plane Principal Stress:
x B y 0 xy B
1
x y
2
x y
2
2
xy
2
1 1.37 MPa Ans
2
x y
2
x y
2
2
xy
2
2 197.81 MPa Ans
Problem 9-41
The bolt is fixed to its support at C. If a force of 90 N is applied to the wrench to tighten it, determine
the principal stresses developed in the bolt shank at point A. Represent the results on an element located
at this point. The shank has a diameter of 6 mm.
Given: do 6mm a 150mm L 50mm
P 90N
Solution:
Internal Force and Moment : At section AB:
Mx P L
Ty P a
Section Property :
I
64
do
4 J
32
do
4 cA 0.5do
Normal Stress: A
Mx cA
I A
212.21 MPa
Shear Stress : A
Ty cA
J A
318.31 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
x A y 0 xy A
1
x y
2
x y
2
2
xy
2
1 441.63MPa Ans
2
x y
2
x y
2
2
xy
2
2 229.42 MPa Ans
Orientation of Principal Stress:
tan 2 p
2 xy
x y
= p
1
2
atan
2 xy
x y
'p p 90deg
p 35.783 deg 'p 54.217 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'
x y
2
x y
2
cos 2 p xy sin 2 p
x' 441.63 MPa
Therefore, p1 p p1 35.78 deg Ans
p2 'p p2 54.22 deg Ans
Problem 9-42
Solve Prob. 9-41 for point B.
Given: do 6mm a 150mm L 50mm
P 90N
Solution:
Internal Force and Moment : At section AB:
Mx P L
Ty P a
Vz P
Section Property : ro 0.5do
I
64
do
4 J
32
do
4
QB
4ro
3
ro
2
2
Normal Stress: cB 0 B
Mx cB
I B
0 MPa
Shear Stress : bB do cB ro
B
Vz QB
I bB
Ty cB
J B 314.07 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
x B y 0 xy B
1
x y
2
x y
2
2
xy
2
1 314.07 MPa Ans
2
x y
2
x y
2
2
xy
2
2 314.07 MPa Ans
Orientation of Principal Stress:
)2tan( Ptan 2 p
2 xy
x y
= p1 45deg Ans
p2 p1 90deg p2 45 deg Ans
Problem 9-43
The beam has a rectangular cross section and is subjected to the loadings shown. Determine the
principal stresses that are developed at point A and point B, which are located just to the left of the
20-kN load. Show the results on elements located at these points.
Given: b 100mm d 200mm
F 20kN P 10kN
L 4m
Solution:
Support Reactions : By symmetry, R1=R ; R2= R
+ Fy=0; 2R F 0= R 0.5F
+ Fx=0; H1 P 0= H1 P
Internal Force and Moment : At Section A-B:
+ Fx=0; H1 N 0= N H1
+ Fy=0; R V 0= V R
+ O=0; M R 0.5L( ) 0= M 0.5R L
Section Property :
A b d I
1
12
b d3
QA 0 (since A' = 0) QB b 0.5d( ) 0.25d( )
Normal Stress: N
A
M c
I
=
cA 0.5d A
N
A
M cA
I A
30.5 MPa
cB 0 B
N
A
M cB
I B
0.5 MPa
Shear Stress :
Since QA = 0, A 0
B
V QB
I b B
0.75 MPa
In-plane Principal Stress: Applying Eq. 9-5
At A: xA A yA 0 xy 0
Since no shear stress acts upon the element,
A1 yA A1 0 MPa Ans
A2 xA A2 30.50 MPa Ans
At B: xB B yB 0 xy B
B1
xB yB
2
xB yB
2
2
B
2
B1 0.541 MPa Ans
B2
xB yB
2
xB yB
2
2
B
2
B2 1.041 MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2 p
2 B
xB yB
= p
1
2
atan
2 B
xB yB
p 35.783 deg
'p p 90deg 'p 54.217 deg
Use Eq. 9-1 to determine the principal plane of 1 and 2.
x'_B
xB yB
2
xB yB
2
cos 2 p B sin 2 p
x'_B 1.04 MPa
Therefore, p1 'p p1 54.22 deg Ans
p2 p p2 35.78 deg Ans
Problem 9-44
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at = 15
rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.
If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located on
the surface of the shaft.
Given: do 250mm L 0.75m
F 1230kN P 900kW 15
rad
s
Solution:
Internal Force and Moment : As shown on FBD
To
P
To 60.00 kN m
N F
Section Property :
A
do
2
4
A 49087.39 mm2
J
do
4
32
J 383495196.97 mm4
Normal Stress:
a
N
A a
25.06 MPa
Shear Stress : cmax 0.5 do
o
To cmax
J o
19.56 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
x a y 0 xy o
1
x y
2
x y
2
2
xy
2
1 10.70 MPa Ans
2
x y
2
x y
2
2
xy
2
2 35.75 MPa Ans
Problem 9-45
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at = 15
rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.
If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any point
located on the surface of the shaft.
Given: do 250mm L 0.75m
F 1230kN P 900kW 15
rad
s
Solution:
Internal Force and Moment : As shown on FBD
To
P
To 60.00 kN m
N F
Section Property :
A
do
2
4
A 49087.39 mm2
J
do
4
32
J 383495196.97 mm4
Normal Stress:
a
N
A a
25.06 MPa
Shear Stress : cmax 0.5 do
o
To cmax
J o
19.56 MPa
Maximum In-plane Shear Stress: Applying Eq. 9-7
x a y 0 xy o
max
x y
2
2
xy
2
max 23.2 MPa Ans
L 250mmGiven: do 75mm di 68mm
At section AB:
P 100N a 300mm
Solution:
Internal Force and Moment :
Ans
Tx P a
My P L
Vz P
Section Property : ro 0.5do ri 0.5di
I
64
do
4 di
4 J
32
do
4 di
4
QAz
4ro
3
ro
2
2
4ri
3
ri
2
2
Normal Stress: cA 0 A
My cA
I A
0 MPa
Shear Stress : bA do di cA ro
A
Vz QAz
I bA
Tx cA
J A 0.863 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
x A z 0 xz A
1
x z
2
x z
2
2
xz
2
1 0.863 MPa Ans
2
x z
2
x z
2
2
xz
2
2 0.863 MPa
Problem 9-46
The steel pipe has an inner diameter of 68mm and an outer diameter of 75 mm. If it is fixed at C and
subjected to the horizontal 100-N force acting on the handle of the pipe wrench at its end, determine
the principal stresses in the pipe at point A which is located on the surface of the pipe.
L 250mmGiven: do 75mm di 68mm
At section AB:
P 100N a 300mm
Solution:
Internal Force and Moment :
Ans
Tx P a
My P L
Vz P
Section Property : ro 0.5do ri 0.5di
I
64
do
4 di
4 J
32
do
4 di
4
QBz 0 (Since A'=0)
Normal Stress: cB ro B
My cB
I
B 1.862 MPa
Shear Stress : cB ro B
Tx cB
J
B 1.117 MPa
In-plane Principal Stresses: Applying Eq. 9-5,
x B z 0 xz B
1
x z
2
x z
2
2
xz
2
1 2.385 MPa Ans
2
x z
2
x z
2
2
xz
2
2 0.523 MPa
Problem 9-47
Solve Prob. 9-46 for point B, which is located on the surface of the pipe.
Problem 9-48
The cantilevered beam is subjected to the load at its end. Determine the principal stresses in the beam at
points A and B.
Given: b 120mm h 150mm P 15kN L 1.2m
yA 45mm zA 60mm y
4
5
z
3
5yB 75mm zB 20mm
Solution:
Internal Force and Moment : At Section A-B:
+ Fz0; Vz P z 0= Vz P z
+ Fy=0; Vy P y 0= Vy P y
Mz P y L
My P z L
Section Property :
Iz
1
12
b h3 Iz 33.75 10
6 m4
Iy
1
12
h b3 Iy 21.60 10
6 m4
QA.y b 0.5h yA yA 0.5 0.5h yA QA.y 216000 mm
3
QB.z h 0.5b zB zB 0.5 0.5b zB QB.z 240000 mm
3
QA.z 0 (since A' = 0)
QB.y 0 (since A' = 0)
Normal Stress:
A
Mz yA
Iz
My zA
Iy
A 10.8 MPa
B
Mz yB
Iz
My zB
Iy
B 42.0 MPa
Shear Stress : V Q
I t
=
A
Vy QA.y
Iz b
A 0.640 MPa
B
Vz QB.z
Iy h
B 0.667 MPa
In-plane Principal Stress: Applying Eq. 9-5
At A: xA A yA 0 xy 0
A1
xA yA
2
xA yA
2
2
A
2
A1 0.0378 MPa Ans
A2
xA yA
2
xA yA
2
2
A
2
A2 10.84 MPa Ans
At B: xB B zB 0 xz B
B1
xB zB
2
xB zB
2
2
B
2
B1 42.01 MPa Ans
B2
xB zB
2
xB zB
2
2
B
2
B2 0.0106 MPa Ans
Problem 9-49
The box beam is subjected to the loading shown. Determine the principal stresses in the beam at points
A and B.
Given: bo 200mm bi 150mm
do 200mm di 150mm
L1 0.9m L2 1.5m
P1 4kN P2 6kN
Solution: L L1 2L2
Support Reactions : Given
+ Fy=0; R1 R2 P1 P2 0=
+ R2=0;
Ans
Guess R1 1kN R2 1kN
R1
R2
Find R1 R2
R1
R2
8.2
1.8
kN
Internal Force and Moment : At section A-B:
N 0
V P1 R1
M P1 L1 0.5L2 R1 0.5L2
Section Property :
I
1
12
bo do
3 bi di
3
QA 0 QB 0 (Since A'=0)
For point A: A 0
cA 0.5 do A
M cA
I A
0.494 MPa
1 A 1 0.494 MPa Ans
2 0 2 0.000 MPa Ans
For point B: B 0
cB 0.5 di B
M cB
I B
0.37 MPa
1 0 1 0.000 MPa Ans
2 B 2 0.370 MPa
P1 L R1 L L1 P2 L2 0=
Problem 9-50
A bar has a circular cross section with a diameter of 25 mm. It is subjected to a torque and a bending
moment. At the point of maximum bending stress the principal stresses are 140MPa and -70 MPa.
Determine the torque and the bending moment.
Given: do 6mm
1 140MPa 2 70MPa
Solution:
In-plane Principal Stresses: Applying Eq. 9-5,
Given y 0
1
x y
2
x y
2
2
xy
2= (1)
2
x y
2
x y
2
2
xy
2= (2)
Solving Eqs. (1) and (2):
Guess x 2MPa xy 1MPa
x
xy
Find x xy
x
xy
70.00
98.99
MPa
Section Property : ro 0.5do
I
64
do
4 J
32
do
4
Normal Stress: M c
I
=
c ro M
x I
c
M 1.484 N m Ans
Shear Stress: T c
J
=
c ro T
xy J
c
T 4.199 N m Ans
Problem 9-51
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point A.
Also calculate the maximum in-plane shear stress at this point.
Unit Used: kPa 1000Pa
Given: b 100mm h 200mm Px 0.5kN
yA 100mm zA 0mm Py 0.8kN
My 0.04kN m Mz 0.03kN m
Solution:
Section Property :
A b h A 20000 mm2
Iz
1
12
b h3 Iz 66.67 10
6 m4
Iy
1
12
h b3 Iy 16.67 10
6 m4
QA.y 0 (since A' = 0)
Normal Stress:
A
Px
A
Mz yA
Iz
My zA
Iy
A 20.0 kPa
Shear Stress : Since QA = 0, A 0
In-plane Principal Stress:
x A y 0 xy 0
Since no shear stress acts upon the element,
1 y 1 0 kPa Ans
2 x 2 20 kPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
x y
2
2
xy
2
max 10 kPa Ans
Problem 9-52
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point B.
Also calculate the maximum in-plane shear stress at this point.
Unit Used: kPa 1000Pa
Given: b 100mm h 200mm Px 0.5kN
yB 0mm zB 50mm Py 0.8kN
My 0.04kN m Mz 0.03kN m
Solution:
Section Property :
A b h A 20000 mm2
Iz
1
12
b h3 Iz 66.67 10
6 m4
Iy
1
12
h b3 Iy 16.67 10
6 m4
QB.y b 0.5h( ) 0.25h( )
Normal Stress:
B
Px
A
Mz yB
Iz
My zB
Iy
B 145.0 kPa
Shear Stress :
B
Py QB.y
Iz b
B 60.00 kPa
In-plane Principal Stress:
xB B yB 0 xy B
B1
xB yB
2
xB yB
2
2
B
2
B1 166.6 kPa Ans
B2
xB yB
2
xB yB
2
2
B
2
B2 21.61 kPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
xB yB
2
2
xy
2
max 94.11 kPa Ans
Problem 9-53
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point C.
Also calculate the maximum in-plane shear stress at this point.
Unit Used: kPa 1000Pa
Given: b 100mm h 200mm Px 0.5kN
yC 50mm zC 0mm Py 0.8kN
My 0.04kN m Mz 0.03kN m
Solution:
Section Property :
A b h A 20000 mm2
Iz
1
12
b h3 Iz 66.67 10
6 m4
Iy
1
12
h b3 Iy 16.67 10
6 m4
QC.y b 0.5h yC yC 0.5 0.5h yC QC.y 375000 mm
3
Normal Stress:
C
Px
A
Mz yC
Iz
My zC
Iy
C 47.5 kPa
Shear Stress :
C
Py QC.y
Iz b
C 45.00 kPa
In-plane Principal Stress:
xC C yC 0 xy C
C1
xC yC
2
xC yC
2
2
C
2
C1 74.63 kPa Ans
C2
xC yC
2
xC yC
2
2
C
2
C2 27.13 kPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
max
xC yC
2
2
xy
2
max 50.88 kPa Ans
Problem 9-54
The beam has a rectangular cross section and is subjected to the loads shown. Write a computer
program that can be used to determine the principal stresses at points A, B, C, and D. Show an
application of the program using the values h = 300 mm, b = 200 mm, Nx = 2 kN, Vy = 1.5 kN, Vz = 0,
My = 0, and Mz = -225 kN·m.
Problem 9-55
The member has a rectangular cross section and is subjected to the loading shown. Write a computer
program that can be used to determine the principal stresses at points A, B, and C. Show an application
of the program using the values b = 150 mm, h = 200 mm, P = 1.5 kN, x = 75 mm, z = -50 mm, Vx =
300 N, and Vz = 600 N.
Problem 9-56
Solve Prob. 9-4 using Mohr's circle.
Given: x 0.650MPa y 0.400MPa
' 60deg xy 0MPa
Solution: 90deg ' 30.00 deg
Center :
c
x y
2 c
0.125 MPa
Radius :
R x c R 0.525 MPa
Coordinates:
A x 0 B y 0 C c 0
Stresses: 
x' c R cos 2 x' 0.387 MPa Ans
x'y' R sin 2 x'y' 0.455 MPa Ans
Problem 9-57
Solve Prob. 9-2 using Mohr's circle.
Given: x 5MPa y 3MPa xy 8MPa ' 40deg
Solution:
Center :
c
x y
2 c
4 MPa
Radius :
R x c
2
xy
2 R 8.062 MPa
Angles: 90deg ' 130 deg
atan
xy
x c
82.875 deg
180deg 2 2.875 deg
Stresses: 
x' c R cos x' 4.052 MPa Ans
x'y' R sin x'y' 0.404 MPa Ans
Problem 9-58
Solve Prob. 9-3 using Mohr's circle.
Given: x 0.350MPa y 0.200MPa
' 50deg xy 0MPa
Solution: 90deg ' 140 deg
Center :
c
x y
2 c
0.075 MPa
Radius :
R x c
2
xy
2 R 0.275 MPa
Coordinates:
A x 0 C c 0
Angles:
360deg 2 80 deg
Stresses: (represented by coordinates of point P)
x' c R cos x' 0.123 MPa Ans
x'y' R sin x'y' 0.271 MPa Ans
Problem 9-59
Solve Prob. 9-10 using Mohr's circle.
Given: x 0MPa y 0.300MPa
30deg xy 0.950MPa
Solution:
Center :
c
x y
2 c
0.15 MPa
Radius :
R x c
2
xy
2 R 0.962 MPa
Angles:
atan
xy
x c
81.027 deg
2 21.027 deg
Stresses: 
x' c R cos x' 0.748 MPa Ans
y' c R cos y' 1.048 MPa Ans
x'y' R sin x'y' 0.345 MPa Ans
Problem 9-60
Solve Prob. 9-6 using Mohr's circle.
Given: x 90MPa y 50MPa xy 35MPa
' 60deg
Solution:
Center :
c
x y
2 c
70 MPa
Radius :
R x c
2
xy
2 R 40.311 MPa
Angles: 90deg '
atan
xy
x c
60.255 deg
180deg 2 59.745 deg
Stresses: 
x' c R cos x' 49.689 MPa Ans
x'y' R sin x'y' 34.821 MPa Ans
Problem 9-61
Solve Prob. 9-11 using Mohr's circle.
Given: x 0.300MPa y 0MPa
60deg xy 0.120MPa
Solution:
Center :
c
x y
2 c
0.15 MPa
Radius :
R x c
2
xy
2 R 0.1921 MPa
Coordinates:
A x xy C c 0
Angles:
atan
xy
x c
38.660 deg
180deg 2 21.340 deg
Stresses: (represented by coordinates of points P and Q)
x' c R cos x' 0.0289 MPa Ans
y' c R cos y' 0.329 MPa Ans
x'y' R sin x'y' 0.0699 MPa Ans
Problem 9-62
Solve Prob. 9-13 using Mohr's circle.
Given: x 45MPa y 60MPa xy 30MPa
Solution:
Center :
c
x y
2 c
7.5 MPa
Radius :
R x c
2
xy
2 R 60.467 MPa
Coordinates:
A x xy C c 0
Stresses: 
1 c R 1 52.97 MPa Ans
2 c R 2 67.97 MPa Ans
max R max 60.47 MPa Ans
avg c avg 7.5 MPa Ans
Angles:
p1
1
2
atan
xy
x c
p1 14.872 deg (Counter-clockwise) Ans
2 s1 90deg 2 p1=
s1 45deg p1
s1 30.128 deg (Clockwise) Ans
Problem 9-63
Solve Prob. 9-14 using Mohr's circle.
Given: x 180MPa y 0MPa xy 150MPa
Solution:
Center :
c
x y
2 c
90 MPa
Radius :
R x c
2
xy
2 R 174.929 MPa
Coordinates:
A x xy B y xy C c 0
Stresses: 
1 c R 1 264.93 MPa Ans
2 c R 2 84.93 MPa Ans
max R max 174.93 MPa Ans
avg c avg 90 MPa Ans
Angles:
p
1
2
atan
xy
x c
p 29.518 deg (Clockwise) Ans
2 s 90deg 2 p=
s 45deg p
s 15.482 deg (Counter-clockwise) Ans
Problem 9-64
Solve Prob. 9-16 using Mohr's circle.
Given: x 200MPa y 250MPa xy 175MPa
Solution:
Center :
c
x y
2 c
25 MPa
Radius :
R x c
2
xy
2 R 285.044 MPa
Coordinates:
A x xy B y xy C c 0
Stresses: 
1 c R 1 310.04 MPa Ans
2 c R 2 260.04 MPa Ans
max R max 285.04 MPa Ans
avg c avg 25 MPa Ans
Angles:
p
1
2
atan
xy
x c
p 18.937 deg (Clockwise) Ans
2 s 90deg 2 p=
s 45deg p
s 26.063 deg (Counter-clockwise) Ans
Problem 9-65
Solve Prob. 9-15 using Mohr's circle.
Given: x 30MPa y 0MPa xy 12MPa
Solution:
Center :
c
x y
2 c
15 MPa
Radius :
R x c
2
xy
2 R 19.209MPa
Coordinates:
A x xy C c 0
Stresses: 
1 c R 1 4.21 MPa Ans
2 c R 2 34.21 MPa Ans
max R max 19.21 MPa Ans
avg c avg 15 MPa Ans
Angles:
p2
1
2
atan
xy
x c
p2 19.330 deg (Countr-clockwise) Ans
2 s2 2 p2 90deg=
s2 p2 45deg
s2 64.330 deg (Countr-clockwise) Ans
Problem 9-66
Determine the equivalent state of stress if an element is oriented 20° clockwise from the element
shown. Show the result on the element.
Given: x 3MPa y 2MPa
20deg xy 4MPa
Solution:
Center :
c
x y
2 c
0.5 MPa
Radius :
R x c
2
xy
2 R 4.717 MPa
Coordinates:
A x xy C c 0
Angles:
atan
xy
x c
57.995 deg
2 17.995 deg
Stresses: (represented by coordinates of points P and Q)
x' c R cos x' 4.986 MPa Ans
y' c R cos y' 3.986 MPa Ans
x'y' R sin x'y' 1.457 MPa Ans
Problem 9-67
Determine the equivalent state of stress if an element is oriented 60° counterclockwise from the
element shown.
Given: x 0.750MPa y 0.800MPa
60deg xy 0.450MPa
Solution:
Center :
c
x y
2 c
0.025 MPa
Radius :
R x c
2
xy
2 R 0.8962 MPa
Coordinates:
A x xy B y xy C c 0
Angles:
atan
xy
x c
30.141 deg
2 89.859 deg
Stresses: 
x' c R cos x' 0.0228 MPa Ans
y' c R cos y' 0.0272 MPa Ans
x'y' R sin x'y' 0.896 MPa Ans
Problem 9-68
Determine the equivalent state of stress if an element is oriented 30° clockwise from the element
shown.
Given: x 350MPa y 230MPa
30deg xy 480MPa
Solution:
Center :
c
x y
2 c
290 MPa
Radius :
R x c
2
xy
2 R 483.7355 MPa
Coordinates:
A x xy B y xy C c 0
Angles:
atan
xy
x c
82.875 deg
2 22.875 deg
Stresses: 
x' c R cos x' 735.6922 MPa Ans
y' c R cos y' 155.6922 MPa Ans
x'y' R sin x'y' 188.038 MPa Ans
Problem 9-69
Determine the equivalent state of stress if an element is oriented 30° clockwise from the element
shown. Show the result on the element.
Given: x 7MPa y 8MPa
30deg xy 15MPa
Solution:
Center :
c
x y
2 c
7.5 MPa
Radius :
R x c
2
xy
2 R 15.0083 MPa
Coordinates:
A x xy C c 0
Angles:
atan
xy
x c
88.091 deg
180deg 2 31.909 deg
Stresses: (represented by coordinates of points P and Q)
x' c R cos x' 5.240 MPa Ans
y' c R cos y' 20.240 MPa Ans
x'y' R sin x'y' 7.933 MPa Ans
Problem 9-70
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: x 350MPa y 200MPa xy 500MPa
Solution:
Center :
c
x y
2 c
75 MPa
Radius :
R x c
2
xy
2 R 570.636 MPa
Coordinates:
A x xy C c 0
a) Principal Stresses:
1 c R 1 645.64 MPa Ans
2 c R 2 495.64 MPa Ans
Angles:
p1
1
2
atan
xy
x c
p1 30.595 deg (Counter-clockwise) Ans
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
max R max 570.64 MPa Ans
avg c avg 75 MPa Ans
2 s 90deg 2 p1=
s 45deg p1
s 14.405 deg (Clockwise) Ans
Problem 9-71
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: x 10MPa y 80MPa xy 60MPa
Solution:
Center :
c
x y
2 c
45 MPa
Radius :
R x c
2
xy
2 R 69.462 MPa
Coordinates:
A x xy C c 0
a) Principal Stresses:
1 c R 1 114.46 MPa Ans
2 c R 2 24.46 MPa Ans
Angles:
p2
1
2
atan
xy
x c
p1 90deg p2
p1 60.128 deg (Clockwise) Ans
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
max R max 69.46 MPa Ans
avg c avg 45 MPa Ans
2 s2 90deg 2 p2=
s2 45deg p2
s2 15.128 deg (Clockwise) Ans
Problem 9-72
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: x 0MPa y 50MPa xy 30MPa
Solution:
Center :
c
x y
2 c
25 MPa
Radius :
R x c
2
xy
2 R 39.051 MPa
Coordinates:
A x xy B y xy C c 0
a) Principal Stresses:
1 c R 1 64.05 MPa Ans
2 c R 2 14.05 MPa Ans
Angles:
p
1
2
atan
xy
x c
p 25.097 deg (Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
max R max 39.05 MPa Ans
avg c avg 25 MPa Ans
2 s2 90deg 2 p=
s2 45deg p
s2 19.903 deg (Counter-clockwise)
Problem 9-73
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: x 12MPa y 8MPa xy 4MPa
Solution:
Center :
c
x y
2 c
10 MPa
Radius :
R x c
2
xy
2 R 4.472 MPa
Coordinates:
A x xy B y xy C c 0
a) Principal Stresses:
1 c R 1 5.53 MPa Ans
2 c R 2 14.47 MPa Ans
Angles:
p
1
2
atan
xy
x c
p 31.717 deg (Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
max R max 4.47 MPa Ans
avg c avg 10 MPa Ans
2 s2 90deg 2 p=
s2 45deg p
s2 13.283 deg (Counter-clockwise)
Problem 9-74
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: x 45MPa y 30MPa xy 50MPa
Solution:
Center :
c
x y
2 c
37.5 MPa
Radius :
R x c
2
xy
2 R 50.559 MPa
Coordinates:
A x xy B y xy C c 0
a) Principal Stresses:
1 c R 1 88.06 MPa Ans
2 c R 2 13.06 MPa Ans
Angles:
p
1
2
atan
xy
x c
p 40.735 deg (Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
max R max 50.56 MPa Ans
avg c avg 37.5 MPa Ans
2 s2 90deg 2 p=
s2 45deg p
s2 4.265 deg (Counter-clockwise)
Problem 9-75
The square steel plate has a thickness of 12 mm and is subjected to the edge loading shown. Determine
the principal stresses developed in the steel.
Given: x 0MPa y 0MPa qxy 3.2
kN
m
L 100mm t 12mm
Solution:
xy
qxy
t xy
0.267 MPa
Center :
c
x y
2 c
0 MPa
Radius :
R x c
2
xy
2 R 0.267 MPa
Coordinates:
A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 0.267 MPa Ans
2 c R 2 0.267 MPa Ans
1 115.60 MPa
Given: x 105MPa y 0MPa xy 35MPa
Solution:
Center :
c
x y
2 c
52.5 MPa
Radius :
R x c
2
xy
2 R 63.097 MPa
Coordinates:
A x xy C c 0
a) In-planePrincipal Stresses:
The coordinates of points B and D represent 1 and , respectively.
1 c R
Ans
Ans
2 c R 2 10.60 MPa Ans
Angles:
p
1
2
atan
xy
x c
p 16.845 deg (Clockwise)
b) Maximum In-plane Shear Stresses:
Represented by the coordinates of point E on the circle.
max R max 63.10 MPa Ans
2 s 90deg 2 p=
s 45deg p
s 28.155 deg (Counter-clockwise)
Problem 9-76
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Problem 9-77
Draw Mohr's circle that describes each of the following states of stress.
Given: x 30MPa y 30MPa
xy 0MPa
Solution:
Cases (a) and (b) are the same.
Center :
c
x y
2 c
0 MPa
Radius :
R x c
2
xy
2 R 30 MPa
Coordinates:
A x xy C c 0
Problem 9-78
Draw Mohr's circle that describes each of the following states of stress.
a) Given: x 0.8MPa xy 0MPa
y 0.6MPa
Solution:
Center : c
x y
2
c 0.1 MPa
Radius : R x c
2
xy
2
R 0.7 MPa
Coordinates: A x xy C c 0
b) Given: x 0MPa xy 0MPa
y 2MPa
Solution:
Center : c
x y
2
c 1 MPa
Radius : R x c
2
xy
2
R 1 MPa
Coordinates: A x xy C c 0
c) Given: x 0MPa xy 20MPa
y 0MPa
Solution:
Center : c
x y
2
c 0 MPa
Radius : R x c
2
xy
2
R 20 MPa
Coordinates: A x xy C c 0Problem 9-79
A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting
state of stress with reference to an element oriented as shown on the bottom.
Unit used: kPa 1000Pa
Given: x_a 50kPa y_a 0 xy_a 0
y_b 18kPa x_b 0 xy_b 45kPa
a 30deg b 50deg
Solution:
For element a:
Center : c_a
x_a y_a
2 c_a
25 kPa
Radius : Ra x_a c_a
2
xy_a
2 Ra 25 kPa
Coordinates: A x_a xy_a B x_a xy_a C c_a 0
'x_a c_a Ra cos 2 a
'y_a c_a Ra cos 2 a
'xy_a Ra sin 2 a
For element b: b 90deg b
Center : c_b
x_b y_b
2 c_b
9 kPa
Radius : Rb x_b c_b
2
xy_b
2 Rb 45.891 kPa
Coordinates: A x_b xy_b B x_a xy_b C c_b 0
Angles : 
atan
xy_b
x_b c_b
78.69 deg
b 2 b
b 1.310 deg
'x_b c_b Rb cos b
'y_b c_b Rb cos b
'xy_b Rb sin b
Resultants:
'x 'x_a 'x_b 'x 74.38 kPa Ans
'y 'y_a 'y_b 'y 42.38 kPa Ans
'xy 'xy_a 'xy_b 'xy 22.70 kPa Ans
Problem 9-80
Mohr's circle for the state of stress in Fig. 9-15a is shown in Fig. 9-15b. Show that finding the
coordinates of point P( x' , x'y' ) on the circle gives the same value as the stress-transformation Eqs.
9-1 and 9-2.
Problem 9-81
The cantilevered rectangular bar is subjected to the force of 25 kN. Determine the principal stresses at
point A.
Given: b 80mm d 160mm cA 40mm
P 25kN L 0.4m bA 40mm
rv
3
5
rh
4
5Solution:
Internal Force and Moment : At Section A-B:
+ Fx=0; P rh N 0= N P rh
+ Fy=0; P rv V 0= V P rv
+ O=0; M P rv L 0= M P rv L
Section Property : dA 0.5d cA
A b d I
1
12
b d3
QA b dA 0.5dA cA
Normal Stress:
A
N
A
M cA
I A
10.352 MPa
Shear Stress :
A
V QA
I b A
1.318 MPa
Construction of Mohr's Circle :
x A y 0MPa xy A
Center : c
x y
2 c
5.176 MPa
Radius : R x c
2
xy
2
R 5.341 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 10.52 MPa Ans
2 c R 2 0.165 MPa Ans
Problem 9-82
Solve Prob. 9-81 for the principal stresses at point B.
Given: b 80mm d 160mm cB 25mm
P 25kN L 0.4m bB 25mm
rv
3
5
rh
4
5Solution:
Internal Force and Moment : At Section A-B:
+ Fx=0; P rh N 0= N P rh
+ Fy=0; P rv V 0= V P rv
+ O=0; M P rv L 0= M P rv L
Section Property : dB 0.5d cB
A b d I
1
12
b d3
QB b dB 0.5dB cB
Normal Stress:
B
N
A
M cB
I B
3.931 MPa
Shear Stress :
B
V QB
I b B
1.586 MPa
Construction of Mohr's Circle :
x B y 0MPa xy B
Center : c
x y
2 c
1.965 MPa
Radius : R x c
2
xy
2
R 2.526 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 0.560 MPa Ans
2 c R 2 4.491 MPa Ans
Problem 9-83
The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at
B. If a man having a weight of 1500 N (~150 kg) stands in the center of the tread, determine the
principal stresses developed in the supporting truck on the cross section at point C. The stairs move at
constant velocity.
Given: b 12mm d 50mm cC 0mm
W 1.5kN hC 150mm 30deg
vA 450mm hA 375mm hB 150mm
Solution:
Support Reactions :
+ A=0; W hA By hB 0= By
W hA
hB
Bx By tan
Internal Force and Moment : At Section C:
+ Fx=0; Bx V 0= V Bx
+ Fy=0; By N 0= N By
+ O=0; Bx hC M 0= M Bx hC
Section Property : dC 0.5d
A b d I
1
12
b d3 QC b dC 0.5dC
Normal Stress:
C
N
A
M cC
I C
6.250 MPa
Shear Stress :
C
V QC
I b C
5.413 MPa
Construction of Mohr's Circle :
x 0MPa y C xy C
Center : c
x y
2 c
3.125 MPa
Radius : R x c
2
xy
2 R 6.25 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 3.125 MPa Ans
2 c R 2 9.375 MPa Ans
Problem 9-84
The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not
rotate while subjected to a force of 400 N, determine the principal stresses in the material on the cross
section at point C.
Given: b 7.5mm d 20mm cC 5mm
P 0.4kN L 100mm
Solution:
Internal Force and Moment : At Section C:
+ Fy=0; V P 0= V P
+ O=0; M P L 0= M P L
Section Property : dC 0.5d cC
A b d I
1
12
b d3
QC b dC cC 0.5dC
Normal Stress:
C
M cC
I C
40.000 MPa
Shear Stress :
C
V QC
I b C
3.000 MPa
Construction of Mohr's Circle :
x C y 0MPa xy C
Center : c
x y
2 c
20 MPa
Radius : R x c
2
xy
2 R 20.224 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 40.224 MPa Ans
2 c R 2 0.224 MPa Ans
Problem 9-85
The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses at
point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an
angle of 30° with the horizontal as shown.
Unit Used: kPa 1000Pa
Given: b 100mm a 1m w 0.2
kN
mh 200mm L 2.5m
hD 75mm 60deg
Solution:
Support Reactions: Due to symmetry, B=C=R
+ Fy=0; 2R w L 0= R 0.5w L
Internal Force and Moment : At Section D:
+ Fy=0; V R w a 0=
V R w a
+ O=0; M R a w a 0.5a( ) 0=
M R a w a 0.5a( )
Section Property : cD 0.5h hD
A b h I
1
12
b h3
QD b hD cD 0.5hD
Normal Stress: D
M cD
I D
56.25 kPa
Shear Stress : D
V QD
I b D
3.516 kPa
Construction of Mohr's Circle :
x D y 0MPa xy D
Center : c
x y
2 c
28.125 kPa
Radius : R x c
2
xy
2 R 28.344 kPa
Coordinates: A x xy C c 0
Angles:
atan
xy
x c
7.125 deg
180deg 2 52.875 deg
Stresses on the Rotated Element: (represented by coordinates of point P)
x' c R cos x' 11.02 kPa Ans
x'y' R sin x'y' 22.60 kPa Ans
Problem 9-86
The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses at
point E that act perpendicular and parallel, respectively, to the grains. The grains at this point make an
angle of 60° with the horizontal as shown.
Unit Used: kPa 1000Pa
Given: b 50mm w 0.2
kN
mh 100mm
L 2.5m 60deg
Solution:
Support Reactions: Due to symmetry, B=C=R
+ Fy=0; 2R w L 0= R 0.5w L
Internal Force: At Section E:
+ Fy=0; N R 0= N R
Section Property :
A b h A 5000 mm2
Normal Stress: E
N
A E
50 kPa
Shear Stress : D 0
Construction of Mohr's Circle :
x E y 0MPa xy D
Center : c
x y
2 c
25 kPa
Radius : R x c
2
xy
2 R 25 kPa
Coordinates: A x xy C c 0
Angles:
atan
xy
x c
0.000 deg
180deg 2 60.000 deg
Stresses on the Rotated Element: (represented by coordinates of point P)
x' c R cos x' 12.50 kPa Ans
x'y' R sin x'y' 21.65 kPa Ans
Problem 9-87
The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal
stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the
results on elements located at these points.
Given: do 15mm P 0.6kN
a 50mm L 0.1m
Solution:
Internal Force: At Section AB:
N P N 0.600 kN
M P a M 0.030 kN m
Section Property :
A
4
do
2 A 176.71 mm2
I
64
do
4 I 0 m2 mm2
QA 0 (since A' = 0)
QB 0 (since A' = 0)
Normal Stress:
cA 0.5do A
N
A
M cA
I A
87.15 MPa
cB 0.5do B
N
A
M cB
I B
93.94 MPa
Shear Stress :
A 0 (since QA = 0)
B 0 (since QB = 0)
For A:
Construction of Mohr's Circle :
x A y 0MPa xy A
Center : c
x y
2 c
43.573 MPa
Radius : R x c
2
xy
2 R 43.573 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and A represent 1 and , respectively.
1 c R 1 0 MPa Ans
2 c R 2 87.15 MPa Ans
Maximum In-plane Shear Stresses: 
Represented by the coordinates of point E on the circle.max R max 43.57 MPa Ans
2 s 90deg= s 45deg (Counter-clockwise) Ans
For B:
Construction of Mohr's Circle :
x B y 0MPa xy B
Center : c
x y
2 c
46.968 MPa
Radius : R x c
2
xy
2 R 46.968 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points A and B represent 1 and , respectively.
1 c R 1 93.94 MPa Ans
2 c R 2 0 MPa Ans
Maximum In-plane Shear Stresses: 
Represented by the coordinates of point E on the circle.
max R max 46.97 MPa Ans
2 s 90deg= s 45deg (Clockwise) Ans
Problem 9-88
Draw the three Mohr's circles that describe each of the following states of stress.
a) max 6MPa int 0
min 0
b) max 50MPa int 0
min 40MPa
c) Unit used: kPa 1000Pa
max 600kPa int 200kPa
min 100kPa
d) max 0 int 7MPa
min 9MPa
e) max 30MPa int 30MPa
min 30MPa
Problem 9-89
Draw the three Mohr's circles that describe each of the following states of stress.
a) 1 15MPa 2 0
3 15MPa
max 15MPa
b) 1 65MPa 2 65MPa
3 65MPa
max 65MPa
Problem 9-90
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: x 100MPa y 90MPa z 80MPa
xy 0MPa yz 40MPa xz 0MPa
Solution:
Construction of Mohr's Circle in y-z Plane :
Center : c
y z
2 c
5 MPa
Radius : R y c
2
yz
2 R 93.941 MPa
Coordinates: A y yz C c 0
In-plane Principal Stresses: 
The coordinates of points A and B represent 1 and , respectively.
1 c R 1 98.941 MPa
2 c R 2 88.941 MPa
Construction of Three Circles :
From the results obtained above,
max 1 int 2 min x
max 98.94 MPa Ans
int 88.94 MPa Ans
min 100.00 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
abs.max
max min
2
abs.max 99.47 MPa Ans
Problem 9-91
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: x 0MPa y 7MPa z 0MPa
xy 0MPa yz 50MPa xz 5MPa
Solution:
Construction of Mohr's Circle in x-z Plane :
Center : c
x z
2 c
0 MPa
Radius : R x c
2
xz
2 R 5 MPa
Coordinates: A y xz C c 0
In-plane Principal Stresses: 
The coordinates of points A and B represent 1 and , respectively.
1 c R 1 5 MPa
2 c R 2 5 MPa
Construction of Three Circles :
From the results obtained above,
max y int 1 min 2
max 7 MPa Ans
int 5 MPa Ans
min 5 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2
max 6 MPa Ans
Problem 9-92
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: x 150MPa y 0MPa z 90MPa
xy 0MPa yz 80MPa xz 0MPa
Solution:
Construction of Mohr's Circle in y-z Plane :
Center : c
y z
2 c
45 MPa
Radius : R z c
2
yz
2 R 91.788 MPa
Coordinates: A y yz B z yz C c 0
In-plane Principal Stresses: 
The coordinates of points A and B represent 1 and , respectively.
1 c R 1 136.788 MPa
2 c R 2 46.788 MPa
Construction of Three Circles :
From the results obtained above,
max x int 1 min 2
max 150.00 MPa Ans
int 136.79 MPa Ans
min 46.79 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2
max 98.39 MPa Ans
Problem 9-93
The principal stresses acting at a point in a body are shown. Draw the three Mohr's circles that
describe this state of stress, and find the maximum in-plane shear stresses and associated average
normal stresses for the x-y, y-z, and x-z planes. For each case, show the results on the element oriented
in the appropriate direction.
Given: x 40MPa y 40MPa z 40MPa
xy 0MPa yz 0MPa xz 0MPa
Solution:
Construction of Three Circles :
max x int y min z
max 40.00 MPa
int 40.00 MPa
min 40.00 MPa
For x-y Plane :
avg
x y
2 avg
0 MPa Ans
max
x y
2 max
40 MPa Ans
For y-z Plane :
'avg
y z
2
'avg 40 MPa Ans
'max
y z
2
'max 0 MPa Ans
For x-y Plane :
''avg
x z
2
''avg 0 MPa Ans
''max
x z
2
''max 40 MPa Ans
Problem 9-94
Consider the general case of plane stress as shown. Write a computer program that will show a plot of
the three Mohr's circles for the element, and will also calculate the maximum in-plane shear stress and
the absolute maximum shear stress.
Problem 9-95
The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the
principal stresses acting at points A and B and the absolute maximum shear stress.
Given: ro 25mm L 450mm
P 0.8kN To 0.045kN m Mo 0.3kN m
Solution: ro
Internal Force and Moment : At Section AB:
Vy P Tx To Mz Mo P L
Section Property :
J
2
ro
4
A ro
2 Iz 4
ro
4
QA 0 (Since A'=0) QB
4 ro
3
0.5A( )
Normal Stress: cA ro cB 0
A
Mz cA
Iz
A 4.889 MPa
B
Mz cB
Iz
B 0 MPa
Shear Stress : bB 2 ro
A
Tx
J A 1.833 MPa
B
Vy QB
Iz bB
Tx
J B
1.290 MPa
For Point A:
Construction of Mohr's Circle:
x A z 0MPa xz A
Center : c
x z
2 c
2.445 MPa
Radius : R z c
2
xz
2 R 3.056 MPa
Coordinates: A z xz C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 5.500 MPa
2 c R 2 0.611 MPa
Three Mohr's Circles: y 0
From the results obtained above,
max 1 int y min 2
max 5.50 MPa Ans
int 0.00 MPa Ans
min 0.611 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2 max
3.06 MPa Ans
For Point B:
Construction of Mohr's Circle:
x B z 0MPa xz B
Center : c
x z
2 c
0 MPa
Radius : R z c
2
xz
2 R 1.290 MPa
Coordinates: A z xz C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 1.290 MPa
2 c R 2 1.290 MPa
Three Mohr's Circles: y 0
From the results obtained above,
max 1 int y min 2
max 1.29 MPa Ans
int 0.00 MPa Ans
min 1.29 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2 max
1.29 MPa Ans
Problem 9-96
The bolt is fixed to its support at C. If a force of 90 kN is applied to the wrench to tighten it, determine
the principal stresses and the absolute maximum shear stress developed in the bolt shank at point A.
Represent the results on an element located at this point. The shank has a diameter of 6 mm.
Given: do 6mm a 50mm
P 90N L 150mm
Solution: 0.5do
Internal Force and Moment : At Section AB:
Vy P Mz P a Tx P L
Section Property :
A
4
do
2 Iz 64
do
4 J
32
do
4
QA 0 (Since A'=0)
Normal Stress: cA 0.5do
A
Mz cA
Iz
A 212.21 MPa
Shear Stress :
A
Tx
J A
318.31 MPa
Construction of Mohr's Circle in x-z Plane :
x A z 0MPa xz A
Center : c
x z
2 c
106.1 MPa
Radius : R z c
2
xz
2 R 335.53 MPa
Coordinates: A z xz C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 441.631 MPa
2 c R 2 229.425 MPa
Angles:
p2
1
2
atan
xz
z c
p2 35.78 deg
2 p1 180deg 2 p2= p1 90deg p2 p1 54.22 deg (Clockwise)
Construction of Three Circles : y 0
From the results obtained above,
max 1 int y min 2
max 441.63 MPa Ans
int 0.00 MPa Ans
min 229.42 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2
max 335.53 MPa Ans
And the orientation is,
s
1
2
atan
z c
xz
s 9.22 deg
Problem 9-97
Solve Prob. 9-96 for point B.
Given: do 6mm a 50mm
P 90N L 150mm
Solution: 0.5do
Internal Force and Moment : At Section AB:
Vy P Mz P a Tx P L
Section Property :
A
4
do
2 Iz 64
do
4 J
32
do
4
QB
4
3
A
2
Normal Stress: cB 0
B
Mz cB
Iz
B 0 MPa
Shear Stress : bB do
B
Vy QB
Iz bB
Tx
J B314.07 MPa
Construction of Mohr's Circle in x-z Plane :
x B z 0MPa xz B
Center : c
x z
2 c
0 MPa
Radius : R z c
2
xz
2 R 314.07 MPa
Coordinates: A z xz C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 314.07 MPa
2 c R 2 314.07 MPa
Angles:
p1
1
2
90deg( ) p1 45.00 deg (Clockwise)
Construction of Three Circles : y 0
From the results obtained above,
max 1 int y min 2
max 314.07 MPa Ans
int 0.00 MPa Ans
min 314.07 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2
max 314.07 MPa Ans
Problem 9-98
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: x 90MPa y 70MPa z 120MPa
xy 30MPa yz 0MPa xz 0MPa
Solution:
Construction of Mohr's Circle in x-y Plane :
Center : c
x y
2 c
10 MPa
Radius : R x c
2
xy
2 R 85.44 MPa
Coordinates: A y xy B x xy C c 0
In-plane Principal Stresses: 
The coordinates of points representing 1 and , respectively, are
1 c R 1 75.440 MPa
2 c R 2 95.440 MPa
Construction of Three Circles :
From the results obtained above,
max 1 int 2 min z
max 75.44 MPa Ans
int 95.44 MPa Ans
min 120.00 MPa Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
max
max min
2
max 97.72 MPa Ans
Problem 9-99
The cylindrical pressure vessel has an inner radius of 1.25 m and a wall thickness of 15 mm. It is made
from steel plates that are welded along a 45° seam with the horizontal. Determine the normal and shear
stress components along this seam if the vessel is subjected to an internal pressure of 3 MPa.
Given: t 15mm ri 1250mm p 3MPa
45deg
Solution:
Hoop Stress :
ri
t
83.33
Since > 10. then thin-wall analysis can be used.
1
p ri
t 1
250 MPa
Longitudinal Stress :
2
p ri
2 t 2
125 MPa
Construction of Mohr's Circle:
x 2 y 1 xy 0
Center : c
x y
2 c
187.5 MPa
Radius : R y c
2
xy
2 R 62.5 MPa
Coordinates: A x xz C c 0
Stresses: (represented by coordinates of point P on the circle)
x' c x' 187.5 MPa Ans
x'y' R x'y' 62.5 MPa Ans
Problem 9-100
Determine the equivalent state of stress if an element is oriented 40° clockwise from the element
shown. Use Mohr's circle.
Given: x 6MPa y 10MPa
40deg xy 0MPa
Solution:
Center :
c
x y
2 c
2 MPa
Radius :
R x c R 8 MPa
Coordinates:
A x 0 B y 0 C c 0
Stresses: 
x' c R cos 2 x' 0.611 MPa Ans
x'y' R sin 2 x'y' 7.878 MPa Ans
y' c R cos 2 y' 3.389 MPa Ans
Problem 9-101
The internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consist
of an axial force of 12.5 kN, a bending moment of 1.2 kN·m, and a torsional moment of 2.25 kN·m.
Determine the principal stresses at point A. Also compute the maximum in-plane shear stress at this
point.
Given: do 150mm N 12.5kN
Mz 1.2kN m Tx 2.25kN m
Solution: 0.5do
Section Property :
A
4
do
2 Iz 64
do
4 J
32
do
4
Normal Stress: cA
A
N
A
Mz cA
Iz
A 4.33 MPa
Shear Stress :
A
Tx
J A
3.4 MPa
Construction of Mohr's Circle in x-y Plane :
x A y 0MPa xy A
Center : c
x y
2 c
2.16 MPa
Radius : R x c
2
xy
2 R 4.03 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 1.862 MPa
2 c R 2 6.191 MPa
Maximum In-plane Shear Stress : 
Represented by the coordinates of point E on the circle.
max R max 4.03 MPa Ans
Problem 9-102
The internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consist
of an axial force of 12.5 kN, a bending moment of 1.2 kN·m, and a torsional moment of 2.25 kN·m.
Determine the principal stresses at point B. Also compute the maximum in-plane shear stress at this
point.
Given: do 150mm N 12.5kN
Mz 1.2kN m Tx 2.25kN m
Solution: 0.5do
Section Property :
A
4
do
2 Iz 64
do
4 J
32
do
4
Normal Stress: cB 0
B
N
A
Mz cB
Iz
B 0.707 MPa
Shear Stress :
B
Tx
J B
3.395 MPa
Construction of Mohr's Circle in x-y Plane :
x B y 0MPa xy B
Center : c
x y
2 c
0.354 MPa
Radius : R x c
2
xy
2 R 3.414 MPa
Coordinates: A x xy C c 0
In-plane Principal Stresses: 
The coordinates of points B and D represent 1 and , respectively.
1 c R 1 3.060 MPa
2 c R 2 3.767 MPa
Maximum In-plane Shear Stress : 
Represented by the coordinates of point E on the circle.
max R max 3.41 MPa Ans
Problem 9-103
Determine the equivalent state of stress on an element if it is oriented 30° clockwise from the element
shown. Use the stress-transformation equations.
Given: x 0MPa y 0.300MPa
30deg xy 0.950MPa
Solution:
Normal Stress:
x'
x y
2
x y
2
cos 2 xy sin 2
x' 0.898 MPa Ans
y'
x y
2
x y
2
cos 2 xy sin 2
y' 0.598 MPa Ans
Shear Stress:
x'y'
x y
2
sin 2 xy cos 2
x'y' 0.605 MPa Ans
Problem 9-104
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB.
Ans
Given: x 50MPa y 100MPa
' 30deg xy 28MPa
Solution:
Construction of Mohr's Circle: 90deg '
Center : c
x y
2 c
75 MPa
Radius : R y c
2
xy
2 R 37.54 MPa
Coordinates: A x xz C c 0
Angles:
atan
xy
x c
48.240 deg
360deg 2 71.760 deg
Stresses: (represented by coordinates of point P on the circle)
x' c R cos x' 63.25 MPa Ans
x'y' R sin x'y' 35.65 MPa