Análise de Tensão em Elementos Triangulares

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PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited 
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without permission. 
 
 
PROBLEM 7.1 
For the given state of stress, determine the normal and shearing stresses exerted on the 
oblique face of the shaded triangular element shown. Use a method of analysis based on 
the equilibrium of that element, as was done in the derivations of Sec. 7.2. 
 
SOLUTION 
 
 Stresses Areas Forces 
 0: 15 sin 30 cos 30 15 cos 30 sin 30 + 10 cos 30 cos 30 0σ= − ° ° − ° ° ° =F A A A A 
 230 sin 30 cos 30 10 cos 30σ = ° ° − ° 
 5.49 ksiσ =  
 0: 15 sin 30 sin 30 15 cos 30 cos 30 10 cos 30 sin 30 0τΣ = + ° ° − ° ° − ° ° =F A A A A 
 2 215(cos 30 sin 30 ) + 10 cos 30 sin 30 τ = ° − ° ° ° 
 11.83 ksiτ = 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
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PROBLEM 7.2 
For the given state of stress, determine the normal and shearing stresses 
exerted on the oblique face of the shaded triangular element shown. Use a 
method of analysis based on the equilibrium of that element, as was done in the 
derivations of Sec. 7.2. 
 
SOLUTION 
 
 
 
 
 
 Stresses Areas Forces 
 0:Σ =F 
 80 cos55 cos55 40 sin 55 sin 55 0σ − ° ° + ° ° =A A A 
 2 280 cos 55 40sin 55σ = ° − ° 0.521 MPaσ = − 
 0:Σ =F 
 80 cos 55 sin 55 40 sin 55 cos 55τ − ° ° − ° °A A A 
 120 cos 55 sin 55τ = ° ° 56.4 MPaτ = 

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without permission. 
 
 
PROBLEM 7.3 
For the given state of stress, determine the normal and shearing stresses exerted 
on the oblique face of the shaded triangular element shown. Use a method of 
analysis based on the equilibrium of that element, as was done in the derivations 
of Sec. 7.2. 
 
SOLUTION 




 Stresses Areas Forces 
 0:Σ =F 
 5 cos60 sin 60 6 sin 60 sin 60 5 sin 60 cos60 0σ + ° ° − ° ° + ° ° =A A A A 
 26sin 60 10cos60 sin 60σ = ° − ° ° 0.1699 ksiσ =  
 0:Σ =F 
 5 cos 60 cos60 6 sin 60 cos60 5 sin 60 sin 60 0τ + ° ° − ° ° − ° ° =A A A A 
 2 25(sin 60 cos 60 ) 6sin 60 cos60τ = ° − ° + ° ° 5.10 ksiτ =  
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
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without permission. 
 
 
PROBLEM 7.4 
For the given state of stress, determine the normal and shearing stresses 
exerted on the oblique face of the shaded triangular element shown. Use a 
method of analysis based on the equilibrium of that element, as was done in the 
derivations of Sec. 7.2. 
 
SOLUTION 
 
 Stresses Areas Forces 
 0: 18 cos 15 sin 15
45 cos 15 cos 15 27 sin 15 sin 15
+ 18 sin 15 cos 15 0
σΣ = + ° °
+ ° ° − ° °
° ° =
F A A
A A
A
 
 
2
2
18 cos 15 sin 15 45 cos 15
27sin 15 18 sin 15 cos 15
σ = − ° ° − °
+ ° − ° °
 
 49.2 MPaσ = −  
 0: 18 cos 15 cos 15
45 cos 15 sin 15
27 sin 15 cos 15
18 sin 15 sin 15 0
τΣ = + ° °
− ° °
− ° °
− ° ° =
F A A
A
A
A
 
 2 218(cos 15 sin 15 ) (45 27)cos 15 sin 15τ = − ° − ° + + ° ° 
 2.41 MPaτ = 
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PROBLEM 7.5 
For the given state of stress, determine (a) the principal planes, (b) the principal 
stresses. 
 
SOLUTION 
 60 MPa 40 MPa 35 MPax y xyσ σ τ= − = − = 
(a) 
2 (2) (35)
tan 2 3.50
60 40
xy
p
x y
τ
θ
σ σ
= = = −
− − +
 
 2 74.05pθ = − ° 37.0 , 53.0pθ = − ° °  
(b) 
2
2
max, min
2
2
2 2
60 40 60 40
(35)
2 2
50 36.4 MPa
x y x y
xy
σ σ σ σ
σ τ
+ − 
= ± + 
 
− − − + = ± + 
 
= − ±
 
 max 13.60 MPaσ = −  
 min 86.4 MPaσ = − 
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PROBLEM 7.6 
For the given state of stress, determine (a) the principal planes, (b) the principal 
stresses. 
 
SOLUTION 
 10MPa 50 MPa 15MPax y xyσ σ τ= = = − 
(a) 
2 2( 15)
tan 2 = 0.750
10 50
xy
p
x y
τ
θ
σ σ
−= − =
− −
 
 2 36.8699pθ = ° 18.4 , 108.4pθ = ° ° 
(b) 
2
2
max, min
2
2
+
2 2
10 50 10 50
( 15)
2 2
σ σ σ σ
σ τ
+ − 
= ±  
 
+ − = ± + − 
 
x y x y
xy 
 30 25= ±  max 55.0 ksiσ = 
 min 5.00 ksiσ = 

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PROBLEM 7.7 
For the given state of stress, determine (a) the principal planes, (b) the principal 
stresses. 
 
SOLUTION 
 4 ksi 12 ksi 15 ksix y xyσ σ τ= = − = − 
(a) 
2 (2)( 15)
tan 2 1.875
4 12
xy
p
x y
τ
θ
σ σ
−= = = −
− +
 
 2 61.93pθ = − ° 31.0 , 59.0pθ = − ° °  
(b) 
2
2
max, min
2
2
2 2
4 12 4 12
15
2 2
4 17
x y x y
xy
σ σ σ σ
σ τ
+ − 
= ± + 
 
− + = ± + 
 
= − ±
 
 max 13.00 ksiσ =  
 min 21.0 ksiσ = − 
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PROBLEM 7.8 
For the given state of stress, determine (a) the principal planes, (b) the principal 
stresses. 
 
SOLUTION 
 8 ksi 12 ksi 5 ksix y xyσ σ τ= − = = 
(a) 
2 2(5)
tan 2 = 0.5
8 12
xy
p
x y
τ
θ
σ σ
= = −
− − −
 
 2 26.5651pθ = − ° 13.3 , 76.7pθ = − ° ° 
(b) 
2
2
max, min
2
2
+
2 2
8 12 8 12
(5)
2 2
x y x y
xy
σ σ σ σ
σ τ
+ − 
= ±  
 
− + − − = ± + 
 

 2 11.1803= ±  max13.18 ksiσ = 
 min 9.18 ksiσ = − 

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PROBLEM 7.9 
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 
 60 MPa 40 MPa 35 MPax y xyσ σ τ= − = − = 
(a) 
60 40
tan 2 0.2857
2 (2)(35)
x y
s
xy
σ σ
θ
τ
− − += − = − = 
 2 15.95sθ = ° 8.0 , 98.0sθ = ° °  
(b) 
2
2
max 2
x y
xy
σ σ
τ τ
− 
= + 
 
 
 
2
260 40 (35)
2
− + = + 
 
 max 36.4 MPaτ =  
(c) ave
60 40
2 2
x yσ σσ σ
+ − −′ = = = 50.0 MPaσ ′ = − 
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without permission. 
 
 
PROBLEM 7.10 
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 
 10 MPa 50 MPa 15 MPax y xyσ σ τ= = = − 
(a) 
10 50
tan 2 1.33333
2 2( 15)
x y
s
xy
σ σ
θ
τ
− −= − = − = −
−
 
 2 53.130sθ = − ° 26.6 , 63.4sθ = − ° ° 
(b) 
2
2
max + 2
x y
xy
σ σ
τ τ
− 
=  
 


2
210 50 ( 15)
2
− = + − 
 
 max 25.0 MPaτ = 
(c) ave 2
σ σ
σ σ
+
′ = = x y 
 
10 50
2
+= 30.0 MPaσ′ = 

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without permission. 
 
 
PROBLEM 7.11 
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 
 4 ksi 12 ksi 15 ksix y xyσ σ τ= = − = − 
(a) 
4 12
tan 2 0.53333
2 (2)( 15)
x y
s
xy
σ σ
θ
τ
− += − = − =
−
 
 2 28.07sθ = ° 14.0 , 104.0sθ = ° °  
(b) 
2
2
max 2
x y
xy
σ σ
τ τ
− 
= + 
 

 
2
24 12 ( 15)
2
+ = + − 
 
 max 17.00 ksiτ =  
(c) ave
4 12
2
σ σ −′ = = 4.00 ksiσ ′ = − 
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PROBLEM 7.12 
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 
 8 ksi 12 ksi 5 ksix y xyσ σ τ= − = = 
(a) 
8 12
tan 2 2.0
2 2(5)
x y
s
xy
σ σ
θ
τ
− − −= − = − = + 
 2 63.435sθ = ° 31.7 , 121.7sθ = ° ° 
(b) 
2
2
max 2
x y
xy
σ σ
τ τ
− 
= + 
 


2
28 12 (5)
2
− − = + 
 
 max 11.18 ksiτ =  
(c) ave 2
σ σ
σ σ
+
′ = = x y 
 
8 12
2
− += 2.00 ksiσ′ = 

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without permission. 
 
 
PROBLEM 7.13 
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 
 
0 8 ksi 5 ksi
4 ksi 4 ksi
2 2
cos 2 + sin 2
2 2
 sin 2 + cos 2
2
cos 2 sin 2
2 2
x y xy
x y x y
x y x y
x xy
x y
x y xy
x y x y
y xy
σ σ τ
σ σ σ σ
σ σ σ σ
σ θ τ θ
σ σ
τ θ τ θ
σ σ σ σ
σ θ τ θ
′
′ ′
′
= = =
+ −
= = −
+ −
= +
−
= −
+ −
= − −
 
(a) 25 2 50θ θ= − ° = − ° 
 4 4 cos ( 50°) + 5 sin ( 50°)xσ ′ = − − − 2.40 ksixσ ′ = −  
 4 sin ( 50 ) 5 cos ( 50 )x yτ ′ ′ = − ° + − ° 0.15 ksix yτ ′ ′ =  
 4 4 cos ( 50 ) 5 sin ( 50)yσ ′ = + − ° − − 10.40 ksiyσ ′ =  
(b) 10 2 20θ θ= ° = ° 
 4 4 cos (20°) + 5 sin (20°)xσ ′ = − 1.95 ksixσ ′ =  
 4 sin (20°) + 5 cos (20°)x yτ ′ ′ = 6.07 ksix yτ ′ ′ =  
 4 4 cos (20°) 5 cos (20°)yσ ′ = + − 6.05 ksiyσ ′ = 
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without permission. 
 
 
PROBLEM 7.14 
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 
 
60 MPa 90 MPa 30 MPa
15 MPa 75 MPa
2 2
cos 2 + sin 2
2 2
 sin 2 + cos 2
2
cos 2 sin 2
2 2
x y xy
x y x y
x y x y
x xy
x y
xy xy
x y x y
y xy
σ σ τ
σ σ σ σ
σ σ σ σ
σ θ τ θ
σ σ
τ θ τ θ
σ σ σ σ
σ θ τ θ
′
′
= − = =
+ −
= = −
+ −
= +
−
= −
+ −
= − −
 
(a) 25 2 50θ θ= − ° = − ° 
 15 75 cos ( 50 ) 30 sin ( 50 )xσ ′ = − − ° + − ° 56.2 MPaxσ ′ = −  
 75 sin ( 50 ) 30 cos ( 50 )x yτ ′ ′ = + − ° + − ° 38.2 MPax yτ ′ ′ = −  
 15 75 cos ( 50 ) 30 sin ( 50 )yσ ′ = + − ° − − ° 86.2 MPayσ ′ =  
(b) 10 2 20θ θ= ° = ° 
 15 75 cos (20°) + 30 sin (20°)xσ ′ = − 45.2 MPaxσ ′ = −  
 75 sin (20°) + 30 cos (20°)x yτ ′ ′ = + 53.8 MPax yτ ′ ′ =  
 15 75 cos (20°) 30 sin (20°)yσ ′ = + − 75.2 MPayσ ′ = 
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without permission. 
 
PROBLEM 7.15 
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 
 
8 ksi 12 ksi 6 ksi
2 ksi 10 ksi
2 2
cos 2 + sin 2
2 2
 sin 2 + cos 2
2
cos 2 sin 2
2 2
x y xy
x y x y
x y x y
x xy
x y
x y xy
x y x y
y xy
σ σ τ
σ σ σ σ
σ σ σ σ
σ θ τ θ
σ σ
τ θ τ θ
σ σ σ σ
σ θ τ θ
′
′ ′
′
= = − = −
+ −
= − =
+ −
= +
−
= −
+ −
= − − 
(a) 25 2 50θ θ= − ° = − ° 
 2 10 cos ( 50 ) 6 sin ( 50 )xσ ′ = − + − ° − − ° 9.02 ksi xσ ′ =  
 10 sin ( 50 ) 6 cos ( 50 )x yτ ′ ′ = − − ° − − ° 3.80 ksix yτ ′ ′ =  
 2 10 cos ( 50 ) 6 sin ( 50 )yσ ′ = − − − ° + − ° 13.02 ksiyσ ′ = −  
(b) 10 2 20θ θ= ° = ° 
 2 10 cos (20°) 6 sin (20°)xσ ′ = − + − 5.34ksixσ ′ = 
 10 sin (20°) 6 cos (20°)x yτ ′ ′ = − −  9.06 ksix yτ ′ ′ = − 
 2 10 cos (20°) + 6 sin (20°)yσ ′ = − − 9.34 ksiyσ ′ = − 
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without permission. 
 
 
PROBLEM 7.16 
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 
 
0 80 MPa 50 MPa
40 MPa 40 MPa
2 2
cos 2 sin 2
2 2
sin 2 + cos 2
2
cos 2 sin 2
2 2
x y xy
x y x y
x y x y
x xy
x y
x y xy
x y x y
y xy
σ σ τ
σ σ σ σ
σ σ σ σ
σ θ τ θ
σ σ
τ θ τ θ
σ σ σ σ
σ θ τ θ
′
′ ′
′
= = − = −
+ −
= − =
+ −
= + +
−
= −
+ −
= − −
 
(a) 25θ = − ° 2 50θ = − ° 
 40 40 cos ( 50 ) 50 sin ( 50°)xσ ′ = − + − ° − − 24.0 MPaxσ ′ =  
 40 sin ( 50°) 50 cos ( 50 )x yτ ′ ′ = − − − − ° 1.5 MPax yτ ′ ′ = −  
 40 40 cos ( 50 ) 50 sin ( 50 )yσ ′ = − − − ° + − ° 104.0 MPayσ ′ = − 
(b) 10 2 20θ θ= ° = ° 
 40 40 cos (20°) 50 sin (20°)xσ ′ = − + − 19.5 MPaxσ ′ = −  
 40 sin (20°) 50 cos (20°)x yτ ′ ′ = − − 60.7 MPax yτ ′ ′ = −  
 40 40 cos (20°) + 50 sin (20°)yσ ′ = − − 60.5 MPayσ ′ = − 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
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without permission. 
 
 
PROBLEM 7.17 
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 
 
4 MPa 1.6 MPa 0
15 2 30
x y xyσ σ τ
θ θ
= − = − =
= − ° = − °
 
(a) sin 2 + cos 2
2
x y
x y xy
σ σ
τ θ τ θ′ ′
−
= − 
 
4 ( 1.6)
 sin ( 30 ) 0
2
− − −= − − ° + 0.600 MPax yτ ′ ′ = − 
(b) cos 2 + sin 2
2 2
x y x y
x xy
σ σ σ σ
σ θ τ θ′
+ −
= + 
 
4 ( 1.6) 4 ( 1.6)
cos ( 30 ) 0
2 2
− + − − − −= + − ° + 3.84 MPaxσ ′ = − 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
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without permission. 
 
 
PROBLEM 7.18 
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 parallel to the grain, (b) the 
normal stress perpendicular to the grain. 
 
SOLUTION 

0 0 400 psi
15 2 30
x y xyσ σ τ
θ θ
= = =
= − ° = − °

(a) sin 2 + cos 2
2
x y
x y xy
σ σ
τ θ τ θ′ ′
−
= − 
 0 400cos( 30 )= + − °  346 psix yτ ′ ′ = 
(b) cos 2 + sin 2
2 2
x y x y
x xy
σ σ σ σ
σ θ τ θ′
+ −
= + 
 0 0 400sin( 30 )= + + − °  200 psixσ ′ = − 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 
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without permission. 
 
 
PROBLEM 7.19 
A steel pipe of 12-in. outer diameter is fabricated from 1
4
-in. -thick plate by 
welding along a helix that forms an angle of 22.5° with a plane perpendicular 
to the axis of the pipe. Knowing that a 40-kip axial force P and an 80-kip in.⋅ 
torque T, each directed as shown, are applied to the pipe, determine σ and τ 
in directions, respectively, normal and tangential to the weld. 
 
SOLUTION 
 ( )
( )
2 2 2
1 2
2 2 2 2 2
2 1
4 4 4 4 4
2 1
1
12 in., 6 in., 0.25 in.
2
5.75 in.
(6 5.75 ) 9.2284 in
(6 5.75 ) 318.67 in
2 2
d c d t
c c t
A c c
J c c
π π
π π
= = = =
= − =
= − = − =
= − = − =
 
Stresses: 
 2
40
4.3344 ksi
9.2284
(80)(6)
1.5063 ksi
318.67
0, 4.3344 ksi, 1.5063 ksix y xy
P
A
Tc
J
σ
τ
σ σ τ
= −
= − = −
=
= =
= = − =
 
Choose the x′ and y′ axes, respectively, tangential and normal to the weld. 
Then and 22.5w y w x yσ σ τ τ θ′ ′ ′= = = ° 
 
cos 2 sin 2
2 2
( 4.3344) [ ( 4.3344)]
cos 45 1.5063 sin 45°
2 2
4.76 ksi
x y x y
y xy
σ σ σ σ
σ θ τ θ′
+ −
= − −
− − −= − ° −
= − 4.76 ksiwσ = −  
 
sin 2 cos 2
2
[ ( 4.3344)]
sin 45 1.5063 cos 45
2
0.467 ksi
x y
x y xy
σ σ
τ θ τ θ′ ′
−
= − +
− −= − ° + °
= − 0.467 ksiwτ = − 
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PROBLEM 7.20 
Two members of uniform cross section 50 80 mm× are glued together along 
plane a-a that forms an angle of 25° with the horizontal. Knowing that the allowable 
stresses for the glued joint are 800 kPaσ = and 600 kPa,τ = determine the largest 
centric load P that can be applied. 
 
SOLUTION 
For plane a-a, 65 .θ = ° 
 
 
2 2 2
3 3 3
3
2 2
2 2
3
0, 0,
 cos sin 2 sin cos 0 sin 65 0
(50 10 )(80 10 )(800 10 )
3.90 10 N
sin 65 sin 65
( )sin cos (cos sin ) sin 65 cos65 0
(50 10 )(8
sin 65 cos 65
x xy y
x y xy
x y xy
P
A
P
A
A
P
P
A
A
P
σ τ σ
σ σ θ σ θ τ θ θ
σ
τ σ σ θ θ τ θ θ
τ
− −
−
= = =
= + + = + ° +
× × ×= = = ×
° °
= − − + − = ° ° +
×= =
° °
3 3
30 10 )(600 10 ) 6.27 10 N
sin 65 cos65
−× × = ×
° °
 
Allowable value of P is the smaller one. 3.90 kNP = 
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PROBLEM 7.21 
Two steel plates of uniform cross section 10 80mm× are welded 
together as shown. Knowing that centric 100-kN forces are applied to 
the welded plates and that 25β = ° , determine (a) the in-plane shearing 
stress parallel to the weld, (b) the normal stress perpendicular to the 
weld. 
 
SOLUTION 
 Area of weld: 
 
3 3
6 2
(10 10 )(80 10 )
cos 25
882.7 10 m
wA
− −
−
× ×=
°
= ×
 
 
 
(a) 0: 100sin 25 0 42.26 kN = − ° = =s s sF F F 
 
3
6
6
42.26 10
47.9 10 Pa
882.7 10
s
w
w
F
A
τ −
×= = = ×
×
 47.9 MPawτ =  
(b) 0: 100cos 25 0 90.63 kN = − ° = =n n nF F F 
 
3
6
6
90.63 10
102.7 10 Pa
882.7 10
n
w
w
F
A
σ −
×= = = ×
×
 102.7 MPawσ =  
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PROBLEM 7.22 
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 the in-plane shearing stress parallel to the 
weld is 30 MPa, determine (a) the angle β, (b) the corresponding normal 
stress perpendicular to the weld. 
 
SOLUTION 
 Area of weld: 
 
3 3
6
2
(10 10 )(80 10 )
cos
800 10
m
cos
wA β
β
− −
−
× ×=
×=
 
 
(a) 3
3
6 6
6
0: 100sin 0 100sin kN 100 10 sin N
100 10 sin
30 10 125 10 sin cos
800 10 / cos
β β β
βτ β β
β−
 = − = = = ×
×= × = = ×
×
s s s
s
w
w
F F F
F
A
 
 
6
6
1 30 10
sin cos sin 2 0.240
2 125 10
β β β ×= = =
×
 14.34β = °  
(b) 
6
6 2
0: 100cos 0 100cos14.34 96.88 kN
800 10
825.74 10 m
cos14.34
β
−
−
 = − = = ° =
×= = ×
n n n
w
F F F
A


3
6
6
96.88 10
117.3 10 Pa
825.74 10
n
w
F
A
σ −
×= = = ×
×
 117.3 MPaσ =  
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PROBLEM 7.23 
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 
Equivalent force-couple system at center of shaft in section at point H: 
 
400 lb (400)(6) 2400 lb in
(400)(2) 800 lb in
V M
T
= = = ⋅
= = ⋅
 
Shaft cross section: 
4 4 4
1
1 in. 0.5 in.
2
1
0.098175 in 0.049087 in
2 2
d c d
J c I J
π
= = =
= = = =
 
Torsion: 3
(800)(0.5)
4.074 10 psi 4.074 ksi
0.098175
Tc
J
τ = = = × = 
Bending: 3
(2400)(0.5)
24.446 10 psi 24.446 ksi
0.049087
Mc
I
σ = = = × = 
Transverse shear: Stress at point H is zero. 
 
ave
2
2 2 2
24.446 ksi, 0, 4.074 ksi
1
( ) 12.223 ksi
2
(12.223) (4.074)
2
12.884 ksi
x y xy
x y
x y
xyR
σ σ τ
σ σ σ
σ σ
τ
= = =
= + =
− 
= + = + 
 
=
 
 avea Rσ σ= + 25.1 ksiaσ =  
 aveb Rσ σ= − 0.661 ksibσ = − 
 max Rτ =  max 12.88 ksiτ =  

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PROBLEM 7.24 
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 as 
on top of the 34 -in. diameter shaft. 
 
SOLUTION 
Equivalent force-couple system at center of shaft in section at point H: 
 
24 lb (24)(6) 144 lb in
(24)(10) 240 lb in
V M
T
= = = ⋅
= = ⋅
 
Shaft cross section: 
4 4 4
1
0.75 in., 0.375 in.
2
1
0.031063 in 0.015532 in
2 2
d c d
J c I J
π
= = =
= = = =
 
Torsion: 3
(240)(0.375)
2.897 10 psi 2.897 ksi
0.031063
Tc
J
τ = = = × = 
Bending: 3
(144)(0.375)
3.477 10 psi 3.477 ksi
0.015532
Mc
I
σ = = = × = 
Transverse shear: At point H, the stress due to transverse shear is zero. 
Resultant stresses: 
ave
2
2 2 2
3.477 ksi, 0, 2.897 ksi
1
( ) 1.738 ksi
2
1.738 2.897 3.378 ksi
2
x y xy
x y
x y
xyR
σ σ τ
σ σ σ
σ σ
τ
= = =
= + =
− 
= + = + = 
 
 
 avea Rσ σ= + 5.12 ksiaσ = 
 aveb Rσ σ= − 1.640 ksibσ = − 
 max Rτ =  max 3.38 ksiτ = 
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PROBLEM 7.25 
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 
at point K. 
 
SOLUTION 
 
( )4 4 6 4
6 4
6 4
102
51 mm 45 mm
2 2
4.1855 10 mm
2
4.1855 10 m
1
2.0927 10 m
2
o
o i o
o i
d
r r r t
J r r
I J
π
−
−
= = = = − =
= − = ×
= ×
= = ×
 
Force-couple system at center of tube in the plane containing points H and K: 
 
3
3 3
3 3
10 kN
10 10 N
(10 10 )(200 10 )
2000 N m
(10 10 )(150 10 )
1500 N m
x
y
z
F
M
M
−
−
=
= ×
= × ×
= ⋅
= − × ×
= − ⋅
 
Torsion: At point K, place local x-axis in negative global z-direction. 
 
3
3
6
6
2000 N m
51 10 m
(2000)(51 10 )
4.1855 10
24.37 10 Pa
24.37 MPa
y
o
xy
T M
c r
Tc
J
τ
−
−
= = ⋅
= = ×
×= =
×
= ×
= 
 
 
 
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PROBLEM 7.25 (Continued) 
 
Transverse shear: Stress due to transverse shear xV F= is zero at point K. 
Bending: 
 
3
6
6
| | (1500)(51 10 )
| | 36.56 10 Pa 36.56 MPa
2.0927 10
z
y
M c
I
σ
−
−
×= = = × =
×
 
Point K lies on compression side of neutral axis: 
 36.56 MPayσ = − 
Total stresses at point K: 
 ave
2
2
0, 36.56 MPa, 24.37 MPa
1
( ) 18.28 MPa
2
30.46 MPa
2
x y xy
x y
x y
xyR
σ σ τ
σ σ σ
σ σ
τ
= = − =
= + = −
− 
= + = 
 
 
 max ave 18.28 30.46Rσ σ= + = − + max 12.18 MPaσ =  
 min ave 18.28 30.46Rσ σ= − = − − min 48.7 MPaσ = −  
 max Rτ =  max 30.5 MPaτ = 
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PROBLEM 7.26 
The axle of an automobile is acted upon by the forces and couple 
shown. Knowing that the diameter of the solid axle is 32 mm, 
determine (a) the principal planes and principal stresses at point H 
located on top of the axle, (b) the maximum shearing stress at the 
same point. 
 
SOLUTION 
 3
1 1
(32) 16 mm 16 10 m
2 2
c d −= = = = × 
Torsion: 6
3 3 3
2 2(350 N m)
54.399 10 Pa 54.399 MPa
(16 10 m)
Tc T
J c
τ
π π −
⋅= = = = × =
×
 
Bending: 4 3 4 9 4
3
3
6
9
(16 10 ) 51.472 10 m
4 4
(0.15m)(3 10 N) 450 N m
(450)(16 10 )
139.882 10 Pa 139.882 MPa
51.472 10
I c
M
My
I
π π
σ
− −
−
−
= = × = ×
= × = ⋅
×= − = − = − × = −
×
 
 
Top view: Stresses: 
 
 
 
 
 
ave
2
2 2 2
139.882 MPa 0 54.399 MPa
1 1
( ) ( 139.882 0) 69.941 MPa
2 2
( 69.941) ( 54.399) 88.606 MPa
2
x y xy
x y
x y
xyR
σ σ τ
σ σ σ
σ σ
τ
= − = = −
= + = − + = −
− 
= + = − + − = 
 
 
(a) max ave 69.941 88.606Rσ σ= + = − + max 18.67 MPaσ =  
 min ave 69.941 88.606Rσ σ= − = − − min 158.5 MPaσ = −  
 
 
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PROBLEM 7.26 (Continued) 
 
 
2 (2)( 54.399)
tan 2 0.77778 2 37.88
139.882
xy
p p
x y
τ
θ θ
σ σ
−= = = = °
− −
 
 
 
 18.9 and 108.9°pθ = ° 
 
 
(b) max 88.6 MPaRτ = = max 88.6 MPaτ =  
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PROBLEM 7.27 
For the state of plane stress shown, determine (a) the largest value of xyτ for which 
the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the 
corresponding principal stresses. 
 
SOLUTION 
 
2 2
2 2
max
2 2
10 ksi, 8 ksi, ?
10 ( 8)
9 12 ksi
x y xy
x y
xy xy
xy
R
z z
σ σ τ
σ σ
τ τ τ
τ
= = − =
−  − − = = + = +   
  
= + =
 
(a) 2 212 9xyτ = − 7.94 ksixyτ =  
(b) ave
1
( ) 1 ksi
2 x y
σ σ σ= + = 
 ave 1 12 13 ksia Rσ σ= + = + = 13.00 ksiaσ =  
 ave 1 12 11 ksib Rσ σ= − = − = − 11.00 ksibσ = −  
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PROBLEM 7.28 
For the state of plane stress shown, determine the largest value of yσ for which 
the maximum in-plane shearing stress is equal to or less than 75 MPa. 
 
SOLUTION 
 60 MPa, ?, 20 MPax y xyσ σ τ= = = 
Let 
 .
2
x yu
σ σ−
= 
Then 
 
2 2
2 2 2 2
2
75 MPa
75 20 72.284 MPa
2 60 (2)(72.284) 84.6 MPa or 205 MPa
y x
xy
xy
y x
u
R u
u R
u
σ σ
τ
τ
σ σ
= −
= + =
= ± − = ± − =
= − = = −
 
Largest value of yσ is required. 205 MPayσ = 
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PROBLEM 7.29 
Determine the range of values of xσ for which the maximum in-plane shearing stress 
is equal to or less than 10 ksi. 
 
SOLUTION 
 ?, 15 ksi, 8 ksix y xyσ σ τ= = = 
Let 
2 2
max
2 2 2 2
2
2
10 ksi
10 8 6 ksi
2 15 (2)(6) 27 ksi or 3 ksi
x y
x y
xy
z
xy
x y
u u
R u
u R
u
σ σ
σ σ
τ τ
τ
σ σ
−
= = +
= + = =
= ± − = ± − = ±
= + = ± =
 
Allowable range: 3 ksi 27 ksixσ≤ ≤ 
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PROBLEM 7.30 
For the state of plane stress shown, determine (a) the value of xyτ for which the 
in-plane shearing stress parallel to the weld is zero, (b) the corresponding 
principal stresses. 
 
SOLUTION 
 12 MPa, 2 MPa, ?x y xyσ σ τ= = = 
Since 0,x yτ ′ ′ = x ′-direction is a principal direction. 
 
15
2
tan 2
p
xy
p
x y
θ
τ
θ
σ σ
= − °
=
−
 
(a) 
1 1
( ) tan 2 (12 2) tan( 30 )
2 2xy x y p
τ σ σ θ= − = − − ° 2.89 MPaxyτ = −  
 
2
2 2 2
ave
5 2.89 5.7735 MPa
2
1
( ) 7 MPa
2
x y
xy
x y
R
σ σ
τ
σ σ σ
− 
= + = + = 
 
= + =
 
(b) ave 7 5.7735a Rσ σ= + = + 12.77 MPaaσ =  
 ave 7 5.7735b Rσ σ= − = − 1.226 MPabσ = 
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PROBLEM 7.31 
Solve Probs. 7.5 and 7.9, using Mohr’s circle. 
PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the 
principal planes, (b) the principal stresses. 
PROBLEM 7.9 through 7.12 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
x y
σ
σ
τ
σ σ
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
σ τ
σ τ
σ
− = − −
= −
= −
 
(a) 
35
tan 3.500
10
GX
CG
β = = = 
 
74.05
1
37.03
2b
β
θ β
= °
= − = − ° 37.0bθ = − °  
 
180 105.95
1
52.97
2a
α β
θ α
= ° − = °
= = ° 53.0aθ = °  
 
2 2 2 210 35 36.4 MPaR CG GX= + = + = 
(b) min ave 50 36.4Rσ σ= − = − − min 86.4 MPaσ = −  
 max ave 50 36.4Rσ σ= + = − + max 13.6 MPaσ = −  
(a′) 45 7.97d Bθ θ= + ° = ° 8.0dθ = °  
 45 97.97e Aθ θ= + ° = ° 98.0eθ = °  
 max 36.4 MPaRτ = = max 36.4 MPaτ =  
(b′) ave 50 MPaσ σ′ = = − 50.0 MPaσ ′ = − 
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PROBLEM 7.32 
Solve Probs 7.7 and 7.11, using Mohr’s circle. 
PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the 
principal planes, (b) the principal stresses. 
PROBLEM 7.9 through 7.12 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
4 ksi,
12 ksi,
15 ksi
4 ksi
2
x
y
xy
x y
σ
σ
τ
σ σ
σ
=
= −
= −
+
= = −
 
Plotted points for Mohr’s circle: 
 
ave
: ( , ) (4 ksi, 15 ksi)
: ( , ) ( 12 ksi, 15 ksi)
: ( , 0) ( 4 ksi, 0)
x xy
y xy
X
Y
C
σ τ
σ τ
σ
− =
= − −
= −
 
(a) 
15
tan 1.875
8
FX
CF
α = = = 
 
61.93
1
30.96
2a
α
θ α
= °
= − = − ° 31.0aθ = − °  
 
180 118.07
1
59.04
2b
β α
θ β
= ° − = °
= = ° 59.0bθ = °  
 2 2 2 2( ) ( ) (8) (15) 17 ksiR CF FX= + = + = 
(b) max ave 4 17a Rσ σ σ= = + = − + max 13.00 ksiσ =  
 min min ave 4 17Rσ σ σ= = − = − − min 21.0 ksiσ = −  
(a′) 45 14.04d aθ θ= + ° = ° 14.0dθ = °  
 45 104.04e bθ θ= + ° = ° 104.0eθ = °  
 max Rτ = max 17.00 ksiτ =  
(b′) aveσ σ′ = 4.00 ksiσ ′ = −  
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PROBLEM 7.33 
Solve Problem 7.10, using Mohr’s circle. 
PROBLEM 7.9 through 7.12 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
10 MPa 50 MPa 15 MPa
1 1
( ) (10 50) 30 MPa
2 2
x y xy
x y
σ σ τ
σ σ σ
= = = −
= + = + =
 
Plotted points for Mohr’s circle: 
 
,
ave
: ( ) (10 MPa, 15 MPa)
: ( , ) (50 MPa, 15 MPa)
: ( , 0) (30 MPa, 0)
x xy
y xy
X
Y
C
σ τ
σ τ
σ
− =
= −
=
 
 
15
tan 0.75
20
36.87
1
18.43
2b
FX
FC
α
α
θ α
= = =
= °
= = °
 
 
(a) 45d bθ θ= − ° 26.6dθ = − °  
 45e bθ θ= + ° 63.4eθ = °  
(b) 
2 2 2 220 15 25 MPaR CF FX= + = + = 
 max (in-plane)τ = R max (in-plane) 25.0 MPaτ =  
(c) aveσ σ′ = 30.0 MPaσ′ =  
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PROBLEM 7.34 
Solve Prob. 7.12, using Mohr’s circle. 
PROBLEM 7.9 through 7.12 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 
 8 ksi 12 ksi 5 ksix y xyσ σ τ= − = = 
Plotted points for Mohr’s circle: 
 
ave
: ( , ) ( 8 ksi, 5 ksi)
: ( , ) (12 ksi, 5 ksi)
: ( , 0) (2 ksi, 0)
x xy
y xy
X
Y
C
σ τ
σ τ
σ
− = − −
=
=
 
 
5
tan 0.5
10
26.565
180 153.435
1
76.718
2a
FX
FC
α
α
β α
θ β
= = =
= °
= − = °
= = °
 
(a) 45d aθ θ= + ° 121.7dθ = °  
 45e aθ θ= − ° 31.7eθ = °  
(b) 
2 2 2 210 5 11.1803 ksiR CF FX= + = + = 
 max (in-plane)τ = R max (in-plane) 11.18 ksiτ =  
(c) aveσ σ′ = 2.00 ksiσ′ =  
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PROBLEM 7.35 
Solve Prob. 7.13, using Mohr’s circle. 
PROBLEM 7.13 through 7.16 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
x y
σ
σ
τ
σ σ
σ
=
=
=
+
= =
 
Plotted points for Mohr’s circle: 
 
: (0, 5 ksi)
: (8 ksi, 5 ksi)
: (4 ksi, 0)
X
Y
C
−
 
 
2 2 2 2
5
tan 2 1.25
4
2 51.34
4 5 6.40 ksi
p
p
FX
FC
R FC FX
θ
θ
= = =
= °
= + = + =
 
(a) 25θ = ° . 2 50θ = ° 
 51.34 50 1.34ϕ = ° − ° = ° 
 ave cosx Rσ σ ϕ′ = − 2.40 ksixσ ′ = − 
 sinx y Rτ ϕ′ ′ = 0.15 ksix yτ ′ ′ = 
 ave cosy Rσ σ ϕ′ = + 10.40 ksiyσ ′ =  
(b) 10θ = ° . 2 20θ = ° 
 51.34 20 71.34ϕ = ° + ° = ° 
 ave cosx Rσ σ ϕ′ = − 1.95 ksixσ ′ =  
 sinx y Rτ ϕ′ ′ = 6.07 ksix yτ ′ ′ =  
 ave cosy Rσ σ ϕ′ = +  6.05 ksiyσ ′ =  
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PROBLEM 7.36 
Solve Prob 7.14, using Mohr’s circle. 
PROBLEM 7.13 through 7.16 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
x y
σ
σ
τ
σ σ
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 = = =p
FX
FC
 
 2 21.80 10.90θ θ= ° = °p P 
 
2 2 2 275 30 80.78 MPaR FC FX= + = + = 
 
(a) 25θ = ° . 2 50θ = ° 
 2 2 50 21.80 28.20Pϕ θ θ= − = ° − ° = ° 
 ave cosσ σ ϕ′ = −x R 56.2 MPaσ ′ = −x 
 sinτ ϕ′ ′ = −x y R 38.2 MPaτ ′ ′ = −x y 
 ave cosσ σ ϕ′ = +y R 86.2 MPayσ ′ =  
 
 
 
 
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PROBLEM 7.36 (Continued) 
 
 
(b) 10θ = ° . 2 20θ = ° 
 2 2 21.80 20 41.80ϕ θ θ= + = ° + ° = °p 
 ave cosσ σ ϕ′ = −x R 45.2 MPaσ ′ = −x  
 sinτ ϕ′ ′ =x y R 53.8 MPaτ ′ ′ =x y  
 ave cosσ σ ϕ′ = +y R  75.2 MPaσ ′ =y  
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PROBLEM 7.37 
Solve Prob. 7.15, using Mohr’s circle. 
PROBLEM 7.13 through 7.16 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
8 ksi,
12 ksi,
6 ksi
2 ksi
2
x
y
xy
x y
σ
σ
τ
σ σ
σ
=
= −
= −
+
= = −
 
Plotted points for Mohr’s circle: 
 
: (8 ksi, 6 ksi)
: ( 12 ksi, 6 ksi)
: ( 2 ksi, 0)
X
Y
C
− −
−
 
 
2 2 2 2
6
tan 2 0.6
10
2 30.96
10 6 11.66 ksi
θ
θ
= = =
= °
= + = + =
p
p
FX
CF
R CF FX
 
(a) 25θ = ° . 2 50θ = ° 
 50 30.96 19.04ϕ = ° − ° = ° 
 ave cosx Rσ σ ϕ′ = + 9.02 ksixσ ′ = 
 sinx y Rτ ϕ′ ′ = 3.80 ksix yτ ′ ′ = 
 ave cosy Rσ σ ϕ′ = − 13.02 ksiyσ ′ = −  
(b) 10θ = ° . 2 20θ = ° 
 30.96 20 50.96ϕ = ° + ° = ° 
 ave cosx Rσ σ ϕ′ = + 5.34 ksixσ ′ =  
 sinx y Rτ ϕ′ ′ = − 9.06 ksix yτ ′ ′ = −  
 ave cosy Rσ σ ϕ′ = − 9.34 ksiyσ ′ = −  
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PROBLEM 7.38 
Solve Prob. 7.16, using Mohr’s circle. 
PROBLEM 7.13 through 7.16 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
x y
σ
σ
τ
σ σ
σ
=
= −
= −
+
= = −
 
Plotted points for Mohr’s circle: 
 
: (0, 50 MPa)
: ( 80 MPa, 50 MPa)
: ( 40 MPa, 0)
X
Y
C
− −
−
 
 
2 2 2 2
50
tan 2 1.25
40
2 51.34
40 50
64.03 MPa
p
p
FX
CF
R CF FX
θ
θ
= = =
= °
= + = +
=
 
(a) 25θ = ° . 2 50θ = ° 
 51.34 50 1.34ϕ = ° − ° = ° 
 ave cosx Rσ σ ϕ′ = + 24.0 MPaxσ ′ = 
 sinx y Rτ ϕ′′ = − 1.5 MPax yτ ′ ′ = −  
 ave cosy Rσ σ ϕ′ = − 104.0 MPayσ ′ = −  
(b) 10θ = ° . 2 20θ = ° 
 51.34 20 71.34ϕ = ° + ° = ° 
 ave cosx Rσ σ ϕ′ = + 19.5 MPaxσ ′ = −  
 sinx y Rτ ϕ′ ′ = − 60.7 MPax yτ ′ ′ = −  
 ave cosy Rσ σ ϕ′ = − 60.5 MPayσ ′ = −  
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PROBLEM 7.39 
Solve Prob. 7.17, using Mohr’s circle. 
PROBLEM 7.17 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
x y
σ σ τ
σ σ
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
σ τ
σ τ
σ
− = −
= −
= −
 
 
15 . 2 30
1.2MPa 1.2MPaCX R
θ θ= − ° = − °
= =
 
(a) sin 30 sin 30 1.2sin 30x y CX Rτ ′ ′ ′= − ° = − ° = − ° 0.600 MPax yτ ′ ′ = −  
(b) ave cos30 2.8 1.2cos30x CXσ σ′ ′= − ° = − − ° 3.84 MPaxσ ′ = −  
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PROBLEM 7.40 
Solve Prob. 7.18, using Mohr’s circle. 
PROBLEM 7.18 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 parallel 
to the grain, (b) the normal stress perpendicular to the grain. 
 
SOLUTION 
 
ave
0, 400psi
0
2
x y xy
x y
σ σ τ
σ σ
σ
= = =
+
= =
 
 
Points: 
 
ave
: ( , ) (0, 400 psi)
: ( , ) (0, 400 psi)
: ( , 0) (0, 0)
x xy
y xy
X
Y
C
σ τ
σ τ
σ
− = −
=
=
 
15 ,θ = − ° 2 30
400 psiCX R
θ = − °
= =
 
(a) cos 30 400 cos30x y Rτ ′ ′ = ° = ° 346 psix yτ ′ ′ =  
(b) ave sin 30 400 sin 30x Rσ σ′ = − ° = − ° 200 psixσ ′ = − 
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PROBLEM 7.41 
Solve Prob. 7.19, using Mohr’s circle. 
PROBLEM 7.19 A steel pipe of 12-in. outer diameter is fabricated from 
1
4 -in.-thick plate by welding along a helix which forms an angle of 22.5° with 
a plane perpendicular to the axis of the pipe. Knowing that a 40-kip axial force 
P and an 80-kip ⋅ in. torque T, each directed as shown, are applied to the pipe, 
determine σ and τ in directions, respectively, normal and tangential to the 
weld. 
 
SOLUTION 
 
 
 
 
 
 
 ( )
( )
2 2 2
1 2
2 2 2 2 2
2 1
4 4 4 4 4
2 1
1
12 in. 6 in., 0.25 in.
2
5.75 in.
(6 5.75 ) 9.2284 in
(6 5.75 ) 318.67 in
2 2
d c d t
c c t
A c c
J c c
π π
π π
= = = =
= − =
= − = − =
= − = − =
 
Stresses: 
 2
40
4.3344 ksi
9.2284
(80)(6)
1.5063 ksi
318.67
0, 4.3344 ksi, 1.5063 ksix y xy
P
A
Tc
J
σ
τ
σ σ τ
= − = − = −
= − = − =
= = − =
 
Draw the Mohr’s circle. 
 
: (0, 1.5063 ksi)
: ( 4.3344 ksi,1.5063 ksi)
: ( 2.1672 ksi, 0)
X
Y
C
−
−
−
 
 
2 2
1.5063
tan 0.69504 34.8
2.1672
(2)(22.5 ) 10.8
(2.1672) (1.5063) 2.6393 ksiR
α α
β α
= = = °
= ° − = °
= + =
 
 2.1672 2.6393 cos 10.8wσ = − − ° 4.76 ksiwσ = −  
 2.6393sin10.2wτ = − ° 0.467 ksiwτ = − 
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PROBLEM 7.42 
Solve Prob. 7.20, using Mohr’s circle. 
PROBLEM 7.20 Two members of uniform cross section 50 80× mm are glued 
together along plane a-a that forms an angle of 25° with the horizontal. Knowing 
that the allowable stresses for the glued joint are 800σ = kPa and 600τ = kPa, 
determine the largest centric load P that can be applied. 
 
SOLUTION 
 
 3 3
3 2
0
0
/
(50 10 )(80 10 )
4 10 m
(1 cos50 )
2
2
1 cos50
x
xy
y P A
A
P
A
A
P
σ
τ
σ
σ
σ
− −
−
=
=
=
= × ×
= ×
= + °
=
+ °
 
 
3 3
3
(2)(4 10 )(800 10 )
1 cos50
3.90 10 N
P
P
−× ×≤
+ °
≤ ×
 
 
3 3
32 (2)(4 10 )(600 10 )sin 50 6.27 10 N
2 sin 50 sin 50
P A
P
A
ττ
−× ×= ° = ≤ = ×
° °
 
Choosing the smaller value, 3.90 kNP = 
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PROBLEM 7.43 
Solve Prob. 7.21, using Mohr’s circle. 
PROBLEM 7.21 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 25β = ° , determine (a) the in-
plane shearing stress parallel to the weld, (b) the normal stress 
perpendicular to the weld. 
 
SOLUTION 
 
3
6
3) 3)
100 10
125 10 Pa 125 MPa
(10 10 (80 10
x
P
A
σ − −
×= = = × =
× ×
 
 0 0y xyσ τ= = 
 From Mohr’s circle: 
 (a) 62.5sin 50wτ = ° 47.9 MPawτ =  
 (b) 62.5 62.5cos50wσ = + ° 
 102.7 MPawσ = 
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PROBLEM 7.44 
Solve Prob. 7.22, using Mohr’s circle. 
PROBLEM 7.22 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 the in-plane shearing stress parallel 
to the weld is 30 MPa, determine (a) the angle β, (b) the corresponding 
normal stress perpendicular to the weld. 
 
SOLUTION 
 
3
6
3 3
100 10
125 10 Pa 125 MPa
(10 10 )(80 10 )
x
P
A
σ − −
×= = = × =
× ×
 
 0 0y xyσ τ= = 
 From Mohr’s circle: 
 (a) 
30
sin 2 0.48
62.5
β = = 14.3β = °  
 (b) 62.5 62.5cos 2σ β= + 
 117.3 MPaσ = 
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PROBLEM 7.45 
Solve Prob. 7.23, using Mohr’s circle.PROBLEM 7.23 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 
Equivalent force-couple system at center of shaft in section at point H: 
 400 lb (400)(6) 2400 lb inV M= = = ⋅ 
 (400)(2) 800 lb inT = = ⋅ 
Shaft cross section 
1
1 in. 0.5 in.
2
d c d= = = 
 4 4 4
1
0.098175 in 0.049087 in
2 2
J c I J
π= = = = 
Torsion: 3
(800)(0.5)
4.074 10 psi 4.074 ksi
0.098175
Tc
J
τ = = = × = 
Bending: 3
(2400)(0.5)
24.446 10 psi 24.446 ksi
0.049087
Mc
I
σ = = = × = 
Transverse shear: Stress at point H is zero. 
Resultant stresses: 
ave
2
2
2 2
24.446 ksi, 0, 4.074 ksi
1
( ) 12.223 ksi
2
2
(12.223) (4.074) 12.884 ksi
x y xy
x y
x y
xyR
σ σ τ
σ σ σ
σ σ
τ
= = =
= + =
− 
= + 
 
= + =
 
 avea Rσ σ= + 25.107 ksiaσ =  
 aveb Rσ σ= − 0.661 ksibσ = − 
 max Rτ =  max 12.88 ksiτ =  

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PROBLEM 7.46 
Solve Prob. 7.24 using Mohr’s circle. 
PROBLEM 7.24 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 as on top of the 34 -in.-diameter shaft. 
 
 
SOLUTION 
Equivalent force-couple system at center of shaft in section at point H: 
 24 lb (24)(6) 144 lb inV M= = = ⋅ 
 (24)(10) 240 lb inT = = ⋅ 
Shaft cross section: 
1
0.75 in. 0.375 in.
2
d c d= = = 
 4 4 4
1
0.031063 in 0.015532 in
2 2
J c I J
π= = = = 
Torsion: 3
(240)(0.375)
2.897 10 psi 2.897 ksi
0.031063
Tc
J
τ = = = × = 
Bending: 3
(144)(0.375)
3.477 10 psi 3.477 ksi
0.015532
Mc
I
σ = = = × = 
Transverse shear: At point H, stress due to transverse shear is zero. 
Resultant stresses: 
ave
2
2
2 2
3.477 ksi, 0, 2.897 ksi
1
( ) 1.738 ksi
2
2
1.738 2.897 3.378 ksi
x y xy
x y
x y
xyR
σ σ τ
σ σ σ
σ σ
τ
= = =
= + =
− 
= + 
 
= + =
 
 avea Rσ σ= + 5.116 ksiaσ =  
 aveb Rσ σ= − 1.640 ksibσ = − 
 max Rτ =  max 3.378 ksiτ =  

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PROBLEM 7.47 
Solve Prob. 7.25, using Mohr’s circle. 
PROBLEM 7.25 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 at point K. 
 
 
SOLUTION 
 
102
51 mm 45 mm
2 2
o
o i o
d
r r r t= = = = − = 
 ( )4 4 6 4 6 44.1855 10 mm 4.1855 10 m
2 o i
J r r
π −= − = × = × 
 6 4
1
2.0927 10 m
2
I J −= = × 
Force-couple system at center of tube in the plane containing points H and K: 
 
3
3 3
3 3
10 10 N
(10 10 )(200 10 ) 2000 N m
(10 10 )(150 10 ) 1500 N m
x
y
z
F
M
M
−
−
= ×
= × × = ⋅
= − × × = − ⋅
 
Torsion: 
3
3
6
2000 N m
51 10 m
(2000)(51 10 )
24.37 MPa
4.1855 10
y
o
xy
T M
c r
Tc
J
τ
−
−
−
= = ⋅
= = ×
×= = =
×
 
Note that the local x-axis is taken along a negative global z-direction. 
Transverse shear: Stress due to xV F= is zero at point K. 
Bending: 
3
6
(1500)(51 10 )
36.56 MPa
2.0927 10
z
y
M c
I
σ
−
−
×= = =
×
 
 Point K lies on compression side of neutral axis. 36.56 MPayσ = − 
 
 
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PROBLEM 7.47 (Continued) 
 
Total stresses at point K: 0, 36.56 MPa, 24.37 MPax y xyσ σ τ= = − = 
 ave
1
( ) 18.28 MPa
2 x y
σ σ σ= + = − 
 
2
2 30.46 MPa
2
x y
xyR
σ σ
τ
− 
= + = 
 
 
 max ave 18.28 30.46Rσ σ= + = − + 
 max 12.18 MPaσ = 
 min ave 18.28 30.46Rσ σ= − = − − 
 min 48.74 MPaσ = −  
 max Rτ = max 30.46 MPaτ =  

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PROBLEM 7.48 
Solve Prob. 7.26, using Mohr’s circle. 
PROBLEM 7.26 The axle of an automobile is acted upon by the 
forces and couple shown. Knowing that the diameter of the solid 
axle is 32 mm, determine (a) the principal planes and principal 
stresses at point H located on top of the axle, (b) the maximum 
shearing stress at the same point. 
 
SOLUTION 
 3
1 1
(32) 16 mm 16 10 m
2 2
c d −= = = = × 
Torsion: 
3
2Tc T
J c
τ
π
= = 
 6
3 3
2(350 N m)
54.399 10 Pa 54.399 MPa
(16 10 m)
τ
π −
⋅= = × =
×
 
Bending: 4 3 4 9 4(16 10 ) 51.472 10 m
4 4
I c
π π − −= = × = × 
 3(0.15m)(3 10 N) 450 N mM = × = ⋅ 
 
3
6
9
(450)(16 10 )
139.882 10 Pa 139.882 MPa
51.472 10
My
I
σ
−
−
×= − = − = − × = −
×
 
Top view Stresses 
 
 
 
 139.882 MPa, 0, 54.399 MPax y xyσ σ τ= − = = − 
Plotted points: : ( 139.882, 54.399); : (0, 54.399); : ( 69.941, 0)X Y C− − − 
 
ave
2
2
2
2
1
( ) 69.941 MPa
2
2
139.882
(54.399) 88.606 MPa
2
x y
x y
xyR
σ σ σ
σ σ
τ
= + = −
− 
= + 
 
− = + = 
 
 
 
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PROBLEM 7.48 (Continued) 
 
 
2 (2)( 54.399)
tan 2
139.882
0.77778
xy
p
x y
τ
θ
σ σ
−= =
− −
=
 
(a) 18.9aθ = ° , 108.9bθ = °  
 ave 69.941 88.606a Rσ σ= − = − − 158.5 MPaaσ = −  
 ave 69.941 88.606b Rσ σ= + = − + 18.67 MPabσ =  
(b) max Rτ =  max 88.6 MPaτ = 
 
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PROBLEM 7.49 
Solve Prob. 7.27, using Mohr’s circle. 
PROBLEM 7.27 For the state of plane stress shown, determine (a) the largest 
value of xyτ for which the maximum in-plane shearing stress is equal to or less 
than 12 ksi, (b) the corresponding principal stresses. 
 
SOLUTION 
The center of the Mohr’s circle lies at point C with coordinates 
 ( )10 8, 0 , 0 (1 ksi, 0).2 2x yσ σ+  −= =   
The radius of thecircle is max(in-plane) 12 ksi.τ = 
 
The stress point ( , )x xyσ τ− lies along the line 1 2X X of the Mohr circle diagram. The extreme points with 
12 ksiR ≤ are 1X and 2.X 
(a) The largest allowable value of xyτ is obtained from triangle CDX : 
 
2 2 2 2
1 2 1DX DX CX CD= = − 
 2 212 9xyτ = − 7.94 ksixyτ =  
(b) The principal stresses are 1 12aσ = + 13.00 ksiaσ =  
 1 12bσ = − 11.00 ksibσ = − 
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PROBLEM 7.50 
Solve Prob. 7.28, using Mohr’s circle. 
PROBLEM 7.28 For the state of plane stress shown, determine the largest value 
of yσ for which the maximum in-plane shearing stress is equal to or less than 
75 MPa. 
 
SOLUTION 
 60 MPa, ?, 20 MPax y xyσ σ τ= = = 
 
Given: 
 
max
2 2 2 2
75 MPa
2 150 MPa
(2)( ) 40 MPa
150 40 144.6 MPa
xy
R
XY R
DY
XD XY DY
τ
τ
= =
= =
= =
= − = − =
 
 60 144.6y x XDσ σ= + = + 205 MPayσ = 
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PROBLEM 7.51 
Solve Prob. 7.29, using Mohr’s circle. 
PROBLEM 7.29 Determine the range of values of xσ for which the maximum in-
plane shearing stress is equal to or less than 10 ksi. 
 
SOLUTION 
For the Mohr’s circle, point Y lies at (15 ksi, 8 ksi). The radius of limiting circles is 10 ksi.R = 
Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one. 
 1
2
10 ksi
10 ksi
C Y
C Y
=
=
 
 
Noting right triangles 1C DY and 2 ,C DY 
 
2 2 2 2 2 2
1 1 1 18 10 6 ksiC D DY C Y C D C D+ = + = = 
Coordinates of point C1 are (0, 15 6) (0, 9 ksi).− = 
Likewise, coordinates of point C2 are = (0, 15 6) (0, 21 ksi).+ = 
Coordinates of point X1: (9 6, 8) (3 ksi, 8 ksi)− − = − 
Coordinates of point X2: (21 6, 8) (27 ksi, 8 ksi)+ − = − 
The point ( , )x xyσ τ− must lie on the line X1 X2. 
Thus, 3 ksi 27 ksixσ≤ ≤  
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PROBLEM 7.52 
Solve Prob. 7.30, using Mohr’s circle. 
PROBLEM 7.30 For the state of plane stress shown, determine (a) the value of 
xyτ for which the in-plane shearing stress parallel to the weld is zero, (b) the 
corresponding principal stresses. 
 
SOLUTION 
Point X of Mohr’s circle must lie on X X′ ′′ so that 12 MPa.xσ = Likewise, point Y lies on line Y Y′ ′′ so 
that 2 MPa.yσ = The coordinates of C are 
 2 12 , 0 (7 MPa, 0).
2
+ = 
 
Counterclockwise rotation through 150° brings line CX to CB, where 0.τ = 
 
12 2
sec 30 sec 30 5.77 MPa
2 2
x yR
σ σ− −= ° = ° = 
(a) tan 30
2
x y
xy
σ σ
τ
−
= − ° 
 
12 2
tan 30
2
−= − ° 2.89 MPaxyτ = −  
(b) ave 7 5.77a Rσ σ= + = + 12.77 MPaaσ =  
 ave 7 5.77b Rσ σ= − = − 1.23 MPabσ = 
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PROBLEM 7.53 
Solve Problem 7.30, using Mohr’s circle and assuming that the weld forms an 
angle of 60° with the horizontal. 
PROBLEM 7.30 For the state of plane stress shown, determine (a) the value of 
xyτ for which the in-plane shearing stress parallel to the weld is zero, (b) the 
corresponding principal stresses. 
 
SOLUTION 
 Locate point C at 
12 2
7 MPa
2
σ += = with 0τ = . 
 Angle 120XCB = ° 
 
12 2
2 2
5 MPa
x yσ σ− −=
=
 
 
 
 
 
 
 
 
 
5sec60
10 MPa
R = °
=
 
 5 tan 60xyτ = − ° 
 8.66 MPaxyτ = −  
 avea Rσ σ= + 
 7 10= + 17 MPaaσ =  
 aveb Rσ σ= − 
 7 10= − 3 MPabσ = −  
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PROBLEM 7.54 
Determine the principal planes and the 
principal stresses for the state of plane stress 
resulting from the superposition of the two 
states of stress shown. 
 
SOLUTION 
Mohr’s circle for 1st stress state. 
 4 ksixσ = 
 4 ksiyσ = − 
 0xyτ = 
 
Resultant stresses: 
 4 4 8 ksixσ = + = 
 4 7 3 ksiyσ = − + = 
 6 0 6 ksixyτ = + = 
 ave
1
( ) 5.5 ksi
2 x y
σ σ σ= + = 
 
2 (2)(6)
tan 2 2.4
5
xy
p
x y
τ
θ
σ σ
= = =
−
 
 2 67.38pθ = ° 33.69aθ = °  
 123.69bθ = ° 

2
2
2 2
2
2.5 6 6.5 ksi
x y
xyR
σ σ
τ
− 
= + 
 
= + =

 avea Rσ σ= +  12 ksiaσ = 
 aveb Rσ σ= −  1 ksibσ = −  
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PROBLEM 7.55 
Determine the principal planes and the principal 
stresses for the state of plane stress resulting from 
the superposition of the two states of stress shown. 
 
SOLUTION 
Mohr’s circle for 2nd stress state: 
 
20 20 cos 60°
30 MPa
20 20 cos 60°
10 MPa
20 sin 60°
17.32 MPa
x
y
xy
σ
σ
τ
= +
=
= −
=
=
=
 
Resultant stresses: 
 35 30 65 MPaxσ = + = 
 25 10 35 MPayσ = + = 
 0 17.32 17.32 MPaxyτ = + = 
 ave
1 1
( ) (65 35) 50 MPa
2 2x y
σ σ σ= + = + = 
 
2 (2)(17.32)
tan 2 1.1547
65 35
xy
p
x y
τ
θ
σ σ
= = =
− −
 
 2 49.11 ,pθ = ° 24.6 , 114.6a bθ θ= ° = °  
 
2
2 22.91 MPa
2
x y
xyR
σ σ
τ
− 
= + = 
 
 
 avea Rσ σ= + 72.91 MPaaσ = 
 aveb Rσ σ= −  27.09 MPabσ = 
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PROBLEM 7.56 
Determine the principal planes and the principal 
stresses for the state of plane stress resulting from the 
superposition of the two states of stress shown. 
 
SOLUTION 
Mohr’s circle for 2nd stress state: 
 
0 0
0 0
0
1 1
 cos 2
2 2
1 1
 cos 2
2 2
1
 sin 2
2
x
y
xy
σ σ σ θ
σ σ σ θ
τ σ θ
= +
= −
=
 
Resultant stresses: 
 
0 0 0 0 0
0 0 00
0 0
ave 0
0
0 0 
1 1 3 1
 cos 2 cos 2
2 2 2 2
1 1 1 1
0 cos 2 cos 2
2 2 2 2
1 1
0 sin 2 sin 2
2 2
1
( )
2
2 sin 2
tan 2
cos 2
sin 2
tan
1 cos 2
x
y
xy
x y
xy
p
x y
σ σ σ σ θ σ σ θ
σ σ σ θ σ σ θ
τ σ θ σ θ
σ σ σ σ
τ σ θθ
σ σ σ σ θ
θ θ
θ
= + + = +
= + − = −
= + =
= + =
= =
− +
= =
+
 
 
1
2p
θ θ=  
 
2 2 2
2
0 0 0
2 2
0 0 0
1 1 1
 cos 2 sin 2
2 2 2 2
1 2
1 2 cos 2 + cos 2 sin 2 1 cos 2 |cos |
2 2
x y
xyR
σ σ
τ σ σ θ σ θ
σ θ θ θ σ θ σ θ
−     = + = + +     
    
= + + = + = 
 avea Rσ σ= + 0 0 cos aσ σ σ θ= + 
 aveb Rσ σ= − 0 0 cos bσ σ σ θ= − 
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PROBLEM 7.57 
Determine the principal planes and the principal stresses 
for the state of plane stress resulting from the 
superposition of the two states of stress shown. 
 
SOLUTION 
 Mohr’s circle for 2nd state of stress: 
 
0
0
0
x
y
x y
σ
σ
τ τ
′
′
′ ′
=
=
=
 
 
0 0 0 0
0 0
3 3
sin 60 sin 60
2 2
1
cos60
2
σ τ τ σ τ τ
τ τ τ
= − ° = − = ° =
= ° =
x y
xy
 
Resultant stresses: 
 
0 0 0 0
0 0 0
3 3 3 3
0 0
2 2 2 2
1 3
2 2
σ τ τ σ τ τ
τ τ τ τ
= − = − = + =
= + =
x y
xy
 
 
ave
22 2
2
0 0 0
1
( ) 0
2
3 3
3
2 2 2
x y
x y
xyR
σ σ σ
σ σ
τ τ τ τ
= + =
 −   = + = + =          
 
 
3
(2)
2 2
tan 2 3
3
xy
p
x y
τ
θ
σ σ
 
 
 = = = −
− −
 
 2 60pθ = − ° 30 60b aθ θ= − ° = °  
 avea Rσ σ= + 03aσ τ=  
 aveb Rσ σ= − 03bσ τ= − 
 
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PROBLEM 7.58 
For the state of stress shown, determine the 
range of values of θ for which the magnitude of 
the shearing stress x yτ ′ ′ is equal to or less than 
8 ksi 
 
SOLUTION 
 
ave
2
2
2 2
16 ksi, 0
6 ksi
1
( ) 8 ksi
2
2
( 8) (6) 10 ksi
2 (2) (6)
tan 2 0.75
16
2 36.870
18.435
x y
xy
x y
x y
xy
xy
p
x y
p
b
R
σ σ
τ
σ σ σ
σ σ
τ
τ
θ
σ σ
θ
θ
= − =
=
= + = −
− 
= + 
 
= − + =
= = = −
− −
= − °
= − °
 
8 ksix yτ ′ ′ ≤ for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle ϕ is 
calculated from 
 
8
sin 2 8 sin 2 0.8
10
R ϕ ϕ= = = 
 
2 53.130 26.565
18.435 26.565 45
18.435 26.565 8.13
90 45
90 98.13
k b
k b
u h
v k
ϕ ϕ
θ θ ϕ
θ θ ϕ
θ θ
θ θ
= ° = °
= − = − ° − ° = − °
= + = − + ° = °
= + ° = °
= + ° = °
 
Permissible range of θ : h kθ θ θ≤ ≤ 45 8.13θ− ° ≤ ≤ ° 
 θ θ θ≤ ≤u v 45 98.13θ° ≤ ≤ ° 
 Also, 135 188.13 and 225 278.13θ θ° ≤ ≤ ° ° ≤ ≤ ° 
 
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PROBLEM 7.59 
For the state of stress shown, determine the 
range of values of θ for which the normal 
stress xσ ′ is equal to or less than 50 MPa. 
 
SOLUTION 
 
ave
2
2
2 2
90 MPa, 0
60 MPa
1
( ) 45 MPa
2
2
45 60 75 MPa
2 (2) ( 60) 4
tan 2
90 3
2 53.13
26.565
x y
xy
x y
x y
xy
xy
p
x y
p
a
R
σ σ
τ
σ σ σ
σ σ
τ
τ
θ
σ σ
θ
θ
= =
= −
= + =
− 
= + 
 
= + =
−= = = −
−
= − °
= − °
 
 
50 MPaxσ ′ ≤ for states of stress corresponding to the arc HBK of Mohr’s circle. From the circle, 
 cos 2 50 45 5 MPaR ϕ = − = 
 
5
cos 2 0.066667
75
ϕ = = 
 
2 86.177 43.089
26.565 43.089 16.524
2 2 360 4 32.524 360 172.355 220.169
110.085
h a
k h
k
ϕ ϕ
θ θ ϕ
θ θ ϕ
θ
= ° = °
= + = − ° + ° = °
= + ° − = ° + ° − ° = °
= °
 
Permissible range of θ : h kθ θ θ≤ ≤ 
 16.524 110.085θ° ≤ ≤ °  
 Also, 196.524 290.085θ° ≤ ≤ ° 
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PROBLEM 7.60 
For the state of stress shown, determine the 
range of values of θ for which the normal stress 
xσ ′ is equal to or less than 100 MPa. 
 
SOLUTION 
 
ave
2
2
2 2
90 MPa, 0
60 MPa
1
( ) 45 MPa
2
2
45 60 75 MPa
2 (2) ( 60) 4
tan 2
90 3
2 53.13
26.565
x y
xy
x y
x y
xy
xy
p
x y
p
a
R
σ σ
τ
σ σ σ
σ σ
τ
τ
θ
σ σ
θ
θ
= =
= −
= + =
− 
= + 
 
= + =
−= = = −
−
= − °
= − °
 
100 MPaxσ ′ ≤ for states of stress corresponding to arc HBK of Mohr’s circle. From the circle, 
 cos2 100 45 55 MPaR ϕ = − = 
 
55
cos 2 0.73333
75
ϕ = = 
 
2 42.833 21.417
26.565 21.417 5.15
2 2 360 4 10.297 360 85.666 264.037
132.02
h a
k h
k
ϕ ϕ
θ θ ϕ
θ θ ϕ
θ
= ° = °
= + = − ° + ° = − °
= + ° − = − ° + ° − ° = °
= °
 
Permissible range of θ is h kθ θ θ≤ ≤ 
 5.15 132.02θ− ° ≤ ≤ °  
 Also, 174.85 312.02θ° ≤ ≤ ° 
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PROBLEM 7.61 
For the element shown, determine the range of values of xyτ for which the 
maximum tensile stress is equal to or less than 60 MPa. 
 
SOLUTION 
 
ave
20 MPa 120 MPa
1
( ) 70 MPa
2
x y
x y
σ σ
σ σ σ
= − = −
= + = −
 
Set max ave
max ave
60 MPa
130 MPa
R
R
σ σ
σ σ
= = +
= − =
 
But 
 
2
2
2
2
2 2
2
2
130 50
120 MPa
x x
xy
x x
xy
R
R
σ σ τ
σ στ
− = + 
 
− = −  
 
= −
=
 
Range of :xyτ 120 MPa 120 MPaxyτ− ≤ ≤ 
 
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PROBLEM 7.62 
For the element shown, determine the range of values of xyτ for which the 
maximum in-plane shearing stress is equal to or less than 150 MPa. 
 
SOLUTION 
 
20MPa 120 MPa
1 ( ) 50 MPa
2
x y
x y
σ σ
σ σ
= − = −
− =
 
Set max (in-plane) 150 MPaRτ = = 
But 
2
2
2
2
2 2
2
2
150 50
141.4 MPa
x y
xy
x y
xy
R
R
σ σ
τ
σ σ
τ
− 
= + 
 
− 
= −  
 
= −
=
 
Range of :xyτ 141.4 MPa 141.4 MPaxyτ− ≤ ≤ 
 
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