2 Hibbeler - Resistências dos Materiais - 7ªed - Soluções

Prévia do material em texto

0 1 2 3 4
200
0
200
400
600
800
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-22
Draw the shear and moment diagrams for the compound beam.The three segments are connected by
pins at B and E.
Given: a 2m� b 1m� F 3kN� 
LBE a 2 b˜�� 
w 0.8
kN
m
� 
LAB a b�� 
Solution: L 3a 4 b˜�� 
 Equilibrium :
 
Consider segment AB:
60B=0; A a b�( )˜ F b( )˜� 0= A b
a b� F˜� A 1.00 kN 
+ 6Fy=0; A B� F� 0= B F A�� B 2.00 kN 
Consider segment BE:
By symmetry, E B= D C=
+ 6Fy=0; 2C 2B� w a 2 b˜�( )˜� 0= C B w
2
a 2 b˜�( )˜�� C 3.60 kN 
D C� D 3.60 kN 
E B� E 2.00 kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a 2˜�(��� 
x4 a 2 b˜�( ) 1.01 a 2 b˜�( )˜� 2a 2 b˜�( )��� 
x5 2a 2 b˜�( ) 1.01 2a 2 b˜�( )˜� 2a 3 b˜�( )��� 
x6 2a 3 b˜�( ) 1.01 2a 3 b˜�( )˜� 2a 4 b˜�( )��� 
x7 2a 4 b˜�( ) 1.01 2a 4 b˜�( )˜� 3a 4 b˜�( )��� 
V1 x1� � A 1kN˜� V2 x2� � A F�( ) 1kN˜� V3 x3� � A F� w x3 LAB�� �˜�ª¬ º¼ 1kN˜� 
V4 x4� � A F� C� w x4 LAB�� �˜�ª¬ º¼ 1kN˜� V5 x5� � A F� C� D� w x5 LAB�� �˜�ª¬ º¼ ˜� 
V6 x6� � A F� C� D� w LBE� �˜�ª¬ º¼ 1kN˜� V7 x7� � A 2F� C� D� w LBE� �˜�ª¬ º¼ 1kN˜� 
M1 x1� � A x1˜kN m˜� M2 x2� � A x2� �˜ F x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � B� x3 LAB�� �˜ 0.5w x3 LAB�� �2˜�ª¬ º¼ 1kN m˜˜� 
M4 x4� � B� x4 LAB�� �˜ 0.5w x4 LAB�� �2˜� C x4 LAB� b�� �˜�ª¬ º¼ 1kN m˜˜� 
M5 x5� � B� x5 LAB�� �˜ 0.5w x5 LAB�� �2˜� C x5 LAB� b�� �˜� D x5 LAB� b� a�� �˜�ª¬ º¼˜� 
M6 x6� � E x6 LAB� LBE�� �˜ª¬ º¼ 1kN m˜˜� 
M7 x7� � E x7 LAB� LBE�� �˜ F x7 LAB� LBE� b�� �˜�ª¬ º¼ 1kN m˜˜� 
0 2 4 6 8 10
4
2
0
2
4
Distance (m)
S
h
ea
r 
(k
N
)
V1 x1� �
V2 x2� �
V3 x3� �
V4 x4� �
V5 x5� �
V6 x6� �
V7 x7� �
x1 x2� x3� x4� x5� x6� x7�
0 2 4 6 8 10
3
2
1
0
1
2
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
M4 x4� �
M5 x5� �
M6 x6� �
M7 x7� �
x1 x2� x3� x4� x5� x6� x7�
Problem 6-23
Draw the shear and moment diagrams for the beam.
Given: a 1.5m� Mo 30kN m˜� 
w 30
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; w� a˜ A� w a˜� B� 0=
 60B=0; Mo w a˜( ) 2.5a( )˜� A 2a( )˜� w a˜( ) 0.5a( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
57.50
32.50
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a��� x3 2a 1.01 2˜ a� 3a��� 
V1 x1� � w� x1˜kN� V2 x2� � w a˜( )� A�[ ] 1kN˜� 
V3 x3� � w a˜( )� A� w x3 2a�� �˜�ª¬ º¼ 1kN˜� 
M1 x1� � Mo 0.5w x1
2˜�
kN m˜� M2 x2� � Mo w a˜( ) x2 0.5a�� �˜� A x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � Mo w a˜( ) x3 0.5a�� �˜� A x3 a�� �˜� 0.5w x3 2a�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3 4
40
20
0
20
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 1 2 3 4
10
0
10
20
30
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-24
The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed
loading on the beam over its 0.6-m length. Draw the shear and moment diagrams for the beam if it
supports a uniform loading of 30 kN/m.
Given: a 0.3m� c 0.6m� 
b 2.4m� 
w 30
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A w b˜� qB� � c˜� 0=
 60A=0; w b˜( ) a 0.5b�( )˜ qB c˜� � a b� 0.5c�( )˜� 0=
Guess A 1kN� qB 1
kN
m
� 
A
qB
§¨
©
·
¹ Find A qB�� �� A 36.00 kN qB 60.00
kN
m
 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a b� c�( )��� 
V1 x1� � AkN� V2 x2� � A w x2 a�� �˜�ª¬ º¼ 1kN˜� 
V3 x3� � A w b˜� qB x3 a� b�� �˜�ª¬ º¼ 1kN˜� 
M1 x1� � A x1˜kN m˜� M2 x2� � A x2� �˜ 0.5w x2 a�� �2˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � A x3� �˜ w b˜( ) x3 a� 0.5 b˜�� �˜� 0.5qB x3 a� b�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 0.5 1 1.5 2 2.5 3
40
20
0
20
40
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 0.5 1 1.5 2 2.5 3
0
10
20
30
40
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-25
Draw the shear and moment diagrams for the beam. The two segments are joined together at B.
Given: a 0.9m� P 40kN� 
b 1.5m� 
w 50
kN
m
� 
c 2.4m� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A P� w c˜� C� 0=
 60B=0; w c˜( ) 0.5c( )˜ C c( )˜� 0=
Guess A 1kN� C 1kN� 
A
C
§¨
©
·
¹ Find A C�( )� 
A
C
§¨
©
·
¹
100
60
§¨
©
·
¹ kN 
MA P a˜ C w c˜�( ) a b�( )˜�� MA 180 kN m˜ 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a b� c�( )��� 
V1 x1� � AkN� V2 x2� � A P�( ) 1kN˜� V3 x3� � A P� w x3 a� b�� �˜�ª¬ º¼ 1kN˜� 
M1 x1� � MA� A x1˜�kN m˜� M2 x2� � MA� A x2˜� P x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � MA� A x3˜� P x3 a�� �˜� 0.5w x3 a� b�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3 4
50
0
50
100
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 1 2 3 4
200
100
0
100
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-26
Consider the general problem of a cantilevered beam subjected to n concentrated loads and a constant
distributed loading w. Write a computer program that can be used to determine the internal shear and
moment at any specified location x along the beam, and plot the shear and moment diagrams for the
beam. Show an application of the program using the values P1 = 4 kN, d1 = 2 m, w = 800 N/m, a1 = 2
m, a2 = 4 m, L = 4 m.
Problem 6-27
Determine the placement distance a of the roller support so that the largest absolute value of the
moment is a minimum. Draw the shear and moment diagrams for this condition.
Problem 6-28
Draw the shear and moment diagrams for the rod. Only vertical reactions occur at its ends A and B.
Given: a 900mm� A 360N� 
B 720N� 
w' 2.4
kN
m
� 
Solution:
+ 6Fy=0; A 0.5w'
xo
a
§¨
©
·
¹˜ xo˜� 0=
xo
A a˜
0.5w'
� 
xo 519.62 mm 
 60 Mmax A xo˜ 0.5w'
xo
a
§¨
©
·
¹˜ xo˜
xo
3
§¨
©
·
¹˜�� 
Mmax 124.71 N m˜ 
x 0 0.01 a˜� a��� V x( ) A 0.5w' x
a
§¨
©
·
¹˜ x˜�
ª«¬
º»¼
1
N
˜� M x( ) A x˜ 0.5w' x
a
§¨
©
·
¹˜ x˜
x
3
§¨
©
·
¹˜�
ª«¬
º»¼
1
N m˜� 
0 0.2 0.4 0.6 0.8
500
0
Distance (m)
S
h
ea
r 
(N
)
V x( )
x
0 0.2 0.4 0.6 0.8
0
50
100
Distance m)
M
o
m
en
t 
(N
-m
)
M x( )
x
Problem 6-29
Draw the shear and moment diagrams for the beam.
Given: Set L 1m� wo 1
kN
m
� 
a
L
3
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A 2 0.5wo� � a˜� wo a˜� B� 0=
 60B=0; A 3 a˜( )˜ 0.5 wo˜ a˜� � 2a a3�§¨© ·¹˜� wo a˜� � 1.5a( )˜� 0.5 wo˜ a˜� � 2a3§¨© ·¹˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.33
0.33
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a( )��� x3 2a( ) 1.01 2a( )˜� 3a(��� 
V1 x1� � A wo2
x1
a
§¨
©
·
¹˜ x1˜�
ª«¬
º»¼
1
kN
˜� V2 x2� � A 0.5 wo˜ a˜� wo x2 a�� �˜�ª¬ º¼ 1kN˜� 
V3 x3� � A 0.5 wo˜ a˜� wo a˜� wo x3 2a�� �˜ 1 0.5 x3 2a�a˜�§¨©
·
¹˜�
ª«¬
º»¼
1
kN
˜� 
0 0.2 0.4 0.6 0.8
0.4
0.2
0
0.2
0.4
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
M1 x1� � A x1˜ wo2
x1
a
§¨
©
·
¹˜ x1˜
x1
3
˜�ª«¬
º»¼
1
N m˜˜� 
M2 x2� � A x2˜ wo a˜2 x2 2a3�§¨© ·¹˜� 0.5wo x2 a�� �2˜�
ª«¬
º»¼
1
N m˜˜� 
M'3 x3� � wo2 x3 2 a˜�� �2˜ 1
x3 2 a˜�
a
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜� 
M3 x3� � A x3˜ wo a˜2 x3 2 a˜3�§¨© ·¹˜� wo a˜� � x3 1.5 a˜�� �˜� M'3 x3� ��
ª«¬
º»¼
1
N m˜˜� 
0 0.2 0.4 0.6 0.8
0
20
40
60
80
100120
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-30
Draw the shear and moment diagrams for the beam.
Set: L 1m� wo 1
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� 0.5 wo˜ L˜� 0=
 60B=0; A 2 L˜
3
˜
wo
2
L˜§¨©
·
¹
L
3
§¨
©
·
¹˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.25
0.25
§¨
©
·
¹ kN 
Let a
L
3
� 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 3a��� 
V1 x1� � wo2�
x1
L
˜ x1� �˜ª«¬
º»¼
1
kN
˜� V2 x2� � A wo2
x2
L
˜ x2� �˜�ª«¬
º»¼
1
kN
˜� 
M1 x1� � wo2�
x1
L
˜ x1� �˜ x13§¨©
·
¹˜
ª«¬
º»¼
1
kN m˜˜� 
M2 x2� � A x2 a�� �˜ wo2
x2
L
˜ x2� �˜ x23§¨©
·
¹˜�
ª«¬
º»¼
1
kN m˜˜� 
0 0.2 0.4 0.6 0.8
0.2
0
0.2
Distance (m)
S
h
ea
r 
 W
o
 (
k
N
)
V1 x1� �
V2 x2� �
x1 x2�
0 0.2 0.4 0.6 0.8
0
0.02
0.04
Distance (m)
M
o
m
en
t 
W
o
*
L
*
L
 (
k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Problem 6-31
The T-beam is subjected to the loading shown. Draw the shear and moment diagrams.
Given: a 2m� P 10kN� 
b 3m� 
w 3
kN
m
� 
c 3m� 
Solution:
 Equilibrium : Given
+ 6Fy=0; P� A� w c˜� B� 0=
 60B=0; P� a b� c�( )˜ A b c�( )˜� w c˜( ) 0.5c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
15.58
3.42
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a b� c�( )��� 
V1 x1� � P�kN� V2 x2� � P� A�( ) 1kN˜� V3 x3� � P� A� w x3 a� b�� �˜�ª¬ º¼ 1kN˜� 
M1 x1� � P� x1˜kN m˜� M2 x2� � P� x2˜ A x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � P� x3˜ A x3 a�� �˜� 0.5 w˜ x3 a� b�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3 4 5 6 7 8
10
5
0
5
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 2 4 6 8
25
20
15
10
5
0
5
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-32
The ski supports the 900-N (~90-kg) weight of the man. If the snow loading on its bottom surface is
trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the
ski.
Given: a 0.5m� P 900N� 
Solution:
 Equilibrium :
+ 6Fy=0; P� w
2
2a 4a�( )˜� 0=
w
P
3a
� w 600 N
m
 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a��� 
x3 2a 1.01 2a( )˜� 3a��� x4 3a 1.01 3a( )˜� 4a��� 
V1 x1� � w2
x1
a
§¨
©
·
¹˜ x1˜
ª«¬
º»¼
1
N
˜� 
V2 x2� � 0.5 w˜ a˜ w x2 a�� �˜�ª¬ º¼ 1N˜� 
V3 x3� � 0.5 w˜ a˜ w a˜� P� w x3 2a�� �˜�ª¬ º¼ 1N˜� 
V4 x4� � 0.5 w˜ a˜ w a˜� P� w a� w x4 3a�� �˜ 1 0.5 x4 3a�a˜�§¨©
·
¹˜�
ª«¬
º»¼
1
N
˜� 
0 0.5 1 1.5 2
400
200
0
200
400
Distance (m)
S
h
ea
r 
(N
)
V1 x1� �
V2 x2� �
V3 x3� �
V4 x4� �
x1 x2� x3� x4�
M1 x1� � w2
x1
a
§¨
©
·
¹˜ x1˜
x1
3
˜ª«¬
º»¼
1
N m˜˜� 
M2 x2� � w a˜2 x2 2a3�§¨© ·¹˜ w2 x2 a�� �2˜�ª«¬ º»¼ 1N m˜˜� 
M3 x3� � w a˜2 x3 2a3�§¨© ·¹˜ w2 x3 a�� �2˜� P x3 2a�� �˜�ª«¬ º»¼ 1N m˜˜� 
M'4 x4� � w2 x4 3 a˜�� �2˜ 1
x4 3 a˜�
a
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜� 
M4 x4� � w a˜2 x4 2 a˜3�§¨© ·¹˜ 2w a˜( ) x4 2 a˜�� �˜� P x4 2a�� �˜� M'4 x4� ��ª«¬ º»¼ 1N m˜˜� 
0 0.5 1 1.5 2
0
50
100
150
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
M4 x4� �
x1 x2� x3� x4�
Problem 6-33
Draw the shear and moment diagrams for the beam.
Given: L 9m� 
wo 50
kN
m
� 
a 0.5L� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A 2 0.5wo� � a˜� B� 0=
 60B=0; A L˜ 0.5 wo˜ a˜� � a a3�§¨© ·¹˜� 0.5 wo˜ a˜� � 2a3§¨© ·¹˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
112.50
112.50
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a( )��� 
V1 x1� � A wo x1˜ 1 0.5 x1a˜�§¨©
·
¹˜�
ª«¬
º»¼
1
kN
˜� 
V2 x2� � A 0.5 wo˜ a˜� wo2
x2 a�
a
§¨
©
·
¹˜ x2 a�� �˜�
ª«¬
º»¼
1
kN
˜� 
M1 x1� � A x1˜ wo2 x12˜ 1
x1
a
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜�
ª«¬
º»¼
1
kN m˜˜� 
M'2 x2� � wo2
x2 a�
a
§¨
©
·
¹˜ x2 a�� �˜
x2 a�
3
§¨
©
·
¹˜� 
M2 x2� � A x2˜ wo a˜2 x2 a3�§¨© ·¹˜� M'2 x2� ��
ª«¬
º»¼
1
kN m˜˜� 
0 5
100
0
100
Distance (m)
S
h
ea
r 
(k
N
)
V1 x1� �
V2 x2� �
x1 x2�
0 5
0
100
200
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Problem 6-34
Draw the shear and moment diagrams for the wood beam, and determine the shear and moment
throughout the beam as functions of x.
Given: a 1m� P 1kN� 
b 1.5m� 
w 2
kN
m
� 
c 1m� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A w b˜� B� 2P� 0=
 60B=0; P� a b�( )˜ A b˜� w b˜( ) 0.5b( )˜� P c˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
2.50
2.50
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a b� c�( )��� 
V1 x1� � P�kN� V2 x2� � P� A� w x2 a�� �˜�ª¬ º¼ 1kN˜� V3 x3� � P� A� w b˜� B�( ) 1kN˜� 
M1 x1� � P� x1˜kN m˜� M2 x2� � P� x2˜ A x2 a�� �˜� 0.5w x2 a�� �2˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � P� x3˜ A x3 a�� �˜� w b˜( ) x3 a� 0.5 b˜�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
0 0.5 1 1.5 2 2.5 3 3.5
2
1
0
1
2
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 0.5 1 1.5 2 2.5 3 3.5
1
0.8
0.6
0.4
0.2
0
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-35
The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kN/m
caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw
the shear and moment diagrams for the pin.
Given: L 100mm� w 0.4 kN
m
� 
a 0.2L� 
Solution:
 Equilibrium :
+ 6Fy=0; 2 0.5wo� � a˜ w 3a( )˜� 0=
wo 3w� wo 1.20
kN
m
 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 4a( )��� x3 4a( ) 1.01 4a( )˜� L��� 
V1 x1� � wo2
x1
a
§¨
©
·
¹˜ x1˜
ª«¬
º»¼
1
N
˜� V2 x2� � wo2 a˜ w x2 a�� �˜�ª«¬
º»¼
1
N
˜� 
V3 x3� � wo2 a˜ w 3a( )˜� wo x3 4a�� �˜ 1 0.5
x3 4a�
a
˜�§¨©
·
¹˜�
ª«¬
º»¼
1
N
˜� 
0 0.02 0.04 0.06 0.08
10
0
10
Distance (m)
S
h
ea
r 
(N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
M1 x1� � wo2
x1
a
§¨
©
·
¹˜ x1˜
x1
3
˜ª«¬
º»¼
1
N m˜˜� 
M2 x2� � wo a˜2 x2 2a3�§¨© ·¹˜ 0.5w x2 a�� �2˜�
ª«¬
º»¼
1
N m˜˜� 
M'3 x3� � wo2 x3 4 a˜�� �2˜ 1
x3 4 a˜�
a
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜� 
M3 x3� � wo a˜2 x3 2 a˜3�§¨© ·¹˜ w 3a( )˜ x3 2.5 a˜�� �˜�
ª«¬
º»¼ M'3 x3� ��
ª«¬
º»¼
1
N m˜˜� 
0 0.02 0.04 0.06 0.08
0
0.1
0.2
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-36
Draw the shear and moment diagrams for the beam.
Given: a 3.6m� b 1.8m� 
MA 2.25kN m˜� w 45
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� 0.5w b˜� 0=
 60B=0; MA A a˜� 0.5w b˜( )
b
3
§¨
©
·
¹˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
7.38�
47.88
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
V1 x1� � AkN� V2 x2� � A B� w x2 a�� �˜ 1 0.5 x2 a�b˜�§¨©
·
¹˜�
ª«¬
º»¼
1
kN
˜� 
M1 x1� � MA A x1˜�� � 1kN m˜˜� 
M2 x2� � MA A x2˜� B x2 a�� �˜� w2 x2 a�� �2˜ 1
x2 a�
b
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜�
ª«¬
º»¼
1
kN m˜˜� 
0 1 2 3 4 5
20
0
20
40
Distance (m)
S
h
ea
r 
(k
N
)
V1 x1� �
V2 x2� �
x1 x2�
0 1 2 3 4 5
30
2010
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Problem 6-37
The compound beam consists of two segments that are pinned together at B. Draw the shear and
moment diagrams if it supports the distributed loading shown.
Set: L 1m� a L
3
� w 1 kN
m
� 
Solution:
Consider segment AB.
 6MB=0; A 2a( )˜ w
2
2a
L
§¨
©
·
¹˜ 2 a˜( )˜
2a
3
§¨
©
·
¹˜� 0=
A
2w L˜
27
� A 0.0741 kN 
Consider whole beam ABC.
MC A L˜
w L˜
2
L
3
§¨
©
·
¹˜�=
MC
5� w L2˜
54
� 
MC 0.09259� kN m˜ 
x1 0 0.01 2a( )˜� 2a( )��� 
V x( ) A
w
2
x
L
§¨
©
·
¹˜ x˜�
ª«¬
º»¼
1
kN
˜� M x( ) A x˜ w
2
x
L
§¨
©
·
¹˜ x˜
x
3
§¨
©
·
¹˜�
ª«¬
º»¼
1
kN m˜� 
0 0.5 1
0.4
0.2
0
0.2
Distance (m)
S
h
ea
r 
W
*
L
 (
k
N
)
V x( )
x
0 0.5 1
0.1
0.05
0
Distance m)
M
o
m
en
t 
 w
*
L
*
L
 
(k
N
-m
)
M x( )
x
Problem 6-38
Draw the shear and moment diagrams for the beam.
Given: L 3m� wo 12
kN
m
� w1 18
kN
m
� 
Solution:
 Equilibrium :
+ 6Fy=0; B wo w1�� � L2˜� 
 60B=0; MB wo L˜� � L2˜ w1 wo�� � L2˜ L3§¨© ·¹˜�� 
B 45.00 kN MB 63.00 kN m˜ 
w' w1 wo�� 
x1 0 0.01 L˜� L��� 
V x( ) wo� x˜
w'
2
x
L
§¨
©
·
¹˜ x˜�
ª«¬
º»¼
1
kN
˜� M x( ) wo� x˜
x
2
§¨
©
·
¹˜
w'
2
x
L
§¨
©
·
¹˜ x˜
x
3
§¨
©
·
¹˜�
ª«¬
º»¼
1
kN m˜˜� 
0 1 2 3
40
20
0
Distance (m)
S
h
ea
r 
(k
N
)
V x( )
x
0 1 2 3
50
0
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M x( )
x
Problem 6-39
Draw the shear and moment diagrams for the beam and determine the shear and moment as functions
of x.
Given: a 3m� wo 200
N
m
� w1 400
N
m
� 
Solution: L 2a� w' w1 wo�� 
 Equilibrium : Given
+ 6Fy=0; A wo w1�� � a2˜� B� 0=
+ 60B=0; A 2 a˜( )˜ wo a˜� � a2˜� w' a2˜ a3§¨© ·¹˜� 0=
Guess A 1N� B 1N� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
200
700
§¨
©
·
¹ N 
Shear and Moment Functions :
For 0 < x < 3m,
V A� V 200 N Ans
M A x˜= M 200x( ) N˜ m˜= Ans
For 3m < x < 6m,
V' A wo x a�( )˜�
w'
2
x a�
a
§¨
©
·
¹˜ x a�( )˜�=
V' 500
100
3
x
2˜�§¨©
·
¹ N˜= Ans
M' A x˜ wo x a�( )˜
x a�
2
˜� w'
2
x a�
a
§¨
©
·
¹˜ x a�( )˜
x a�
3
§¨
©
·
¹˜�=
M' 600 500x� 100
9
x
3˜�§¨©
·
¹ N˜ m˜= Ans
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� L��� 
V1 x1� � AN� V2 x2� � A wo x2 a�� �˜� w'2
x2 a�
a
§¨
©
·
¹˜ x2 a�� �˜�
ª«¬
º»¼
1
N
˜� 
M1 x1� � A x1˜� � 1N m˜˜� 
M2 x2� � A x2˜ wo x2 a�� �˜ x2 a�2˜� w'2
x2 a�
L
§¨
©
·
¹˜ x2 a�� �˜
x2 a�
3
§¨
©
·
¹˜�
ª«¬
º»¼
1
N m˜˜� 
0 2 4 6
500
0
500
Distance (m)
S
h
ea
r 
(N
)
V1 x1� �
V2 x2� �
x1 x2�
0 2 4 6
0
200
400
600
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Problem 6-40
Determine the placement distance a of the roller support so that the largest absolute value of the
moment is a minimum. Draw the shear and moment diagrams for this condition.
Solution:
 Equilibrium :
6Fy=0; A 2P� B� 0=
 
+
60A=0; P L
2
˜ B a˜� P L˜� 0=
+ B
3 L˜
2a
P˜= A 4a 3L�
2a
P˜=
Internal Moment :
For positive moment, Mmax A
L
2
˜=
For negative moment, Mmin P� L a�( )˜=
When Mmax Mmin=
A
L
2
˜ P L a�( )˜= 4a 3L�
2a
P˜§¨©
·
¹
L
2
˜ P L a�( )˜=
4a 3L�( ) L˜ 4a L a�( )=
a
3
2
L= Ans
Set: L 1m� P 1kN� 
a
3
2
L� A 4a 3L�
2a
P˜� B 3 L˜
2a
P˜� 
a' 0.5L� b' a a'�� c' L a�� 
x1 0 0.01 a'˜� a'��� x2 a' 1.01 a'˜� a' b'�( )��� x3 a' b'�( ) 1.01 a' b'�( )˜� L��� 
V1 x1� � A 1kN˜� V2 x2� � A P�( ) 1kN˜� V3 x3� � A P� B�( ) 1kN˜� 
M1 x1� � A x1˜kN m˜� M2 x2� � A x2� �˜ P x2 a'�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � A x3� �˜ P x3 a'�� �˜� B x3 a'� b'�� �˜�ª¬ º¼ 1kN m˜˜� 
0 0.2 0.4 0.6 0.8
1
0.5
0
0.5
1
Distance (m)
S
h
ea
r 
P
 (
k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 0.2 0.4 0.6 0.8
0.1
0
0.1
Distance (m)
M
o
m
en
t 
P
*
L
 (
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-41
Draw the shear and moment diagrams for the beam.
Given: a 2m� w0 8
kN
m
� 
Solution: unit
kN
m
3
� w unit( ) 2˜ x2=
Wx
0
x
xwµ´¶ d=
Wx unit( )
0
x
x2x
2µ´¶ d
§¨
©¨
·
¹
˜= Wx unit( )
2
3
˜ x3˜=
Wa unit( )
2
3
˜ a3˜=
+ 6Fy=0; A
0
a
xwµ´¶ d� 0= A unit( )
0
a
x2x
2µ´¶ d˜� A 5.33 kN 
xc
0
a
xw x˜µ´¶ d
A
= xc
unit
A
0
a
x2 x
3˜µ´¶ d� xc 1.500 m 
MA A xc� �˜� MA 8.00 kN m˜ 
x 0 0.01 a˜� a��� V x( ) A unit( ) 2
3
˜ x3�ª«¬
º»¼
1
kN
˜� 
M x( ) MA� A x˜� unit( )
2
3
˜ x3˜ x˜ 1
xc
a
�§¨©
·
¹˜�
ª«¬
º»¼
1
kN m˜˜� 
0 0.5 1 1.5 2
0
2
4
6
Distance (m)
S
h
ea
r 
(k
N
)
V x( )
x
0 0.5 1 1.5 2
8
6
4
2
0
Distance m)
M
o
m
en
t 
(k
N
-m
)
M x( )
x
Problem 6-42
The truck is to be used to transport the concrete column. If the column has a uniform weight of w
(force/length), determine the equal placement a of the supports from the ends so that the absolute
maximum bending moment in the column is as small as possible. Also, draw the shear and moment
diagrams for the column.
Solution:
 Support Reactions: By symmetry, A=B=R
+ 6Fy=0; 2R wL� 0=
R 0.5w L˜=
Internal Moment :
For negative moment, Mmin 0.5� w a2˜=
For positive moment, Mmax w
L
2
˜§¨©
·
¹
L
4
˜ R L
2
a�§¨©
·
¹�=(at mid-span)
Mmax
w L˜
8
4a L�( )˜=
For optimal minimum: Mmax Mmin=
w L˜
8
4a L�( )˜ 1
2
� w˜ a2˜= 4a L�( ) L˜ 4� a2=
Let D a
L
= D2 D� 0.25� 0=
D 1
2
1� 12 4 0.25�( )˜��ª¬ º¼˜� 
D 0.2071 
a 0.2071L= Ans
Set: L 1m� w 1 kN
m
� 
a DL� R 0.5w L˜� b L 2a�� 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� L��� 
V1 x1� � w� x1˜kN� V2 x2� � R w x2˜�� � 1kN˜� V3 x3� � 2R w x3˜�� � 1kN˜� 
M1 x1� � w�2 x12˜§¨© ·¹ 1kN m˜˜� M2 x2� � R x2 a�� �˜ w2 x22˜�ª«¬ º»¼ 1kN m˜˜� 
M3 x3� � R x3 a�� �˜ R x3 a� b�� �˜� w2 x32˜�ª«¬ º»¼ 1kN m˜˜� 
0 0.2 0.4 0.6 0.8
0.4
0.2
0
0.2
0.4
Distance (m)
S
h
ea
r 
 w
*
L
 (
k
N
)
V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 0.2 0.4 0.6 0.8
0.04
0.02
0
0.02
0.04
Distance (m)
M
o
m
en
t 
 w
*
L
*
L
 (
k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-43
A member having the dimensions shown is to be used to resist an internal bending moment of M = 2
kN·m. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b)
about the y axis. Sketch the stress distribution for each case.
Given: d 120mm� b 60mm� 
Mz 2kN m˜� My 2kN m˜� 
Solution:
Iz
1
12
b˜ d3˜� Iy
1
12
d˜ b3˜� 
Maximum Stress: V M c
I
˜=
(a) About the z axis
ymax
d
2
� Vmax Mz� � ymaxIz˜� 
Vmax 13.89 MPa Ans
(b) About the y axis
zmax
b
2
� Vmax My� � zmaxIy˜� 
Vmax 27.78 MPa Ans
Problem 6-44
The steel rod having a diameter of 20 mm is subjected to an internal moment of M = 300 N·m.
Determine the stress created at points A and B. Also, sketch a threedimensional view of the stress
distribution acting over the cross section.
Given: d 20mm� M 300N m˜� 
T 45deg� 
Solution: I
S
4
d
2
§¨
©
·
¹
4
˜� V M y
I
˜=
yA
d
2
� VA M
yA
I
˜� 
VA 381.97 MPa Ans
yB
d
2
§¨
©
·
¹ sin T� �˜� VB M
yB
I˜� 
VB 270.09 MPa Ans
Problem 6-45
The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by th
stresses acting on both the top and bottom boards,A and B, of the beam.
Given: bf 200mm� tf 25mm� 
tw 25mm� dw 150mm� 
Solution: D dw 2tf�� 
I
1
12
bf D
3˜ bf 2tw�� � dw3˜�ª¬ º¼˜� 
Bending Stress: V M c
I
˜=
Set M 1kN m˜� 
co 0.5D� Vo M
co
I
˜� Vo 1.097143 MPa 
ci 0.5dw� Vi M
ci
I
˜� Vi 0.822857 MPa 
Resultant Force and Moment: For board A or B.
F
1
2
Vo Vi�� �˜ bf˜ tf˜� F 4.800 kN 
Centroid of force: Vi bf tf˜� �˜ tf2˜ 12 Vo Vi�� �˜ bf tf˜� �˜
tf
3
˜� F yc˜=
yc
1
F
Vi bf tf˜� �˜ tf2˜ 12 Vo Vi�� �˜ bf tf˜� �˜
tf
3
˜�ª«¬
º»¼� 
yc 11.905 mm 
M' F D 2yc�� �˜� 
M' 0.8457 kN m˜ 
Hence, %M
M'
M
100˜� 
%M 84.57 Ans
Problem 6-46
Determine the moment M that should be applied to the beam in order to create a compressive stress at
point D of VD = 30 MPa. Also sketch the stress distribution acting over the cross section and compute
the maximum stress developed in the beam.
Given: bf 200mm� tf 25mm� dw 150mm� 
tw 25mm� VD 30MPa� 
Solution: D dw 2tf�� 
I
1
12
bf D
3˜ bf 2tw�� � dw3˜�ª¬ º¼˜� 
I 91145833.33 mm
4 
Bending Stress: V M c
I
˜=
cD 0.5dw� VD M
cD
I
˜= M
VD I˜
cD
� 
M 36.46 kN m˜ Ans
cmax 0.5D� Vmax M
cmax
I
˜� 
Vmax 40.00 MPa Ans
Problem 6-47
The slab of marble, which can be assumed a linear elastic brittle material, has a specific weight of 24
kN/m3 and a thickness of 20 mm. Calculate the maximum bending stress in the slab if it is supported
(a) on its side and (b) on its edges. If the fracture stress is V f = 1.5 MPa, explain the consequences of
supporting the slab in each position. 
Given: t 20mm� L 1.5m� d 0.5m� 
J 24 kN
m
3
� Vf 1.5MPa� 
Solution: w J d˜ t˜� w 0.24 kN
m
 
Mmax
1
8
w˜ L2˜� 
Is
1
12
t˜ d3˜� Ie
1
12
d˜ t3˜� 
Maximum Stress: V M c
I
˜=
(a) Supported on its side
c1
d
2
� Vmax Mmax� � c1Is˜� 
Vmax 0.081 MPa Ans
(b) Supported on its edges
c2
t
2
� Vmax Mmax� � c2Ie˜� 
Vmax 2.025 MPa Ans
> Vf = 1.5 MPa
The marble slab will break if it is supported as in case (b).
Problem 6-48
The slab of marble, which can be assumed a linear elastic brittle material, has a specific weight of 24
kN/m3. If it is supported on its edges as shown in (b), determine the minimum thickness it should have
without causing it to break.The fracture stress is V f = 1.5 MPa. 
Given: L 1.5m� d 0.5m� 
J 24 kN
m
3
� Vf 1.5MPa� 
Solution: w J d˜ t˜=
Mmax
1
8
w˜ L2˜= Mmax
1
8
J d˜ t˜� �˜ L2˜=
Maximum Stress: c
t
2
= Ie
1
12
d˜ t3˜=
Vmax Mmax
c
Ie
˜= Vmax Mmax
6
d t
2˜
˜=
Thus, Vmax
1
8
J d˜ t˜� �˜ L2˜ 6
d t
2˜
˜=
t
3 J˜ L2˜
4 Vf˜
� 
t 27 mm Ans
Problem 6-49
A beam has the cross section shown. If it is made of steel that has an allowable stress of Vallow = 170
MPa, determine the largest internal moment the beam can resist if the moment is applied (a) about the z
axis, (b) about the y axis.
Given: bf 120mm� tf 5mm� d 120mm� 
tw 5mm� Vallow 170MPa� 
Solution: D d 2tf�� 
Iz
1
12
bf D
3˜ bf tw�� � d3˜�ª¬ º¼˜� 
Iy 2
1
12
tf˜ bf3˜§¨©
·
¹
1
12
d˜ tw3˜�� 
Bending Stress: Vallow M
c
I
˜=
(a) About the z axis
cz
D
2
� Mz Vallow� � Izcz˜� 
Mz 14.15 kN m˜ Ans
(b) About the y axis
cy
bf
2
� My Vallow� � Iycy˜� 
My 4.08 kN m˜ Ans
Problem 6-50
Two considerations have been proposed for the design of a beam. Determine which one will support a
moment of with the least amount of M = 150 kN·m bending stress. What is that stress? By what
percentage is it more effective?
Given: bf 200mm� dw 300mm� 
tf.a 15mm� tw.a 30mm� 
tf.b 30mm� tw.b 15mm� 
M 150kN m˜� 
Solution:
Section Property:
For section (a):
Da dw 2tf.a�� Ia
1
12
bf Da
3˜ bf tw.a�� � dw3˜�ª¬ º¼˜� 
For section (b):
Db dw 2tf.b�� Ib
1
12
bf Db
3˜ bf tw.b�� � dw3˜�ª¬ º¼˜� 
Maximum Bending Stress: V M c
I
˜=
For section (a):
cmax 0.5Da� Vmax M
cmax
Ia
˜� 
Vmax 114.35 MPa 
For section (b):
c'max 0.5Db� V'max M
c'max
Ib
˜� 
V'max 74.72 MPa Ans
By comparison, section (b) will have the least amount of bending stress. 
%eff
Vmax V'max�
V'max
100˜� 
%eff 53.03 Ans
Problem 6-51
The aluminum machine part is subjected to a moment of Determine the bending stress M = 75 N·m.
created at points B and C on the cross section. Sketch the results on a volume element located at each
of these points.
Given: bf 80mm� tf 10mm� 
tw 10mm� dw 40mm� 
M 75N m˜� 
Solution: D dw tf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf 2 dw tw˜� � 0.5dw tf�� �˜�
bf tf˜ 2dw tw˜�
� 
yc 17.50 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
tw dw
3˜˜ dw tw˜� � yc 0.5dw tf�� ��ª¬ º¼2˜�� 
I If 2Iw�� 
Bending Stress: V M c
I
˜=
At B: cB yc� VB M
cB
I
˜� VB 3.612 MPa Ans
At C: cC yc tf�� VC M
cC
I
˜� VC 1.548 MPa Ans
Problem 6-52
The aluminum machine part is subjected to a moment of M = 75 N·m. Determine the maximum tensil
and compressive bending stresses in the part.
Given: bf 80mm� tf 10mm� 
tw 10mm� dw 40mm� 
M 75N m˜� 
Solution: D dw tf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf 2 dw tw˜� � 0.5dw tf�� �˜�
bf tf˜ 2dw tw˜�
� 
yc 17.50 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
tw dw
3˜˜ dw tw˜� � yc 0.5dw tf�� ��ª¬ º¼2˜�� 
I If 2Iw�� 
Bending Stress: V M c
I
˜=
For compression:
cc yc� Vc_max M
cc
I
˜� Vc_max 3.612 MPa Ans
For tension:
ct D yc�� Vt_max M
ct
I
˜� Vt_max 6.709 MPa Ans
Problem 6-53
A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on
the cross section is M = 450 N·m, determine the resultant force the bending stress produces on the top
board A and on the side board B.
Given: bf 240mm� tf 15mm� 
tw 20mm� dw 200mm� 
M 450N m˜� 
Solution: D dw 2tf�� 
Iy
1
12
D bf
3˜ dw bf 2tw�� �3˜�ª¬ º¼˜� 
Bending Stress: V M c
I
˜=
co 0.5bf� Vo M
co
Iy
˜� Vo 0.410251 MPa 
ci 0.5bf tw�� Vi M
ci
Iy
˜� Vi 0.341876 MPa 
Resultant Force : For board A or B.
FA
Vo
2
bf
2
tf˜
§¨
©
·
¹˜
Vo
2
bf
2
tf˜
§¨
©
·
¹˜�� FA 0 kN Ans
FB
1
2
Vo Vi�� �˜ dw˜ tw˜� FB 1.504 kN Ans
Problem 6-54
The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment
M = 8 kN·m, determine the bending stress acting at points A and B, and show the results acting on
volume elements located at these points.
Given: b'f 50mm� tf 20mm� 
tw 20mm� dw 220mm� 
M 8kN m˜� 
Solution:
I
1
12
tw dw
3˜˜ 2 1
12
b'f˜ tf3˜§¨©
·
¹�� 
I 17813333.33 mm
4 
Bending Stress: V M c
I
˜=
At A: cA 0.5dw� VA M
cA
I
˜� VA 49.401 MPa Ans
At B: cB 0.5tf� VB M
cB
I
˜� VB 4.491 MPa Ans
Problem 6-55
The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the moment
M = 8 kN·m, determine the maximum bending stress in the beam, and sketch a three-dimensional view
of the stress distribution acting over the entire cross-sectional area.
Given: b'f 50mm� tf 20mm� 
tw 20mm� dw 220mm� 
M 8kN m˜� 
Solution:
I
1
12
tw dw
3˜˜ 2 1
12
b'f˜ tf3˜§¨©
·
¹�� 
I 17813333.33 mm
4 
Bending Stress: V M c
I
˜=
cmax 0.5dw� Vmax M
cmax
I
˜� Vmax 49.401 MPa Ans
At B:
cB 0.5tf� VB M
cB
I
˜� VB 4.491 MPa 
Problem 6-56
The beam is made from three boards nailed together as shown. If the moment acting on the cross
section is M = 1.5 kN·m, determinethe maximum bending stress in the beam. Sketch a
three-dimensional view of the stress distribution acting over the cross section.
Given: bf 250mm� b'f 150mm� tf 38mm� 
tw 25mm� d 300mm� 
M 1.5kN m˜� 
Solution: D d 2tf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf d tw˜� � 0.5d tf�� �˜� b'f tf˜� � D 0.5tf�� �˜�
bf tf˜ d tw˜� b'f tf˜�
� 
yc 159.71 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
tw˜ d3˜ d tw˜� � yc 0.5d tf�� ��ª¬ º¼2˜�� 
I'f
1
12
b'f˜ tf3˜ b'f tf˜� � yc D 0.5tf�� ��ª¬ º¼2˜�� 
I If Iw� I'f�� 
Bending Stress: V M c
I
˜=
At B: cmax D yc�� Vmax M
cmax
I
˜� Vmax 0.684 MPa Ans
At A: cA cmax tf�� VA M
cA
I
˜� VA 0.564 MPa 
At C: cC yc tf�� VC M
cC
I
˜� VC 0.385 MPa 
At D: cD yc� VD M
cD
I
˜� VD 0.505 MPa 
Problem 6-57
Determine the resultant force the bending stresses produce on the top board A of the beam if M = 1.5
kN·m.
Given: bf 250mm� b'f 150mm� tf 38mm� 
tw 25mm� d 300mm� 
M 1.5kN m˜� 
Solution: D d 2tf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf d tw˜� � 0.5d tf�� �˜� b'f tf˜� � D 0.5tf�� �˜�
bf tf˜ d tw˜� b'f tf˜�
� 
yc 159.71 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
tw˜ d3˜ d tw˜� � yc 0.5d tf�� ��ª¬ º¼2˜�� 
I'f
1
12
b'f˜ tf3˜ b'f tf˜� � yc D 0.5tf�� ��ª¬ º¼2˜�� 
I If Iw� I'f�� 
Bending Stress: V M c
I
˜=
At C: cC yc tf�� VC M
cC
I
˜� VC 0.385 MPa 
At D: cD yc� VD M
cD
I
˜� VD 0.505 MPa 
The resultant Force: For top board A
F 0.5 VC VD�� � bf tf˜� �˜� F 4.23 kN Ans
Problem 6-58
The control level is used on a riding lawn mower. Determine the maximum bending stress in the lever
at section a-a if a force of 100 N is applied to the handle. The lever is supported by a pin at A and a
wire at B. Section a-a is square, 6 mm by 6 mm.
Given: L 50mm� b 6mm� 
d 6mm� F 100N� 
Solution: I
1
12
b d
3˜� �˜� 
M F L˜� 
M 5.00 N m˜ 
Bending Stress: V M c
I
˜=
c
d
2
� Vmax M
c
I
˜� 
Vmax 138.89 MPa Ans
Problem 6-59
Determine the largest bending stress developed in the member if it is subjected to an internal bending
moment of M = 40 kN·m.
Given: bf 100mm� tf 10mm� rf 30mm� 
tw 10mm� dw 180mm� 
M 40kN m˜� 
Solution: D dw tf� 2rf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf dw tw˜� � 0.5dw tf�� �˜� S rf2˜§© ·¹ rf dw� tf�� �˜�
bf tf˜ dw tw˜� S rf2˜�
� 
yc 143.41 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
tw˜ dw3˜ dw tw˜� � yc 0.5dw tf�� ��ª¬ º¼2˜�� 
I'f
S
4
rf
4˜ S rf2˜§© ·¹ yc rf dw� tf�� ��ª¬ º¼2˜�� 
I If Iw� I'f�� 
Bending Stress: V M c
I
˜=
Maximum stress occurs at the bottom fibre.
cmax yc� Vmax M
cmax
I
˜� Vmax 128.51 MPa Ans
Problem 6-60
The tapered casting supports the loading shown. Determine the bending stress at points A and B. The
cross section at section a-a is given in the figure.
Given: La 250mm� P 750N� 
Lb 375mm� Lc 125mm� 
b 100mm� t 25mm� 
d 75mm� 
Solution:
 Equilibrium :
 60C=0; F1 2Lb Lc�� �˜ P Lc Lb�� �˜� P Lb� �˜� 0=
F1
P Lc Lb�� �˜ P Lb� �˜�
2 Lb˜ Lc�
� 
F1 750.00 N 
 Section a-a : D d 2t�� 
M F1 La˜� M 187.50 N m˜ 
I
1
12
b D
3˜ b d3˜�� �˜� 
Bending Stress: V M c
I
˜=
cA
D
2
� VA M
cA
I
˜� VA 0.918 MPa Ans
cB
d
2
� VB M
cB
I
˜� VB 0.551 MPa Ans
Problem 6-61
If the shaft in Prob. 6-1 has a diameter of 100 mm, determine the absolute maximum bending stress in
the shaft.
Given: a 250mm� b 800mm� 
F 24kN� do 100mm� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� F� 0=
 60A=0; F� a˜ B b˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
31.50
7.50�
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
M1 x1� � F� x1˜kN m˜� M2 x2� � F� x2� �˜ A x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
0 0.5 1
5
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Max. Moment : unit kN m˜� 
M1 a( ) 6.00� 
Bending Stress: M M1 a( ) unit˜� I
S do4˜
64
� c'
do
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 61.12 MPa Ans
Problem 6-62
If the shaft in Prob. 6-3 has a diameter of 40 mm, determine the absolute maximum bending stress in
the shaft.
Given: a 350mm� b 500mm� c 375mm� 
d 300mm� do 40mm� 
B 400N� C 550N� E 175N� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A D� B� C� E� 0=
 60D=0; A a b� c�( )˜ B b c�( )˜� C c˜� E d˜� 0=
Guess A 1N� D 1N� 
A
D
§¨
©
·
¹ Find A D�( )� 
A
D
§¨
©
·
¹
411.22
713.78
§¨
©
·
¹ N 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
x4 a b� c� 1.01 a b� c�( )˜� a b� c� d���� 
M1 x1� � A x1˜N m˜� M2 x2� � A x2� �˜ B x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � A x3� �˜ B x3 a�� �˜� C x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
M4 x4� � A x4� �˜ B x4 a�� �˜� C x4 a� b�� �˜� D x4 a� b� c�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.2 0.4 0.6 0.8 1 1.2 1.4
100
0
100
200
Distance (m)
M
o
m
en
t 
(N
-m
) M1 x1� �
M2 x2� �
M3 x3� �
M4 x4� �
x1 x2� x3� x4�
Max. Moment : unit N m˜� 
M3 a b�( ) 149.54 
Bending Stress: M M3 a b�( ) unit˜� I
S do4˜
64
� c'
do
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 23.8 MPa Ans
Problem 6-63
If the shaft in Prob. 6-6 has a diameter of 50 mm, determine the absolute maximum bending stress in
the shaft.
Given: a 125mm� b 600mm� c 75mm� 
F1 0.8kN� F2 1.5kN� do 50mm� 
Solution: L a b� c�� 
 Equilibrium : Given
+ 6Fy=0; A F1� F2� B� 0=
 60B=0; A L( )˜ F1 b c�( )˜� F2 c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.8156
1.4844
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � A x1˜N m˜� M2 x2� � A x2� �˜ F1 x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � A x3� �˜ F1 x3 a�� �˜� F2 x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.2 0.4 0.6
0
50
100
150
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit N m˜� 
M2 a b�( ) 111.33 
Bending Stress: M M2 a b�( ) unit˜� I
S do4˜
64
� c'
do
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 9.072 MPa Ans
Problem 6-64
If the shaft in Prob. 6-8 has a diameter of 30 mm and thickness of 10 mm, determine the absolute
maximum bending stress in the shaft.
Given: a 400mm� h 80mm� F 5kN� 
do 30mm� t 10mm� 
Solution: Given di do 2t�� 
 Equilibrium :
+ 6Fy=0; A C� 0=
 60C=0; A a˜ F h˜� 0=
Guess A 1N� C 1N� 
A
C
§¨
©
·
¹ Find A C�( )� 
A
C
§¨
©
·
¹
1.00�
1.00
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� M1 x1� � A x1˜N m˜� 
0 0.2 0
600
400
200
0
Distane (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
x1
Max. Moment : unit N m˜� 
M1 a( ) 400.00� 
Bending Stress: M M1 a( ) unit˜� I
S
64
do
4
di
4�§© ·¹˜� c'
do
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 152.8 MPa Ans
Problem 6-65
If the beam ACB in Prob. 6-9 has a square cross section, 150 mm by 150 mm, determine the absolute
maximum bending stress in the beam.
Given: a 1m� b 1m� 
c 1m� d 0.25m� 
ao 150mm� 
F1 75kN� F2 100kN� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A F1� B� 0=
 60C=0; A a b� c�( )˜ F1 b c�( )˜� F2 d( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
58.33
16.67
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � A x1˜kN m˜� M2 x2� � A x2� �˜ F1 x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � A x3� �˜ F1 x3 a�� �˜� F2 d˜�ª¬ º¼ 1kN m˜˜� 
0 0.5 1 1.5 2 2.5 3
0
50
100
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2�x3�
Max. Moment : unit kN m˜� 
M1 a( ) 58.33 
Bending Stress: M M1 a( ) unit˜� I
ao
4
12
� c'
ao
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 103.7 MPa Ans
Problem 6-66
If the crane boom ABC in Prob. 6-10 has a rectangular cross section with a base of 60 mm, determine
its required height h to the nearest multiples of 5 mm if the allowable bending stress is Vallow = 170 MPa.
Given: a 0.9m� b 1.5m� bo 60mm� 
c 1.2m� W 6kN� Vallow 170MPa� 
Solution: d a
2
c
2�� 
v
c
d
� h a
d
� 
 Equilibrium : Given
+
 
6Fy=0; Ay� B v˜� W� 0=
60A=0; B� v˜( ) a˜ W a b�( )˜� 0=
+ 6Fx=0; Ax B h˜� 0=
Guess Ax 1kN� Ay 1kN� B 1kN� 
Ax
Ay
B
§¨
¨¨
©
·
¸
¹
Find Ax Ay� B�� �� 
Ax
Ay
B
§¨
¨¨
©
·
¸
¹
12
10
20
§¨
©¨
·
¹
kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
M1 x1� � Ay� x1˜kN m˜� M2 x2� � Ay� x2� �˜ B v˜( ) x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
0 1 2
10
5
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Max. Moment : unit kN m˜� 
M1 a( ) 9.00� 
Bending Stress: M M1 a( ) unit˜� 
I
bo ho
3˜
12
= c'
ho
2
=
V M c'
I
˜=
ho
6 M
bo Vallow� �˜� ho 72.76 mm 
Use ho = 75mm Ans
Problem 6-67
If the crane boom ABC in Prob. 6-10 has a rectangular cross section with a base of 50 mm and a
height of 75 mm, determine the absolute maximum bending stress in the boom.
Given: a 0.9m� b 1.5m� bo 50mm� 
c 1.2m� W 6kN� ho 75mm� 
Solution: d a
2
c
2�� 
v
c
d
� h a
d
� 
 Equilibrium : Given
+
 
6Fy=0; Ay� B v˜� W� 0=
60A=0; B� v˜( ) a˜ W a b�( )˜� 0=
+ 6Fx=0; Ax B h˜� 0=
Guess Ax 1kN� Ay 1kN� B 1kN� 
Ax
Ay
B
§¨
¨¨
©
·
¸
¹
Find Ax Ay� B�� �� 
Ax
Ay
B
§¨
¨¨
©
·
¸
¹
12
10
20
§¨
©¨
·
¹
kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
M1 x1� � Ay� x1˜kN m˜� M2 x2� � Ay� x2� �˜ B v˜( ) x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
0 1 2
10
5
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Max. Moment : unit kN m˜� 
M1 a( ) 9.00� 
Bending Stress: M M1 a( ) unit˜� 
I
bo ho
3˜
12
� c'
ho
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� 
Vmax 192 MPa Ans
Problem 6-68
Determine the absolute maximum bending stress in the beam in Prob. 6-24. The cross section is
rectangular with a base of 75 mm and height of 100 mm.
Given: a 0.3m� b 2.4m� c 0.6m� 
bo 75mm� ho 100mm� w 30
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A w b˜� qB� � c˜� 0=
 60A=0; w b˜( ) a 0.5b�( )˜ qB c˜� � a b� 0.5c�( )˜� 0=
Guess A 1kN� qB 1
kN
m
� 
A
qB
§¨
©
·
¹ Find A qB�� �� A 36.00 kN qB 60.00
kN
m
 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a b� c�( )��� 
M1 x1� � A x1˜kN m˜� M2 x2� � A x2� �˜ 0.5w x2 a�� �2˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � A x3� �˜ w b˜( ) x3 a� 0.5 b˜�� �˜� 0.5qB x3 a� b�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3
0
20
40
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit kN m˜� 
b' 0.5 b˜� 
M2 a b'�( ) 32.40 
Bending Stress:
M M2 a b'�( ) unit˜� 
I
bo ho
3˜
12
� c'
ho
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� 
Vmax 259.2 MPa Ans
Problem 6-69
Determine the absolute maximum bending stress in the beam in Prob. 6-25. Each segment has a
rectangular cross section with a base of 100 mm and height of 200 mm.
Given: a 0.9m� b 1.5m� c 2.4m� 
bo 100mm� ho 200mm� 
P 40kN� w 50 kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A P� w c˜� C� 0=
 60B=0; w c˜( ) 0.5c( )˜ C c( )˜� 0=
Guess A 1kN� C 1kN� 
A
C
§¨
©
·
¹ Find A C�( )� 
A
C
§¨
©
·
¹
100
60
§¨
©
·
¹ kN 
MA P a˜ C w c˜�( ) a b�( )˜�� MA 180 kN m˜ 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� a b� c�( )��� 
M1 x1� � MA� A x1˜�kN m˜� M2 x2� � MA� A x2˜� P x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � MA� A x3˜� P x3 a�� �˜� 0.5w x3 a� b�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 2 4
200
0
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit kN m˜� 
c' 0.5 c˜� 
M3 a b� c'�( ) 36.00 
Bending Stress:
M M2 a b� c'�( ) unit˜� 
I
bo ho
3˜
12
� co
ho
2
� 
V M c'
I
˜= Vmax M
co
I
˜� 
Vmax 108 MPa Ans
Problem 6-70
Determine the absolute maximum bending stress in the 20-mm-diameter pin in Prob. 6-35.
Given: L 100mm� w 0.4 kN
m
� 
a 0.2L� do 20mm� 
Solution:
 Equilibrium :
+ 6Fy=0; 2 0.5wo� � a˜ w 3a( )˜� 0=
wo 3w� wo 1.20
kN
m
 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 4a( )��� x3 4a( ) 1.01 4a( )˜� L��� 
M1 x1� � wo2
x1
a
§¨
©
·
¹˜ x1˜
x1
3
˜ª«¬
º»¼
1
N m˜˜� M2 x2� �
wo a˜
2
x2
2a
3
�§¨©
·
¹˜ 0.5w x2 a�� �2˜�
ª«¬
º»¼
1
N m˜˜� 
M'3 x3� � wo2 x3 4 a˜�� �2˜ 1
x3 4 a˜�
a
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜� 
M3 x3� � wo a˜2 x3 2 a˜3�§¨© ·¹˜ w 3a( )˜ x3 2.5 a˜�� �˜�
ª«¬
º»¼ M'3 x3� ��
ª«¬
º»¼
1
N m˜˜� 
0 0.02 0.04 0.06 0.08
0
0.1
0.2
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit N m˜� 
M2 .5L( ) 0.260 
Bending Stress: M M2 .5L( ) unit˜� I
S
64
do
4˜� c'
do
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 0.331 MPa Ans
Problem 6-71
The member has a cross section with the dimensions shown. Determine the largest internal moment M
that can be applied without exceeding allowable tensile and compressive stresses of (V t )allow = 150
MPa and (V�c )allow = 100 MPa, respectively.
Given: bf 100mm� tf 10mm� rf 30mm� 
tw 10mm� dw 180mm� 
Vt.allow 150MPa� Vc.allow 100MPa� 
Solution: D dw tf� 2rf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf dw tw˜� � 0.5dw tf�� �˜� S rf2˜§© ·¹ rf dw� tf�� �˜�
bf tf˜ dw tw˜� S rf2˜�
� 
yc 143.41 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
tw˜ dw3˜ dw tw˜� � yc 0.5dw tf�� ��ª¬ º¼2˜�� 
I'f
S
4
rf
4˜ S rf2˜§© ·¹ yc rf dw� tf�� ��ª¬ º¼2˜�� 
I If Iw� I'f�� I 44639608.23 mm4 
Maximum Bending Stress: V M c
I
˜=
Assume failure due to tensile stress.
ct.max yc� Vt.max M
ct.max
I
˜= Mt
Vt.allow I˜
ct.max
� Mt 46.69 kN m˜ 
Assume failure due to compressive stress.
cc.max D yc�� Vc.max M
cc.max
I
˜= Mc
Vc.allow I˜
cc.max
� Mc 41.88 kN m˜ 
Mallow min Mt Mc�� �� 
Mallow 41.88 kN m˜ Ans
Problem 6-72
Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to th
concentrated forces. The sleeve bearings at A and B support only vertical forces.
Given: a 0.8m� b 1.2m� c 0.6m� 
F1 0.6kN� F2 0.4kN� 
do 30mm� 
Solution: L a b� c�� 
 Equilibrium : Given
+
 
6Fy=0; A F1� F2� B� 0=
60B=0; A b( )˜ F1 a b�( )˜� F2 c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.8
0.2
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � F1� x1˜N m˜� M2 x2� � F1� x2� �˜ A x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � F1� x3� �˜ A x3 a�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.5 1 1.5 2 2.5
600
400
200
0
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit N m˜� 
M1 a( ) 480.000� 
Bending Stress: M M1 a( ) unit˜� I
S
64
do
4˜� c'
do
2
� 
V M c'
I
˜= Vmax M
c'
I
˜� Vmax 181.1 MPa Ans
Problem 6-73
Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces.
The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is Vallow
160 MPa.
Given: a 0.8m� b 1.2m� c 0.6m� 
F1 0.6kN� F2 0.4kN�Vallow 160MPa� 
Solution: L a b� c�� 
 Equilibrium : Given
+
 
6Fy=0; A F1� F2� B� 0=
60B=0; A b( )˜ F1 a b�( )˜� F2 c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.8
0.2
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � F1� x1˜N m˜� M2 x2� � F1� x2� �˜ A x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � F1� x3� �˜ A x3 a�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.5 1 1.5 2 2.5
600
400
200
0
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit N m˜� 
M1 a( ) 480.000� 
Bending Stress: M M1 a( ) unit˜� I
S
64
do
4˜= c'
do
2
=
V M c'
I
˜= Vallow M
32
S do3˜
˜= do
3
32 M
S Vallow
� do 31.26 mm Ans
Problem 6-74
Determine the absolute maximum bending stress in the 40-mm-diameter shaft which is subjected to the
concentrated forces. The sleeve bearings at A and B support only vertical forces.
Given: a 300mm� b 450mm� 
c 375mm� do 40mm� 
F1 2kN� F2 1.5kN� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A F1� B� F2� 0=
 60B=0; A a b�( )˜ F1 b( )˜� F2 c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.45
3.05
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
V1 x1� � A 1kN˜� V2 x2� � A F1�� � 1kN˜� V3 x3� � A F1� B�� � 1kN˜� 
M1 x1� � A x1˜N m˜� M2 x2� � A x2� �˜ F1 x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � A x3� �˜ F1 x3 a�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.2 0.4 0.6 0.8 1
2
1
0
1
2
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 0.2 0.4 0.6 0.8 1
1000
500
0
500
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit N m˜� 
M3 a b�( ) 562.50� 
Bending Stress:
M M2 a b�( ) unit˜� I
S do4˜
64
� co
do
2
� 
V M c'
I
˜= Vmax M
co
I
˜� 
Vmax 89.52 MPa Ans
Problem 6-75
Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces.
The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is Vallow =
150 MPa.
Given: a 300mm� b 450mm� 
c 375mm� Vallow 150MPa� 
F1 2kN� F2 1.5kN� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A F1� B� F2� 0=
 60B=0; A a b�( )˜ F1 b( )˜� F2 c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.45
3.05
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � A x1˜N m˜� M2 x2� � A x2� �˜ F1 x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � A x3� �˜ F1 x3 a�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.5 1
1000
500
0
500
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit N m˜� 
M3 a b�( ) 562.50� 
Bending Stress:
M M3 a b�( )� unit˜� 
I
S do4˜
64
= co
do
2
=
V M
co
I
˜=
do
3
32M
S Vallow˜
� 
do 33.68 mm Ans
Problem 6-76
The bolster or main supporting girder of a truck body is subjected to the uniform distributed load.
Determine the bending stress at points A and B.
Given: L1 2.4m� L2 3.6m� 
b 150mm� tf 20mm� 
d 300mm� tw 12mm� 
w 25
kN
m
� 
Solution:
By symmetry : F1 R= F2 R=
 Equilibrium :
+ 6Fy=0; w� L1 L2�� �˜ 2R� 0=
R 0.5 w˜ L1 L2�� �˜� 
 
MAB R L1˜ 0.5w L12˜�� 
Section properties : D d 2 tf˜�� 
I
1
12
b D
3˜ b tw�� � d3˜�ª¬ º¼˜� 
Bending Stress: V M c
I
˜=
cB
d
2
� VB MAB
cB
I
˜� 
VB 89.6 MPa Ans
cA
d
2
tf�� VA MAB
cA
I
˜� 
VA 77.65 MPa Ans
Problem 6-77
A portion of the femur can be modeled as a tube having an inner diameter of 9.5 mm and an outer
diameter of 32 mm. Determine the maximum elastic static force P that can be applied to its center
without causing failure. Assume the bone to be roller supported at its ends. The V-H diagram for the
bone mass is shown and is the same in tension as in compression.
Given: L1 100mm� L2 100mm� 
di 9.5mm� do 32mm� 
He 0.02
mm
mm
� Ve 8.75MPa� 
Hr 0.06
mm
mm
� Vr 16.1MPa� 
Solution:
By symmetry : R 0.5P=
Mmax R L1˜= Mmax 0.5P L1˜=
Section properties :
I
S
64
do
4
di
4�§© ·¹˜� 
Bending Stress: V M c
I
˜= c
do
2
=
M
2V I˜
do
=
Vmax Ve=Requires:
P
2Ve I˜
0.5 L1˜� � do˜� 
P 558.6 N Ans
Problem 6-78
If the beam in Prob. 6-20 has a rectangular cross section with a width of 200 mm and a height of 400
mm, determine the absolute maximum bending stress in the beam.
Given: a 2.4m� b 1.2m� 
P1 50kN� 
w 30
kN
m
� M2 60kN m˜� P2 40kN� 
bo 200mm� do 400mm� 
Solution:
 Equilibrium :
+
A w a˜ P1� P2�� A 162 kN 6Fy=0;
 MA w a˜( ) 0.5a( )˜ P1 a˜� P2 a b�( )˜� M2�� 60A=0;
MA 410.40 kN m˜ 
 As indicated in the moment diagram, the maximum moment is MA.
Section properties :
I
1
12
bo˜ do3˜� 
Bending Stress: V M c
I
˜=
co
do
2
� Vmax MA
co
I
˜� 
Vmax 76.95 MPa Ans
Problem 6-79
If the shaft has a diameter of 37.5 mm, determine the absolute maximum bending stress in the shaft.
Given: a 450mm� b 600mm� c 300mm� 
do 37.5mm� F1 1000N� F2 750N� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A 2F1� B� 2F2� 0=
 60B=0; 2� F1 a b�( )˜ A b( )˜� 2F2 c( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
2.75
0.75
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � 2� F1 x1˜kN m˜� M2 x2� � 2� F1 x2� �˜ A x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � 2� F1 x3� �˜ A x3 a�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
0 0.2 0.4 0.6 0.8 1 1.2
1
0.5
0
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit kN m˜� 
M1 a( ) 0.900� 
Bending Stress:
Mmax M1 a( ) unit˜� I
S do4˜
64
� co
do
2
� 
V M c'
I
˜= Vmax Mmax
co
I
˜� Vmax 173.84 MPa Ans
Problem 6-80
If the beam has a square cross section of 225 mm on each side, determine the absolute maximum
bending stress in the beam.
Given: a 2.5m� b 2.5m� 
P 6kN� w 15 kN
m
� 
bo 225mm� do 225mm� 
Solution:
 Equilibrium :
+
A w a˜ P�� A 43.5 kN 6Fy=0;
 MA w a˜( ) 0.5a( )˜ P a b�( )˜�� 60A=0;
MA 76.88 kN m˜ 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
M1 x1� � MA� A x1˜� 0.5w x12˜�§© ·¹ 1kN m˜˜� 
M2 x2� � MA� A x2˜� w a˜( ) x2 0.5 a˜�� �˜�ª¬ º¼ 1kN m˜˜� 
0 2 4
50
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
As indicated in the moment diagram, the
maximum moment is MA.
Section properties :
I
1
12
bo˜ do3˜� 
Bending Stress: V M c
I
˜=
co
do
2
� Vmax MA
co
I
˜� 
Vmax 40.49 MPa Ans
Problem 6-81
The beam is subjected to the load P at its center. Determine the placement a of the supports so that th
absolute maximum bending stress in the beam is as large as possible. What is this stress?
Solution:
 Equilibrium : By symmetry, A=B=R
+ 6Fy=0; 2R P� 0=
R 0.5P=
Max. Moment : 
Mmax R 0.5L a�( )˜=
Mmax 0.5P 0.5L a�( )˜=
For the largest Mmax require, a 0� Ans
Mmax
P L˜
4
=
Bending Stress: V M c'
I
˜=
I
b d
3˜
12
= c'
d
2
= Vmax Mmax
c'
I
˜=
Vmax
P L˜
2
3
b d
2˜
˜= Ans
Problem 6-82
If the beam in Prob. 6-23 has a cross section as shown, determine the absolute maximum bending
stress in the beam.
Given: Mo 30kN m˜� w 30
kN
m
� 
a 1.5m� b 100mm� tf 12mm� 
d 168mm� tw 6mm� 
Solution:
 Equilibrium : Given
+ 6Fy=0; w� a˜ A� w a˜� B� 0=
 60B=0; Mo w a˜( ) 2.5a( )˜� A 2a( )˜� w a˜( ) 0.5a()˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
57.50
32.50
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a��� x3 2a 1.01 2˜ a� 3a��� 
M1 x1� � Mo 0.5w x1
2˜�
kN m˜� M2 x2� � Mo w a˜( ) x2 0.5a�� �˜� A x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � Mo w a˜( ) x3 0.5a�� �˜� A x3 a�� �˜� 0.5w x3 2a�� �2˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3 4
0
20
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
As indicated in the moment
diagram, the maximum
moment is Mo.
Section properties : D d 2 tf˜�� I
1
12
b D
3˜ b tw�� � d3˜�ª¬ º¼˜� 
Bending Stress: V M c
I
˜=
co
D
2
� Vmax Mo
co
I
˜� Vmax 131.87 MPa Ans
Problem 6-83
The pin is used to connect the three links together. Due to wear, the load is distributed over the top and
bottom of the pin as shown on the free-body diagram. If the diameter of the pin is 10 mm, determine
the maximum bending stress on the cross-sectional area at the center section a-a. For the solution it is
first necessary to determine the load intensities w1 and w2 .
Given: P 2kN� 
a 25mm� b 37.5mm� 
do 10mm� 
Solution:
 Maa P�
a
3
b
2
�§¨©
·
¹˜ P
b
2
§¨
©
·
¹˜�� 
Maa 16.6667� N m˜ 
Bending Stress:
I
S do4˜
64
� co
do
2
� 
V M c'
I
˜= Vmax Maa
co
I
˜� 
Vmax 169.77 MPa Ans
Problem 6-84
A shaft is made of a polymer having an elliptical cross-section. If it resists an internal moment of M =
50 N·m determine the maximum bending stress developed in the material (a) using the flexure formul
where Iz = 1/4 S (0.08 m)(0.04 m)3, (b) using integration. Sketch a three-dimensional view of the
stress distribution acting over the cross-sectional area.
Given: a 80mm� b 40mm� 
Mz 50N m˜� 
Solution:
a) Using the flexure formula, cmax b� 
Iz
S a˜ b3˜
4
� Iz 4021238.60 mm4 
Vmax Mz
cmax
Iz
˜� 
Vmax 0.497 MPa Ans
b) Using integration,
y
2
b
2
z
2
a
2
� 1= z a
b
b
2
y
2�˜=
Izo
A
Ay
2µ´¶ d
§¨
©¨
·
¹
= Izo
b�
b
yy
2
2z( )˜µ´¶ d=
Izo
b�
b
y2y
2 a
b
b
2
y
2�˜§¨©
·
¹˜
µ´
µ¶
d� 
Izo 4021631.98 mm
4 
Bending Stress: V'max Mz
cmax
Izo
˜� 
V'max 0.497 MPa Ans
Problem 6-85
Solve Prob. 6-84 if the moment is M = 50 N·m, applied about the y axis instead of the z axis. Here
Iy = 1/4 S (0.04 m)(0.08 m)3.
Given: a 80mm� b 40mm� 
My 50N m˜� 
Solution:
a) Using the flexure formula, cmax a� 
Iy
S b˜ a3˜
4
� Iy 16084954.39 mm4 
Vmax My
cmax
Iy
˜� 
Vmax 0.249 MPa Ans
b) Using integration,
y
2
b
2
z
2
a
2
� 1= y b
a
a
2
z
2�˜=
Iyo
A
Az
2µ´¶ d
§¨
©¨
·
¹
= Iyo
b�
b
zz
2
2y( )˜µ´¶ d=
Iyo
a�
a
z2z
2 b
a
a
2
z
2�˜§¨©
·
¹˜
µ´
µ¶
d� 
Iyo 16086527.94 mm
4 
Bending Stress: V'max My
cmax
Iyo
˜� 
V'max 0.249 MPa Ans
Problem 6-86
The simply supported beam is made from four 16-mm-diameter rods, which are bundled as shown.
Determine the maximum bending stress in the beam due to the loading shown.
Given: L1 0.5m� L2 1.5m� 
do 16mm� P 400N� 
Solution:
By symmetry : F1 R= F2 R=
 Equilibrium :
+ 6Fy=0; 2� P 2 R˜� 0= R P� 
 
Mmax R L1 0.5L2�� �˜ P 0.5L2� �˜�� 
Section properties : A
S
4
do
2˜� 
I 4
S
64
do
4˜ A
do
2
§¨
©
·
¹
2
˜�
ª«¬
º»¼˜� 
Bending Stress: V M c
I
˜=
cmax do� Vmax Mmax
cmax
I
˜� 
Vmax 49.74 MPa Ans
Problem 6-87
Solve Prob. 6-86 if the bundle is rotated 45° and set on the supports.
Given: L1 0.5m� L2 1.5m� 
do 16mm� P 400N� 
Solution:
By symmetry : F1 R= F2 R=
 Equilibrium :
+ 6Fy=0; 2� P 2 R˜� 0= R P� 
 
Mmax R L1 0.5L2�� �˜ P 0.5L2� �˜�� 
Section properties : A
S
4
do
2˜� d' 0.5 do2 do2�˜� 
I 2
S
64
do
4˜§¨©
·
¹˜ 2
S
64
do
4˜ A d'2˜�§¨©
·
¹˜�� 
Bending Stress: V M c
I
˜=
cmax d'
do
2
�� Vmax Mmax
cmax
I
˜� 
Vmax 60.04 MPa Ans
Problem 6-88
The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed load
w0 that it can support so that the maximum bending stress in the beam does not exceed V max = 150
MPa.
Given: L1 4m� L2 4m� 
b 200mm� t 8mm� 
d 250mm� Vmax 150MPa� 
Solution:
By symmetry : F1 R= F2 R=
 Equilibrium :
+ 6Fy=0; 0.5� wo L1 L2�� �˜ 2R� 0=
R 0.25 wo˜ L1 L2�� �˜=
 
Mmax R L1˜ 0.5wo L1˜� � L13˜�=
Mmax 0.25 wo˜ L1 L2�� �˜ L1˜ 0.5wo L1˜� � L13˜�=
Section properties : D d 2 t˜�� 
I
1
12
b D
3˜ b t�( ) d3˜�ª¬ º¼˜� 
Bending Stress: V M c
I
˜= cmax
D
2
� 
Vmax 0.25 wo˜ L1 L2�� �˜ L1˜ 0.5wo L1˜� � L13˜�ª«¬
º»¼
D
2I
˜=
wo
Vmax 2I( )˜
D
1
0.25 L1 L2�� �˜ L1˜ 0.5L1� � L13˜�
˜� 
wo 13.47
kN
m
 Ans
Problem 6-89
The steel beam has the cross-sectional area shown. If w0 = 10 kN/m, determine the maximum bending
stress in the beam.
Given: L1 4m� L2 4m� 
b 200mm� t 8mm� 
d 250mm� 
wo 10
kN
m
� 
Solution:
By symmetry : F1 R= F2 R=
 Equilibrium :
+ 6Fy=0; 0.5� wo L1 L2�� �˜ 2R� 0=
R 0.25 wo˜ L1 L2�� �˜� 
 
Mmax R L1˜ 0.5wo L1˜� � L13˜�� 
Section properties : D d 2 t˜�� 
I
1
12
b D
3˜ b t�( ) d3˜�ª¬ º¼˜� 
Bending Stress: V M c
I
˜=
cmax
D
2
� Vmax Mmax
cmax
I
˜� 
Vmax 111.38 MPa Ans
Problem 6-90
The beam has a rectangular cross section as shown. Determine the largest load P that can be
supported on its overhanging ends so that the bending stress in the beam does not exceed V max = 10
MPa.
Given: a 0.5m� bo 50mm� 
Vallow 10MPa� do 100mm� 
Solution:
By symmetry : A R= B R=
 Equilibrium :
+ 6Fy=0; 2� P 2 R˜� 0= R P=
 
Mmax P� 1.5a( )˜ R 0.5a( )˜�= Mmax P� a˜=
Section properties :
I
1
12
bo do
3˜§© ·¹˜� 
Bending Stress: V M c
I
˜= cmax
do
2
� 
V P a˜( )
do
2I
˜= P
Vallow 2I( )˜
a do˜
� P 1.67 kN Ans
R P� 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a��� x3 2a 1.01 2a( )˜� 3a��� 
M1 x1� � P� x1˜kN m˜� M2 x2� � P� x2� �˜ R x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � P� x3� �˜ R x3 a�� �˜� R x3 2a�� �˜�ª¬ º¼ 1kN m˜˜� 
0 0.5 1 1.5
1
0.5
0
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-91
The beam has the rectangular cross section shown. If P = 1.5 kN, determine the maximum bending
stress in the beam. Sketch the stress distribution acting over the cross section.
Given: a 0.5m� bo 50mm� 
P 1.5kN� do 100mm� 
Solution:
By symmetry : A R= B R=
 Equilibrium :
+ 6Fy=0; 2� P 2 R˜� 0= R P� 
 
Mmax P� 1.5a( )˜ R 0.5a( )˜�= Mmax P� a˜� Mmax 0.75� kN m˜ 
Section properties :
I
1
12
bo do
3˜§© ·¹˜� 
Bending Stress:
cmax
do
2
� Vmax Mmax
cmax
I
˜� Vmax 9.00 MPa Ans
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a��� x3 2a 1.01 2a( )˜� 3a��� 
M1 x1� � P� x1˜kN m˜� M2 x2� � P� x2� �˜ R x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � P� x3� �˜ R x3 a�� �˜� R x3 2a�� �˜�ª¬ º¼ 1kN m˜˜� 
0 0.5 1 1.5
1
0.5
0
Distance (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-92
The beam is subjected to the loading shown. If its cross-sectional dimension a = 180 mm, determine
the absolute maximum bending stress in the beam.
Given: L1 2m� L2 1m� 
P 60kN� a 180mm� 
w 40
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� P� w L1˜� 0=
 60A=0; P L1 L2�� �˜ B L1˜� 0.5w L12˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
10.00
130.00
§¨
©
·
¹ kN 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2���� 
M1 x1� � A x1˜ 0.5w x12˜�§© ·¹ 1kN m˜˜� 
M2 x2� � A x2˜ w L1˜� � x2 0.5 L1˜�� �˜� B x2 L1�� �˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 360
40
20
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Max. Moment : unit kN m˜� 
M1 L1� � 60.000� 
Mmax M1 L1� �� unit˜� 
Mmax 60.00 kN m˜ 
Section properties :
yc
6 yi

Ai˜� �˜
6 Ai� �˜=
yc
a
a
3
˜§¨©
·
¹
a
6
˜ a
2
2a
3
˜§¨©
·
¹
2a
3
§¨
©
·
¹˜�
a
a
3
˜§¨©
·
¹
a
2
2a
3
˜§¨©
·
¹�
� yc 75.00 mm 
Iw
1
12
a
2
§¨
©
·
¹˜
2a
3
§¨
©
·
¹
3
˜ a
2
2a
3
˜§¨©
·
¹ yc
2a
3
§¨
©
·
¹�
ª«¬
º»¼
2
˜�� 
If
1
12
a( )˜ a
3
§¨
©
·
¹
3
˜ a a
3
˜§¨©
·
¹ yc
a
6
§¨
©
·
¹�
ª«¬
º»¼
2
˜�� 
I If Iw�� 
Bending Stress: V M c
I
˜= cmax a yc�� 
Vmax Mmax� � cmaxI˜� 
Vmax 105.11 MPa Ans
Problem 6-93
The beam is subjected to the loading shown. Determine its required cross-sectional dimension a, if the
allowable bending stress for the material is Vallo w = 150 MPa.
Given: L1 2m� L2 1m� 
P 60kN� Vallow 150MPa� 
w 40
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� P� w L1˜� 0=
 60A=0; P L1 L2�� �˜ B L1˜� 0.5w L12˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
10.00
130.00
§¨
©
·
¹ kN 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2���� 
M1 x1� � A x1˜ 0.5w x12˜�§© ·¹ 1kN m˜˜� 
M2 x2� � A x2˜ w L1˜� � x2 0.5 L1˜�� �˜� B x2 L1�� �˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3
60
40
20
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Max. Moment : unit kN m˜� 
M1 L1� � 60.000� 
Mmax M1 L1� �� unit˜� 
Mmax 60.00 kN m˜ 
Section properties :
yc
6 yi

Ai˜� �˜
6 Ai� �˜=
yc
a
a
3
˜§¨©
·
¹
a
6
˜ a
2
2a
3
˜§¨©
·
¹
2a
3
§¨
©
·
¹˜�
a
a
3
˜§¨©
·
¹
a
2
2a
3
˜§¨©
·
¹�
= yc
5a
12
=
Iw
1
12
a
2
§¨
©
·
¹˜
2a
3
§¨
©
·
¹
3
˜ a
2
2a
3
˜§¨©
·
¹ yc
2a
3
§¨
©
·
¹�
ª«¬
º»¼
2
˜�= Iw
43a
4
1296
=
If
1
12
a( )˜ a
3
§¨
©
·
¹
3
˜ a a
3
˜§¨©
·
¹ yc
a
6
§¨
©
·
¹�
ª«¬
º»¼
2
˜�= If
31a
4
1296
=
I If Iw�= I
37a
4
648
=
Bending Stress: V M c
I
˜= cmax a yc�= cmax
7a
12
=
I Mmax� � cmaxVallow˜=
a
3
648Mmax
7
12
§¨
©
·
¹˜
37Vallow
� 
a 159.88 mm Ans
Problem 6-94
The wing spar ABD of a light plane is made from 2014T6 aluminum and has a cross-sectional area of
1000 mm2, a depth of 80 mm, and a moment of inertia about its neutral axis of 1.662 (106) mm4.
Determine the absolute maximum bending stress in the spar if the anticipated loading is to be as shown.
Assume A, B, and C are pins. Connection is made along the central longitudinal axis of the spar.
Given: a 1m� b 2m� 
A 1000m
2� wo 15
kN
m
� 
I 1.662 10
6� �˜ mm4� 
do 80mm� 
Solution: L a b�� 
 Equilibrium : Given
+ 6Fy=0; A B� 0.5wo L( )˜� 0=
 60A=0; B a˜ 0.5wo L˜� � L3§¨© ·¹˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
0.00�
22.50
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
M1 x1� � A x1˜ wo2 x12˜ 1
x1
L
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜�
ª«¬
º»¼
1
kN m˜˜� 
M2 x2� � A x2˜ B x2 a�� �˜�ª¬ º¼ wo2 x22˜ 1
x2
L
§¨
©
·
¹
1
3
˜�ª«¬
º»¼˜�
ª«¬
º»¼
1
kN m˜˜� 
0 1 2 3
0
5
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Max. Moment : unit kN m˜� 
M1 a( ) 6.667 
Mmax M1 a( ) unit˜� 
Bending Stress:
V M c
I
˜= cmax
do
2
� 
Vmax Mmax
cmax
I
˜� 
Vmax 160.45 MPa Ans
( < VY = 414 MPa)
Problem 6-95
The boat has a weight of 11.5 kN and a center of gravity at G. If it rests on the trailer at the smooth
contact A and can be considered pinned at B, determine the absolute maximum bending stress
developed in the main strut of the trailer. Consider the strut to be a box-beam having the dimensions
shown and pinned at C.
Given: a 0.9m� b 1.8m� 
c 1.2m� d 0.3m� 
bo 45mm� do 75mm� 
bi 38mm� di 45mm� 
W 11.5kN� 
Solution:
 Equilibrium (for boat) : Given
+ 6Fy=0; A W� B� 0=
 60B=0; A a b�( )˜ W b d�( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
6.389
5.111
§¨
©
·
¹ kN 
 Equilibrium (for assembly) : Given
+ 6Fy=0; D A� B� C� 0=
 60C=0; A� a b� c�( )˜ D b c�( )˜� B c˜� 0=
Guess C 1kN� D 1kN� 
C
D
§¨
©
·
¹ Find C D�( )� 
C
D
§¨
©
·
¹
1.15
10.35
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
M1 x1� � A� x1˜kN m˜� M2 x2� � A� x2� �˜ D x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � A� x3� �˜ D x3 a�� �˜� B x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3
5
0
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Max. Moment : unit kN m˜� 
M1 a( ) 5.750� 
Mmax M1 a( )� unit˜� 
Section properties :
I
1
12
bo do
3˜ bi di3˜�§© ·¹˜� 
Bending Stress:
V M c
I
˜= cmax
do
2
� 
Vmax Mmax
cmax
I
˜� 
Vmax 166.7 MPa Ans
Problem 6-96
The beam supports the load of 25 kN. Determine the absolute maximum bending stress in the beam if
the sides of its triangular cross section are a = 150 mm.
Given: a 150mm� L 0.6m� P 25kN� 
Solution:
Mmax P L˜� 
Section Property :
I
1
36
a˜ a sin 60deg( )˜( )3˜� 
Maximum Bending Stress: V M c
I
˜=
cmax
2
3
a˜ sin 60deg( )˜� 
Vmax Mmax� � cmaxI˜� 
Vmax 142.2 MPa Ans
Problem 6-97
The beam supports the load of 25 kN. Determine the required size a of the sides of its triangular cross
section if the allowable bending stress is V allow = 126 MPa.
Given: L 0.6m� P 25kN� Vallow 126MPa� 
Solution:
Mmax P L˜� 
Section Property :
I
1
36
a˜ a sin 60deg( )˜( )3˜=
Maximum Bending Stress: V M c
I
˜=
cmax
2
3
a˜ sin 60deg( )˜=
V Mmax� � cmaxI˜=
I Mmax� � cmaxVallow˜=
a
24
sin 60deg( )
2
Mmax
Vallow
˜§¨
©
·
¹
1
3
� 
a 156.2 mm Ans
Problem 6-98
The wood beam is subjected to the uniform load of w = 3 kN/m. If the allowable bending stress for the
material is V allow = 10 MPa, determine the required dimension b of its cross section. Assume the
support at A is a pin and B is a roller.
Given: L1 2m� Vallow 10MPa� 
L2 1m� w 3
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� w L1˜� 0=
 60A=0; B� L1 L2�� �˜ 0.5w L12˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
4.00
2.00
§¨
©
·
¹ kN 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2���� 
M1 x1� � A x1˜ 0.5w x12˜�§© ·¹ 1kN m˜˜� 
M2 x2� � A x2˜ w L1˜� � x2 0.5 L1˜�� �˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3
0
2
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
As indicated in the moment diagram, the
maximum moment occurs in L1 such that
V1=0:
V1 A w xc� �˜�=
xc
A
w
� xc 1.333 m 
Max. Moment : unit kN m˜� 
M1 xc� � 2.667 
Mmax M1 xc� � unit˜� 
Mmax 2.667 kN m˜ 
Section properties : I
1
12
b˜ 1.5 b˜( )3˜=
Bending Stress: V M
co
I
˜= co
1.5 b˜
2
=
I Mmax
co
Vallow
˜=
b
6 Mmax˜
2.25 Vallow˜
§¨
©
·
¹
1
3
� b 89.3 mm Ans
Problem 6-99
The wood beam has a rectangular cross section in the proportion shown. Determine its required
dimension b if the allowable bending stress is V allow = 10 MPa.
Given: L1 2m� Vallow 10MPa� 
L2 2m� w 0.5
kN
m
� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� w L1˜� 0=
 60A=0; B� L1 L2�� �˜ 0.5w L12˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ FindA B�( )� 
A
B
§¨
©
·
¹
0.75
0.25
§¨
©
·
¹ kN 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2���� 
M1 x1� � A x1˜ 0.5w x12˜�§© ·¹ 1kN m˜˜� 
M2 x2� � A x2˜ w L1˜� � x2 0.5 L1˜�� �˜�ª¬ º¼ 1kN m˜˜� 
0 2
0
0.5
1
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
As indicated in the moment diagram, the
maximum moment occurs in L1 such that
V1=0:
V1 A w xc� �˜�=
xc
A
w
� xc 1.500 m 
Max. Moment : unit kN m˜� 
M1 xc� � 0.5625 
Mmax M1 xc� � unit˜� 
Mmax 0.5625 kN m˜ 
Section properties : I
1
12
b˜ 1.5 b˜( )3˜=
Bending Stress: V M
co
I
˜= co
1.5 b˜
2
=
I Mmax
co
Vallow
˜=
b
3
6 Mmax˜
2.25 Vallow˜
� b 53.1 mm Ans
Problem 6-100
A beam is made of a material that has a modulus of elasticity in compression different from that given
for tension. Determine the location c of the neutral axis, and derive an expression for the maximum
tensile stress in the beam having the dimensions shown if it is subjected to the bending moment M.
Problem 6-101
The beam has a rectangular cross section and is subjected to a bending moment M. If the material
from which it is made has a different modulus of elasticity for tension and compression as shown,
determine the location c of the neutral axis and the maximum compressive stress in the beam.
Problem 6-102
The box beam is subjected to a bending moment of M = 25 kN·m directed as shown. Determine the
maximum bending stress in the beam and the orientation of the neutral axis.
Given: ay 3�� az 4� ar 5� M 25kN m˜� 
bo 150mm� do 150mm� 
bi 100mm� di 100mm� 
Solution:
Internal Moment Components :
My
ay
ar
M˜� Mz
az
ar
M˜� 
Section Property :
Iy
1
12
do bo
3˜ di bi3˜�§© ·¹˜� Iz 112 bo do
3˜ bi di3˜�§© ·¹˜� 
Maximum Bending Stress: By inspection, maximum bending stress occurs at B and D.
V
Mz y˜
Iz
�
My z˜
Iy
�=
At B : yB 0.5do� zB 0.5bo� 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�� 
VB 77.5� MPa (C) Ans
At D : yD 0.5� do� zD 0.5� bo� 
VD
Mz yD˜
Iz
�
My zD˜
Iy
�� 
VD 77.5 MPa (T) Ans
Orientation of Neutral Axis : tan D� � Iz
Iy
tan T� �˜=
T atan
ay
az
§¨
©
·
¹
� D atan
Iz
Iy
tan T� �˜§¨
©
·
¹
� D 36.87� deg Ans
y' bo tan D� �˜� y' 112.50� mm 
Problem 6-103
Determine the maximum magnitude of the bending moment M so that the bending stress in the member
does not exceed 100 MPa.
Given: ay 3�� az 4� ar 5� Vallow 100MPa� 
bo 150mm� do 150mm� 
bi 100mm� di 100mm� 
Solution:
Internal Moment Components :
My
ay
ar
M˜= Mz
az
ar
M˜=
Section Property :
Iy
1
12
do bo
3˜ di bi3˜�§© ·¹˜� Iz 112 bo do
3˜ bi di3˜�§© ·¹˜� 
Maximum Bending Stress: By inspection, maximum bending stress occurs at B and D.
Apply the flexure formula for biaxial bending at either point B or D. V
Mz y˜
Iz
�
My z˜
Iy
�=
At B : yB 0.5do� zB 0.5bo� 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�=
Vallow
az
ar
M˜§¨
©
·
¹
�
yB
Iz
˜
ay
ar
M˜§¨
©
·
¹
zB
Iy
˜�=
M
Vallow
az
ar
�
yB
Iz
˜
ay
ar
§¨
©
·
¹
zB
Iy
˜�
� 
M 32.24 kN m˜ Ans
Problem 6-104
The beam has a rectangular cross section. If it is subjected to a bending moment of M = 3500 N·m
directed as shown, determine the maximum bending stress in the beam and the orientation of the
neutral axis.
Given: M 3.5kN m˜� T' 30deg� 
b 150mm� d 300mm� 
Solution: T 180deg T'�� �� T 150 deg 
Internal Moment Components :
My M sin T� �˜� Mz M cos T� �˜� 
Section Property :
Iy
1
12
d˜ b3˜� Iz
1
12
b˜ d3˜� 
Maximum Bending Stress: V
Mz y˜
Iz
�
My z˜
Iy
�=
At A : yA 0.5d� zA 0.5b� 
VA
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 2.903 MPa (T) Ans
At B : yB 0.5� d� zB 0.5� b� 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 2.903� MPa (C) Ans
At C : yC 0.5d� zC 0.5� b� 
VC
Mz yC˜
Iz
�
My zC˜
Iy
�� VC 0.208� MPa (C)
At D : yD 0.5� d� zD 0.5b� 
VD
Mz yD˜
Iz
�
My zD˜
Iy
�� VD 0.208 MPa (T)
Orientation of Neutral Axis : tan D� � Iz
Iy
tan T� �˜=
D atan
Iz
Iy
tan T� �˜§¨
©
·
¹
� D 66.59� deg Ans
z' 0.5 b˜ 0.5d
tan D� ��� z' 10.05 mm 
Problem 6-105
The T-beam is subjected to a bending moment of M = 15 kN·m. directed as shown. Determine the
maximum bending stress in the beam and the orientation of the neutral axis. The location of the
centroid, C, must be determined.
Given: M 15kN m˜� T' 60deg� 
bf 300mm� tf 50mm� 
tw 50mm� dw 200mm� 
Solution: T 180deg T'�� �� T 120 deg 
Internal Moment Components :
My M sin T� �˜� Mz M cos T� �˜� 
Section Property :
Iy
1
12
tf bf
3˜ dw tw3˜�§© ·¹˜� Iy 110416666.67 mm4 
yc
6 yi

Ai˜� �˜
6 Ai� �˜= yc
bf tf˜� � 0.5tf� �˜ tw dw˜� � 0.5dw tf�� �˜�
bf tf˜� � tw dw˜� ��� yc 75.00 mm 
Iz
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜� 112 tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�ª«¬ º»¼�� 
Iz 130208333.33 mm
4 
Maximum Bending Stress: V
Mz y˜
Iz
�
My z˜
Iy
�=
At A : yA yc� zA 0.5bf� 
VA
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 21.97 MPa (T) Ans
At B : yB yc� zB 0.5� bf� 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 13.33� MPa (C)
At D : yD tf dw� yc�� ��� zD 0.5� tw� 
VD
Mz yD˜
Iz
�
My zD˜
Iy
�� VD 13.02� MPa (C)
Orientation of Neutral Axis : tan D� � Iz
Iy
tan T� �˜=
D atan
Iz
Iy
tan T� �˜§¨
©
·
¹
� D 63.91� deg Ans
z' 0.5bf
yc
tan D� ��� z' 186.72 mm 
Problem 6-106
If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of
M = 520 N·m and is directed as shown, determine the bending stress at points A and B. The location y
of the centroid C of the strut's cross-sectional area must be determined.Also, specify the orientation o
the neutral axis.
T' atan 5
12
§¨
©
·
¹� Given: M 520N m˜� 
bf 400mm� tf 20mm� 
tw 20mm� dw 180mm� 
Solution: D dw tf�� T 180deg T'�� �� 
Internal Moment Components :
My M sin T� �˜� Mz M cos T� �˜� 
Section Property :
Iy
1
12
D bf
3˜ dw bf 2tw�� �3˜�ª¬ º¼˜� Iy 366826666.67 mm4 
yc
6 yi

Ai˜� �˜
6 Ai� �˜= yc
bf tf˜� � 0.5tf� �˜ 2 tw dw˜� � 0.5dw tf�� �˜�
bf tf˜� � 2 tw dw˜� ��� yc 57.37 mm 
Iz
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜� 2 112 tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�ª«¬ º»¼˜�� 
Iz 57601403.51 mm
4 
Maximum Bending Stress: V
Mz y˜
Iz
�
My z˜
Iy
�=
At A : yA yc D�� zA 0.5� bf� 
VA
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 1.298� MPa (C) Ans
At B : yB yc� zB 0.5bf� 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 0.587 MPa (T)
Orientation of Neutral Axis : tan D� � Iz
Iy
tan T� �˜=
D atan
Iz
Iy
tan T� �˜§¨
©
·
¹
� D 3.74� deg Ans
Problem 6-107
The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of
M = 520 N·m and is directed as shown. Determine the maximum bending stress in the strut. The
location y of the centroid C of the strut's cross-sectional area must be determined.Also, specify the
orientation of the neutral axis.
T' atan 5
12
§¨
©
·
¹� Given: M 520N m˜� 
bf 400mm� tf 20mm� 
tw 20mm� dw 180mm� 
Solution: D dw tf�� T 180deg T'�� �� 
Internal Moment Components :
My M sin T� �˜� Mz M cos T� �˜� 
Section Property :
Iy
1
12
D bf
3˜ dw bf 2tw�� �3˜�ª¬ º¼˜� Iy 366826666.67 mm4 
yc
6 yi

Ai˜� �˜
6 Ai� �˜= yc
bf tf˜� � 0.5tf� �˜ 2 tw dw˜� � 0.5dw tf�� �˜�
bf tf˜� � 2 tw dw˜� ��� yc 57.37 mm 
Iz
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜� 2 112 tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�ª«¬ º»¼˜�� 
Iz 57601403.51 mm
4 
Maximum Bending Stress: V
Mz y˜
Iz
�
My z˜
Iy
�=
By inspection, the maximumbending stress can occur at either point A or B.
At A : yA yc D�� zA 0.5� bf� 
VA
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 1.298� MPa (C) Ans
At B : yB yc� zB 0.5bf� 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 0.587 MPa (T)
Orientation of Neutral Axis : tan D� � Iz
Iy
tan T� �˜=
D atan
Iz
Iy
tan T� �˜§¨
©
·
¹
� D 3.74� deg Ans
Problem 6-108
The 30-mm-diameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shown
It is supported on two journal bearings at A and B which offer no resistance to axial loading.
Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft.
Determine the maximum bending stress developed in the shaft.
Given: a 1m� Fy 150N� Fz 400N� do 30mm� 
Solution:
 Equilibrium : In x-y plane. Given
+ 6Fy=0; Ay By� 2Fy� 0=
 60A=0; 2� Fy a˜ By 2a( )˜� 0=
Guess Ay 1N� By 1N� 
Ay
By
§¨
©¨
·
¹
Find Ay By�� �� Ay
By
§¨
©¨
·
¹
0.45
0.15�
§¨
©
·
¹ kN 
 Equilibrium : In x-z plane, by symmetry: Az = Bz = Rz. 
+ 6Fz=0; 2Rz 2Fz� 0=
Rz Fz� Rz 400 N 
Internal Moment Components: 
The shaft is subjected to two bending moment components My and Mz. The moment disgram
for each component is drawn.
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 2a( )��� x3 2a( ) 1.01 2a( )˜� 3a��� 
Mz1 x1� � 2� Fy x1˜N m˜� Mz2 x2� � 2� Fy x2� �˜ Ay x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
Mz3 x3� � 2� Fy x3� �˜ Ay x3 a�� �˜�ª¬ º¼ 1N m˜˜� 
My1 x1� � 0� My2 x2� � Rz x2 a�� �˜N m˜� My3 x3� � Rz x3 a�� �˜ 2Fz x3 2a�� �˜�ª¬ º¼ 1N˜˜� 
0 1 2 3
400
200
0
Distane (m)
M
o
m
en
t 
(N
-m
)
Mz1 x1� �
Mz2 x2� �
Mz3 x3� �
x1 x2� x3�
0 1 2
0
500
Distane (m)
M
o
m
en
t 
(N
-m
)
My2 x2� �
My3 x3� �
x2 x3�
Maximum Bending Stress: unit N m˜� 
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My
2 + Mz
2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at E (x=2a).
Mmax Mz2 2a( )
2
My2 2a( )
2� unit˜� Mmax 427.20 N m˜ 
Hence,
cmax
do
2
� I S
64
do
4˜� Vmax Mmax
cmax
I
˜� 
Vmax 161.2 MPa Ans
M1 x1� � Mz1 x1� �2 My1 x1� �2�� 
M2 x2� � Mz2 x2� �2 My2 x2� �2�� 
M3 x3� � Mz3 x3� �2 My3 x3� �2�� 
0 0.5 1 1.5 2 2.5 3
0
200
400
Distane (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-109
The shaft is subjected to the vertical and horizontal loadings of two pulleys D and E as shown. It is
supported on two journal bearings at A and B which offer no resistance to axial loading.
Furthermore,the coupling to the motor at C can be assumed not to offer any support to the shaft.
Determine the required diameter d of the shaft if the allowable bending stress for the material is V allow
= 180 MPa.
Given: a 1m� Fy 150N� Fz 400N� Vallow 180MPa� 
Solution:
 Equilibrium : In x-y plane. Given
+ 6Fy=0; Ay By� 2Fy� 0=
 60A=0; 2� Fy a˜ By 2a( )˜� 0=
Guess Ay 1N� By 1N� 
Ay
By
§¨
©¨
·
¹
Find Ay By�� �� Ay
By
§¨
©¨
·
¹
0.45
0.15�
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� 3a��� 
Mz1 x1� � 2� Fy x1˜N m˜� Mz2 x2� � 2� Fy x2� �˜ Ay x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
 Equilibrium : In x-z plane, by symmetry: Az = Bz = Rz. 
+ 6Fz=0; 2Rz 2Fz� 0=
Rz Fz� Rz 400 N 
x'1 a 1.01 a˜� 2a( )��� x'2 2a( ) 1.01 2a( )˜� 3a��� 
My1 x'1� � Rz x'1 a�� �˜ N m˜� My2 x'2� � Rz x'2 a�� �˜ 2Fz x'2 2a�� �˜�ª¬ º¼ 1N m˜˜� 
Internal Moment Components: 
The shaft is subjected to two bending moment components My and Mz. The moment disgram
for each component is drawn.
0 1 2 3
400
200
0
Distane (m)
M
o
m
en
t 
(N
-m
)
Mz1 x1� �
Mz2 x2� �
x1 x2�
0 1 2
0
500
Distane (m)
M
o
m
en
t 
(N
-m
)
My1 x'1� �
My2 x'2� �
x'1 x'2�
Maximum Bending Stress: unit N m˜� 
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My
2 + Mz
2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at E (x=2a).
Mmax Mz2 2a( )
2
My1 2a( )
2� unit˜� Mmax 427.20 N m˜ 
Hence,
cmax
do
2
= I
S
64
do
4˜= Vallow Mmax
cmax
I
˜=
S
64
do
4˜
Mmax
Vallow
§¨
©
·
¹
do
2
˜=
do
3
32Mmax
S Vallow˜
� 
do 28.91 mm Ans
Problem 6-110
The board is used as a simply supported floor joist. If a bending moment of M = 1.2 kN·m is applied
3° from the z axis, determine the stress developed in the board at the corner A. Compare this stress
with that developed by the same moment applied along the z axis (T = 0°). What is the angle a for the
neutral axis when T = 3°? Comment: Normally, floor boards would be nailed to the top of the beam so
that T = 0° (nearly) and the high stress due to misalignment would not occur.
Given: M 1.2kN m˜� T 3deg� 
b 50mm� d 150mm� 
Solution:
Internal Moment Components :
My M� sin T� �˜� Mz M cos T� �˜� 
yc 0.5d� Section Property :
Iy
1
12
d˜ b3˜� Iz
1
12
b˜ d3˜� 
Maximum Bending Stress: V
Mz y˜
Iz
�
My z˜
Iy
�=
At A : yA yc�� zA 0.5� b� 
VA
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 7.40 MPa (T) Ans
Orientation of Neutral Axis : tan D� � Iz
Iy
tan T� �˜=
D atan
Iz
Iy
tan T� �˜§¨
©
·
¹
� D 25.25 deg Ans
When T = 0 : M'y 0� M'z M� 
V'A
M'z yA˜
Iz
�
M'y zA˜
Iy
�� V'A 6.40 MPa (T) Ans
Problem 6-111
Consider the general case of a prismatic beam subjected to bending-moment components My and Mz,
as shown, when the x, y, z axes pass through the centroid of the cross section. If the material is
linear-elastic, the normal stress in the beam is a linear function of position such that V = a + by + cz.
Using the equilibrium conditions , determine the
constants a, b, and c, and show that the normal stress can be determined from the equation 
V = [- (Mz Iy + My Iyz) y + (My Iz + Mz Iyz) z] / (Iy Iz - Iyz2), where the moments and products of
inertia are defined in Appendix A.
³³³ � AzAyA dAyıM,dAzıM,dAı0
Given: Linear function: Vx a by� z�=
Solution:
Equilibrium Conditios :
³³³
³³
�� 
�� 
AAA
AA
x
dAzcdAybdAa0
dAcz)by(a0,dAı0
³³³
³³
�� 
�� 
A
2
AA
y
A
y
A
xy
dAzcdAyzbdAzaM
dAcz)by(azM,dAzıM
³³³
³³
��� 
�� 
AA
2
A
z
A
z
A
xz
dAyzcdAybdAyaM
dAcz)by(ay-M,dAyı-M
³³ AA 0dAzdAyThe integrals are defined in Appendix A. Note that
0c0bdAa0
A
˜�˜� ³Thus, 
yyzy IcIb0aM ˜�˜�˜ 
yzyz IcIb0aM ˜�˜�˜� 
0)A (since0a z Solving for a, b and c : Ans
2
yzzyyzyzy I-IID where)IMI(M
D
1
-b ˜ ˜�˜ 
Ans
)IMI(M
D
1
c yzzzy ˜�˜ Ans
Bending Stress:
 z)IMI(M
D
1
 y)()IMI(M
D
1ı zyyzzyzyyzx ˜˜�˜��˜˜�˜ QED
In matrix form,
V 1
D
Mz My� � Iy
Iyz
Iyz
Iz
§¨
©¨
·
¹
˜ y�
z
§¨
©
·
¹˜=
Problem 6-112
The 65-mm-diameter steel shaft is subjected to the two loads that act in the directions shown. If the
journal bearings at A and B do not exert an axial force on the shaft, determine the absolute maximum
bending stress developed in the shaft.
Given: a 1.25m� b 1m� F 4kN� 
L 2a b�� do 65mm� T 30deg� 
Solution:
 Equilibrium : In x-y plane, by symmetry: Ay = By = Ry. 
+ 6Fy=0; 2Ry 2F cos T� �˜� 0=
Ry F cos T� �˜� Ry 3.464 kN 
 Equilibrium :
 
In x-z plane, by anti-symmetry: Az = -Bz = Rz. 
60B=0; Az L˜ F sin T� �˜ b( )˜� 0=
Rz F sin T� �˜ bL§¨©
·
¹˜� Rz 0.571 kN 
Internal Moment Components: 
The shaft is subjected to two bending moment components My and Mz. The moment disgram
for each component is drawn.
x1 0 0.01 a˜� a���x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� L��� 
Mz1 x1� � Ry x1˜kN m˜� Mz2 x2� � Ry x2� �˜ F cos T� �˜ x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
Mz3 x3� � Ry x3� �˜ F cos T� �˜ x3 a�� �˜� F cos T� �˜ x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
My1 x1� � Rz x1˜kN m˜� My2 x2� � Rz x2� �˜ F sin T� �˜ x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
My3 x3� � Rz x3� �˜ F sin T� �˜ x3 a�� �˜� F sin T� �˜ x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
0 2
0
5
Distane (m)
M
o
m
en
t 
(k
N
-m
)
Mz1 x1� �
Mz2 x2� �
Mz3 x3� �
x1 x2� x3�
0 2
1
0
1
Distane (m)
M
o
m
en
t 
(k
N
-m
)
My1 x1� �
My2 x2� �
My3 x3� �
x1 x2� x3�
Maximum Bending Stress: unit kN m˜� 
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My
2 + Mz
2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at x=a.
Mmax Mz1 a( )
2
My1 a( )
2� unit˜� Mmax 4.389 kN m˜ 
Hence,
cmax
do
2
� I S
64
do
4˜� Vmax Mmax
cmax
I
˜� 
Vmax 162.8 MPa Ans
M1 x1� � Mz1 x1� �2 My1 x1� �2�� 
M2 x2� � Mz2 x2� �2 My2 x2� �2�� 
M3 x3� � Mz3 x3� �2 My3 x3� �2�� 
0 1 2 3
0
5
Distane (m)
M
o
m
en
t 
(N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-113
The steel shaft is subjected to the two loads that act in the directions shown. If the journal bearings at
A and B do not exert an axial force on the shaft, determine the required diameter of the shaft if the
allowable bending stress is V allow = 180 MPa.
Given: a 1.25m� b 1m� F 4kN� 
L 2a b�� T 30deg� Vallow 180MPa� 
Solution:
 Equilibrium : In x-y plane, by symmetry: Ay = By = Ry. 
+ 6Fy=0; 2Ry 2F cos T� �˜� 0=
Ry F cos T� �˜� Ry 3.464 kN 
 Equilibrium :
 
In x-z plane, by anti-symmetry: Az = -Bz = Rz. 
60B=0; Az L˜ F sin T� �˜ b( )˜� 0=
Rz F sin T� �˜ bL§¨©
·
¹˜� Rz 0.571 kN 
Internal Moment Components: 
The shaft is subjected to two bending moment components My and Mz. The moment disgram
for each component is drawn.
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b�( )��� x3 a b�( ) 1.01 a b�( )˜� L��� 
Mz1 x1� � Ry x1˜kN m˜� Mz2 x2� � Ry x2� �˜ F cos T� �˜ x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
Mz3 x3� � Ry x3� �˜ F cos T� �˜ x3 a�� �˜� F cos T� �˜ x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
My1 x1� � Rz x1˜kN m˜� My2 x2� � Rz x2� �˜ F sin T� �˜ x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
My3 x3� � Rz x3� �˜ F sin T� �˜ x3 a�� �˜� F sin T� �˜ x3 a� b�� �˜�ª¬ º¼ 1kN m˜˜� 
0 2
0
5
Distane (m)
M
o
m
en
t 
(k
N
-m
)
Mz1 x1� �
Mz2 x2� �
Mz3 x3� �
x1 x2� x3�
0 2
1
0
1
Distane (m)
M
o
m
en
t 
(k
N
-m
)
My1 x1� �
My2 x2� �
My3 x3� �
x1 x2� x3�
Maximum Bending Stress: unit kN m˜� 
Since all the axes through the circle's center for circular shaft are principal axes, then the
resultant moment M = (My
2 + Mz
2)0.5 can be used to determine the maximum bending stress.
The maximum bending stress moment occurs at x=a.
Mmax Mz1 a( )
2
My1 a( )
2� unit˜� Mmax 4.389 kN m˜ 
Hence,
cmax
do
2
= I
S
64
do
4˜= Vallow Mmax
cmax
I
˜=
S
64
do
4˜
Mmax
Vallow
§¨
©
·
¹
do
2
˜=
do
3
32Mmax
S Vallow˜
� 
do 62.86 mm Ans
Problem 6-114
Using the techniques outlined in Appendix A, Example A.4 or A.5, the Z section has principal mome
of inertia of Iy = 0.060(10
-3) m4 and Iz = 0.471(10
-3) m4, computed about the principal axes of inertia y
and z, respectively. If the section is subjected to an internal moment of M = 250 N·m directed
horizontally as shown, determine the stress produced at point A. Solve the problem using Eq. 6-17.
Given: M 250N m˜� T 32.9deg� 
bf 300mm� tf 50mm� 
tw 50mm� dw 150mm� 
Iy 0.060 10
3�� � m4� Iz 0.471 10 3�� � m4� 
Solution:
Internal Moment Components :
My M cos T� �˜� Mz M sin T� �˜� 
Coordinates of Point A :
y'A 0.5bf� z'A dw 0.5tf�� ��� 
yA
zA
§¨
©¨
·
¹
cos T� �
sin T� �
sin T� ��
cos T� �
§¨
©
·
¹
y'A
z'A
§¨
©¨
·
¹
˜� 
yA
zA
§¨
©¨
·
¹
221.00
65.46�
§¨
©
·
¹ mm 
Bending Stress:
VA
Mz yA˜
Iz
�
My zA˜
Iy
�§¨
©
·
¹
� 
VA 0.293� MPa (C) Ans
Problem 6-115
Solve Prob. 6-114 using the equation developed in Prob. 6-111.
Given: M 250N m˜� T 32.9deg� 
bf 300mm� tf 50mm� 
tw 50mm� dw 150mm� 
Iy 0.060 10
3�� � m4� Iz 0.471 10 3�� � m4� 
Solution:
Internal Moment Components :
My' M� Mz' 0� 
Section Property : cz.w 0.5dw 0.5tf�� cy.w 0.5bf 0.5tw�� 
Iy'
1
12
bf˜ tf3˜ 2
1
12
tw˜ dw3˜ tw dw˜ cz.w2˜�§¨©
·
¹�� Iy' 181.25 10
6�u m4 
Iz'
1
12
tf˜ bf3˜ 2
1
12
dw˜ tw3˜ dw tw˜ cy.w2˜�§¨©
·
¹�� Iz' 350.00 10
6�u m4 
Iy'z' tw dw˜ cz.w� �˜ cy.w�� �˜ tw dw˜ cz.w�� �˜ cy.w� �˜�� Iy'z' 187.50� 10 6�u m4 
Coordinates of Point A :
y'A 0.5bf� z'A dw 0.5tf�� ��� 
y'A 150 mm z'A 175� mm 
Bending Stress: Using formula developed in Prob. 6-111.
D Iy' Iz'˜ Iy'z'2�� 
VA
1
D
Mz' My'� � Iy'
Iy'z'
Iy'z'
Iz'
§¨
©¨
·
¹
˜
y'A�
z'A
§¨
©¨
·
¹
˜� 
VA 0.293� MPa (C) Ans
Problem 6-116
Using the techniques outlined in Appendix A, Example A.4 or A.5, the Z section has principal mome
of inertia of Iy = 0.060(10
-3) m4 and Iz = 0.471(10
-3) m4, computed about the principal axes of inertia y
and z, respectively. If the section is subjected to an internal moment of M = 250 N·m directed
horizontally as shown, determine the stress produced at point B. Solve the problem using Eq. 6-17.
Given: M 250N m˜� T 32.9deg� 
bf 300mm� tf 50mm� 
tw 50mm� dw 150mm� 
Iy 0.060 10
3�� � m4� Iz 0.471 10 3�� � m4� 
Solution:
Internal Moment Components :
My M cos T� �˜� Mz M sin T� �˜� 
Coordinates of Point B :
y'B 0.5� bf� z'B dw 0.5tf�� 
yB
zB
§¨
©¨
·
¹
cos T� �
sin T� �
sin T� ��
cos T� �
§¨
©
·
¹
y'B
z'B
§¨
©¨
·
¹
˜� 
yB
zB
§¨
©¨
·
¹
221.00�
65.46
§¨
©
·
¹ mm 
Bending Stress:
VB
Mz yB˜
Iz
�
My zB˜
Iy
�§¨
©
·
¹
� 
VB 0.293 MPa (T) Ans
Problem 6-117
For the section, Iy' = 31.7110-62 m4, Iz' = 114(10
-6) m4, Iy'z' = 15.1(10
-6) m4. Using the techniques
outlined in Appendix A, the member's cross-sectional area has principal moments of inertia of Iy =
29.0(10-6) m4 and Iz = 117(10
-6) m4, computed about the principal axes of inertia y and z, respectively
If the section is subjected to a moment of M = 2500 N·m directed as shown, determine
the stress produced at point A, using Eq. 6-17.
Given: M 2500N m˜� T' 10.10deg� 
h1 80mm� h2 140mm� 
b1 60mm� b2 60mm� 
Iy 29.0 10
6�� � m4� Iz 117 10 6�� � m4� 
Solution: T T'�� 
Internal Moment Components :
My M sin T� �˜� Mz M cos T� �˜� 
Coordinates of Point A :
y'A h2�� z'A b2�� 
yA
zA
§¨
©¨
·
¹
cos T� �
sin T� �
sin T� ��
cos T� �
§¨
©
·
¹
y'A
z'A
§¨
©¨
·
¹
˜� 
yA
zA
§¨
©¨
·
¹
148.35�
34.52�
§¨
©
·
¹ mm 
Bending Stress:
VA
Mz yA˜
Iz
�
My zA˜
Iy
�§¨
©
·
¹
� 
VA 2.599 MPa (T) Ans
Problem 6-118
Solve Prob. 6-117 using the equation developed in Prob. 6-111.
Given: M 2500N m˜� T' 10.10deg� 
h1 80mm� h2 140mm� 
b1 60mm� b2 60mm� 
Iy' 31.7 10
6�� � m4� Iz' 114 10 6�� � m4� 
Iy'z' 15.1 10
6�� � m4� 
Solution: T T'�� 
Internal Moment Components :
My' 0� Mz' M� 
Coordinates of Point A :
y'A h2�� z'A b2�� 
yA
zA
§¨
©¨
·
¹
cos T� �
sin T� �
sin T� ��
cos T� �
§¨
©
·
¹
y'A
z'A
§¨
©¨
·
¹
˜� 
yA
zA
§¨
©¨
·
¹
148.35�
34.52�
§¨
©
·
¹ mm 
Bending Stress: Using formula developed in Prob. 6-111.
D Iy' Iz'˜ Iy'z'2�� 
VA
1
D
Mz' My'� � Iy'
Iy'z'
Iy'z'
Iz'
§¨
©¨
·
¹
˜
y'A�
z'A
§¨
©¨·
¹
˜� 
VA 2.608 MPa (T) Ans
Problem 6-119
The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). Determine the
dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two
metals. What maximum moment will this beam support if the allowable bending stress for the
aluminum is (V allow) al = 128 MPa and for the brass (V allow) br = 35 MPa?
Given: b 150mm� hal 50mm� 
Eal 68.9GPa� Ebr 101GPa� 
Val_allow 128MPa� Vbr_allow 35MPa� 
Solution:
Section Property : n
Eal
Ebr
� n 0.682178 
Abr b h˜= A'al n b˜( ) hal˜� 
yc
6yi Ai˜
6A= yc
A'al 0.5 hal˜� �˜ Abr hal 0.5 h˜�� �˜�
A'al Abr�
=
Given yc hal� 
hal
A'al 0.5 hal˜� �˜ b h˜ hal 0.5 h˜�� �˜�
A'al b h˜�
=
Guess h 10mm� 
h Find h( )� h 41.30 mm Ans
Abr b h˜� 
Ibr
1
12
b˜ h3˜ Abr 0.5h( )2˜�� 
I'al
1
12
n b˜( )˜ hal3˜ A'al yc 0.5hal�� �2˜�� 
I Ibr I'al�� I 7785108.17 mm4 
Allowable Bending Stress: V M y˜
I
=
Assume failure of red brass:
ybr h� Mbr
Vbr_allow� � I˜
ybr
� Mbr 6.60 kN m˜ 
Assume failure of aluminum:
yal hal� Mal
Val_allow� � I˜
n yal˜
� Mal 29.22 kN m˜ 
Mallow min Mbr Mal�� �� 
Mallow 6.60 kN m˜ Ans
Problem 6-120
The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 4
mm, determine the maximum moment that can be applied to the beam if the allowable bending stress
for the aluminum is (V allow) al = 128 MPa and for the brass (V allow) br = 35 MPa?
Given: b 150mm� hal 50mm� hbr 40mm� 
Eal 68.9GPa� Ebr 101GPa� 
Val_allow 128MPa� Vbr_allow 35MPa� 
Solution:
Section Property : n
Eal
Ebr
� n 0.682178 
Abr b hbr˜� A'al n b˜( ) hal˜� 
yc
6yi Ai˜
6A= yc
A'al 0.5 hal˜� �˜ Abr hal 0.5 hbr˜�� �˜�
A'al Abr�
� yc 49.289 mm 
Ibr
1
12
b˜ hbr3˜ Abr hal 0.5hbr� yc�� �2˜�� 
I'al
1
12
n b˜( )˜ hal3˜ A'al yc 0.5hal�� �2˜�� 
I Ibr I'al�� I 7457987.63 mm4 
Allowable Bending Stress: V M y˜
I
=
Assume failure of red brass:
ybr hal hbr� yc�� Mbr
Vbr_allow� � I˜
ybr
� Mbr 6.41 kN m˜ 
Assume failure of aluminum:
yal yc� Mal
Val_allow� � I˜
n yal˜
� Mal 28.39 kN m˜ 
Mallow min Mbr Mal�� �� 
Mallow 6.41 kN m˜ Ans
Problem 6-121
A wood beam is reinforced with steel straps at its top and bottom as shown. Determine the maximum
bending stress developed in the wood and steel if the beam is subjected to a bending moment of M = 
kN·m. Sketch the stress distribution acting over the cross section. Take Ew = 11 GPa, Est = 200 GPa.
Given: b 200mm� hw 300mm� tst 20mm� 
Ew 11GPa� Est 200GPa� 
M 5kN m˜� 
Solution:
Section Property :
n
Est
Ew
� n 18.181818 
Aw b hw˜� A'st n b˜( ) tst˜� 
I'st
1
12
n b˜( )˜ tst3˜ A'st 0.5hw 0.5tst�� �2˜�� 
I
1
12
b˜ hw3˜ 2I'st�� I 4178484848.48 mm4 
Maximum Bending Stress: V M y˜
I
=
For wood beam,
yw 0.5hw� Vw.max
M yw˜
I
� Vw.max 0.179 MPa Ans
For steel straps,
yst 0.5hw tst�� Vst.max
n M˜ yst˜
I
� Vst.max 3.699 MPa Ans
At y'st 0.5hw� V'st
n M˜ y'st˜
I
� V'st 3.263 MPa 
Problem 6-122
The Douglas Fir beam is reinforced with A-36 steel straps at its center and sides. Determine the
maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz =
10 kN·m. Sketch the stress distribution acting over the cross section.
Given: bst 12mm� d 150mm� 
bw 50mm� Mz 10kN m˜� 
Ew 13.1GPa� Est 200GPa� 
Solution:
Section Property : n
Ew
Est
� n 0.0655 
Iz
1
12
3bst� � d3˜ n 2bw� �˜ d3˜�ª¬ º¼˜� 
Maximum Bending Stress: V
Mz y˜
Iz
=
ymax 0.5d� 
Vst
Mz ymax˜
Iz
� Vst 62.7 MPa Ans
Vw n Vst� �˜� Vw 4.10 MPa Ans
Problem 6-123
The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and
in the wood if the beam is subjected to a moment of M = 1.2 kN·m. Est = 200 GPa, Ew = 12GPa.
Given: bst 399mm� dst 100mm� tst 12mm� 
bw 375mm� dw 88mm� M 1.2kN m˜� 
Ew 12GPa� Est 200GPa� 
Solution: df dw� 
n
Ew
Est
� n 0.06 
Section Property :
As1 2tst df˜� As2 bst tst˜� A'w n bw˜ dw˜� 
yc
A'w As1�� � 0.5dw tst�� �˜ As2 0.5tst� �˜�
A'w As1� As2�
� yc 29.04 mm 
Iw
1
12
n bw˜ dw3˜§© ·¹˜ A'w 0.5dw tst� yc�� �2˜�� 
Is1
1
12
2tst df
3˜§© ·¹˜ As1 0.5dw tst� yc�� �2˜�� 
Is2
1
12
bst tst
3˜§© ·¹˜ As2 0.5tst yc�� �2˜�� 
I Iw Is1� Is2�� 
Maximum Bending Stress: V M c˜
I
=
cmax dst yc�� 
Vst
M cmax˜
I
� Vst 10.37 MPa Ans
Vw n Vst� �˜� Vw 0.62 MPa Ans
Problem 6-124
The Douglas Fir beam is reinforced with A-36 steel straps at its sides. Determine the maximum stress
developed in the wood and steel if the beam is subjected to a bending moment of Mz = 4 kN·m. Sketc
the stress distribution acting over the cross section.
Given: bst 15mm� d 350mm� 
bw 200mm� Mz 4kN m˜� 
Ew 13.1GPa� Est 200GPa� 
Solution:
Section Property : n
Ew
Est
� n 0.0655 
Iz
1
12
2bst� � d3˜ n bw˜ d3˜�ª¬ º¼˜� 
Maximum Bending Stress: V
Mz y˜
Iz
=
ymax 0.5d� 
Vst
Mz ymax˜
Iz
� Vst 4.546 MPa Ans
Vw n Vst� �˜� Vw 0.298 MPa Ans
Problem 6-125
The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the cross
section shown. If it is subjected to a moment of M = 6.5 kN·m, determine the maximum stress in the
brass and steel. Also, what is the stress in each material at the seam where they are bonded together?
Given: b 125mm� hst 100mm� hbr 100mm� 
Est 200GPa� Ebr 101GPa� 
M 6.5kN m˜� 
Solution:
Section Property : n
Ebr
Est
� n 0.505 
Ast b hst˜� A'br n b˜( ) hbr˜� 
yc
6yi Ai˜
6A= yc
A'br 0.5 hbr˜� �˜ Ast hbr 0.5 hst˜�� �˜�
A'br Ast�
� yc 116.445 mm 
Ist
1
12
b˜ hst3˜ Ast hbr 0.5hst� yc�� �2˜�� 
I'br
1
12
n b˜( )˜ hbr3˜ A'br yc 0.5hbr�� �2˜�� 
I Ist I'br�� I 57620604.93 mm4 
Maximum Bending Stress: V M y˜
I
=
For steel,
yst hbr hst� yc�� Vst.max
M yst˜
I
� Vst.max 9.426 MPa Ans
For red brass,
ybr yc� Vbr.max
n M˜ ybr˜
I
� Vbr.max 6.634 MPa Ans
Bending Stress at the Seam:
yseam yc hbr�� V'st
M yseam˜
I
� V'st 1.855 MPa Ans
V'br n V'st˜� V'br 0.937 MPa Ans
Problem 6-126
The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the cross
section shown. If the allowable bending stress for the steel is (V allow)st = 180 MPa and for the brass
(V allow)br = 60 MPa, determine the maximum moment M that can be applied to the beam.
Given: b 125mm� hst 100mm� hbr 100mm� 
Est 200GPa� Ebr 101GPa� 
Vst_allow 180MPa� Vbr_allow 60MPa� 
Solution:
Section Property : n
Ebr
Est
� n 0.505 
Ast b hst˜� A'br n b˜( ) hbr˜� 
yc
6yi Ai˜
6A= yc
A'br 0.5 hbr˜� �˜ Ast hbr 0.5 hst˜�� �˜�
A'br Ast�
� yc 116.445 mm 
Ist
1
12
b˜ hst3˜ Ast hbr 0.5hst� yc�� �2˜�� 
I'br
1
12
n b˜( )˜ hbr3˜ A'br yc 0.5hbr�� �2˜�� 
I Ist I'br�� I 57620604.93 mm4 
Allowable Bending Stress: V M y˜
I
=
Assume failure of red brass:
ybr yc� Mbr
Vbr_allow� � I˜
n ybr˜
� Mbr 58.79 kN m˜ 
Assume failure of steel:
yst hbr hst� yc�� Mst
Vst_allow� � I˜
yst
� Mst 124.13 kN m˜ 
Mallow min Mbr Mst�� �� 
Mallow 58.79 kN m˜ Ans
Problem 6-127
The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress
for the steel is (V st) allow = 280 MPa and the allowable compressive stress for the concrete is (V conc)
allow = 21 MPa, determine the maximum moment M that can be applied to the section. Assume the
concrete cannot support a tensile stress. Est = 200 GPa, Econc = 26.5 GPa.
Given: bf 550mm� df 100mm� 
bw 150mm� dw 450mm� 
dr 25mm� hr 50mm� 
Econc 26.5GPa� Est 200GPa� 
Vc_allow 21MPa� Vst_allow 280MPa� 
Solution:
n
Est
Econc
� 
Section Property : n 7.54717 
A'stn 2˜ S 0.5dr� �2˜� 
Given A'st dw h'w� hr�� �˜ bf df˜ 0.5df h'w�� �˜� bw h'w˜ 0.5h'w� �˜� 0=
Guess h'w 10mm� 
h'w Find h'w� �� h'w 3.41 mm 
Ist A'st dw h'w� hr�� �2˜� 
If
1
12
bf df
3˜§© ·¹˜ bf df˜� � 0.5df h'w�� �2˜�� 
I'w
1
12
bw h'w
3˜§© ·¹˜ bw h'w˜� � 0.5h'w� �2˜�� 
I Ist If� I'w�� 
V M y˜
I
=
Assume concrete fails:
ymax df h'w�� Mconc
Vc_allow� � I˜
ymax
� Mconc 277.83 kN m˜ 
Assume steel fails:
yst dw h'w� hr�� Mst
Vst_allow� � I˜
n yst˜
� 
Mst 127.98 kN m˜ 
Mallow min Mconc Mst�� �� 
Mallow 127.98 kN m˜ Ans
Problem 6-128
Determine the maximum uniform distributed load w0 that can be supported by the reinforced concrete
beam if the allowable tensile stress for the steel is (V st) allow = 200 MPa and the allowable compressive
stress for the concrete is (V conc) allow = 20 MPa. Assume the concrete cannot support a tensile stress.
Take Est = 200 GPa, Econc = 25 GPa.
Given: b 250mm� d 500mm� 
dr 16mm� hr 50mm� 
Econc 25GPa� Est 200GPa� 
Vc_allow 20MPa� L 2.5m� 
Vst_allow 200MPa� 
Solution:
n
Est
Econc
� 
Section Property : n 8 
A'st n 2˜ S 0.5dr� �2˜� 
Given A'st d h'� hr�� �˜ b h'˜ 0.5h'( )˜� 0=
Guess h' 10mm� 
h' Find h'( )� h' 95.51 mm 
I A'st d h'� hr�� �2˜ 112 b h'3˜� �˜ b h'˜( ) 0.5h'( )2˜�ª«¬ º»¼�� 
V M y˜
I
=
Assume concrete fails:
ymax h'� Mconc
Vc_allow� � I˜
ymax
� Mconc 99.85 kN m˜ 
Assume steel fails:
yst d h'� hr�� Mst
Vst_allow� � I˜
n yst˜
� Mst 33.63 kN m˜ 
Thus, steel fails first.
Maximum moment ocurs over the middle support: Mmax
wo L
2˜
2
=
wo
2 Mst˜
L
2
� 
wo 10.76
kN
m
 Ans
Problem 6-129
A bimetallic strip is made from pieces of 2014-T6 aluminum and C83400 red brass, having the cross
section shown. A temperature increase causes its neutral surface to be bent into a circular arc having a
radius of 400 mm. Determine the moment that must be acting on its cross section due to the thermal
stress. Take Eal = 74 GPa, Ebr = 102 GPa.
Given: bbr 6mm� dbr 2mm� 
bal 6mm� dal 2mm� 
Eal 74GPa� U 400mm� 
Ebr 102GPa� 
Solution: Transform the section to brass
Section Property : n
Eal
Ebr
� n 0.72549 
A'al n bal˜ dal˜� Abr bbr dbr˜� 
yc
A'al 0.5dal� �˜ Abr 0.5dbr dal�� �˜�
A'al Abr�
� yc 2.16 mm 
Ial
1
12
n bal˜ dal3˜§© ·¹˜ A'al 0.5dal yc�� �2˜�� 
Ibr
1
12
bbr dbr
3˜§© ·¹˜ Abr 0.5dbr dal� yc�� �2˜�� 
I Ial Ibr�� 
Maximum Bending Stress:
1
U
M
E I˜=
M
Ebr� � I˜
U� 
M 6.91 N m˜ Ans
Problem 6-130
The fork is used as part of a nosewheel assembly for an airplane. If the maximum wheel reaction at the
end of the fork is 4.5 kN, determine the maximum bending stress in the curved portion of the fork at
section a-a. There the cross-sectional area is circular, having a diameter of 50 mm.
Given: rc 250mm� a 150mm� F 4.5kN� 
T 30deg� do 50mm� 
Solution:
Internal Moment :
 60C=0; M F a rc sin T� �˜�� �˜� 0=
M F a rc sin T� �˜�� �˜� 
M 112.5 N m˜ 
Section Property : ro 0.5do� 
A S ro2˜� 
³ AA_r rdAȈI IA_r 2 S˜ rc rc2 ro2��§© ·¹˜=
R
A
IA_r
= R
S ro2˜
2 S˜ rc rc2 ro2��§© ·¹˜
� 
R 249.373 mm 
Bending Stress: V M R r�( )˜
A r˜ rc R�� �˜=
rA rc ro�� VA
M R rA�� �˜
A rA˜ rc R�� �˜� VA 9.91 MPa 
rB rc ro�� VB
M R rB�� �˜
A rB˜ rc R�� �˜� VB 8.52� MPa 
Vmax max VA VB�� �� 
Vmax 9.91 MPa (T) Ans
Given: bf 75mm� b'f 150mm� tf 10mm� 
Problem 6-131
Determine the greatest magnitude of the applied forces P if the allowable bending stress is (V allow)c =
50 MPa in compression and (V allow)t = 120 MPa in tension.
tw 10mm� dw 150mm� ri 250mm� 
Vc.allow 50� MPa� Vt.allow 120MPa� 
Solution:
Internal Moment :
M = P(dw+tf ) kN-m is positive since it tends
to increase the beam's radius of curvature.
Section Property : re ri dw� 2tf�� 
A bf tf˜ dw tw˜� b'f tf˜�� A 3750 mm2 
r
 6 ri

Ai˜� �˜
6 Ai� �˜= rc
b'f tf˜� � ri 0.5tf�� �˜ dw tw˜� � ri 0.5dw� tf�� �˜� bf tf˜� � re 0.5tf�� �˜�
A
� 
rc 319.000 mm 
³ AA_r rdAȈI IA_r b'f ln
ri tf�
ri
§¨
©
·
¹
˜ tw ln
re tf�
ri tf�
§¨
©
·
¹
˜� bf ln
re
re tf�
§¨
©
·
¹
˜�� 
IA_r 12.245 mm 
R
A
IA_r
� R 306.243 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜= M
V A˜ r˜ rc R�� �˜
R r�=
Assume tension failure.
r ri� P dw tf�� �˜ Vt.allow A˜ r˜ rc R�� �˜R r�= P
Vt.allow A˜ r˜ rc R�� �˜
dw tf�� � R r�( )˜� 
P 159.48 kN 
Assume compression failure.
r' re� P' dw tf�� �˜ Vc.allow A˜ r'˜ rc R�� �˜R r'�= P'
Vc.allow A˜ r'˜ rc R�� �˜
dw tf�� � R r'�( )˜� 
P' 55.2 kN 
Pallow min P P'�( )� Pallow 55.20 kN Ans
Given: bf 75mm� b'f 150mm� tf 10mm� 
Problem 6-132
If P = 6 kN, determine the maximum tensile and compressive bending stresses in the beam.
tw 10mm� dw 150mm� ri 250mm� 
P 6kN� 
Solution:
Internal Moment :
M P dw tf�� �˜� M 0.960 kN m˜ 
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re ri dw� 2tf�� 
A bf tf˜ dw tw˜� b'f tf˜�� A 3750 mm2 
r
 6 ri

Ai˜� �˜
6 Ai� �˜= rc
b'f tf˜� � ri 0.5tf�� �˜ dw tw˜� � ri 0.5dw� tf�� �˜� bf tf˜� � re 0.5tf�� �˜�
A
� 
rc 319.000 mm 
³ AA_r rdAȈI IA_r b'f ln
ri tf�
ri
§¨
©
·
¹
˜ tw ln
re tf�
ri tf�
§¨
©
·
¹
˜� bf ln
re
re tf�
§¨
©
·
¹
˜�� 
IA_r 12.245 mm 
R
A
IA_r
� R 306.243 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
Maximum tensile stress: r ri� 
Vt_max
M R r�( )˜
A r˜ rc R�� �˜� Vt_max 4.51 MPa (T) Ans
Maximum compressive stress: r re� 
Vc_max
M R r�( )˜
A r˜ rc R�� �˜� Vc_max 5.44� MPa (C) Ans
Problem 6-133
The curved beam is subjected to a bending moment of M = 900 N·m as shown. Determine the stress
at points A and B, and show the stress on a volume element located at each of these points.
Given: bf 100mm� tf 20mm� ri 400mm� 
tw 15mm� dw 150mm� 
M 900� N m˜� T 30deg� 
Solution:
Internal Moment :
M is negative since it tends to decrease the
beam's radius of curvature.
Section Property : re ri dw� tf�� 
A bf tf˜ dw tw˜�� A 4250 mm2 
r
 6 ri

Ai˜� �˜
6 Ai� �˜= rc
dw tw˜� � ri 0.5dw�� �˜ bf tf˜� � re 0.5tf�� �˜�
A
� 
rc 515.000 mm 
³ AA_r rdAȈI IA_r tw ln
re tf�
ri
§¨
©
·
¹
˜ bf ln
re
re tf�
§¨
©
·
¹
˜�� 
IA_r 8.349 mm 
R
A
IA_r
� R 509.067 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
At A: rA re� VA
M R rA�� �˜
A rA˜ rc R�� �˜� VA 3.82 MPa (T) Ans
At B: rB ri� VB
M R rB�� �˜
A rB˜ rc R�� �˜� VB 9.73� MPa (C) Ans
Problem 6-134
The curved beam is subjected to a bending moment of M = 900 N·m. Determine the stress at point C.
Given: bf 100mm� tf 20mm� ri 400mm� 
tw 15mm� dw 150mm� 
M 900� N m˜� T 30deg� 
Solution:
Internal Moment :
M is negative since it tends to decrease the
beam's radius of curvature.
Section Property : re ri dw� tf�� 
A bf tf˜ dw tw˜�� A 4250 mm2 
r
 6 ri

Ai˜� �˜
6 Ai� �˜= rc
dw tw˜� � ri 0.5dw�� �˜ bf tf˜� � re 0.5tf�� �˜�
A
� 
rc 515.000 mm 
³ AA_r rdAȈI IA_r tw ln
re tf�
ri
§¨
©
·
¹
˜ bf ln
re
re tf�
§¨
©
·
¹
˜�� 
IA_r 8.349 mm 
R
A
IA_r
� R 509.067 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
At C: rC re tf�� VC
M R rC�� �˜
A rC˜ rc R�� �˜� VC 2.66 MPa (T) Ans
Problem 6-135
The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple
as shown, determine the maximum tensile and compressive stress acting at section a-a. Sketch the
stress distribution on the section in three dimensions.
Given: b 50mm� h 75mm� ri 162.5mm� 
P 250N� T 60deg� a 150mm� 
Solution:
Internal Moment :
M P a sin T� �˜ h cos T� �˜�� �˜� 
M 41.851 N m˜ 
M is positive since it tends to increase the
beam's radius of curvature.Section Property : re ri h�� 
A b h˜� A 3750 mm2 
rc ri 0.5 h˜�� rc 200 mm 
³ AA_r rdAȈI IA_r b ln
re
ri
§¨
©
·
¹
˜� IA_r 18.974 mm 
R
A
IA_r
� R 197.634 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
At A: rA re� VA
M R rA�� �˜
A rA˜ rc R�� �˜� VA 0.792� MPa (C) Ans
At B: rB ri� VB
M R rB�� �˜
A rB˜ rc R�� �˜� VB 1.020 MPa (T) Ans
Problem 6-136
The circular spring clamp produces a compressive force of 3 N on the plates. Determine the maximum
bending stress produced in the spring at A. The spring has a rectangular cross section as shown.
Given: b 20mm� h 10mm� ri 200mm� 
P 3N� a 220mm� 
Solution:
Internal Moment :
M P a˜� 
M 0.660 N m˜ 
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re ri h�� 
A b h˜� A 200 mm2 
rc ri 0.5 h˜�� rc 205 mm 
³ AA_r rdAȈI IA_r b ln
re
ri
§¨
©
·
¹
˜� IA_r 0.976 mm 
R
A
IA_r
� R 204.959 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
Maximum tensile stress: rA ri� 
Vt_max
M R rA�� �˜
A rA˜ rc R�� �˜� Vt_max 2.01 MPa (T) Ans
Maximum compressive stress: r'A re� 
Vc_max
M R r'A�� �˜
A r'A˜ rc R�� �˜� Vc_max 1.95� MPa (C) Ans
Problem 6-137
Determine the maximum compressive force the spring clamp can exert on the plates if the allowable
bending stress for the clamp is V allow = 4 MPa.
Given: b 20mm� h 10mm� ri 200mm� a 220mm� 
Vt.allow 4MPa� Vc.allow 4� MPa� 
Solution:
Section Property : re ri h�� 
A b h˜� A 200 mm2 
rc ri 0.5 h˜�� rc 205 mm 
³ AA_r rdAȈI IA_r b ln
re
ri
§¨
©
·
¹
˜� IA_r 0.976 mm 
R
A
IA_r
� R 204.959 mm 
Internal Moment :
Mmax P a R�( )˜=
M is positive since it tends to increase the
beam's radius of curvature.
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜= M
V A˜ r˜ rc R�� �˜
R r�=
Assume tension failure.
r ri� P a R�( )˜
Vt.allow A˜ r˜ rc R�� �˜
R r�= P
Vt.allow A˜ r˜ rc R�� �˜
a R�( ) R r�( )˜� 
P 3.087 N 
Assume compression failure.
r' re� P' a R�( )˜
Vc.allow A˜ r'˜ rc R�� �˜
R r'�= P'
Vc.allow A˜ r'˜ rc R�� �˜
a R�( ) R r'�( )˜� 
P' 3.189 N 
Pallow min P P'�( )� Pallow 3.087 N Ans
Problem 6-138
While in flight, the curved rib on the jet plane is subjected to an anticipated moment of M = 16 N·m a
the section. Determine the maximum bending stress in the rib at this section, and sketch a two-
dimensional view of the stress distribution.
Given: bf 30mm� tf 5mm� 
tw 5mm� dw 20mm� 
M 16N m˜� ri 600mm� 
Solution:
Internal Moment :
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re ri dw� 2tf�� 
A 2bf tf˜ dw tw˜�� A 400 mm2 
r
 6 ri

Ai˜� �˜
6 Ai� �˜= rc
bf tf˜� � ri 0.5tf�� �˜ dw tw˜� � ri 0.5dw� tf�� �˜� bf tf˜� � re 0.5tf�� �˜�
A
� 
rc 615.000 mm 
³ AA_r rdAȈI IA_r bf ln
ri tf�
ri
§¨
©
·
¹
˜ tw ln
re tf�
ri tf�
§¨
©
·
¹
˜� bf ln
re
re tf�
§¨
©
·
¹
˜�� 
IA_r 0.651 mm 
R
A
IA_r
� 
R 614.793 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
Maximum tensile stress: r ri� 
Vt_max
M R r�( )˜
A r˜ rc R�� �˜� Vt_max 4.77 MPa (T) Ans
Maximum compressive stress: r' re� 
Vc_max
M R r'�( )˜
A r'˜ rc R�� �˜� Vc_max 4.67� MPa (C) Ans
Problem 6-139
The steel rod has a circular cross section. If it is gripped at its ends and a couple moment of M = 1.5
N·m is developed at each grip, determine the stress acting at points A and B and at the centroid C.
Given: rci 50mm� rce 75mm� 
ro 12mm� M 1.5N m˜� 
Solution:
Internal Moment :
M = 1.5Nm is positive since it tends to
increase the beam's radius of curvature.
Section Property :
A S ro2˜� rc 0.5 rci rce�� �� 
³ AA_r rdAȈI IA_r 2 S˜ rc rc2 ro2��§© ·¹˜� 
R
A
IA_r
� R 61.919 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
rA rci� VA
M R rA�� �˜
A rA˜ rc R�� �˜� VA 1.3594 MPa (T) Ans
rB rce� VB
M R rB�� �˜
A rB˜ rc R�� �˜� VB 0.9947� MPa (C) Ans
rC rc� VC
M R rC�� �˜
A rC˜ rc R�� �˜� VC 0.0531� MPa (C) Ans
Problem 6-140
The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple
as shown, determine the maximum tensile and compressive stresses acting at section a-a. Sketch the
stress distribution on the section in three dimensions.
Given: b 50mm� h 75mm� 
ri 100mm� P 250N� 
d' 50mm� d 150mm� 
Solution:
Internal Moment :
M P d˜� 
M 37.5 N m˜ 
M is positive since it tends to increase the
beam's radius of curvature.
Section Property : re ri h�� 
A b h˜� A 3750 mm2 
rc ri 0.5 h˜�� rc 137.5 mm 
³ AA_r rdAȈI IA_r b ln
re
ri
§¨
©
·
¹
˜� IA_r 27.981 mm 
R
A
IA_r
� R 134.021 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜=
Maximum tensile stress: r ri� 
Vt_max
M R r�( )˜
A r˜ rc R�� �˜� Vt_max 0.978 MPa (T) Ans
Maximum compressive stress: r' re� 
Vc_max
M R r'�( )˜
A r'˜ rc R�� �˜� Vc_max 0.673� MPa (C) Ans
Problem 6-141
The member has an elliptical cross section. If it is subjected to a moment of M = 50 N·m, determine
the stress at points A and B. Is the stress at point A', which is located on the member near the wall, the
same as that at A? Explain.
Given: ri 100mm� La 150mm� Lb 75mm� 
M 50N m˜� 
Solution:
Internal Moment :
M is positive since it tends to increase the
member's radius of curvature.
Section Property : ao 0.5La� bo 0.5Lb� 
re ri La�� 
A S ao˜ bo˜� A 8835.73 mm2 
rc ri ao�� rc 175 mm 
³ AA_r rdAȈI IA_r
2S bo˜
ao
rc rc
2
ao
2��§© ·¹˜=
R
A
IA_r
= R
S ao˜ bo˜ ao� �˜
2 S˜ bo˜ rc rc2 ao2��§© ·¹˜
� 
R 166.557 mm 
Bending Stress: V M R r�( )˜
A r˜ rc R�� �˜=
rA ri� VA
M R rA�� �˜
A rA˜ rc R�� �˜� VA 0.446 MPa (T) Ans
rB re� VB
M R rB�� �˜
A rB˜ rc R�� �˜� VB 0.224� MPa (C) Ans
No. The stress at point A' is not the same as that at A,
because of localized stress concentration. Ans
Problem 6-142
The member has an elliptical cross section. If the allowable bending stress is V allow = 125 MPa,
determine the maximum moment M that can be applied to the member.
Given: ri 100mm� La 150mm� Lb 75mm� 
Vt.allow 125MPa� Vc.allow 125� MPa� 
Solution:
Internal Moment :
M is positive since it tends to increase the
member's radius of curvature.
Section Property : ao 0.5La� bo 0.5Lb� 
re ri La�� 
A S ao˜ bo˜� A 8835.73 mm2 
rc ri ao�� rc 175 mm 
³ AA_r rdAȈI IA_r
2S bo˜
ao
rc rc
2
ao
2��§© ·¹˜=
R
A
IA_r
= R
S ao˜ bo˜ ao� �˜
2 S˜ bo˜ rc rc2 ao2��§© ·¹˜
� 
R 166.557 mm 
Normal Stress: V M R r�( )˜
A r˜ rc R�� �˜= M
V A˜ r˜ rc R�� �˜
R r�=
Assume tension failure.
r ri� M
Vt.allow A˜ r˜ rc R�� �˜
R r�= M
Vt.allow A˜ r˜ rc R�� �˜
R r�� 
M 14.01 kN m˜ 
Assume compression failure.
r' re� M'
Vc.allow A˜ r'˜ rc R�� �˜
R r'�= M'
Vc.allow A˜ r'˜ rc R�� �˜
R r'�� 
M' 27.94 kN m˜ 
Mallow min M M'�( )� Mallow 14.01 kN m˜ Ans
Problem 6-143
The bar has a thickness of 6.25 mm and is made of a material having an allowable bending stress of
V allow = 126 MPa. Determine the maximum moment M that can be applied.
Given: t 6.25mm� r 6.25mm� 
w 100mm� h 25mm� 
Vallow 126MPa� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 8138.02 mm4 
Stress Concentration Factor :
w
h
4 r
h
0.25 
From Fig. 6-48, K 1.45� 
Maximum Moment : V K M c˜
I
˜=
c 0.5 h˜� Mmax
Vallow� � I˜
K c˜� 
Mmax 56.57 N m˜ Ans
Problem 6-144
The bar has a thickness of 12.5 mm and is subjected to a moment of 90 N·m. Determine the maximum
bending stress in the bar.
Given: t 12.5mm� r 6.25mm� 
w 100mm� h 25mm� 
M 90N m˜� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 16276.04 mm4 
Stress Concentration Factor :
w
h
4 r
h
0.25 
From Fig. 6-48, K 1.45� 
MaximumBending Stress :
c 0.5 h˜� Vmax K
M c˜
I
˜� 
Vmax 100.2 MPa Ans
Problem 6-145
The bar is subjected to a moment of M = 40 N·m. Determine the smallest radius r of the fillets so that
an allowable bending stress of V allow = 124 MPa is not exceeded.
Given: w 80mm� h 20mm� t 7mm� 
M 40N m˜� Vallow 124MPa� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 4666.67 mm4 
Allowable Bending Stress : V K M c˜
I
˜=
c 0.5 h˜� K
Vallow I˜
M c˜� 
K 1.45 
Stress Concentration Factor :
From Fig. 6-48, with K 1.45 w
h
4 
then,
r
h
0.25=
r 0.25 h˜� 
r 5.00 mm Ans
Problem 6-146
The bar is subjected to a moment of M = 17.5 N · m. If r = 5 mm, determine the maximum bending
stress in the material.
Given: w 80mm� h 20mm� 
t 7mm� r 5mm� 
M 17.5N m˜� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 4666.67 mm4 
Stress Concentration Factor :
w
h
4 r
h
0.25 
From Fig. 6-48, K 1.45� 
Maximum Bending Stress :
c 0.5 h˜� Vmax K
M c˜
I
˜� 
Vmax 54.4 MPa Ans
Problem 6-147
The bar is subjected to a moment of M = 20 N·m. Determine the maximum bending stress in the bar
and sketch, approximately, how the stress varies over the critical section.
Given: w 30mm� h 10mm� 
t 5mm� r 1.5mm� 
M 20N m˜� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 416.67 mm4 
Stress Concentration Factor :
w
h
3 r
h
0.15 
From Fig. 6-48, K 1.6� 
Maximum Bending Stress :
c 0.5 h˜� Vmax K
M c˜
I
˜� 
Vmax 384 MPa Ans
Problem 6-148
The allowable bending stress for the bar is V allow = 175 MPa. Determine the maximum moment M
that can be applied to the bar.
Given: w 30mm� h 10mm� 
t 5mm� r 1.5mm� 
Vallow 175MPa� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 416.67 mm4 
Stress Concentration Factor :
w
h
3 r
h
0.15 
From Fig. 6-48, K 1.6� 
Maximum Moment : V K M c˜
I
˜=
c 0.5 h˜� M
Vallow� � I˜
K c˜� 
M 9.11 N m˜ Ans
Problem 6-149
Determine the maximum bending stress developed in the bar if it is subjected to the couples shown.
The bar has a thickness of 6 mm.
Given: t 6mm� w 108mm� 
h1 72mm� h2 36mm� 
r1 7.2mm� r2 27mm� 
M1 20N m˜� M2 7.5N m˜� 
Mw 12.5N m˜� 
Solution:
Section Property:
For the larger section 1: For the smaller section 2:
I1
1
12
t˜ h13˜� I1 186624 mm4 I2
1
12
t˜ h23˜� I2 23328 mm4 
Stress Concentration Factor :
For the larger section 1: For the smaller section 2:
w
h1
1.5 
r1
h1
0.1 w
h2
3 
r2
h2
0.75 
From Fig. 6-48, K1 1.755� From Fig. 6-48, K2 1.15� 
Maximum Moment : V K M c˜
I
˜=
For the larger section 1: For the smaller section 2:
c1 0.5 h1˜� V1 K1
M1 c1˜
I1
˜� c2 0.5 h2˜� V2 K2
M2 c2˜
I2
˜� 
V1 6.77 MPa V2 6.66 MPa 
Vmax max V1 V2�� �� 
Vmax 6.77 MPa Ans
Problem 6-150
Determine the length L of the center portion of the bar so that the maximum bending stress at A, B,
and C is the same.The bar has a thickness of 10 mm.
Given: w 60mm� t 10mm� 
h 40mm� r 7mm� 
a 200mm� P 350N� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 53333.33 mm4 Support Reaction : By symmetry, A =B = R
+ 6Fy=0; 2R P� 0=
Stress Concentration Factor :
R 0.5P� R 175 N 
w
h
1.5 r
h
0.175 
 Internal Moment : 
MA R a˜� MA 35.00 N m˜ From Fig. 6-48, K 1.5� 
MB MA� 
Maximum Bending Stresses at A and B :
MC R a 0.5L�( )˜=
c 0.5 h˜� VA.max K
MA c˜
I
˜� 
VA.max 19.688 MPa 
VB.max VA.max� 
At Section C-C:
VC.max VA.max� Require,
I'
1
12
t˜ w3˜� I' 180000 mm4 
Maximum Bending Stress :
c' 0.5 w˜� VC.max
MC c'˜
I'
=
VA.max
R a 0.5L�( )˜ c'˜
I'
=
L
2VA.max I'˜
R c'˜ 2a�
§¨
©
·
¹� 
L 950 mm Ans
Problem 6-151
If the radius of each notch on the plate is r = 10 mm, determine the largest moment M that can be
applied. The allowable bending stress for the material is V allow = 180 MPa.
Given: w 165mm� h 125mm� 
t 20mm� r 10mm� 
Vallow 180MPa� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 3255208.33 mm4 
Stress Concentration Factor :
b 0.5 w h�( )˜� 
b
r
2 r
h
0.08 
From Fig. 6-50, K 2.1� 
Maximum Moment : V K M c˜
I
˜=
c 0.5 h˜� M
Vallow� � I˜
K c˜� 
M 4.464 kN m˜ Ans
Problem 6-152
The stepped bar has a thickness of 15 mm. Determine the maximum moment that can be applied to its
ends if it is made of a material having an allowable bending stress of V allow = 200 MPa.
Given: w 45mm� t 15mm� 
h1 30mm� h2 10mm� 
r1 3mm� r2 6mm� 
Vallow 200MPa� 
Solution:
Section Property:
For the larger section 1: For the smaller section 2:
I1
1
12
t˜ h13˜� I1 33750 mm4 I2
1
12
t˜ h23˜� I2 1250 mm4 
Stress Concentration Factor :
For the larger section 1: For the smaller section 2:
w
h1
1.5 
r1
h1
0.1 
h1
h2
3 
r2
h2
0.6 
From Fig. 6-48, K1 1.75� From Fig. 6-48, K2 1.2� 
Maximum Moment : V K M c˜
I
˜=
For the larger section 1: For the smaller section 2:
c1 0.5 h1˜� M1
Vallow� � I1˜
K1 c1˜
� c2 0.5 h2˜� M2
Vallow� � I2˜
K2 c2˜
� 
M1 257.14 N m˜ M2 41.67 N m˜ 
Mallow min M1 M2�� �� 
Mallow 41.67 N m˜ Ans
Problem 6-153
The bar has a thickness of 12.5 mm and is made of a material having an allowable bending stress of
V allow = 140 MPa. Determine the maximum moment M that can be applied.
Given: t 12.5mm� r 7.5mm� 
w 150mm� h 50mm� 
Vallow 140MPa� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 130208.33 mm4 
Stress Concentration Factor :
w
h
3 r
h
0.15 
From Fig. 6-48, K 1.6� 
Maximum Moment : V K M c˜
I
˜=
c 0.5 h˜� M
Vallow� � I˜
K c˜� 
M 455.73 N m˜ Ans
Problem 6-154
The bar has a thickness of 12.5 mm and is subjected to a moment of 900 N·m. Determine the
maximum bending stress in the bar.
Given: t 12.5mm� r 7.5mm� 
w 150mm� h 50mm� 
M 900N m˜� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 130208.33 mm4 
Stress Concentration Factor :
w
h
3 r
h
0.15 
From Fig. 6-48, K 1.6� 
Maximum Bending Stress :
c 0.5 h˜� Vmax K
M c˜
I
˜� 
Vmax 276.5 MPa Ans
Problem 6-155
The simply supported notched bar is subjected to two forces P. Determine the largest magnitude of P
that can be applied without causing the material to yield. The material is A-36 steel. Each notch has a
radius of r = 3 mm.
Given: t 12mm� r 3mm� 
w 42mm� h 30mm� 
a 500mm� VY 250MPa� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 27000.00 mm4 
 Support Reaction : By symmstry, R1=R2=R
+
Stress Concentration Factor : 6Fy=0; 2R 2P� 0= R P=
b 0.5 w h�( )˜� 
 Internal Moment : At mid-span,
b
r
2 r
h
0.1 
MC R a˜=
From Fig. 6-50, K 1.92� MC P a˜=
Maximum Moment : VY K
MC c˜
I
˜=
c 0.5 h˜� 
MC
VY� � I˜
K c˜=
P a˜
VY� � I˜
K c˜=
P
VY� � I˜
a K˜ c˜� 
P 468.75 N Ans
Problem 6-156
The simply supported notched bar is subjected to the two loads, each having a magnitude of P = 500
N. Determine the maximum bending stress developed in the bar, and sketch the bending-stress
distribution acting over the cross section at the center of the bar. Each notch has a radius of r = 3 mm.
Given: t 12mm� r 3mm� 
w 42mm� h 30mm� 
a 500mm� P 500N� 
Solution:
Section Property:
I
1
12
t˜ h3˜� I 27000.00 mm4 
 Support Reaction : By symmstry, R1=R2=R
+
Stress Concentration Factor : 6Fy=0; 2R 2P� 0= R P� 
b 0.5 w h�( )˜� 
 Internal Moment : At mid-span,
b
r
2 r
h
0.1 
MC R a˜� 
From Fig. 6-50, K 1.92� MC 250 N m˜ 
Maximum Bending Stress :
c 0.5 h˜� Vmax K
MC c˜
I
˜� 
Vmax 266.7 MPa Ans
Problem 6-157
A rectangular A-36 steel bar has a width of 25 mm and height of 75 mm. Determine the moment
applied about the horizontal axis that will cause halfthe bar to yield.
Given: b 25mm� VY 250MPa� 
d 75mm� 
Solution: de 0.5d� dp 0.5d� 
Elastic-plastic Moment:
M VY b
de
2
˜§¨©
·
¹˜
2de
3
˜ VY b
dp
2
˜§¨©
·
¹˜ de
dp
2
�§¨©
·
¹˜�� 
M 9.52 kN m˜ Ans
Problem 6-158
The box beam is made of an elastic perfectly plastic material for which VY = 250 MPa. Determine the
residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then
released.
Given: bo 200mm� do 200mm� 
bi 150mm� di 150mm� 
VY 250MPa� 
Solution:
Section Property:
tb 0.5 do di�� �� td 0.5 bo bi�� �� 
I
1
12
bo do
3˜ bi di3˜�§© ·¹˜� I 91145833.33 mm4 
Plastic Moment:
Mp VY bo tb˜� �˜ do tb�� �˜ VY 2td di2˜§¨©
·
¹˜
di
2
§¨
©
·
¹˜�� 
Mp 289062.50 N m˜ 
Modulus of Rupture:
The modulus of rupture Vr can be determined using the flexure formula
with the application of reverse plastic moment Mp.
c 0.5do� Vr
Mp c˜
I
� Vr 317.14 MPa 
Residul Bending Stress:
V't Vr VY�� V't 67.14 MPa Ans
V'b Vr VY�� V'b 67.14 MPa Ans
Problem 6-159
The box beam is made of an elastic plastic material for which VY = 250 MPa. Determine the residual
stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.
Given: bf 200mm� dw 200mm� 
tf 15mm� tw 20mm� 
VY 250MPa� 
Solution:
Section Property:
D dw 2tf�� I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 82783333.33 mm
4 
Plastic Moment:
Mp VY bf tf˜� �˜ D tf�� �˜ VY tw dw2˜§¨©
·
¹˜
dw
2
§¨
©
·
¹˜�� 
Mp 211.25 kN m˜ 
Modulus of Rupture:
The modulus of rupture Vr can be determined using the flexure formula
with the application of reverse plastic moment Mp.
c 0.5D� Vr
Mp c˜
I
� Vr 293.46 MPa 
Residul Bending Stress:
V't Vr VY�� V't 43.46 MPa Ans
V'b Vr VY�� V'b 43.46 MPa Ans
Problem 6-160
Determine the plastic section modulus and the shape factor of the beam's cross section.
Set a mm� 
Given: bf 2a� tf a� 
dw 2a� tw a� 
Solution:
Section Property :
A bf tf˜ dw tw˜�� 
yc
6 yi

Ai˜� �˜
6 Ai� �˜= yc
bf tf˜� � 0.5tf� �˜ dw tw˜� � 0.5dw tf�� �˜�
A
� 
yc 1.25 a 
I
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜� 112 tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�ª«¬ º»¼�� 
I 3.08 a
4 
Maximum Elastic Moment :
c dw tf� yc�� c 1.75 a 
VY
MY c˜
I
=
MY
VY
I
c
=
I
c
§¨
©
·
¹ 1.7619 a
3 
MY 1.7619 a
3� � VY˜=
Plastic Moment :
0dAı
A
 ³ VY tw˜ d( )˜ VY tw˜ dw d�� �˜� VY bf tf˜� �˜� 0=
d
tw dw˜ bf tf˜�
2tw
� d 2 a 
darm 0.5tf dw� 0.5d�� darm 1.50 a 
Mp VY tw˜ d( )˜ darm˜=
Mp
VY
tw d( )˜ darm˜= tw d( )˜ darm˜ 3.00 a3 
Mp 3.00 a
3� � VY˜=
Shape Factor : k
Mp
MY
= k
3.00 a
3
1.7619 a
3
� k 1.70 Ans
Plastic Section Modulus : z
Mp
VY
= z 3.00 a
3
= Ans
Problem 6-161
The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and th
plastic moment that can be applied to the cross section. Take a = 50 mmand VY = 230 MPa.
Given: a 50mm� bf 2a� tf a� 
VY 230MPa� dw 2a� tw a� 
Solution:
Section Property :
A bf tf˜ dw tw˜�� 
yc
6 yi

Ai˜� �˜
6 Ai� �˜= yc
bf tf˜� � 0.5tf� �˜ dw tw˜� � 0.5dw tf�� �˜�
A
� 
yc 62.5 mm 
I
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜� 112 tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�ª«¬ º»¼�� 
I 19270833.33 mm
4 
Maximum Elastic Moment :
c dw tf� yc�� c 87.50 mm 
VY
MY c˜
I
= MY
VY I˜
c
� MY 50.65 kN m˜ Ans
Plastic Moment :
0dAı
A
 ³ VY tw˜ d( )˜ VY tw˜ dw d�� �˜� VY bf tf˜� �˜� 0=
d
tw dw˜ bf tf˜�
2tw
� d 100 mm 
darm 0.5tf dw� 0.5d�� darm 75.00 mm 
Mp VY tw˜ d( )˜ darm˜� 
Mp 86.25 kN m˜ Ans
Problem 6-162
The rod has a circular cross section. If it is made of an elastic plastic material, determine the shape
factor and the plastic section modulus Z.
Set r mm� 
Solution:
Section Property :
A Sr2� I S
4
r
4� �� 
Maximum Elastic Moment :
c r� VY
MY c˜
I
=
MY
VY
I
c
=
I
c
§¨
©
·
¹ 0.7854 r
3 
MY 0.7854 r
3� � VY˜=
Plastic Moment :
darm 2
4r
3S
§¨
©
·
¹� 
Mp VY
A
2
§¨
©
·
¹˜ darm˜=
Mp
VY
A
2
§¨
©
·
¹ darm˜=
A
2
§¨
©
·
¹ darm˜ 1.3333 r
3 
Mp 1.3333 r
3� � VY˜=
Shape Factor : k
Mp
MY
= k
1.3333 r
3
0.7854 r
3
� k 1.70 Ans
Plastic Section Modulus : z
Mp
VY
= z 1.333 r
3
= Ans
Problem 6-163
The rod has a circular cross section. If it is made of an elastic plastic material, determine the maximum
elastic moment and plastic moment that can be applied to the cross section.Take r = 75 mm, VY = 250
MPa.
Given: r 75mm� VY 250MPa� 
Solution:
Section Property :
A Sr2� I S
4
r
4� �� 
Maximum Elastic Moment :
c r� VY
MY c˜
I
= MY
VY I˜
c
� 
MY 82.83 kN m˜ Ans
Plastic Moment :
darm 2
4r
3S
§¨
©
·
¹� 
Mp VY
A
2
§¨
©
·
¹˜ darm˜� Mp 140.63 kN m˜ Ans
Problem 6-164
Determine the plastic section modulus and the shape factor of the cross section.
Set a mm� 
Given: bf 3a� tf a� 
dw 3a� tw a� 
Solution:
Section Property :
A bf tf˜ dw tf�� � tw˜�� A 5 a2 
I
1
12
bf˜ tf3˜
1
12
tw˜ dw3 tf3�§© ·¹˜�� I 2.41667 a4 
Maximum Elastic Moment :
c 0.5dw� c 1.5 a 
VY
MY c˜
I
=
MY
VY
I
c
=
I
c
§¨
©
·
¹ 1.61111 a
3 
MY 1.61111 a
3� � VY˜=
Plastic Moment :
dw.arm tf 0.5 dw tf�� ��� dw.arm 2.00 a 
df.arm 0.5tf� df.arm 0.50 a 
Mp VY bf
tf
2
˜§¨©
·
¹˜ df.arm˜ VY tw˜
dw tf�
2
§¨
©
·
¹˜ dw.arm˜�=
Mp
VY
bf
tf
2
˜§¨©
·
¹ df.arm˜ tw
dw tf�
2
§¨
©
·
¹˜ dw.arm˜�=
bf
tf
2
˜§¨©
·
¹ df.arm˜ tw
dw tf�
2
§¨
©
·
¹˜ dw.arm˜� 2.75 a
3 
Mp 2.75 a
3� � VY˜=
Shape Factor : k
Mp
MY
= k
2.75 a
3
1.61111 a
3
� k 1.71 Ans
Plastic Section Modulus : z
Mp
VY
= z 2.75 a
3
= Ans
Problem 6-165
The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and th
plastic moment that can be applied to the cross section. Take a = 50 mm and VY = 250 MPa.
Given: a 50mm� VY 250MPa� 
bf 3a� tf a� 
dw 3a� tw a� 
Solution:
Section Property :
A bf tf˜ dw tf�� � tw˜�� A 5 a2 
I
1
12
bf˜ tf3˜
1
12
tw˜ dw3 tf3�§© ·¹˜�� I 2.41667 a4 
Maximum Elastic Moment :
c 0.5dw� c 1.5 a 
VY
MY c˜
I
= MY
VY I˜
c
� MY 50.35 kN m˜ Ans
Plastic Moment :
dw.arm tf 0.5 dw tf�� ��� dw.arm 2.00 a 
df.arm 0.5tf� df.arm 0.50 a 
Mp VY bf
tf
2
˜§¨©
·
¹˜ df.arm˜ VY tw˜
dw tf�
2
§¨
©
·
¹˜ dw.arm˜�� 
Mp 85.94 kN m˜ Ans
Problem 6-166
The beam is made of an elastic perfectly plastic material. Determine the plastic moment Mp that can be
supported by a beam having the cross section shown. VY = 210 MPa.
Given: ro 50mm� tw 25mm� VY 210MPa� 
ri 25mm� dw 250mm� 
Solution:
 Plastic Moment :
A1 S ro2 ri2�§© ·¹˜� darm1 dw 2ro�� 
A2 tw 0.5dw� �� darm2 0.5dw� 
Mp VY A1˜� � darm1˜ VY A2˜� � darm2˜�� 
Mp 515 kN m˜ Ans
Problem 6-167
Determine the plastic moment Mp that can be supported by a beam having the cross section shown.
VY = 210 MPa.
Given: ro 50mm� tw 25mm� VY 210MPa� 
ri 25mm� dw 250mm� 
Solution:
A1 S ro2 ri2�§© ·¹˜=
A2 tw dw d'�� �=
A3 tw d'=
0dAı
A
 ³ VY A1˜ VY A2˜� VY A3˜� 0=
A1 A2� A3� 0=
S ro2 ri2�§© ·¹˜ tw dw d'�� �� tw d'� 0=
d'
S ro2 ri2�§© ·¹˜ tw dw� ��
2tw
� 
d' 242.81 mm 
 Plastic Moment :
A1 S ro2 ri2�§© ·¹˜� darm1 ro dw d'�� ��� 
A2 tw dw d'�� �� darm2 0.5 dw d'�� �� 
A3 tw d'� darm3 0.5d'� 
Mp VY A1˜� � darm1˜ VY A2˜� � darm2˜� VY A3˜� � darm3˜�� 
Mp 225.6 kN m˜ Ans
Problem 6-168
Determine the plastic section modulus and the shape factor for the member having thetubular cross
section.
Set d mm� 
Solution:
Section Property :
A
S
4
2d( )
2
d
2�ª¬ º¼� A 2.35619 d2 
I
S
64
2d( )
4
d
4�ª¬ º¼� I 0.73631 d4 
Maximum Elastic Moment :
c d� VY
MY c˜
I
=
MY
VY
I
c
=
I
c
§¨
©
·
¹ 0.73631 d
3 
MY 0.73631 d
3� � VY˜=
Plastic Moment :
yc
6 yi

Ai˜� �˜
6 Ai� �˜= yc
0.5
S
4
§¨
©
·
¹ 2d( )
2 4d
3S
§¨
©
·
¹˜ 0.5
S
4
§¨
©
·
¹ d
2 1
2
˜ 4d
3S
§¨
©
·
¹�
0.5A
� 
yc 0.49515 d 
darm 2yc� 
Mp VY
A
2
§¨
©
·
¹˜ darm˜=
Mp
VY
A
2
§¨
©
·
¹ darm˜=
A
2
§¨
©
·
¹ darm˜ 1.16667 d
3 
Mp 1.16667 d
3� � VY˜=
Shape Factor : k
Mp
MY
= k
1.16667 d
3
0.73631 d
3
� k 1.58 Ans
Plastic Section Modulus : z
Mp
VY
= z 1.16667 d
3
= Ans
Problem 6-169
Determine the plastic section modulus and the shape factor for the member.
Solution: Set b mm� h mm� 
Section Property :
A
1
2
b h˜( )= I 1
36
b h
3˜� �=
Maximum Elastic Moment :
c
2
3
h˜� VY
MY c˜
I
=
MY
VY
I
c
=
I
c
§¨
©
·
¹
1
24
b˜ h2˜=
MY
1
24
b˜ h2˜§¨©
·
¹ VY˜=Plastic Moment :
From the geometry, b'
d
h
b˜=
A'
1
2
b'˜ d˜= Atrp
1
2
b b'�( )˜ h d�( )˜=
A'
1
2
d
h
b˜§¨©
·
¹˜ d˜= Atrp
1
2
b
d
h
b˜�§¨©
·
¹˜ h d�( )˜=
0dAı
A
 ³ VY A'� �˜ VY Atrp� �˜� 0=
A' Atrp=
1
2
d
h
b˜§¨©
·
¹˜ d˜
1
2
b
d
h
b˜�§¨©
·
¹˜ h d�( )˜= d
h
2
= b'
b
2
=
Note: The centroid of a trapezoidal area was used
in the calculation.
hc
h d�
3
2 b'˜ b�
b' b�˜=
darm
1
3
d˜ h d�( ) hc�ª¬ º¼�=
darm
1
3
d˜ h d�( ) b' 2 b˜�
3 b' b�( )�=
darm
1
3
h
2
˜ h h
2
�§¨©
·
¹
b 2 2 b˜�
3 b 2 b˜�� �˜�= darm
4 2 2�� � h˜
6
=
Mp VY
A
2
§¨
©
·
¹˜ darm˜=
Mp
VY
A
2
§¨
©
·
¹ darm˜=
A
2
darm˜
2 2�
6
b˜ h2˜=
Mp
2 2�
6
b˜ h2˜§¨©
·
¹ VY˜=
Shape Factor : k
Mp
MY
= k
2 2�
6
b˜ h2˜
1
24
b˜ h2˜
� k 2.34 Ans
Plastic Section Modulus : z
Mp
VY
= z
2 2�
6
b˜ h2˜= Ans
Problem 6-170
The member is made of elastic perfectly plastic material for which VY = 230 MPa. Determine the
maximum elastic moment and the plastic moment that can be applied to the cross section.Take b = 50
mm and h = 80 mm.
Given: b 50mm� h 80mm� VY 230MPa� 
Solution:
Section Property :
A
1
2
b h˜( )� A 2000 mm2 
I
1
36
b h
3˜� �� I 711111.11 mm4 
Maximum Elastic Moment :
c
2
3
h˜� VY
MY c˜
I
= MY
VY I˜
c
� MY 3.07 kN m˜ Ans
Plastic Moment :
From the geometry, b'
d
h
b˜= A'
1
2
b'˜ d˜= Atrap
1
2
b b'�( )˜ h d�( )˜=
A'
1
2
d
h
b˜§¨©
·
¹˜ d˜= Atrap
1
2
b
d
h
b˜�§¨©
·
¹˜ h d�( )˜=
0dAı
A
 ³ VY A'� �˜ VY Atrap� �˜� 0= A' Atrap=
1
2
d
h
b˜§¨©
·
¹˜ d˜
1
2
b
d
h
b˜�§¨©
·
¹˜ h d�( )˜= d
h
2
= b'
b
2
=
Note: The centroid of a trapezoidal area was used
in the calculation.
hc
h d�
3
2 b'˜ b�
b' b�˜=
darm
1
3
d˜ h d�( ) hc�ª¬ º¼�=
darm
1
3
d˜ h d�( ) b' 2 b˜�
3 b' b�( )�=
darm
1
3
h
2
˜ h h
2
�§¨©
·
¹
b 2 2 b˜�
3 b 2 b˜�� �˜�=
darm
4 2 2�� � h˜
6
� darm 31.24 mm 
Mp VY
A
2
§¨
©
·
¹˜ darm˜� 
Mp 7.19 kN m˜ Ans
Problem 6-171
The wide-flange member is made from an elasticplastic material. Determine the shape factor and the
plastic section modulus Z.
Set b mm� h mm� t mm� 
Given: bf b� tf t� 
D h� tw t� 
Solution: dw D 2 tf˜�= dw h 2t�� 
Section Property :
A bf D˜ bf tw�� � dw˜�= A b h˜ b t�( ) h 2t�( )˜�� 
I
1
12
bf˜ D3˜
1
12
bf tw�� �˜ dw3˜�=
I
1
12
b˜ h3˜ 1
12
b t�( )˜ h 2t�( )3˜�� 
Maximum Elastic Moment :
c 0.5D� c 0.5h� 
VY
MY c˜
I
=
MY
VY
I
c
=
I
c
1
6
b˜ h2˜ 1
6 h˜ b t�( )˜ h 2t�( )
3˜�=
MY
1
6 h˜ b h
3˜ b t�( ) h 2 t˜�( )3˜�ª¬ º¼˜ VY˜=
Plastic Moment :
dw.arm 0.5dw= dw.arm
h 2t�
2
=
df.arm D tf�= df.arm h t�=
Mp VY bf tf˜� �˜ df.arm˜ VY tw˜ dw2§¨©
·
¹˜ dw.arm˜�=
Mp
VY
b t˜( ) h t�( )˜ t h 2t�
2
§¨
©
·
¹˜
h 2t�
2
˜�=
Mp b t˜( ) h t�( )˜
t
4
h 2t�( )2˜�ª«¬
º»¼ VY˜=
Shape Factor : k
Mp
MY
= k
b t˜( ) h t�( )˜ t
4
h 2t�( )2˜�
1
6 h˜ b h
3˜ b t�( ) h 2 t˜�( )3˜�ª¬ º¼˜
=
k
3 h˜
2
4b t˜ h t�( )˜ t h 2t�( )2˜�
b h
3˜ b t�( ) h 2 t˜�( )3˜�
ª««¬
º»»¼
=
Ans
Plastic Section Modulus : z
Mp
VY
= z b t˜( ) h t�( )˜ t
4
h 2t�( )2˜�=
Ans
Problem 6-172
The beam is made of an elastic-plastic material for which VY = 200 MPa. If the largest moment in the
beam occurs within the center section a-a, determine the magnitude of each force P that causes this
moment to be (a) the largest elastic moment and (b) the largest plastic moment.
Given: a 2m� b 100mm� h 200mm� 
VY 200MPa� 
Solution:
Section Property :
A b h˜� A 20000 mm2 
I
1
12
b˜ h3˜� I 66666666.67 mm4 
a) Maximum Elastic Moment :
c 0.5 h˜� MY P a˜= VY
MY c˜
I
=
VY
P a˜ c˜
I
= P
VY I˜
a c˜� 
P 66.67 kN Ans
b) Plastic Moment :
darm
h
2
� Mp VY
A
2
§¨
©
·
¹˜ darm˜� Mp 200.00 kN m˜ 
P'
Mp
a
� P' 100.00 kN Ans
Problem 6-173
The beam is made of phenolic, a structural plastic, that has the stress-strain curve shown. If a portion
of the curve can be represented by the equation V = (5(106)H )1/2 MPa, determine the magnitude of w
the distributed load that can be applied to the beam without causing the maximum strain in its fibers a
the critical section to exceed Hmax = 0.005 mm/mm.
Given: b 150mm� h 150mm� 
V2 5 106� � H=
L 2m� Hmax 0.005
mm
mm
� 
Solution:
Stress-strain Relationship : unit MPa� 
When Hmax = 0.005, 
Vmax unit 5 106� � Hmax˜� 
Vmax 158.11 MPa 
Resultant Internal Forces :
The resultant internal forces T and C can be evaluated from the volume of
stress block which is a paraboloid, T = C.
T
2
3
Vmax˜
b h˜
2
§¨
©
·
¹˜� T 1185.85 kN 
darm 2
3
5
h
2
§¨
©
·
¹˜
ª«¬
º»¼˜� darm 90 mm 
 Maximum Internal Moment :
Mmax T darm˜� Mmax 106.73 kN m˜ 
By observation, the maximum moment occurs over the middle support.
Mmax w L˜= w
Mmax
L
� 
w 53.36 m
kN
m
 Ans
Problem 6-174
The box beam is made from an elastic plastic material for which VY = 175 MPa. Determine the
intensity of the distributed load w0 that will cause the moment to be (a) the largest elastic moment and
(b) the largest plastic moment.
Given: bo 200mm� do 400mm� 
bi 150mm� di 300mm� 
L 3m� VY 175MPa� 
Solution:
 Support Reaction : By symmstry, R1=R2=R
+ 6Fy=0; 2R 2 0.5 wo˜ L˜� �� 0=
R 0.5 wo˜ L˜=
 Maximum Moment :
M R L˜ L
3
0.5 wo˜ L˜� ��= M wo L
2˜
3
=
a) Elastic Analysis : V M c˜
I
=
c 0.5do� I
1
12
bo do
3˜ bi di3˜�§© ·¹˜� 
MY
VY I˜
c
� MY 638.02 kN m˜ 
wo
3 MY˜
L
2
� wo 212.67
kN
m
 Ans
b) Plastic Analysis : tw 0.5 bo bi�� �� tf 0.5 do di�� �� 
Af bo tf˜� darm1 do tf�� 
Aw di tw˜� darm2 0.5di� 
Mp VY Af˜� � darm1˜ VY Aw˜� � darm2˜�� 
Mp 809.4 kN m˜ 
w'o
3 Mp˜
L
2
� w'o 269.79
kN
m
 Ans
Problem 6-175
The beam is made of a polyester that has the stress-strain curve shown. If the curve can be
represented by the equation V = [140 tan-1(15H )] MPa, where tan-1(15H�) is in radians, determine the
magnitude of the force P that can be applied to the beam without causing the maximum strain in its
fibers at the critical section to exceed Hmax = 0.003 mm/mm.
Given: b 50mm� h 100mm� 
V 140 atan 15H� �=
L 2.4m� Hmax 0.003
mm
mm
� 
Solution:
 Support Reaction : By symmstry, R1=R2=R+ 6Fy=0; 2R 2P� 0=
R P=
 Maximum Moment :
M R L˜= M P L˜=
Stress-strain Relationship : unit MPa� 
The bending stress can bs expressed
in terms of y using
H
Hmax
0.5h
y˜=
V 140 atan 15
Hmax
0.5h
y˜§¨©
·
¹ unit˜=
When Hmax = 0.003, ymax 0.5h� 
Vmax 140 atan 15
Hmax
0.5h
ymax˜
§¨
©
·
¹˜ unit˜� 
Vmax 6.30 MPa 
³ A dAıyMResultant Internal Moment :
M 2
0
0.5h
yy 140 atan 15
Hmax
0.5h
y˜§¨©
·
¹ unit˜
§¨
©
·
¹˜ b˜
µ´
µ¶
d� 
M 524.79 N m˜ 
P
M
L
� P 218.66 N Ans
Problem 6-176
The stress-strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut
made of this material is subjected to bending, determine the moment resisted by the strut if the
maximum stress reaches a value of (a) VA and (b) VB.
Given: b 50mm� d 75mm� 
VA 980MPa� VB 1260MPa� 
HA 0.01
mm
mm
� HB 0.04
mm
mm
� 
Solution:
 Maximum Elastic Moment :
Since the stress is linearly related to strain up
to point A, the flexure formula can be applied.
V M c˜
I
=
c 0.5d� I b d3˜� 
MY
VA I˜
c
� MY 551.25 kN m˜ Ans
 UItimate Moment :
yA
HA
HB
0.5d( )˜� h 0.5d yA�� 
C1 T1= T1
VA VB�
2
b h˜( )˜� 
C2 T2= T2
VA
2
b yA˜� �˜� 
Note: The centroid of a trapezoidal area was used
in the calculation of moment.
hc
h
3
2VB VA�
VB VA�
˜� 
darm1 2 yA hc�� �� 
darm2
2
3
2yA� �� 
M T1� � darm1˜ T2� � darm2˜�� 
M 78.54 kN m˜ Ans
Ʌ�
not zero.
Problem 6-177
A beam is made from polypropylene plastic and has a stress-strain diagram that can be approximated
by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of H = 0.02
mm/mm, determine the maximum moment M.
Given: b 30mm� h 100mm� 
V 10 4 H MPa=
Hmax 0.02
mm
mm
� 
Solution:
Stress-strain Relationship : unit MPa� 
The bending stress can bs expressed
in terms of y using
H
Hmax
0.5h
y˜=
V 10
4 Hmax
0.5h
y˜ unit˜=
³ A dAıyMResultant Internal Moment :
M 2
0
0.5h
yy 10
4 Hmax
0.5h
y˜ unit˜
§¨
©¨
·
¹˜ b˜
µ´
µ
µ¶
d� 
M 0.251 kN m˜ Ans
Problem 6-178
The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by the
straight line segments shown. Assuming that this diagram is the same for both tension and
compression, determine the moment the bar will support if the maximum strain at the top and bottom
fibers of the beam is Hmax = 0.03.
Given: b 75mm� d 100mm� 
VA 420MPa� VB 560MPa� 
HA 0.006
mm
mm
� HB 0.025
mm
mm
� 
VC 630MPa� HC 0.05
mm
mm
� 
Hmax 0.03
mm
mm
� 
Solution:
 Maximum Stress :
Vmax VB�
Hmax HB�
VC VB�
HC HB�
= Vmax
VC VB�
HC HB�
Hmax HB�� �˜ VB�� Vmax 574 MPa 
 Maximum Moment :
yA
HA
Hmax
0.5d( )˜� yB
HB
Hmax
0.5d( )˜� h1 0.5d yB�� h2 yB yA�� 
C1 T1= T1
VB Vmax�
2
b h1˜� �˜� 
C2 T2= T2
VA VB�
2
b h2˜� �˜� 
C3 T3= T3
VA
2
b yA˜� �˜� 
Note: The centroid of a trapezoidal area was used
in the calculation of moment.
hc1
h1
3
2Vmax VB�
Vmax VB�
˜� hc2
h2
3
2VB VA�
VB VA�
˜� 
darm1 2 yB hc1�� �� 
darm2 2 yA hc2�� �� 
darm3
2
3
2yA� �� 
M T1� � darm1˜ T2� � darm2˜� T3� � darm3˜�� 
M 96.48 kN m˜ Ans
Problem 6-179
The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by the
straight line segments shown. Assuming that this diagram is the same for both tension and
compression, determine the moment the bar will support if the maximum strain at the top and bottom
fibers of the beam is Hmax = 0.05.
Given: b 75mm� d 100mm� 
VA 420MPa� VB 560MPa� 
HA 0.006
mm
mm
� HB 0.025
mm
mm
� 
VC 630MPa� HC 0.05
mm
mm
� 
Solution: Hmax HC� 
 Stress-strain Relationship :
V1
H
VA
HA
= V1
VA
HA
H˜=
V2 VA�
H HA�
VB VA�
HB HA�
= V2
VB VA�
HB HA�
H HA�� �˜ VA�=
V3 VB�
H HB�
VC VB�
HC HB�
= V3
VC VB�
HC HB�
H HB�� �˜ VB�=
yA
HA
Hmax
0.5d( )˜� yB
HB
Hmax
0.5d( )˜� 
Strain : H
Hmax
0.5d
y˜=
V1
VA
HA
Hmax
0.5d
y˜§¨©
·
¹˜= for 0 < y < yA
V2
VB VA�
HB HA�
Hmax
0.5d
y˜ HA�
§¨
©
·
¹˜ VA�= for yA < y < yB
V3
VC VB�
HC HB�
Hmax
0.5d
y˜ HB�
§¨
©
·
¹˜ VB�= for yB < y < 0.5d
³ A dAıyMResultant Moment :
M1 2
0
yA
yy
VA
HA
Hmax
0.5d
y˜§¨©
·
¹˜
ª«¬
º»¼
˜ b˜µ´µ¶
d� M1 0.76 kN m˜ 
M2 2
yA
yB
yy
VB VA�
HB HA�
Hmax
0.5d
y˜ HA�
§¨
©
·
¹˜ VA�
ª«¬
º»¼
˜ b˜µ´µ
µ¶
d� M2 22.28 kN m˜ 
M3 2
yA
yB
yy
VC VB�
HC HB�
Hmax
0.5d
y˜ HB�
§¨
©
·
¹˜ VB�
ª«¬
º»¼
˜ b˜µ´µ
µ¶
d� M3 23.8 kN m˜ 
M M1 M2� M3�� 
M 46.84 kN m˜ Ans
Note: The solution can also be obtained from stress blocks as in Prob, 6-178
Problem 6-180
The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectly
plastic in compression. Determine the maximum bending moment M that can be supported by the bea
so that the compressive material at the outer edge starts to yield.
Solution:
A1 a d˜= C
1
2
VY A1˜=
A2 a h d�( )= T VY A2˜=
0dAı
A
 ³ C T� 0=
1
2
VY A1˜ VY A2˜� 0=
1
2
A1 A2� 0=
1
2
a d˜ a h d�( )� 0=
d
2 h˜
3
=
 Plastic Moment :
darm
2
3
d
1
2
h d�( )�= darm
2
3
2 h˜
3
1
2
h
2 h˜
3
�§¨©
·
¹�= darm
11 h˜
18
=
Mp VY A2˜� � darm˜=
Mp VY a˜ h d�( )˜
11 h˜
18
˜=
Mp
11 a˜ h2˜
54
VY˜= Ans
Problem 6-181
The plexiglass bar has a stress-strain curve that can be approximated by the straight-line segments
shown. Determine the largest moment M that can be applied to the bar before it fails.
Given: b 20mm� h 20mm� 
Vt1 40MPa� Vt2 60MPa� 
Ht1 0.02
mm
mm
� Ht2 0.04
mm
mm
� 
Vc1 80� MPa� Vc2 100� MPa� 
Hc1 0.04�
mm
mm
� Hc2 0.06�
mm
mm
� 
Solution:
Assume failure due to tension and Hc Hc1�
A1 b h d�( )= C
1
2
Vc A1˜=
A2
1
2
b˜ d˜= T1
1
2
Vt1 A2˜=
A3
1
2
b˜ d˜= T2
1
2
Vt1 Vt2�� � A3˜=
0dAı
A
 ³ C T1� T2� 0=
1
2
Vc b h d�( )[ ]˜
1
2
Vt1
1
2
b˜ d˜§¨©
·
¹˜�
1
2
Vt1 Vt2�� � 12 b˜ d˜§¨© ·¹˜� 0=
Vc
h
d
1�§¨©
·
¹˜ Vt.1 0.5Vt.2�=
Try Vc 74.833MPa� then d
h Vc˜
Vt1 0.5Vt2� Vc�
� d 10.334 mm 
Vc Vc1�
Check :
From the strain diagram, Hc
h d�
d
Ht2˜� Hc 0.037417
mm
mm
 O.K! Hc Hc1�
From the V-H diagram, V
Hc
Hc1
Vc1˜� V 74.833 MPa O.K! Close to assumed value.
Hence,
C
1
2
Vc b h d�( )[ ]˜� C 7.2336 kN 
T1
1
2
Vt1
1
2
b˜ d˜§¨©
·
¹˜� T1 2.0667 kN 
T2
1
2
Vt1 Vt2�� � 12 b˜ d˜§¨© ·¹˜� T2 5.1668 kN 
 Ultimate Moment :
darm1
2
3
h d�( )� darm1 6.4442 mm 
darm2
2
3
d
2
§¨
©
·
¹� darm2 3.4446 mm 
Note: The centroid of a trapezoidal area was used
in the calculation.
hc
0.5d
3
2Vt1 Vt2�
Vt1 Vt2�
˜� hc 2.4112 mm 
darm3 d hc�� darm3 7.9225 mm 
Mult C darm1˜ T1 darm2˜� T2 darm3˜�� 
Mult 94.67 N m˜ Ans
Problem 6-182
The beam is made from three boards nailed together as shown. If the moment acting on the cross
section is M = 650 N·m, determine the resultant force the bending stress produces on the top board.
Given: bf 290mm� tf 15mm� 
tw 20mm� dw 125mm� 
M 650N m˜� 
Solution: D dw tf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf 2 dw tw˜� � 0.5dw tf�� �˜�
bf tf˜ 2dw tw˜�
� 
yc 44.933 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
2tw� �˜ dw3˜ 2tw dw˜� � yc 0.5dw tf�� ��ª¬ º¼2˜�� 
I If Iw�� I 17990374.89 mm4 
Bending Stress: V M c
I
˜=
At B: cB yc tf�� VB M
cB
I
˜� VB 1.0815 MPa 
At A: cA yc� VA M
cA
I
˜� VA 1.6235 MPa 
ResultantForce: For the top board.
F 0.5 VA VB�� � bf tf˜� �˜� F 5.883 kN Ans
Problem 6-183
The beam is made from three boards nailed together as shown. Determine the maximum tensile and
compressive stresses in the beam.
Given: bf 290mm� tf 15mm� 
tw 20mm� dw 125mm� 
M 650N m˜� 
Solution: D dw tf�� 
y
 6 yi

Ai˜� �˜
6 Ai� �˜=
yc
bf tf˜� � 0.5˜ tf 2 dw tw˜� � 0.5dw tf�� �˜�
bf tf˜ 2dw tw˜�
� 
yc 44.933 mm 
If
1
12
bf˜ tf3˜ bf tf˜� � yc 0.5tf�� �2˜�� 
Iw
1
12
2tw� �˜ dw3˜ 2tw dw˜� � yc 0.5dw tf�� ��ª¬ º¼2˜�� 
I If Iw�� I 17990374.89 mm4 
Maximum Bending Stress: V M c
I
˜=
For compression:
cc yc� Vc_max M
cc
I
˜� Vc_max 1.623 MPa Ans
For tension:
ct D yc�� Vt_max M
ct
I
˜� Vt_max 3.435 MPa Ans
Problem 6-184
Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as
functions of x, where 0 xd < 1.8 m.
Given: a 2.4m� b 1.2m� 
P 40kN� 
w 30
kN
m
� 
M 75kN m˜� 
Solution:
 Equilibrium :
+
A w a˜ P�� A 112 kN 6Fy=0;
 MA w a˜( ) 0.5a( )˜ P a b�( )˜� M�� 60A=0;
MA 305.40 kN m˜ 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� 
V1 x1� � A w x1˜�� � 1kN˜� V2 x2� � A w a˜�( ) 1kN˜� Ans
M1 x1� � MA� A x1˜� 0.5w x12˜�§© ·¹ 1kN m˜˜� Ans
M2 x2� � MA� A x2˜� w a˜( ) x2 0.5 a˜�� �˜� M�ª¬ º¼ 1kN m˜˜� 
0 1 2 3
0
50
100
150
Distance (m)
S
h
ea
r 
(k
N
)
V1 x1� �
V2 x2� �
x1 x2�
0 1 2 3
400
200
0
Distane (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
x1 x2�
Problem 6-185
Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced by
equivalent loadings at point C on the axis of the beam.
Given: a 1.2m� b 1.2m� 
c 1.2m� d 0.3m� 
F1 75kN� F2 100kN� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A F1� B� 0=
 60C=0; A a b� c�( )˜ F1 b c�( )˜� F2 d( )˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
58.33
16.67
§¨
©
·
¹ kN 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
V1 x1� � A 1kN˜� V2 x2� � A F1�� � 1kN˜� V3 x3� � A F1�� � 1kN˜� 
M1 x1� � A x1˜kN m˜� M2 x2� � A x2� �˜ F1 x2 a�� �˜�ª¬ º¼ 1kN m˜˜� 
M3 x3� � A x3� �˜ F1 x3 a�� �˜� F2 d˜�ª¬ º¼ 1kN m˜˜� 
0 1 2 3
0
50
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
0 1 2 3
0
50
100
Distance (m)
M
o
m
en
t 
(k
N
-m
)
M1 x1� �
M2 x2� �
M3 x3� �
x1 x2� x3�
Problem 6-186
Determine the plastic section modulus and the shape factor for the wide-flange beam.
Given: bf 180mm� tf 20mm� 
dw 180mm� tw 30mm� 
Solution: D dw 2 tf˜�� 
Section Property :
A bf D˜ bf tw�� � dw˜�� A 12600 mm2 
I
1
12
bf˜ D3˜
1
12
bf tw�� �˜ dw3˜�ª«¬ º»¼� I 86820000 mm4 
Set VY MPa� 
Maximum Elastic Moment :
c 0.5D� VY
MY c˜
I
= MY
I
c
§¨
©
·
¹ VY˜� 
MY
VY
789272.73 mm
3 
Plastic Moment :
dw.arm 0.5dw� dw.arm 90 mm 
df.arm D tf�� df.arm 200 mm 
Mp bf tf˜� � df.arm˜ tw dw2§¨©
·
¹˜ dw.arm˜�
ª«¬
º»¼ VY˜� 
Mp
VY
963000.00 mm
3 
Plastic Section Modulus : z
Mp
VY
� z 963 10 6�u m3 Ans
Shape Factor : k
Mp
MY
� k 1.22 Ans
Problem 6-187
Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt,
gear, and flywheel. The bearings at A and B exert only vertical reactions on the shaft.
Given: a 200mm� b 400mm� 
c 300mm� d 200mm� 
C 450N� D 300� N� 
E 150N� 
Solution:
 Equilibrium : Given
+ 6Fy=0; A B� C� D� E� 0=
 60B=0; A a b� c�( )˜ C b c�( )˜� D c˜� E d˜� 0=
Guess A 1N� B 1N� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
216.67
83.33
§¨
©
·
¹ N 
x1 0 0.01 a˜� a��� x2 a 1.01 a˜� a b���� x3 a b� 1.01 a b�( )˜� a b� c���� 
x4 a b� c� 1.01 a b� c�( )˜� a b� c� d���� 
V1 x1� � A 1N˜� V2 x2� � A C�( ) 1N˜� V3 x3� � A C� D�( ) 1N˜� 
V4 x4� � A C� D� B�( ) 1N˜� 
M1 x1� � A x1˜N m˜� M2 x2� � A x2� �˜ C x2 a�� �˜�ª¬ º¼ 1N m˜˜� 
M3 x3� � A x3� �˜ C x3 a�� �˜� D x3 a� b�� �˜�ª¬ º¼ 1N m˜˜� 
M4 x4� � A x4� �˜ C x4 a�� �˜� D x4 a� b�� �˜� B x4 a� b� c�� �˜�ª¬ º¼ 1N m˜˜� 
0 0.2 0.4 0.6 0.8 1
400
200
0
200
400
Distance (m)
S
h
ea
r 
(N
)
V1 x1� �
V2 x2� �
V3 x3� �
V4 x4� �
x1 x2� x3� x4�
0 0.2 0.4 0.6 0.8 1
50
0
50
Distance (m)
M
o
m
en
t 
(N
-m
) M1 x1� �
M2 x2� �
M3 x3� �
M4 x4� �
x1 x2� x3� x4�
Problem 6-188
The beam is constructed from four pieces of wood, glued together as shown. If the internal bending
moment is M = 120 kN·m, determine the maximum bending stress in the beam. Sketch a
three-dimensional view of the stress distribution acting over the cross section.
Given: bo 300mm� do 300mm� 
bi 250mm� di 250mm� 
M 120kN m˜� 
Solution:
Section Property :
I
1
12
bo do
3˜ bi di3˜�§© ·¹˜� 
Maximum Bending Stress: V M c˜
I
=
cmax 0.5do� 
Vmax
M cmax˜
I
� 
Vmax 51.51 MPa Ans
Problem 6-189
The beam is constructed from four pieces of wood, glued together as shown. If the internal bending
moment is M = 120 kN·m, determine the resultant force the bending moment exerts on the top and
bottom boards of the beam.
Given: bo 300mm� do 300mm� 
bi 250mm� di 250mm� 
M 120kN m˜� 
Solution:
Section Property : t 0.5 do di�� �� 
I
1
12
bo do
3˜ bi di3˜�§© ·¹˜� 
Maximum Bending Stress: V M c˜
I
=
co 0.5do� Vo
M co˜
I
� Vo 51.505 MPa 
ci 0.5di� Vi
M ci˜
I
� Vi 42.921 MPa 
Resultant Force :
F
1
2
Vo Vi�� �˜ t bo˜� �˜� 
F 354.10 kN Ans
Problem 6-190
For the section, Iy =31.7(10
-6) m4, Iz = 114(10
-6) m4, Iyz = 15.1(10
-6) m4. Using the techniques outline
in Appendix A, the member's cross-sectional area has principal moments of inertia of Iy' = 29(10
-6) m
and Iz' = 117(10
-6) m4, computed about the principal axes of inertia y' and z', respectively. If the sectio
is subjected to a moment of M = 2 kN·m directed as shown, determine the stress produced at point A,
(a) using Eq. 6-11 and (b) using the equation developed in Prob. 6-111.
Given: M 2000N m˜� T 10.10deg� 
b1 80mm� b2 140mm� 
h1 60mm� h2 60mm� 
Iy 31.7 10
6�� � m4� Iz 114 10 6�� � m4� 
Iyz 15.1 10
6�� � m4� 
Iy' 29.0 10
6�� � m4� Iz' 117 10 6�� � m4� 
Solution: T' T�� 
Coordinates of Point A : yA b2� zA h2� 
y'A
z'A
§¨
©¨
·
¹
cos T'� �
sin T'� �
sin T'� ��
cos T'� �
§¨
©
·
¹
yA
zA
§¨
©¨
·
¹
˜� 
y'A
z'A
§¨
©¨
·
¹
148.35
34.52
§¨
©
·
¹ mm 
a) Using Eq. 6-11
Internal Moment Components :
My' M sin T� �˜� Mz' M cos T� �˜� 
Bending Stress:
VA
Mz' y'A˜
Iz'
�
My' z'A˜
Iy'
�§¨
©
·
¹
� VA 2.079� MPa (C) Ans
b) Using the equation developed in Prob. 6-111
Internal Moment Components :
My 0� Mz M� 
Bending Stress: Using formula developed in Prob. 6-111.
D Iy Iz˜ Iyz2�� 
VA
1
D
Mz My� � Iy
Iyz
Iyz
Iz
§¨
©¨
·
¹
˜
yA�
zA
§¨
©¨
·
¹
˜� 
VA 2.086� MPa (C) Ans
Problem 6-191
The strut has a square cross section a by a and is subjected to the bending moment M applied at an
angle T as shown. Determine the maximum bending stress in terms of a, M, and T. What angle T wil
give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.
Solution:
Internal Moment Components :
My M� sin T� �˜= Mz M� cos T� �˜=
Section Property :
Iy
a
4
12
= Iz
a
4
12
=
Maximum Bending Stress:
By inspection, maximum bending stress occurs at A (and B).
At A : yA 0.5a= zA 0.5� a=
V
Mz y˜
Iz
�
My z˜
Iy
�=
V 12� M cos T� �˜ 0.5a( )˜
a
4
� 12� M sin T� �˜ 0.5� a( )˜
a
4
�=
V 6 M˜
a
3
cosT� � sin T� ��� �˜= Ans
TV
d
d
6 M˜
a
3
sin T� �� cos T� ��� �˜=
TV
d
d
0= sin T� �� cos T� �� 0=
tan T� � 1= T 45deg� Ans
Orientation of Neutral Axis : T' T�� 
tan D� � Iz
Iy
tan T'� �˜= Iz
Iy
1=
D atan 1 tan T'� �˜� �� 
D 45.00� deg Ans
Problem 7-1
If the beam is subjected to a shear of V = 15 kN, determine the web's shear stress at A and B. Indicate
the shear-stress components on a volume element located at these points. Set w = 125 mm. Show that
the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182(10
-3) m4.
Given: bf 200mm� b'f 125mm� tf 30mm� 
tw 25mm� dw 250mm� 
V 15kN� 
Solution:
Section Property : D dw 2tf�� 
A bf tf˜ dw tw˜� b'f tf˜�� A 16000 mm2 
y
 6 yi

Ai˜� �˜
6 Ai� �˜= yc
b'f tf˜� � 0.5tf� �˜ dw tw˜� � 0.5dw tf�� �˜� bf tf˜� � D 0.5tf�� �˜�
A
� 
yc 174.69 mm 
I'f
1
12
b'f˜ tf3˜ b'f tf˜� � 0.5tf yc�� �2˜�� 
Iw
1
12
tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�� 
If
1
12
bf˜ tf3˜ bf tf˜� � D 0.5tf� yc�� �2˜�� 
I If Iw� I'f�� I 218.18 10 6�u m4 
Q 6 yi
˜ A1˜= QA D 0.5tf� yc�� � bf tf˜� �˜� QA 721875.00 mm3 
QB yc 0.5tf�� � b'f tf˜� �˜� QB 598828.12 mm3 
Shear Stress: W V Q˜
I t˜=
WA
V QA˜
I tw˜
� WA 1.99 MPa Ans
WB
V QB˜
I tw˜
� WB 1.65 MPa Ans
Problem 7-2
If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in
the beam. Set w = 200 mm.
Given: bf 200mm� tf 30mm� 
tw 25mm� dw 250mm� 
V 30kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 268.65 10
6�u m4 
Q 6 yi
˜ A1˜= Qmax bf tf˜� � D2
tf
2
�§¨©
·
¹˜
dw
2
tw˜
§¨
©
·
¹
dw
4
§¨
©
·
¹˜�� 
Qmax 1035312.50 mm
3 
Shear Stress: W V Q˜
I t˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I tw˜
� Wmax 4.62 MPa Ans
Problem 7-3
If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the
web of the beam. Set w = 200 mm.
Given: bf 200mm� dw 250mm� 
tf 30mm� tw 25mm� 
V 30kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 268.65 10
6�u m4 
Af 0.5D y�( ) bf˜= yf 0.5 0.5D y�( ) y�=
yf 0.5 0.5D y�( )=
Qf Af yf˜=
Qf 0.5 0.5D y�( ) bf˜ 0.5D y�( )˜=
Qf 0.5bf 0.25D
2
y
2�� �=
Shear Stress: W V Q˜
I t˜=
Wf
V
I bf˜
§¨
©
·
¹
Qf˜=
Wf
V
I bf˜
§¨
©
·
¹
0.5bf 0.25D
2
y
2�� �ª¬ º¼˜=
³ A 0dAV WResultant Shear Force:
For the flange,
Vf
V
I bf˜ 0.5dw
0.5D
y0.5 bf˜ 0.25 D2˜ y2�� �˜ª¬ º¼ bf˜µ´¶ d
ª««¬
º»»¼
˜� 
Vf 1.46 kN 
Vw V 2Vf�� 
Vw 27.09 kN Ans
Problem 7-4
If the wide-flange beam is subjected to a shear of V = 125 kN, determine the maximum shear stress in
the beam.
Given: bf 200mm� dw 250mm� 
tf 25mm� tw 25mm� 
V 125kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
Qmax bf tf˜� � D2
tf
2
�§¨©
·
¹˜
dw
2
tw˜
§¨
©
·
¹
dw
4
§¨
©
·
¹˜�� 
Maximum Shear Stress: W V Q˜
I t˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I tw˜
� 
Wmax 19.87 MPa Ans
Problem 7-5
If the wide-flange beam is subjected to a shear of V = 125 kN, determine the shear force resisted by
the web of the beam.
Given: bf 200mm� dw 250mm� 
tf 25mm� tw 25mm� 
V 125kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
A1 bf tf˜= y1c 0.5 D tf�� �=
A2 0.5dw y�� � tw˜= y2c 0.5 0.5dw y�� � y�=
y2c 0.5 0.5dw y�� �=
Qw A1 y1c� �˜ A2 y2c� �˜�=
Qw 0.5bf tf˜ D tf�� �˜ 0.5 0.5dw y�� � tw˜ 0.5dw. y�� �˜�=
Qw 0.5bf tf˜ D tf�� �˜ 0.5tw 0.25dw2 y2�§© ·¹�=
Shear Stress: W V Q˜
I t˜=
Ww
V
I tw˜
§¨
©
·
¹
Qw˜=
Ww
V
I tw˜
§¨
©
·
¹
0.5 bf˜ tf˜ D tf�� �˜ 0.5 tw˜ 0.25 dw2˜ y2�§© ·¹˜�ª¬ º¼˜=
³ A 0dAV WResultant Shear Force: For the web.
Vw
0.5� dw
0.5dw
y
V
I tw˜
§¨
©
·
¹
0.5 bf˜ tf˜ D tf�� �˜ 0.5 tw˜ 0.25 dw2˜ y2�§© ·¹˜�ª¬ º¼˜ tw˜µ´µ¶
d� 
Vw 115.04 kN Ans
Problem 7-6
The beam has a rectangular cross section and is made of wood having an allowable shear stress of
Wallow = 11.2 MPa. If it is subjected to a shear of V = 20 kN, determine the smallest dimension a of its
bottom and 1.5a of its sides.
Given: V 20kN� Wallow 11.2MPa� 
Solution:
Section Property :
I
1
12
a˜ 1.5a( )3˜= t a=
Qmax
1.5a
2
a˜§¨©
·
¹
1.5a
4
§¨
©
·
¹˜=
Allowablwe Shear Stress: W V Q˜
I t˜=
I t˜
V Qmax˜
Wallow
=
1
12
a˜ 1.5a( )3˜ a˜ VWallow
§¨
©
·
¹
1.5a
2
a˜§¨©
·
¹˜
1.5a
4
§¨
©
·
¹˜=
a
V
Wallow
� 
a 42.26 mm Ans
Problem 7-7
The beam has a rectangular cross section and is made of wood. If it is subjected to a shear of V = 20
kN, and a = 250 mm, determine the maximum shear stress and plot the shearstress variation over the
cross section. Sketch the result in three dimensions.
Given: a 250mm� V 20kN� 
Solution:
Section Property : b a� d 1.5a� 
I
1
12
b˜ d3˜� 
Qmax
d
2
b˜§¨©
·
¹
d
4
§¨
©
·
¹˜� 
Maximum Shear Stress: W V Q˜
I b˜=
Maximum shear stress occurs at the point where
the neutral axis passes through the section.
Wmax
V Qmax˜
I b˜� 
Wmax 0.320 MPa Ans
Problem 7-8
Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.
Given: bf 120mm� tf 12mm� 
tw 80mm� dw 60mm� 
V 20kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 5.21 10
6�u m4 
Q 6 yi
˜ A1˜= Qmax bf tf˜� � D2
tf
2
�§¨©
·
¹˜
dw
2
tw˜
§¨
©
·
¹
dw
4
§¨
©
·
¹˜�� 
Qmax 87840.00 mm
3 
Shear Stress: W V Q˜
I t˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I tw˜
� 
Wmax 4.22 MPa Ans
Problem 7-9
Determine the maximum shear force V that the strut can support if the allowable shear stress for the
material is Wallow = 40 MPa.
Given: bf 120mm� tf 12mm� 
tw 80mm� dw 60mm� 
Wallow 40MPa� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 5.21 10
6�u m4 
Q 6 yi
˜ A1˜= Qmax bf tf˜� � D2
tf
2
�§¨©
·
¹˜
dw
2
tw˜
§¨
©
·
¹
dw
4
§¨
©
·
¹˜�� 
Qmax 87840.00 mm
3 
Shear Stress: W V Q˜
I t˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wallow
V Qmax˜
I tw˜
= V
I tw˜ Wallow˜
Qmax
� 
V 189.69 kN Ans
Problem 7-10
Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a
shear force of V = 15 kN.
Given: bf 120mm� tf 12mm� 
tw 80mm� dw 60mm� 
V 15kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 5.21 10
6�u m4 
Q 6 yi
˜ A1˜=
QA bf tf˜� � D2
tf
2
�§¨©
·
¹˜� QA 51840.00 mm
3 
Qmax bf tf˜� � D2
tf
2
�§¨©
·
¹˜
dw
2
tw˜
§¨
©
·
¹
dw
4
§¨
©
·
¹˜�� 
Qmax 87840.00 mm
3 
Shear Stress: W V Q˜
I t˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I tw˜
� Wmax 3.16 MPa Ans
Ww_A
V QA˜
I tw˜
� Ww_A 1.87 MPa Ans
Wf_A
V QA˜
I bf˜
� Wf_A 1.24 MPa Ans
Problem 7-11
If the pipe is subjected to a shear of V = 75 kN, determine the maximum shear stress in the pipe.
Given: ro 60mm� ri 50mm� V 75kN� 
Solution:
Section Property : t ro ri�� 
I
S
4
ro
4
ri
4�§© ·¹˜� 
Qmax
4ro
3S
S ro2˜
2
§¨
©
·
¹˜
4ri3S
S ri2˜
2
§¨
©
·
¹˜�� 
Maximum Shear Stress: W V Q˜
I b˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I 2t( )˜� 
Wmax 43.17 MPa Ans
Problem 7-12
The strut is subjected to a vertical shear of V = 130 kN. Plot the intensity of the shear-stress
distribution acting over the cross-sectional area, and compute the resultant shear force developed in the
vertical segment AB.
Given: bf 350mm� tf 50mm� 
tw 50mm� dw 350mm� 
V 130kN� 
Solution:
Section Property : a 0.5 bf tw�� �� 
I
1
12
bf tf
3˜ tw dw3˜� tw tf3˜�§© ·¹˜� 
I 181.77 10
6�u m4 
Q 6 yi
˜ A1˜=
QC a tw˜� � a2
tf
2
�§¨©
·
¹˜� QC 750000 mm
3 
QD a tw˜� � a2
tf
2
�§¨©
·
¹˜
tf
2
bf˜
§¨
©
·
¹
tf
4
§¨
©
·
¹˜�� QD 859375 mm
3 
W V Q˜
I t˜=Shear Stress:
Ww_C
V QC˜
I tw˜
� Wf_C
V QC˜
I bf˜
� WD
V QD˜
I bf˜
� 
Ww_C 10.73 MPa Wf_C 1.53 MPa WD 1.76 MPa 
³ A 0dAV WResultant Shear Force:
Aw 0.5dw y�� � tw˜= yw 0.5 0.5dw y�� � y�=
yw 0.5 0.5dw y�� �=
Qw Aw yw˜=
Qw 0.5 0.5dw y�� � tw˜ 0.5dw y�� �˜=
Q 0.5tw 0.25dw
2
y
2�§© ·¹=
Ww
V
I tw˜
§¨
©
·
¹
Qw˜= Ww
V
I tw˜
§¨
©
·
¹
0.5tw 0.25dw
2
y
2�§© ·¹ª¬ º¼˜=
VAB
V
I tw˜ 0.5tf
0.5dw
y0.5 tw˜ 0.25 dw2˜ y2�§© ·¹˜ª¬ º¼ tw˜µ´µ¶ d
ª««¬
º»»¼
˜� 
VAB 50.29 kN Ans
Problem 7-13
The steel rod has a radius of 30 mm. If it is subjected to a shear of V = 25 kN, determine the
maximum shear stress.
Given: r 30mm� V 25kN� 
Solution:
Section Property :
I
S
4
r
4˜� 
Qmax
4r
3S
S r2˜
2
§¨
©
·
¹˜� 
Maximum Shear Stress: W V Q˜
I b˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I 2r( )˜� 
Wmax 11.79 MPa Ans
Problem 7-14
If the T-beam is subjected to a vertical shear of V = 60 kN, determine the maximum shear stress in the
beam. Also, compute the shear-stress jump at the flange-web junction AB. Sketch the variation of the
shear-stress intensity over the entire cross section.
Given: bf 300mm� dw 150mm� 
tf 75mm� tw 100mm� 
V 60kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� ��
bf tf˜ dw tw˜�
� yc 82.50 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� 
Qmax tw D yc�� �˜ D yc�2˜� 
QAB bf tf˜� � yc 0.5tf�� �˜� 
Shear Stress: W V Q˜
I b˜=
Wmax
V Qmax˜
I tw˜
� Wmax 3.993 MPa Ans
Wf_AB
V QAB˜
I bf˜
� Wf_AB 1.327 MPa Ans
Ww_AB
V QAB˜
I tw˜
� Ww_AB 3.982 MPa Ans
Problem 7-15
If the T-beam is subjected to a vertical shear of V = 60 kN, determine the vertical shear force resisted
by the flange.
Given: bf 300mm� dw 150mm� 
tf 75mm� tw 100mm� 
V 60kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� ��
bf tf˜ dw tw˜�
� yc 82.50 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� 
A'f yc y�� � bf˜= yfc 0.5 yc y�� � y�=
yfc 0.5 yc y�� �=
Q A'f yfc˜=
Q 0.5 yc y�� � bf˜ yc y�� �=
Q 0.5bf yc
2
y
2�§© ·¹=
Shear Stress: W V Q˜
I b˜=
Wf
V
I bf˜
§¨
©
·
¹
Q˜=
Wf
V
I bf˜
§¨
©
·
¹
0.5bf yc
2
y
2�§© ·¹ª¬ º¼˜=
³ A 0dAV WResultant Shear Force: For the flange. yo yc tf�� 
Vf
yo
yc
y
V
I bf˜
§¨
©
·
¹
0.5bf yc
2
y
2�§© ·¹ª¬ º¼˜ bf˜µ´µ¶
d� 
Vf 19.08 kN Ans
Problem 7-16
The T-beam is subjected to the loading shown. Determine the maximum transverse shear stress in the
beam at the critical section.
Given: L1 2m� L2 2m� L3 3m� 
bf 100mm� dw 100mm� 
tf 20mm� tw 20mm� 
P 20kN� w 8 kN
m
� 
Solution: L L1 L2� L3�� 
Support Reaction :
 Equilibrium : Given
+ 6Fy=0; A P� w L3˜� B� 0=
 60B=0; A L P L2 L3�� �˜� w L3˜� � 0.5L3� �˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
19.43
24.57
§¨
©
·
¹ kN 
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� ��
bf tf˜ dw tw˜�
� yc 40.00 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� I 5333333.33 mm4 
Qmax tw D yc�� �˜ D yc�2˜� Qmax 64000 mm3 
Maximum Shear Stress: W V Q˜
I b˜= Vmax B� 
Wmax
Vmax Qmax˜
I( ) tw˜
� 
Wmax 14.74 MPa Ans
Shear Force Diagram:
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � AkN� V2 x2� � A P�( ) 1kN˜� V3 x3� � A P� w x3 L1� L2�� �˜�ª¬ º¼ 1kN˜� 
0 2 4 6
20
0
20
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Problem 7-17
Determine the largest end forces P that the member can support if the allowable shear stress is Wallow =
70 MPa. The supports at A and B only exert vertical reactions on the beam.
Given: L1 1m� Wallow 70MPa� 
L2 2m� 
w 3
kN
m
� 
L3 1m� 
do 100mm� bo 160mm� 
di 60mm� bi 80mm� 
Solution:
Section Property :
yc
0.5do bo do˜� � 0.5di� � bi di˜� ��
bo do˜ bi di˜�
� yc 58.57 mm 
I1
1
12
bo˜ do3˜ bo do˜� � 0.5do yc�� �2˜�� 
I2
1
12
bi˜ di3˜ bi di˜� � 0.5di yc�� �2˜�� I I1 I2�� 
Qmax A'f yfc˜= A'f yc bo bi�� �˜= y'c 0.5yc= Qmax 0.5yc2 bo bi�� �u=
Maximum Shear Stress: W V Q˜
I b˜= Vmax P=
Wallow
P
I bo bi�� �˜ª«¬
º»¼
0.5yc
2
bo bi�� �ª¬ º¼˜=
Wallow
P
I
§¨
©
·
¹ 0.5yc
2§© ·¹˜=
P
I
0.5yc
2
§¨
©
·
¹
Wallow� �˜� 
P 373.42 kN Ans
Shear Force Diagram: L L1 L2� L3�� 
 Equilibrium : Given
+ 6Fy=0; A w L2˜� B� 2P� 0=
 60B=0; P� L1 L2�� �˜ A L2˜� w L2˜� � 0.5L2� �˜� P L3˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
376.42
376.42
§¨
©
·
¹ kN 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � P�kN� V2 x2� � P� A� w x2 L1�� �˜�ª¬ º¼ 1kN˜� V3 x3� � P� A� w L2˜� B�� � 1kN˜� 
0 1 2 3 4
0
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Problem 7-18
If the force P = 4 kN, determine the maximum shear stress in the beam at the critical section. The
supports at A and B only exert vertical reactions on the beam.
Given: L1 1m� P 4kN� 
L2 2m� 
w 3
kN
m
� 
L3 1m� 
do 100mm� bo 160mm� 
di 60mm� bi 80mm� 
Solution:
Section Property :
yc
0.5do bo do˜� � 0.5di� � bi di˜� ��
bo do˜ bi di˜�
� yc 58.57 mm 
I1
1
12
bo˜ do3˜ bo do˜� � 0.5do yc�� �2˜�� 
I2
1
12
bi˜ di3˜ bi di˜� � 0.5di yc�� �2˜�� I I1 I2�� 
Qmax A'f yfc˜= A'f yc bo bi�� �˜= y'c 0.5yc= Qmax 0.5yc2 bo bi�� �u=
Maximum Shear Stress: W V Q˜
I b˜= Vmax P=
Wmax
P
I bo bi�� �˜ª«¬
º»¼
0.5yc
2
bo bi�� �ª¬ º¼˜=
Wmax
P
I
§¨
©
·
¹ 0.5yc
2§© ·¹˜� 
Wmax 0.750 MPa Ans
Shear Force Diagram: L L1 L2� L3�� 
 Equilibrium : Given
+ 6Fy=0; A w L2˜� B� 2P� 0=
 60B=0; P� L1 L2�� �˜ A L2˜� w L2˜� � 0.5L2� �˜� P L3˜� 0=
Guess A 1kN� B 1kN� 
A
B
§¨
©
·
¹ Find A B�( )� 
A
B
§¨
©
·
¹
7.00
7.00
§¨
©
·
¹ kN 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � P�kN� V2 x2� � P� A� w x2 L1�� �˜�ª¬ º¼ 1kN˜� V3 x3� � P� A� w L2˜� B�� � 1kN˜� 
0 1 2 3 4
10
0
10
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Problem 7-19
Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is
the maximum shear stress greater than the average shear stress acting over the cross section?
Problem 7-20
Develop an expression for the average vertical componentof shear stress acting on the horizontal plane
through the shaft, located a distance y from the neutral axis.
Problem 7-21
Railroad ties must be designed to resist large shear loadings. If the tie is subjected to the 150-kN rail
loadings and the gravel bed exerts a distributed reaction as shown, determine the intensity w for
equilibrium, and find the maximum shear stress in the tie.
Given: L1 0.45m� P 150kN� 
L2 0.90m� 
w 3
kN
m
� 
L3 0.45m� 
d 150mm� b 200mm� 
Solution:
 Equilibrium :
+ 6Fy=0; 0.5w L1˜ w L2˜� 0.5w L3˜� 2P� 0=
w
2P
0.5L1 L2� 0.5L3�
� 
w 222.22
kN
m
 
Section Property : I
1
12
b˜ d3˜� Qmax 0.5 b˜ d˜( ) 0.25˜ d� 
Shear Force Diagram: L L1 L2� L3�� 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � 0.5 x1˜ w x1˜L1
§¨
©
·
¹
1
kN
˜� V2 x2� � 0.5w L1˜ P� w x2 L1�� �˜�ª¬ º¼ 1kN˜� 
V3 x3� � 0.5 w˜ L1˜ 2 P˜� w L2˜� w x3 L1� L2�� �˜ 1 x3 L1� L2�2L3�
§¨
©
·
¹
˜�ª«¬
º»¼
1
kN
˜� 
0 0.5 1 1.5
100
0
100
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Maximum Shear Stress:
W V Q˜
I b˜=
Vmax V2 L1� � kN˜� 
Wmax
Vmax
I b˜
§¨
©
·
¹ Qmax˜� 
Wmax 5 MPa Ans
Problem 7-22
The beam is subjected to a uniform load w. Determine the placement a of the supports so that the shear
stress in the beam is as small as possible. What is this stress?
Set w
kN
m
� a m� L 5m� 
Given: L1 a� L3 a� 
L2 L 2a�� 
Solution:
 Equilibrium : By equilibrium, A = B = R
+
6Fy=0; 2R w L˜� 0=
R 0.5w L˜� 
Shear Force Diagram:
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � w x1˜� � 1kN˜� V2 x2� � w x2˜ R�� � 1kN˜� V3 x3� � w x3˜ 2 R˜�� � 1kN˜� 
0 1 2 3 4 5
0
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Require,
V1 a( ) V2 a( )�=
w a˜ w a˜ R�( )�=
2w a˜ 0.5w L˜=
a 0.25L=
Vmax w a˜= Vmax 0.25w L˜� 
Section Property : I
1
12
b˜ d3˜= Qmax b
d
2
˜§¨©
·
¹
d
4
˜=
Maximum Shear Stress: W V Q˜
I b˜=
Wmax
Vmax
I b˜
§¨
©
·
¹ Qmax˜= Wmax
0.25w L˜
1
12
b˜ d3˜ b˜
§¨
©¨
·
¹
b
d
2
˜§¨©
·
¹
d
4
˜ª«¬
º»¼˜=
Wmax
3w L˜
8 b˜ d˜= Ans
Problem 7-23
The timber beam is to be notched at its ends as shown. If it is to support the loading shown, determine
the smallest depth d of the beam at the notch if the allowable shear stress is Wallow = 3 MPa. The beam
has a width of 200 mm. 
Given: L1 1.2m� P1 12.5kN� 
L2 1.8m� P2 25.0kN� 
L3 1.8m� P3 12.5kN� 
L4 1.2m� b 200mm� 
Wallow 3MPa� 
Solution:
 Equilibrium : By symmetry, R1=R , R2=R 
+ 6Fy=0; P1 P2� P3� 2R� 0=
R 0.5 P1 P2� P3�� �� R 25.00 kN 
Section Property : I
1
12
b˜ d3˜=
Qmax 0.5 b˜ d˜( ) 0.25˜ d= Qmax 0.125 b˜ d2˜=
Maximum Shear Stress:
W V Q˜
I b˜= Vmax R� 
Wallow
R 0.125b d
2˜� �˜
1
12
b˜ d3˜§¨©
·
¹ b˜
=
d
12 0.125( )˜ R˜
Wallow� � b˜� 
d 62.5 mm Ans
Problem 7-24
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the shear stress developed in the glued joints at section a-a. The supports at
C and D exert only vertical reactions on the beam.
Given: bf 150mm� dw 200mm� 
tf 40mm� tw 50mm� 
P 25kN� 
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ 6Fy=0; 3P 2R� 0=
R 1.5P� 
R 37.50 kN 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜ dw3˜�� 
QA bf tf˜� � 0.5D 0.5tf�� �˜� 
QB QA� 
Shear Stress: W V Q˜
I b˜= Vaa R P�� 
WA
Vaa QA˜
I tw˜
� WA 0.747 MPa Ans
WB
Vaa QB˜
I tw˜
� WB 0.747 MPa Ans
Problem 7-25
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the maximum shear stress developed in the glued joints. The supports at C
and D exert only vertical reactions on the beam.
Given: bf 150mm� dw 200mm� 
tf 40mm� tw 50mm� 
P 25kN� 
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ 6Fy=0; 3P 2R� 0=
R 1.5P� 
R 37.50 kN 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜ dw3˜�� 
QA bf tf˜� � 0.5D 0.5tf�� �˜� 
QB QA� 
Shear Stress: W V Q˜
I b˜= Vmax R� 
WA
Vmax QA˜
I tw˜
� WA 2.24 MPa Ans
WB
Vmax QB˜
I tw˜
� WB 2.24 MPa Ans
Problem 7-26
The beam is made from three boards glued together at the seams A and B. If it is subjected to the
loading shown, determine the maximum vertical shear force resisted by the top flange of the beam. The
supports at C and D exert only vertical reactions on the beam.
Given: bf 150mm� dw 200mm� 
tf 40mm� tw 50mm� P 25kN� 
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ 6Fy=0; 3P 2R� 0=
R 1.5P� 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜ dw3˜�� 
yc 0.5D� 
A'f yc y�� � bf˜= yfc 0.5 yc y�� � y�=
yfc 0.5 yc y�� �=
Q A'f yfc˜=
Q 0.5 yc y�� � bf˜ yc y�� �=
Q 0.5bf yc
2
y
2�§© ·¹=
Shear Stress in flange: W V Q˜
I b˜= V R=
W R
I bf˜
§¨
©
·
¹
Q˜=
W R
I
§¨
©
·
¹ 0.5 yc
2
y
2�§© ·¹ª¬ º¼˜=
Resultant Shear Force: For the flange. yo yc tf�� Vf
A
AWµ´¶ d=
Vf
yo
yc
y
R
I
§¨
©
·
¹ 0.5 yc
2
y
2�§© ·¹ª¬ º¼˜ bf˜µ´µ¶
d� 
Vf 2.36 kN Ans
Problem 7-27
Determine the shear stress at points B and C located on the web of the fiberglass beam.
Given: bf 100mm� dw 150mm� 
tf 18mm� tw 12mm� 
L1 2m� L2 0.6m� 
L3 2m� 
wo 2.5
kN
m
� w1 3
kN
m
� 
Solution: L L1 L2� L3�� 
 Equilibrium : Given
+ 6Fy=0; A wo L1˜� 0.5 w1˜ L3˜� D� 0=
 60D=0; A L˜ wo L1˜ L 0.5L1�� �˜� 0.5 w1˜ L3˜ 2L33§¨©
·
¹˜� 0=
Guess A 1kN� D 1kN� 
A
D
§¨
©
·
¹ Find A D�( )� 
A
D
§¨
©
·
¹
4.783
3.217
§¨
©
·
¹ kN 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜ dw3˜�� 
QB bf tf˜� � 0.5D 0.5tf�� �˜� 
QC QB� 
Shear Stress: W V Q˜
I b˜= VBC A wo 0.5L1� �˜�� 
WB
VBC QB˜
I tw˜
� WB 0.572 MPa Ans
WC
VBC QC˜
I tw˜
� WC 0.572 MPa Ans
Problem 7-28
Determine the maximum shear stress acting in the fiberglass beam at the critical section.
Given: bf 100mm� dw 150mm� 
tf 18mm� tw 12mm� 
L1 2m� L2 0.6m� 
L3 2m� 
wo 2.5
kN
m
� w1 3
kN
m
� 
Solution: L L1 L2� L3�� 
 Equilibrium : Given
+ 6Fy=0; A wo L1˜� 0.5 w1˜ L3˜� D� 0=
 60D=0; A L˜ wo L1˜ L 0.5L1�� �˜� 0.5 w1˜ L3˜ 2L33§¨©
·
¹˜� 0=
Guess A 1kN� D 1kN� 
A
D
§¨
©
·
¹ Find A D�( )� 
A
D
§¨
©
·
¹
4.783
3.217
§¨
©
·
¹ kN 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜ dw3˜�� 
Qmax bf tf˜� � 0.5D 0.5tf�� �˜ 0.5dw tw˜� � 0.25dw� �˜�� 
Shear Stress: W V Q˜
I b˜= Vmax A� 
Wmax
Vmax Qmax˜
I tw˜
� Wmax 1.467 MPa Ans
Shear Force Diagram: L L1 L2� L3�� 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � A wo x1˜�� � 1kN˜� V2 x2� � A wo L1˜�� � 1kN˜� 
V3 x3� � A wo L1˜� w1 x3 L1� L2�� �˜ 1 x3 L1� L2�2L3�
§¨
©
·
¹
˜�ª«¬
º»¼
1
kN
˜� 
0 1 2 3 4
5
0
5
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Problem 7-29
The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected to
the loading shown, determine the shear stress developed in the glued joints at the critical section. The
supports at C and D exert only vertical reactions on the beam.
Given: bf 200mm� dw 200mm� 
tf 50mm� tw 50mm� 
wo 3
kN
m
� L 2.5m� 
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ 6Fy=0; wo L˜ 2R� 0=
R 0.5 wo L˜� �� 
R 3.75 kN 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜dw3˜�� 
QA bf tf˜� � 0.5D 0.5tf�� �˜� 
QB QA� 
Shear Stress: W V Q˜
I b˜= Vmax R� 
WA
Vmax QA˜
I tw˜
� WA 0.225 MPa Ans
WB
Vmax QB˜
I tw˜
� WB 0.225 MPa Ans
Problem 7-30
The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected to
the loading shown, determine the vertical shear force resisted by the top flange of the beam at the
critical section. The supports at C and D exert only vertical reactions on the beam.
Given: bf 200mm� dw 200mm� 
tf 50mm� tw 50mm� 
L 2.5m� 
wo 3
kN
m
� 
Solution:
 Equilibrium : By symmetry, RC=R , RD=R 
+ 6Fy=0; wo L˜ 2R� 0=
R 0.5 wo L˜� �� 
R 3.75 kN 
Section Property : D dw 2tf�� 
I
1
12
bf˜ D3˜
1
12
tw˜ dw3˜�� 
yc 0.5D� 
A'f yc y�� � bf˜= yfc 0.5 yc y�� � y�=
yfc 0.5 yc y�� �=
Q A'f yfc˜=
Q 0.5 yc y�� � bf˜ yc y�� �=
Q 0.5bf yc
2
y
2�§© ·¹=
Shear Stress in flange: W V Q˜
I b˜= Vmax R=
W R
I bf˜
§¨
©
·
¹
Q˜=
W R
I
§¨
©
·
¹ 0.5 yc
2
y
2�§© ·¹ª¬ º¼˜=
Resultant Shear Force: For the flange. yo yc tf�� Vf
A
AWµ´¶ d=
Vf
yo
yc
y
R
I
§¨
©
·
¹ 0.5 yc
2
y
2�§© ·¹ª¬ º¼˜ bf˜µ´µ¶
d� 
Vf 0.30 kN Ans
Problem 7-31
Determine the variation of the shear stress over the cross section of a hollow rivet. What is the
maximum shear stress in the rivet? Also, show that if then W�max = 2( V/A ).01 rr o
Problem 7-32
The beam has a square cross section and is subjected to the shear force V. Sketch the shear-stress
distribution over the cross section and specify the maximum shear stress. Also, from the neutral axis,
locate where a crack along the member will first start to appear due to shear.
Problem 7-33
Write a computer program that can be used to determine the maximum shear stress in the beam that
has the cross section shown, and is subjected to a specified constant distributed load w and
concentrated force P. Show an application of the program using the values L = 4 m, a = 2 m, P = 1.5
kN, d1 = 0, d2 = 2 m, w = 400 N/m, t1 = 15 mm, t2 = 20 mm, b = 50 mm, and h = 150 mm.
Problem 7-34
The beam has a rectangular cross section and is subjected to a load P that is just large enough to
develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic-plastic, then at a
distance x < L the moment M = P x creates a region of plastic yielding with an associated elastic core
having a height 2y'. This situation has been described by Eq. 6-30 and the moment M is distributed
over the cross section as shown in Fig. 6-54e. Prove that the maximum shear stress developed in the
beam is given by W�max = 3/2 (P/A'),where A' = 2y'b, the cross-sectional area of the elastic core.
Problem 7-35
The beam in Fig. 6-54f is subjected to a fully plastic moment Mp. Prove that the longitudinal and
transverse shear stresses in the beam are zero. Hint: Consider an element of the beam as shown in Fig.
7-4d.
Problem 7-36
The beam is constructed from two boards fastened together at the top and bottom with two rows of
nails spaced every 150 mm. If each nail can support a 2.5-kN shear force, determine the maximum
shear force V that can be applied to the beam.
Given: b 150mm� d1 50mm� 
d2 50mm� sn 150mm� 
Fallow 2.5kN� 
Solution:
Section Property : D d1 d2�� 
I
1
12
b˜ D3˜� 
Q b d1˜� � 0.5d1� �˜� 
Shear Flow : q
V Q˜
I
=
There are two rows of nails. Hence, the allowable shear flow is qallow
2Fallow
sn
� 
2Fallow
sn
Vmax Q˜
I
=
Vmax
2 I˜ Fallow
Q sn˜
� 
Vmax 2.222 kN Ans
Problem 7-37
The beam is constructed from two boards fastened together at the top and bottom with two rows of
nails spaced every 150 mm. If an internal shear force of V = 3 kN is applied to the boards, determine
the shear force resisted by each nail.
Given: b 150mm� d1 50mm� 
d2 50mm� sn 150mm� 
V 3kN� 
Solution:
Section Property : D d1 d2�� 
I
1
12
b˜ D3˜� 
Q b d1˜� � 0.5d1� �˜� 
Shear Flow : q
V Q˜
I
� q 45.00 kN
m
 
There are two rows of nails. Hence, the shear force resisted by each nail is F
q
2
sn˜� 
F 3.37 kN Ans
Problem 7-38
A beam is constructed from five boards bolted together as shown. Determine the maximum shear
force developed in each bolt if the bolts are spaced s = 250 mm apart and the applied shear is V = 35
kN.
Given: d1 250mm� d2 350mm� 
t 25mm� a 100mm� 
s 250mm� V 35kN� 
Solution: a' d2 d1 a�� ��� a' 200 mm 
h a' 0.5d1�� h 325 mm 
Section Property :
yc
2 d1 t˜� � 0.5d1 a'�� �˜ 3 d2 t˜� � 0.5d2� �˜�
2 d1 t˜� � 3 d2 t˜� ��� 
yc 223.39 mm 
I1
1
12
2t( )˜ d13˜ 2t d1˜� � 0.5d1 a'� yc�� �2˜�� 
I2
1
12
3 t˜( )˜ d23˜ 3 t˜ d2˜� � 0.5d2 yc�� �2˜�� 
I I1 I2�� I 523597110.22 mm4 
Q d1 2t( )˜ h yc�� �˜� Q 1270161.29 mm3 
Shear Flow :
q
V Q˜
I
� q 84.90 kN
m
 
There are four planes on the bolt. Hencs, the shear
force resisted by each shear plane of the bolt is
F
q
4
s˜§¨©
·
¹� F 5.31 kN Ans
Problem 7-39
A beam is constructed from five boards bolted together as shown. Determine the maximum spacing s
of the bolts if they can each resist a shear of 20 kN and the applied shear is V = 45 kN.
Given: d1 250mm� d2 350mm� 
t 25mm� a 100mm� 
V 45kN� Fallow 20kN� 
Solution: a' d2 d1 a�� ��� a' 200 mm 
h a' 0.5d1�� h 325 mm 
Section Property :
yc
2 d1 t˜� � 0.5d1 a'�� �˜ 3 d2 t˜� � 0.5d2� �˜�
2 d1 t˜� � 3 d2 t˜� ��� 
yc 223.39 mm 
I1
1
12
2t( )˜ d13˜ 2t d1˜� � 0.5d1 a'� yc�� �2˜�� 
I2
1
12
3 t˜( )˜ d23˜ 3 t˜ d2˜� � 0.5d2 yc�� �2˜�� 
I I1 I2�� I 523597110.22 mm4 
Q d1 2t( )˜ h yc�� �˜� Q 1270161.29 mm3 
Shear Flow : q
V Q˜
I
� q 109.16 kN
m
 
Since there are four planes on the bolt, the allowable shear flow is qallow
4Fallow
s
=
s
4Fallow
q
� 
s 732.9 mm Ans
Problem 7-40
The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the
nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2
mm.
Given: bf 250mm� dw 150mm� 
t 30mm� a 100mm� 
s 100mm� V 800N� 
do 2mm� 
Solution: h' 0.5t� h' 15 mm 
Section Property :
yc
bf t˜ 0.5t( )˜ dw 2t( )˜ 0.5dw� �˜�
bf t˜ 2 dw t˜� ��� 
yc 47.73 mm 
I1
1
12
bf˜ t3˜ bf t˜� � 0.5t yc�� �2˜�� 
I2
1
12
2 t˜( )˜ dw3˜ 2 t˜ dw˜� � 0.5dw yc�� �2˜�� 
I I1 I2�� I 32164772.73 mm4 
Q bf t˜ yc h'�� �˜� Q 245454.55 mm3 
Shear Flow :
q
V Q˜
I
� q 6.105 kN
m
 
F q s˜� F 0.6105 kN 
Since each side of the beam resists this shear force, then
Ao
S
4
do
2˜� Wavg
F
2Ao
� 
Wavg 97.16 MPa Ans
Problem 7-41
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of
150 mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine
the maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.
Given: bf 75mm� dw 75mm� t 12mm� 
dp 150mm� dg 50mm� 
V 250kN� Fallow 75kN� 
Solution:
Section Property : D 2dw 2t� dg�� 
IT
1
12
bf˜ D3˜
1
12
t˜ dg3˜�
1
12
bf t�� �˜ D 2t�( )3˜�� 
IP
1
12
2t( )˜ dp3˜� 
I IT IP�� 
Q bf t˜� � 0.5D 0.5t�( )˜ t dw˜� � 0.5dw 0.5dg�� �˜�� 
Shear Flow : q
V Q˜
I
=
Since there are two shear planes on the bolt, the allowable shear flow is q
2F
sn
=
2Fallow
sn
V Q˜
I
=
sn
2 I˜ Fallow
V Q˜� 
sn 138.0 mm Ans
Problem 7-42
The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of
150 mm and a thickness of 12 mm. If the bolts are spaced at s = 200 mm, determine the maximum
shear force V that can be applied to the cross section. Each bolt can resist a shear force of 75 kN.Given: bf 75mm� dw 75mm� t 12mm� 
dp 150mm� dg 50mm� 
sn 200mm� Fallow 75kN� 
Solution:
Section Property : D 2dw 2t� dg�� 
IT
1
12
bf˜ D3˜
1
12
t˜ dg3˜�
1
12
bf t�� �˜ D 2t�( )3˜�� 
IP
1
12
2t( )˜ dp3˜� 
I IT IP�� 
q
V Q˜
I
Q bf t˜� � 0.5D 0.5t�( )˜ t dw˜� � 0.5dw 0.5dg�� �˜�� 
Shear Flow : =
Since there are two shear planes on the bolt, the allowable shear flow is q
2F
sn
=
2Fallow
sn
V Q˜
I
=
V
2 I˜ Fallow
sn Q˜
� 
V 172.5 kN Ans
Problem 7-43
The double-web girder is constructed from two plywood sheets that are secured to wood members at
its top and bottom. If each fastener can support 3 kN in single shear, determine the required spacing s
of the fasteners needed to support the loading P = 15 kN. Assume A is pinned and B is a roller.
Given: bi 150mm� di 250mm� 
t 12mm� do 350mm� 
P 15kN� Fallow 3kN� 
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ 6Fy=0; 2R P� 0=
R 0.5P� 
Section Property : bo bi 2t�� d' 0.5 do di�� �� 
I
1
12
bo˜ do3˜
1
12
bi˜ di3˜�� 
Q bi d'˜� � 0.5do 0.5d'�� �˜� 
Shear Flow : q
V Q˜
I
= Vmax R� 
Since there are two shear planes on the bolt, the allowable shear flow is q
2F
sn
=
2Fallow
sn
Vmax Q˜
I
=
sn
2 I˜ Fallow
Vmax Q˜
� 
sn 303.2 mm Ans
Problem 7-44
The double-web girder is constructed from two plywood sheets that are secured to wood members at
its top and bottom. The allowable bending stress for the wood is Vallow = 56 MPa and the allowable
shear stress is Wallow = 21 MPa. If the fasteners are spaced s = 150 mm and each fastener can support
3 kN in single shear, determine the maximum load P that can be applied to the beam. 
Given: bi 150mm� di 250mm� 
t 12mm� do 350mm� 
sn 150mm� Vallow 56MPa� 
Fallow 3kN� Wallow 21MPa� 
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ 6Fy=0; 2R P� 0=
R 0.5P=
Section Property : bo bi 2t�� 
d' 0.5 do di�� �� 
I
1
12
bo˜ do3˜
1
12
bi˜ di3˜�� 
Q bi d'˜� � 0.5do 0.5d'�� �˜� 
Shear Flow : q
V Q˜
I
= Vmax R= q
2F
sn
=
Since there are two shear planes on the bolt, the allowable shear flow is
2Fallow
sn
Vmax Q˜
I
=
2Fallow
sn
0.5P Q˜
I
=
P
4 I˜ Fallow
sn Q˜
� 
P 30.32 kN Ans
Problem 7-45
The beam is made from three polystyrene strips that are glued together as shown. If the glue has a
shear strength of 80 kPa, determine the maximum load P that can be applied without causing the glue
to lose its bond.
Given: bf 30mm� tf 40mm� 
tw 20mm� dw 60mm� 
Wallow 0.080MPa� 
Solution:
 Equilibrium : By equilibrium, A = B = R
+
6Fy=0; 2R P� 2 1
4
P˜§¨©
·
¹� 0=
R 0.75P=
Maximum Shear : Vmax R= Vmax
3
4
P=
Section Property : D dw 2tf�� 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� 
I 6.68 10
6�u m4 
Q 6 yi
˜ A1˜= Q bf tf˜� � D2
tf
2
�§¨©
·
¹˜� Q 60000 mm
3 
Shear Stress: W V Q˜
I t˜=
Wallow
Vmax Q˜
I tw˜
=
Vmax
Wallow I˜ tw˜
Q
=
3
4
P
Wallow I˜ tw˜
Q
=
P
4Wallow I˜ tw˜
3Q
� 
P 0.238 kN Ans
Problem 7-46
The beam is made from four boards nailed together as shown. If the nails can each support a shear
force of 500 N., determine their required spacings s' and s if the beam is subjected to a shear of V =
3.5 kN.
Given: bf 250mm� dw 250mm� 
tf 25mm� tw 40mm� 
tb 25mm� db 75mm� 
V 3.5kN� Fallow 0.5kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� �� 0.5db 2tb db˜� ��
bf tf˜ dw tw˜� 2 tb db˜� ��� 
yc 85.94 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I3
1
12
2tb� �˜ db3˜ 2tb db˜� � 0.5db yc�� �2˜�� 
I I1 I2� I3�� 
QC tb db˜� � yc 0.5db�� �˜� 
QD dw tw˜� � D yc� 0.5dw�� �˜� 
Shear Flow : q
V Q˜
I
=
The allowable shear flow at points C and D are : qC
F
sn
= qD
F
s'n
=
Fallow
sn
V QC˜
I
=
Fallow
s'n
V QD˜
I
=
sn
I Fallow
V QC˜
� s'n
I Fallow
V QD˜
� 
sn 216.6 mm s'n 30.7 mm Ans
Problem 7-47
The beam is fabricated from two equivalent channels and two plates. Each plate has a height of 150
mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine the
maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.
Given: bf 300mm� dw 88mm� t 12mm� 
dp 150mm� dg 50mm� 
V 250kN� Fallow 75kN� 
Solution:
Section Property : D 2dw 2t� dg�� 
IU
1
12
bf˜ D3˜
1
12
2t( )˜ dg3˜�
1
12
bf 2t�� �˜ D 2t�( )3˜�� 
IP
1
12
2t( )˜ dp3˜� 
I IU IP�� 
Q bf t˜� � 0.5D 0.5t�( )˜ 2t dw˜� � 0.5dw 0.5dg�� �˜�� 
Shear Flow : q
V Q˜
I
=
Since there are two rows of bolts, the allowable shear flow is q
2F
sn
=
2Fallow
sn
V Q˜
I
=
sn
2 I˜ Fallow
V Q˜� 
sn 137.6 mm Ans
Problem 7-48
A built-up timber beam is made from n boards, each having a rectangular cross section. Write a
computer program that can be used to determine the maximum shear stress in the beam when it is
subjected to any shear V. Show an application of the program using a cross section that is in
the form of a “T” and a box.
Problem 7-49
The timber T-beam is subjected to a load consisting of n concentrated forces Pn ., If the allowable
shear Vnail for each of the nails is known, write a computer program that will specify the nail spacing
between each load. Show an application of the program using the values L = 4.5 m, a1 = 1.2 m, P1 = 3
kN, a2 = 2.4 m, P2 = 7.5 kN, b1 = 37.5 mm, h1 = 250 mm, b2 = 200 mm, h2 = 25 mm, and Vnail = 1
kN.
Problem 7-50
The strut is constructed from three pieces of plastic that are glued together as shown. If the allowable
shear stress for the plastic is Wallow = 5.6 MPa and each glue joint can withstand 50 kN/m, determine
the largest allowable distributed loading w that can be applied to the strut.
Given: L1 1m� Wallow 5.6MPa� 
L2 2m� qallow 50
kN
m
� 
L3 1m� 
bf 74mm� tf 25mm� 
dw 75mm� tw 12mm� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � 2tw dw˜� ��
bf tf˜ 2tw dw˜�
� 
yc 37.16 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
2tw� �˜ dw3˜ 2tw dw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� 
Qmax 2tw� � D yc�� �˜ D yc�2˜� 
QA bf tf˜� � yc 0.5tf�� �˜� 
Allowable Shear Stress: W V Q˜
I t˜= Vmax w L1˜=
Wallow
w L1˜
I 2tw� �˜
ª«¬
º»¼
2tw� � D yc�� �˜ D yc�2˜ª«¬
º»¼˜=
Wallow
w L1˜
2I
D yc�� �2˜= w 2I
L1 D yc�� �2˜ Wallow� �˜� w 9.13
kN
m
 
Shear Flow : Assume the beam fails at the glue joint and the allowable shear flow is 2 qallow˜
2qallow
V Q˜
I
= 2qallow
w L1˜� � QA˜
I
=
w
2 I˜ qallow
L1 QA˜
� w 7.06 kN
m
 (Controls !) Ans
Shear Force Diagram: L L1 L2� L3�� 
 Equilibrium : By symmetry, R1 R= R2 R=
+ 6Fy=0; 2R w L˜� 0=
R 0.5w L˜� 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � w� x1˜kN� V2 x2� � w� x2˜ R�� � 1kN˜� V3 x3� � w� x3˜ 2R�� � 1kN˜� 
0 1 2 3 4
10
0
10
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Problem 7-51
The strut is constructed from three pieces of plastic that are glued together as shown. If the distributed
load w = 3 kN/m, determine the shear flow that must be resisted by each glue joint.
Given: L1 1m� 
L2 2m� w 3 kN
m
� 
L3 1m� 
bf 74mm� tf 25mm� 
dw 75mm� tw 12mm� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � 2tw dw˜� ��
bf tf˜ 2tw dw˜�
� 
yc 37.16 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
2tw� �˜ dw3˜ 2tw dw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� 
QA bf tf˜� � yc 0.5tf�� �˜� 
ShearFlow : Since there are two glue joints, hence 2q
V Q˜
I
=
Vmax w L1˜� 
q
Vmax QA˜
2I
� 
q 21.24
kN
m
 Ans
Shear Force Diagram: L L1 L2� L3�� 
 Equilibrium : By symmetry, R1 R= R2 R=
+ 6Fy=0; 2R w L˜� 0=
R 0.5w L˜� 
x1 0 0.01 L1˜� L1��� x2 L1 1.01 L1˜� L1 L2�� ���� x3 L1 L2�� � 1.01 L1 L2�� �˜� L( )��� 
V1 x1� � w� x1˜kN� V2 x2� � w� x2˜ R�� � 1kN˜� V3 x3� � w� x3˜ 2R�� � 1kN˜� 
0 1 2 3 4
10
0
10
Distance (m)
S
h
ea
r 
(k
N
) V1 x1� �
V2 x2� �
V3 x3� �
x1 x2� x3�
Problem 7-52
The beam is subjected to the loading shown, where P = 7 kN. Determine the average shear stress
developed in the nails within region AB of the beam. The nails are located on each side of the beam and
are spaced 100 mm apart. Each nail has a diameter of 5 mm.
Given: bf 250mm� dw 150mm� 
t 30mm� a 2m� 
s 100mm� do 5mm� 
P' 3kN� P 7kN� 
Solution:
Section Property :
I
1
12
bf 2 t˜�� � dw3˜ bf dw 2 t˜�� �3˜�ª¬ º¼˜� 
I 72000000 mm
4 
Q bf t˜ 0.5dw 0.5t�� �˜� 
Q 450000 mm
3 
Maximum Shear : Vmax P' P�� Vmax 10 kN 
Shear Flow :
q
Vmax Q˜
I
� q 62.500 kN
m
 
There are two rows of nails. Hence, the sher force
resisted by each nail is
F
q
2
s˜� F 3.125 kN 
Ao
S
4
do
2˜� Wavg
F
Ao
� 
Wavg 159.2 MPa Ans
Problem 7-53
The beam is constructed from four boards which are nailed together. If the nails are on both sides of
the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the
end of the beam.
Given: bf 250mm� dw 150mm� 
t 30mm� a 2m� 
s 100mm� P' 3kN� 
Fallow 3kN� 
Solution:
Section Property :
I
1
12
bf 2 t˜�� � dw3˜ bf dw 2 t˜�� �3˜�ª¬ º¼˜� 
I 72000000 mm
4 
Q bf t˜ 0.5dw 0.5t�� �˜� 
Q 450000 mm
3 
Maximum Shear : Vmax P' P�=
There are two rows of nails. Hence, the allowable sher flow is
qallow
2Fallow
s
� 
qallow 60.00
kN
m
 
Shear Flow :
qallow
Vmax Q˜
I
= qallow
P' P�( ) Q˜
I
=
P
qallow I˜
Q
§¨
©
·
¹ P'�� 
P 6.60 kN Ans
Problem 7-54
The member consists of two plastic channel strips 12 mm thick, bonded together at A and B. If the
glue can support an allowable shear stress of Wallow = 4.2 MPa, determine the maximum intensity w0 of
the triangular distributed loading that can be applied to the member based on the strength of the glue.
Given: bo 150mm� t 12mm� 
do 150mm� L 4m� 
Wallow 4.2MPa� 
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ 6Fy=0; 2R 0.5wo L˜� 0=
R 0.25wo L˜=
Section Property : bi bo 2t�� di do 2t�� 
I
1
12
bo˜ do3˜
1
12
bi˜ di3˜�� 
Q bo t˜� � 0.5do 0.5t�� �˜ 2 t˜ 0.5 di˜� �˜ª¬ º¼ 0.5di� �˜�� 
Shear Flow : q
V Q˜
I
= Vmax R=
Since there are two planes of glue, the allowable shear flow is 2t Wallow˜
2t( ) Wallow˜
Vmax Q˜
I
=
2 t˜( )Wallow
0.25wo L˜� � Q˜
I
=
wo
8t I˜ Wallow
L Q˜� 
wo 9.73
kN
m
 Ans
Problem 7-55
The member consists of two plastic channel strips 12 mm thick, glued together at A and B. If the
distributed load has a maximum intensity of w0 = 50 kN/m, determine the maximum shear stress
resisted by the glue.
Given: bo 150mm� do 150mm� L 4m� 
t 12mm� wo 50
kN
m
� 
Solution:
 Equilibrium : By symmetry, A=R , B=R 
+ 6Fy=0; 2R 0.5wo L˜� 0=
R 0.25wo L˜� 
Section Property : bi bo 2t�� di do 2t�� 
I
1
12
bo˜ do3˜
1
12
bi˜ di3˜�� 
Q bo t˜� � 0.5do 0.5t�� �˜ 2 t˜ 0.5 di˜� �˜ª¬ º¼ 0.5di� �˜�� 
Allowable Shear Stress: W V Q˜
I b˜= Vmax R� 
Wmax
Vmax Q˜
I 2t( )˜� 
Wmax 21.58 MPa Ans
Problem 7-56
A shear force of V = 18 kN is applied to the symmetric box girder. Determine the shear flow at A and
B.
Given: bf 125mm� dw 300mm� t 10mm� 
dm 200mm� dg 30mm� V 18kN� 
Solution:
Section Property :
I1
1
12
bf˜ t3˜ bf t˜� � 0.5dw 0.5t�� �2˜�� 
I2
1
12
bf˜ t3˜ bf t˜� � 0.5t 0.5dm�� �2˜�� 
I3
1
12
t˜ dw3˜� 
I 2I1 2I2� 2I3�� I 125166666.67 mm4 
QA bf t˜� � 0.5dw 0.5t�� �˜� QA 181250 mm3 
QB bf t˜� � 0.5t 0.5dm�� �˜� QB 131250 mm3 
Shear Flow :
qA
1
2
V QA˜
I
� qA 13.03
kN
m
 
Ans
qB
1
2
V QB˜
I
� qB 9.44
kN
m
 Ans
Problem 7-57
A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at C.
Given: bf 125mm� dw 300mm� t 10mm� 
dm 200mm� dg 30mm� V 18kN� 
Solution:
Section Property :
I1
1
12
bf˜ t3˜ bf t˜� � 0.5dw 0.5t�� �2˜�� 
I2
1
12
bf˜ t3˜ bf t˜� � 0.5t 0.5dm�� �2˜�� 
I3
1
12
t˜ dw3˜� 
I 2I1 2I2� 2I3�� 
I 125166666.67 mm
4 
QC bf t˜� � 0.5dw 0.5t�� �˜ bf t˜� � 0.5t 0.5dm�� �˜� 2 0.5dw t˜� � 0.25dw� �˜�� 
QC 537500 mm
3 
Shear Flow :
qC
1
2
V QC˜
I
� qC 38.65
kN
m
 Ans
Problem 7-58
The channel is subjected to a shear of V = 75 kN. Determine the shear flow developed at point A.
Given: bf 400mm� dw 200mm� 
tf 30mm� tw 30mm� 
V 75kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � 2tw dw˜� ��
bf tf˜ 2tw dw˜�
� 
yc 72.50 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
2tw� �˜ dw3˜ 2tw dw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� 
I 120250000 mm
4 
QA bf tf˜� � yc 0.5tf�� �˜� 
QA 690000 mm
3 
Shear Flow :
qA
1
2
V QA˜
I
� qA 215.2
kN
m
 Ans
Problem 7-59
The channel is subjected to a shear of V = 75 kN. Determine the maximum shear flow in the channel.
Given: bf 400mm� dw 200mm� 
tf 30mm� tw 30mm� 
V 75kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � 2tw dw˜� ��
bf tf˜ 2tw dw˜�
� 
yc 72.50 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
2tw� �˜ dw3˜ 2tw dw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� 
I 120250000 mm
4 
Qmax tw D yc�� �˜ 12˜ D yc�� �˜� 
Qmax 372093.75 mm
3 
Shear Flow :
qmax
V Qmax˜
I
� qmax 232.1
kN
m
 Ans
Problem 7-60
The beam supports a vertical shear of V = 35 kN. Determine the resultant force developed in segment
AB of the beam.
Given: bf 125mm� dw 250mm� 
tf 12mm� tw 12mm� 
V 35kN� 
Solution:
Section Property : D dw 2tf�� 
I
1
12
2tf� �˜ bf3˜ 112 dw˜ tw3˜�� 
Q A'f yfc˜=
Q tf 0.5bf y�� �˜ 0.5 0.5bf y�� � y�ª¬ º¼˜=
Q 0.5tf 0.25bf
2
y
2�§© ·¹=
Shear Flow: q
V Q˜
I
=
q
V tf˜
2I
0.25bf
2
y
2�§© ·¹˜=
Resultant Shear Force: For AB: yo 0.5tw� VAB
A
yqµ´¶ d=
VAB
yo
0.5bf
y
V tf˜
2I
0.25bf
2
y
2�§© ·¹˜µ´µ¶
d� 
VAB 7.43 kN Ans
Problem 7-61
The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V
= 150 N, determine the shear flow at points A and B.
Given: bf 60mm� b'f 80mm� 
tf 10mm� dw 40mm� 
tw 10mm� V 150N� 
Solution:
Section Property : D dw 2tf�� 
A bf tf˜ 2dw tw˜� b'f tf˜�� 
A 2200 mm
2 
y
 6 yi

Ai˜� �˜
6 Ai� �˜= yc
b'f tf˜� � 0.5tf� �˜ 2 dw tw˜� � 0.5dw tf�� �˜� bf tf˜� � D 0.5tf�� �˜�
A
� 
yc 27.73 mm 
I'f
1
12
b'f˜ tf3˜ b'f tf˜� � 0.5tf yc�� �2˜�� 
Iw
1
12
tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�� 
If
1
12
bf˜ tf3˜ bf tf˜� � D 0.5tf� yc�� �2˜�� 
I If 2Iw� I'f�� 
I 981.9697 10
9�u m4 
QA 0.5b'f tf˜� � yc 0.5tf�� �˜� QA 9090.91 mm3 
QB bf tf˜� � D 0.5tf� yc�� �˜� QB 16363.64 mm3 
Shear Flow :
qA
V QA˜
I
� qA 1.39
kN
m
 
Ans
qB
1
2
V QB˜
I
� qB 1.25
kN
m
 Ans
Problem 7-62
The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V
= 150 N, determine the maximum shear flow in the strut.
Given: bf 60mm� b'f 80mm� 
tf 10mm� dw 40mm� 
tw 10mm� V 150N� 
Solution:
Section Property : D dw 2tf�� 
A bf tf˜ 2dw tw˜� b'f tf˜�� 
A 2200 mm
2 
y
 6 yi

Ai˜� �˜
6 Ai� �˜= yc
b'f tf˜� � 0.5tf� �˜ 2 dw tw˜�� 0.5dw tf�� �˜� bf tf˜� � D 0.5tf�� �˜�
A
� 
yc 27.73 mm 
I'f
1
12
b'f˜ tf3˜ b'f tf˜� � 0.5tf yc�� �2˜�� 
Iw
1
12
tw˜ dw3˜ tw dw˜� � 0.5dw tf� yc�� �2˜�� 
If
1
12
bf˜ tf3˜ bf tf˜� � D 0.5tf� yc�� �2˜�� 
I If 2Iw� I'f�� 
I 981.9697 10
9�u m4 
Qmax bf tf˜� � D 0.5tf� yc�� �˜ 2 tw˜ D yc� tf�� �˜ 12˜ D yc� tf�� �˜�� 
Qmax 21324.38 mm
3 
Shear Flow :
qmax
1
2
V Qmax˜
I
� qmax 1.63
kN
m
 Ans
Problem 7-63
The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks.
Given: L 125mm� t 6mm� 
T 45deg� V 10kN� 
Solution:
Section Property :
h L cos T� �˜� b t
sin T� �� 
I
1
12
2b( )˜ h3˜� 
Q A' y'c˜=
Q t
0.5h y�
sin T� �§¨©
·
¹˜ 0.5 0.5h y�( ) y�[ ]˜=
Q
t
2 sin T� � 0.25h
2
y
2�� �=
Shear Flow: q
V Q˜
I
=
q
V t˜
2I sin T� �˜ 0.25h
2
y
2�� �˜=
At y = 0, q qmax=
qmax
V t˜
2I sin T� �˜ 0.25h
2� �˜� 
qmax 84.85
kN
m
 Ans
Problem 7-64
The beam is subjected to a shear force of V = 25 kN. Determine the shear flow at points A and B.
Given: bf 274mm� tf 12mm� 
b'f 250mm� t'f 12mm� 
dw 200mm� tw 12mm� 
d'w 50mm� V 25kN� 
Solution:
Section Property : D dw tf�� 
D' D d'w�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � 2tw dw˜� �� D' 0.5t'f�� � b'f t'f˜� ��
bf tf˜ 2tw dw˜� b'f t'f˜�
� 
yc 92.47 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
2tw� �˜ dw3˜ 2tw dw˜� � 0.5dw tf� yc�� �2˜�� 
I3
1
12
b'f˜ t'f3˜ b'f t'f˜� � D' 0.5t'f� yc�� �2˜�� 
I I1 I2� I3�� 
QA bf tf˜� � yc 0.5tf�� �˜� 
QB b'f t'f˜� � D' 0.5t'f� yc�� �˜� 
Shear Flow :
qA
V QA˜
2I
� qA 65.09
kN
m
 Ans
qB
V QB˜
2I
� qB 43.63
kN
m
 Ans
Problem 7-65
The beam is constructed from four plates and is subjected to a shear force of V = 25 kN. Determine
the maximum shear flow in the cross section.
Given: bf 274mm� tf 12mm� 
b'f 250mm� t'f 12mm� 
dw 200mm� tw 12mm� 
d'w 50mm� V 25kN� 
Solution:
Section Property : D dw tf�� 
D' D d'w�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � 2tw dw˜� �� D' 0.5t'f�� � b'f t'f˜� ��
bf tf˜ 2tw dw˜� b'f t'f˜�
� 
yc 92.47 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
2tw� �˜ dw3˜ 2tw dw˜� � 0.5dw tf� yc�� �2˜�� 
I3
1
12
b'f˜ t'f3˜ b'f t'f˜� � D' 0.5t'f� yc�� �2˜�� 
I I1 I2� I3�� 
Qmax bf tf˜� � yc 0.5tf�� �˜ 2tw yc tf�2§¨©
·
¹˜ yc tf�� �˜�� 
Maximum Shear Flow :
qmax
V Qmax˜
2I
� qmax 82.88
kN
m
 Ans
Problem 7-66
A shear force of V = 18 kN is applied to the box girder. Determine the position d of the stiffener plates
BE and FG so that the shear flow at A is twice as great as the shear flow at B. Use the centerline
dimensions for the calculation. All plates are 10 mm thick.
Given: t 10mm� bf 135mm t�� 
V 18kN� dw t 290mm�� 
Solution:
Section Property :
QA bf t˜� � 0.5dw 0.5t�� �˜� QA 181250 mm3 
QB bf t˜� � d( )˜=
Shear Flow :
qA
1
2
V QA˜
I
= qB
1
2
V QB˜
I
=
Require, qA 2qB=
1
2
V QA˜
I
2
1
2
V QB˜
I
§¨
©
·
¹˜=
QA 2QB=
QA 2 bf t˜� � d( )˜=
d
QA
2bf t˜
� 
d 72.50 mm Ans
Problem 7-67
The pipe is subjected to a shear force of V = 40 kN. Determine the shear flow in the pipe at points A
and B.
Given: ri 150mm� t 5mm� 
V 40kN� 
Solution:
Section Property : ro ri t�� 
I
S
4
ro
4
ri
4�§© ·¹˜� 
Since a' -> 0, then QA 0� 
QB
4ro
3S
S ro2˜
2
§¨
©
·
¹
4ri
3S
S ri2˜
2
§¨
©
·
¹�� 
Shear Flow :
qA
V QA˜
I
� qA 0.00
kN
m
 Ans
qB
V QB˜
2I
� qB 83.48
kN
m
 Ans
Problem 7-68
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown where b2 > b1. The member segments have the same thickness t.
Problem 7-69
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.The member segments have the same thickness t.
Problem 7-70
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-71
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-72
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-73
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Problem 7-74
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown. The member segments have the same thickness t.
Given: dT 150mm� dv 150mm� 
T 30deg� 
Solution: Set t 1mm� 
Section Property :
hT dT sin T� �˜� bT t
sin T� �� 
IT
1
12
bT˜ hT3˜ dT t˜� � dv hT�2§¨©
·
¹
2
˜�� 
Iv
1
12
t˜ dv3˜� 
I 2IT Iv�� 
y'c 0.5dv hT� 0.5x sin T� �˜�=
Q A' y'c˜= Q x t˜( ) 0.5dv hT� 0.5x sin T� �˜�� �˜=
Shear Flow Resultant:
q
V Q˜
I
= V P=
q
P
x t˜
I
0.5dv hT� 0.5x sin T� �˜�� �˜=
F1
P
0
dT
x
x t˜
I
0.5dv hT� 0.5x sin T� �˜�� �˜µ´µ¶ d=
Shear Center: Summing moment about point A
P e˜ F1 dv˜ cos T� �˜=
e
F1
P
dv˜ cos T� �˜=
e
0
dT
xdv cos T� �˜ x t˜I§¨©
·
¹˜ 0.5dv hT� 0.5x sin T� �˜�� �˜µ´µ¶ d� 
e 43.30 mm Ans
Problem 7-75
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its
side.
Given: a 100mm� b 100mm� 
Solution:
Set t mm� P kN� 
Section Property :
h 2a� 
I
1
12
2t( )˜ h3˜ 2 b t˜( ) a2˜�� I 3.3333 a3t 
Q1 y t˜( )
1
2
˜ y( )= Q1
t
2
y
2
=
Q2 a t˜( )
1
2
˜ a( ) x t˜( ) a( )˜�= Q1
a t˜
2
a 2x�( )=
Shear Flow Resultant:
q1
V Q1˜
I
= q1
P t˜ y2˜
2I
=
q2
V Q2˜
I
= q2
P a t˜( )˜ a 2x�( )˜
2I
=
Fw
0
a
yq1µ´¶ d= Fw
0
a
y
P t˜ y2˜
2I
µ´
µ¶
d� Fw 0.05 P 
Ff
0
b
xq2µ´¶ d= Ff
0
b
x
P a t˜( )˜ a 2x�( )˜
2I
µ´
µ¶
d� Ff 0.3 P 
Shear Center: Summing moment about point A
P e˜ 2Fw b˜ Ff h˜�=
e
1
P
2Fw b˜ Ff h˜�� �˜� 
e 70 mm Ans
Problem 7-76
Determine the location e of the shear center, point O, for the thin-walled member having a slit along its
side. Each element has aconstant thickness t.
Given: a mm� b a� 
Solution:
Set t mm� P kN� 
Section Property :
h 2a� 
I
1
12
2t( )˜ h3˜ 2 b t˜( ) a2˜�� I 3.3333 a3t 
Q1 y t˜( )
1
2
˜ y( )= Q1
t
2
y
2
=
Q2 a t˜( )
1
2
˜ a( ) x t˜( ) a( )˜�= Q1
a t˜
2
a 2x�( )=
Shear Flow Resultant:
q1
V Q1˜
I
= q1
P t˜ y2˜
2I
=
q2
V Q2˜
I
= q2
P a t˜( )˜ a 2x�( )˜
2I
=
Fw
0
a
yq1µ´¶ d= Fw
0
a
y
P t˜ y2˜
2I
µ´
µ¶
d� 
Fw 0.05 P 
Ff
0
b
xq2µ´¶ d= Ff
0
b
x
P a t˜( )˜ a 2x�( )˜
2I
µ´
µ¶
d� 
Ff 0.3 P 
Shear Center: Summing moment about point A
P e˜ 2Fw b˜ Ff h˜�=
e
1
P
2Fw b˜ Ff h˜�� �˜� 
e 0.7 a Ans
Problem 7-77
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Given: a mm� T 60deg� 
Solution:
Set t mm� P kN� 
Section Property :
b a sin T� �˜� b 0.86603 a 
t'
t
cos T� �� t' 2 t h
a
2
� 
I
1
12
t( )˜ a3˜ 1
12
t'( ) a
3˜�� I 0.25 a3t 
Q1 y' t'˜( )
1
2
˜ y'( )= Q1
t'
2
y'
2
=
Q2 h t'˜( )
1
2
˜ h( ) h y�( ) t˜ y 1
2
h y�( )�ª«¬
º»¼�=
Q2
12
t'h
2
t h
2
y
2�� ��ª¬ º¼=
Shear Flow Resultant:
q1
V Q1˜
I
= q1
P t'˜ y'2˜
2I
=
q2
V Q2˜
I
= q2
P t'h
2
t h
2
y
2�� ��ª¬ º¼˜
2I
=
F'
0
h
yq1µ´¶ d= F'
0
h
y'
P t'˜ y'2˜
2I
µ´
µ¶
d� F' 0.1667 P 
F
0
h
yq2µ´¶ d= F
0
h
y
P t'h
2
t h
2
y
2�� ��ª¬ º¼˜
2I
µ´
µ¶
d� F 0.6667 P 
Shear Center: Summing moment about point A
P e˜ 2 F b˜ F' 0˜�( )=
e
1
P
2F b˜( )˜� 
e 1.1547 a Ans
Problem 7-78
If the angle has a thickness of 3 mm, a height h = 100 mm, and it is subjected to a shear of V = 50 N,
determine the shear flow at point A and the maximum shear flow in the angle.
Given: h 100mm� t 3mm� 
T 45deg� V 50N� 
Solution:
Section Property :
t'
t
sin T� �� t' 4.2426 mm 
I
1
12
2t'( )˜ h3˜� 
Q A' y'c˜=
Q t' 0.5h y�( )˜ 1
2
0.5h y�( ) y�ª«¬
º»¼˜=
Q
t'
2
0.25h
2
y
2�� �=
Shear Flow: q
V Q˜
I
=
q
V t'˜
2I
0.25h
2
y
2�� �˜=
At A, yA 0.5 h˜� qA 0� Ans
At y = 0, q qmax=
qmax
V t'˜
2I
0.25h
2� �˜� 
qmax 375
N
m
 Ans
Problem 7-79
The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.
Indicate numerical values at all peaks. The thickness is 6 mm and the legs (AB) are 125 mm.
Given: L 125mm� t 6mm� 
T 45deg� V 10kN� 
Solution:
Section Property :
h L cos T� �˜� t' t
sin T� �� t' 8.4853 mm 
I
1
12
2t'( )˜ h3˜� 
Q A' y'c˜=
Q t' 0.5h y�( )˜ 1
2
0.5h y�( ) y�ª«¬
º»¼˜=
Q
t'
2
0.25h
2
y
2�� �=
Shear Flow: q
V Q˜
I
=
q
V t'˜
2I
0.25h
2
y
2�� �˜= Ans
At y = 0, q qmax=
qmax
V t'˜
2I
0.25h
2� �˜� 
qmax 84.85
kN
m
 Ans
Problem 7-80
Determine the placement e for the force P so that the beam bends downward without twisting. Take
h = 200 mm.
Given: h1 100mm� bf 300mm� 
h2 200mm� 
Solution:
Set t mm� P kN� 
Section Property :
I
1
12
bf t
3˜ t h13˜� t h23˜�§© ·¹˜� 
Qw2 t 0.5h2 y�� �˜ y 12 0.5h2 y�� ��ª«¬ º»¼˜=
Qw2
t
2
0.25h2
2
y
2�§© ·¹˜=
Shear Flow Resultant:
qw2
V Qw2˜
I
= qw2
P t˜ 0.25h22 y2�§© ·¹˜
2I
=
Fw2
0.5� h2
0.5h2
xqw2
µ´
µ¶
d=
Fw2
0.5� h2
0.5h2
y
P t˜ 0.25h22 y2�§© ·¹˜
2I
µ´
µ
µ¶
d� 
Fw2 0.8889 P 
Shear Center: Summing moment about point A
P e˜ Fw2 bf t�� �˜=
e
1
P
Fw2 bf t�� �˜ª¬ º¼˜� 
e 267.5 mm Ans
Problem 7-81
A force P is applied to the web of the beam as shown. If e = 250 mm, determine the height h of the
right flange so that the beam will deflect downward without twisting. The member segments have the
same thickness t.
Given: h1 100mm� bf 300mm� 
e 250mm� 
Solution:
Set t mm� P kN� 
Shear Center: Summing moment about point A
P e˜ Fw2 bf t�� �˜= Fw2 ebf t� P˜� 
Assume bf+t equal to bf : Fw2
e
bf
P˜� 
Section Property : Assume bf t
3 negligible.
I
1
12
t h1
3˜ t h23˜�§© ·¹˜=
Qw2 t 0.5h2 y�� �˜ y 12 0.5h2 y�� ��ª«¬ º»¼˜=
Qw2
t
2
0.25h2
2
y
2�§© ·¹˜=
Shear Flow Resultant:
qw2
V Qw2˜
I
= qw2
P t˜ 0.25h22 y2�§© ·¹˜
2I
=
Fw2
0.5� h2
0.5h2
xqw2
µ´
µ¶
d=
e
bf
P˜
0.5� h2
0.5h2
y
6P t˜ 0.25h22 y2�§© ·¹˜
t h1
3˜ t h23˜�
µ´
µ
µ¶
d=
Given
e h1
3
h2
3�§© ·¹˜
6bf 0.5� h2
0.5h2
y0.25h2
2
y
2�§© ·¹µ´µ¶ d=
Guess h2 10mm� h2 Find h2� �� 
h2 171.0 mm Ans
Problem 7-82
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Problem 7-83
Determine the location e of the shear center, point O, for the tube having a slit along its length.
Problem 7-84
The beam is fabricated from four boards nailed together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are uniformly spaced s = 75 mm. The beam is
subjected to a shear of V = 22.5 kN.
Given: bf 250mm� dw 300mm� 
tf 25mm� tw 25mm� 
tb 25mm� db 100mm� 
V 22.5kN� sn 75mm� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� �� 0.5db 2tb db˜� ��
bf tf˜ dw tw˜� 2 tb db˜� ��� 
yc 87.50 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I3
1
12
2tb� �˜ db3˜ 2tb db˜� � 0.5db yc�� �2˜�� 
I I1 I2� I3�� 
QC tb db˜� � yc 0.5db�� �˜� 
QD dw tw˜� � D yc� 0.5dw�� �˜� 
Shear Flow : q
V Q˜
I
=
The allowable shear flow at points C and D are : qC
FC
sn
= qD
FD
sn
=
FC
sn
V QC˜
I
=
FD
sn
V QD˜
I
=
FC
V QC˜ sn˜
I
� FD
V QD˜ sn˜
I
� 
FC 0.987 kN FD 6.906 kN Ans
Problem 7-85
The beam is constructed from four boards glued together at their seams. If the glue can withstand 15
kN/m, what is the maximum vertical shear V that the beam can support?
Given: bf 100mm� dw 249mm� 
tf 12mm� tw 12mm� di 75mm� 
qallow 15
kN
m
� 
Solution:
Section Property :
I
2
12
tw˜ dw3˜
2
12
bf˜ tf3˜� 2bf tf˜
di tf�
2
§¨
©
·
¹
2
˜�� 
Q bf tf˜� � di tf�2˜� 
Shear Flow : Since there are two glue joints, hence 2q
V Q˜
I
=
Vmax
2I qallow˜
Q
� 
Vmax 20.37 kN Ans
Problem 7-86
Solve Prob. 7-85 if the beam is rotated 90° from the position shown.
Given: bf 100mm� dw 249mm� 
tf 12mm� tw 12mm� di 75mm� 
qallow 15
kN
m
� 
Solution:
Section Property :
I
2
12
tf˜ bf3˜
2
12
dw˜ tw3˜� 2dw tw˜
bf tw�
2
§¨
©
·
¹
2
˜�� 
Q dw tw˜� � bf tw�2˜� 
Shear Flow : Since there are two glue joints, hence 2q
V Q˜
I
=
Vmax
2I qallow˜
Q
� 
Vmax 3.731 kN Ans
Problem 7-87
The member is subjected to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C.
The thickness of each thin-walled segment is 15 mm.
Given: bf 200mm� dw 300mm� 
tf 15mm� tw 15mm� 
tb 15mm� db 115mm� 
V 2kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� �� 0.5db 2tb db˜� ��
bf tf˜ dw tw˜� 2 tb db˜� ��� 
yc 87.98 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I3
1
12
2tb� �˜ db3˜ 2tb db˜� � 0.5db yc�� �2˜�� 
I I1 I2� I3�� 
I 86939045.38 mm
4 
QA 0� 
QB tb db˜� � yc 0.5db�� �˜� QB 52577.05 mm3 
QC QB tf 0.5bf 0.5tw�� �˜ yc 0.5tf�� �˜�� QC 164242.29 mm3 
Shear Flow : q
V Q˜
I
=
qA 0� Ans
qB
V QB˜
I
� qB 1.210
kN
m
 Ans
qC
V QC˜
I
� qC 3.778
kN
m
 Ans
Problem 7-88
The member is subjected to a shear force of V = 2 kN. Determine the maximum shear flow in the
member. All segments of the cross section are 15 mm thick.
Given: bf 200mm� dw 300mm� 
tf 15mm� tw 15mm� 
tb 15mm� db 115mm� 
V 2kN� 
Solution:
Section Property : D dw tf�� 
yc
0.5tf bf tf˜� � 0.5dw tf�� � dw tw˜� �� 0.5db 2tb db˜� ��
bf tf˜ dw tw˜� 2 tb db˜� ��� 
yc 87.98 mm 
I1
1
12
bf˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I3
1
12
2tb� �˜ db3˜ 2tb db˜� � 0.5db yc�� �2˜�� 
I I1 I2� I3�� 
I 86939045.38 mm
4 
Qmax tw D yc�� �˜ 12˜ D yc�� �˜� 
Qmax 386537.47 mm
3 
Shear Flow : q
V Q˜
I
=
Maximum shear flow occurs at the point where the neutral axis passes through the section.
qmax
V Qmax˜
I
� qmax 8.892
kN
m
 Ans
Problem 7-89
The beam is made from three thin plates welded together as shown. If it is subjected to a shear of
V = 48 kN, determine the shear flow at points A and B. Also, calculate the maximum shear stress in
the beam.
Given: bf 215mm� tf 15mm� 
tw 15mm� dw 315mm� 
V 48kN� h 200mm� 
Solution:
Section Property : a 0.5 bf tw�� �� 
yc
bf tw�� � tf˜ª¬ º¼ h 0.5tf�� � dw tw˜� � 0.5dw���
bf tw�� � tf˜ dw tw˜�� 
yc 176.92 mm 
If
1
12
bf tw�� �˜ tf3˜ bf tw�� � tf˜ h 0.5 tf˜� yc�� �2˜�� 
Iw
1
12
tw˜ dw3˜ tw dw˜� � yc 0.5dw�� �2˜�� 
I If Iw�� I 43.71347 10 6�u m4 
QA a tw˜� � dw yc� 0.5a�� �˜� QA 132123.79 mm3 
QB a tf˜� � h 0.5 tf˜� yc�� �˜� QB 45873.79 mm3 
Qmax tw yc� �˜ 12˜ yc� �˜� Qmax 234748.45 mm3 
Shear Flow : q
V Q˜
I
=
qA
V QA˜
I
� qA 145.1
kN
m
 Ans
qB
V QB˜
I
� qB 50.37
kN
m
 Ans
Maximum Shear Stress: W V Q˜
I b˜=
Maximum shear stress occurs at the point where the neutral axis passes through the section.
Wmax
V Qmax˜
I tw� �˜� Wmax 17.18 MPa Ans
Problem 7-90
A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjected
to a shear force of V = 1.25 kN, determine the shear stress at points A and C. Indicate the results on
volume elements located at these points.
Given: bf 100mm� dw 50mm� b'f 25mm� 
t 6mm� V 1.25kN� 
Solution:
Section Property : D dw 2t�� 
yc
0.5t 2b'f t˜� � 0.5dw t�� � 2dw t˜� �� D 0.5t�( ) bf t˜� ��
2b'f t˜ 2dw t˜� bf t˜�
� 
yc 36.60 mm 
I1
1
12
b'f˜ t3˜ b'f t˜� � 0.5t yc�� �2˜�� 
I2
1
12
t˜ dw3˜ dw t˜� � 0.5dw t� yc�� �2˜�� 
I3
1
12
bf˜ t3˜ bf t˜� � D 0.5t� yc�� �2˜�� 
I 2I1 2I2� I3�� 
QA b'f t˜� � yc 0.5t�� �˜� 
QC b'f t˜� � yc 0.5t�� �˜ dw t˜� � 0.5dw t� yc�˜� 0.5bf t˜� � D 0.5t� yc�� �˜�� 
QC 0.00 mm
3 (since A' = 0)
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I t˜� WA 1.335 MPa Ans
WC
V QC˜
I t˜� WC 0 MPa Ans
Problem 7-91
A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjected
to a shear force of V = 1.25 kN, determine the shear stress at point B.
Given: bf 100mm� dw 50mm� b'f 25mm� 
t 6mm� V 1.25kN� 
Solution:
Section Property : D dw 2t�� 
yc
0.5t 2b'f t˜� � 0.5dw t�� � 2dw t˜� �� D 0.5t�( ) bf t˜� ��
2b'f t˜ 2dw t˜� bf t˜�
� 
yc 36.60 mm 
I1
1
12
b'f˜ t3˜ b'f t˜� � 0.5t yc�� �2˜�� 
I2
1
12
t˜ dw3˜ dw t˜� � 0.5dw t� yc�� �2˜�� 
I3
1
12
bf˜ t3˜ bf t˜� � D 0.5t� yc�� �2˜�� 
I 2I1 2I2� I3�� 
QB bf t˜� � D 0.5t� yc�� �˜� 
Shear Stress : W V Q˜
I b˜=
WB
V QB˜
I 2t( )˜� WB 1.781 MPa Ans
Problem 7-92
Determine the location e of the shear center, point O, for the thin-walled member having the cross
section shown.
Problem 7-93
Sketch the intensity of the shear-stress distribution acting over the beam's cross-sectional area, and
determine the resultant shear force acting on the segment AB. The shear acting at the section is V =
175 kN. Show that INA = 340.82(10
6) mm4.
Given: b1 200mm� d1 200mm� V 175kN� 
b2 50mm� d2 150mm� 
Solution:
Section Property : D d1 d2�� 
yc
0.5d1 b1 d1˜� � 0.5d2 d1�� � b2 d2˜� ��
b1 d1˜ b2 d2˜�
� yc 127.63 mm 
y'c D yc�� 
I1
1
12
b1˜ d13˜ b1 d1˜� � 0.5d1 yc�� �2˜�� 
I2
1
12
b2˜ d23˜ b2 d2˜� � D 0.5d2� yc�� �2˜�� 
I I1 I2�� I 340.82 106u mm4 (Q.E.D)
A'1 yc y1�� � b1˜= y1c 0.5 yc y1�� � y1�= y1c 0.5 yc y1�� �=
Q1 A'1 y'1c˜= Q1 0.5 yc y1�� � b1˜ yc y1�� �= Q1 0.5b1 yc2 y12�§© ·¹=
A'2 y'c y2�� � b2˜= y2c 0.5 y'c y2�� � y2�= y2c 0.5 y'c y2�� �=
Q2 A'2 y'2c˜= Q2 0.5 y'c y2�� � b2˜ y'c y2�� �= Q2 0.5b2 y'c2 y22�§© ·¹=
Shear Stress : W V Q˜
I b˜= WCB
V
I
§¨
©
·
¹ 0.5 yc
2
y1
2�§© ·¹ª¬ º¼˜=
W1B
V
I
§¨
©
·
¹ 0.5yc
2§© ·¹˜� At B: y1 0=
W1B 4.18 MPa 
At C: y1 yc d1�� W1C
V
I
§¨
©
·
¹ 0.5 yc
2
y1
2�§© ·¹ª¬ º¼˜� 
W1C 2.84 MPa 
WAB
V
INA
§¨
©
·
¹
0.5 y'c
2
y2
2�§© ·¹ª¬ º¼˜=
At C: y2 y'c d2�� W2C
V
I
§¨
©
·
¹ 0.5 y'c
2
y2
2�§© ·¹ª¬ º¼˜� 
W2C 11.35 MPa 
Resultant Shear Force: For segment AB. yo y'c d2�� VAB
A
AWABµ´¶ d=
VAB
yo
y'c
y
V
I
§¨
©
·
¹ 0.5 y'c
2
y
2�§© ·¹ª¬ º¼˜ b2˜µ´µ¶
d� 
VAB 49.78 kN Ans
Problem 8-1
A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p =
300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.
Given: r 1.5m� p 0.3MPa� Vallow 12MPa� 
Solution:
Normal Stress : Vallow
p r˜
2 t˜=
t
p r˜
2 Vallow˜
� 
t 18.75 mm Ans
Problem 8-2
A pressurized spherical tank is to be made of 125mm-thick steel. If it is subjected to an internal
pressure of p = 1.4 MPa, determine its outer radius if the maximum normal stress is not to exceed 105
MPa.
Given: t 125mm� p 1.4MPa� Vallow 105MPa� 
Solution:
Normal Stress : V p r˜
2 t˜=
ri
2 t˜ Vallow
p
� ri 18.750 m 
ro ri t�� ro 18.875 m Ans
Problem 8-3
The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stress
in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 0.5 MPa.
The wall has a thickness of 6 mm and the inner diameter of the cylinder is 200 mm.
Given: t 6mm� p 0.5MPa� 
ri 200mm� 
Solution:
Case (a) :
Hoop Stress : V1
p ri˜
t
� V1 16.67 MPa Ans
Normal Stress : V2 0� Ans
Case (b) :
Hoop Stress : V1
p ri˜
t
� V1 16.67 MPa Ans
Normal Stress : V2
p ri˜
2 t˜� V2 8.33 MPa Ans
Problem 8-4
The tank of the air compressor is subjected to an internal pressure of 0.63 MPa. If the internal diameter
of the tank is 550 mm, and the wall thickness is 6 mm, determine the stress components acting at point
A. Draw a volume element of the material at this point, and show the results on the element.
Given: t 6mm� p 0.63MPa� 
di 550mm� 
Solution: ri 0.5di� 
Hoop Stress : D
ri
t
� D 45.83 
Since D > 10. then thin-wall analysis can be used.
V1
p ri˜
t
� V1 28.88 MPa Ans
Longitudinal Stress :
V2
p ri˜
2 t˜� V2 14.44 MPa Ans
Problem 8-5
The open-ended pipe has a wall thickness of 2 mm and an internal diameter of 40 mm. Calculate the
pressure that ice exerted on the interior wall of the pipe to cause it to burst in the manner shown. The
maximum stress that the material can support at freezing temperatures is Vmax = 360 MPa. Show the
stress acting on a small element of material just before the pipe fails.
Given: t 2mm� Vallow 360MPa� 
di 40mm� 
Solution: ri 0.5di� 
Hoop Stress : D
ri
t
� D 10.00 
Since D > 10. then thin-wall analysis can be used.
V1 Vallow� 
V1
p ri˜
t
= p
Vallow t˜
ri
� 
p 36.0 MPa Ans
Longitudinal Stress :
Since the pipe is open at both neds, then
V2 0� Ans
Problem 8-6
The open-ended polyvinyl chloride pipe has an inner diameter of 100 mm and thickness of 5 mm. If it
carries flowing water at 0.42 MPa pressure, determine the state of stress in the walls of the pipe.
Given: t 5mm� p 0.42MPa� 
di 100mm� 
Solution: ri 0.5di� 
Hoop Stress : V1
p ri˜
t
� V1 4.2 MPa Ans
Normal Stress : V2 0� Ans
There is no stress componenet in the longitudinal direction since pipe has open ends.
Problem 8-7
If the flow of water within the pipe in Prob. 8-6 is stopped due to the closing of a valve, determine the
state of stress in the walls of the pipe. Neglect the weight of the water. Assume the supports only exert
vertical forces on the pipe.
Given: t 5mm� p 0.42MPa� 
di 100mm� 
Solution: ri 0.5di� 
Hoop Stress : V1
p ri˜
t
� V1 4.2 MPa Ans
Normal Stress : V2
p ri˜
2 t˜� V2 2.1 MPa Ans
Problem 8-8
The A-36-steel band is 50 mm wide and is secured around the smooth rigid cylinder. If the bolts are
tightened so that the tension in them is 2 kN, determine the normal stress in the band, the pressure
exerted on the cylinder, and the distance half the band stretches.
Given: t 3mm� b 50mm� 
r 200mm� F 2kN� 
E 200GPa� 
Solution: rb r 0.5t�� Lb S rb˜� 
Tensile Stress in the Band :
V1
F
b t˜� V1 13.33 MPa Ans
Hoop Stress : V p r˜
t
=
p
tV1
rb
� p 0.199 MPa Ans
Stectch:
G H1 Lb˜= H1
V1
E
=
G
V1 Lb˜
E
� G 0.0422 mm Ans
Problem 8-9
The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is then
subjected to a nonlinear temperature drop of 'T = 12 sin2T °C, where T is in radians, determine the
circumferential stress in the band.
Unit used: qC deg� 
Given: t 0.4mm� b 25mm� 
r' 250mm� E 193GPa� 
'T 12 sin T� �2˜= D 17 10 6�� �˜ 1qC� 
Solution:
Compatibility: Since the band is fitted to a rigid cylinder
(which does not deform under load), then
GF GT� 0=
P 2S r˜� �˜
A E˜
0
2S
TD 'T� �˜ r˜µ´¶ d� 0=
2S r˜
E
P
A
§¨
©
·
¹˜ 12 0
2S
TD sin T� �2˜ r˜µ´¶ d� 0=
However,
P
A
Vc=
2S r˜
E
Vc˜ 12 D˜ r˜
0
2S
Tsin T� �2µ´¶ d=
Vc
6 D˜ E
S 0
2S
Tsin T� �2µ´¶ d
§¨
©¨
·
¹
˜ qC˜� 
Vc 19.69 MPa Ans
Problem 8-10
The barrel is filled to the top with water. Determine the distance s that the top hoop should be placed
from the bottom hoop so that the tensile force in each hoop is the same. Also, what is the force in each
hoop? The barrel has an inner diameter of 1.2 m. Neglect its wall thickness. Assume that only the
hoops resist the water pressure. Note: Water develops pressure in the barrel according to Pascal's law,
p = (0.01z) MPa, where z is the depth from the surface of the water in meter.
Given: d 1.2m� p 0.01z MPa˜=
h 2.4m� h' 0.6m� 
Solution: r 0.5d� 
P
0
h
zp 2r( )˜µ´¶ d=
P 2r
0
h
z0.01zµ´¶ d
§¨
©¨
·
¹
˜ MPa
m
˜� 
P 34.56 kN 
 Equilibrium for the Steel Hoop :
6Fy=0; P 4F� 0= F 0.25P� F 8.64 kN 
60Base=0; P h
3
˜ 2F h'˜� 2F h' s�( )˜� 0=
s
P h˜
6F
h'� h'�� 
s 400 mm Ans
Problem 8-11
A wood pipe having an inner diameter of 0.9 m is bound together using steel hoops having a
cross-sectional area of 125 mm2. If the allowable stress for the hoops is Vallow = 84 MPa, determine
their maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressure
of 28 kPa. Assume each hoop supports the pressure loading acting along the length s of the pipe.
Given: d 0.9m� p 28 10 3�� �˜ MPa˜� 
As 125mm
2� Vallow 84MPa� 
Solution: r 0.5d� 
P p 2r s˜( )˜= F Vallow As� �˜=
 Equilibrium for the Steel Hoop :
From the FBD,
+ 6Fy=0; P 2F� 0=
p 2r s˜( )˜ 2Vallow As� �˜� 0=
s
Vallow As� �˜
p r˜� 
s 833.33 mm Ans
Problem 8-12
A boiler is constructed of 8-mm thick steel plates that are fastened together at their ends using a butt
joint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm
apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress
in the boiler's plate apart from the seam, (b) the circumferential stress in the outer cover plate along the
rivet line a-a, and (c) the shear stress in the rivets.
Given: to 8mm� ri 750mm� p 1.35MPa� 
tc 8mm� db 10mm� s 50mm� 
Solution:
a) Hoop Stress :
V1
p ri˜
to
� V1 126.6 MPa Ans
b) Hoop Stress in cover plate along line a-a :
Consider a width of s (mm), Fo (in boiler plate) = F'o (in cover plates)
V1 s to˜� �˜ V'1 s db�� �˜ 2 tc˜� �˜=
V'1
V1 s to˜� �˜
s db�� � 2 tc˜� �˜� 
V'1 79.1 MPa Ans
c) Shear Stress in Rivet : rb 0.5db� 
From the FBD,
+ 6Fy=0; Fb V1 s to˜� �˜� 0=
Fb V1 s to˜� �˜� 
Wavg
1
2
Fb
S rb2˜
§¨
©¨
·
¹
˜� 
Wavg 322.3 MPa Ans
Problem 8-13
The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a
pressure p. Determine the change in the internal radius of the ring after this pressure is applied. The
modulus of elasticity for the ring is E.
Solution:
 Equilibrium for the Ring :
+ 6Fy=0; 2P 2p ri˜ w˜� 0=
P p ri˜ w˜=
Hoop Stress and Strain for the Ring :
V1
P
A
= V1
p ri˜ w˜
ro ri�� � w˜= V1
p ri˜
ro ri�
=
Using Hooke's Law,
H1
V1
E
= H1
p ri˜
ro ri�� � E˜= (1)
However,
H1
2 S˜ ri
1
˜ 2 S˜ ri˜�
2 S˜ r˜= H1
ri
1
ri�
r
= H1
Gri
r
=
Then, from Eq.(1)
Gri
r
p ri˜
ro ri�� � E˜=
Gri
p ri
2˜
ro ri�� � E˜= Ans
Problem 8-14
A closed-ended pressure vessel is fabricated by cross-winding glass filaments over a mandrel, so that
the wall thickness t of the vessel is composed entirely of filament and an epoxy binder as shown in the
figure. Consider a segment of the vessel of width w and wrapped at an angle T. If the vessel is subjected
to an internal pressure p, show that the force in the segment is FT = V0 wt , where V0 is the stress in the
filaments. Also, show that the stresses in the hoop and longitudinal directions are Vh = V0 sin2 T��and Vl
= V0 cos2 T�, respectively. At what angle T (optimum winding angle) would the filaments have to be
wound so that the hoop and longitudinal stresses are equivalent?
Problem 8-15
The steel bracket is used to connect the ends of two cables. If the allowable normal stress for the steel
is Vallow = 168 MPa, determine the largest tensile force P that can be applied to the cables. The bracket
has a thickness of 12 mm and a width of 18 mm.
Given: b 18mm� a 50mm� 
t 12mm� Vallow 168MPa� 
Solution:
Internal Force and Moment :
ao a 0.5 b˜�� 
N P=
M P ao˜=
Section Property :
A t b˜� I 1
12
t˜ b3˜� 
Alowable Normal Stress: V N
A
M c˜
I
�=
The maximum normal stress occurs at the bottom of the steel bracket.
cmax 0.5 b˜� Vallow
P
A
P ao˜ cmax˜
I
�=
P
Vallow
1
A
ao cmax˜
I
�
� 
P 1.756 kN Ans
Problem 8-16
The steel bracket is used to connect the ends of two cables. If the applied force P = 2.5 kN, determine
the maximum normal stress in the bracket. The bracket has a thickness of 12 mm and a width of 18
mm.
Given: b 18mm� a 50mm� 
t 12mm� P 2.5kN� 
Solution:
Internal Force and Moment :
ao a 0.5 b˜�� 
N P� 
M P ao˜� 
Section Property :
A t b˜� I 1
12
t˜ b3˜� 
Alowable Normal Stress: V N
A
M c˜
I
�=
The maximum normal stress occurs at the bottom of the steel bracket.
cmax 0.5 b˜� Vmax
N
A
M cmax˜
I
�� 
Vmax 239.2 MPa Ans
Problem 8-17
The joint is subjected to a force of 1.25 kN as shown. Sketch the normal-stress distribution acting over
section a-a if the member has a rectangular cross section of width 12 mm and thickness 18 mm.
Given: b 18mm� ah 50mm� av 32mm� 
t 12mm� P 1.25kN� 
v 3� h 4� r 5� 
Solution:
Internal Force and Moment :
+ 6Fx=0; P h
r
§¨
©
·
¹˜ N� 0= N P
h
r
§¨
©
·
¹˜� 
+ 6Fy=0; V P v
r
§¨
©
·
¹˜� 0= V P
v
r
§¨
©
·
¹˜� 
+ 60A=0; M P h
r
§¨
©
·
¹˜ av˜� P
v
r
§¨
©
·
¹˜ ah˜� 0=
M P
v
r
§¨
©
·
¹˜ ah˜ P
h
r
§¨
©
·
¹˜ av˜�� 
Section Property :
A b t˜� I 1
12
b˜ t3˜� 
Normal Stress: V N
A
M c˜
I
�=
ctop 0.5 t˜� Vtop
N
A
M ctop˜
I
�� Vtop 17.36 MPa (T) Ans
cbot 0.5� t˜� Vbot
N
A
M cbot˜
I
�� Vbot 8.10� MPa (C) Ans
Location of zero stress:
Vtop
Vbot
yo
t yo�
=
yo
t Vtop˜
Vbot Vtop�
� 
yo 8.18 mm 
Problem 8-18
The joint is subjected to a force of 1.25 kN as shown. Determine the state of stress at points A and B,
and sketch the results on differential elements located at these points. The member has a rectangular
cross-sectional area of width 12 mm and thickness 18 mm.
Given: b 18mm� ah 50mm� av 32mm� 
t 12mm� P 1.25kN� 
v 3� h 4� r 5� 
Solution:
Internal Force and Moment :
+ 6Fx=0; P h
r
§¨
©
·
¹˜ N� 0= N P
h
r
§¨
©
·
¹˜� 
+ 6Fy=0; V P v
r
§¨
©
·
¹˜� 0= V P
v
r
§¨
©
·
¹˜� 
+ 60A=0; M P h
r
§¨
©
·
¹˜ av˜� P
v
r
§¨
©
·
¹˜ ah˜� 0=
M P
v
r
§¨
©
·
¹˜ ah˜ P
h
r
§¨
©
·
¹˜ av˜�� 
Section Property:
A b t˜� I 1
12
b˜ t3˜� 
QA 0.5 t˜ b˜( ) 0.25t( )˜� 
QB 0� (since A' = 0)
Normal Stress: V N
A
M c˜
I
�=
cA 0� VA
N
A
M cA˜
I
�� VA 4.63 MPa (T) Ans
cB 0.5� t˜� VB
N
A
M cB˜
I
�� VB 8.10� MPa (C) Ans
Shear Stress : W V Q˜
I b˜=
WA
V QA˜
I b˜� WA 5.21 MPa Ans
WB
V QB˜
I b˜� WB 0 MPa Ans
Problem 8-19
The coping saw has an adjustable blade that is tightened with a tension of 40 N. Determine the state of
stress in the frame at points A and B.
Given: t 3mm� a 100mm� 
h 8mm� P 40N� 
Solution:
Internal Force and Moment :
At A: NA P�� MA P a˜� 
At B: NB 0� MB P 0.5a( )˜� 
Section Property :
A t h˜� I 1
12
t˜ h3˜� 
State of Stress: V N
A
M c˜
I
�=
At A:
cA 0.5 h˜� VA
NA
A
MA cA˜
I
�� 
VA 123.3 MPa (T) Ans
At B:
cB 0.5 h˜� VB
NB
A
MB cB˜
I
�� 
VB 62.5 MPa (T) Ans
Problem 8-20
Determine the maximum and minimum normal stress in the bracket at section a when the load is
applied at x = 0.
Given: a 30mm� b 20mm� 
xe 0.5a� P 4kN� 
Solution:
Internal Force and Moment :
N P�� 
M P xe˜� M 0.060 kN m˜ 
Section Property :
A a b˜� A 600 mm2 
I
1
12
b˜ a3˜� I 45000 mm4 
Normal Stress: V N
A
M c˜
I
�=
cmax 0.5a� Vt
N
A
M cmax˜
I
�� Vt 13.33 MPa (T)
cmin 0.5� a˜� Vc
N
A
M cmin˜
I
�� Vc 26.67� MPa (C)
Vmax max Vt Vc�� �� Vmax 26.67 MPa Ans
Vmin min Vt Vc�� �� Vmin 13.33 MPa Ans
Problem 8-21
Determine the maximum and minimum normal stress in the bracket at section a when the load is
applied at x = 50 mm.
Given: a 30mm� b 20mm� 
P 4� kN� xe 0.5a 50mm�� 
Solution:
Internal Force and Moment :
N P� 
M P xe˜� M 0.140 kN m˜ 
Section Property :
A a b˜� A 600 mm2 
I
1
12
b˜ a3˜� I 45000 mm4 
Normal Stress: V N
A
M c˜
I
�=
cmax 0.5a� Vt
N
A
M cmax˜
I
�� Vt 40.00 MPa (T)
cmin 0.5� a˜� Vc
N
A
M cmin˜
I
�� Vc 53.33� MPa (C)
Vmax max Vt Vc�� �� Vmax 53.33 MPa Ans
Vmin min Vt Vc�� �� Vmin 40.00 MPa Ans
Problem 8-22
The vertical force P acts on the bottom of the plate having a negligible weight. Determine the maximum
distance d to the edge of the plate at which it can be applied so that it produces no compressive
stresses on the plate at section a-a. The plate has a thickness of 10 mm and P acts along the centerline
of this thickness.
Given: t 10mm� b 150mm� 
Solution:
Internal Force and Moment :
N P=
xe d 0.5b�= M P xe˜=
Section Property :
A t b˜� A 1500 mm2 
I
1
12
t˜ b3˜� I 2812500 mm4 
Normal Stress: Require Vmin 0� 
I
Mc
A
Nı r 
cmin 0.5� b˜� 0
P
A
P xe˜� � cmin˜
I
�=
0
P
A
P d 0.5b�( )˜ 0.5� b˜( )˜
I
�=
d
2I
A b˜
b
2
�� 
d 100 mm Ans
Problem 8-23
The vertical force P = 600 N acts on the bottom of the plate having a negligible weight. The plate has a
thickness of 10 mm and P acts along the centerline of this thickness such that d = 100 mm. Plot the
distribution of normal stress acting along section a-a.
Given: t 10mm� b 150mm� 
d 100mm� P 0.6kN� 
Solution:
Internal Force and Moment :
N P� 
xe d 0.5b�� M P xe˜� M 0.015 kN m˜ 
Section Property :
A t b˜� A 1500 mm2 
I
1
12
t˜ b3˜� I 2812500 mm4 
Normal Stress: V N
A
M c˜
I
�=
cA 0.5� b� VA
N
A
M cA˜
I
�� VA 0 MPa (C) Ans
cB 0.5 b˜� VB
N
A
M cB˜
I
�� VB 0.800 MPa (T) Ans
Problem 8-24
The gondola and passengers have a weight of 7.5 kN and center of gravity at G. The suspender arm
AE has a square cross-sectional area of 38 mm by 38 mm, and is pin connected at its ends A and E.
Determine the largest tensile stress developed in regions AB and DC of the arm.
Given: b 38mm� d 38mm� ah 375mm� 
L1 1.2m� L2 1.65m� W 7.5kN� 
Solution:
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
Segment AB :
NAB W� MAB 0� 
Maximum Normal Stress: V N
A
M c˜
I
�=
VAB
NAB
A
� VAB 5.19 MPa (T) Ans
Segment DC :
NDC W� MDC W ah˜� 
Maximum Normal Stress: V N
A
M c˜
I
�=
cmax 0.5d� VDC
NDC
A
MDC cmax˜
I
�� 
VDC 312.73 MPa (T) Ans
Problem 8-25
The stepped support is subjected to the bearing load of 50 kN. Determine the maximum and minimum
compressive stress in the material.
Given: a 100mm� b 100mm� 
P 50kN� xe 0.5a 30mm�� 
Solution:
Internal Force and Moment :
N P�� 
M P xe˜� M 1.000 kN m˜ 
Section Property :
For the bottom portion of the stepped support.
A a b˜� A 10000 mm2 
I
1
12
b˜ a3˜� I 8333333.33 mm4 
Normal Stress: V N
A
M c˜
I
�=
cmax 0.5a� Vt
N
A
M cmax˜
I
�� Vt 1 MPa (T)
cmin 0.5� a˜� Vc
N
A
M cmin˜
I
�� Vc 11.00� MPa (C)
Vc_max max 0 Vc�� �� Vc_max 11 MPa Ans
Vc_min min 0 Vc�� �� Vc_min 0 MPa Ans
Problem 8-26
The bar has a diameter of 40 mm. If it is subjected to a force of 800 N as shown, determine the stress
components that act at point A and show the results on a volume element located at this point.
Given: do 40mm� a 200mm� 
P 0.8kN� T 30deg� 
Solution:
Internal Force and Moment : At section AB.
N P sin T� �˜� N 0.400 kN 
V P cos T� �˜� V 0.693 kN 
M V a˜� M 0.1386 kN m˜ 
Section Property : ro 0.5do� 
A S ro2˜� A 1256.64 mm2 
I
S
4
ro
4˜� 
I 125663.71 mm
4 
QA
4ro
3S
A
2
§¨
©
·
¹˜� QA 5333.33 mm
3 
Normal Stress: V N
A
M c˜
I
�=
cA 0� VA
N
A
M cA˜
I
�� 
VA 0.318 MPa (T) Ans
Shear Stress : W V Q˜
I b˜=
WA
V QA˜
I do˜
� WA 0.735 MPa Ans
Problem 8-27
Solve Prob. 8-26 for point B.
Given: do 40mm� a 200mm� 
P 0.8kN� T 30deg� 
Solution:
Internal Force and Moment : At section AB.
N P sin T� �˜� N 0.400 kN 
V P cos T� �˜� V 0.693 kN 
M V a˜� M 0.1386 kN m˜ 
Section Property : ro 0.5do� 
A S ro2˜� A 1256.64 mm2 
I
S
4
ro
4˜� 
I 125663.71 mm
4 
QB 0� (since A'=0 )
Normal Stress: V N
A
M c˜
I
�=
cB ro�� VB
N
A
M cB˜
I
�� 
VB 21.73� MPa (C) Ans
Shear Stress : W V Q˜
I b˜=
WA
V QB˜
I do˜
� WA 0 MPa Ans
Problem 8-28
Since concrete can support little or no tension, this problem can be avoided by using wires or rods to
prestress the concrete once it is formed. Consider the simply supported beam shown, which has a
rectangular cross section of 450 mm by 300 mm. If concrete has a specific weight of 24 kN/m3,
determine the required tension in rod AB, which runs through the beam so that no tensile stress is
developed in the concrete at its center section a-a. Neglect the size of the rod and any deflection of the
beam.
Given: b 300mm� d 450mm� 
d' 400mm� J 24 kN
m
3
� 
L 2.4m� 
Solution: a d d'�� 
w J b˜ d˜� 
Support Reactions : By symmetry, RA= R ; RA= R
+ 2R w L˜� 0= R 0.5w L˜� 
Internal Force and Moment :
+ 6Fx=0; T N� 0= N T=
+ 60O=0; M T 0.5d a�( )˜� R 0.5L( )˜� 0.5w L˜( ) 0.25L( )˜� 0=
M R 0.25 L˜( )˜ T 0.5 d˜ a�( )˜�=
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
Normal Stress: Va
N
A
M c˜
I
�=
Requires Va= 0 0 T�
A
M ca˜
I
�=
ca 0.5d� 0
T�
A
R 0.25 L˜( )˜ T 0.5 d˜ a�( )˜�[ ] ca˜
I
�=
T
R 0.25 L˜( )˜
0.5 d˜ a�( ) I
A ca˜
�
� 
T 9.331 kN Ans
Problem 8-29
Solve Prob. 8-28 if the rod has a diameter of 12 mm. Use the transformed area method discussed in
Sec. 6.6. Est = 200 GPa, Ec = 25 GPa.
Given: b 300mm� d 450mm� 
d' 400mm� do 12mm� 
Est 200GPa� Ec 25GPa� 
L 2.4m� J 24 kN
m
3
� 
Solution: a d d'�� 
w J b˜ d˜� 
Support Reactions : By symmetry, RA= R ; RA= R
+ 2R w L˜� 0= R 0.5w L˜� 
Internal Force and Moment :
+ 6Fx=0; T N� 0= N T=
+ 60O=0; M T d yc� a�� �˜� R 0.5L( )˜� 0.5w L˜( ) 0.25L( )˜� 0=
M R 0.25 L˜( )˜ T d yc� a�� �˜�=
Section Property :
n
Est
Ec
� A'conc n 1�( )
S
4
do
2˜§¨©
·¹˜� 
A b d˜ A'conc�� 
yc
b d˜ 0.5d( )˜ A'conc d'˜�
A
� 
I
1
12
b˜ d3˜ b d˜ 0.5d yc�� �2˜� A'conc d' yc�� �2˜�� 
Normal Stress: Va
N
A
M c˜
I
�=
Requires Va= 0 0 T�
A
M ca˜
I
�=
ca d yc�� 0
T�
A
R 0.25 L˜( )˜ T d yc� a�� �˜�ª¬ º¼ ca˜
I
�=
T
R 0.25 L˜( )˜
d yc� a�� � IA ca˜�
� 
T 9.343 kN Ans
Problem 8-30
The block is subjected to the two axial loads shown. Determine the normal stress developed at points A
and B. Neglect the weight of the block.
Given: b 50mm� d 75mm� 
P1 250N� P2 500N� 
Solution:
Internal Force and Moment :
+ 6Fx=0; N� P1� P2� 0= N P1� P2�� 
+ 60z=0; Mz P1 0.5d( )˜� P2 0.5d( )˜� 0=
Mz 0.5 d˜ P2 P1�� �˜� 
+ 60y=0; My P1 0.5b( )˜� P2 0.5b( )˜� 0=
My 0.5 b˜ P2 P1�� �˜� 
Section Property :
A b d˜� Iz
1
12
b˜ d3˜� Iy
1
12
d˜ b3˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At A: yA 0.5d� zA 0.5b� 
VA
N
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 0.200� MPa (C) Ans
At B: yB 0.5d� zB 0.5� b� 
VB
N
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 0.600� MPa (C) Ans
Solution:
(T)
Given: b 50mm� d 75mm� 
P1 250N� P2 500N� 
60z=0;
Internal Force and Moment :
+ 6Fx=0; N� P1� P2� 0= N P1� P2�� 
+
Ans
Mz P1 0.5d( )˜� P2 0.5d( )˜� 0=
Mz 0.5 d˜ P2 P1�� �˜� 
+ 60y=0; My P1 0.5b( )˜� P2 0.5b( )˜� 0=
My 0.5 b˜ P2 P1�� �˜� 
Section Property :
A b d˜� Iz
1
12
b˜ d3˜� Iy
1
12
d˜ b3˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At A: yA 0.5d� zA 0.5b� 
VA
N
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 0.200� MPa (C) Ans
At B: yB 0.5d� zB 0.5� b� 
VB
N
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 0.600� MPa (C) Ans
At C: yC 0.5� d� zC 0.5� b� 
VC
N
A
Mz yC˜
Iz
�
My zC˜
Iy
�� VC 0.200� MPa (C) Ans
At D: yD 0.5� d� zD 0.5b� 
VD
N
A
Mz yD˜
Iz
�
My zD˜
Iy
�� VD 0.200 MPa 
Problem 8-31
The block is subjected to the two axial loads shown. Sketch the normal stress distribution acting over
the cross section at section a-a. Neglect the weight of the block.
Problem 8-32
A bar having a square cross section of 30 mm by 30 mm is 2 m long and is held upward. If it has a
mass of 5 kg/m, determine the largest angle T, measured from the vertical, at which it can be
supported before it is subjected to a tensile stress near the grip.
Given: b 30mm� L 2m� 
t 30mm� mo 5
kg
m
� 
Solution: W mo g˜ L˜� W 0.0981 kN 
Internal Force and Moment :
 + 6Fy=0; N� W cos T� �˜� 0= N W� cos T� �˜=
+
60O=0; M W sin T� �˜ 0.5 L˜( )˜� 0= M 0.5W L˜ sin T� �˜=
Section Property :
A b t˜� I 1
12
b˜ t3˜� 
Normal Stress: Require Vmax 0� 
I
Mc
A
Nı r 
cmax 0.5 b˜� 0
N
A
M cmax˜
I
�=
0
W� cos T� �˜
b t˜
0.5W L˜ sin T� �˜ 0.5 b˜( )˜
1
12
b˜ t3˜
�=
tan T� � t2
3 L˜ b˜=
T atan t
2
3 L˜ b˜
§¨
©
·
¹� 
T 0.00500 rad T 0.286 deg Ans
Problem 8-33
Solve Prob. 8-32 if the bar has a circular cross section of 30-mm diameter.
Given: do 30mm� L 2m� 
mo 5
kg
m
� 
Solution: W mo g˜ L˜� W 0.0981 kN 
Internal Force and Moment :
 + 6Fy=0; N� W cos T� �˜� 0= N W� cos T� �˜=
+
60O=0; M W sin T� �˜ 0.5 L˜( )˜� 0= M 0.5W L˜ sin T� �˜=
Section Property : ro 0.5do� 
A S ro2˜� A 706.86 mm2 
I
S
4
ro
4˜� I 39760.78 mm4 
Normal Stress: Require Vmax 0� 
I
Mc
A
Nı r 
cmax ro� 0
N
A
M cmax˜
I
�=
0
W� cos T� �˜
A
0.5W L˜ sin T� �˜ ro� �˜
I
�=
tan T� � 2I
A L˜ ro˜
=
T atan 2I
A L˜ ro˜
§¨
©
·
¹
� 
T 0.00375 rad T 0.215 deg Ans
Problem 8-34
The wide-flange beam is subjected to the loading shown. Determine the stress components at points A
and B and show the results on a volume element at each of these points. Use the shear formula to
compute the shear stress.
Given: b 100mm� d 150mm� 
t 12mm� d'B 50mm� 
P1 2.5kN� P2 12.5kN� 
P3 15kN� L1 0.5m� 
L2 1m� L3 1.5m� 
Solution: L 3 L1˜ L2� L3�� 
Support Reactions : Given
+ 6Fy=0; R1 R2� P1� P2� P3� 0=
+ 60R2=0; R1� L˜ P1 L L1�� �˜� P2 L2 L3�� �˜� P3 L3˜� 0=
Guess R1 1N� R2 1N� 
R1
R2
§¨
©¨
·
¹
Find R1 R2�� �� R1
R2
§¨
©¨
·
¹
15.63
14.38
§¨
©
·
¹ kN 
At Section A-B:
M R1 2L1� �˜� V R1 P1�� 
Section Property : D d 2t�� 
A 2 b˜ t˜ d t˜�� I 1
12
b˜ D3˜ 1
12
QA 0� (since A' = 0)
Normal Stress: V M y˜
I
=
yA 0.5D� VA
M� yA˜
I
� VA 70.98� MPa (C) Ans
yB 0.5d d'B�� VB
M yB˜
I
� VB 20.4 MPa (T) Ans
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I b˜� WA 0.00 MPa Ans
WB
V QB˜
I t˜� WB 7.265 MPa Ans
b t�( )˜ d3˜�� 
QB b t˜ 0.5D 0.5t�( )˜ d'B t˜ 0.5D t� 0.5d'B�� �˜�� 
Problem 8-35
The cantilevered beam is used to support the load of 8 kN. Determine the state of stress at points A
and B, and sketch the results on differential elements located at each of these points.
Given: bo 100mm� d 100mm� 
t 10mm� L 3m� 
dA 25mm� dB 45mm� 
P 8kN� 
Solution:
Internal Force and Moment : At Section A-B:
V P� V 8 kN 
M P L˜� M 24 kN m˜ 
Section Property : bi bo 2t�� 
A 2 d˜ t˜ bi t˜�� A 2800 mm2 
I
1
12
bi˜ t3˜
1
12
2t( )˜ d3˜�� I 1673333.33 mm4 
QA dA t˜� � 0.5d 0.5dA�� �˜� QA 9375 mm3 
QB dB t˜� � 0.5d 0.5dB�� �˜� QB 12375 mm3 
Normal Stress: V M y˜
I
=
yA 0.5d dA�� VA
M yA˜
I
� VA 358.6 MPa (T) Ans
yB 0.5d dB�� VB
M yB˜
I
� VB 71.7 MPa (T) Ans
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I t˜� WA 4.48 MPa Ans
WB
V QB˜
I t˜� WB 5.92 MPa Ans
Problem 8-36
The cylinder of negligible weight rests on a smooth floor. Determine the eccentric distance ey at which
the load can be placed so that the normal stress at point A is zero.
Solution:
Internal Force and Moment :
V P�=
M P ey� �˜=
Section Property :
A S r2˜=
I
S
4
r
4˜=
Normal Stress: Require Vmax 0� 
I
Mc
A
Nı r 
cA r= 0
N
A
M cr˜
I
�=
0
P�
S r2˜
4P ey� �˜ r˜
S r4˜
�=
ey
r
4
= Ans
Problem 8-37
The beam supports the loading shown. Determine the state of stress at points E and F at section a-a,
and represent the results on a differential volume element located at each of these points.
Given: bf 150mm� dw 200mm� 
tf 10mm� tw 15mm� 
a 1m� b 2m� 
yB 3.3m� 
w 10
kN
m
� 
yD 0.3m� 
Solution: L 2 a˜ b�� 
Support Reactions : Given
Dx
L
yB yD�
§¨
©
·
¹
Dy˜= (1)
+ 6Fy=0; Cy Dy� w 2a( )˜� 0= (2)
+ 60C=0; w� 2a( )˜ a˜ Dy L( )˜� Dx yD� �˜� 0= (3)
Solving Eqs.(1), (2) and (3). Guess Cy 1N� Dx 1N� Dy 1N� 
Cy
Dx
Dy
§¨
¨¨
©
·
¸
¹
Find Cy Dx� Dy�� �� 
Cy
Dx
Dy
§¨
¨¨
©
·
¸
¹
15.4545
6.0606
4.5455
§¨
©¨
·
¹
kN 
Internal Force and Moment : At Section a-a:
N Dx�� N 6.0606� kN 
V w a˜ Dy�� V 5.4545 kN 
M Dy� a b�( )˜ Dx yD˜� w a˜ 0.5a( )˜�� M 10.4545� kN m˜ 
Section Property : do dw 2tf�� 
A 2 bf˜ tf˜ dw tw˜�� A 6000 mm2 
I
1
12
bf˜ do3˜
1
12
bf tw�� �˜ dw3˜�� I 43100000 mm4 
QE bf tf˜ 0.5do 0.5tf�� �˜ 0.5dw tw˜ 0.25dw� �˜�� QE 232500 mm3 
QF 0� (since A' = 0)
I
Mc
A
Nı r Normal Stress:
cE 0� VE
N
A
M cE˜
I
�� VE 1.01� MPa (C) Ans
cF 0.5 do˜� VF
N
A
M cF˜
I
�� VF 27.69� MPa (C) Ans
Shear Stress : W V Q˜
I t˜=
WE
V QE˜
I tw˜
� WE 1.96 MPa Ans
WF
V QF˜
I bf˜
� WF 0 MPa Ans
Problem 8-38
The metal link is subjected to the axial force of P = 7 kN. Its original cross section is to be altered by
cutting a circular groove into one side. Determine the distance a the groove can penetrate into the cross
section so that the tensile stress does not exceed Vallow = 175 MPa. Offer a better way to remove this
depth of material from the cross section and calculate the tensile stress for this case. Neglect the
effects of stressconcentration.
Given: h 80mm� t 25mm� 
P 7kN� Vallow 175MPa� 
Solution:
Internal Force and Moment : At narrow section.
+ 6Fx=0; P N� 0= N P� 
+ 60base=0; M N h a�
2
˜� P h
2
˜� 0= M 0.5P a˜=
Section Property :
A h a�( ) t˜= I t
12
h a�( )3˜=
Normal Stress: Require Vmax Vallow� 
cmax 0.5 d a�( )=
Vmax
N
A
M cmax˜
I
�=
Vallow
P
h a�( ) t˜
0.5P a˜ 0.5 h a�( )[ ]˜
t
12
h a�( )3˜
�=
Given Vallow t˜� � h a�( )2˜ P h a�( )˜� 3P a˜� 0= (1)
Solving Eq.(1),. Guess a 1mm� 
a Find a( )� 
a 61.94 mm Ans
Better way: To remove material equally from both sides such that M=0.
A h a�( ) t˜� 
V'max
N
A
0�� 
V'max 15.50 MPa Ans
Problem 8-39
Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN. Indicate
the result as a differential volume element.
Given: bf 150mm� dw 200mm� 
tf 20mm� tw 15mm� 
a 2m� b 0.75m� c 1m� 
hG 375mm� rG 250mm� P 4kN� 
Solution: L a b� c�� 
Support Reactions : Given
+ 6Fx=0; Cx P� 0= Cx P� 
+ 60D=0; P hg rG�� �˜ a˜ Cy L( )˜� 0= Cy hG rG�L P� 
Internal Force and Moment : At Section A-B:
N Cx� N 4 kN 
V Cy� V 0.6667 kN 
M Cy c˜� M 0.6667 kN m˜ 
Section Property : do dw 2tf�� 
A 2 bf˜ tf˜ dw tw˜�� A 9000 mm2 
I
1
12
bf˜ do3˜
1
12
bf tw�� �˜ dw3˜�� I 82800000 mm4 
QA bf tf˜ 0.5do 0.5tf�� �˜ 0.5dw tw˜ 0.25dw� �˜�� QA 405000 mm3 
I
Mc
A
Nı r Normal Stress:
cA 0� VA
N
A
M cA˜
I
�� 
VA 0.444 MPa (T) Ans
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I tw˜
� WA 0.217 MPa Ans
Problem 8-40
Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN. Indicate
the result as a differential volume element.
Given: bf 150mm� dw 200mm� 
tf 20mm� tw 15mm� 
a 2m� b 0.75m� c 1m� 
hG 375mm� rG 250mm� P 4kN� 
Solution: L a b� c�� 
Support Reactions : Given
+ 6Fx=0; Cx P� 0= Cx P� 
+ 60D=0; P hg rG�� �˜ a˜ Cy L( )˜� 0= Cy hG rG�L P� 
Internal Force and Moment : At Section A-B:
N Cx� N 4 kN 
V Cy� V 0.6667 kN 
M Cy c˜� M 0.6667 kN m˜ 
Section Property : do dw 2tf�� 
A 2 bf˜ tf˜ dw tw˜�� A 9000 mm2 
I
1
12
bf˜ do3˜
1
12
bf tw�� �˜ dw3˜�� I 82800000 mm4 
QB 0� (since A' = 0)
I
Mc
A
Nı r Normal Stress:
cB 0.5� do� VB
N
A
M cB˜
I
�� 
VB 0.522� MPa (C) Ans
Shear Stress : W V Q˜
I t˜=
WB
V QB˜
I tw˜
� WB 0 MPa Ans
Problem 8-41
The bearing pin supports the load of 3.5 kN. Determine the stress components in the support member
at point A. The support is 12 mm thick.
Given: b 12mm� d 18mm� W 3.5kN� 
L1 50mm� L2 75mm� a 32mm� 
T 30deg� 
Solution:
Internal Force and Moment :
6Fx=0; N W cos T� �˜� 0= N W cos T� �˜� 
6Fy=0; V W sin T� �˜� 0= V W sin T� �˜� 
+ 60=0; M W a L1 sin T� �˜�� �˜� 0=
M W a L1 sin T� �˜�� �˜� 
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
QA 0� (since A' = 0)
Normal Stress: V N
A
M y˜
I
�=
At A: yA 0.5� d� 
VA
N
A
M yA˜
I
�� 
VA 23.78� MPa (C) Ans
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I b˜� 
WA 0 MPa Ans
Problem 8-42
The bearing pin supports the load of 3.5 kN. Determine the stress components in the support member
at point B. The support is 12 mm thick.
Given: b 12mm� d 18mm� W 3.5kN� 
L1 50mm� L2 75mm� a 32mm� 
T 30deg� 
Solution:
Internal Force and Moment :
6Fx=0; N W cos T� �˜� 0= N W cos T� �˜� 
6Fy=0; V W sin T� �˜� 0= V W sin T� �˜� 
+ 60=0; M W a L1 sin T� �˜�� �˜� 0=
M W a L1 sin T� �˜�� �˜� 
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
QB 0� (since A' = 0)
Normal Stress: V N
A
M y˜
I
�=
At B: yB 0.5d� 
VB
N
A
M yB˜
I
�� 
VB 51.84 MPa (T) Ans
Shear Stress : W V Q˜
I t˜=
WB
V QB˜
I b˜� 
WB 0 MPa Ans
p 8
kN
m
2
� 
Given: b 3.6m� h 1.8m� 
ro 75mm� ri 68mm� 
W Nx� 0=
W 7.5kN� d 0.9m� 
Solution: P p b˜ h˜� 
Internal Force and Moment :
6Fx=0;
Ans
Nx W�� 
6Fy=0; Vy P� 0= Vy P� 
6Fz=0; Vz 0� 
60x=0; Mx P 0.5b( )˜� 0= Mx 0.5 P˜ b˜� 
60y=0; My W 0.5b( )˜� 0= My 0.5 W˜ b˜� 
60z=0; Mz P 0.5h d�( )˜� 0= Mz P� 0.5h d�( )˜� 
Nx 7.5� kN Vy 51.84 kN Vz 0 kN 
Mx 93.312 kN m˜ My 13.5 kN m˜ Mz 93.312� kN m˜ 
Section Property : A S ro2 ri2�§© ·¹˜� 
Iy
S
4
ro
4
ri
4�§© ·¹˜� Iz Iy� 
QC_y
4ro
3S
S
2
ro
2˜§¨©
·
¹˜
4ri
3S
S
2
ri
2˜§¨©
·
¹˜�� 
QD_z QC_y� QD_y 0� 
QC_z 0� J S
2
ro
4
ri
4�§© ·¹˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At C: yC 0� zC ri� 
VC
N
A
Mz yC˜
Iz
�
My zC˜
Iy
�� VC 113.9 MPa (T) Ans
At D: yD ro� zD 0� 
VD
N
A
Mz yD˜
Iz
�
My zD˜
Iy
�� VD 868.5 MPa (T)
Problem 8-43
The uniform sign has a weight of 7.5 kN and is supported by the pipe AB, which has an inner radius of
68 mm and an outer radius of 75 mm. If the face of the sign is subjected to a uniform wind pressure
of p = 8 kN/m2, determine the state of stress at points C and D. Show the results on a differential
volume element located at each of these points. Neglect the thickness of the sign, and assume that it is
supported along the outside edge of the pipe.
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wt
T U˜
J
=
At C: t 2 ro ri�� �˜� U ri� 
Wv_y
Vy QC_y˜
Iz t˜
� Wt
Mx U˜
J
� 
WC_xy Wv_y Wt�� WC_xy 360.8� MPa Ans
WC_xz 0� Ans
WC_yz 0� Ans
At D: t 2 ro ri�� �˜� U' ro� 
Wv_z
Vz QD_z˜
Iy t˜
� W't
Mx U'˜
J
� 
WD_xz Wv_z W't�� WD_xz 434.3 MPa Ans
WD_xy 0� Ans
WD_yz 0� Ans
Problem 8-44
Solve Prob. 8-43 for points E and F.
Given: b 3.6m� h 1.8m� 
ro 75mm� ri 68mm� 
p 8
kN
m
2
� 
W 7.5kN� d 0.9m� 
Solution: P p b˜ h˜� 
Internal Force and Moment :
6Fx=0; W Nx� 0= Nx W�� 
6Fy=0; Vy P� 0= Vy P� 
6Fz=0; Vz 0� 
60x=0; Mx P 0.5b( )˜� 0= Mx 0.5 P˜ b˜� 
60y=0; My W 0.5b( )˜� 0= My 0.5 W˜ b˜� 
60z=0; Mz P 0.5h d�( )˜� 0= Mz P� 0.5h d�( )˜� 
Nx 7.5� kN Vy 51.84 kN Vz 0 kN 
Mx 93.312 kN m˜ My 13.5 kN m˜ Mz 93.312� kN m˜ 
Section Property : A S ro2 ri2�§© ·¹˜� 
Iy
S
4
ro
4
ri
4�§© ·¹˜� Iz Iy� 
QF_y
4ro
3S
S
2
ro
2˜§¨©
·
¹˜
4ri
3S
S
2
ri
2˜§¨©
·
¹˜�� 
QE_z QF_y� QE_y 0� 
QF_z 0� J S
2
ro
4
ri
4�§© ·¹˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At F: yF 0� zF ro�� 
VF
N
A
Mz yF˜
Iz
�
My zF˜
Iy
�� VF 125.7� MPa (C) Ans
At E: yE ro�� zE 0� 
VE
N
A
Mz yE˜
Iz
�
My zE˜
Iy
�� VE 868.5� MPa (C) Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wt
T U˜
J
=
At F: t 2 ro ri�� �˜� U ro� 
Wv_y
Vy QF_y˜
Iz t˜
� Wt
Mx U˜
J
� 
WF_xy Wv_y Wt�� WF_xy 467.2 MPa Ans
WF_xz 0� Ans
WF_yz 0� Ans
At E: t 2 ro ri�� �˜� U' ro� 
Wv_z
Vz QE_z˜
Iy t˜
� W't
Mx U'˜
J
� 
WE_xz Wv_z W't�� WE_xz 434.3� MPa Ans
WE_xy 0� Ans
WE_yz 0� Ans
Problem 8-45
The bar has a diameter of 40 mm. If it is subjected to the two force components at its end as shown,
determine the state of stress at point A and show the results on a differential volume element located at
this point.
Given: a 100mm� b 150mm� 
do 40mm� Py 0.3kN� 
Pz 0.5� kN� 
Solution:
Internal Force and Moment :
6Fx=0; Nx 0� 
6Fy=0; Vy Py� 0= Vy Py�� 
6Fz=0; Vz Pz� 0= Vz Pz�� 
60x=0; Mx 0� 
60y=0; My Pz b�( )˜� 0= My Pz� b˜� 
60z=0; Mz Py b�( )˜� 0= Mz Py b˜� 
Nx 0 kN Vy 0.3� kN Vz 0.5 kN 
Mx 0 kN m˜ My 0.075 kN m˜ Mz 0.045 kN m˜ 
Section Property : ro 0.5do� 
A S ro2˜� J
S
2
ro
4˜�Iy
S
64
do
4˜� Iz Iy� 
QA_z 0� QA_y
4ro
3S
A
2
§¨
©
·
¹˜� 
Normal Stress: V
Nx
A
Mz y˜
Iz
�
My z˜
Iy
�=
At A: yA 0� zA ro� 
VA
Nx
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 11.9 MPa (T) Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wt
T U˜
J
=
At A: ty do� U ro� 
Wv_y
Vy QA_y˜
Iz ty˜
� Wt
Mx U˜
J
� 
Wv_y 0.318� MPa Wt 0 MPa 
WA_xy Wv_y Wt�� WA_xy 0.318� MPa Ans
WA_xz 0� Ans
WA_yz 0� Ans
Problem 8-46
Solve Prob. 8-45 for point B.
Given: a 100mm� b 150mm� 
do 40mm� Py 0.3kN� 
Pz 0.5� kN� 
Solution:
Internal Force and Moment :
6Fx=0; Nx 0� 
6Fy=0; Vy Py� 0= Vy Py�� 
6Fz=0; Vz Pz� 0= Vz Pz�� 
60x=0; Mx 0� 
60y=0; My Pz b�( )˜� 0= My Pz� b˜� 
60z=0; Mz Py b�( )˜� 0= Mz Py b˜� 
Nx 0 kN Vy 0.3� kN Vz 0.5 kN 
Mx 0 kN m˜ My 0.075 kN m˜ Mz 0.045 kN m˜ 
Section Property : ro 0.5do� 
A S ro2˜� J
S
2
ro
4˜� 
Iy
S
64
do
4˜� Iz Iy� 
QB_y 0� QB_z
4ro
3S
A
2
§¨
©
·
¹˜� 
Normal Stress: V
Nx
A
Mz y˜
Iz
�
My z˜
Iy
�=
At B: yB ro� zB 0� 
VB
Nx
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 7.16� MPa (C) Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wt
T U˜
J
=
At B: tz do� U ro� 
Wv_z
Vz QB_z˜
Iy tz˜
� Wt
Mx U˜
J
� 
Wv_z 0.531 MPa Wt 0 MPa 
WB_xz Wv_z Wt�� WB_xz 0.531 MPa Ans
WB_xy 0� Ans
WB_yz 0� Ans
Problem 8-47
The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mg
and center of mass at G. If the pipe has an outer diameter of 70 mm and a wall thickness of 10
mm,determine the state of stress acting at point C. Show the results on a differential volume element
located at this point. Neglect the weight of the pipe.
Given: do 70mm� t 10mm� 
hA 75mm� L 3m� 
Mo 3000kg� T 45deg� 
Solution: W Mo g˜� W 29.42 kN 
Equilibrium :
+ 6Fy=0; P W� 0= P W� 
Support Reactions : By symmetry, Ay = By = R
P 2R� 0=
R 0.5� P� R 14.71� kN 
Also, Ax = -Bx and tan T� � Ay
Ax
=
Ax
R
tan T� �� Ax 14.71� kN 
Internal Force and Moment : At Section C-D:
N Ax� N 14.71� kN 
V R 0.5W�� V 0 kN 
M V 0.5 L˜( )˜ Ax hA� �˜�� M 1.103� kN m˜ 
Section Property : di do 2t�� 
A
S
4
do
2
di
2�§© ·¹˜� A 1884.96 mm2 
I
S
64
do
4
di
4�§© ·¹˜� I 871791.96 mm4 
I
Mc
A
Nı r Normal Stress:
cC 0.5do� VC
N
A
M cC˜
I
�� VC 52.10� MPa (C) Ans
Shear Stress : W V Q˜
I t˜=
WC 0� (since V = 0) Ans
Problem 8-48
The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mg
and center of mass at G. If the pipe has an outer diameter of 70 mm and a wall thickness of 10
mm,determine the state of stress acting at point D. Show the results on a differential volume element
located at this point. Neglect the weight of the pipe.
Given: do 70mm� t 10mm� 
hA 75mm� L 3m� 
Mo 3000kg� T 45deg� 
Solution: W Mo g˜� W 29.42 kN 
Equilibrium :
+ 6Fy=0; P W� 0= P W� 
Support Reactions : By symmetry, Ay = By = R
P 2R� 0=
R 0.5� P� R 14.71� kN 
Also, Ax = -Bx and tan T� � Ay
Ax
=
Ax
R
tan T� �� Ax 14.71� kN 
Internal Force and Moment : At Section C-D:
N Ax� N 14.71� kN 
V R 0.5W�� V 0 kN 
M V 0.5 L˜( )˜ Ax hA� �˜�� M 1.103� kN m˜ 
Section Property : di do 2t�� 
A
S
4
do
2
di
2�§© ·¹˜� A 1884.96 mm2 
I
S
64
do
4
di
4�§© ·¹˜� I 871791.96 mm4 
I
Mc
A
Nı r Normal Stress:
cD 0� VD
N
A
M cD˜
I
�� VD 7.80� MPa (C) Ans
Shear Stress : W V Q˜
I t˜=
WD 0� (since V = 0) Ans
Problem 8-49
The sign is subjected to the uniform wind loading. Determine the stress components at points A and B
on the 100-mm-diameter supporting post. Show the results on a volume element located at each of
these points.
Given: do 100mm� a 3m� po 1500Pa� 
bo 2m� ho 1m� 
Solution:
Px po� bo˜ ho˜� Px 3.00� kN 
Internal Force and Moment :
6Fx=0; Vx Px� 0= Vx Px�� 
6Fy=0; Vy 0� 
6Fz=0; Nz 0� 
60x=0; Mx 0� 
60y=0; My Px a 0.5ho�� �˜� 0= My Px� a 0.5ho�� �˜� 
60z=0; Mz Px 0.5bo� �˜� Mz Px 0.5bo� �˜� 
Vx 3 kN Vy 0 kN Nz 0 kN 
Mx 0 kN m˜ My 10.5 kN m˜ Mz 3� kN m˜ 
Section Property : ro 0.5 do˜� 
A S ro2§© ·¹˜� A 7853.98 mm2 
I
S
4
ro
4§© ·¹˜� I 4908738.52 mm4 
J
S
2
ro
4§© ·¹˜� J 9817477.04 mm4 
QA 0� (since A' = 0)
QB
4ro
3S
A
2
§¨
©
·
¹˜� 
At A: xA ro� At B: xB 0� 
UA ro� UB ro� 
Normal Stress:
I
Mc
A
Nı r VA
Nz
A
My xA˜
I
�� VB
Nz
A
My xB˜
I
�� 
VA 107.0 MPa (T) Ans VB 0.0 MPa Ans
Shear Stress :
W T U˜
J
V Q˜
I t˜�= WA
Mz UA˜
J
� (since QA = 0) WB
Mz UB˜
J
Vx QB˜
I do˜
�� 
WA 15.28 MPa Ans WB 14.77 MPa Ans
Problem 8-50
The sign is subjected to the uniform wind loading. Determine the stress components at points C and D
on the 100-mm-diameter supporting post. Show the results on a volume element located at each of
these points.
Given: do 100mm� a 3m� po 1500Pa� 
bo 2m� ho 1m� 
Solution:
Px po� bo˜ ho˜� Px 3.00� kN 
Internal Force and Moment :
6Fx=0; Vx Px� 0= Vx Px�� 
6Fy=0; Vy 0� 
6Fz=0; Nz 0� 
60x=0; Mx 0� 
60y=0; My Px a 0.5ho�� �˜� 0= My Px� a 0.5ho�� �˜� 
60z=0; Mz Px 0.5bo� �˜� Mz Px 0.5bo� �˜� 
Vx 3 kN Vy 0 kN Nz 0 kN 
Mx 0 kN m˜ My 10.5 kN m˜ Mz 3� kN m˜ 
Section Property : ro 0.5 do˜� 
A S ro2§© ·¹˜� A 7853.98 mm2 
I
S
4
ro
4§© ·¹˜� I 4908738.52 mm4 
J
S
2
ro
4§© ·¹˜� J 9817477.04 mm4 
QC 0� (since A' = 0)
QD
4ro
3S
A
2
§¨
©
·
¹˜� 
At C: xC ro�� At D: xD 0� 
UC ro� UD ro� 
Normal Stress:
I
Mc
A
Nı r VC
Nz
A
My xC˜
I
�� VD
Nz
A
My xD˜
I
�� 
VC 107.0� MPa (C) Ans VD 0 MPa Ans
Shear Stress :
W T U˜
J
V Q˜
I t˜�= WC
Mz UC˜
J
� (since QC = 0) WD
Mz� UD˜
J
Vx QD˜
I do˜
�� 
WC 15.28 MPa Ans WD 15.79 MPa Ans
Problem 8-51
The 18-mm-diameter shaft is subjected to the loading shown. Determine the stress components at
point A. Sketch the results on a volume element located at this point. The journal bearing at C can exert
only force components Cy and Cz on the shaft, and the thrust bearing at D can exert force components
Dx, Dy, and Dz on the shaft.
Given: do 18mm� L 500mm� 
ax 50mm� ay 200mm� 
P 600N� a 250mm� 
Solution: ro 0.5do� 
Support Reactions at C :
Tx 0� My 0� Mz 0� 
Cx 0� Cy 0� Cz P� 
Internal Force and Moment at A :
N 0� Vy 0� Vz Cz� 
Tx 0� My Cz� a˜� Mz 0� 
Section Property :
Iy
S
4
ro
4˜� Iz Iy� A S ro2˜� 
QA 0� J
S
2
ro
4˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At A: yA 0� zA ro� 
VA
N
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 262.0� MPa (C) Ans
Shear Stress : W V Q˜
I b˜=
At A: b 0� WA
Vz QA˜
Iy b˜
� WA 0.00 Ans
Problem 8-52
Solve Prob. 8-51 for the stress components at point B.
Given: do 18mm� L 500mm� 
ax 50mm� ay 200mm� 
P 600N� a 250mm� 
Solution: ro 0.5do� 
Support Reactions at C :
Tx 0� My 0� Mz 0� 
Cx 0� Cy 0� Cz P� 
Internal Force and Moment at B :
N 0� Vy 0� Vz Cz� 
Tx 0� My Cz� a˜� Mz 0� 
Section Property :
Iy
S
4
ro
4˜� Iz Iy� A S ro2˜� 
J
S
2
ro
4˜� 
QB
4ro
3S
S
2
ro
2˜§¨©
·
¹˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At B: yB ro� zB 0� 
VA
N
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VA 0.0 MPa Ans
Shear Stress : W V Q˜
I b˜=
At B: b 2 ro˜� WB
Vz QB˜
Iy b˜
� WB 3.14 MPa AnsProblem 8-53
The solid rod is subjected to the loading shown. Determine the state of stress developed in the material
at point A, and show the results on a differential volume element at this point.
Given: ro 30mm� a 150mm� 
Px 10� kN� Py 1kN� 
Pz 15kN� Tx 0.2kN m˜� 
Solution:
Internal Force and Moment : At Section A
6Fx=0; Nx Px� 0= Nx Px�� 
6Fy=0; Vy 0� 
Nx 10 kN 6Fz=0; Vz 0� 
Vy 0 kN 60x=0; Mx Tx� 0= Mx Tx�� 
Vz 0 kN 60y=0; My 0� 
Mx 0.2� kN m˜ 60z=0; Mz Px ro�� �˜� 0= Mz Px� ro˜� 
My 0 kN m˜ 
Section Property : U ro� Mz 0.3 kN m˜ 
A S ro2§© ·¹˜� A 2827.43 mm2 
I
S
4
ro
4§© ·¹˜� I 636172.51 mm4 
J
S
2
ro
4§© ·¹˜� J 1272345.02 mm4 
QA_z
4ro
3S
A
2
§¨
©
·
¹˜� QA_y 0� (since A' = 0)
y
y
z
zx
I
zM
I
yM
A
Nı �� Normal Stress:
yA ro�� zA 0� VA
Nx
A
Mz yA˜
I
�
My zA˜
I
�� VA 17.7 MPa (T) Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wv_z 0� (since Vz= 0)
Wt
T U˜
J
= Wt
Mx U˜
J
� Wt 4.72� MPa 
Wxz Wv_z Wt�� Wxz 4.716 MPa Ans
Wxy 0� Ans
Wyz 0� Ans
Problem 8-54
The solid rod is subjected to the loading shown. Determine the state of stress at point B, and show the
results on a differential volume element at this point.
Given: ro 30mm� a 150mm� 
Px 10� kN� Py 10kN� 
Pz 15kN� Tx 0.2kN m˜� 
Solution:
Internal Force and Moment : At Section B
6Fx=0; Nx Px� 0= Nx Px�� 
6Fy=0; Vy Py� 0= Vy Py�� 
6Fz=0; Vz 0� Nx 10 kN 
60x=0; Mx Py ro˜� Tx� 0= Mx Py ro˜ Tx�� Vy 10� kN 
60y=0; My 0� Vz 0 kN 
60z=0; Mz Px ro�� �˜� Py a�( )˜� 0= Mz Px� ro˜ Py a˜�� Mx 0.1 kN m˜ 
My 0 kN m˜ Section Property : U ro� 
Mz 1.8 kN m˜ 
A S ro2§© ·¹˜� A 2827.43 mm2 
I
S
4
ro
4§© ·¹˜� I 636172.51 mm4 
J
S
2
ro
4§© ·¹˜� J 1272345.02 mm4 
QB_z
4ro
3S
A
2
§¨
©
·
¹˜� QB_y 0� (since A' = 0)
y
y
z
zx
I
zM
I
yM
A
Nı �� Normal Stress:
yB ro� zB 0� VB
Nx
A
Mz yB˜
I
�
My zB˜
I
�� VB 81.3� MPa (C) Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wv_z 0� (since Vz= 0)
Wt
T U˜
J
= Wt
Mx U˜
J
� Wt 2.36 MPa 
Wxz Wv_z Wt�� Wxz 2.358 MPa Ans
Wxy 0� Ans
Wyz 0� Ans
Problem 8-55
The solid rod is subjected to the loading shown. Determine the state of stress at point C, and show the
results on a differential volume element at this point.
Given: ro 30mm� a 150mm� 
Px 10� kN� Py 10kN� 
Pz 15kN� Tx 0.2kN m˜� 
Solution:
Internal Force and Moment : At Section A
6Fx=0; Nx Px� 0= Nx Px�� 
6Fy=0; Vy Py� 0= Vy Py�� 
6Fz=0; Vz Pz� 0= Vz Pz�� Nx 10 kN 60x=0; Mx Py ro˜� Pz ro˜� Tx� 0= Mx Py ro˜ Pz ro˜� Tx�� Vy 10� kN 60y=0; My Pz a�( )˜� 0= My Pz� a˜� Vz 15� kN 60z=0; Mz Px ro�� �˜� Py 3a�( )˜� 0= Mz Px� ro˜ 3Py a˜�� Mx 0.35� kN m˜ 
Section Property : U ro� My 2.25� kN m˜ 
Mz 4.8 kN m˜ A S ro2§© ·¹˜� A 2827.43 mm2 
I
S
4
ro
4§© ·¹˜� I 636172.51 mm4 
J
S
2
ro
4§© ·¹˜� J 1272345.02 mm4 
QC_y
4ro
3S
A
2
§¨
©
·
¹˜� QC_z 0� (since A' = 0)
y
y
z
zx
I
zM
I
yM
A
Nı �� Normal Stress:
yC 0� zC ro� VC
Nx
A
Mz yC˜
I
�
My zC˜
I
�� VC 102.6� MPa (C) Ans
Shear Stress :
The transverse shear stress in the z and y directions and the torsional shear stress can be
obtained using the shear formula for Wv and the torsional formula for Wt respectively.
Wv
V Q˜
I t˜= Wv_y
Vy QC_y˜
I 2ro� �˜� Wv_y 4.72� MPa 
Wt
T U˜
J
= Wt
Mx U˜
J
� Wt 8.25� MPa 
Wxy Wv_y Wt�� Wxy 3.54 MPa Ans
Wxz 0� Ans
Wyz 0� Ans
Problem 8-56
The 25-mm-diameter rod is subjected to the loads shown. Determine the state of stress at point A, and
show the results on a differential element located at this point.
Given: ax 200� mm� az 75mm� do 25mm� 
Px 375� N� Py 400� N� Pz 500N� 
Solution: ro 0.5do� 
Internal Force and Moment :
6Fx=0; Nx Px� 0= Nx Px�� 
6Fy=0; Vy Py� 0= Vy Py�� 
6Fz=0; Vz Pz� 0= Vz Pz�� 
60x=0; Tx Py az� �˜� 0= Tx Py az� �˜� 
60y=0; My Pz ax� �˜� Px az� �˜� 0= My Pz ax� �˜ Px az� �˜�� 
60z=0; Mz Py ax� �˜� 0= Mz Py� ax� �˜� 
Section Property : A S ro2˜� 
Iy
S
4
ro
4§© ·¹˜� Iz Iy� 
QA_y 0� QA_z
4ro
3S
S
2
ro
2˜§¨©
·
¹˜� 
J
S
2
ro
4§© ·¹˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At A: yA ro� zA 0� 
VA
Nx
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 52.9 MPa (T) Ans
Shear Stress : Wv
V Q˜
I b˜= Wt
T U˜
J
=
At A: bz 2 ro˜� U ro� 
Wv_z
Vz QA_z˜
Iy bz˜
� Wt
Tx U˜
J
� 
WA_xz Wv_z Wt�� WA_xz 11.14 MPa Ans
WA_xy 0� Ans
Problem 8-57
The 25-mm-diameter rod is subjected to the loads shown. Determine the state of stress at point B, and
show the results on a differential element located at this point.
Given: ax 200� mm� az 75mm� do 25mm� 
Px 375� N� Py 400� N� Pz 500N� 
Solution: ro 0.5do� 
Internal Force and Moment :
6Fx=0; Nx Px� 0= Nx Px�� 
6Fy=0; Vy Py� 0= Vy Py�� 
6Fz=0; Vz Pz� 0= Vz Pz�� 
60x=0; Tx Py az� �˜� 0= Tx Py az� �˜� 
60y=0; My Pz ax� �˜� Px az� �˜� 0= My Pz ax� �˜ Px az� �˜�� 
60z=0; Mz Py ax� �˜� 0= Mz Py� ax� �˜� 
Section Property : A S ro2˜� 
Iy
S
4
ro
4§© ·¹˜� Iz Iy� 
QB_z 0� QB_y
4ro
3S
S
2
ro
2˜§¨©
·
¹˜� 
J
S
2
ro
4§© ·¹˜� 
Normal Stress: V N
A
Mz y˜
Iz
�
My z˜
Iy
�=
At B: yB 0� zB ro� 
VB
Nx
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 46.1� MPa (C) Ans
Shear Stress : Wv
V Q˜
I b˜= Wt
T U˜
J
=
At B: by 2 ro˜� U ro� 
Wv_y
Vy QB_y˜
Iz by˜
� Wt
Tx U˜
J
� 
WB_xy Wv_y Wt�� WB_xy 10.86 MPa Ans
WB_xz 0� Ans
Problem 8-58
The crane boom is subjected to the load of 2.5 kN. Determine the state of stress at points A and B.
Show the results on a differential volume element located at each of these points.
Given: b 75mm� d 76mm� t 12mm� 
ah 1.5m� av 2.4m� P 2.5kN� 
v 4� h 3� r 5� 
Solution:
Internal Force and Moment :
+ 6Fx=0; V� P h
r
§¨
©
·
¹˜� 0= V P
h
r
§¨
©
·
¹˜� 
+ 6Fy=0; N� P v
r
§¨
©
·
¹˜� 0= N P�
v
r
§¨
©
·
¹˜� 
+ 60O=0; M P h
r
§¨
©
·
¹˜ av˜� P
v
r
§¨
©
·
¹˜ ah˜� 0=
M P
v
r
§¨
©
·
¹˜ ah˜ P
h
r
§¨
©
·
¹˜ av˜�� 
Section Property : D d 2t�� 
A b D˜ t d˜�� I 1
12
b˜ D3˜ 1
12
b t�( )˜ d3˜�� 
QA 0� QB 0� (since A' = 0)
Normal Stress: V N
A
M c˜
I
�=
cA 0.5D� VA
N
A
M cA˜
I
�� VA 83.34 MPa (T) Ans
cB 0.5� D� VB
N
A
M cB˜
I
�� VB 83.95� MPa (C) Ans
Shear Stress : W V Q˜
I b˜=
WA
V QA˜
I b˜� WA 0 MPa Ans
WB
V QB˜
I b˜� WB 0 MPa Ans
Problem 8-59
The masonry pier is subjected to the 800-kN load. Determine the equation of the line y = f (x) along
which the load can be placed without causing a tensile stress in the pier. Neglect the weight of the pier.
Given: a 1.5m� b 2.25m� Pz 800� kN� 
xA a�� yA b�� 
Solution:
Section Property :
A 2a( ) 2 b˜( )˜� A 13.5 m2 
Ix
1
12
2a( )˜ 2 b˜( )3� Ix 22.78125 m4 
Iy
1
12
2b( )˜ 2 a˜( )3� Iy 10.125 m4 
Force and Moment :
Mx Pz y˜=
My Pz� x˜=
Normal Stress: Require VA 0� 
VA
Pz
A
Mx yA˜
Ix
�
My xA˜
Iy
�=
0
Pz
A
Pz y˜� � yA˜
Ix
�
Pz x˜� � xA˜
Iy
�=
0
1
A
yA
Ix
§¨
©
·
¹
y˜�
xA
Iy
§¨
©
·
¹
x˜�=
0
1
4a b˜
3
4a b
2˜
§¨
©
·
¹
y˜� 3
4b a
2˜
§¨
©
·
¹
x˜�=
y
b
3
b
a
x˜�=
y 0.75 1.5 x˜�= Ans
Problem 8-60
The masonry pier is subjected to the 800-kN load. If x= 0.25 m and y = 0.5 m, determine the normal
stress at each corner A, B, C, D (not shown) and plot the stress distribution over the cross section.
Neglect the weight of the pier.
Unit Used: kPa 10
3
Pa� 
Given: a 1.5m� b 2.25m� Pz 800� kN� 
xA a�� xB a� xC a� xD a�� 
yA b�� yB b�� yC b� yD b� 
x 0.25m� y 0.5m� 
Solution:
Section Property :
A 2a( ) 2 b˜( )˜� A 13.5 m2 
Ix
1
12
2a( )˜ 2 b˜( )3� Ix 22.78125 m4 
Iy
1
12
2b( )˜ 2 a˜( )3� Iy 10.125 m4 
Force and Moment :
Mx Pz y˜� Mx 400� kN m˜ 
My Pz� x˜� My 200 kN m˜ 
Normal Stress: V
Pz
A
Mx y˜
Ix
�
My x˜
Iy
�=
VA
Pz
A
Mx yA˜
Ix
�
My xA˜
Iy
�� VA 9.877 kPa (T) Ans
VB
Pz
A
Mx yB˜
Ix
�
My xB˜
Iy
�� VB 49.38� kPa (C) Ans
VC
Pz
A
Mx yC˜
Ix
�
My xC˜
Iy
�� VC 128.4� kPa (C) Ans
VD
Pz
A
Mx yD˜
Ix
�
My xD˜
Iy
�� VD 69.1� kPa (C) Ans
Problem 8-61
The symmetrically loaded spreader bar is used to lift the 10-kN (~1-tonne) tank. Determine the state of
stress at points A and B, and indicate the results on a differential volume elements.
Given: b 25mm� d 50mm� a 0.45m� 
L 1.2m� W 10kN� T 30deg� 
Solution:
Support Reactions :
+ 6Fy=0; W� 2F cos T� �˜� 0= F W
2 cos T� �˜� 
Internal Force and Moment :
+ 6Fx=0; F sin T� �˜ N� 0= N F sin T� �˜� 
+ 6Fy=0; V F cos T� �˜� 0= V F cos T� �˜� 
+ 60B=0; M F cos T� �˜ a˜� 0= M F cos T� �˜ a˜� 
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
QB 0.5 d˜ b˜( ) 0.25d( )˜� 
QA 0� (since A' = 0)
Normal Stress: V N
A
M c˜
I
�=
cA 0.5d� VA
N
A
M cA˜
I
�� VA 218.31 MPa (T) Ans
cB 0� VB
N
A
M cB˜
I
�� VB 2.31 MPa (T) Ans
Shear Stress : W V Q˜
I b˜=
WA
V QA˜
I b˜� WA 0 MPa Ans
WB
V QB˜
I b˜� WB 6.00 MPa Ans
Problem 8-62
A post having the dimensions shown is subjected to the bearing load P. Specify the region to which
this load can be applied without causing tensile stress to be developed at points A, B, C, and D.
Problem 8-63
The man has a mass of 100 kg and center of mass at G. If he holds himself in the position shown,
determine the maximum tensile and compressive stress developed in the curved bar at section a-a. He
is supported uniformly by two bars, each having a diameter of 25 mm. Assume the floor is smooth.
Given: Ri 150mm� do 25mm� mo 100kg� 
e 0.3m� a 0.35m� b 1m� 
Solution:
Equilibrium: For the man.
+ 60toe=0; mo g˜� � b˜ 2P a b�( )˜� 0=
P mo g˜� � b2 a b�( )˜� 
P 0.3632 kN 
Section Property : ro 0.5do� rc Ri ro�� 
A S ro2˜� A 490.87 mm2 
³ AA_r rdAȈI IA_r 2S rc rc2 ro2��§© ·¹˜� 
IA_r 3.0252 mm 
R
A
IA_r
� R 162.259 mm 
Internal Force and Moment :
As shown on BFBD. The internal moment must be
computed about the neutral axis. M is negative since it
tends to decrease the bar's radius of curvature.
N P�� N 0.3632� kN 
M P� R e�( )˜� M 0.16790� kN m˜ 
Maximum Normal Stress:
For tensile stress, Vt
N
A
M R r2�� �˜
A r1˜ rc R�� �˜�=
r2 rc ro�� Vt
N
A
M R r2�� �˜
A r2˜ rc R�� �˜�� Vt 102.7 MPa (T) Ans
For compressive stress, Vc
N
A
M R r1�� �˜
A r1˜ rc R�� �˜�=
r1 rc ro�� Vc
N
A
M R r1�� �˜
A r1˜ rc R�� �˜�� Vc 116.9� MPa (C) Ans
Problem 8-64
The block is subjected to the three axial loads shown. Determine the normal stress developed at points
A and B. Neglect the weight of the block.
Given: b 100mm� b' 50mm� 
d 75mm� d' 125mm� 
P1 500N� P2 1250N� P3 250N� 
Solution: B b 2b'�� D d 2d'�� 
Internal Force and Moment :
+ 6Fz=0; N� P1� P2� P3� 0=
N P1 P2� P3�� ��� 
+ 60x=0; Mx P1 0.5d( )˜� P2 0.5d( )˜� P3 0.5D( )˜� 0=
Mx P1 0.5 d˜( )˜ P2 0.5 d˜( )˜�ª¬ º¼ P3 0.5 D˜( )˜�� 
+ 60y=0; My P1 0.5B( )˜� P2 0.5B( )˜� P3 0.5 b˜( )˜� 0=
My P1� 0.5 B˜( )˜ P2 0.5 B˜( )˜� P3 0.5 b˜( )˜�� 
Section Property :
A B D˜ 4b' d'˜�� Ix
1
12
b˜ D3˜ 2
12
b'˜ d3˜�� 
Iy
1
12
d˜ B3˜ 2
12
d'˜ b3˜�� 
Normal Stress: V N
A
Mx y˜
Ix
�
My x˜
Iy
�=
At A: xA 0.5B� yA 0.5� d� 
VA
N
A
Mx yA˜
Ix
�
My xA˜
Iy
�� VA 0.1703� MPa (C) Ans
At B: xB 0.5b� yB 0.5� D� 
VB
N
A
Mx yB˜
Ix
�
My xB˜
Iy
�� VB 0.0977� MPa (C) Ans
Problem 8-65
If P = 15 kN, plot the distribution of stress acting over the cross section a-a of the offset link.
Given: ho 50mm� P 15kN� 
to 10mm� a 30mm� 
Solution:
Section Property :
A ho to˜� A 500 mm2 
I
1
12
to˜ ho3� I 104166.67 mm4 
Moment :
M P a 0.5ho�� �˜� M 0.825 kN m˜ 
Normal Stress:
I
Mc
A
Nı r VA
P
A
M 0.5ho� �˜
I
�� VA 228 MPa (T) Ans
VB
P
A
M 0.5ho� �˜
I
�� VB 168� MPa (C) Ans
y
VA
ho y�
VB
= y ho
VA
VA VB�
˜� 
y 28.79 mm 
Problem 8-66
Determine the magnitude of the load P that will cause a maximum normal stress Vmax = 200 MPa of in
the link at section a-a.
Given: ho 50mm� a 30mm� 
to 10mm� Vallow 200MPa� 
Solution:
Section Property :
A ho to˜� A 500 mm2 
I
1
12
to˜ ho3� I 104166.67 mm4 
Moment :
M P a 0.5ho�� �˜=
Normal Stress: The maximum normal stress occurs at A.
I
Mc
A
Nı r VA
P
A
M 0.5ho� �˜
I
�=
Vallow
P
A
P a 0.5ho�� �˜ 0.5ho� �˜
I
�=
P Vallow A˜� � II A a 0.5ho�� �˜ 0.5ho� �˜�˜� 
P 13.16 kN Ans
Problem 8-67
Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a
radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall
of the cylinder.
Given: t 2mm� P 2kN� 
ri 45mm� 
Solution:
A Sri2� p
P
A
� 
p 0.3144 MPa 
Hoop Stress : D
ri
t
� D 22.50 
Since D > 10. then thin-wall analysis can be used.
V1
p ri˜
t
� V1 7.07 MPa Ans
Longitudinal Stress :
V2 0� Ans
The pressure p is supported by the surface of the
pistons in the longitudinal direction.
Problem 8-68
Determine the maximum force P that can be exerted on each of the two pistons so that the
circumferential stress component in the cylinder does not exceed 3 MPa. Each piston has a radius of
45 mm and the cylinder has a wall thickness of 2 mm.
Given: t 2mm� Vallow 3MPa� 
ri 45mm� 
Solution:
A Sri2� p
P
A
=
Hoop Stress : D
ri
t
� D 22.50 
Since D > 10. then thin-wall analysis can be used.
V1
p ri˜
t
= Vallow
P ri˜
A t˜=
P
Vallow
ri
A˜ t˜� 
P 0.848 kN Ans
Problem 8-69
The screw of the clamp exerts a compressive force of 2.5 kN on the wood blocks. Determine the
maximum normal stress developed along section a-a. The cross section there is rectangular, 18 mm by
12 mm.
Given: b 12mm� d 18mm� 
a 100mm� P 2.5kN� 
Solution:
Internal Force and Moment :
N P� 
M P a˜� 
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
Normal Stress: V N
A
M c˜
I
�=
cmax 0.5d� Vmax
N
A
M cmax˜
I
�� 
Vmax 397.4 MPa (T) Ans
Problem 8-70
The wall hanger has a thickness of 6 mm and is used to support the vertical reactions of the beam that
is loaded as shown. If the load is transferred uniformly to each strap of the hanger, determine the state
of stress at points C and D of the strap at B. Assume the vertical reaction F at this end acts in the
center and on the edge of the bracket as shown.
Given: t 6mm� d 50mm� P 50kN� 
L1 1.2m� L2 1.8m� w 30
kN
m
� 
Solution: L L1 L2�� 
 Support Reactions : Given
+ 6Fy=0; FA P� w L2˜� FB� 0=
 60B=0; FA L˜ P L 0.5L1�� �˜� w L2˜ 0.5L2� �˜� 0=
Guess FA 1kN� FB 1kN� 
FA
FB
§¨
©¨
·
¹
Find FA FB�� �� 
FA
FB
§¨
©¨
·
¹
56.20
47.80
§¨
©
·
¹ kN 
Section Property :
A 2t d˜� I 2
12
t˜ d3˜� 
At Section CD: P FB� M P 0.5d( )˜� V 0� 
Stresses: V P
A
M y˜
I
�= W V Q˜
I b˜=
At C: yC 0� VC
P
A
M yC˜
I
�� 
VC 79.67 MPa (T) Ans
WC 0� Ans
At D:yD 0.5� d� 
VD
P
A
M yD˜
I
�� 
VD 159.33� MPa (C) Ans
WD 0� Ans
Problem 8-71
The support is subjected to the compressive load P. Determine the absolute maximum and minimum
normal stress acting in the material.
Problem 8-72
The support has a circular cross section with a radius that increases linearly with depth. If it is
subjected to the compressive load P, determine the maximum and minimum normal stress acting in the
material.
Problem 8-73
The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter
of 1.5 m and a wall thickness of 18 mm. If the largest normal stress is not to exceed 150 MPa,
determine the maximum pressure the tank can sustain. Also, compute the number of bolts required to
attach the cap to the tank if each bolt has a diameter of 20 mm. The allowable stress for the bolts is
(Vallow)b = 180 MPa.
Given: t 18mm� di 1.5m� Vallow 150MPa� 
db 20mm� Vb.allow 180MPa� 
Solution: ri 0.5di� 
Hoop Stress : D
ri
t
� D 41.67 
Since D > 10. then thin-wall analysis can be used.
V1
p ri˜
t
= Vallow
p ri˜
t
=
p
Vallow t˜
ri
� 
p 3.6 MPa Ans
Force Equilibrium for the Cap : A Sri2� 
+ 6Fy=0; p A˜ Fb� 0=
Fb p A˜� 
Fb 6361.73 kN 
Allowable Normal Stress for Bolts : Ab
S
4
db
2� 
Vb.allow
Fb
n Ab˜
=
n
Fb
Vb.allow� � Ab˜� 
n 112.5 
Use n 113� Ans
Problem 8-74
The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameter
of 1.5 m and a wall thickness of 18 mm. If the pressure in the tank is p = 1.20 MPa, determine the
force in the 16 bolts that are used to attach the cap to the tank. Also, specify the state of stress in the
wall of the tank.
Given: t 18mm� di 1.5m� p 1.20MPa� 
n 16� 
Solution: ri 0.5di� 
Hoop Stress : D
ri
t
� D 41.67 
Since D > 10. then thin-wall analysis can be used.
V1
p ri˜
t
� 
V1 50 MPa Ans
Longitudinal Stress :
V2
p ri˜
2 t˜� 
V2 25 MPa Ans
Force Equilibrium for the Cap : A Sri2� 
+ 6Fy=0; p A˜ 16 Fb˜� 0=
Fb
p A˜
16
� 
Fb 132.5 kN Ans
Problem 8-75
The crowbar is used to pull out the nail at A. If a force of 40 N is required, determine the stress
components in the bar at points D and E. Show the results on a differential volume element located at
each of these points. The bar has a circular cross section with a diameter of 12 mm. No slipping
occurs at B.
Given: do 12mm� dA 60mm� hA 75mm� 
a 125mm� dP 300mm� 
F 40N� hP 300mm� 
Solution: rp dP
2
hP
2�� 
Support Reactions :
+ 60B=0; F hA˜ P rp� �˜� 0=
P
F hA˜
rp
� 
Internal Force and Moment :
+ 6Fx=0; N 0� 
+ 6Fy=0; V P� 0= V P� 
+ 60O=0; M P a˜� 0= M P a˜� 
Section Property : ro 0.5do� 
A S ro2˜� I
S
4
ro
4˜� 
QE
4ro
3S 0.5S ro
2˜§© ·¹˜� 
QD 0� (since A' = 0)
Normal Stress: V N
A
M c˜
I
�=
cD ro� VD
N
A
M cD˜
I
�� VD 5.21 MPa (T) Ans
cE 0� VE
N
A
M cE˜
I
�� VE 0.00 MPa Ans
Shear Stress : W V Q˜
I b˜=
bE 2 ro˜� WE
V QE˜
I bE˜
� WE 0.0834 MPa Ans
WD 0� Ans
Problem 8-76
The screw of the clamp exerts a compressive force of 2.5 kN on the wood blocks. Sketch the stress
distribution along section a-a of the clamp. The cross section there is rectangular, 18 mm by 12 mm.
Given: b 12mm� d 18mm� 
a 100mm� P 2.5kN� 
Solution:
Internal Force and Moment :
N P� 
M P a˜� 
Section Property :
A b d˜� I 1
12
b˜ d3˜� 
Normal Stress: V N
A
M c˜
I
�=
cmax 0.5d� Vmax
N
A
M cmax˜
I
�� 
Vmax 397.4 MPa (T) Ans
cmin 0.5d� Vmin
N
A
M cmin˜
I
�� 
Vmin 374.2� MPa (C) Ans
y
d y�
Vmin
Vmax
=
y
d Vmin
Vmin Vmax�
� 
y 8.73 mm 
Problem 8-77
The clamp is made from members AB and AC, which are pin connected at A. If the compressive force
at C and B is 180 N, determine the state of stress at point F, and indicate the results on a differential
volume element. The screw DE is subjected only to a tensile force along its axis.
Given: h 15mm� t 15mm� P 180N� 
a 30mm� b 40mm� 
Solution:
Support Reactions :
+ 60O=0; P b a�( )˜ FDE a( )˜� 0=
FDE
b a�
a
P˜� 
FDE 0.420 kN 
Internal Force and Moment :
+ 6Fy=0; N' 0� 
+ 6Fx=0; V FDE� P� 0= V P FDE�� 
+ 60O=0; M P b 0.5a�( )˜� FDE 0.5a( )˜� 0= M FDE 0.5a( )˜ P b 0.5a�( )˜�� 
M 3.60� N m˜ 
Section Property:
A h t˜� A 225 mm2 
I
1
12
t˜ h3˜� I 4218.75 mm4 
QF 0� (since A' = 0)
Normal Stress: V N'
A
M c˜
I
�=
cF 0.5 h˜� VF
N'
A
M cF˜
I
�� 
VF 6.40� MPa (C) Ans
Shear Stress : W V Q˜
I b˜=
WF
V QF˜
I t˜� WF 0 MPa Ans
Problem 8-78
The eye is subjected to the force of 250 N. Determine the maximum tensile and compressive stresses
at section a-a. The cross section is circular and has a diameter of 6 mm. Use the curved-beam formula
to compute the bending stress.
Given: Ri 30mm� do 6mm� P 0.250kN� 
Solution: ro 0.5do� rc Ri ro�� 
Section Property :
A S ro2˜� A 28.27 mm2 
³ AA_r rdAȈI IA_r 2S rc rc2 ro2��§© ·¹˜� 
IA_r 0.8586 mm 
R
A
IA_r
� R 32.932 mm 
Internal Force and Moment :
As shown on BFBD. The internal moment must be
computed about the neutral axis. M is positive since it
tends to increase the beam's radius of curvature.
N P� 
M P R˜� 
Maximum Normal Stress:
For tensile stress, Vt
N
A
M R r1�� �˜
A r1˜ rc R�� �˜�=
r1 rc ro�� Vt
N
A
M R r1�� �˜
A r1˜ rc R�� �˜�� 
Vt 425.3 MPa (T) Ans
For compressive stress, Vc
N
A
M R r2�� �˜
A r1˜ rc R�� �˜�=
r2 rc ro�� Vc
N
A
M R r2�� �˜
A r2˜ rc R�� �˜�� 
Vc 354.4� MPa (C) Ans
Problem 8-79
Solve Prob. 8-78 if the cross section is square, having dimensions of 6 mm by 6 mm.
Given: Ri 30mm� do 6mm� P 0.250kN� 
Solution: ro 0.5do� rc Ri ro�� 
Section Property :
A do
2� A 36.00 mm2 
³ AA_r rdAȈI IA_r do� � ln
rc ro�
rc ro�
§¨
©
·
¹
˜� 
IA_r 1.0939 mm 
R
A
IA_r
� 
R 32.91 mm 
Internal Force and Moment :
As shown on BFBD. The internal moment must be
computed about the neutral axis. M is positive since it
tends to increase the beam's radius of curvature.
N P� 
M P R˜� 
Maximum Normal Stress:
For tensile stress, Vt
N
A
M R r1�� �˜
A r1˜ rc R�� �˜�=
r1 rc ro�� Vt
N
A
M R r1�� �˜
A r1˜ rc R�� �˜�� 
Vt 250.2 MPa (T) Ans
For compressive stress, Vc
N
A
M R r2�� �˜
A r1˜ rc R�� �˜�=
r2 rc ro�� Vc
N
A
M R r2�� �˜
A r2˜ rc R�� �˜�� 
Vc 208.4� MPa (C) Ans
Problem 9-1
Prove that the sum of the normal stresses Vx + Vy = Vx' + Vy' is constant. See Figs. 9-2a and 9-2b.
Solution:
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�= (1)
Vy'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�= (2)
(1) + (2) :
LHS Vx' Vy'�=
RHS
Vx Vy�
2
Vx Vy�
2
�= RHS Vx Vy�=
Hence,
Vx' Vy'� Vx Vy�= (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: Vx 3MPa� Vy 5MPa� Wxy 8� MPa� I 40deg� 
Solution: Set 'A m2� T 180deg I�� 
Force Equilibrium: For the sectioned element,
'Ax 'A cos I� �˜� 'Ay 'A sin I� �˜� 
Fxx Vx 'Ax˜� Fxy Wxy 'Ax˜� 
Fyy Vy 'Ay˜� Fyx Wxy 'Ay˜� 
Given
+ 6Fx'=0; 'Fx' Fxy sin T� �˜� Fxx cos T� �˜� Fyx cos T� �˜� Fyy sin T� �˜� 0=
+ 6Fy'=0; 'Fy' Fxy cos T� �˜� Fxx sin T� �˜� Fyx sin T� �˜� Fyy cos T� �˜� 0=
Guess 'Fx' 1kN� 'Fy' 1kN� 
'Fx'
'Fy'
§¨
©¨
·
¹
Find 'Fx' 'Fy'�� �� 'Fx''Fy'
§¨
©¨
·
¹
4052.11�
404.38�
§¨
©
·
¹ kN 
Normal and Shear Stress: V
0A
F
A
§¨
©
·
¹limo=
Vx''Fx'
'A
� Vx' 4.052� MPa Ans
Wx'y'
'Fy'
'A
� Wx'y' 0.404� MPa Ans
The negative signs indicate that the sense of Vx' and Wx'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: Vx 0.200� MPa� Vy 0.350MPa� 
I 50deg� Wxy 0MPa� 
Solution: Set 'A m2� T 180deg I�� 
Force Equilibrium: For the sectioned element,
'Ax 'A cos I� �˜� 'Ay 'A sin I� �˜� 
Fxx Vx 'Ax˜� Fxy Wxy 'Ax˜� Fxy 0.00 
Fyy Vy 'Ay˜� Fyx Wxy 'Ay˜� Fyx 0.00 
Given
+ 6Fx'=0; 'Fx' Fxy sin T� �˜� Fxx cos T� �˜� Fyx cos T� �˜� Fyy sin T� �˜� 0=
+ 6Fy'=0; 'Fy' Fxy cos T� �˜� Fxx sin T� �˜� Fyx sin I� �˜� Fyy cos T� �˜� 0=
Guess 'Fx' 1kN� 'Fy' 1kN� 
'Fx'
'Fy'
§¨
©¨
·
¹
Find 'Fx' 'Fy'�� �� 'Fx''Fy'
§¨
©¨
·
¹
122.75
270.82
§¨
©
·
¹ kN 
Normal and Shear Stress: V
0A
F
A
§¨
©
·
¹limo=
Vx'
'Fx'
'A
� Vx' 0.123 MPa Ans
Wx'y'
'Fy'
'A
� Wx'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: Vx 0.650� MPa� Vy 0.400MPa� 
I 60deg� Wxy 0MPa� 
Solution: Set 'A m2� T 90deg I�� ��� 
Force Equilibrium: For the sectioned element,
'Ax 'A sin I� �˜� 'Ay 'A cos I� �˜� 
Fxx Vx 'Ax˜� Fxy Wxy 'Ax˜� Fxy 0.00 
Fyy Vy 'Ay˜� Fyx Wxy 'Ay˜� Fyx 0.00 
Given
 + 6Fx'=0; 'Fx' Fxy sin T� �˜� Fxx cos T� �˜� Fyx cos T� �˜� Fyy sin T� �˜� 0=
+ 6Fy'=0; 'Fy' Fxy cos T� �˜� Fxx sin T� �˜� Fyx sin T� �˜� Fyy cos T� �˜� 0=
Guess 'Fx' 1kN� 'Fy' 1kN� 
'Fx'
'Fy'
§¨
©¨
·
¹
Find 'Fx' 'Fy'�� �� 'Fx''Fy'
§¨
©¨
·
¹
387.50�
454.66
§¨
©
·
¹ kN 
Normal and Shear Stress: V
0A
F
A
§¨
©
·
¹limo=
Vx'
'Fx'
'A
� Vx' 0.387� MPa Ans
Wx'y'
'Fy'
'A
� Wx'y' 0.455 MPa Ans
The negative signs indicate that the sense of Vx' 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: Vx 60� MPa� Vy 50� MPa� 
I 30deg� Wxy 28MPa� 
Solution: Set 'A m2� T 180deg I�� 
Force Equilibrium: For the sectioned element,
'Ax 'A cos I� �˜� 'Ay 'A sin I� �˜� 
Fxx Vx 'Ax˜� Fxy Wxy 'Ax˜� Fxy 24248.71 kN 
Fyy Vy 'Ay˜� Fyx Wxy 'Ay˜� Fyx 14000.00 kN 
Given
+ 6Fx'=0; 'Fx' Fxy sin T� �˜� Fxx cos T� �˜� Fyx cos T� �˜� Fyy sin T� �˜� 0=
 + 6Fy'=0; 'Fy' Fxy cos T� �˜� Fxx sin T� �˜� Fyx sin T� �˜� Fyy cos T� �˜� 0=
Guess 'Fx' 1kN� 'Fy' 1kN� 
'Fx'
'Fy'
§¨
©¨
·
¹
Find 'Fx' 'Fy'�� �� 'Fx''Fy'
§¨
©¨
·
¹
33251.29�
18330.13
§¨
©
·
¹ kN 
Normal and Shear Stress: V
0A
F
A
§¨
©
·
¹limo=
Vx'
'Fx'
'A
� Vx' 33.251� MPa Ans
Wx'y'
'Fy'
'A
� Wx'y' 18.330 MPa Ans
The negative signs indicate that the sense of Vx' 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: Vx 90MPa� Vy 50MPa� 
Wxy 35� MPa� I 60deg� 
Solution: Set 'A m2� T 90deg I�� ��� 
Force Equilibrium: For the sectioned element,
'Ax 'A sin I� �˜� 'Ay 'A cos I� �˜� 
Fxx Vx 'Ax˜� Fxy Wxy 'Ax˜� 
Fyy Vy 'Ay˜� Fyx Wxy 'Ay˜� 
Given
+ 6Fx'=0; 'Fx' Fxy sin T� �˜� Fxx cos T� �˜� Fyx cos T� �˜� Fyy sin T� �˜� 0=
 + 6Fy'=0; 'Fy' Fxy cos T� �˜� Fxx sin T� �˜� Fyx sin T� �˜� Fyy cos T� �˜� 0=
Guess 'Fx' 1kN� 'Fy' 1kN� 
'Fx'
'Fy'
§¨
©¨
·
¹
Find 'Fx' 'Fy'�� �� 'Fx''Fy'
§¨
©¨
·
¹
49689.11
34820.51�
§¨
©
·
¹ kN 
Normal and Shear Stress: V
0A
F
A
§¨
©
·
¹limo=
Vx'
'Fx'
'A
� Vx' 49.69 MPa Ans
Wx'y'
'Fy'
'A
� Wx'y' 34.82� MPa Ans
The negative signs indicate that the sense of Wx'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: Vx 5MPa� Vy 3MPa� Wxy 8MPa� I 40deg� 
Solution: T 90deg I�� 
Normal Stress:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 4.05� MPa Ans
The negative signs indicate that the sense of Vx' is a compressive stress.
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'y' 0.404� MPa Ans
The negative signs indicate that the sense of Wx'y' is in the -y' direction.
Problem 9-8
Solve Prob. 9-4 using the stress-transformation equations developed in Sec. 9.2.
Given: Vx 0.650� MPa� Vy 0.400MPa� 
I 60deg� Wxy 0MPa� 
Solution: T 90deg I�� 
Normal Stress:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 0.387� MPa Ans
The negative signs indicate that the sense of Vx' is a compressive stress.
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'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: Vx 90MPa� Vy 50MPa� 
Wxy 35� MPa� I 60deg� 
Solution: T 90deg I�� ��� 
Normal Stress:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 49.69 MPa Ans
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'y' 34.82� MPa Ans
The negative signs indicate that the sense of Wx'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: Vx 0kPa� Vy 300� kPa� 
T 30deg� Wxy 950kPa� 
Solution:
Normal Stress:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 747.7 kPa Ans
Vy'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vy' 1047.7� kPa Ans
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'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: Vx 0.300MPa� Vy 0MPa� 
T 60� deg� Wxy 0.120MPa� 
Solution:
Normal Stress:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 0.0289� MPa Ans
Vy'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vy' 0.329 MPa Ans
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'y' 0.0699 MPa Ans
Problem 9-12
Solve Prob. 9-6 using the stress-transformation equations.
Given: Vx 90MPa� Vy 50MPa� 
I 60deg� Wxy 35� MPa� 
Solution: T 90deg I�� ��� 
Normal Stress:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 49.69 MPa Ans
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'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: Vx 45MPa� Vy 60� MPa� Wxy 30MPa� 
Solution:
(a) Principal Stress:
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 52.97 MPa 
Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 67.97� MPa 
Ans
Orientation of Principal Stress:
tan 2Tp� � 2WxyVx Vy�= Tp
1
2
atan
2Wxy
Vx Vy�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 14.87 deg T'p 75.13� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2Tp� �˜� Wxy sin 2Tp� �˜�� 
Vx' 52.97 MPa 
Therefore,Tp1 Tp� Tp1 14.87 deg Ans
Tp2 T'p� Tp2 75.13� deg Ans
(b)
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 60.47 MPa Ans
Vavg
Vx Vy�
2
� Vavg 7.50� MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2Ts� � Vx Vy�
2Wxy
�= Ts
1
2
atan
Vx Vy�
2Wxy
�§¨
©
·
¹
� T's Ts 90deg�� 
Ts 30.13� deg T's 59.87 deg 
By observation, in order to preserve equilibrium along AB, Wmax 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: Vx 180MPa� Vy 0MPa� Wxy 150� MPa� 
Solution:
(a) Principal Stress:
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 264.93 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 84.93� MPa Ans
Orientation of Principal Stress:
tan 2Tp� � 2WxyVx Vy�= Tp
1
2
atan
2Wxy
Vx Vy�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 29.52� deg T'p 60.48 deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2Tp� �˜� Wxy sin 2Tp� �˜�� 
Vx' 264.93 MPa 
Therefore, Tp1 Tp� Tp1 29.52� deg Ans
Tp2 T'p� Tp2 60.48 deg Ans
(b)
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 174.93 MPa Ans
Vavg
Vx Vy�
2
� Vavg 90.00 MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2Ts� � Vx Vy�
2Wxy
�= Ts
1
2
atan
Vx Vy�
2Wxy
�§¨
©
·
¹
� T's Ts 90deg�� 
Ts 15.48 deg T's 74.52� deg 
By observation, in order to preserve equilibrium along AB, Wmax 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: Vx 30� MPa� Vy 0MPa� Wxy 12� MPa� 
Solution:
(a) Principal Stress:
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 4.21 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 34.21� MPa Ans
Orientation of Principal Stress:
tan 2Tp� � 2WxyVx Vy�= Tp
1
2
atan
2Wxy
Vx Vy�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 19.33 deg T'p 70.67� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2Tp� �˜� Wxy sin 2Tp� �˜�� 
Vx' 34.21� MPa 
Therefore, Tp1 T'p� Tp1 70.67� deg Ans
Tp2 Tp� Tp2 19.33 deg Ans
(b)
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 19.21 MPa Ans
Vavg
Vx Vy�
2
� Vavg 15.00� MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2Ts� � Vx Vy�
2Wxy
�= Ts
1
2
atan
Vx Vy�
2Wxy
�§¨
©
·
¹
� T's Ts 90deg�� 
Ts 25.67� deg T's 64.33 deg 
By observation, in order to preserve equilibrium along AB, Wmax 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: Vx 200� MPa� Vy 250MPa� Wxy 175MPa� 
Solution:
(a) Principal Stress:
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 310.04 MPa 
Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 260.04� MPa 
Ans
Orientation of Principal Stress:
tan 2Tp� � 2WxyVx Vy�= Tp
1
2
atan
2Wxy
Vx Vy�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 18.94� deg T'p 71.06 deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2Tp� �˜� Wxy sin 2Tp� �˜�� 
Vx' 260.04� MPa 
Therefore, Tp1 T'p� Tp1 71.06 deg Ans
Tp2 Tp� Tp2 18.94� deg Ans
(b)
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 285.04 MPa Ans
Vavg
Vx Vy�
2
� Vavg 25.00 MPa Ans
Orientation of Maximum In-plane Shear Stress:
tan 2Ts� � Vx Vy�
2Wxy
�= Ts
1
2
atan
Vx Vy�
2Wxy
�§¨
©
·
¹
� T's Ts 90deg�� 
Ts 26.06 deg T's 63.94� deg 
By observation, in order to preserve equilibrium along AB, Wmax 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) Vx'_a 200� MPa� Wx'y'_a 0MPa� 
Vy'_a 350� MPa� Ta 30� deg� 
(a) Vx'_b 0� Wx'y'_b 58MPa� 
Vy'_b 0� Tb 25deg� 
Solution:
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 of the text.
For element (a):
Vx_a
Vx'_a Vy'_a�
2
Vx'_a Vy'_a�
2
cos 2Ta� �˜� Wx'y'_a sin 2Ta� �˜�§¨©
·
¹� Vx_a 237.50� MPa 
Vy_a
Vx'_a Vy'_a�
2
Vx'_a Vy'_a�
2
cos 2Ta� �˜� Wx'y'_a sin 2Ta� �˜�§¨©
·
¹� Vy_a 312.50� MPa 
Wxy_a
Vx'_a Vy'_a�
2
� sin 2Ta� �˜ Wx'y'_a cos 2Ta� �˜�§¨©
·
¹� Wxy_a 64.95 MPa 
For element (b):
Vx_b
Vx'_b Vy'_b�
2
Vx'_b Vy'_b�
2
cos 2Tb� �˜� Wx'y'_b sin 2Tb� �˜�§¨©
·
¹� Vx_b 44.43 MPa 
Vy_b
Vx'_b Vy'_b�
2
Vx'_b Vy'_b�
2
cos 2Tb� �˜� Wx'y'_b sin 2Tb� �˜�§¨©
·
¹� Vy_b 44.43� MPa 
Wxy_b
Vx'_b Vy'_b�
2
� sin 2Tb� �˜ Wx'y'_b cos 2Tb� �˜�§¨©
·
¹� Wxy_b 37.28 MPa 
Combining the stress componenets of two elements yields
Vx Vx_a Vx_b�� Vx 193.1� MPa Ans
Vy Vy_a Vy_b�� Vy 356.9� MPa Ans
Wxy Wxy_a Wxy_b�� Wxy 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.
Vx 0MPa� Vy 0MPa� 
Wxy
q
t
� Wxy 0.333 MPa 
In-plane Principal Stress: Apply Eq. 9-5.
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 0.333 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 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.
Vx
qx
t
� Vx 3.00 MPa 
Vy
qy
t
� Vy 4.00 MPa 
Wxy 0� 
Maximum In-plane Shear Stress: Apply Eq. 9-7.
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 0.500 MPa Ans
Average Normal Stress: Apply Eq. 9-8.
Vavg
Vx Vy�
2
� Vavg 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: Va 80MPa� Vb 60MPa� 
T 45deg� E 60deg� 
Solution:
Vx Vb sin E� �˜� 
Wxy Vb cos E� �˜� 
Given
Va
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�=
Wa
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�=
Guess Vy 1MPa� Wa 1MPa� 
Vy
Wa
§¨
©¨
·
¹
Find Vy Wa�� �� VyWa
§¨
©¨
·
¹
48.04
1.96�
§¨
©
·
¹ MPa Ans
Principal Stress:
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 80.06 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 19.94 MPa Ans
Problem 9-21
The stress acting on two planes at a point is indicated. Determine the normal stress Vb and the principal
stresses at the point.
Given: Va 4MPa� Wx'y' 2� MPa� 
Ibb 45deg� E 60deg� 
Solution:
Stress Transformation Equations :
Applying Eqs. 9-3 and 9-1 with T Ibb� 90deg�� 
Vy Va sin E� �˜� Wxy Va cos E� �˜� Vx' Vb=
Given
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxysin 2T� �˜�=
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�=
Guess Vx 1MPa� Vx' 1MPa� 
Vx
Vx'
§¨
©¨
·
¹
Find Vx Vx'�� �� VxVx'
§¨
©¨
·
¹
7.464
7.464
§¨
©
·
¹ MPa 
Vb Vx'� 
Vb 7.464 MPa Ans
In-plane Principal Stress: Applying Eq. 9-5,
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 8.29 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 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 :
+ 60O=0; E' L˜ P a b�( )˜� 0=
E'
P a b�( )˜
L
� E' 24 kN 
Internal Force and Moment : At Section A-B:
+ 6Fx=0; E V� 0= V E'�� 
+ 60O=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 mm3 
QA 0� (since A' = 0)
Normal Stress: V M c˜
I
=
cA 0.5� h� VA
M cA˜
I
� VA 192.00� MPa 
cB 0� VB
M cB˜
I
� VB 0 MPa 
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I t˜� WA 0.00 MPa 
WB
V QB˜
I t˜� WB 24.00� MPa 
In-plane Principal Stress:
At A: VxA VA� VyA 0� Wxy WA� 
Since no shear stress acts upon the element,
VA1 VyA� VA1 0 MPa Ans
VA2 VxA� VA2 192� MPa Ans
At B: VxB VB� VyB 0� Wxy WB� 
2
2
2.1
22
xy
yxyx WVVVVV �¸¸¹
·
¨¨©
§ �r� 
VB1 Wxy�� VB1 24 MPa Ans
VB2 Wxy� VB2 24� MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2Tp� � 2WBVxB VyB��= Tp
1
2
90deg( )� T'p Tp 90deg�� 
Tp 45 deg T'p 45� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'_B
VxB VyB�
2
VxB VyB�
2
cos 2Tp� �˜� WB sin 2Tp� �˜�� 
Vx'_B 24.00� MPa 
Therefore, Tp1 T'p� Tp1 45.00� deg Ans
Tp2 Tp� Tp2 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 :
+ 60E=0; P 2a( )˜ R L˜� 0=
R
2P a˜
L
� R 16 kN 
Internal Force and Moment : At Section C-D:
+ 6Fx=0; R V� 0= V R� 
+ 60O=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 mm3 
QC 0� (since A' = 0)
Normal Stress: V M c˜
I
=
cC 0.5h� VC
M cC˜
I
� VC 256.00 MPa 
cD 0.5� h˜ hD�� VD
M cD˜
I
� VD 153.6� MPa 
Shear Stress : W V Q˜
I t˜=
WC
V QC˜
I t˜� WC 0.00 MPa 
WD
V QD˜
I t˜� WD 10.24 MPa 
In-plane Principal Stress:
At C: VxC 0� VyC VC� Wxy WC� 
Since no shear stress acts upon the element,
VC1 VyC� VC1 256 MPa Ans
VC2 VxC� VC2 0 MPa Ans
At D: VxD 0� VyD VD� Wxy WD� 
VD1
VxD VyD�
2
VxD VyD�
2
§¨
©
·
¹
2
WD2��� VD1 0.680 MPa Ans
VD2
VxD VyD�
2
VxD VyD�
2
§¨
©
·
¹
2
WD2��� VD2 154.28� MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point D,
tan 2Tp� � 2WDVxD VyD�= Tp
1
2
atan
2WD
VxD VyD�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 3.797 deg T'p 86.203� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'_D
VxD VyD�
2
VxD VyD�
2
cos 2Tp� �˜� WD sin 2Tp� �˜�� 
Vx'_D 0.680 MPa 
Therefore, Tp1 Tp� Tp1 3.80 deg Ans
Tp2 T'p� Tp2 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� I 20deg� 
h 60mm� t 25mm� 
Solution:
Vx
P
h t˜� Vx 0.1667 MPa 
Vy 0� 
Wxy 0� 
T 90deg I�� 
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� Vx' 19.50 kPa Ans
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� Wx'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
Vx = 2.8 MPa, determine the necessary compressive stress Vy that will cause failure.
Given: Vx 2.8MPa� Wxy 0MPa� 
Tgrain 58deg� Wx'y' 3.85MPa� 
Solution: T Tgrain 90deg�� 
Shear Stress:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�=
Wx'y'
Vx Vy�
2
� sin 2T� �˜=
Vy
2Wx'y'
sin 2T� � Vx�� 
Vy 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:
+ 6Fy=0; V w a˜� 0= V w a˜� 
+ 60A=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˜ tf3˜ bf tf˜� � 0.5tf yc�� �2˜�� 
I2
1
12
tw˜ dw3˜ dw tw˜� � 0.5dw tf� yc�� �2˜�� 
I I1 I2�� I 16562500.00 mm4 
QA 0� (since A' = 0)
QB hB tw˜� � D yc� 0.5 hB˜�� �� QB 92500 mm3 
Normal Stress: V M c˜
I
=
cA yc� VA
M cA˜
I
� VA 152.15 MPa 
cB D yc� hB�� ��� VB M cB˜I� VB 195.62� MPa 
Shear Stress : W V Q˜
I t˜= WA
V QA˜
I bf˜
� WA 0.00 MPa 
WB
V QB˜
I tw˜
� WB 6.702 MPa 
In-plane Principal Stress:
At A: VxA VA� VyA 0� Wxy WA� 
Since no shear stress acts upon the element,
VA1 VxA� VA1 152.15 MPa Ans
VA2 VyA� VA2 0 MPa Ans
At B: VxB VB� VyB 0� Wxy WB� 
VB1
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB1 0.229 MPa Ans
VB2
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB2 195.852� MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2Tp� � 2WBVxB VyB��= Tp
1
2
atan
2� WB
VxB VyB�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 1.96 deg T'p 88.04� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'_B
VxB VyB�
2
VxB VyB�
2
cos 2Tp� �˜� WB sin 2Tp� �˜�� 
Vx'_B 194.94� MPa 
Therefore, Tp1 T'p� Tp1 88.04� deg Ans
Tp2 Tp� Tp2 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:
+ 6Fx=0; N P� 0= N P� 
+ 60O=0; M P a( )˜� 0= M P a˜� 
Section Property :
A
S do2˜
4
� A 176.71 mm2 
I
S do4˜
64
� I 2485.05 mm4 
I
Mc
A
N r 2.1VNormal Stress:
cA 0.5do� VA
N
A
M cA˜
I
�� VA 87.15� MPa 
cB 0.5do� VB
N
A
M cB˜
I
�� VB 93.94 MPa 
In-plane Principal Stress:
At A: VxA VA� VyA 0� Wxy 0� 
Since no shear stress acts upon the element,
VA1 VyA� VA1 0 MPa Ans
VA2 VxA� VA2 87.15� MPa Ans
At B: VxB VB� VyB 0� Wxy 0� 
Since no shear stress acts upon the element,
VB1 VxB� VB1 93.94 MPa Ans
VB2 VyB� VB2 0 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7
WA.max
VxA VyA�
2
§¨
©
·
¹
2
Wxy2�� WA.max 43.6 MPa Ans
WB.max
VxB VyB�
2
§¨
©
·
¹
2
Wxy2�� WB.max 47.0 MPa Ans
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
f )ș2tan( S_A Ts_A 45deg� Anstan 2Ts_A� � VxA VyA�
2Wxy
�=
AnsT's_A Ts_A 90deg�� T's_A 45� deg Ans
f )ș2tan( S_Btan 2Ts_B� � VxB VyB�
2Wxy
�= Ts_B 45� deg� Ans
AnsT's_B Ts_B 90deg�� T's_B 45 deg Ans
By observation, in order to preserve equilibrium along AB, Wmax has to act in thedirection shown in the figure.
Average Normal Stress: Applying Eq. 9-8
Vavg_A
VxA VyA�
2
� Vavg_A 43.57� MPa 
Vavg_B
VxB VyB�
2
� Vavg_B 46.97 MPa 
Problem 9-28
The simply supported beam is subjected to the traction stress W0 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
+ 6Fy=0; 2R F� 0= R 0.5F� 
+ 6Fx=0; H1 P� 0= H1 P� 
Internal Force and Moment : At Section A-B:
+ 6Fx=0; H1 N� 0= N H1�� 
+ 6Fy=0; R V� 0= V R�� 
+ 60O=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: V N
A
M c˜
I
�=
cA 0.5� d� VA
N
A
M cA˜
I
�� VA 0.942� MPa 
cB 0.5d� VB
N
A
M cB˜
I
�� VB 0.764 MPa 
Shear Stress : Since QA = QB = 0, WA 0� WB 0� 
In-plane Principal Stress:
At A: VxA VA� VyA 0� Wxy 0� 
Since no shear stress acts upon the element,
VA1 VyA� VA1 0 MPa Ans
VA2 VxA� VA2 0.942� MPa Ans
At B: VxB VB� VyB 0� Wxy 0� 
Since no shear stress acts upon the element,
VB1 VxB� VB1 0.764 MPa Ans
VB2 VyB� VB2 0 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax_A
VxA VyA�
2
§¨
©
·
¹
2
Wxy2�� Wmax_A 0.471 MPa Ans
Wmax_B
VxB VyB�
2
§¨
©
·
¹
2
Wxy2�� Wmax_B 0.382 MPa Ans
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
f )ș2tan( S_A Ts_A 45deg� Anstan 2Ts_A� � VxA VyA�
2Wxy
�=
AnsT's_A Ts_A 90deg�� T's_A 45� deg Ans
f )ș2tan( S_Btan 2Ts_B� � VxB VyB�
2Wxy
�= Ts_B 45� deg� Ans
AnsT's_B Ts_B 90deg�� T's_B 45 deg Ans
By observation, in order to preserve equilibrium along AB, Wmax has to act in the
direction shown in the figure.
Average Normal Stress: Applying Eq. 9-8,
Vavg_A
VxA VyA�
2
� Vavg_A 0.471� MPa Ans
Vavg_B
VxB VyB�
2
� Vavg_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� T 30deg� 
a 3m� 
w 8
kN
m
� 
Solution:
Internal Force and Moment : At Section A-B:
+ 6Fx=0; P cos T� �˜ N� 0= N P cos T� �˜� 
+ 6Fy=0; V P sin T� �˜� w a˜� 0=
V P sin T� �˜ w a˜�� 
+ 60O=0; M P sin T� �˜ a( )˜� w a˜( ) 0.5a( )˜� 0=
M P a˜ sin T� �˜ 0.5w a2˜�� 
Section Property : D dw 2tf�� 
A bf D˜ bf tw�� � dw˜�� A 6000 mm2 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� I 50.80 10 6�u m4 
QA bf tf˜� � D2
tf
2
�§¨©
·
¹˜� QA 210000 mm
3 QB QA� 
Normal Stress: V N
A
M c˜
I
�=
cA 0.5dw� VA
N
A
M cA˜
I
�� VA 148.293 MPa 
cB 0.5� dw� VB
N
A
M cB˜
I
�� VB 141.077� MPa 
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I tw˜
� WA 15.09 MPa 
WB
V QB˜
I tw˜
� WB 15.09 MPa 
In-plane Principal Stress:
At A: VxA VA� VyA 0� Wxy WA� 
VA1
VxA VyA�
2
VxA VyA�
2
§¨
©
·
¹
2
WA2��� VA1 149.8 MPa Ans
VA2
VxA VyA�
2
VxA VyA�
2
§¨
©
·
¹
2
WA2��� VA2 1.52� MPa Ans
At B: VxB VB� VyB 0� Wxy WB� 
VB1
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB1 1.60 MPa Ans
VB2
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB2 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� 
T 30deg� P 10N� 
Solution:
Section Property : di do 2t�� 
A
S
4
do
2
di
2�§© ·¹˜� A 91.11 mm2 
Normal Stress:
Vx
P
A
� Vx 109.76 kPa 
Vy 0� 
Wxy 0� 
Shear stress along the seam:
Wx'y'
Vx Vy�
2
� sin 2T� �˜ Wxy cos 2T� �˜�� 
Wx'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� 
T 30deg� P 10N� 
Solution:
Section Property : di do 2t�� 
A
S
4
do
2
di
2�§© ·¹˜� A 91.11 mm2 
Normal Stress:
Vx
P
A
� Vx 109.76 kPa 
Vy 0� 
Wxy 0� 
Normal stress perpendicular to the seam:
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2T� �˜� Wxy sin 2T� �˜�� 
Vx' 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:
6Fx=0; N P� w L˜� 0= N P� w L˜�� 
60x=0; T Mx� 0= T Mx� 
Section Property : di do 2t�� 
A
S
4
do
2
di
2�§© ·¹˜� J S32 do
4
di
4�§© ·¹˜� 
Normal Stress: V N
A
� V 9.457� MPa 
Shear Stress :
c 0.5do� W
T c˜
J
� W 28.850 MPa 
a) In-plane Principal Stresses:
Vx 0� Vy V� Wxy W� 
for any point on the shaft's surface. Applying Eq. 9-5,
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 24.51 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 33.96� MPa Ans
b) Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 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 500� kN� My 30� kN m˜� 
Py 800� kN� Mz 40kN m˜� 
Solution:
Section Property : D dw 2tf�� 
A bf D˜ bf tw�� � dw˜�� A 30000 mm2 
Iz
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� Iz 350.00 10 6�u m4 
Iy
1
12
2tf bf
3˜ dw tw3˜�§© ·¹˜� Iy 68.75 10 6�u m4 
QA 0� (since A' = 0)
Normal Stress: yA 0.5D� zA 0.5bf� 
VA
Px
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 77.446� MPa 
Shear Stress : Since QA = 0, WA 0� 
In-plane Principal Stress:
Vx VA� Vy 0� Wxy 0� 
Since no shear stress acts upon the element,
V1 Vy� V1 0 MPa Ans
V2 Vx� V2 77.45� MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 38.72 MPa Ans
Problem 9-37
Solve Prob. 9-36 for point B.
Given: bf 200mm� tf 50mm� 
tw 50mm� dw 200mm� 
Px 500� kN� My 30� kN m˜� 
Py 800� kN� Mz 40kN m˜� 
Solution:
Section Property : D dw 2tf�� 
A bf D˜ bf tw�� � dw˜�� A 30000 mm2 
Iz
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� Iz 350.00 10 6�u m4 
Iy
1
12
2tf bf
3˜ dw tw3˜�§© ·¹˜� Iy 68.75 10 6�u m4 
QB 0� (since A' = 0)
Normal Stress: yB 0.5� D�zB 0.5� bf� 
VB
Px
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 44.113 MPa 
Shear Stress : Since QB = 0, WB 0� 
In-plane Principal Stress:
Vx VB� Vy 0� Wxy 0� 
Since no shear stress acts upon the element,
V1 Vx� V1 44.113 MPa Ans
V2 Vy� V2 0.00 MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 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 500� kN� My 30� kN m˜� 
Py 800� kN� Mz 40kN m˜� 
Solution:
Section Property : D dw 2tf�� 
A bf D˜ bf tw�� � dw˜�� A 30000 mm2 
Iz
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� Iz 350.00 10 6�u m4 
Iy
1
12
2tf bf
3˜ dw tw3˜�§© ·¹˜� Iy 68.75 10 6�u m4 
QC bf tf˜� � D2
tf
2
�§¨©
·
¹˜� QC 1250000 mm
3 
Normal Stress: yC 0.5� dw� zC 0� 
VC
Px
A
Mz yC˜
Iz
�
My zC˜
Iy
�� VC 5.238� MPa 
Shear Stress : W V Q˜
I t˜=
WC
Py QC˜
Iz tw˜
� WC 57.14� MPa 
In-plane Principal Stress:
Vx VC� Vy 0� Wxy WC� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 54.58 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 59.82� MPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 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:
+ 6Fy=0; V P� 0= V P� 
+ 60O=0; M P a( )˜� 0= M P a˜� 
Section Property : D dw 2tf�� 
A bf D˜ bf tw�� � dw˜�� A 7300 mm2 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� I 95.45 10 6�u m4 
QA bf tf˜� � D2
tf
2
�§¨©
·
¹˜� QA 314400 mm
3 
Normal Stress: V M c˜
I
=
cA 0.5dw� VA
M cA˜
I
� VA 196.435 MPa 
Shear Stress : W V Q˜
I t˜=
WA
V QA˜
I tw˜
� WA 16.47 MPa 
In-plane Principal Stress:
Vx VA� Vy 0� Wxy WA� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 197.81 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 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:
+ 6Fy=0; V P� 0= V P� 
+ 60O=0; M P a( )˜� 0= M P a˜� 
Section Property : D dw 2tf�� 
A bf D˜ bf tw�� � dw˜�� A 7300 mm2 
I
1
12
bf D
3˜ bf tw�� � dw3˜�ª¬ º¼˜� I 95.45 10 6�u m4 
QB bf tf˜� � D2
tf
2
�§¨©
·
¹˜� QB 314400 mm
3 
Normal Stress: V M c˜
I
=
cB 0.5� dw� VB
M cB˜
I
� VB 196.435� MPa 
Shear Stress : W V Q˜
I t˜=
WB
V QB˜
I tw˜
� WB 16.47 MPa 
In-plane Principal Stress:
Vx VB� Vy 0� Wxy WB� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 1.37 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 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
S
64
do
4˜� J S
32
do
4˜� cA 0.5do� 
Normal Stress: VA
Mx cA˜
I
� VA 212.21 MPa 
Shear Stress : WA
Ty cA˜
J
� WA 318.31 MPa 
In-plane Principal Stresses: Applying Eq. 9-5,
Vx VA� Vy 0� Wxy WA� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 441.63 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 229.42� MPa Ans
Orientation of Principal Stress:
tan 2Tp� � 2WxyVx Vy�= Tp
1
2
atan
2Wxy
Vx Vy�
§¨
©
·
¹
� T'p Tp 90deg�� 
Tp 35.783 deg T'p 54.217� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'
Vx Vy�
2
Vx Vy�
2
cos 2Tp� �˜� Wxy sin 2Tp� �˜�� 
Vx' 441.63 MPa 
Therefore, Tp1 Tp� Tp1 35.78 deg Ans
Tp2 T'p� Tp2 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
S
64
do
4˜� J S
32
do
4˜� 
QB
4ro
3S
S ro2˜
2
§¨
©
·
¹˜� 
Normal Stress: cBV 0� VB
Mx cBV˜
I
� VB 0 MPa 
Shear Stress : bB do� cBW ro� 
WB
Vz QB˜
I bB˜
Ty cBW˜
J
�� WB 314.07� MPa 
In-plane Principal Stresses: Applying Eq. 9-5,
Vx VB� Vy 0� Wxy WB� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 314.07 MPa Ans
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 314.07� MPa Ans
Orientation of Principal Stress:
f )ș2tan( Ptan 2Tp� � 2WxyVx Vy�= Tp1 45deg� Ans
Tp2 Tp1 90deg�� Tp2 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
+ 6Fy=0; 2R F� 0= R 0.5F� 
+ 6Fx=0; H1 P� 0= H1 P� 
Internal Force and Moment : At Section A-B:
+ 6Fx=0; H1 N� 0= N H1�� 
+ 6Fy=0; R V� 0= V R�� 
+ 60O=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: V N
A
M c˜
I
�=
cA 0.5� d� VA
N
A
M cA˜
I
�� VA 30.5� MPa 
cB 0� VB
N
A
M cB˜
I
�� VB 0.5� MPa 
Shear Stress :
Since QA = 0, WA 0� 
WB
V QB˜
I b˜� WB 0.75� MPa 
In-plane Principal Stress: Applying Eq. 9-5
At A: VxA VA� VyA 0� Wxy 0� 
Since no shear stress acts upon the element,
VA1 VyA� VA1 0 MPa Ans
VA2 VxA� VA2 30.50� MPa Ans
At B: VxB VB� VyB 0� Wxy WB� 
VB1
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB1 0.541 MPa Ans
VB2
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB2 1.041� MPa Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2Tp� � 2WBVxB VyB�= Tp
1
2
atan
2WB
VxB VyB�
§¨
©
·
¹
� Tp 35.783 deg 
T'p Tp 90deg�� T'p 54.217� deg 
Use Eq. 9-1 to determine the principal plane of V1 and V2.
Vx'_B
VxB VyB�
2
VxB VyB�
2
cos 2Tp� �˜� WB sin 2Tp� �˜�� 
Vx'_B 1.04� MPa 
Therefore, Tp1 T'p� Tp1 54.22� deg Ans
Tp2 Tp� Tp2 35.78 deg Ans
Problem 9-44
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at Z = 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� Z 15 rad
s
� 
Solution:
Internal Force and Moment : As shown on FBD
To
P
Z� To 60.00 kN m˜ 
N F�� 
Section Property :
A
S do2˜
4
� A 49087.39 mm2 
J
S do4˜
32
� J 383495196.97 mm4 
Normal Stress:
Va
N
A
� Va 25.06� MPa 
Shear Stress : cmax 0.5 do˜� 
Wo
To cmax˜
J
� Wo 19.56 MPa 
In-plane Principal Stresses: Applying Eq. 9-5,
Vx Va� Vy 0� Wxy Wo� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V1 10.70 MPa Ans
V2
Vx Vy�2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��� V2 35.75� MPa Ans
Problem 9-45
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at Z = 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� Z 15 rad
s
� 
Solution:
Internal Force and Moment : As shown on FBD
To
P
Z� To 60.00 kN m˜ 
N F�� 
Section Property :
A
S do2˜
4
� A 49087.39 mm2 
J
S do4˜
32
� J 383495196.97 mm4 
Normal Stress:
Va
N
A
� Va 25.06� MPa 
Shear Stress : cmax 0.5 do˜� 
Wo
To cmax˜
J
� Wo 19.56 MPa 
Maximum In-plane Shear Stress: Applying Eq. 9-7
Vx Va� Vy 0� Wxy Wo� 
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 23.2 MPa Ans
L 250mm� Given: 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
S
64
do
4
di
4�§© ·¹˜� J S32 do
4
di
4�§© ·¹˜� 
QAz
4ro
3S
S ro2˜
2
§¨
©
·
¹˜
4ri
3S
S ri2˜
2
§¨
©
·
¹˜�� 
Normal Stress: cAV 0� VA
My cAV˜
I
� VA 0 MPa 
Shear Stress : bA do di�� cAW ro� 
WA
Vz QAz˜
I bA˜
Tx cAW˜
J
�� WA 0.863� MPa 
In-plane Principal Stresses: Applying Eq. 9-5,
Vx VA� Vz 0� Wxz WA� 
V1
Vx Vz�
2
Vx Vz�
2
§¨
©
·
¹
2
Wxz2��� V1 0.863 MPa Ans
V2
Vx Vz�
2
Vx Vz�
2
§¨
©
·
¹
2
Wxz2��� V2 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 250mm� Given: 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
S
64
do
4
di
4�§© ·¹˜� J S32 do
4
di
4�§© ·¹˜� 
QBz 0� (Since A'=0)
Normal Stress: cBV ro� VB
My cBV˜
I
� 
VB 1.862 MPa 
Shear Stress : cBW ro� WB
Tx cBW˜
J
�� 
WB 1.117� MPa 
In-plane Principal Stresses: Applying Eq. 9-5,
Vx VB� Vz 0� Wxz WB� 
V1
Vx Vz�
2
Vx Vz�
2
§¨
©
·
¹
2
Wxz2��� V1 2.385 MPa Ans
V2
Vx Vz�
2
Vx Vz�
2
§¨
©
·
¹
2
Wxz2��� V2 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� yT
4�
5
� zT
3
5
� 
yB 75mm� zB 20� mm� 
Solution:
Internal Force and Moment : At Section A-B:
+ 6Fz0; Vz P zT˜� 0= Vz P� zT˜� 
+ 6Fy=0; Vy P yT˜� 0= Vy P� yT˜� 
Mz P yT˜ L˜� 
My P� zT˜ L˜� 
Section Property :
Iz
1
12
b˜ h3˜� Iz 33.75 10 6�u m4 
Iy
1
12
h˜ b3˜� Iy 21.60 10 6�u m4 
QA.y b 0.5h yA�� �˜ yA 0.5 0.5h yA�� ��ª¬ º¼˜� QA.y 216000 mm3 
QB.z h 0.5b zB�� �˜ zB 0.5 0.5b zB�� ��ª¬ º¼˜� QB.z 240000 mm3 
QA.z 0� (since A' = 0)
QB.y 0� (since A' = 0)
Normal Stress:
VA
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 10.8� MPa 
VB
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 42.0 MPa 
Shear Stress : W V Q˜
I t˜=
WA
Vy QA.y˜
Iz b˜
� WA 0.640 MPa 
WB
Vz QB.z˜
Iy h˜
� WB 0.667� MPa 
In-plane Principal Stress: Applying Eq. 9-5
At A: VxA VA� VyA 0� Wxy 0� 
VA1
VxA VyA�
2
VxA VyA�
2
§¨
©
·
¹
2
WA2��� VA1 0.0378 MPa Ans
VA2
VxA VyA�
2
VxA VyA�
2
§¨
©
·
¹
2
WA2��� VA2 10.84� MPa Ans
At B: VxB VB� VzB 0� Wxz WB� 
VB1
VxB VzB�
2
VxB VzB�
2
§¨
©
·
¹
2
WB2��� VB1 42.01 MPa Ans
VB2
VxB VzB�
2
VxB VzB�
2
§¨
©
·
¹
2
WB2��� VB2 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
+ 6Fy=0; R1 R2� P1� P2� 0=
+ 60R2=0;
Ans
Guess R1 1kN� R2 1kN� 
R1
R2
§¨
©¨
·
¹
Find R1 R2�� �� R1
R2
§¨
©¨
·
¹
8.2
1.8
§¨
©
·
¹ 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 di3˜�§© ·¹˜� 
QA 0� QB 0� (Since A'=0)
For point A: WA 0� 
cA 0.5 do˜� VA
M cA˜
I
� VA 0.494 MPa 
V1 VA� V1 0.494 MPa Ans
V2 0� V2 0.000 MPa Ans
For point B: WB 0� 
cB 0.5� di˜� VB
M cB˜
I
� VB 0.37� MPa 
V1 0� V1 0.000 MPa Ans
V2 VB� V2 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 140 MPa and -70 MPa.
Determine the torque and the bending moment.
Given: do 6mm� 
V1 140MPa� V2 70� MPa� 
Solution:
In-plane Principal Stresses: Applying Eq. 9-5,
Given Vy 0� 
V1
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��= (1)
V2
Vx Vy�
2
Vx Vy�
2
§¨
©
·
¹
2
Wxy2��= (2)
Solving Eqs. (1) and (2):
Guess Vx 2MPa� Wxy 1MPa� 
Vx
Wxy
§¨
©¨
·
¹
Find Vx Wxy�� �� VxWxy
§¨
©¨
·
¹
70.00
98.99
§¨
©
·
¹ MPa 
Section Property : ro 0.5do� 
I
S
64
do
4˜� J S
32
do
4˜� 
Normal Stress: V M c˜
I
=
c ro� M
Vx I˜
c
� M 1.484 N m˜ Ans
Shear Stress: W T c˜
J
=
c ro� T
Wxy 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 0.04� kN m˜� Mz 0.03kN m˜� 
Solution:
Section Property :
A b h˜� A 20000 mm2 
Iz
1
12
b˜ h3˜� Iz 66.67 10 6�u m4 
Iy
1
12
h˜ b3˜� Iy 16.67 10 6�u m4 
QA.y 0� (since A' = 0)
Normal Stress:
VA
Px
A
Mz yA˜
Iz
�
My zA˜
Iy
�� VA 20.0� kPa 
Shear Stress : Since QA = 0, WA 0� 
In-plane Principal Stress:
Vx VA� Vy 0� Wxy 0� 
Since no shear stress acts upon the element,
V1 Vy� V1 0 kPa Ans
V2 Vx� V2 20� kPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
Vx Vy�
2
§¨
©
·
¹
2
Wxy2�� Wmax 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 50� mm� Py 0.8kN� 
My 0.04� kN m˜� Mz 0.03kN m˜� 
Solution:
Section Property :
A b h˜� A 20000 mm2 
Iz
1
12
b˜ h3˜� Iz 66.67 10 6�u m4 
Iy
1
12
h˜ b3˜� Iy 16.67 10 6�u m4 
QB.y b 0.5h( )˜ 0.25h( )˜� 
Normal Stress:
VB
Px
A
Mz yB˜
Iz
�
My zB˜
Iy
�� VB 145.0 kPa 
Shear Stress :
WB
Py QB.y˜
Iz b˜
� WB 60.00 kPa 
In-plane Principal Stress:
VxB VB� VyB 0� Wxy WB� 
VB1
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2
WB2��� VB1 166.6 kPa Ans
VB2
VxB VyB�
2
VxB VyB�
2
§¨
©
·
¹
2WB2��� VB2 21.61� kPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
VxB VyB�
2
§¨
©
·
¹
2
Wxy2�� Wmax 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 50� mm� zC 0mm� Py 0.8kN� 
My 0.04� kN m˜� Mz 0.03kN m˜� 
Solution:
Section Property :
A b h˜� A 20000 mm2 
Iz
1
12
b˜ h3˜� Iz 66.67 10 6�u m4 
Iy
1
12
h˜ b3˜� Iy 16.67 10 6�u m4 
QC.y b 0.5h yC�� �˜ yC 0.5 0.5h yC�� ��ª¬ º¼˜� QC.y 375000 mm3 
Normal Stress:
VC
Px
A
Mz yC˜
Iz
�
My zC˜
Iy
�� VC 47.5 kPa 
Shear Stress :
WC
Py QC.y˜
Iz b˜
� WC 45.00 kPa 
In-plane Principal Stress:
VxC VC� VyC 0� Wxy WC� 
VC1
VxC VyC�
2
VxC VyC�
2
§¨
©
·
¹
2
WC2��� VC1 74.63 kPa Ans
VC2
VxC VyC�
2
VxC VyC�
2
§¨
©
·
¹
2
WC2��� VC2 27.13� kPa Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
Wmax
VxC VyC�
2
§¨
©
·
¹
2
Wxy2�� Wmax 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: Vx 0.650� MPa� Vy 0.400MPa� 
I' 60deg� Wxy 0MPa� 
Solution: T 90deg I'�� T 30.00 deg 
Center :
Vc
Vx Vy�
2
� Vc 0.125� MPa 
Radius :
R Vx Vc�� R 0.525 MPa 
Coordinates:
A Vx 0�� � B Vy 0�� � C Vc 0�� �
Stresses: 
Vx' Vc R cos 2T� �˜�� Vx' 0.387� MPa Ans
Wx'y' R sin 2T� �˜� Wx'y' 0.455 MPa Ans
Problem 9-57
Solve Prob. 9-2 using Mohr's circle.
Given: Vx 5MPa� Vy 3MPa� Wxy 8MPa� I' 40deg� 
Solution:
Center :
Vc
Vx Vy�
2
� Vc 4 MPa 
Radius :
R Vx Vc�� �2 Wxy2�� R 8.062 MPa 
Angles: T 90deg I'�� T 130 deg 
I atan
Wxy
Vx Vc�
§¨
©
·
¹
� I 82.875 deg 
D 180deg 2T I�� ��� D 2.875 deg 
Stresses: 
Vx' Vc R cos D� �˜�� Vx' 4.052� MPa Ans
Wx'y' R� sin D� �˜� Wx'y' 0.404� MPa Ans
Problem 9-58
Solve Prob. 9-3 using Mohr's circle.
Given: Vx 0.350MPa� Vy 0.200� MPa� 
I' 50deg� Wxy 0MPa� 
Solution: T 90deg I'�� T 140 deg 
Center :
Vc
Vx Vy�
2
� Vc 0.075 MPa 
Radius :
R Vx Vc�� �2 Wxy2�� R 0.275 MPa 
Coordinates:
A Vx 0�� � C Vc 0�� �
Angles:
D 360deg 2T�� D 80 deg 
Stresses: (represented by coordinates of point P)
Vx' Vc R cos D� �˜�� Vx' 0.123 MPa Ans
Wx'y' R sin D� �˜� Wx'y' 0.271 MPa Ans
Problem 9-59
Solve Prob. 9-10 using Mohr's circle.
Given: Vx 0MPa� Vy 0.300� MPa� 
T 30deg� Wxy 0.950MPa� 
Solution:
Center :
Vc
Vx Vy�
2
� Vc 0.15� MPa 
Radius :
R Vx Vc�� �2 Wxy2�� R 0.962 MPa 
Angles:
I atan
Wxy
Vx Vc�
§¨
©
·
¹
� I 81.027 deg 
D 2T I�� D 21.027� deg 
Stresses: 
Vx' Vc R cos D� �˜�� Vx' 0.748 MPa Ans
Vy' Vc R cos D� �˜�� Vy' 1.048� MPa Ans
Wx'y' R� sin D� �˜� Wx'y' 0.345 MPa Ans
Problem 9-60
Solve Prob. 9-6 using Mohr's circle.
Given: Vx 90MPa� Vy 50MPa� Wxy 35� MPa� 
I' 60deg� 
Solution:
Center :
Vc
Vx Vy�
2
� Vc 70 MPa 
Radius :
R Vx Vc�� �2 Wxy2�� R 40.311 MPa 
Angles: T 90deg I'�� 
I atan
Wxy
Vx Vc�
§¨
©
·
¹
� I 60.255� deg 
D 180deg 2T I�� ��� D 59.745 deg 
Stresses: 
Vx' Vc R cos D� �˜�� Vx' 49.689 MPa Ans
Wx'y' R� sin D� �˜� Wx'y' 34.821� MPa Ans