Philpot_MoM_2nd_ch07-11_ISM

Prévia do material em texto

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7.1 For the cantilever beam and loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 Fig. P7.1 
Solution 
Beam equilibrium: 
 
0
0
0
2
0
0
 0
2
2
y y
y
A A
A
F A w L
A w L
L
M M w L
w L
M
 
 
 
Section a-a: 
 
0
0 0
0
0 ( )
0yF w L w x V
V w L w x w L x 
 
 
2
0
0 0
2 2
0 0 2 20
00 ( )
2
0
2 2
2 2
a a
w L x
M w Lx w x M
w L w x
M
w
L x wL Lxw x 
 
 
(b) Shear-force and bending-moment diagrams 
 
 
 
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7.2 For the simply supported beam shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 
 Fig. P7.2 
Solution 
Beam equilibrium: 
 
0
( ) 0
and
y y y
A y
y y
F A C P
M Pa C a b
Pa Pb
C A
a b a b
 
 
Section a-a: 
For the interval 0 ≤ x < a: 
 
0
0
y y
a a y
Pb
V
a b
Pb
M x
a b
Pb
F A V V
a b
Pb
M A x M x M
a b
 
 
 
Section b-b: 
For the interval a ≤ x < b: 
 
0y y
Pb
F A P V P V
a b
Pa
V
a b
 
 
 
(
( )
) ( ) 0b b y
Pb
M A x P x a M x P x a M
a
Pb
b
M x P x a
a b
 
 
 
 
 
 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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(b) Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.3 For the cantilever beam and loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 Fig. P7.3 
Solution 
Beam equilibrium: 
 
0
 0
2 2
 
2 2
y a b y
y a b
C a b C
C a b
F w a w b C
C w a w b
a b
M w a b w b M
a b
M w a b w b
 
 
 
 
Section a-a: 
For the interval 0 ≤ x < a: 
 
2
2
0
0
2
y a
a a
a
a
a
V w xF w x V
x
M w x M
w x
M 
 
 
Section b-b: 
For the interval a ≤ x < b: 
 
2
2 2
0
0
2 2
y a b
b b a
b
b
a
b
aV w a w x a
F w a w x a V
a
w x aa
M w a x
x a
M w a x w x a M
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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(b) Shear-force and bending-moment diagrams 
 
 
 
 
 
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7.4 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 
 Fig. P7.4 
Solution 
Beam equilibrium: 
 
2
2
( ) 0
2 2
( 2 )
2( )
( ) 0
2 2
(2 )
2( )
C a b y
a b
y
A a b y
a b
y
a b
M w a b w b A a b
w a a b w b
A
a b
a b
M w a w b a C a b
w a w b a b
C
a b
 
 
 
Section a-a: 
For the interval 0 ≤ x < a: 
 
2
2 22
0
0
2
2
( 2 )
2( )
( 2 )
2 2( )
y y a a y
a a y a
a b
a
y
a ba a
F A w x V V w x A
x
M A x w x M
w x
M
w a a b w b
w x
a b
w x w a a b w b
A
a b
x x
 
 
 
Section b-b: 
For the interval a ≤ x < b: 
 
2( 2 )
( )
2( ) 2( )
( ) 0
( )
a b
a
y a b
a b
b
y
y
F A w a w x a V
V A w a w x
w a a b w b
w a w x a
a b a b
a
 
 
 22 2( 2 )
2
0
2 2
2 22( ) 2(2 )
b b y a b
ay a b
a b
b
a x a
M A x w a x w x a M
x aa
M A x w
x aaw a a b w b
x w a x w
a b a
a x w
b
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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(b) Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.5 For the cantilever beam and loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 Fig. P7.5 
Solution 
Beam equilibrium: 
 
0 0
2
0 0
0
2 2
 0
2 3 6
y y y
B B B
w L w L
F B B
w L L w L
M M M
 
 
 
Section a-a: 
 
0
2
0
3
0
0 02
0
2 3
2
6
y
a a
w xw x x
F V
L
w x x x
M M
L
V
L
w x
M
L
 
 
 
(b) Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.6 For the simply supported beam shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Determine the location and the magnitude of the 
maximum bending moment. 
 
 Fig. P7.6 
Solution 
Beam equilibrium: 
 
0 0
0 0
0
2 3 6
2
0
2 3 3
B y y
A y y
w L L w L
M A L A
w L L w L
M B L B
 
 
 
Section a-a: 
 
2
0 0 0
3
0 0
2
0 0
3
0 0
0
6 2
6
0
2 6 2
0
2
6
3 6 6
y y
a a y
w x x w L w x
F A V V
L L
w x x x w Lx w x
M A x M M
L
w L w x
V
L
w x w Lx
M
L
L
 
 
(b) Shear-force and bending-moment diagrams 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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(c) Location and magnitude of maximum bending moment: 
The maximum bending moment is located at the location where V = 0. Therefore, the maximum 
bending moment occurs at: 
 
2
0 0
2 2
0 0
0
6
0.577350
2
2 6 3 3
w L w x
V
L
w x w L L
L
L
L
x Ans. 
 
Substitute this value of x into the bending moment equation to determine the moment magnitude: 
 
3
0 0
3
0 2
0
0
max
6 6
(0.577350 )
0.06415
(0.577
6
0
350 )
6
w x w Lx
M
L
w L w L L
M
L
w L Ans. 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.7 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Report the maximum bending moment and its 
location. 
 
 Fig. P7.7 
Solution 
Beam equilibrium: 
(50 kN)(3 m) (75 kN)(6 m) (10 m) 0
60 kN
50 kN 75 kN 0
65 kN
A y
y
y y y
y
M D
D
F A D
A
 
 
Section a-a: 
For the interval 0 ≤ x < 3 m: 
 65 kN 0
(6
65 kN
(65 5 kNk ) N) 0
y y
a a y
V
M
F A V V
M A x M x M x
 
 
Section b-b: 
For the interval 3 m ≤ x < 6 m: 
 50 kN 65 kN 50 k
15 k
0
N
Ny yF A V
V
V 
 
(50 kN)( 3 m)
(65 kN) (50 kN)( 3 m
(15 kN) 150 kN m
) 0
-
b b yM A x x
M x
M
x x M
 
 
 
Section c-c: 
For the interval 6 m ≤ x < 10 m: 
 
50 kN 75 kN
65 kN 50 kN 75 kN 0
60 kN
y yF A V
V
V
 
 
 
 
(50 kN)( 3 m) (75 kN)( 6 m)
(65 kN) (50 kN)( 3 m) (75 kN)(
(60 
6
kN) 600 kN-m
 m) 0
c c yM A x x x M
x x x M
M x
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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 (b) Shear-force and bending-moment diagrams 
 
 
 
 
 
(c) Maximum bending moment 
 and its location 
 
 Mmax = 240 kN-m @ x = 6 m 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.8 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Report the maximum positive bending moment, 
the maximum negative bending moment, and their 
respective locations. 
 
 Fig. P7.8 
Solution 
Beam equilibrium: 
(20 kN)(2 m) (60 kN)(6 m) (8 m) 0
40 kN
20 kN 60 kN 0
40 kN
B y
y
y y y
y
M D
D
F B D
B
 
 
Section a-a: 
For the interval 0 ≤ x < 2 m: 
 2020 kN 0
(20 k
 kN
N (20 kN)) 0
y
a a
F
MM M
VV
xx
 
 
Section b-b: 
For the interval 2 m ≤ x < 8 m: 
 
20 kN
(20 kN) 80 kN-m
20 kN 20 kN 40 kN 0
(20 kN) ( 2 m)
(20 kN) (40 kN)( 2 m) 0
y y
b b y
F B V V
M x B x M
x M
x
x
V
M
 
 
 
 
 
 
 
 
 
 
 
 
 
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Section c-c: 
For the interval 8 m ≤ x < 10 m: 
 
20 kN 60 kN
20 kN 40 kN 60 k
40 kN
N 0
y yF B
V
V
V
 
 
 
 
(20 kN) ( 2 m) (60 kN)( 8 m)
(20 kN) (40 kN)( 2 m) (60 kN)( 8 m)
(40 kN) 400 kN-
0
m
c c yM x B x x
M
M
x x M
x
x
 
 
 (b) Shear-force and bending-moment diagrams 
 
 
 
(c) Maximum bending moment 
 and its location 
 
 Mmax-positive = 80 kN-m @ x = 8 m 
 
 Mmax-negative = –40 kN-m @ x = 2 m 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.9 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Report the maximum positivebending moment, 
the maximum negative bending moment, and their 
respective locations. 
 
 Fig. P7.9 
Solution 
Beam equilibrium: 
 
(7 kips/ft)(30 ft)(15 ft) (21 ft) 0
150 kips
(7 kips/ft)(30 ft) 0
60 kips
C y
y
y y y
y
M B
B
F B C
C
 
 
Section a-a: 
For the interval 0 ≤ x < 9 ft: 
 
2
(7 kips/ft) 0
(7 kips/ft)( )
(7 kips/ft)
(7 kips/f )
2
0
2
t
y
a a
F x V
x
V x
xM x MM
 
 
 
Section b-b: 
For the interval 9 ft ≤ x < 30 ft: 
 (7 kips/ft) (7 kips/ft) 150 kips 0
(7 kips/ft) 150 kips
y yF x B V x V
V x
 
2
(7 kips/ft)( ) ( 9 ft)
2
(7 kips/ft)( ) (150 kips)( 9 ft) 0
2
(7 kips/ft)
(150 kips) 1,350 kips
2
b b y
x
M x B
M x
x M
x
x x M
x
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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 (b) Shear-force and bending-moment diagrams 
 
 
(c) Maximum bending moment 
 and its location 
 
 Mmax-positive = 257.14 kip-ft @ x = 21.43 ft 
 
 Mmax-negative = –283.50 kip-ft @ x = 9 ft 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.10 For the cantilever beam and loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 Fig. P7.10 
Solution 
Beam equilibrium: 
 
(4 kips/ft)(8 ft) 0
32 kips
(4 kips/ft)(8 ft)(12 ft) 0
384 kip-ft
y y
y
C C
C
F C
C
M M
M
 
 
Section a-a: 
For the interval 0 ≤ x < 8 ft: 
 
2
(4 kips/ft)
4 kips
(4 kips/f
/f
t) 0
(4 kips/ft
t
2
)( ) 0
2
y
a a
F x V
x
M x M
V x
M x
 
 
Section b-b: 
For the interval 8 ft ≤ x < 16 ft: 
 
(4 kips/ft)(8 ft) 0
(4 kips/ft)(8 ft)( 4 ft
32 kips
(32 kips) 128 ki f
0
p t
)
-
y
b b
F V
M x M
V
M x
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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 (b) Shear-force and bending-moment diagrams 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.11 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Report the maximum bending moment and its 
location. 
 
 Fig. P7.11 
Solution 
Beam equilibrium: 
 
(42 kips)(10 ft) (6 kips/ft)(20 ft)(20 ft)
(30 ft) 0
94 kips
A
y
y
M
C
C
 
 42 kips (6 kips/ft)(20 ft) 0
68 kips
y y y
y
F A C
A
 
 
Section a-a: 
For the interval 0 ≤ x < 10 ft: 
 68 kips
(
68 kips 0
(68 kips 6) 0 8 kips)
y y
a a y
F A V V
M A x M Mx
V
M x
 
 
Section b-b: 
For the interval 10 ft ≤ x < 30 ft: 
 
42 kips (6 kips/ft)( 10 ft)
68 kips 42 kips (6 kips/ft)( 10 
(6 kips
ft)
/ft) 86 
0
kips
y yF A x V
x V
V x
 
 
 
2
2
10 ft
(42 kips)( 10 ft) (6 kips/ft)( 10 ft)
2
6 kips/ft
(68 kips) (42 kips)( 10 ft) ( 10 f
3 86 120 kip
t
2
-
)
ft
0
b b y
x
M A x x x M
x x x
M
M
x x
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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 (b) Shear-force and bending-moment diagrams 
 
 
(c) Maximum bending moment 
 and its location 
 
 Mmax = 736.33 kip-ft @ x = 14.33 ft 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.12 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Report the maximum positive bending moment, 
the maximum negative bending moment, and their 
respective locations. 
 
 Fig. P7.12 
Solution 
Beam equilibrium: 
 
180 kN-m (9 m) (36 kN)(12 m) 0
28 kN
36 kN 0
8 kN
A y
y
y y y
y
M C
C
F A C
A
 
 
Section a-a: 
For the interval 0 ≤ x < 4 m: 
 8 kN 0
(
8 kN
(8 kN)8 kN) 0
y y
a a y
V
M
F A V V
M xA x M x M
 
 
Section b-b: 
For the interval 4 m ≤ x < 9 m: 
 
8 kN
(8 kN) 180 kN-m
8 kN 0
180 kN-m
(8 kN) 180 kN-m 0
y y
x y
F A V V
M A x M
M
x
V
x
M 
 
 
Section c-c: 
For the interval 9 m ≤ x < 12 m: 
 8 kN 28 kN 0
36 kN
y y yF A C V V
V
 
 
 
( 9 m) 180 kN-m
(8 kN) (28 kN)( 9 
(36 kN) 432
m) 180 kN
 kN
-m 0
-m
c c y yM A x C
M
x M
x
x
x M
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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 (b) Shear-force and bending-moment diagrams 
 
 
(c) Maximum bending moment 
 and its location 
 
 Mmax-positive = 32 kN-m @ x = 4 m 
 
 Mmax-negative = –148 kN-m @ x = 4 m 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.13 For the cantilever beam and loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 
 Fig. P7.13 
Solution 
Beam equilibrium: 
 
 
Section a-a: 
For the interval 0 ≤ x < 8 ft: 
 
 
Section b-b: 
For the interval 8 ft ≤ x < 14 ft: 
 
 
 
 
 
 
 
 
 
 
 
 
 
5 kips/ft 6 ft 0
 30 kips
120 kip-ft 5 kips/ft 6 ft 3 ft 0
 210 kip-ft
y y
y
C C
C
F C
C
M M
M
-
0
120 kip-
0 kips
120 kip- tft 0 f
y
a a
VF V
MM M
-
2
5 kips/ft 8 ft 0
 
8 ft
120 kip-ft 5 kips/
5 40
ft 8 ft =0
2
 
 kips
2.5 40 280 kip-f t
y
b b
F x V
x
M x M
V x
M x x
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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(b) Shear-force and bending-moment diagrams: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.14 For the cantilever beam and loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
 
 Fig. P7.14 
Solution 
Beam equilibrium: 
 
Section a-a: 
For the interval 0 ≤ x < 9 ft: 
 
Section b-b: 
For the interval 9 ft ≤ x < 14 ft: 
 
 
1
6 kips/ft 9 ft 17 kips 0
2
 44 kips
1
6 kips/ft 9 ft 6 ft
2
 17 kips 14 ft 0
 400 kip-ft
y y
y
A A
A
F A
A
M M
M
2
1 6 1 6
44 kips 0
2 9 2 9
 
1 6
2 9 3
1 6
 400 kip-ft 44 kips 0
2
44 ki s
9
p
3
3
y y
a a A y
x x
F A x V x V
x x
M M A x x M
x x
x x
x
V
M
3
 44 400 kip-ft
9
 
x
M x
1
6 kips/ft 9 ft
2
1
 44 kips 6 kips/ft 9 ft
17 ki
0
2
 ps
y yF A V
V
V
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
-
1
6 kips/ft 9 ft 6 ft
2
1
 400 kip-ft 6 kips/ft 9 ft 6 ft 044
1
 kips
2
 7 238 kip-f t
b b A yM M A x x M
M x
x x M
 
 
 
(b) Shear-force and bending-moment diagrams: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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7.15 For the simply supported beam subjected to the 
loading shown, 
(a) Derive equations for the shear force V and the 
bending moment M for any location in the beam. 
(Place the origin at point A.) 
(b) Plot the shear-force and bending-moment 
diagrams for the beam using the derived functions. 
(c) Report the maximum positive bending moment, 
the maximum negative bending moment, and their 
respective locations. 
 
 Fig. P7.15 
Solution 
Beam equilibrium: 
 
Section a-a: 
For the interval 0 ≤ x < 13 ft: 
 
Section b-b: 
For the interval 13 ft ≤ x < 17 ft: 
 
2
61.03 kips 07 kips/ft 7 kips/ft
 
250 kip-ft7 kips/ft
2
 250 kip-ft 061.03 kips 7
7 61.03 kips
3.5 61.
 kips/ft
2
 3 0
y y
b b y
F A x V x V
x
M A x x M
x
x x M
V x
M x 250 kip-ftx
 
 
 
250 kip-ft 7 kips/ft 25 ft 12.5 ft
17 ft 0
 113.97 kips
7 kips/ft 25 ft
 113.97 kips 7 kips/ft 25 ft 0
 61.03 kips
A
y
y
y y y
y
y
M
C
C
F A C
A
A
2
61.03 kips 07 kips/ft 7 kips/ft
 
7 kips/ft
2
 061.03 kips 7 
7 61.03 kips
3.5 61.03 
kips/ft
2
 kip- t f 
y y
a a y
F A x V x V
x
M A x x
V x
M
M
x
x
x
M
x
x
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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Section c-c: 
For the interval 17 ft ≤ x < 25 ft: 
7 kips/ft
 61.03 kips 113.97 kips 07 kips/ft
 
( 17 ft) 7 kips/ft
2
250 kip-ft
 (17 ft)175 kips 113.97 kips 7 kips/ft
2
7 175 kips
y y y
c c y y
F A C x V
x V
x
M A x C x x
M
x
x
V
x
x
2
250 kip-ft 0
 3.5 175 2,187.5 kip -ftM x x
M 
 
 
 
 
(b) Shear-force and bending-moment diagrams: 
 
 
(c) Maximum bending moment and its 
location 
 
 Mmax-positive = 266.04 kip-ft 
 @ x = 8.72 ft 
 
 Mmax-nagative = –224.0 kip-ft 
 @ x = 17 ft 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.16 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.16 
Solution 
Beam equilibrium: 
 
 
 
Shear-force and bending-moment diagrams: 
 
 
28 kips 4 ft 42 kips 8 ft 14 ft 032.00 kips
34 kips 56 kips
 32.00 kips 28 kips 42 kips 0
 38.00 kips
A y
y
y y y
y
y
M D
D
F A D
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.17 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.17 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
35 kN 4 m 45 kN 8 m
 15 kN 14 m 10 m 0
 71 kN
35 kN 45 kN 15 kN
 71 kN 35 kN 45 kN 15 kN 0
 24 kN
A
y
y
y y y
y
y
M
D
D
F A D
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.18 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.18 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
 
15 kips 25 kips 0
 10 kips
15 kips 9 ft 25 kips 3 ft 0
 60 kip-ft
y y
y
C C
C
F C
C
M M
M
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.19 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 
 Fig. P7.19 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
12 ft 6 ft 18 ft 010 kips/ft
 40 kips
10 kips/ft 12 ft
 40 kips 10 kips/ft 12 ft 0
 80 kips
A y
y
y y y
y
y
M C
C
F A C
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.20 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.20 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
4.5 kips/ft 12 ft 9 ft 12 ft 0
 40.50 kips
4.5 kips/ft 12 ft
 40.50 kips 4.5 kips/ft 12 ft 0
 13.50 kips
A y
y
y y y
y
y
M C
C
F A C
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.21 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.21 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
 
40 kN/m 3 m 50 kN 0
 70 kN
40 kN/m 3 m 1.5 m
(50 kN)(3 m) 60 kN-m 0
 90 kN-m
y y
y
A A
A
F A
A
M M
M
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.22 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.22 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
 
28 kips 9 kips/ft 5 ft 0
 17 kips
28 kips 8 ft 9 kips/ft 5 ft 2.5 ft 0
 111.5 kip-ft
y y
y
C C
C
F C
C
M M
M
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.23 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 
 Fig. P7.23 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
10 kips/ft 9 ft 4.5 ft 6 ft 20 ft 0
 47.25 kips
10 kips/ft 9 ft 0
 42.75 kips
A y
y
y y y
y
M D
D
F A D
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation ofthis work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
7.24 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.24 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
 
4.5 kips/ft 7 ft 7 ft
 38 kips 14.5 ft 18 ft 0
 18.36 kip
4.5 kips/ft 7 ft 38 kips
 18.36 kips 4.5 kips/ft 7 ft 38 kips 0
 11.86 kips
A
y
y
y y y
y
y
M
E
E
F A E
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.25 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.25 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
60 kN 2.5 m 45 kN/m 5 m 7.5 m
 9 m 0
 156.67 kN
60 kN 45 kN/m 4 m
 156.67 kN 60 kN 45 kN/m 4 m 0
 83.33 kN
A
y
y
y y y
y
y
M
D
D
F A D
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.26 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 
 Fig. P7.26 
Solution 
Beam equilibrium: 
 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
10 kips 0
 10 kips
10 kips 10 ft 60 kip-ft 0
 40 kip-ft
y y
y
C C
C
F C
C
M M
M
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.27 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 
 Fig. P7.27 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
2 kN 11 kN 0
 13 kN
50 kN-m 2 kN 3.5 m
 11 kN 6 m 0
 23 kN-m
y y
y
A A
A
F A
A
M M
M
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.28 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.28 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
66 kN-m 96 kN-m 12 m 0
 13.50 kN
13.50 kN 0
 13.50 kN
A y
y
y y y y
y
M D
D
F A D A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.29 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 Fig. P7.29 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
80 kN-m 25 kN/m 6 m 3 m
 50 kN-m 6 m 0
 70 kN
25 kN/m 6 m
 70 kN 25 kN/m 6 m 0
 80 kN
A
y
y
y y y
y
y
M
B
B
F A B
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.30 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments (both 
positive and negative) along with their respective 
locations. Clearly differentiate straight-line and 
curved portions of the diagrams. 
 
 Fig. P7.30 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams: 
 
 
 
 
25 kN-m 15 kN 8 m
 7 kN/m 3 m 13.5 m 12 m 0
 31.54 kN
15 kN 7 kN/m 3 m
 31.54 kN 15 kN 7 kN/m 3 m 0
 4.46 kN
A
y
y
y y y
y
y
M
D
D
F A D
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
7.31 Draw the shear-force and bending-moment 
diagram for the beam shown in Fig. P7.31. Assume 
the upward reaction providedby the ground to be 
uniformly distributed. Label all significant points 
on each diagram. Determine the maximum value of 
(a) the internal shear force and (b) the internal 
bending moment. 
 Fig. P7.31 
Solution 
Beam equilibrium: 
 2 kips/ft 8 ft 25 kips 25 kips 16 ft 0
 4.125 kips/ft
yF w
w
 
 
 
 
 
 
Shear-force and bending-moment diagrams 
 
 
(a) Maximum value of internal shear force: 
 V = 16.50 kips @ x = 4 ft Ans. 
 V = −16.50 kips @ x = 12 ft Ans. 
 
(b) Maximum value of internal bending 
moment: 
 M = 33 kip-ft @ x = 4 ft, 12 ft Ans. 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.32 Draw the shear-force and bending-moment 
diagram for the beam shown in Fig. P7.32. Assume 
the upward reaction provided by the ground to be 
uniformly distributed. Label all significant points 
on each diagram. Determine the maximum value of 
(a) the internal shear force and (b) the internal 
bending moment. 
 Fig. P7.32 
Solution 
Beam equilibrium: 
 
40 kN/m 1 m 50 kN
 40 kN/m 1 m 4 m 0
 32.5 kN/m
yF
w
w
 
 
 
 
 
 
Shear-force and bending-moment diagrams 
 
 
(a) Maximum value of internal shear force: 
 V = ±25 kN @ x = 2 m Ans. 
 
(b) Maximum value of internal bending moment: 
 M = −4.62 kN-m @ x = 1.23 m 
 M = −4.62 kN-m @ x = 2.77 m 
 Mmax = 5.00 kN-m Ans. 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.33 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Additionally: 
(a) Determine V and M in the beam at a point 
located 0.75 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of C. 
 
 Fig. P7.33 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 3.75 m, 
 V = 177.5 kN Ans. 
 M = 1,167 kN-m Ans. 
 
(b) At x = 13.75 m, 
 V = −323 kN Ans. 
 M = 442 kN-m Ans. 
 
 
 
 
 
125 kN 3 m 50 kN/m 12 m 9 m
 15 m 0
 385.00 kN
125 kN 50 kN/m 12 m
 385.00 kN 125 kN 50 kN/m 12 m 0
 340.00 kN
A
y
y
y y y
y
y
M
C
C
F A C
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.34 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Additionally: 
(a) Determine V and M in the beam at a point 
located 0.75 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of C. 
 
 Fig. P7.34 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 3.75 m, 
 V = 85.5 kN Ans. 
 M = 30.4 kN-m Ans. 
 
(b) At x = 7.75 m, 
 V = −74.5 kN Ans. 
 M = 52.4 kN-m Ans. 
 
 
 
 
15 kN 3 m 40 kN/m 6 m 3 m
 18 kN 10 m 6 m 0
 142.50 kN
15 kN 40 kN/m 6 m 18 kN
 142.5 kN 15 kN 40 kN/m 6 m 18 kN 0
 130.50 kN
B
y
y
y y y
y
y
M
C
C
F B C
B
B
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.35 Use the graphical method to construct the shear-
force and bending-moment diagrams for the beam 
shown. Label all significant points on each diagram 
and identify the maximum moments along with their 
respective locations. Additionally: 
(a) Determine V and M in the beam at a point located 
0.75 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of C. 
 
 Fig. P7.35 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 3.75 m, 
 V = 91.3 kN Ans. 
 M = 199.2 kN-m Ans. 
 
(b) At x = 6.75 m, 
 V = −103.8 kN Ans. 
 M = 180.5 kN-m Ans. 
 
 
 
25 kN/m 3 m 1.5 m
 65 kN/m 5 m 2.5 m 5 m 0
 185.00 kN
25 kN/m 3 m 65 kN/m 5 m
 185.00 kN 25 kN/m 3 m
65 kN/m 5 m 0
 65.00 kN
B
y
y
y y y
y
y
M
C
C
F B C
B
B
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.36 Use the graphical method to construct the shear-
force and bending-moment diagrams for the beam 
shown. Label all significant points on each diagram 
and identify the maximum moments along with their 
respective locations. Additionally: 
(a) Determine V and M in the beam at a point located 
0.75 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of C. 
 
 Fig. P7.36 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 3.25 m, 
 V = −21.3 kN Ans. 
 M = 16.09 kN-m Ans. 
 
(b) At x = 4.75 m, 
 V = 31.25 kN Ans. 
 M = 23.6 kN-m Ans. 
 
 
 
 
75 kN 60 kN 35 kN/m 6 m 0
 75.00 kN
75 kN 6 m 60 kN 3.5 m 120 kN-m
 35 kN/m 6 m 3 m 0
 90.00 kN-m
y y
y
C
C
C
F C
C
M
M
M
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.37 Use the graphical method to construct the shear-
force and bending-moment diagrams for the beam 
shown. Label all significant points on each diagram 
and identify the maximum moments along with their 
respective locations. Additionally: 
(a) Determine V and M in the beam at a point located 
1.50 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of D. 
 
 Fig. P7.37 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 4.5 m, 
 V = 93.7 kN Ans. 
 M = 23.9 kN-m Ans. 
 
(b) At x = 10.75 m, 
 V = −125.1 kN Ans. 
 M = 75.7 kN-m Ans. 
 
 
 
 
52 kN 3 m 35 kN/m 9 m 4.5 m
 150 kN-m 36 kN 12 m 9 m 0
 204.83 kN
52 kN 35 kN/m 9 m 36 kN
 204.83 kN 52 kN 35 kN/m 9 m 36 kN 0
 198.17 kN
B
y
y
y y y
y
y
M
D
D
F B D
B
B
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.38 Use the graphical method to construct the shear-
force and bending-moment diagrams for the beam 
shown. Label all significant points on each diagram 
and identify the maximum moments along with their 
respective locations. Additionally: 
(a) Determine V and M in the beam at a point located 
1.50 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of D. 
 
 Fig. P7.38 
Solution 
Beam equilibrium: 
Shear-force and bending-moment diagrams 
 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 5.0 m, 
 V = 54.0 kN Ans. 
 M = −44.0 kN-m Ans. 
 
(b) At x = 12.25 m, 
 V = −47.3 kN Ans. 
 M = −49.5 kN-m Ans. 
 
 
 
25 kN/m 3.5 m 1.75 m
 25 kN/m 12.5 m 6.25 m 80 kN 5.5 m
 20 kN 12.5 m 10 m 0
 161.00 kN
25 kN/m 16 m 80 kN 20 kN
 161.00 kN 25 kN/m 16 m
80 kN 20 kN 0
 
B
y
y
y y y
y
M
D
D
F B D
B
 179.00 kNyB
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.39 Use the graphical method to construct the shear-
force and bending-moment diagrams for the beam 
shown. Label all significant points on each diagram 
and identify the maximum moments along with their 
respective locations. Additionally: 
(a) Determine V and M in the beam at a point located 
1.50 m to the right of B. 
(b) Determine V and M in the beam at a point 
located 1.25 m to the left of D. 
 
 Fig. P7.39 
Solution 
Beam equilibrium: 
Shear-force and bending-moment diagrams 
 
Shear force V and bending moment M 
at specific locations: 
 
(a) At x = 3.50 m, 
 V = 285 kN Ans. 
 M = 63.8 kN-m Ans. 
 
(b) At x = 7.75 m, 
 V = −190.0 kN Ans. 
 M = 331 kN-m Ans. 
 
 
 
160 kN 2 m 50 kN/m 2 m 1 m
 50 kN/m 2 m 1 m
 120 kN/m 5 m 4.5 m 7 m 0
 340 kN
160 kN 50 kN/m 4 m
 120 kN/m 5 m
 340 kN 160 kN 50 kN/m 4 m
 
B
y
y
y y y
y
M
D
D
F B D
B
 120 kN/m 5 m 0
 620 kNyB
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.40 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.40 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
 
 
225 kN-m 120 kN/m 4 m 2 m
 60 kN/m 2.5 m 8.75 m 7.5 m 0
 273 kN
120 kN/m 4 m 60 kN/m 2.5 m
 273 kN 120 kN/m 4 m
 60 kN/m 2.5 m 0
 357 kN
A
y
y
y y y
y
y
M
C
C
F A C
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.41 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.41 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
 
 
25 kip-ft 5 kips/ft 3 ft 1.5 ft
 5 kips/ft 5 ft 2.5 ft 25 kips 10 ft
 15 ft 0
 17.67 kips
5 kips/ft 8 ft 25 kips
 17.67 kips 5 kips/ft 8 ft 25 kips 0
 
B
y
y
y y y
y
M
E
E
F B E
B
 47.33 kipsyB
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.42 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.42 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
 
 
35 kip-ft 8 kips/ft 9 ft 4.5 ft
 17 kips 12 ft 9 ft 0
 62.56 kips
8 kips/ft 9 ft 17 kips
 62.56 kips 8 kips/ft 9 ft 17 kips 0
 26.44 kips
B
y
y
y y y
y
y
M
C
C
F B C
B
B
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permissionof the copyright owner is unlawful. 
7.43 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.43 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
6 kips/ft 30 ft 15 ft 60 kips 10 ft
 60 kips 20 ft 3 kips/ft 10 ft 35 ft
 90 kip-ft 30 ft 0
 62.00 kips
6 kips/ft 30 ft 10 ft3 kips/ft
 60 kips 60 k
A
y
y
y y y
M
D
D
F A D
ips
 62.00 kips 6 kips/ft 30 ft 10 ft3 kips/ft
 60 kips 60 kips 0
 28.00 kips
y
y
A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.44 Use the graphical method to construct the shear-
force and bending-moment diagrams for the beam 
shown. Label all significant points on each diagram 
and identify the maximum moments along with their 
respective locations. Clearly differentiate straight-line 
and curved portions of the diagrams. 
 
 Fig. P7.44 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
 
 
5 kips 5 ft 2 kips/ft 20 ft 10 ft 25 kip-ft
 15 kips 8 ft 10 kips 23 ft 20 ft 0
 23 kips
5 kips 2 kips/ft 20 ft 15 kips 10 kips
 23 kips 5 kips 2 kips/ft 20 ft
 
B
y
y
y y y
y
M
D
D
F B D
B
 15 kips 10 kips 0
 17 kipsyB
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.45 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.45 
Solution 
Beam equilibrium: 
 
Shear-force and bending-moment diagrams 
 
 
 
 
50 kN/m 2 m 1 m 20 kN 2 m
 25 kN/m 3 m 3.5 m 50 kN 5 m 0
 47.50 kN-m
50 kN/m 2 m 20 kN
 25 kN/m 3 m 50 kN 0
 55 kN
A A
A
y y
y
M M
M
F A
A
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.46 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.46 
Solution 
Beam equilibrium: 
 
20 kips 15 ft 6 kips/ft 8 ft 11 ft
 12 kips/ft 7 ft 3.5 ft
 70 kips 7 ft 0
 32.00 kip-ft
20 kips 6 kips/ft 8 ft 70 kips
 12 kips/ft 7 ft 0
 
C
C
C
y
y
M
M
M
F
C
C 42.00 kipsy
 
 
 
Shear-force and bending-moment diagrams 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.47 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.47 
Solution 
Beam equilibrium: 
 
4,000 lb-ft 800 lb/ft 4 ft 2 ft
 9,000 lb-ft 600 lb/ft 10 ft 10 ft
 3,600 lb 10 ft 15 ft 0
 5,640 lb
B
y
y
M
E
E
 
 
800 lb/ft 4 ft 600 lb/ft 10 ft 3,600 lb
 5,640 lb 800 lb/ft 4 ft 600 lb/ft 10 ft 3,600 lb 0
 7,160 lb
y y y
y
y
F B E
B
B
 
 
 
Shear-force and bending-moment diagrams 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.48 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.48 
Solution 
Beam equilibrium: 
 
400 kN-m 500 kN-m 600 kN-m
 60 kN 2 m 1 m 120 kN/m 4 m 4 m
 60 kN/m 2 m 7 m 150 kN 6 m 8 m 0
 530 kN
B
y
y
M
E
E
 
 
60 kN 2 m 120 kN/m 4 m
 60 kN 2 m 150 kN
 530 kN 60 kN 2 m 120 kN/m 4 m
 60 kN 2 m 150 kN
 340 kN
y y y
y
y
F B E
B
B
 
 
Shear-force and bending-moment diagrams 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.49 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.49 
Solution 
Beam equilibrium: 
 
 
 
Consider free-body diagram of DE: 
 
55 kN/m 3 m 1.5 m 3 m 0
 82.5 kN
55 kN/m 3 m
 82.5 kN 55 kN/m 3 m 0
 82.5 kN
D y
y
y y y
y
y
M E
E
F D E
D
D
 
 
 
Consider free-body diagram of ABCD: 
 
60 kN-m 75 kN/m 5 m 2.5 m
 100 kN 2.5 m 5 m 3.5 m
 60 kN-m 75 kN/m 5 m 2.5 m
 100 kN 2.5m 82.5 kN 5 m 3.5 m 0
 440 kN
75 kN/m 5 m 100 kN
 
A
y y
y
y
y y y y
M
D C
C
C
F A C D
A 440 kN 75 kN/m 5 m 100 kN 82.5 kN 0
 117.5 kN
y
yA
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.50 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.50 
Solution 
Beam equilibrium: 
 
 
 
Consider free-body diagram of ABC: 
 
500 lb/ft 10 ft 5 ft 15 ft 0
 1,666.67 lb
500 lb/ft 10 ft
 1,666.67 lb 500 lb/ft 10 ft 0
 3,333.33 lb
A y
y
y y y
y
y
M C
C
F A C
A
A
 
 
 
Consider free-body diagram of CDE: 
 
1,200 lb
 1,666.67 lb 1,200 lb 0
 2,866.67 lb
10 ft 1, 200 lb 8 ft
 1,666.67 lb 10 ft 1,200 lb 8 ft 0
 26,266.67 lb-ft
y y y
y
y
E y E
E
E
F C E
E
E
M C M
M
M
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
7.51 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.51 
Solution 
Beam equilibrium: 
 
1
2
1
2
1
2
70 kN/m 7 m (3.5 m)
70 kN/m 3 m 8 m
55 kN 10 m 7 m 0
 443.57 kN
70 kN/m 7 m
70 kN/m 3 m 55 kN
443.57 kN 70 kN/m 7 m
70 kN/m 3 m 55 kN 0
 206.43 kN
A
y
y
y y y
y
y
M
B
B
F A B
A
A
 
Shear-force and bending-moment diagrams 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
7.52 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 Fig. P7.52 
Solution 
Beam equilibrium: 
 
1
2
1
2
1
2
8 ft
6 kips/ft 8 ft
3
4 kips/ft 15 ft (14.5 ft) 17 ft 0
 47.41 kips
6 kips/ft 8 ft
4 kips/ft 15 ft
47.41 kips 6 kips/ft 8 ft
4 kips/ft 15 ft 0
 36.59 k
B
y
y
y y y
y
y
M
D
D
F B D
B
B ips
 
 
Shear-force and bending-moment diagrams 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
7.53 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.53 
Solution 
Beam equilibrium: 
 
1
2
1
2
(9 kips)(4 ft)
4 kips/ft 9 ft (13 ft) 0
 270.00 kip-ft
(9 kips) 4 kips/ft 9 ft 0
27.00 kips
A
A
A
y y
y
M
M
M
F A
A
 
 
 
Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
7.54 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 
 Fig. P7.54 
Solution 
Beam equilibrium: 
 
1
2
1
2
(25 kN)(6 m)
30 kN/m 3 m 3 m 0
 285.00 kN-m
25 kN 30 kN/m 3 m 0
 70 kN
D
D
D
y y
y
M
M
M
F D
D
 
 
 
Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.55 Use the graphical method to construct the 
shear-force and bending-moment diagrams for the 
beam shown. Label all significant points on each 
diagram and identify the maximum moments along 
with their respective locations. Clearly differentiate 
straight-line and curved portions of the diagrams. 
 Fig. P7.55 
Solution 
Beam equilibrium: 
1
2
1
2
1
2
(6 kips/ft)(22 ft)(11 ft)
2(8 ft)
(9 kips/ft)(8 ft) 22 ft (22 ft) 0
3
 110.73 kips(6 kips/ft)(22 ft) (9 kips/ft)(8 ft)
(110.73 kips) (6 kips/ft)(22 ft) (9 k
A
y
y
y y y
y
M
B
B
F A B
A ips/ft)(8 ft) 0
 57.27 kipsyA
 
Shear-force and bending-moment diagrams 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.56 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.56 
Solution 
Beam equilibrium: 
 
(180 lb)(2 ft) (450 lb)(6 ft) (9 ft) 0
 340 lb
180 lb 450 lb 0
 290 lb
A y
y
y y y
y
M D
D
F A D
A
 
 
Load function w(x): 
 
1 1 1 1
( ) 290 lb 0 ft 180 lb 2 ft 450 lb 6 ft 340 lb 9 ftw x x x x x
 Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 0 0 0
( ) 290 lb 0 ft 180 lb 2 ft 450 lb 6 ft 340 lb 9 ftV x x x x x
 Ans. 
 
1 1 1 1
( ) 290 lb 0 ft 180 lb 2 ft 450 lb 6 ft 340 lb 9 ftM x x x x x
 Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.57 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.57 
Solution 
Beam equilibrium: 
 
(10 kN)(2.5 m) (35 kN)(3 m) (5 m) 0
 16 kN
10 kN 35 kN 0
 29 kN
B y
y
y y y
y
M D
D
F B D
B
 
 
Load function w(x): 
 
1 1 1 1
( ) 10 kN 0 m 29 kN 2.5 m 35 kN 5.5 m 16 kN 7.5 mw x x x x x
 Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 0 0 0
( ) 10 kN 0 m 29 kN 2.5 m 35 kN 5.5 m 16 kN 7.5 mV x x x x x
 Ans. 
 
1 1 1 1
( ) 10 kN 0 m 29 kN 2.5 m 35 kN 5.5 m 16 kN 7.5 mM x x x x x
 Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.58 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.58 
Solution 
Beam equilibrium: 
 
(30 kN)(3 m) (20 kN)(7 m)
(15 kN)(15 m) (10 m) 0
 45.5 kN
30 kN 20 kN 15 kN 0
 19.5 kN
B
y
y
y y y
y
M
D
D
F A D
A
 
Load function w(x): 
 
1 1 1
1 1
( ) 19.5 kN 0 m 30 kN 3 m 20 kN 7 m
45.5 kN 10 m 15 kN 15 m
w x x x x
x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 0 0
0 0
( ) 19.5 kN 0 m 30 kN 3 m 20 kN 7 m
45.5 kN 10 m 15 kN 15 m
V x x x x
x x Ans. 
 
1 1 1
1 1
( ) 19.5 kN 0 m 30 kN 3 m 20 kN 7 m
45.5 kN 10 m 15 kN 15 m
M x x x x
x x Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.59 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.59 
Solution 
Beam equilibrium: 
 
5 kN 0
 5 kN
(5 kN)(6 m) 20 kN-m 0
 10 kN-m
y y
y
C C
C
F C
C
M M
M
 
 
Load function w(x): 
 
1 2 1 2
( ) 5 kN 0 m 20 kN-m 3 m 5 kN 6 m 10 kN-m 6 mw x x x x x
 Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 0 1
( ) 5 kN 0 m 20 kN-m 3 m 5 kN 6 m 10 kN-m 6 mV x x x x x
 Ans. 
 
1 0 1 0
( ) 5 kN 0 m 20 kN-m 3 m 5 kN 6 m 10 kN-m 6 mM x x x x x
 Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.60 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.60 
Solution 
Beam equilibrium: 
 
(35 kN/m)(2 m) 0
 70 kN
(35 kN/m)(2 m)(4 m) 0
 280 kN-m
y y
y
A A
A
F A
A
M M
M
 
 
Load function w(x): 
 
1 2 0 0
( ) 70 kN 0 m 280 kN-m 0 m 35 kN/m 3 m 35 kN/m 5 mw x x x x x
 Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 1 1
( ) 70 kN 0 m 280 kN-m 0 m 35 kN/m 3 m 35 kN/m 5 mV x x x x x
 Ans. 
 
1 0 2 235 kN/m 35 kN/m
( ) 70 kN 0 m 280 kN-m 0 m 3 m 5 m
2 2
M x x x x x
 Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.61 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.61 
Solution 
Beam equilibrium: 
(25 kN)(4 m)(2 m) (32 kN)(6 m)(8 m) 0
 49 kN
(25 kN)(4 m) 32 kN 0
 83 kN
A y
y
y y y
y
M D
D
F A D
A
 
 
Load function w(x): 
 
1 0 0
1 1
( ) 83 kN 0 m 25 kN/m 0 m 25 kN/m 4 m
32 kN 6 m 49 kN 8 m
w x x x x
x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 1
0 0
( ) 83 kN 0 m 25 kN/m 0 m 25 kN/m 4 m
32 kN 6 m 49 kN 8 m
V x x x x
x x Ans. 
 
1 2 2
1 1
25 kN/m 25 kN/m
( ) 83 kN 0 m 0 m 4 m
2 2
32 kN 6 m 49 kN 8 m
M x x x x
x x Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.62 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.62 
Solution 
Beam equilibrium: 
 
(3,000 lb)(5 ft) 8,000 lb-ft
(800 lb)(7 ft)(12.5 ft) (20 ft) 0
 3,850 lb
3,000 lb (800 lb)(7 ft) 0
 4,750 lb
A
y
y
y y y
y
M
E
E
F A E
A
 
Load function w(x): 
 
1 1 2
0 0 1
( ) 4,750 lb 0 ft 3,000 lb 5 ft 8,000 lb-ft 5 ft
800 lb/ft 9 ft 800 lb/ft 16 ft 3,850 lb 20 ft
w x x x x
x x x Ans. 
Shear-force function V(x) and bending-moment function M(x): 
 
0 0 1
1 1 0
( ) 4,750 lb 0 ft 3,000 lb 5 ft 8,000 lb-ft 5 ft
800 lb/ft 9 ft 800 lb/ft 16 ft 3,850 lb 20 ft
V x x x x
x x x Ans. 
 
1 1 0
2 2 1
( ) 4,750 lb 0 ft 3,000 lb 5 ft 8,000 lb-ft 5 ft
800 lb/ft 800 lb/ft
9 ft 16 ft 3,850 lb 20 ft
2 2
M x x x x
x x x Ans. 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.63 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.63 
Solution 
Beam equilibrium: 
 
(800 lb/ft)(12 ft) (800 lb)(6 ft) 0
 14,400 lb
(800 lb/ft)(12 ft)(6 ft)
(800 lb)(6 ft)(21 ft) 0
 158,400 lb-ft
y y
y
A
A
A
F A
A
M
M
M
 
 
Load function w(x): 
 1 2 0
0 0 0
( ) 14,400 lb 0 ft 158,400 lb-ft 0 ft 800 lb-ft 0 ft
800 lb/ft 12 ft 800 lb/ft 18 ft 800 lb/ft 24 ft
w x x x x
x x x
 Ans. 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 1
1 1 1
( ) 14,400 lb 0 ft 158,400 lb-ft 0 ft 800 lb-ft 0 ft
800 lb/ft 12 ft 800 lb/ft 18 ft 800 lb/ft 24 ft
V x x x x
x x x Ans. 
 
1 0 2
2 2 2
800 lb-ft
( ) 14,400 lb 0 ft 158,400 lb-ft 0 ft 0 ft
2
800 lb/ft 800 lb/ft 800 lb/ft
12 ft 18 ft 24 ft
2 2 2
M x x x x
x x x Ans. 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.64 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.64 
Solution 
Beam equilibrium: 
12 kN-m (18 kN/m)(2 m)(2 m) (5 m) 0
 12 kN
(18 kN/m)(2 m) 0
 24 kN
A y
y
y y y
y
M D
D
F A D
A
 
Load function w(x): 
 
1 2 0
0 1
( ) 24 kN 0 m 12 kN-m 0 m 18 kN/m 1 m
18 kN/m 3 m 12 kN 5 m
w x x x x
x x Ans. 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 1
1 0
( ) 24 kN 0 m 12 kN-m 0 m 18 kN/m 1 m
18 kN/m 3 m 12 kN 5 m
V x x x x
x x Ans. 
 
1 0 2
2 1
18 kN/m
( ) 24 kN 0 m 12 kN-m 0 m 1 m
2
18 kN/m
3 m 12 kN 5 m
2
M x x x x
x x Ans. 
Shear-force and bending-moment diagrams: 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.65 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.65 
Solution 
Beam equilibrium: 
1
2
1
2
1
2
(6 kips/ft)(22 ft)(11 ft)
2(8 ft)
(9 kips/ft)(8 ft) 22 ft (22 ft) 0
3
 110.73 kips
(6 kips/ft)(22 ft) (9 kips/ft)(8 ft)
(110.73 kips) (6 kips/ft)(22 ft) (9 k
A
y
y
y y y
y
M
B
B
F A B
A ips/ft)(8 ft) 0
 57.27 kipsyA
 
 Load function w(x): 
 
1 0 1 0
1 1 0
( ) 57.27 kips 0 ft 6 kips/ft 0 ft 110.73 kips 22 ft 6 kips/ft 22 ft
9 kips/ft 9 kips/ft
22 ft 30 ft 9 kips/ft 30 ft
8 ft 8 ft
w x x x x x
x x xAns. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 0 1
2 2 1
( ) 57.27 kips 0 ft 6 kips/ft 0 ft 110.73 kips 22 ft 6 kips/ft 22 ft
9 kips/ft 9 kips/ft
22 ft 30 ft 9 kips/ft 30 ft
2(8 ft) 2(8 ft)
V x x x x x
x x x Ans. 
 
1 2 1 2
3 3 2
6 kips/ft 6 kips/ft
( ) 57.27 kips 0 ft 0 ft 110.73 kips 22 ft 22 ft
2 2
9 kips/ft 9 kips/ft 9 kips/ft
22 ft 30 ft 30 ft
6(8 ft) 6(8 ft) 2
M x x x x x
x x x Ans. 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Shear-force and bending-moment diagrams 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.66 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Use V(x) and M(x) to plot the shear-force and 
bending-moment diagrams. 
 
 Fig. P7.66 
Solution 
Beam equilibrium: 
1
2
1
2
(20 kN/m)(3 m) (30 kN/m)(3 m) 0
 105 kN
(20 kN/m)(3 m)(2.5 m)
(30 kN/m)(3 m)(2 m) 0
 240 kN-m
y y
y
C
C
C
F C
C
M
M
M
 
 
Load function w(x): 
 
0 1 0 1
0 1 2
30 kN/m 30 kN/m
( ) 20 kN/m 0 m 0 m 20 kN/m 3 m 3 m
3 m 3 m
30 kN/m 3 m 105 kN 4 m 240 kN-m 4 m
w x x x x x
x x x Ans. 
Shear-force function V(x) and bending-moment function M(x): 
 
1 2 1 2
1 0 1
30 kN/m 30 kN/m
( ) 20 kN/m 0 m 0 m 20 kN/m 3 m 3 m
2(3 m) 2(3 m)
30 kN/m 3 m 105 kN 4 m 240 kN-m 4 m
V x x x x x
x x x Ans. 
 
2 3 2 3
2 1 0
20 kN/m 30 kN/m 20 kN/m 30 kN/m
( ) 0 m 0 m 3 m 3 m
2 6(3 m) 2 6(3 m)
30 kN/m
3 m 105 kN 4 m 240 kN-m 4 m
2
M x x x x x
x x x Ans. 
Shear-force and bending-moment diagrams: 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
7.67 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Determine the maximum bending moment in 
the beam between the two simple supports. 
 
 Fig. P7.67 
Solution 
Beam equilibrium: 1
2
1
2
9 kN-m (18 kN/m)(3 m)(1 m) (3 m) 0
 6 kN
(18 kN/m)(3 m) 0
 21 kN
B y
y
y y y
y
M C
C
F B C
B
 
Load function w(x): 
 
2 1 0
1 1 1
( ) 9 kN-m 0 m 21 kN 1 m 18 kN/m 1 m
18 kN/m 18 kN/m
1 m 4 m 6 kN 4 m
3 m 3 m
w x x x x
x x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
1 0 1
2 2 0
( ) 9 kN-m 0 m 21 kN 1 m 18 kN/m 1 m
18 kN/m 18 kN/m
1 m 4 m 6 kN 4 m
2(3 m) 2(3 m)
V x x x x
x x x Ans. 
 
0 1 2
3 3 1
18 kN/m
( ) 9 kN-m 0 m 21 kN 1 m 1 m
2
18 kN/m 18 kN/m
1 m 4 m 6 kN 4 m
6(3 m) 6(3 m)
M x x x x
x x x Ans. 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Maximum bending moment: 
 Mmax = 5.66 kN-m at x = 2.59 m Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.68 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Determine the maximum bending moment in 
the beam between the two simple supports. 
 
 Fig. P7.68 
Solution 
Beam equilibrium: 1
2
1
2
(5 kips/ft)(9 ft)(6 ft) (14 ft) 0
 9.64 kips
(5 kips/ft)(9 ft) 0
 12.86 kips
A y
y
y y y
y
M C
C
F A C
A
 
 
Load function w(x): 
 
1 1 1
0 1
5 kips/ft 5 kips/ft
( ) 12.86 kips 0 ft 0 ft 9 ft
9 ft 9 ft
5 kips/ft 9 ft 9.64 kips 14 ft
w x x x x
x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 2 2
1 0
5 kips/ft 5 kips/ft
( ) 12.86 kips 0 ft 0 ft 9 ft
2(9 ft) 2(9 ft)
5 kips/ft 9 ft 9.64 kips 14 ft
V x x x x
x x Ans. 
 
1 3 3
2 1
5 kips/ft 5 kips/ft
( ) 12.86 kips 0 ft 0 ft 9 ft
6(9 ft) 6(9 ft)
5 kips/ft
9 ft 9.64 kips 14 ft
2
M x x x x
x x Ans. 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Maximum bending moment: 
 Mmax = 58.3 kip-ft at x = 6.80 ft Ans. 
 
 
Shear-force and bending-moment diagrams 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.69 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Determine the maximum bending moment in 
the beam between the two simple supports. 
 
 Fig. P7.69 
Solution 
Beam equilibrium: 
1
2
1
2
(5 kips/ft)(6 ft)(3 ft)
21 ft
(9 kips/ft)(21 ft) 6 ft (16 ft) 0
3
 82.41 kips
(5 kips/ft)(6 ft) (9 kips/ft)(21 ft) 0
 42.09 kips
A
y
y
y y y
y
M
C
C
F A C
A
 
 
Load function w(x): 
 
1 0 0 0
1 1 1
( ) 42.09 kips 0 ft 5 kips/ft 0 ft 5 kips/ft 6 ft 9 kips/ft 6 ft
9 kips/ft 9 kips/ft
6 ft 82.41 kips 16 ft 27 ft
21 ft 21 ft
w x x x x x
x x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 1 1
2 0 2
( ) 42.09 kips 0 ft 5 kips/ft 0 ft 5 kips/ft 6 ft 9 kips/ft 6 ft
9 kips/ft 9 kips/ft
6 ft 82.41 kips 16 ft 27 ft
2(21 ft) 2(21 ft)
V x x x x x
x x x Ans. 
 
1 2 2 2
3 1 3
5 kips/ft 5 kips/ft 9 kips/ft
( ) 42.09 kips 0 ft 0 ft 6 ft 6 ft
2 2 2
9 kips/ft 9 kips/ft
6 ft 82.41 kips 16 ft 27 ft
6(21 ft) 6(21 ft)
M x x x x x
x x x Ans. 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Maximum bending moment: 
 Mmax = 170.9 kip-ft at x = 7.39 ft Ans. 
 
 
Shear-force and bending-moment diagrams 
 
 
 
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to students enrolled in courses for which the textbookhas been adopted. Any other reproduction or translation of this work beyond that 
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7.70 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Determine the maximum bending moment in 
the beam between the two simple supports. 
 
 Fig. P7.70 
Solution 
Beam equilibrium: 
1
2
1
2
(25 kN/m)(4.0 m)(4.5 m)
2(4.0 m)
(45 kN/m)(4.0 m) 2.5 m (8 m) 0
3
 114.38 kN
(25 kN/m)(4.0 m) (45 kN/m)(4.0 m) 0
 75.63 kN
A
y
y
y y y
y
M
D
D
F A D
A
 
Load function w(x): 
 
1 0 0 1
1 0 1
45 kN/m
( ) 75.63 kN 0 m 25 kN/m 2.5 m 25 kN/m 6.5 m 2.5 m
4.0 m
45 kN/m
6.5 m 45 kN/m 6.5 m 114.38 kN 8 m
4.0 m
w x x x x x
x x xAns. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 1 1 2
2 1 0
45 kN/m
( ) 75.63 kN 0 m 25 kN/m 2.5 m 25 kN/m 6.5 m 2.5 m
2(4.0 m)
45 kN/m
6.5 m 45 kN/m 6.5 m 114.38 kN 8 m
2(4.0 m)
V x x x x x
x x x Ans. 
 
1 2 2 3
3 2 1
25 kN/m 25 kN/m 45 kN/m
( ) 75.63 kN 0 m 2.5 m 6.5 m 2.5 m
2 2 6(4.0 m)
45 kN/m 45 kN/m
6.5 m 6.5 m 114.38 kN 8 m
6(4.0 m) 2
M x x x x x
x x x Ans. 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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Maximum bending moment: 
 Mmax = 275 kN-m at x = 4.57 m Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.71 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Determine the maximum bending moment in 
the beam between the two simple supports. 
 
 Fig. P7.71 
Solution 
Beam equilibrium: 
1
2
1
2
2(7.0 m)
(30 kN/m)(7.0 m)(3.5 m) (40 kN/m)(7.0 m)
3
(50 kN/m)(2.0 m)(1.0 m) (5.5 m) 0
 234.24 kN
(30 kN/m)(7.0 m) (40 kN/m)(7.0 m)
(50 kN/m)(2.0 m) 0
 215.
C
y
y
y y y
y
M
B
B
F B C
C 76 kN
 
Load function w(x): 
 
0 0 1
1 0 1
1 0 0
40 kN/m
( ) 30 kN/m 0 m 40 kN/m 0 m 0 m
7.0 m
40 kN/m
234.24 kN 1.5 m 30 kN/m 7 m 7 m
7.0 m
215.76 kN 7 m 50 kN/m 7.0 m 50 kN/m 9.0 m
w x x x x
x x x
x x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
1 1 2
0 1 2
0 1 1
40 kN/m
( ) 30 kN/m 0 m 40 kN/m 0 m 0 m
2(7.0 m)
40 kN/m
234.24 kN 1.5 m 30 kN/m 7 m 7 m
2(7.0 m)
215.76 kN 7 m 50 kN/m 7.0 m 50 kN/m 9.0 m
V x x x x
x x x
x x x Ans. 
 
2 2 3
1 2 3
1 2 2
30 kN/m 40 kN/m 40 kN/m
( ) 0 m 0 m 0 m
2 2 6(7.0 m)
30 kN/m 40 kN/m
234.24 kN 1.5 m 7 m 7 m
2 6(7.0 m)
50 kN/m 50 kN/m
215.76 kN 7 m 7.0 m 9.0 m
2 2
M x x x x
x x x
x x x Ans. 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Maximum bending moment: 
 Mmax = 86.6 kN-m at x = 4.00 m Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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7.72 For the beam and loading shown, 
(a) Use discontinuity functions to write the 
expression for w(x). Include the beam reactions in 
this expression. 
(b) Integrate w(x) to twice to determine V(x) and 
M(x). 
(c) Determine the maximum bending moment in 
the beam between the two simple supports. 
 
 Fig. P7.72 
Solution 
Beam equilibrium: 
 
1
2
1
2
2(4.5 m)
(60 kN)(1.5 m) (90 kN/m)(4.5 m)
3
(6.5 m) 0
 79.62 kN
60 kN (90 kN/m)(4.5 m) 0
 182.88 kN
B
y
y
y y y
y
M
D
D
F B D
B
 
 
Load function w(x): 
 
1 1 1
1 0 1
90 kN/m
( ) 60 kN 0 m 182.88 kN 1.5 m 1.5 m
4.5 m
90 kN/m
6 m 90 kN/m 6 m 79.62 kN 8 m
4.5 m
w x x x x
x x x Ans. 
 
Shear-force function V(x) and bending-moment function M(x): 
 
0 0 2
2 1 0
90 kN/m
( ) 60 kN 0 m 182.88 kN 1.5 m 1.5 m
2(4.5 m)
90 kN/m
6 m 90 kN/m 6 m 79.62 kN 8 m
2(4.5 m)
V x x x x
x x x Ans. 
 
1 1 3
3 2 1
90 kN/m
( ) 60 kN 0 m 182.88 kN 1.5 m 1.5 m
6(4.5 m)
90 kN/m 90 kN/m
6 m 6 m 79.62 kN 8 m
6(4.5 m) 2
M x x x x
x x x Ans. 
 
 
 
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Maximum bending moment: 
 Mmax = 197.2 kN-m at x = 5.01 m Ans. 
 
Shear-force and bending-moment diagrams: 
 
 
 
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8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E = 
1,900 ksi] planks is bent to a radius of curvature of 40 ft. Determine the maximum bending stress 
developed in the plank. 
Solution 
From Eq. (8.3): 
 
1,900 ksi
( 0.5 in.) 1.979 ksi
(40 ft)(12 in./f
1.979 k
t
si
)
x
E
y        
 Ans. 
 
 
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8.2 A high-strength steel [E = 200 GPa] tube having anoutside diameter of 80 mm and a wall thickness 
of 3 mm is bent into a circular curve having a 52-m radius of curvature. Determine the maximum 
bending stress developed in the tube. 
Solution 
From Eq. (8.3): 
 
200,000 MPa
( 80 mm / 2) 153.846 MPa
(52 m)(1,000 mm/
153.8 MP
m)
ax
E
y        
 Ans.
 
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8.3 A high-strength steel [E = 200 GPa] band saw blade wraps around a pulley that has a diameter of 
450 mm. Determine the maximum bending stress developed in the blade. The blade is 12-mm wide and 
1-mm thick. 
Solution 
The radius of curvature of the band saw blade is: 
 
450 mm 1 mm
225.5 mm
2 2
   
 
From Eq. (8.3): 
 
200,000 MPa
( 0.5 mm) 443.459 MPa
225.5 mm
443 MPax
E
y        
 Ans.
 
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8.4 The boards for a concrete form are to be bent into a circular shape having an inside radius of 10 m. 
What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa? 
Assume that the modulus of elasticity for the wood is 12 GPa. 
Solution 
The radius of curvature of the concrete form is dependent on the board thickness: 
 
10,000 mm
2
t  
 
From Eq. (8.3): 
 12,000 MPa
7 MPa
2
10,000 mm
2
x
E t
y
t
 
 
        
  
 
Solve for t: 
 
12,000 MPa 7 MPa 10,000 mm
2 2
6,000 70,000 3.5
5,996.5
11.67 mm
70,000
t t
t t
t
t
   
       
 

  Ans.
 
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8.5 A beam having a tee-shaped cross section is subjected to equal 12 kN-m bending moments, as 
shown in Fig. P8.5a. The cross-sectional dimensions of the beam are shown in Fig. P8.5b. Determine: 
(a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus 
about the z axis. 
(b) the bending stress at point H. State whether the normal stress at H is tension or compression. 
(c) the maximum bending stress produced in the cross section. State whether the stress is tension or 
compression. 
 
 
Fig. P8.5a Fig. P8.5b 
Solution 
(a) Centroid location in y direction: (reference axis at bottom of tee shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (mm
2
) (mm) (mm
3
) 
top flange 2,500.0 162.5 406,250.0 
stem 3,750.0 75.0 281,250.0 
 6,250.0 mm
2
 
 
687,500.0 mm
3 
 
 3
2
687,500.0 mm
6,250.0 mm
110.0 mm
i i
i
y A
y
A

  

 (measured upward from bottom edge of stem) Ans. 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 130,208.33 52.50 6,890,625.00 7,020,833.33 
stem 7,031,250.00 −35.00 4,593,750.00 11,625,000.00 
Moment of inertia about the z axis (mm
4
) = 18,645,833.33 
 
418,656,000 mmzI 
 Ans. 
 
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Section moduli: 
 
4
3
top
top
4
3
bot
bot
3
18,645,833.33 mm
286,858.974 mm
(175 mm 110 mm)
18,645,833.33 mm
169,507.576 mm
110 mm
169,500 mm
z
z
I
S
c
I
S
c
S
  

  
  Ans. 
 
(b) Bending stress at point H: (y = 175 mm − 25 mm − 110 mm = 40 mm) 
 
4
(12 kN-m)(40 mm)(1,000 N/kN)(1,000 mm/m)
18,654,833.33 mm
25.743 MPa 25.7 MPa (C)
x
z
M y
I
  
 
   Ans. 
 
(c) Maximum bending stress: 
The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the 
cross section is at y = +65 mm, and the bottom of the cross section is at y = −110 mm. The larger 
bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom 
of the cross section. 
 
4
(12 kN-m)( 110 mm)(1,000 N/kN)(1,000 mm/m)
18,654,833.33 mm
70.793 MPa 70.8 MPa (T)
x
z
M y
I
  

 
  Ans. 
 
 
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8.6 A beam is subjected to equal 6.5 kip-ft bending moments, as shown in Fig. P8.6a. The cross-
sectional dimensions of the beam are shown in Fig. P8.6b. Determine: 
(a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus 
about the z axis. 
(b) the bending stress at point H, which is located 2 in. below the z centroidal axis. State whether the 
normal stress at H is tension or compression. 
(c) the maximum bending stress produced in the cross section. State whether the stress is tension or 
compression. 
 
 
Fig. P8.6a Fig. P8.6b 
Solution 
(a) Centroid location in y direction: (reference axis at bottom of shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
left side 8.0 4.0 32.0 
top flange 4.0 7.5 30.0 
right side 8.0 4.0 32.0 
 20.0 in.
2
 
 
94.0 in.
3 
 
 3
2
94.0 i
4.70
n.
20.0 i
 i
n
n
.
.
i i
i
y A
y
A

  

 (measured upward from bottom edge of section) Ans. 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left side 42.667 −0.700 3.920 46.587 
top flange 0.333 2.800 31.360 31.693 
right side 42.667 −0.700 3.920 46.587 
Moment of inertia about the z axis (in.
4
) = 124.867 
 
4124.9 in.zI 
 Ans. 
 
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Section moduli: 
 
4
3
top
top4
3
bot
bot
3
124.867 in.
37.8384 in.
(8 in. 4.7 in.)
124.867 in.
26.5674 in.
4.7 in
26.6 in.
.
z
z
I
S
c
I
S
c
S
  

  
  Ans. 
 
(b) Bending stress at point H: (y = −2 in.) 
 
4
( 6.5 kip-ft)( 2 in.)(12 in./ft)
124.867 in.
1,249 p 1,249 psi ( )i Cs
x
z
M y
I
  
 
 
   Ans. 
 
(c) Maximum bending stress: 
The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the 
cross section is at y = +3.30 in., and the bottom of the cross section is at y = −4.7 in. The larger bending 
stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the 
cross section. 
 
4
( 6.5 kip-ft)( 4.7 in.)(12 in./ft)
124.867 in
2,940 psi (
.
2,935. i C)9 ps
x
z
M y
I
  
 
 
   Ans.
 
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8.7 A beam is subjected to equal 470 N-m bending moments, as shown in Fig. P8.7a. The cross-
sectional dimensions of the beam are shown in Fig. P8.7b. Determine: 
(a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus 
about the z axis. 
(b) the bending stress at point H. State whether the normal stress at H is tension or compression. 
(c) the maximum bending stress produced in the cross section. State whether the stress is tension or 
compression. 
 
Fig. P8.7a Fig. P8.7b 
Solution 
(a) Centroid location in y direction: (reference axis at bottom of U shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (mm
2
) (mm) (mm
3
) 
left side 400.0 25.0 10,000.0 
bottom flange 272.0 4.0 1,088.0 
right side 400.0 25.0 10,000.0 
 1,072.0 mm
2
 
 
21,088.0 mm
3 
 
 3
2
21,088.0 mm
1,072.0 mm
19.67 mm
i i
i
y A
y
A

  

 (measured upward from bottom edge of section) Ans. 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
left side 83,333.33 5.33 11,356.56 94,689.89 
bottom flange 1,450.67 −15.67 66,803.30 68,253.96 
right side 83,333.33 5.33 11,356.56 94,689.89 
Moment of inertia about the z axis (mm
4
) = 257,633.75 
 
4257,600 mmzI 
 Ans. 
 
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Section moduli: 
 
4
3
top
top
4
3
bot
bot
3
257,633.75 mm
8,494.814 mm
(50 mm 19.672 mm)
257,633.75 mm
13,096.708 mm
19.672 
8,495 mm
mm
z
z
I
S
c
I
S
c
S
  

  
  Ans. 
 
(b) Bending stress at point H: (y = 8 mm − 19.672 mm = −11.672 mm) 
 
4
(470 N-m)( 11.672 mm)(1,000 mm/m)
257,633.75 m
21.
m
21 3 MPa.293 (T) MPa
x
z
M y
I
  

 
  Ans. 
 
(c) Maximum bending stress: 
The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the 
cross section is at y = +30.328 mm, and the bottom of the cross section is at y = −19.672 mm. The larger 
bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of 
the cross section. 
 
4
(470 N-m)(30.328 mm)(1,000 mm/m)
257,633.7
55.3 MPa (
5 mm
55.328 C) MPa
x
z
M y
I
  
 
   Ans.
 
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8.8 A beam is subjected to equal 17.5 kip-ft bending moments, as shown in Fig. P8.8a. The cross-
sectional dimensions of the beam are shown in Fig. P8.8b. Determine: 
(a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus 
about the z axis. 
(b) the bending stress at point H. State whether the normal stress at H is tension or compression. 
(c) the bending stress at point K. State whether the normal stress at K is tension or compression. 
(d) the maximum bending stress produced in the cross section. State whether the stress is tension or 
compression. 
 
 
Fig. P8.8a Fig. P8.8b 
Solution 
(a) Centroid location in y direction: (reference axis at bottom of shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
top flange 12.0000 13.0000 156.0000 
web 20.0000 7.0000 140.0000 
bottom flange 20.0000 1.0000 20.0000 
 52.0000 in.
2
 
 
316.0000 in.
3 
 
 3
2
316.0 in.
6.077 in.
52.0
6.08 in.
 in.
i i
i
y A
y
A

   

 (measured upward from bottom edge of bottom 
flange) Ans. 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 4.000 6.923 575.148 579.148 
web 166.667 0.923 17.041 183.708 
bottom flange 6.667 -5.077 515.503 522.170 
Moment of inertia about the z axis (in.
4
) = 1,285.026 
 Ans.
 
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Section Moduli 
4
3
4
3
6.0769 in.
14 in. 6.0769 in. 7.9231 in.
1,285.026 in.
211.460 in.
6.0769 in.
1,285.026 in.
162.188 in.
7.9231 in.
The controlling section modulus is the smaller of the 
bot
top
z
bot
bot
z
top
top
c
c
I
S
c
I
S
c

  
  
  
3
two values; theref
162.2 
ore,
in.S  Ans. 
 
Bending stress at point H: 
From the flexure formula: 
 
4
( 17.5 kip-ft)(7.9231 in. 2 in.)(12 in./ft)
967.9544 psi
1,285.0256 i
968 psi (T)
n.
x
z
M y
I
        Ans. 
 
Bending stress at point K: 
From the flexure formula: 
 
4
( 17.5 kip-ft)( 6.0769 in. 2 in.)(12 in./ft)
666.2543 psi
1,285.026 in
666 psi 
.
(C)x
z
M y
I
          Ans. 
 
Maximum bending stress 
Since ctop > cbot, the maximum bending stress occurs at the top of the flanged shape. From the flexure 
formula: 
 
4
( 17.5 kip-ft)(7.9231 in.)(12 in./ft)
1,294.8 psi
1,285.02
1,295 psi 
6 n.
(T)
i
x
z
M y
I
       Ans. 
 
Also, note that the same maximum bending stress magnitude can be calculated with the section 
modulus: 
 
3
(17.5 kip-ft)(12 in./ft)
1,294.8 psi
162.
1,295 ps
1877 in.
i
 
x
M
S
     Ans. 
The sense of the stress (either tension or compression)would be determined by inspection. 
 
 
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8.9 The cross-sectional dimensions of a beam are 
shown in Fig. P8.9. 
(a) If the bending stress at point K is 43 MPa (C), 
determine the internal bending moment Mz acting 
about the z centroidal axis of the beam. 
(b) Determine the bending stress at point H. State 
whether the normal stress at H is tension or 
compression. 
 
 Fig. P8.9 
Solution 
Centroid location in y direction: (reference axis at bottom of double-tee shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (mm
2
) (mm) (mm
3
) 
top flange 375.0 47.5 17,812.5 
left stem 225.0 22.5 5,062.5 
right stem 225.0 22.5 5,062.5 
 825.0 mm
2
 
 
27,937.5 mm
3 
 3
2
27,937.5 mm
33.864 mm 33.9 mm
825.0 mm
i i
i
y A
y
A

   

 (measured upward from bottom of section) 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 781.250 13.636 69,731.405 70,512.655 
left stem 37,968.750 −11.364 29,054.752 67,023.502 
right stem 37,968.750 −11.364 29,054.752 67,023.502 
Moment of inertia about the z axis (mm
4
) = 204,559.659 
(a) Determine bending moment: 
At point K, y = 50 mm − 5 mm − 33.864 mm = 11.136 mm. The bending stress at K is x = −43 MPa; 
therefore, the bending moment magnitude can be determined from the flexure formula: 
 
2 4( 43 N/mm )(204,559.659 mm )
11.136 mm
789,850.765 N- 790 N-mmm
x
z
x z
M y
I
I
M
y


 

    
  Ans. 
 
(b) Bending stress at point H: 
At point H, y = −33.864 mm. The bending stress is computed with the flexure formula: 
 
4
(789,850.765 N-mm)( 33.864 mm)
130.755 MPa
2
130.8 MPa (
04,559.659 mm
T)x
z
M y
I
       Ans. 
 
 
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8.10 The cross-sectional dimensions of a beam are 
shown in Fig. P8.10. 
(a) If the bending stress at point K is 2,600 psi (T), 
determine the internal bending moment Mz acting 
about the z centroidal axis of the beam. 
(b) Determine the bending stress at point H. State 
whether the normal stress at H is tension or 
compression. 
 
 Fig. P8.10 
Solution 
Centroid location in y direction: (reference axis at bottom of inverted-tee shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
bottom flange 0.56250 0.12500 0.07031 
stem 0.56250 1.37500 0.77344 
 1.12500 in.
2
 
 
0.84375 in.
3 
 3
2
0.84375 in.
0.750 in.
1.1250 in.
i i
i
y A
y
A

  

 (measured upward from bottom edge of section) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
bottom flange 0.00293 −0.62500 0.21973 0.22266 
stem 0.23730 0.62500 0.21973 0.45703 
Moment of inertia about the z axis (in.
4
) = 0.67969 
(a) Determine bending moment: 
At point K, y = 2.50 in. − 0.75 in. = 1.750 in. The bending stress at K is x = +2,600 psi; therefore, the 
bending moment magnitude can be determined from the flexure formula: 
 
4(2,600 psi)(0.67967 in. )
1.750 in.
1,009.820 lb-in. 1,010 lb-in. 84.2 lb-ft
x
z
x z
M y
I
I
M
y


 
    
     Ans. 
 
(b) Bending stress at point H: 
At point H, y = −0.75 in. The bending stress is computed with the flexure formula: 
 
4
( 1,009.820 lb-in.)( 0.75 in.)
1,114.286 psi
0
1,114 psi 
.67969 i
(C)
n.
x
z
M y
I
         Ans. 
 
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8.11 The cross-sectional dimensions of a box-
shaped beam are shown in Fig. P8.11. If the 
maximum allowable bending stress is b = 
15,000 psi, determine the maximum internal 
bending moment Mz magnitude that can be 
applied to the beam. 
 
 Fig. P8.11 
Solution 
Moment of inertia about z axis: 
 3 3
4(3 in.)(2 in.) (2.5 in.)(1 in.) 1.791667 in.
12 12
zI   
 
 
Maximum internal bending moment Mz: 
 
4(15,000 psi)(1.791667 in. )
26,875 lb-in.
1 in.
2,240 lb-ft
z
x
z
x z
M c
I
I
M
c



     Ans. 
 
 
 
 
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8.12 The cross-sectional dimensions of a beam are 
shown in Fig. P8.12. The internal bending moment 
about the z centroidal axis is Mz = +2.70 kip-ft. 
Determine: 
(a) the maximum tension bending stress in the 
beam. 
(b) the maximum compression bending stress in the 
beam. 
 
 Fig. P8.12 
Solution 
Centroid location in y direction: (reference axis at bottom of shape) 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
left stem 2.000 2.000 4.000 
top flange 2.500 3.750 9.375 
right stem 2.000 2.000 4.000 
 6.500 in.
2
 
 
17.375 in.
3 
 3
2
17.375 in.
2.673 in.
6.500 in.
i i
i
y A
y
A

  

 
 (measured upward from bottom edge of section) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left stem 2.66667 −0.67308 0.90607 3.57273 
top flange 0.05208 1.07692 2.89941 2.95149 
right stem 2.66667 −0.67308 0.90607 3.57273 
Moment of inertia about the z axis (in.
4
) = 10.09696 
 
(a) Determine maximum tension bending stress: 
For a positive bending moment, tension bending stresses will be created below the neutral axis. 
Therefore, the maximum tension bending stress will occur at point K (i.e., y = −2.673 in.): 
 
4
(2.70 kip-ft)( 2.673 in.)(12 in./ft)
8.578 ksi
10.09696 in.
8.58 ksi (T)x
z
M y
I
       Ans. 
 
(b) Determine maximum compression bending stress: 
For a positive bending moment, compression bending stresses will be created above the neutral axis. 
Therefore, the maximum compression bending stress will occur at point H (i.e., y = 4 in. − 2.673 in. = 
1.327 in.): 
 
4
(2.70 kip-ft)(1.327 in.)(12 in./ft)
4.258 ksi
10.0969
4.26 ksi
6 in.
 (C)x
z
M y
I
        Ans. 
 
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8.13 The cross-sectional dimensions of a beam are 
shown in Fig. P8.13. 
(a) If the bending stress at point K is 35.0 MPa (T), 
determine the bending stress at point H. State 
whether the normal stress at H is tension or 
compression. 
(b) If the allowable bending stress is b = 165 MPa, 
determine the magnitude of the maximum bending 
moment Mz that can be supported by the beam. 
 
 Fig. P8.13 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi –
y
 d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 540,000.000 160.000 184,320,000.000 184,860,000.000 
web 32,518,666.667 0.000 0.000 32,518,666.667 
bottom flange 540,000.000 −160.000 184,320,000.000 184,860,000.000 
Moment of inertia about the z axis (mm
4
) = 402,238,666.667 
 
(a) At point K, y = −90 mm, and at point H, y = −175 mm. The bending stress at K is x = +35 MPa, and 
the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending 
stress at H can be found from the ratio: 
 
175 mm
(35.0 MPa) 68.056 MPa
90
6
 
8.1 MPa (T)
mm
H K
H K
H
H K
K
y y
y
y
 
 


    

 Ans. 
 
(b) Maximum internal bending moment Mz: 
 
2 4(165 N/mm )(402,238,667 mm )
379,253,600 N-mm
175 m
3 N
m
79 k -m
z
x
z
x z
z
M c
I
I
M
c



     Ans. 
 
 
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8.14 The cross-sectional dimensions of a beam are 
shown in Fig. P8.14. 
(a) If the bending stress at point K is 9.0 MPa (T), 
determine the bending stress at point H. State 
whether the normal stress at H is tension or 
compression. 
(b) If the allowable bending stress is b = 165 MPa, 
determine the magnitude of the maximum bending 
moment Mz that can be supported by the beam. 
 
 Fig. P8.14 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi –
y
 d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
left flange 9,720,000 0 0 9,720,000 
web 31,680 0 0 31,680 
right flange 9,720,000 0 0 9,720,000 
Moment of inertia about the z axis (mm
4
) = 19,471,680 
 
(a) At point K, y = −60 mm, and at point H, y = +90 mm. The bending stress at K is x = +9.0 MPa, and 
the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending 
stress at H can be found from the ratio: 
 
90 mm
(9.0 MP 13.50 a) 13.50 MPa
60 m
MPa (C
m
)
H K
H K
H
H K
K
y y
y
y
 
 

     

 Ans. 
 
(b) Maximum bending moment Mz: 
 
2 4(165 N/mm )(19,471,680 mm )
35,698,080 N-mm
90 mm
35.7 kN-m
z
x
z
x z
M c
I
I
M
c



     Ans. 
 
 
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8.15 The cross-sectional dimensions of a beam are 
shown in Fig. P8.15. The internal bending moment 
about the z centroidal axis is Mz = −1.55 kip-ft. 
Determine: 
(a) the maximum tension bending stress in the beam. 
(b) the maximum compression bending stress in the 
beam. 
 
 Fig. P8.15 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
top flange 8.0 4.5 36.0 
left web 3.0 2.5 7.5 
left bottom flange 3.0 0.5 1.5 
right web 3.0 2.5 7.5 
right bottom flange 3.0 0.5 1.5 
 20.0 in.
2
 
 
54.0 in.
3 
 
 3
2
54.0 in.
2.70 in.
20.0 in.
i i
i
y A
y
A

  

 (measured upward from bottom edge of bottom flange) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 0.6667 1.8000 25.9200 26.5867 
left web 2.2500 −0.2000 0.1200 2.3700 
left bottom flange 0.2500 −2.2000 14.5200 14.7700 
right web 2.2500 −0.2000 0.1200 2.3700 
right bottom flange 0.2500 −2.2000 14.5200 14.7700 
Moment of inertia about the z axis (in.
4
) = 60.8667 
 
(a) Maximum tension bending stress: 
For a negative bending moment, the maximum tension bending stress will occur at the top surface of the 
cross section. From the flexure formula, the bending stress at the top surface is: 
 
4
( 1.55 kip-ft)(5.0 in. 2.70 in.)(12 in./ft)
0.7028 ksi
60.8667 i
703 p
n
si T)
.
 (x
z
M y
I
        Ans. 
 
(b) Maximum compression bending stress: 
The maximum compression bending stress will occur at the bottom surface of the cross section. From 
the flexure formula, the bending stress at the bottom surface is: 
 
4
( 1.55 kip-ft)( 2.70 in.)(12 in./ft)
0.8251 ksi
60.8667 in.
825 psi (C)x
z
M y
I
         Ans. 
 
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8.16 The cross-sectional dimensions of a beam are 
shown in Fig. P8.16. The internal bending moment 
about the z centroidal axis is Mz = +270 lb-ft. 
Determine: 
(a) the maximum tension bending stress in the beam. 
(b) the maximum compression bending stress in the 
beam. 
 
Solution 
Fig. P8.16 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
bottom flange 0.40625 0.06250 0.02539 
left web 0.28125 1.25000 0.35156 
left top flange 0.09375 2.43750 0.22852 
right web 0.28125 1.25000 0.35156 
right top flange 0.09375 2.43750 0.22852 
 1.15625 in.
2
 
 
1.18555 in.
3 
 3
2
1.18555 in.
1.0253 in.
1.15625 in.
i i
i
y A
y
A

  

 (measured upward from bottom edge of bottom flange) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
bottom flange 0.000529 −0.962838 0.376617 0.377146 
left web 0.118652 0.224662 0.014196 0.132848 
left top flange 0.000122 1.412162 0.186956 0.187079 
right web 0.118652 0.224662 0.014196 0.132848 
right top flange 0.000122 1.412162 0.186956 0.187079 
Moment of inertia about the z axis (in.
4
) = 1.016999 
 
(a) Maximum tension bending stress: 
For a positive bending moment of Mz = +270 lb-ft, the maximum tension bending stress will occur at the 
bottom surface of the cross section (i.e., y = −1.0253 in.). From the flexure formula, the bending stress 
at the bottom of the crosssection is: 
 
4
(270 lb-ft)( 1.0253 in.)(12 in./ft)
3,26 3,270 psi6.446 psi
1.016999
(
 i
)
n.
 Tx
z
M y
I
       Ans. 
 
(b) Maximum compression bending stress: 
The maximum compression bending stress will occur at the top surface of the cross section (i.e., y = 2.50 
in. − 1.0253 in. = 1.4747 in.). From the flexure formula, the bending stress at the top of the cross 
section is: 
 
4
(270 lb-ft)(1.4747 in.)(12 in./ft)
4,69 4,700 psi8.164 psi
1.016999 
 
in.
(C)x
z
M y
I
        Ans. 
 
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8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a) having the cross section 
shown in Fig. P8.17b. Determine the maximum tension and compression bending stresses produced in 
segment BC of the beam. 
 
 
Fig. P8.17a Fig. P8.17b 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (mm
2
) (mm) (mm
3
) 
top flange 3,000.0 167.5 502,500.0 
stem 1,440.0 80.0 115,200.0 
 4,440 mm
2
 
 
617,700 mm
3 
 
 3
2
617,700 mm
139.1216 mm
4,440 mm
i i
i
y A
y
A

  

 (measured upward from bottom edge of stem) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 56,250.00 28.38 2,415,997.08 2,472,247.08 
stem 3,072,000.00 −59.12 5,033,327.25 8,105,327.25 
Moment of inertia about the z axis (mm
4
) = 10,577,574.32 
 
 
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Shear-force and bending-moment diagrams: 
 
 
The maximum moment occurs between B and C. The moment magnitude is 12 kN-m. 
 
Maximum tension bending stress: 
For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of 
this cross section. From the flexure formula, the bending stress at the bottom of the tee stem is: 
 
6 4
(12 kN-m)( 139.1216 mm)(1,000 N/kN)(1,000 mm/m)
10.5776 10 m
157.8 MPa )
m
(Tx
z
M y
I
     

 Ans. 
 
Maximum compression bending stress: 
The maximum compression bending stress will occur at the top of the flange: 
 
6 4
(12 kN-m)(175 mm 139.1216 mm)(1,000 N/kN)(1,000 mm/m)
10.5776 10 mm
40.7 MPa 40.7 MPa (C)
x
z
M y
I
  

 

   Ans.
 
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8.18 Two vertical forces are applied to a simply supported beam (Fig. P8.18a) having the cross section 
shown in Fig. P8.18b. Determine the maximum tension and compression bending stresses produced in 
segment BC of the beam. 
 
 
Fig. P8.18a Fig. P8.18b 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
left stem 0.7500 1.5000 1.1250 
bottom flange 0.5000 0.1250 0.0625 
right stem 0.7500 1.5000 1.1250 
 2.000 in.
2
 
 
2.3125 in.
3 
 
 3
2
2.3125 in.
1.1563 in.
2.000 in.
i i
i
y A
y
A

  

 (measured upward from bottom edge of stem) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left stem 0.56250 0.34375 0.08862 0.65112 
bottom flange 0.00260 −1.03125 0.53174 0.53434 
right stem 0.56250 0.34375 0.08862 0.65112 
Moment of inertia about the z axis (in.
4
) = 1.83659 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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Shear-force and bending-moment diagrams: 
 
The maximum moment occurs between B and C. The moment magnitude is 600 lb-ft. 
 
Maximum tension bending stress: 
For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of 
this cross section at y = −1.1563 in. From the flexure formula, the bending stress at the bottom of the U 
shape is: 
 
4
(600 lb-ft)( 1.1563 in.)(12 in./ft)
4,533 4,530 psi (.053 psi
1.83659 in.
T)x
z
M y
I
       Ans. 
 
Maximum compression bending stress: 
The maximum compression bending stress will occur at the top of the U shape, where y = 1.8438 in.: 
 
4
(600 lb-ft)(1.8438 in.)(12 in./ft)
7,228 7,230 psi .265 psi
1
(
.83659 in.
C)x
z
M y
I
        Ans. 
 
 
 
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8.19 A WT230 × 26 standard steel shape is used to support the loads shown on the beam in Fig. P8.19a. 
The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of 
the cross section (Fig. P8.19b). Consider the entire 4-m length of the beam and determine: 
(a) the maximum tension bending stress at any location along the beam, and 
(b) the maximum compression bending stress at any location along the beam. 
 
 
Fig. P8.19a Fig. P8.19b 
Solution 
Section properties 
 From Appendix B: 
6 416.7 10 mmzI
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 13.61 kN-m 
 negative M = −20.00 kN-m 
 
Bending stresses at max positive moment 
 
2
6 4
2
6 4
(13.61 kN-m)(60.7 mm)(1,000)
16.7 10 mm
49.5 MPa (C)
(13.61 kN-m)( 164.3 mm)(1,000)
16.7 10 mm
133.9 MPa (T)
x
x
 
 
Bending stresses at max negative moment 
 
2
6 4
2
6 4
( 20 kN-m)(60.7 mm)(1,000)
16.7 10 mm
72.7 MPa (T)
( 20 kN-m)( 164.3 mm)(1,000)
16.7 10 mm
196.8 MPa (C)
x
x
 
 
(a) Maximum tension bending stress
(b) Maximum compression bending stress
133.9 MPa (T)
196.8 MPa (C)
Ans.
Ans.
 
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8.20 A WT305 × 41 standard steel shape is used to support the loads shown on the beam in Fig. P8.20a. 
The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of 
the cross section (Fig. P8.19b). Consider the entire 10-m length of the beam and determine: 
(a) the maximum tension bending stress at any location along the beam, and 
(b) the maximum compression bending stress at any location along the beam. 
 
 
Fig. P8.20a Fig. P8.20b 
Solution 
Section properties 
 From Appendix B: 
6 448.7 10 mmzI
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 45.84 kN-m 
 negative M = −24.00 kN-m 
 
Bending stresses at max positive moment 
 
2
6 4
2
6 4
(45.84 kN-m)(88.9 mm)(1,000)
48.7 10 mm
83.7 MPa (C)
(45.84 kN-m)( 211.1 mm)(1,000)
48.7 10 mm
198.7 MPa (T)
x
x
 
 
Bending stresses at max negative moment 
 
2
6 4
2
6 4
( 24 kN-m)(88.9 mm)(1,000)
48.7 10 mm
43.8 MPa (T)
( 24 kN-m)( 211.1 mm)(1,000)
48.7 10 mm
104.0 MPa (C)
x
x
 
 
(a) Maximum tension bending stress
(b) Maximum compression bending stress
198.7 MPa (T)
104.0 MPa (C)
Ans.
Ans.
 
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8.21 A steel tee shape is used to support the loads shown on the beam in Fig. P8.21a. The dimensions of 
the shape are shown in Fig. P8.21b. Consider the entire 24-ft length of the beam and determine: 
(a) the maximum tension bending stress at any location along the beam, and 
(b) the maximum compression bending stress at any location along the beam. 
 
 
Fig. P8.21a Fig. P8.21b 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
top flange 24.0000 19.2500 462.0000 
stem 13.8750 9.2500 128.3438 
 37.875 in.
2
 
 
590.3438 in.
3 
 
 
3
2
590.3438 in.
15.5866 in. (from bottom of shape to centroid)
37.8750 in.
4.4134 in. (from top of shape to centroid)
i i
i
y A
y
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 4.5000 3.6634 322.0861 326.5861 
stem 395.7266 −6.3366 557.1219 952.8484 
Moment of inertia about the z axis (in.
4
) = 1,279.4345 
 
 
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Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 100.75 kip-ft 
 negative M = −68.00 kip-ft 
 
Bending stresses at max positive moment 
 
4
4
(100.75 kip-ft)(4.4134 in.)(12 in./ft)
1, 279.4345 in.
4.17 ksi (C)
(100.75 kip-ft)( 15.5866 in.)(12 in./ft)
1, 279.4345 in.
14.73 ksi (T)
x
x
 
 
Bending stresses at max negative moment 
 
4
4
( 68 kip-ft)(4.4134 in.)(12 in./ft)
1,279.4345 in.
2.81 ksi (T)
( 68 kip-ft)( 15.5866 in.)(12 in./ft)
1,279.4345 in.
9.94 ksi (C)
x
x
 
 
(a) Maximum tension bending stress
(b) Maximum compression bending stress
14.73 ksi (T)
9.94 ksi (C)
Ans.
Ans.
 
 
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8.22 A flanged wooden shape is used to support the loads shown on the beam in Fig. P8.22a. The 
dimensions of the shape are shown in Fig. P8.22b. Consider the entire 18-ft length of the beam and 
determine: 
(a) the maximum tension bending stress at any location along the beam, and 
(b) the maximum compression bending stress at any location along the beam. 
 
 
Fig. P8.22a Fig. P8.22b 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
top flange 20.0 11.0 220.0 
web 16.0 6.0 96.0 
bottom flange 12.0 1.0 12.0 
 48.0 in.
2
 
 
328.0 in.
3 
 
 
3
2
328.0 in.
6.8333 in. (from bottom of shape to centroid)
48.0 in.
5.1667 in. (from top of shape to centroid)
i i
i
y A
y
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 6.667 4.167 347.222 353.889 
web 85.333 –0.833 11.111 96.444 
bottom flange 4.000 –5.833 408.333 412.333 
Moment of inertia about the z axis (in.
4
) = 862.667 
 
 
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Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 10,580 lb-ft 
 negative M = −8,400 lb-ft 
 
Bending stresses at max positive moment 
 
4
4
(10,580 lb-ft)(5.1667 in.)(12 in./ft)
862.667 in.
760.4 psi (C)
(10,580 lb-ft)( 6.8333 in.)(12 in./ft)
862.667 in.
1,005.6 psi (T)
x
x
 
 
Bending stresses at max negative moment 
 
4
4
( 8,400 lb-ft)(5.1667 in.)(12 in./ft)
862.667 in.
603.7 psi (T)
( 8,400 lb-ft)( 6.8333 in.)(12 in./ft)
862.667 in.
798.5 psi (C)
x
x
 
 
(a) Maximum tension bending stress
(b) Maximum compression bending stres
1,006 psi (T)
799 psi (C)s
Ans.
Ans.
 
 
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8.23 A channel shape is used to support the loads shown on the beam in Fig. P8.23a. The dimensions of 
the shape are shown in Fig. P8.23b. Consider the entire 12-ft length of the beam and determine: 
(a) the maximum tension bending stress at any location along the beam, and 
(b) the maximum compression bending stress at any location along the beam. 
 
Fig. P8.23a Fig. P8.23b 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
left stem 3.000 3.000 9.000 
top flange 5.500 5.750 31.625 
right stem 3.000 3.000 9.000 
 11.500 in.
2
 
 
49.625 in.
3 
 
 
3
2
49.625in.
4.3152 in. (from bottom of shape to centroid)
11.500 in.
1.6848 in. (from top of shape to centroid)
i i
i
y A
y
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left stem 9.0000 −1.3152 5.1894 14.1894 
top flange 0.1146 1.4348 11.3223 11.4369 
right stem 9.0000 −1.3152 5.1894 14.1894 
Moment of inertia about the z axis (in.
4
) = 39.8157 
 
 
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Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 8,850 lb-ft 
 negative M = −9,839 lb-ft 
 
Bending stresses at max positive moment 
 
4
4
(8,850 lb-ft)(1.6848 in.)(12 in./ft)
39.8157 in.
4,494 psi (C) 4.49 ksi (C)
(8,850 lb-ft)( 4.3152 in.)(12 in./ft)
39.8157 in.
11,510 psi (T) 11.51 ksi (T)
x
x
 
 
Bending stresses at max negative moment 
 
4
4
( 9,839 lb-ft)(1.6848 in.)(12 in./ft)
39.8157 in.
4,996 psi (T) 5.00 ksi (T)
( 9,839 lb-ft)( 4.3152 in.)(12 in./ft)
39.8157 in.
12,796 psi (C) 12.80 ksi (C)
x
x
 
 
(a) Maximum tension bending stress
(b) Maximum compression bending stress
11.51 ksi (T)
12.80 ksi (C)
Ans.
Ans.
 
 
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8.24 A W360 × 72 standard steel shape is used to support the loads shown on the beam in Fig. P8.24a. 
The shape is oriented so that bending occurs about the weak axis as shown in Fig. P8.24b. Consider the 
entire 6-m length of the beam and determine: 
(a) the maximum tension bending stress at any location along the beam, and 
(b) the maximum compression bending stress at any location along the beam. 
 
Fig. P8.24a Fig. P8.24b 
Solution 
Section properties 
 From Appendix B: 
6 421.4 10 mm 204 mmz fI b
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 31.50 kN-m 
 negative M = −25.87 kN-m 
 
Since the shape is symmetric about the z axis, 
the largest bending stresses will occur at the 
location of the largest moment magnitude – 
either positive or negative. In this case, the 
largest bending stresses will occur where the 
moment magnitude is 31.50 kN-m. 
 
Bending stresses at maximum moment 
 
2
6 4
(31.50 kN-m)( 204 mm/2)(1,000)
21.4 10 mm
150.1 MPa (T) and (C)
x
 
 
 
 
(a) Maximum tension bending stress
(b) Maximum compression bending stress
150.1 MPa (T)
150.1 MPa (C)
Ans.
Ans.
 
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8.25 A 1.00-in.-diameter solid steel 
shaft supports loads PA = 180 lb and PC 
= 240 lb as shown in Fig. P8.25. 
Assume L1 = 5 in., L2 = 16 in., and L3 = 
8 in. The bearing at B can be idealized 
as a roller support and the bearing at D 
can be idealized as a pin support. 
Determine the magnitude and location 
of the maximum bending stress in the 
shaft. 
 
 Fig. P8.25 
Solution 
Section properties 
 
4 4 4(1.00 in.) 0.049087 in.
64 64
I D
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 980 lb-in. 
 negative M = −900 lb-in. 
 
Since the circular cross section is symmetric 
about the z axis, the largest bending stresses 
will occur at the location of the largest moment 
magnitude – either positive or negative. In this 
case, the largest bending stresses will occur at 
C, where the moment magnitude is 980 lb-in. 
 
Bending stresses at maximum moment 
 
4
(980 lb-in.)( 1.00 in./2)
0.049087 
9
in.
,980 psi
x
 Ans. 
 
 
 
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8.26 A 30-mm-diameter solid steel 
shaft supports loads PA = 1,400 N and 
PC = 2,600 N as shown in Fig. P8.26. 
Assume L1 = 100 mm, L2 = 200 mm, 
and L3 = 150 mm. The bearing at B can 
be idealized as a roller support and the 
bearing at D can be idealized as a pin 
support. Determine the magnitude and 
location of the maximum bending stress 
in the shaft. 
 
 Fig. P8.26 
Solution 
Section properties 
 
4 4 4(30 mm) 39,760.8 mm
64 64
I D
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 162,857 N-mm 
 negative M = −140,000 N-mm 
 
Since the circular cross section is symmetric 
about the z axis, the largest bending stresses 
will occur at the location of the largest moment 
magnitude – either positive or negative. In this 
case, the largest bending stresses will occur 
where the moment magnitude is 162,857 N-
mm. 
 
Bending stresses at maximum moment 
 
4
(162,857 N-mm)( 30 mm/2)
39,760.8 mm
61.4 MPa
x
 Ans. 
 
 
 
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8.27 A 20-mm-diameter solid steel shaft 
supports loads PA = 500 N, PC = 1,750 
N, and PE = 500 N as shown in Fig. 
P8.27. Assume L1 = 90 mm, L2 = 260 
mm, L3 = 140 mm, and L4 = 160 mm. 
The bearing at B can be idealized as a 
roller support and the bearing at D can 
be idealized as a pin support. Determine 
the magnitude and location of the 
maximum bending stress in the shaft. 
 Fig. P8.27 
Solution 
Section properties 
 
4 4 4(20 mm) 7,853.9816 mm
64 64
zI D
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 91,500 N-mm 
 negative M = −80,000 N-mm 
 
Since the circular cross section is symmetric 
about the z axis, the largest bending stresses 
will occur at the location of the largest moment 
magnitude – either positive or negative. In this 
case, the largest bending stresses will occur at 
C, where the moment magnitude is 91,500 N-
mm. 
 
Bending stresses at maximum moment 
 
4
(91,500 N-mm)( 20 mm/2)
7,853.9816
1
 mm
16.5 MPa
x
 Ans. 
 
 
 
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8.28 A 1.75-in.-diameter solid steel shaft 
supports loads PA = 250 lb, PC = 600 lb, 
and PE = 250 lb as shown in Fig. P8.28. 
Assume L1 = 9 in., L2 = 24 in., L3 = 12 
in., and L4 = 15 in. The bearing at B can 
be idealized as a roller support and the 
bearing at D can be idealized as a pin 
support. Determine the magnitude and 
location of the maximum bending stress 
in the shaft. 
 Fig. P8.28 
Solution 
Section properties 
 
4 4 4(1.75 in.) 0.460386 in.
64 64
I D
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 1,550 lb-in. 
 negative M = −3,750 lb-in. 
 
Since the circular cross section is symmetric 
about the z axis, the largest bending stresses 
will occur at the location of the largest moment 
magnitude – either positive or negative. In this 
case, the largest bending stresses will occur at 
support D, where the moment magnitude is 
3,750 lb-in. 
 
Bending stresses at maximum moment 
 
4
( 3,750 lb-in.)( 1.75 in./2)
0.460386 in.
7,130 psi
x
 Ans. 
 
 
 
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8.29 A HSS12 × 8 × 1/2 standard steel shape 
is used to support the loads shown on the 
beam in Fig. P8.29. The shape is oriented so 
that bending occurs about the strong axis. 
Determine the magnitude and location of the 
maximum bending stress in the beam. 
 
 Fig. P8.29 
Solution 
Section properties 
 From Appendix B: 
4333 in. 12 in.zI d
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 124.59 kip-ft 
 negative M = −72.00 kip-ft 
 
Since the shape is symmetric about the z axis, 
the largest bending stresses will occur at the 
location of the largest moment magnitude – 
either positive or negative. In this case, the 
largest bending stresses will occur at C, where 
the moment magnitude is 124.59 kip-ft. 
 
 
 
 
 
Bending stresses at max moment magnitude 
 
4
(124.59 kip-ft)( 12 in./2)(12 in./ft)
333 in.
26.9 ksix
 Ans.
 
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8.30 A W410 × 60 standard steel shape 
is used to support the loads shown on 
the beam in Fig. P8.30. The shape is 
oriented so that bending occurs about 
the strong axis. Determine the 
magnitude and location of the 
maximum bending stress in the beam. 
 Fig. P8.30 
Solution 
Section properties 
 From Appendix B: 
6 4216 10 mm 406 mmzI d
 
 
Shear-force and bending-moment diagrams 
 
Maximum bending moments 
 positive M = 50 kN-m 
 negative M = −70 kN-m 
 
Since the shape is symmetric about the z axis, 
the largest bending stresses will occur at the 
location of the largest moment magnitude – 
either positive or negative. In this case, the 
largest bending stresses will occur between B 
and C, where the moment magnitude is 70 kN-
m. 
 
 
 
 
 
Bending stresses at max moment magnitude 
 2
6 4
(70 kN-m)( 406 mm/2)(1,000)
216 10 m
65.8 MPa
m
x
 Ans. 
 
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8.31 A solid steel shaft supports loads 
PA = 200 lb and PD = 300 lb as shown 
in Fig. P8.31. Assume L1 = 6 in., 
L2 = 20 in., and L3 = 10 in. The bearing 
at B can be idealized as a roller support 
and the bearing at C can be idealized 
as a pin support. If the allowable 
bending stress is 8 ksi, determine the 
minimum diameter that can be used for 
the shaft. 
 
 Fig. P8.31 
Solution 
Shear-force and bending-moment diagrams 
 
Maximum bending moment magnitude 
 M = 3,000 lb-in. 
 
Minimum required section modulus 
 
33,000 lb-in. 0.375 in.
8,000 psi
x
x
M
S
M
S



 
 
 
 
Section modulus for solid circular section 
 3
32
d
S


 
 
Minimum shaft diameter 
 
3
30.375 in.
1.
32
563 in.
d
d


  Ans. 
 
 
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8.32 A solid steel shaft supports loads 
PA = 500 N and PD = 400 N as shown 
in Fig. P8.32. Assume L1 = 200 mm, 
L2 = 660 mm, and L3 = 340 mm. The 
bearing at B can be idealized as a roller 
support and the bearing at C can be 
idealized as a pin support. If the 
allowable bending stress is 25 MPa, 
determine the minimum diameter that 
can be used for the shaft. 
 
 Fig. P8.32 
Solution 
Shear-force and bending-moment diagrams 
 
Maximum bending moment magnitude 
 M = 136,000 N-mm 
 
Minimum required section modulus 
 
3
2
136,000 N-mm
5,440 mm
25 N/mm
x
x
M
S
M
S



 
  
 
Section modulus for solid circular section 
 3
32
d
S


 
 
Minimum shaft diameter 
 
3
35,440 m
38. m
m
3
1 m
2
d
d


  Ans. 
 
 
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8.33 A simply supported wood beam (Fig. P8.33a) with a span of L = 20 ft supports a uniformly 
distributed load of w = 800 lb/ft. The allowable bending stress of the wood is 1,400 psi. If the aspect 
ratio of the solid rectangular wood beam is specified as h/b = 1.5 (Fig. P8.33b), determine the minimum 
width b that can be used for the beam. 
Fig. P8.33a Fig. P8.33b 
Solution 
Shear-force and bending-moment diagrams 
Also, see Example 7-3 for shear-force and bending-moment diagram development. 
 
 
Maximum bending moment 
 
2 2
max
(800 lb/ft)(20 ft)
8 8
40,000 lb-ft 480,000 lb-in.wL
M  
 
 
 
Minimum required section modulus 
 
3480,000 lb-in. 342.8571 in.
1, 400 psi
x
x
M
S
M
S



 
 
 
 
Section modulus for solid rectangular section 
 3 2/12
/ 2 6
I bh bh
S
c h
  
 
 
 
The aspect ratio of the solid rectangular wood beam is specified as h/b = 1.5; therefore, the section 
modulus can be expressed as: 
 2 2 3
3(1.5 ) 2.25 0.3750
6 6 6
bh b b b
S b   
 
 
Minimum allowable beam width 
 
3 30.3750 342.8571
9.71 in
 
.
in.b
b

  Ans. 
 
The corresponding beam height h is 
 
/ 1.5 1.5 1.5(9.71 in.) 14.57 in.h b h b    
 
 
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8.34 A simply supported wood beam (Fig. P8.34a) with a span of L = 14 ft supports a uniformly 
distributed load of w. The beam width is b = 6 in. and the beam height is h = 10 in. (Fig. P8.34b). The 
allowable bending stress of the wood is 900 psi. Determine the magnitude of the maximum load w that 
may be carried by the beam. 
Fig. P8.34a Fig. P8.34b 
Solution 
Moment of inertia for rectangular cross section about horizontal centroidal axis 
 3 3
4(6 in.)(10 in.) 500 in.
12 12
bh
I   
 
 
Maximum allowable moment 
 4(900 psi)(500 in. )
90,000 lb-in. 7,500 lb-ft
5 in.
x
x
IMc
M
I c
       
 
Shear-force and bending-moment diagrams 
Also, see Example 7-3 for shear-force and bending-moment diagram development. 
 
Determine distributed load intensity 
Equate the moment expression from the bending-
moment diagram to the maximum allowable moment 
that can be applied to the rectangular cross section: 
 2
max 7,500 lb-ft
8
wL
M  
 
 
Solve for the maximum distributed load w that can be 
applied to the 14-ft simple span: 
 
2 2
2
(14 ft)
7,500 lb-ft
8 8
8(7,500 lb-ft)
(14 ft)
306 lb/ft
wL w
w
 
   Ans. 
 
 
 
 
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8.35 A cantilever timber beam (Fig. P8.35a) with a span of L = 2.5 m supports a uniformly distributed 
load of w = 4 kN/m. The allowable bending stress of the wood is 9 MPa. If the aspect ratio of the solid 
rectangular timber is specified as h/b = 0.5 (Fig. P8.35b), determine the minimum width b that can be 
used for the beam. 
Fig. P8.35a Fig. P8.35b 
Solution 
Maximum moment magnitude: 
The maximum bending moment magnitude in the cantilever beam occurs at support A: 
 2 2
6
max
(4 kN/m)(2.5 m)
12.5 kN-m 12.5 10 N-mm
2 2
wL
M     
 
 
Minimum required section modulus 
 
6
6 3
2
12.5 10 N-mm
1.3889 10 mm
9 N/mm
x
x
M
S
M
S



 

   
 
Section modulus for solid rectangular section 
 3 2/12
/ 2 6
I bh bh
S
c h
  
 
 
The aspect ratio of the solid rectangular wood beam is specified as h/b = 0.5; therefore, the section 
modulus can be expressed as: 
 2 2 3
3(0.5 ) 0.25 0.0416667
6 6 6
bh b b b
S b   
 
 
Minimum allowable beam width 
 
3 6 30.0416667 1.3889 10 mm
321.83 mm 322 mm
b
b
 
   Ans. 
 
The corresponding beam height h is 
 
/ 0.5
0.5 0.5(321.83 mm) 161 mm
h b
h b

    
 
 
 
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8.36 A cantilever timber beam (Fig. P8.36a) with a span of L = 3 m supports a uniformly distributed 
load of w. The beam width is b = 300 mm and the beam height is h = 200 mm (Fig. P8.36b). The 
allowable bending stress of the wood is 6 MPa. Determine the magnitude of the maximum load w that 
may be carried by the beam. 
 Fig. P8.36a Fig. P8.36b 
Solution 
Section modulus for solid rectangular section 
 3 2 2
6 3/12 (300 mm)(200 mm) 2 10 mm
/ 2 6 6
I bh bh
S
c h
     
 
 
Maximum allowable bending moment: 
2 6 3 6
allow (6 N/mm )(2 10 mm ) 12 10 N-mmx x
M
M S
S
        
 
Maximum bending moment in cantilever span: 
The maximum bending moment magnitude in the cantilever beam occurs at support A: 
 2
max
2
wL
M 
 
 
Maximum distributed load: 
 
2
allow
6
allow
allow 2 2
2
2 2(12 10 N-mm)
2.67 N/mm
[(3 m)(1,000 mm
2.67 kN/
/ )
m
m ]
wL
M
M
w
L


     Ans. 
 
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8.37 The beam shown in Fig. P8.37 will be 
constructed from a standard steel W-shape 
using an allowable bending stress of 24 ksi. 
(a) Develop a list of five acceptable shapes 
that could be used for this beam. On this list, 
include the most economical W10, W12, 
W14, W16, and W18 shapes. 
(b) Select the most economical W shape for 
this beam. 
 Fig. P8.37 
Solution 
Shear-force and bending-moment diagrams 
 
Maximum bending moment magnitude 
 M = 90 kip-ft 
 
Minimum required section modulus 
 
3(90 kip-ft)(12 in./ft) 45 in.
24 ksi
x
x
M
S
M
S



 
  
 
(a) Acceptable steel W-shapes 
 
3
3
3
3
3
W10 45, 49.1 in.
W12 40, 51.5 in.
W14 34, 48.6 in.
W16 31, 47.2 in.
W18 35, 57.6 in.
S
S
S
S
S
 
 
 
 
 
 
 
(b) Most economical W-shape 
 
W16 31
 Ans. 
 
 
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8.38 The beam shown in Fig. P8.38 will be 
constructed from a standard steel W-shape using 
an allowable bending stress of 165 MPa. 
(a) Develop a list of four acceptable shapes that 
could be used for this beam. Include the most 
economical W360, W410, W460, and W530 
shapes on the list of possibilities. 
(b) Select the most economical W shape for this 
beam. 
 Fig. P8.38 
Solution 
Shear-force and bending-moment diagrams 
 
 
Maximum bending moment magnitude 
 M = 206.630 kN-mMinimum required section modulus 
 
2
3 3
2
(206.63 kN-m)(1,000)
1,252 10 mm
165 N/mm
x
x
M
S
M
S



 
   
 
(a) Acceptable steel W-shapes 
 
3 3
3 3
3 3
3 3
W360 79, 1,270 10 mm
W410 75, 1,330 10 mm
W460 74, 1,460 10 mm
W530 66, 1,340 10 mm
S
S
S
S
  
  
  
  
 
 
(b) Most economical W-shape 
 
W530 66
 Ans. 
 
 
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8.39 The beam shown in Fig. P8.39 will be 
constructed from a standard steel W-shape using 
an allowable bending stress of 165 MPa. 
(a) Develop a list of four acceptable shapes that 
could be used for this beam. Include the most 
economical W360, W410, W460, and W530 
shapes on the list of possibilities. 
(b) Select the most economical W shape for this 
beam. 
 Fig. P8.39 
Solution 
Shear-force and bending-moment diagrams 
 
Maximum bending moment magnitude 
 M = 238.57 kN-m 
 
Minimum required section modulus 
 
2
3 3
2
(238.57 kN-m)(1,000)
1,446 10 mm
165 N/mm
x
x
M
S
M
S



 
   
 
(a) Acceptable steel W-shapes 
 
3 3
3 3
3 3
3 3
W360 101, 1,690 10 mm
W410 85, 1,510 10 mm
W460 74, 1,460 10 mm
W530 74, 1,550 10 mm
S
S
S
S
  
  
  
  
 
 
(b) Most economical W-shape 
 
 or W460 74 W530 74 
 Ans. 
 
 
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8.40 The beam shown in Fig. P8.40 will be constructed from a standard 
steel W-shape using an allowable bending stress of 165 MPa. 
(a) Develop a list of four acceptable shapes that could be used for this 
beam. Include the most economical W310, W360, W410, and W460 
shapes on the list of possibilities. 
(b) Select the most economical W shape for this beam. 
 
 Fig. P8.40 
Solution 
Maximum moment magnitude: 
The maximum bending moment magnitude occurs at the base of the cantilever beam: 
 
max
6
1 1
(15 kN)(3.0 m) (40 kN/m)(3.0 m) (3.0 m)
2 3
105.0 kN-m 105.0 10 N-mm
M  
   
 
Minimum required section modulus 
 
2
3 3
2
(105.0 kN-m)(1,000)
636 10 mm
165 N/mm
x
x
M
S
M
S



     
 
(a) Acceptable steel W-shapes 
 
3 3
3 3
3 3
3 3
W310 60, 844 10 mm
W360 44, 688 10 mm
W410 46.1, 773 10 mm
W460 52, 944 10 mm
S
S
S
S
  
  
  
  
 
 
(b) Most economical W-shape 
 
W360 44
 Ans. 
 
 
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8.41 The beam shown in Fig. P8.41 will be 
constructed from a standard steel HSS-shape 
using an allowable bending stress of 30 ksi. 
(a) Develop a list of three acceptable shapes that 
could be used for this beam. On this list, include 
the most economical HSS8, HSS10, and HSS12 
shapes. 
(b) Select the most economical HSS-shape for this 
beam. 
 
 Fig. P8.41 
Solution 
Shear-force and bending-moment diagrams 
 
Maximum bending moment magnitude 
 M = 45.56 kip-ft 
 
Minimum required section modulus 
 
3(45.56 kip-ft)(12 in./ft) 18.22 in.
30 ksi
x
x
M
S
M
S



 
  
 
(a) Acceptable steel HSS shapes 
 
3
3
3
3
HSS8 none are acceptable
HSS10 4 3 / 8, 20.8 in.
HSS10 6 3 / 8, 27.4 in.
HSS12 6 3 / 8, 35.9 in.
HSS12 8 3 / 8, 43.7 in.
S
S
S
S
  
  
  
  
 
 
(b) Most economical HSS shape 
 
HSS10 4 3 / 8 
 Ans. 
 
 
 
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8.42 A composite beam is fabricated by bolting two 3 in. wide × 12 in. deep timber planks to the sides 
of a 0.50 in. × 12 in. steel plate (Fig. P8.42b). The moduli of elasticity of the timber and the steel are 
1,800 ksi and 30,000 ksi, respectively. The simply supported beam spans a distance of 20 ft and carries 
two concentrated loads P, which are applied at the quarter points of the span (Fig. P8.42a). 
(a) Determine the maximum bending stresses produced in the timber planks and the steel plate if P = 3 
kips. 
(b) Assume that the allowable bending stresses of the timber and the steel are 1,200 psi and 24,000 psi, 
respectively. Determine the largest acceptable magnitude for concentrated loads P. (You may neglect 
the weight of the beam in your calculations.) 
 
 
Fig. P8.42a 
Fig. P8.42b 
 
Solution 
Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is: 
 
2
1
30,000 ksi
16.6667
1,800 ksi
E
n
E
 
Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the 
modular ratio: b2, trans = 16.6667(0.50 in.) = 8.3333 in. Thus, for calculation purposes, the 12 in. × 0.50 
in. steel plate is replaced by a wood board that is 12 in. deep and 8.3333-in. thick. 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
timber (1) 864 0 0 864 
transformed steel plate (2) 1,200 0 0 1,200 
Moment of inertia about the z axis = 2,064 in.
4 
 
 
Maximum bending moment in beam for P = 3 kips 
The maximum bending moment in the simply supported beam with two 3-kip concentrated loads is: 
 
max (3 kips)(5 ft) 15 kip-ft 180 kip-in.M
 
 
Bending stress in timber (1) 
From the flexure formula, the maximum bending stress in timber (1) is: 
 
1 4
(180 kip-in.)( 6 in.)
0.5233 ksi
2,
523 
064 in.
psi
My
I
 Ans. 
 
 
 
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Bending stress in steel plate (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in steel plate (2) is:2 4
(180 kip-in.)( 6 in.)
(16.6667) 8.7209 ksi
2,064 
8,720
n
 p
.
s
i
i
My
n
I
 Ans. 
 
Determine maximum P 
If the allowable bending stress in the timber is 1,200 psi, then the maximum bending moment that may 
be supported by the beam is: 
 4
1
1 max
(1.200 ksi)(2,064 in. )
412.80 kip-in.
6 in.
IMy
M
I y
 
If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be 
supported by the beam is: 
 4
2
2 max
(24.00 ksi)(2,064 in. )
495.36 kip-in.
(16.667)(6 in.)
IMy
n M
I ny
 
Note: The negative signs were omitted in the previous two equations because only the moment 
magnitude is of interest here. 
 
From these two results, the maximum moment that the beam can support is 412.80 kip-in. The 
maximum concentrated load magnitude P that can be supported is found from: 
 
max
max
(5 ft)
412.80 kip-in.
5 ft (5 ft)(12 in./ft)
6.88 kips
M P
M
P Ans. 
 
 
 
 
 
 
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8.43 The cross section of a composite beam that 
consists of 4-mm-thick fiberglass faces bonded to a 20-
mm-thick particleboard core is shown in Fig. P8.43. 
The beam is subjected to a bending moment of 55 N-m 
acting about the z axis. The elastic moduli for the 
fiberglass and the particleboard are 30 GPa and 10 GPa, 
respectively. Determine: 
(a) the maximum bending stresses in the fiberglass 
faces and the particleboard core. 
(b) the stress in the fiberglass at the joint where the two 
materials are bonded together. 
 
 Fig. P8.43 
Solution 
Let the particleboard be denoted as material (1) and the fiberglass as material (2). The modular ratio is: 
 
2
1
30 GPa
3
10 GPa
E
n
E
 
Transform the fiberglass faces into an equivalent amount of particleboard by multiplying their width by 
the modular ratio: b2, trans = 3(50 mm) = 150 mm. Thus, for calculation purposes, the 50 mm × 4 mm 
fiberglass faces are replaced by particleboard faces that are 150-mm wide and 4-mm thick. 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
transformed fiberglass top face 800.00 12.00 86,400.00 87,200.00 
particleboard core 33,333.33 0 0 33,333.33 
transformed fiberglass bot face 800.00 12.00 86,400.00 87,200.00 
Moment of inertia about the z axis = 207,733.33 mm
4 
 
 
Bending stress in particleboard core (1) 
From the flexure formula, the maximum bending stress in the particleboard core is: 
 
1 4
(55 N-m)( 10 mm)(1,000 mm/m)
207,733.33 
2.65 MPa
mm
My
I
 Ans. 
 
Bending stress in fiberglass faces (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in the fiberglass faces (2) is: 
 
2 4
(55 N-m)( 14 mm)(1,000 mm/m)
(3)
207,733.33 mm
11.12 MPa
My
n
I
 Ans. 
 
Bending stress in fiberglass (2) at interface 
At the interface between the particleboard and the fiberglass, y = ±10 mm: 
 
2 4
(55 N-m)( 10 mm)(1,000 mm/m)
(3)
207,733.33 m
7.94
m
 MPa
My
n
I
 Ans. 
 
 
 
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8.44 A composite beam is made of two brass [E = 
100 GPa] plates bonded to an aluminum [E = 75 
GPa] bar, as shown in Fig. P8.44. The beam is 
subjected to a bending moment of 1,750 N-m acting 
about the z axis. Determine: 
(a) the maximum bending stresses in the brass 
plates and the aluminum bar. 
(b) the stress in the brass at the joints where the two 
materials are bonded together. 
 
 Fig. P8.44 
Solution 
Let the aluminum be denoted as material (1) and the brass as material (2). The modular ratio is: 
 
2
1
100 GPa
1.3333
75 GPa
E
n
E
 
Transform the brass plates into an equivalent amount of aluminum by multiplying their width by the 
modular ratio: b2, trans = 1.3333(50 mm) = 66.6666 mm. Thus, for calculation purposes, the 50 mm × 10 
mm brass plates are replaced by aluminum plates that are 66.6666-mm wide and 10-mm thick. 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
transformed top brass plate 5,555.55 20 266,666.40 272,221.95 
aluminum bar 112,500.00 0 0 112,500.00 
transformed bot brass plate 5,555.55 –20 266,666.40 272,221.95 
Moment of inertia about the z axis = 656,943.90 mm
4 
 
 
Bending stress in aluminum bar (1) 
From the flexure formula, the maximum bending stress in the aluminum bar is: 
 
1 4
(1,750 N-m)( 15 mm)(1,000 mm/m)
656,943.90 mm
40.0 MPa
My
I
 Ans. 
 
Maximum bending stress in brass plates (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in the brass plates (2) is: 
 
2 4
(1,750 N-m)( 25 mm)(1,000 mm/m)
(1.3333)
656,943.90 m
88.8 MPa
m
My
n
I
 Ans. 
 
Bending stress in brass plates (2) at interface 
At the interface between the brass plates and the aluminum bar, y = ±15 mm: 
 
2 4
(1,750 N-m)( 15 mm)(1,000 mm/m)
(1.3333)
656,943.90 m
53.3 MPa
m
My
n
I
 Ans. 
 
 
 
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8.45 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite 
beam (Fig. P8.45b). The composite beam is subjected to a bending moment of M = +300 lb-ft about the 
z axis (Fig. P8.45a). Determine: 
(a) the maximum bending stresses in the aluminum and steel bars. 
(b) the stress in the two materials at the joint where they are bonded together. 
 
 
Fig. P8.45a Fig. P8.45b 
Solution 
Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is: 
 
2
1
30,000 ksi
3
10,000 ksi
E
n
E
 
Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the 
modular ratio: b2, trans = 3(2.00 in.) = 6.00 in. Thus, for calculation purposes, the 2.00 in. × 0.75 in. steel 
bar is replaced by an aluminum bar that is 6.00-in. wide and 0.75-in. thick. 
 
Centroid location of the transformed section in the vertical direction 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
aluminum bar (1) 2.00 0.50 1.00 0.25 0.2500 
transformed steel bar (2) 6.00 0.75 4.50 0.875 3.9375 
 5.50 
 
4.1875
 
 
 3
2
4.1875 in.
5.50
0.7614 in.
 in.
i i
i
y Ay
A
 (measured upward from bottom edge of section) 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
aluminum bar (1) 0.02083 –0.5114 0.2615 0.2823 
transformed steel bar (2) 0.2109 0.1136 0.05811 0.2690 
Moment of inertia about the z axis = 0.5514 in.
4 
 
 
 
 
 
 
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(a) Maximum bending stress in aluminum bar (1) 
From the flexure formula, the maximum bending stress in aluminum bar (1) is: 
 
1 4
(300 lb-ft)( 0.7614 in.)(12 in./ft)
0.5514 in
4,970 psi (T)
.
My
I
 Ans. 
 
(a) Maximum bending stress in steel bar (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in steel bar (2) is: 
 
2 4
(300 lb-ft)(1.250 in. 0.7614 in.)(12 in./ft)
(3)
0.5514 i
9,570 psi (C)
n.
My
I
 Ans. 
 
(b) Bending stress in aluminum bar (1) at interface 
 
1 4
(300 lb-ft)(0.50 in. 0.7614 in.)(12 in./ft)
0.5514 in.
1,706 psi (T)
My
I
 Ans. 
 
(b) Bending stress in steel bar (2) at interface 
 
2 4
(300 lb-ft)(0.50 in. 0.7614 in.)(12 in./ft)
(3)
0.5514 in.
5,120 psi (T)
My
I
 Ans. 
 
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8.46 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite 
beam (Fig. P8.46b). The allowable bending stresses for the aluminum and steel bars are 20 ksi and 30 
ksi, respectively. Determine the maximum bending moment M that can be applied to the beam. 
 
 
Fig. P8.46a Fig. P8.46b 
Solution 
Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is: 
 
2
1
30,000 ksi
3
10,000 ksi
E
n
E
 
Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the 
modular ratio: b2, trans = 3(2.00 in.) = 6.00 in. Thus, for calculation purposes, the 2.00 in. × 0.75 in. steel 
bar is replaced by an aluminum bar that is 6.00-in. wide and 0.75-in. thick. 
 
Centroid location of the transformed section in the vertical direction 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
aluminum bar (1) 2.00 0.50 1.00 0.25 0.2500 
transformed steel bar (2) 6.00 0.75 4.50 0.875 3.9375 
 5.50 
 
4.1875
 
 
 3
2
4.1875 in.
5.50
0.7614 in.
 in.
i i
i
y A
y
A
 (measured upward from bottom edge of section) 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
aluminum bar (1) 0.02083 –0.5114 0.2615 0.2823 
transformed steel bar (2) 0.2109 0.1136 0.05811 0.2690 
Moment of inertia about the z axis = 0.5514 in.
4 
 
 
 
 
 
 
 
 
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(a) Maximum bending moment magnitude based on allowable aluminum stress 
Based on an allowable bending stress of 20 ksi for the aluminum, the maximum bending moment 
magnitude that be applied to the cross section is: 
 4
1
1
(20 ksi)(0.5514 in. )
14.484 kip-in.
0.7614 in.
My I
M
I y
 (a) 
 
Maximum bending moment magnitude based on allowable steel stress 
Based on an allowable bending stress of 30 ksi for the steel, the maximum bending moment magnitude 
that be applied to the cross section is: 
 4
2
2
(30 ksi)(0.5514 in. )
11.285 kip-in.
(3)(1.25 in. 0.7614 in.)
My I
n M
I n y
 (b) 
 
Maximum bending moment magnitude 
From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the 
cross section is 
 
max 11.285 kip 940- in l ft. b-M
 Ans. 
 
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8.47 Two steel [E = 30,000 ksi] plates are securely 
attached to a Southern pine [E = 1,800 ksi] timber 
to form a composite beam (Fig. P8.47). The 
allowable bending stress for the steel plates is 
24,000 psi and the allowable bending stress for the 
Southern pine is 1,200 psi. Determine the maximum 
bending moment that can be applied about the 
horizontal axis of the beam. 
 
 Fig. P8.47 
Solution 
Denote the timber as material (1) and denote the steel as material (2). The modular ratio is: 
 
2
1
30,000 ksi
16.6667
1,800 ksi
E
n
E
 
Transform the steel plates into an equivalent amount of timber by multiplying their width by the 
modular ratio: b2, trans = 16.6667(8 in.) = 133.3333 in. Thus, for calculation purposes, the 8 in. × 0.25 in. 
steel plates can be replaced by wood plates that are 133.3333-in. wide and 0.25-in. thick. 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
transformed steel plate at top 0.1736 8.125 2,200.52 2,200.694 
timber (1) 3,413.3333 0 0 3,413.333 
transformed steel plate at bottom 0.1736 –8.125 2,200.52 2,200.694 
Moment of inertia about the z axis = 7,814.72 in.
4 
 
 
(a) Maximum bending moment magnitude based on allowable Southern pine stress 
Based on an allowable bending stress of 1,200 psi for the Southern pine timber, the maximum bending 
moment magnitude that be applied to the cross section is: 
 4
1
1
(1.200 ksi)(7,814.72 in. )
1,172.208 kip-in.
8 in.
IMy
M
I y
 (a) 
 
Maximum bending moment magnitude based on allowable steel stress 
Based on an allowable bending stress of 24,000 psi for the steel plates, the maximum bending moment 
magnitude that be applied to the cross section is: 
 4
2
2
(24 ksi)(7,814.72 in. )
1,364.021 kip-in.
(16.6667)( 8.25 in.)
IMy
n M
I n y
 (b) 
 
Maximum bending moment magnitude 
From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the 
cross section is 
 
max 1,172.208 kip 97.7 -in ki t. p-fM
 Ans. 
 
 
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8.48 A simply supported composite beam 5 m long carries a uniformly distributed load w (Fig. P8.48a). 
The beam is constructed of a Southern pine [E = 12 GPa] timber, 200 mm wide by 360 mm deep, that is 
reinforced on its lower surface by a steel [E = 200 GPa] plate that is 150 mm wide by 12 mm thick (Fig. 
P8.48b). 
(a) Determine the maximum bending stresses produced in the timber and the steel if w = 12 kN/m. 
(b) Assume that the allowable bending stresses of the timber and the steel are 9 MPa and 165 MPa, 
respectively. Determine the largest acceptable magnitude for distributed load w. (You may neglect the 
weight of the beam in your calculations.) 
 
 
 
Fig. P8.48a Fig. P8.48b 
Solution 
Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is: 
 
2
1
200 GPa
16.6667
12 GPa
E
n
E
 
Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the 
modular ratio: b2, trans = 16.6667(150 mm) = 2,500 mm. Thus, for calculation purposes, the 150 mm × 
12 mm steel plate is replaced by a wood board that is 2,500-mm wide and 12-mm thick. 
 
Centroid location of the transformed section in the vertical direction 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
timber (1) 200 360 72,000 192 13,824,000 
transformed steel plate (2) 2,500 12 30,000 6 180,000 
 102,000 
 
14,004,000
 
 
 3
2
14,004,000 mm
102,000 mm
137.294 mm
i i
i
y A
y
A
 (measured upward from bottom edge of section) 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
timber (1) 777,600,000 54.71 215,476,817 993,076,817 
transformed steel plate (2) 360,000 –131.29 517,144,360 517,504,360 
Moment of inertia about the z axis = 
 1,510,581,176 mm
4
 
 = 1.5106 ×10
9
 mm
4 
 
 
 
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Bending moment in beam for w = 12 kN/m 
The bending moment in the simply supported beam with a uniformly distributed load of 12 kN/m is: 
 2 2
6
max
(12 kN/m)(5 m)
37.5 kN-m 37.5 10 N-mm
8 8
wL
M
 
 
Bending stress in timber (1) 
From the flexure formula, the maximum bending stress in timber (1) is: 
 6
1 9 4
(37.5 10 N-mm)(372 mm 137.294 mm)
1.5106 10 mm
5.83 MPa (C)
My
I
 Ans. 
 
Bending stress in steel plate (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in steel plate (2) is: 
 6
2 9 4
(37.5 10 N-mm)( 137.294 mm)
(16.6667)
1.5106 10 m
56.8 MPa (
m
T)
My
I
 Ans. 
 
Determine maximum w 
If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be 
supported by the beam is: 
 2 9 4
61
1 max
(9 N/mm )(1.5106 10 mm )
57.925 10 N-mm
(372 mm 137.294 mm)
IMy
M
I y
 
If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be 
supported by the beam is: 
 2 9 4
62
2 max
(165 N/mm )(1.5106 10 mm )
108.926 10 N-mm
(16.6667)(137.294 mm)
IMy
n M
I ny
 
Note: The negative signs were omitted in the previous two equations because only the moment 
magnitude is of interest here. 
 
From these two results, the maximum moment that the beam can support is 57.925×10
6
 N-mm. The 
maximum distributed load magnitude w that can be supported is found from: 
 
2
max
6
max
2 2
8
8 8(57.925 10 N-mm)(1 m/1000 mm)
18,536 N/m
(5 m
18.54 kN/m
)
wL
M
M
w
L
 Ans. 
 
 
 
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8.49 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material 
bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.49b. The 
elastic modulus of the wood is E = 12 GPa and the elastic modulus of the CFRP is 112 GPa. The simply 
supported beam spans 6 m and carries a concentrated load P at midspan (Fig. P8.49a). 
(a) Determine the maximum bending stresses produced in the timber and the CFRP if P = 4 kN. 
(b) Assume that the allowable bending stresses of the timber and the CFRP are 9 MPa and 1,500 MPa, 
respectively. Determine the largest acceptable magnitude for concentrated load P. (You may neglect the 
weight of the beam in your calculations.) 
 
Fig. P8.49a 
Fig. P8.49b 
Solution 
Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is: 
 
2
1
112 GPa
9.3333
12 GPa
E
n
E
 
Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio: 
b2, trans = 9.3333(40 mm) = 373.33 mm. Thus, for calculation purposes, the 40 mm × 3 mm CFRP is 
replaced by a wood board that is 373.33-mm wide and 3-mm thick. 
 
Centroid location of the transformed section in the vertical direction 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
timber (1) 90 250 22,500 128 2,880,000 
transformed CFRP (2) 373.33 3 1,120 1.5 1,680 
 23,620 
 
2,881,680
 
 3
2
2,881,680 mm
23,620 mm
122.00 mm
i i
i
y A
y
A
 (measured upward from bottom edge of section) 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
timber (1) 117,187,500 6.00 810,000 117,997,500 
transformed CFRP (2) 840 –120.50 16,262,680 16,263,520 
Moment of inertia about the z axis = 
 134,261,020 mm
4
 
 = 134.261 ×10
6
 mm
4 
 
 
 
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Maximum bending moment in beam for P = 4 kN 
The maximum bending moment in the simply supported beam with a concentrated load of 4 kN at 
midspan is: 
 
6
max
(4 kN)(6 m)
6 kN-m 6 10 N-mm
4 4
PL
M
 
 
(a) Bending stress in timber (1) 
From the flexure formula, the maximum bending stress in timber (1) is: 
 6
1 6 4
(6 10 N-mm)(253 mm 122.00 mm)
134.261 1
5.85 MPa (C)
0 mm
My
I
 Ans. 
 
(a) Bending stress in CFRP (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in the CFRP is: 
 6
2 6 4
(6 10 N-mm)( 122.00 mm)
(9.3333)
134.26110 m
50.9 MPa (
m
T)
My
I
 Ans. 
 
(b) Determine maximum P 
If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be 
supported by the beam is: 
 2 6 4
61
1 max
(9 N/mm )(134.261 10 mm )
9.224 10 N-mm
(253 mm 122.00 mm)
IMy
M
I y
 
If the allowable bending stress in the CFRP is 1,500 MPa, then the maximum bending moment that may 
be supported by the beam is: 
 2 6 4
62
2 max
(1,500 N/mm )(134.261 10 mm )
176.867 10 N-mm
(9.3333)(122.00 mm)
IMy
n M
I ny
 
Note: The negative signs were omitted in the previous two equations because only the moment 
magnitude is of interest here. 
 
From these two results, the maximum moment that the beam can support is 9.224×10
6
 N-mm. The 
maximum concentrated load magnitude P that can be supported is found from: 
 
max
6
max
4
4 4(9.224 10 N-mm)(1 m/1000 mm)
6,149 N
(6 m)
6.15 kN
PL
M
M
P
L
 Ans. 
 
 
 
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8.50 Two steel plates, each 4 in. wide and 0.25 in. 
thick, reinforce a wood beam that is 3 in. wide and 
8 in. deep. The steel plates are attached to the 
vertical sides of the wood beam in a position such 
that the composite shape is symmetric about the z 
axis, as shown in the sketch of the beam cross 
section (Fig. P8.50). Determine the maximum 
bending stresses produced in both the wood and the 
steel if a bending moment of Mz = +50 kip-in is 
applied about the z axis. Assume Ewood = 2,000 ksi 
and Esteel = 30,000 ksi. 
 
 Fig. P8.50 
Solution 
Let the wood be denoted as material (1) and the steel plates as material (2). The modular ratio is: 
 
2
1
30,000 ksi
15
2,000 ksi
E
n
E
 
Transform the steel plates (2) into an equivalent amount of wood (1) by multiplying the plate 
thicknesses by the modular ratio: b2, trans = 15(0.25 in.) = 3.75 in. (each). Thus, for calculation purposes, 
each 4 in. × 0.25 in. steel plate is replaced by a wood board that is 4-in. tall and 3.75-in. wide. 
 
Centroid location: Since the transformed section is doubly symmetric, the centroid location is found 
from symmetry. 
 
Moment of inertia about the z centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
wood beam (1) 128 0 0 128 
two transformed steel plates (2) 40 0 0 40 
Moment of inertia about the z axis = 168 in.
4
 
 
Bending stress in wood beam (1) 
From the flexure formula, the maximum bending stress in wood beam (1) is: 
 
1 4
(50 kip-in.)(4 in.)
1
1.190 ksi 1,190 ps
68 in.
iz
z
M c
I
 Ans. 
 
Bending stress in steel plates (2) 
The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the 
maximum bending stress in the steel plates (2) is: 
 
2 4
(50 kip-in.)(2 in.)
(15)
168 in.
8.93 ksi 8,930 psiz
z
M c
n
I
 Ans. 
 
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8.51 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material 
bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.51b. The 
elastic modulus of the wood is 1,700 ksi and the elastic modulus of the CFRP is 23,800 ksi. The simply 
supported beam spans 24 ft and carries two concentrated loads P, which act at the quarter-points of the 
span (Fig. P8.51a). The allowable bending stresses of the timber and the CFRP are 2,400 psi and 
175,000 psi, respectively. Determine the largest acceptable magnitude for the concentrated loads P. 
(You may neglect the weight of the beam in your calculations.) 
 
 
Fig. P8.51a 
Fig. P8.51b 
Solution 
Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is: 
 
2
1
23,800 ksi
14
1,700 ksi
E
n
E
 
Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio: 
b2, trans = 14(3 in.) = 42 in. Thus, for calculation purposes, the 3 in. × 0.125 in. CFRP is replaced by a 
wood board that is 42-in. wide and 0.125-in. thick. 
 
Centroid location of the transformed section in the vertical direction 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
timber (1) 5.5 12 66 6.125 404.25 
transformed CFRP (2) 42.0 0.125 5.25 0.0625 0.3281 
 71.25 
 
404.5781
 
 3
2
404.5781 in.
71.25 in.
5.6783 in.
i i
i
y A
y
A
 (measured upward from bottom edge of section) 
 
Moment of inertia about the horizontal centroidal axis 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
timber (1) 792 0.4467 13.1703 805.170 
transformed CFRP (2) 0.00684 –5.6158 165.5697 165.577 
Moment of inertia about the z axis = 970.747 in.
4 
 
 
 
 
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Determine maximum P 
If the allowable bending stress in the timber is 2,400 psi, then the maximum bending moment that may 
be supported by the beam is: 
 4
1
1 max
(2.40 ksi)(970.747 in. )
361.393 kip-in.
(12.125 in. 5.6783 in.)
IMy
M
I y
 
If the allowable bending stress in the CFRP is 175,000 psi, then the maximum bending moment that may 
be supported by the beam is: 
 4
2
2 max
(175 ksi)(970.747 in. )
2,137 kip-in.
(14)(5.6783 in.)
IMy
n M
I ny
 
Note: The negative signs were omitted in the previous two equations because only the moment 
magnitude is of interest here. 
 
From these two results, the maximum moment that the beam can support is 351.393 kip-in. = 30.116 
kip-ft. The maximum concentrated load magnitude P that can be supported is found from: 
 
max
max
(6 ft)
30.116 kip-ft
6 ft 6 ft
5.02 kips
M P
M
P Ans. 
 
 
 
 
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8.52 A steel pipe assembly supports a 
concentrated load of P = 22 kN as shown in 
Fig. P8.52. The outside diameter of the pipe is 
142 mm and the wall thickness is 6.5 mm. 
Determine the normal stresses produced at 
points H and K. 
Fig. P8.52 
Solution 
Section properties 
 
2 2 2 2 2
4 4 4 4 4
2 142mm 2(6.5 mm) 129 mm
(142 mm) (129 mm) 2,766.958 mm
4 4
(142 mm) (129 mm) 6,364,867 mm
64 64
z
d D t
A D d
I D d
 
 
    
          
          
 
 
Internal forces and moments 
 
22 kN 22,000 N
(22,000 N)(370 mm) 8,140,000 N-mmz
F
M
 
 
 
 
Stresses 
 
axial 2
bending 4
22,000 N
7.951 MPa (C)
2,766.958 mm
(8,140,000 N-mm)(142 mm/2)
90.802 MPa
6,364,867 mm
z
z
F
A
M c
I


  
   
 
 
Normal stress at H 
By inspection, the bending stress at H will be compression; therefore, the normal stress at H is: 
 
7.951 MPa 90.802 MPa 98.753 98.8 MPa (C)MPaH       Ans. 
 
Normal stress at K 
By inspection, the bending stress at K will be tension; therefore, the normal stress at K is: 
 
7.951 MPa 90.802 MPa 82.851 82.9 MPa (T)MPaK       Ans. 
 
 
 
 
 
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8.53 The screw of a clamp exerts a compressive 
force of 350 lb on the wood blocks. Determine the 
normal stresses produced at points H and K. The 
clamp cross-sectional dimensions at the section of 
interest are 1.25 in. by 0.375 in. thick. 
 
Fig. P8.53 
Solution 
Section properties 
 
2
3
4
(0.375 in.)(1.250 in.) 0.468750 in.
(0.375 in.)(1.250 in.)
0.061035 in.
12
z
A
I
 
 
 
 
Internal forces and moments 
 
350 lb
(350 lb)(3.75 in. 1.25 in./2) 1,531.25 lb-in.z
F
M

  
 
 
Stresses 
 
axial 2
bending 4
350 lb
746.667 psi (T)
0.468750 in.
(1,531.25 lb-in.)(1.250 in./2)
15,680.0 psi
0.061035 in.
z
z
F
A
M c
I


  
   
 
 
Normal stress at H 
By inspection, the bending stress at H will be tension; therefore, the normal stress at H is: 
 
746.667 psi 15,680 psi 16,426.667 psi 16,430 psi (T)H     Ans. 
 
Normal stress at K 
By inspection, the bending stress at K will be compression; therefore, the normal stress at K is: 
 
746.667 psi 15,680 psi 14,933.333 ps 14,930 i psi (C)K      Ans. 
 
 
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8.54 Determine the normal stresses produced at points H and K of the pier support shown in Fig. P8.54a. 
 
 
Fig. P8.54a Fig. P8.54b Cross section a–a 
Solution 
Section properties 
 
2
3
9 4
(250 mm)(500 mm) 125,000 mm
(250 mm)(500 mm)
2.60417 10 mm
12
z
A
I
 
  
 
 
Internal forces and moments 
 
250 kN 400 kN 650 kN
(250 kN)(3.25 m) (400 kN)(2.25 m) 87.50 kN-mz
F
M
  
   
 
 
Stresses 
 
axial 2
2
bending 9 4
650,000 N
5.20 MPa (C)
125,000 mm
(87.5 kN-m)(500 mm/2)(1,000)
8.40 MPa
2.60417 10 mm
z
z
F
A
M c
I


  
   

 
 
Normal stress at H 
By inspection, the bending stress at H will be tension; therefore, the normal stress at H is: 
 
5.20 MPa 8.40 MPa 3.20 MP 3.20 MPa (T)aH      Ans. 
 
Normal stress at K 
By inspection, the bending stress at K will be compression; therefore, the normal stress at K is: 
 
5.20 MPa 8.40 MPa 13.60 MPa 13.60 MPa (C)K       Ans. 
 
 
 
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8.55 A tubular steel column CD supports 
horizontal cantilever arm ABC, as shown in Fig. 
P8.55. Column CD has an outside diameter of 
10.75 in. and a wall thickness of 0.365 in. 
Determine the maximum compression stress at 
the base of column CD. 
 
 Fig. P8.55 
Solution 
Section properties 
 
2 2 2 2 2
4 4 4 4 4
2 10.750 in. 2(0.365 in.) 10.020 in.
(10.750 in.) (10.020 in.) 11.908 in.
4 4
(10.750 in.) (10.020 in.) 160.734 in.
64 64
z
d D t
A D d
I D d
 
 
    
          
          
 
 
Internal forces and moments 
 
700 lb 900 lb 1,600 lb
(700 lb)(13 ft) (900 lb)(23 ft) 29,800 lb-ft 357,600 lb-in.
F
M
  
   
 
 
Stresses 
 
axial 2
bending 4
1,600 lb
134.36 psi (C)
11.908 in.
(357,600 lb-in.)(10.75 in./2)
11,958.27 psi
160.734 in.
F
A
M c
I


  
   
 
 
Maximum compression stress at base of column 
 
compression 134.36 psi 11,958.27 psi 12,092.63 12.09 ksi ( psi C)       Ans. 
 
 
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8.56 Determine the normal stresses acting at points H and K for the structure shown in Fig. P8.56a. The 
cross-sectional dimensions of the vertical member are shown in Fig. P8.56b. 
Fig. P8.56a 
 
 
Fig. P8.56b Cross section 
Solution 
Section properties 
 
2
3
4
(4 in.)(8 in.) 32 in.
(4 in.)(8 in.)
170.6667 in.
12
z
A
I
 
 
 
 
Internal forces and moments 
 
1,200 lb 2,800 lb 4,000 lb
(1,200 lb)(12 in. 8 in./2) 19,200 lb-in.z
F
M
  
  
 
 
Stresses 
 
axial 2
bending 4
4,000 lb
125 psi (C)
32 in.
(19,200 lb-in.)(8 in./2)
450 psi
170.6667 in.
z
z
F
A
M c
I


  
   
 
 
Normal stress at H 
By inspection, the bending stress at H will be compression; therefore, the normal stress at H is: 
 
125 psi 450 psi 575 p 575 psi (C)siH       Ans. 
 
Normal stress at K 
By inspection, the bending stress at K will be tension; therefore, the normal stress at K is: 
 
125 psi 450 psi 325 ps 325 psi (T)iK      Ans. 
 
 
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8.57 A W18 × 35 standard steel shape is 
subjected to a tension force P that is applied 
15 in. above the bottom surface of the wide-
flange shape as shown in Fig. P8.57. If the 
tension normal stress of the upper surface of 
the W-shape must be limited to 18 ksi, 
determine the allowable force P that may be 
applied to the member.Fig. P8.57 
Solution 
Section properties (from Appendix B) 
 2
4
Depth 17.7 in.
10.3 in.
510 in.z
d
A
I



 
 
Stresses 
 
axial 2
2
bending 4 4 4
10.3 in.
(15 in. 17.7 in./2)(17.7 in./2) (6.15 in.)(8.85 in.) (54.4275 in. )
510 in. 510 in. 510 in.
z
z
F P
A
M c P P P
I


 

   
 
 
Normal stress on the upper surface of the W-shape 
The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. 
Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given 
by: 
 
2
upper surface 2 4
2 2
2
(54.4275 in. )
10.3 in. 510 in.
(0.097087 in. 0.106721 in. )
(0.203808 in. )
P P
P
P

 

 
 
 
The normal stress on the upper surface of the W-shape must be limited to 18 ksi; therefore, 
 
2
2
(0.203808 in. ) 18 ksi
18 ksi
0.203808 in
88.3 kip
.
s
P
P



   Ans. 
 
 
 
 
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8.58 A WT305 × 41 standard steel shape is 
subjected to a tension force P that is applied 250 
mm above the bottom surface of the tee shape, as 
shown in Fig. P8.58. If the tension normal stress 
of the upper surface of the WT-shape must be 
limited to 150 MPa, determine the allowable 
force P that may be applied to the member. 
 
 
Fig. P8.58 
Solution 
Section properties (from Appendix B) 
 2
6 4
Depth 300 mm
Centroid 88.9 mm (from flange to centroid)
5,230 mm
48.7 10 mmz
d
y
A
I



 
 
 
Stresses 
 
4 2
axial 2
(1.9120 10 mm )
5,230 mm
F P
P
A
     
 
 
bending 6 4
6 4
4 2
(250 mm 88.9 mm)(300 mm 88.9 mm)
48.7 10 mm
(161.1 mm)(211.1 mm)
48.7 10 mm
(6.9832 10 mm )
z
z
M c P
I
P
P

 
 
 



 
 
 
Normal stress on the upper surface of the WT-shape 
The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. 
Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given 
by: 
 
4 2 4 2
upper surface
4 2
(1.9120 10 mm ) (6.9832 10 mm )
(8.8953 10 mm )
P P
P
    
 
   
  
The normal stress on the upper surface of the WT-shape must be limited to 150 MPa; therefore, 
 
4 2
2
4 2
(8.8953 10 mm ) 150 MPa
150 N/mm
168,629 N
8.8953 10 mm
168.6 kN
P
P
 
 
 
   

 Ans. 
 
 
 
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8.59 A pin support consists of a vertical 
plate 60 mm wide by 10 mm thick. The pin 
carries a load of 1,200 N. Determine the 
normal stresses acting at points H and K for 
the structure shown in Fig. P8.59. 
 
 Fig. P8.59 
Solution 
Section properties 
 
2
3
4
(60 mm)(10 mm) 600 mm
(60 mm)(10 mm)
5,000 mm
12
A
I
 
 
 
 
Internal forces and moments 
 
1,200 N
(1,200 N)(30 mm 10 mm/2) 42,000 N-mm
F
M

  
 
 
Stresses 
 
axial 2
bending 4
1,200 N
2.00 MPa (T)
600 mm
(42,000 N-mm)(10 mm/2)
42.00 MPa
5,000 mm
F
A
M c
I


  
   
 
 
Normal stress at H 
By inspection, the bending stress at H will be compression; therefore, the normal stress at H is: 
 
40.0 MPa (C2.00 MPa 42.00 MPa 40.00 ) MPaH      Ans. 
 
Normal stress at K 
By inspection, the bending stress at K will be tension; therefore, the normal stress at K is: 
 
2.00 MPa 42.00 MPa 44.00 MPa 44.0 MPa (T)K     Ans. 
 
 
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8.60 The tee shape shown in Fig. P8.60b is used as a short post to support a load of P = 4,600 lb. The 
load P is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.60a. Determine 
the normal stresses at points H and K, which are located on section a–a. 
 Fig. P8.60a 
 
 
Fig. P8.60b Cross-sectional dimensions 
Solution 
Centroid location in x direction: 
Shape width b height h Area Ai 
xi 
(from left) xi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
flange 12 2 24 1 24 
stem 2 10 20 7 140 
 44 in.
2
 
 
164 in.
3 
 
 
3
2
164 in.
3.7273 in. (from left side to centroid)
44 in.
8.2727 in. (from right side to centroid)
i i
i
x A
x
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = xi – x d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
flange 8 −2.7273 178.5160 186.5160 
stem 166.6667 3.2727 214.2113 380.8790 
Moment of inertia about the z axis (in.
4
) = 567.3940 
 
Internal forces and moments 
 
4,600 lb
(4,600 lb)(5 in. 3.7273 in.) 40,145.455 lb-in.z
F
M

  
 
 
 
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Stresses 
 
axial 2
,bending 4
,bending 4
4,600 lb
104.545 psi
44 in.
(40,145.455 lb-in.)( 3.7273 in.)
263.720 psi
567.3940 in.
(40,145.455 lb-in.)(8.2727 in.)
585.329 psi
567.3940 in.
z
H
z
z
K
z
F
A
M x
I
M x
I




   

   
  
 
 
Normal stress at H 
 
104.545 psi 263.720 psi 368.265 p 368 psi (C)siH       Ans. 
 
Normal stress at K 
 
104.545 psi 585.329 psi 480.784 p 481 psi (si T)K      Ans. 
 
 
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8.61 The tee shape shown in Fig. P8.61b is used as a short post to support a load of P. The load P is 
applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.61a. The tension and 
compression normal stresses in the post must be limited to 1,000 psi and 800 psi, respectively. 
Determine the maximum magnitudeof load P that satisfies both the tension and compression stress 
limits. 
 Fig. P8.61a 
 
 
Fig. P8.61b Cross-sectional dimensions 
Solution 
Centroid location in x direction: 
Shape width b height h Area Ai 
xi 
(from left) xi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
flange 12 2 24 1 24 
stem 2 10 20 7 140 
 44 in.
2
 
 
164 in.
3 
 
 
3
2
164 in.
3.7273 in. (from left side to centroid)
44 in.
8.2727 in. (from right side to centroid)
i i
i
x A
x
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = xi – x d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
flange 8 −2.7273 178.5160 186.5160 
stem 166.6667 3.2727 214.2113 380.8790 
Moment of inertia about the z axis (in.
4
) = 567.3940 
 
Internal forces and moments 
 
(5 in. 3.7273 in.) (8.7273 in.)z
F P
M P P

  
 
 
 
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Stresses 
 
2
axial 2
2
,bending 4
2
,bending 4
(0.022727 in. )
44 in.
(8.7273 in.) ( 3.7273 in.)
(0.057331 in. )
567.3940 in.
(8.7273 in.) (8.2727 in.)
(0.127246 in. )
567.3940 in.
z
H
z
z
K
z
F P
P
A
M x P
P
I
M x P
P
I






    

   
  
 
 
Compression stress limit (at H) 
 
2 2 2
2
(0.022727 in. ) (0.057331 in. ) (0.080058 in. )
(0.080058 in. ) 800 psi
9,992.76 lb
H P P P
P
P
   

    

  
 
Tension stress limit (at K) 
 
2 2 2
2
(0.022727 in. ) (0.127246 in. ) (0.104519 in. )
(0.104519 in. ) 1,000 psi
9,567.64 lb
K P P P
P
P
   

   

  
 
Maximum magnitude of load P 
 
max 9,570 lbP 
 Ans. 
 
 
 
 
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8.62 The tee shape shown in Fig. P8.62b is used as a post that supports a load of P = 25 kN. Note that 
the load P is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.62a. Determine the 
normal stresses at points H and K. 
 
 
Fig. P8.62a Fig. P8.62b Cross-sectional dimensions 
Solution 
Centroid location in x direction: 
Shape width b height h Area Ai 
xi 
(from left) xi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
stem 20 130 2,600 65 169,000 
flange 120 20 2,400 140 336,000 
 5,000 
 
505,000
 
 
 
3
2
505,000 mm
101.0 mm (from left side to centroid)
5,000 mm
49.0 mm (from right side to centroid)
i i
i
x A
x
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = xi – x d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67 
flange 80,000.00 39.0 3,650,400.00 3,730,400.00 
Moment of inertia about the z axis (mm
4
) = 10,761,666.67 
 
Internal forces and moments 
 
25 kN 25,000 N
(25,000 N)(400 mm 49.0 mm) 11,225,000 N-mmz
F
M
 
    
 
 
 
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Stresses 
 
axial 2
,bending 4
,bending 4
25,000 N
5 MPa
5,000 mm
( 11,225,000 N-mm)( 101.0 mm)
105.35 MPa
10,761,666.67 mm
( 11,225,000 N-mm)(49.0 mm)
51.11 MPa
10,761,666.67 mm
z
H
z
z
K
z
F
A
M x
I
M x
I




   
 
  

   
 
 
Normal stress at H 
 
5 MPa 105.35 M 100.4 P MPa (Ta )H     Ans. 
 
Normal stress at K 
 
5 MPa 51.11 M 56.1 MPa (CPa )K    
 Ans. 
 
 
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8.63 The tee shape shown in Fig. P8.63b is used as a post that supports a load of P, which is applied 400 
mm from the flange of the tee shape, as shown in Fig. P8.63a. The tension and compression normal 
stresses in the post must be limited to 165 MPa and 80 MPa, respectively. Determine the maximum 
magnitude of load P that satisfies both the tension and compression stress limits. 
 
 
Fig. P8.63a Fig. P8.63b Cross-sectional dimensions 
Solution 
Centroid location in x direction: 
Shape width b height h Area Ai 
xi 
(from left) xi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
stem 20 130 2,600 65 169,000 
flange 120 20 2,400 140 336,000 
 5,000 
 
505,000
 
 
 
3
2
505,000 mm
101.0 mm (from left side to centroid)
5,000 mm
49.0 mm (from right side to centroid)
i i
i
x A
x
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = xi – x d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67 
flange 80,000.00 39.0 3,650,400.00 3,730,400.00 
Moment of inertia about the z axis (mm
4
) = 10,761,666.67 
 
Internal forces and moments 
 
(400 mm 49.0 mm) (449.0 mm)z
F P
M P P

    
 
 
 
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Stresses 
 
4 2
axial 2
3 2
,bending 4
3 2
,bending 4
(2 10 mm )
5,000 mm
( 449 mm) ( 101.0 mm)
(4.21394 10 mm )
10,761,666.67 mm
( 449 mm) (49.0 mm)
(2.04439 10 mm )
10,761,666.67 mm
z
H
z
z
K
z
F P
P
A
M x P
P
I
M x P
P
I



 
 
 
     
 
   

    
 
 
Tension stress limit (at H) 
 
4 2 3 2
3 2
3 2 2
(2 10 mm ) (4.21394 10 mm )
(4.01394 10 mm )
(4.01394 10 mm ) 165 N/mm
41,106.7 N
H P P
P
P
P
    
 
 
    
 
 
  
 
Compression stress limit (at K) 
 
4 2 3 2 3 2
3 2 2
(2 10 mm ) (2.04439 10 mm ) (2.24439 10 mm )
(2.24439 10 mm ) 80 N/mm
35,644.43 N
K P P P
P
P
      
 
       
 
  
 
Maximum magnitude of load P 
 
max 35.6 kNP 
 Ans. 
 
 
 
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8.64 The tee shape shown in Fig. P8.64b is used as a post that supports a load of P = 25 kN, which is 
applied 400 mm from the flange of the tee shape, as shown in Fig. P8.64a. Determine the magnitudes 
and locations of the maximum tension and compression normal stresses within the vertical portion BC of 
the post. 
 
 
Fig. P8.64a Fig. P8.64b Cross-sectional dimensions 
Solution 
Centroid location in x direction: 
Shape width b height h Area Ai 
xi 
(from left) xi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
stem 20 130 2,600 65 169,000 
flange 120 20 2,400 140 336,000 
 5,000 
 
505,000
 
 
 
3
2
505,000 mm
101.0 mm (from left side to centroid)
5,000 mm
49.0 mm (from right side to centroid)
i i
i
x A
x
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = xi – x d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67 
flange 80,000.00 39.0 3,650,400.00 3,730,400.00 
Moment of inertia about the z axis (mm
4
) = 10,761,666.67 
 
Internal forces and moments 
 
(25 kN)cos35 20.4788 kN 20,478.8 N (vertical component)
(25 kN)sin 35 14.3394 kN 14,339.4 N (horizontal component)
at B (20,478.8 N)(400 mm 49.0 mm) 9,194,981.2 N-mm
at C (20,478.8 N)(400 m
z
z
F
V
M
M
   
   
    
  m 49.0 mm) (14,339.4 N)(1,200 mm) 8,012,298.8 N-mm  
 
 
 
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Normal stress at H at location B 
 
axial 2
,bending 4
20,478.8 N
4.0958 MPa
5,000 mm
( 9,194,981.2 N-mm)( 101.0 mm)
86.2964 MPa
10,761,666.67 mm
4.0958 MPa 86.2964 MPa 82.2 MPa
z
H
z
H
F
A
M x
I




   
 
  
   
 
 
Normal stress at H at location C 
 ,bending 4
(8,012,298.8 N-mm)( 101.0 mm)
75.1967 MPa
10,761,666.67 mm
4.0958 MPa 75.1967 MPa 79.3 MPa
z
H
z
H
M x
I



   
    
 
 
Normal stress at K at location B 
 ,bending 4
( 9,194,981.2 N-mm)(49.0 mm)
41.8666 MPa
10,761,666.67 mm
4.0958 MPa 41.8666 MPa 46.0 MPa
z
K
z
K
M x
I



   
    
 
 
Normal stress at K at location C 
 ,bending 4
(8,012,298.8 N-mm)(49.0 mm)
36.4816 MPa
10,761,666.67 mm
4.0958 MPa 36.4816 MPa 32.4 MPa
z
K
z
K
M x
I


  
   
 
 
Maximum tension stress 
 
max tension 82.2 MP at locaa (T) tion B  Ans. 
 
Maximum compression stress 
 
max compression 79.3 at loMPa cat (C) ion C  Ans. 
 
 
 
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8.65 A beam with a box cross section is subjected to 
a resultant moment magnitude of 2,100 N-m acting 
at the angle shown in Fig. P8.65. Determine: 
(a) the maximum tension and the maximum 
compression bending stresses in the beam. 
(b) the orientation of the neutral axis relative to the 
+z axis. Show its location on a sketch of the cross 
section. 
 
Fig. P8.65 
Solution 
Section properties 
 
3 3
4
3 3
4
(90 mm)(55 mm) (80 mm)(45 mm)
640,312.5 mm
12 12
(55 mm)(90 mm) (45 mm)(80 mm)
1, 421, 250.0 mm
12 12
y
z
I
I
 
 
Moment components 
 (2,100 N-m)sin30 1,050 N-m
(2,100 N-m)cos30 1,818.65 N-m
y
z
M
M
 
 
(a) Maximum bending stresses 
For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. 
Compute normal stress at y = 45 mm, z = 27.5 mm: 
 
4 4
(1,050 N-m)(27.5 mm)(1,000 mm/m) ( 1,818.65 N-m)(45 mm)(1,000 mm/m)
640,312.5 mm 1,421,250.0 mm
45.0952 MPa 57.5827 MPa
102.6 102.7 MPa (T)779 MPa
y z
x
y z
M z M y
I I
 Ans. 
 
Compute normal stress at y = −45 mm, z = −27.5 mm: 
 
4 4
(1,050 N-m)( 27.5 mm)(1,000 mm/m) ( 1,818.65 N-m)( 45 mm)(1,000 mm/m)
640,312.5 mm 1,421,250.0 mm
45.0952 MPa 57.5827 MPa
102 102.7 MPa (C.6779 MPa )
y z
x
y z
M z M y
I I
 Ans. 
 
(b) Orientation of neutral axis 
For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of 
the neutral axis: 
 
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4
4
(1,050 N-m)(1,421,250.0 mm )
tan 1.2815
( 1,818.65 N-m)(640
52.03
,312.5 mm )
(i.e., 52.03 CCW from axis)
y z
z y
M I
M I
z Ans. 
 
 
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8.66 The moment acting on the cross section of the 
T-beam has a magnitude of 22 kip-ft and is oriented 
as shown in Fig. P8.66. Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the orientation of the neutral axis relative to the 
+z axis. Show its location on a sketch of the cross 
section. 
 
Fig. P8.66 
Solution 
Section properties 
Centroid location in y direction: 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
top flange 7.00 1.25 8.7500 8.375 73.28125 
stem 0.75 7.75 5.8125 3.875 22.52344 
 14.5625 
 
95.80469
 
 
 
3
2
95.80469 in.
6.5789 in. (from bottom of shape to centroid)
14.5625 in.
2.4211 in. (from top of shape to centroid)
i i
i
y A
y
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 1.1393 1.7961 28.2273 29.3666 
stem 29.0928 −2.7039 42.4956 71.5884 
Moment of inertia about the z axis (in.
4
) = 100.9550 
 
Moment of inertia about the y axis: 
 3 3
4(1.25 in.)(7.00 in.) (7.75 in.)(0.75 in.) 36.0016 in.
12 12
yI
 
 
Moment components 
 (22 kip-ft)cos55 12.6187 kip-ft 151.4242 kip-in.
(22 kip-ft)sin55 18.0213 kip-ft 216.2561 kip-in.
y
z
M
M
 
 
 
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(a) Bending stress at H 
For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. 
To compute the normal stress at H, use the (y, z) coordinates y = 2.4211 in. and z = −3.5 in.: 
 
4 4
( 151.4242 kip-in.)( 3.50 in.) ( 216.2561 kip-in.)(2.4211 in.)
36.0016 in. 100.9550 in.
14.7211 ksi 5.1862 ksi
19.9 19.91 ksi (T)074 ksi
y z
x
y z
M z M y
I I
 Ans. 
 
(b) Bending stress at K 
To compute the normal stress at K, use the (y, z) coordinates y = −6.5789 in. and z = 0.375 in.: 
 
4 4
( 151.4242 kip-in.)(0.375 in.) ( 216.2561 kip-in.)( 6.5789 in.)
36.0016 in. 1
15.67 ksi (C)
00.9550 in.
1.5773 ksi 14.0927 ksi
15.6700 ksi
y z
x
y z
M z M y
I I
 Ans. 
 
(c) Orientation of neutral axis 
For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of 
the neutral axis: 
 
4
4
( 151.4242 kip-in.)(100.9550 in. )
tan 1.9635
( 216.2561 kip-i
63.0
n.)(36.0016 in. )
(i.e., 63.01 CW from ax1 is)
y z
z y
M I
M I
z Ans. 
 
 
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8.67 A beam with a box cross section is subjected to 
a resultant moment magnitude of 75 kip-in. acting 
at the angle shown in Fig. P8.67. Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the maximum tension and the maximum 
compression bending stresses in the beam. 
(d) the orientation of the neutral axis relative to the 
+z axis. Show its location on a sketch of the cross 
section. 
 
Fig. P8.67 
Solution 
Section properties 
 
3 3
4
3 3
4
(4 in.)(6 in.) (3.25 in.)(5.25 in.)
32.8096 in.
12 12
(6 in.)(4 in.) (5.25 in.)(3.25 in.)
16.9814 in.
12 12
y
z
I
I
 
 
Moment components 
 (75 kip-in.)cos 20 70.4769 kip-in.
(75 kip-in.)sin 20 25.6515 kip-in.
y
z
M
M
 
 
(a) Bending stress at H 
For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. 
To compute the normal stress at H, use the (y, z) coordinates y = −2.0 in. and z = −3.0 in.: 
 
4 4
(70.4769 kip-in.)( 3.0 in.) (25.6515 kip-in.)( 2.0 in.)
32.8096 in. 1
3.42 ksi (C)
6.9814 in.
6.4442 ksi 3.0211 ksi
3.4231 ksi
y z
x
y z
M z M y
I I
 Ans. 
 
(b) Bending stress at K 
To compute the normal stress at K, use the (y, z) coordinates y = 2.0 in. and z = 3.0 in.: 
 
4 4
(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)(2.0 in.)
32.8096 in. 16.9814 in.
6.4442 ksi 3.0211 ksi
3.4231 k 3.42 s ksi (T)i
y z
x
y z
M z M y
I I
 Ans. 
 
 
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(c) Maximum bending stresses 
The maximum tension normal stress occurs at the (y, z) coordinates y = −2.0 in. and z = 3.0 in.: 
 
4 4
(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)( 2.0 in.)
32.8096 in. 16.9814 in.
6.4442 ksi 3.0211 ksi
9.46 9.47 ksi (T53 ksi )
y z
x
y z
M z M y
I I
 Ans. 
 
The maximum compression normal stress occurs at the (y, z) coordinates y = 2.0 in. and z = −3.0 in.: 
 
4 4
(70.4769 kip-in.)( 3.0 in.) (25.6515 kip-in.)(2.0 in.)
32.8096 in. 16.9814 in.
6.4442 ksi 3.0211 ksi
9.4653 k 9.47 ksi (i C)s
y z
x
y z
M z M y
I I
 Ans. 
 
(d) Orientation of neutral axis 
For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of 
the neutral axis: 
 
4
4
(70.4769 kip-in.)(16.9814 in. )
tan 1.4220
(25.6515 kip-in.)
54.88
(32.8096 in. )
(i.e., 54.88 CW from axis)
y z
z y
M I
M I
z Ans. 
 
 
 
 
 
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8.68 The moment acting on the cross section of the 
wide-flange beam has a magnitude of M = 12 kN-m and 
is oriented as shown in Fig. P8.68. Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the orientation of the neutral axis relative to the +z 
axis. Show its location on a sketch of the cross section. 
 
 Fig. P8.68 
Solution 
Section properties 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 59,062.5 97.5 29,944,687.5 30,003,750 
web 4,860,000 0 0 4,860,000 
bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750 
Moment of inertia about the z axis (mm
4
) = 64,867,500 
 
Moment of inertia about the y axis: 
 3 3
4(15 mm)(210 mm) (180 mm)(10 mm)2 23,167,500 mm
12 12
yI
 
 
Moment components 
 6
6
(12 kN-m)sin 35 6.8829 kN-m 6.8829 10 N-mm
(12 kN-m)cos35 9.8298 kN-m 9.8298 10 N-mm
y
z
M
M
 
 
(a) Bending stress at H 
For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses. 
To compute the normal stress at H, use the (y, z) coordinates y = 105 mm and z = −105 mm: 
 
6 6
4 4
(6.8829 10 N-mm)( 105 mm) (9.8298 10 N-mm)(105 mm)
23,167,500 mm 64,
47.1 MPa (C)
867,500 mm
31.1948 MPa 15.9114 MPa
47.1062 MPa
y z
x
y z
M z M y
I I
 Ans. 
 
 
 
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(b) Bending stress at K 
To compute the normal stress at K, use the (y, z) coordinates y = −105 mm and z = 105 mm: 
 
6 6
4 4
(6.8829 10 N-mm)(105 mm) (9.8298 10 N-mm)( 105 mm)
23,167,500 mm 64,867,500 mm
31.1948 MPa 15.9114 MPa
47. 47.1 1062 M MPa (P T)a
y z
x
y z
M z M y
I I
 Ans. 
 
 
(b) Orientation of neutral axis 
For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of 
the neutral axis: 
 
4
4
(6.8829 kN-m)(64,867,500 mm )
tan 1.9605
(9.8298 kN-m)(23,
62.
167,500 mm )
(i.e., 62.98 CW from axis8 )9
y z
z y
M I
M I
z Ans. 
 
 
 
 
 
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8.69 For the cross section shown in Fig. P8.69, 
determine the maximum magnitude of the bending 
moment M so that the bending stress in the wide-
flange shape does not exceed 165 MPa. 
 
 Fig. P8.69 
Solution 
Section properties 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 59,062.5 97.5 29,944,687.5 30,003,750 
web 4,860,000 0 0 4,860,000 
bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750 
Moment of inertia about the z axis (mm
4
) = 64,867,500 
 
Moment of inertia about the y axis: 
 3 3
4(15 mm)(210 mm) (180 mm)(10 mm)2 23,167,500 mm
12 12
yI
 
 
Moment components 
 
sin 35 cos35y zM M M M
 
 
Maximum bending moment magnitude 
The maximum tension bending stress should occur at point K, which has the (y, z) coordinates y = −105 
mm and z = 105 mm: 
 
4 4
sin35 (105 mm) cos35 ( 105 mm)
165 MPa
23,167,500 mm 64,867,500 mm
y z
x
y z
M z M y M M
I I
 
 
 
6 3 6 3 2
6
2.59957 10 mm 1.32595 10 mm 165 N/mm
42.0 kN-42.0327 10 N-mm m
M
M Ans. 
 
 
 
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8.70 The unequal-leg angle is subjected to a bending 
moment of Mz = 20 kip-in. that acts at the orientation 
shown in Fig. P8.70. Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the maximum tension and the maximum 
compression bending stresses in the cross section. 
(d) the orientation of the neutral axis relative to the +z 
axis. Show its location on a sketch of the cross section. 
 
 Fig. P8.70 
Solution 
Section properties 
Centroid location in y direction: 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
upright leg 0.375 4.000 1.5000 2.00 3.00 
bottom leg 2.625 0.375 0.9844 0.1875 0.18457 
 2.4844 
 
3.18457
 
 
3
2
3.18457 in.
1.2818 in.(from bottom of shape to centroid)
2.4844 in.
2.7182 in. (from top of shape to centroid)
i i
i
y A
y
A
 
 
Centroid location in z direction: 
Shape Area Ai 
zi 
(from right edge) zi Ai 
 (in.
2
) (in.) (in.
3
) 
upright leg 1.5000 0.1875 0.2813 
bottom leg 0.9844 1.6875 1.6612 
 2.4844 
 
1.94243
 
 
3
2
1.94243 in.
0.7818 in. (from right edge of shape to centroid)
2.4844 in.
2.2182 in. (from left edge of shape to centroid)
i i
i
z A
z
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
upright leg 2.000 0.7182 0.77372 2.7737 
bottom leg 0.011536 −1.0943 1.17881 1.1903 
Moment of inertia about the z axis (in.
4
) = 3.9640 
 
 
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Moment of inertia about the y axis: 
Shape IC d = zi – z d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
upright leg 0.017578 −0.5943 0.52979 0.5474 
bottom leg 0.565247 0.9057 0.80750 1.3727 
Moment of inertia about the y axis (in.
4
) = 1.9201 
 
Product of inertia about the centroidal axes: 
Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz 
 (in.
4
) (in.) (in.) (in.
2
) (in.
4
) (in.
4
) 
upright leg 0 0.7182 −0.5943 1.5000 −0.6402 −0.6402 
bottom leg 0 −1.0943 0.9057 0.9844 −0.9757 −0.9757 
Product of inertia (in.
4
) = −1.6159 
 
(a) Bending stress at H 
Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the 
bending stresses. Equation (8.22) will be used here. Note that the bending moment component about 
the y axis is zero (i.e., My = 0); therefore, the first term in Eq. (8.22) is eliminated. To compute the 
normal stress at H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.: 
 
2
4 4
4 4 4 2
5
8
(1.9201 in. )(2.7182 in.) ( 1.6159 in. )( 0.4068 in.)
(20 kip-in.)
(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )
4.5619 in.
(20 kip-in.)
5.0001 in.
18.2469 ks
y yz
x z
y z yz
I y I z
M
I I I
18.25 ksii (C)
 Ans. 
 
(b) Bending stress at K 
To compute the normal stress at K, use (y, z) coordinates of y = −0.9068 in. and z = 2.2182 in.: 
 
2
4 4
4 4 4 2
5
8
(1.9201 in. )( 0.9068 in.) ( 1.6159 in. )(2.2182 in.)
(20 kip-in.)
(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )
1.8432 in.
(20 kip-in.)
5.0001 in.
7.3728 ksi
y yz
x z
y z yz
I y I z
M
I I I
7.37 ksi (C)
 Ans. 
 
 
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(d) Orientation of neutral axis 
Since the angle shape has no axis of symmetry, Eq. (8.23) must be used to determine the orientation of 
the neutral axis: 
 
4
4
(20 kip-in.)( 1.6159 in. )
tan 0.8416
(20 kip-in.)(1.9201 in. )
(i.e., 40.08 CCW from axi0 s8 )40.
y z z yz
z y y yz
M I M I
M I M I
z Ans. 
 
 
 
(c) Maximum bending stresses 
Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are 
farthest from the neutral axis are point H and the corner of the angle. The bending stress at H has 
already been computed. To compute the normal stress at the corner of the angle, use (y, z) coordinates 
of y = −1.2818 in. and z = −0.7818 in. 
 
2
4 4
4 4 4 2
5
8
(1.9201 in. )( 1.2818 in.) ( 1.6159 in. )( 0.7818 in.)
(20 kip-in.)
(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )
3.7245 in.
(20 kip-in.)
5.0001 in.
14.8977 ksi
y yz
x z
y z yz
I y I z
M
I I I
14.90 ksi (T)
 
 
Therefore, the maximum compression bending stress is: 
 
18.25 ksi (C)x
 Ans. 
and the maximum tension bending stress is: 
 
14.90 ksi (T)x
 Ans. 
 
 
 
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8.71 For the cross section shown in Fig. P8.71, 
determine the maximum magnitude of the bendingmoment M so that the bending stress in the unequal-
leg angle shape does not exceed 24 ksi. 
 
 Fig. P8.71 
Solution 
Section properties 
Centroid location in y direction: 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
upright leg 0.375 4.000 1.5000 2.00 3.00 
bottom leg 2.625 0.375 0.9844 0.1875 0.18457 
 2.4844 
 
3.18457
 
 
3
2
3.18457 in.
1.2818 in.(from bottom of shape to centroid)
2.4844 in.
2.7182 in. (from top of shape to centroid)
i i
i
y A
y
A
 
 
Centroid location in z direction: 
Shape Area Ai 
zi 
(from right edge) zi Ai 
 (in.
2
) (in.) (in.
3
) 
upright leg 1.5000 0.1875 0.2813 
bottom leg 0.9844 1.6875 1.6612 
 2.4844 
 
1.94243
 
 
3
2
1.94243 in.
0.7818 in. (from right edge of shape to centroid)
2.4844 in.
2.2182 in. (from left edge of shape to centroid)
i i
i
z A
z
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
upright leg 2.000 0.7182 0.77372 2.7737 
bottom leg 0.011536 −1.0943 1.17881 1.1903 
Moment of inertia about the z axis (in.
4
) = 3.9640 
 
 
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Moment of inertia about the y axis: 
Shape IC d = zi – z d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
upright leg 0.017578 −0.5943 0.52979 0.5474 
bottom leg 0.565247 0.9057 0.80750 1.3727 
Moment of inertia about the y axis (in.
4
) = 1.9201 
 
Product of inertia about the centroidal axes: 
Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz 
 (in.
4
) (in.) (in.) (in.
2
) (in.
4
) (in.
4
) 
upright leg 0 0.7182 −0.5943 1.5000 −0.6402 −0.6402 
bottom leg 0 −1.0943 0.9057 0.9844 −0.9757 −0.9757 
Product of inertia (in.
4
) = −1.6159 
 
Orientation of neutral axis 
Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral 
axis from Eq. (8.23) before beginning the stress calculations: 
 
4
4
(20 kip-in.)( 1.6159 in. )
tan 0.8416
(20 kip-in.)(1.9201 in. )
(i.e., 40.08 CCW from axi0 s8 )40.
y z z yz
z y y yz
M I M I
M I M I
z 
 
 
 
 
Allowable moments based on maximum tension and compression bending stresses 
Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are 
farthest from the neutral axis are point H and the corner of the angle. To compute the normal stress at 
H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.: 
 
4 4
2 4 4 4 2
5
3
8
(1.9201 in. )(2.7182 in.) ( 1.6159 in. )( 0.4068 in.)
(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )
4.5619 in.
( 0.9124 in. )
5.0001 in.
y yz
x z z
y z yz
z z
I y I z
M M
I I I
M M 
Therefore, based on an allowable bending stress of 24 ksi at H, the maximum magnitude of Mz is: 
 
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3(0.9124 in. ) 24 ksi
26.3054 kip-in.
z
z
M
M
 (a) 
 
To compute the normal stress at the corner of the angle, use (y, z) coordinates of y = −1.2818 in. and z = 
−0.7818 in. 
 
4 4
2 4 4 4 2
5
3
8
(1.9201 in. )( 1.2818 in.) ( 1.6159 in. )( 0.7818 in.)
(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )
3.7245 in.
(0.7449 in. )
5.0001 in.
y yz
x z z
y z yz
z z
I y I z
M M
I I I
M M 
 
Therefore, based on the bending stress at the corner of the angle, the maximum magnitude of Mz is: 
 
3(0.7449 in. ) 24 ksi
32.2197 kip-in.
z
z
M
M
 (b) 
 
Maximum bending moment Mz 
Compare the results in Eqs. (a) and (b) to find that the maximum bending moment that can be applied to 
the angle shape is: 
 
26.3 kip-in.zM
 Ans. 
 
 
 
 
 
 
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8.72 The moment acting on the cross section of 
the unequal-leg angle has a magnitude of M = 20 
kip-in. and is oriented as shown in Fig. P8.72. 
Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the maximum tension and the maximum 
compression bending stresses in the cross section. 
(d) the orientation of the neutral axis relative to 
the +z axis. Show its location on a sketch of the 
cross section. 
 
 Fig. P8.72 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi – y Area Ai d²A IC + d²A 
 (mm
4
) (mm) (mm
2
) (mm
4
) (mm
4
) 
top flange 130,208.3 112.5 2,500 31,640,625.0 31,770,883.3 
web 10,666,666.7 0 3,200 0 10,666,666.7 
bottom flange 130,208.3 −112.5 2,500 31,640,625.0 31,770,883.3 
Moment of inertia about the z axis (mm
4
) = 74,208,333.3 
 
Moment of inertia about the y axis: 
Shape IC d = zi – z Area Ai d²A IC + d²A 
 (mm
4
) (mm) (mm
2
) (mm
4
) (mm
4
) 
top flange 2,083,333.3 −42.0 2,500 4,410,000 6,493,333.3 
web 68,266.7 0 3,200 0 68,266.7 
bottom flange 2,083,333.3 42.0 2,500 4,410,000 6,493,333.3 
Moment of inertia about the y axis (mm
4
) = 13,054,933.3 
 
Product of inertia about the centroidal axes: 
Shape yc zc Area Ai yc zc Ai Iyz 
 (mm) (mm) (mm
2
) (mm
4
) (mm
4
) 
top flange 112.5 −42.0 2,500 −11,812,500 −11,812,500 
web 0 0 3,200 0 0 
bottom flange −112.5 42.0 2,500 −11,812,500 −11,812,500 
Product of inertia (mm
4
) = −23,625,000 
 
Moment components 
 6
6
(40 kN-m)sin15 10.3528 kN-m 10.3528 10 N-mm
(40 kN-m)cos15 38.6370 kN-m 38.6370 10 N-mm
y
z
M
M
 
 
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(a) Bending stress at H 
Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the 
bending stresses. Equation (8.21) will be used here. 
2 2
6 4 6 4
4 4 4 2
6
( 38.6370 10 N-mm)(13,054,933.3 mm ) ( 10.3528 10 N-mm)( 23,625,000 mm )
(13,054,933.3 mm )(74,208,333.3 mm ) ( 23,625,000 mm )
( 10.3528 10 N-
z y y yz y z z yz
x
y z yz y z yz
M I M I y M I M I z
I I I I I I
y
4 6 4
4 4 4 2
3 3
mm)(74,208,333.3 mm ) ( 38.6370 10 N-mm)( 23,625,000 mm )
(13,054,933.3 mm )(74,208,333.3 mm ) ( 23,625,000 mm )
(0.63271 N/mm ) (0.35197 N/mm )
z
y z 
 
To computethe normal stress at H, use (y, z) coordinates of y = 125 mm and z = −92 mm: 
 
3 3(0.63271 N/mm )(125 mm) (0.35197 N/mm )( 92 mm)
46.7073 MPa 46.7 MPa (T)
x
 Ans. 
 
(b) Bending stress at K 
To compute the normal stress at K, use (y, z) coordinates of y = −125 mm and z = 92 mm: 
 
3 3(0.63271 N/mm )( 125 mm) (0.35197 N/mm )(92 mm)
46.7073 MPa 46.7 MPa (C)
x
 Ans. 
 
(d) Orientation of neutral axis 
Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis 
from Eq. (8.23) to help identify points of maximum stress. 
 
4 4
4 4
tan
( 10.3528 kN-m)(74,208,333.3 mm ) ( 38.6370 kN-m)( 23,625,000 mm )
( 38.6370 kN-m)(13,054,933.3 mm ) ( 10.3528 kN-m)( 23,625,000 mm )
0.55629
(i.e., 29.09 CCW f29.09 rom 
y z z yz
z y y yz
M I M I
M I M I
 axis)z 
 
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(c) Maximum tension and compression bending stresses 
Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are 
farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To 
compute bending stresses at the upper point, use (y, z) coordinates of y = 125 mm and z = 8 mm: 
 
3 3(0.63271 N/mm )(125 mm) (0.35197 N/mm )(8 mm)
81.9045 MPa Maximum tensi81.9 MP on benda (T) ing stress
x
 Ans. 
 
To compute bending stresses at the lower point, use (y, z) coordinates of y = −125 mm and z = −8 mm: 
 
3 3(0.63271 N/mm )( 125 mm) (0.35197 N/mm )( 8 mm)
81.9045 MPa Maximum compre81.9 ssionMPa (C bending stre s) s
x
 Ans. 
 
 
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8.73 The moment acting on the cross section of the 
unequal-leg angle has a magnitude of 14 kN-m and 
is oriented as shown in Fig. P8.73. Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the maximum tension and the maximum 
compression bending stresses in the cross section. 
(d) the orientation of the neutral axis relative to the 
+z axis. Show its location on a sketch of the cross 
section. 
 
 Fig. P8.73 
Solution 
Section properties 
Centroid location in y direction: 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
horizontal leg 150 19 2,850 190.50 542,925.0 
vertical leg 19 181 3,439 90.50 311,229.5 
 6,289 
 
854,154.5
 
 
 
3
2
854,154.5 mm
135.82 mm (from bottom of shape to centroid)
6,289 mm
64.18 mm (from top of shape to centroid)
i i
i
y A
y
A
 
 
Centroid location in z direction: 
Shape Area Ai 
zi 
(from right edge) zi Ai 
 (mm
2
) (mm) (mm
3
) 
horizontal leg 2,850 75.0 213,750.0 
vertical leg 3,439 9.5 32,670.5 
 6,289 
 
246,420.5
 
 
 
3
2
246,420.5 mm
39.18 mm (from right edge of shape to centroid)
6,289 mm
110.82 mm (from left edge of shape to centroid)
i i
i
z A
z
A
 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
horizontal leg 85,737.50 54.68 8,522,088.15 8,607,825.65 
vertical leg 9,388,756.58 −45.32 7,062,503.99 16,451,260.58 
Moment of inertia about the z axis (mm
4
) = 25,059,086.23 
 
 
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Moment of inertia about the y axis: 
Shape IC d = zi – z d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
horizontal leg 5,343,750.00 35.82 3,656,188.87 8,999,938.87 
vertical leg 103,456.58 −29.68 3,029,990.78 3,133,447.36 
Moment of inertia about the y axis (mm
4
) = 12,133,386.23 
 
Product of inertia about the centroidal axes: 
Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz 
 (mm
4
) (mm) (mm) (mm
2
) (mm
4
) (mm
4
) 
horizontal leg 0 54.68 35.82 2,850 5,582,117.16 5,582,117.16 
vertical leg 0 −45.32 −29.68 3,439 4,625,790.65 4,625,790.65 
Product of inertia (mm
4
) = 10,207,907.81 
 
Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the 
bending stresses. Equation (8.21) will be used here. 
2 2
6 4
4 4 4 2
6 4
4
(14 10 N-mm)(12,133,386.23 mm )
(12,133,386.23 mm )(25,059,086.23 mm ) (10,207,907.81 mm )
(14 10 N-mm)(10,207,907.81 mm )
(12,133,386.23 mm )(25
z y y yz y z z yz
x
y z yz y z yz
M I M I y M I M I z
I I I I I I
y
4 4 2
3 3
,059,086.23 mm ) (10,207,907.81 mm )
( 0.84997 N/mm ) (0.71509 N/mm )
z
y z 
 
To compute the normal stress at H, use (y, z) coordinates of y = 45.18 mm and z = 110.82 mm: 
 
3 3( 0.84997 N/mm )(45.18 mm) (0.71509 N/mm )(110.82 mm)
40.8444 MPa 40.8 MPa (T)
x
 Ans. 
 
(b) Bending stress at K 
To compute the normal stress at K, use (y, z) coordinates of y = 64.18 mm and z = −39.18 mm: 
 
3 3( 0.84997 N/mm )(64.18 mm) (0.71509 N/mm )( 39.18 mm)
82.5685 MP 82.6 MPa (a C)
x
 Ans. 
 
(d) Orientation of neutral axis 
Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral 
axis from Eq. (8.23) to help identify points of maximum stress. 
 
4
4
tan
(14 kN-m)(10,207,907.81 mm )
(14 kN-m)(12,133,386.23 mm )
0.84
40.0
131
(i.e., 40.07 CW from axis)7
y z z yz
z y y yz
M I M I
M I M I
z 
 
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(c) Maximum tension and compression bending stresses 
Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are 
farthest from the neutral axis are on the top corner (at K) and on the inside corner of the vertical leg. 
 
To compute bending stresses at the lower point, use (y, z) coordinates of y = −135.82 mm and z = −20.18 
mm: 
 
3 3( 0.84997 N/mm )( 135.82 mm) (0.71509 N/mm )( 20.18 mm)
101.0 MPa (T101.0129 MPa Maximum tension bendi) ng stress
x
 Ans. 
 
The maximum compression bending stress is 
 
3 3( 0.84997 N/mm )(64.18 mm) (0.71509 N/mm )( 39.18 mm)
82.5685 MPa Maximum compre82.6 ssionMPa (C) bending stress
x
 Ans. 
 
 
 
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8.74 The moment acting on the cross section of 
the zee shape has a magnitude of M = 4.75 kip-ft 
and is oriented as shown in Fig. P8.74. Determine: 
(a) the bending stress at point H. 
(b) the bending stress at point K. 
(c) the maximum tension and the maximum 
compression bending stresses in the cross section. 
(d) the orientation of the neutral axis relative to 
the +z axis. Show its location on a sketch of the 
cross section. 
 
 Fig. P8.74 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi – y Area Ai d²A IC + d²A 
 (in.
4
) (in.) (in.
2
) (in.
4
) (in.
4
) 
top flange 0.0260 2.75 1.25 9.4531 9.4792 
web 3.6458 0 1.75 0 3.6458 
bottom flange 0.0260 −2.75 1.25 9.4531 9.4792 
Moment of inertia about the z axis (in.
4
) = 22.6042 
 
Moment of inertia about the y axis: 
Shape IC d = zi – z Area Ai d²A IC + d²A 
 (in.
4
) (in.) (in.
2
) (in.
4
) (in.
4
) 
top flange 0.6510 1.075 1.25 1.4445 2.0956 
web 68,266.7 0 1.75 0 0.0179 
bottom flange 0.6510 −1.075 1.25 1.4445 2.0956 
Moment of inertia about the y axis (in.
4
) = 4.2091 
 
Product of inertia about the centroidal axes: 
Shape yc zc Area Ai yc zc Ai Iyz 
 (in.) (in.) (in.
2
) (in.
4
) (in.
4
) 
top flange 2.75 1.075 1.25 3.6953 3.6953 
web 0 0 1.75 0 0 
bottom flange −2.75 −1.075 1.25 3.6953 3.6953 
Product of inertia (in.
4
) = 7.3906 
 
(a) Bending stress at H 
Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the 
bending stresses. Equation (8.21) will be used here. 
 
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2 2
4 4
4 4 4 2 4 4
( 4.75 kip-ft)(12 in./ft)(4.2091 in. ) ( 4.75 kip-ft)(12 in./ft)(7.3906 in. )
(4.2091 in. )(22.6042 in. ) (7.3906 in. ) (4.2091 in. )(22.6042 in. )
z y y yz y z z yz
x
y z yz y z yz
M I M I y M I M I z
I I I I I I
y
4 2
3 3
(7.3906 in. )
(5.92065 kips/in. ) (10.39584 kips/in. )
z
y z 
 
To compute the normal stress at H, use (y, z) coordinates of y = 3 in. and z = 2.325 in.: 
 
3 3(5.92065 kips/in. )(3 in.) (10.39584 kips/in. )(2.325 in.)
6.4084 ksi 6.41 ksi (C)
x
 Ans. 
 
(b) Bending stress at K 
To compute the normal stress at K, use (y, z) coordinates of y = −2.50 in. and z = −2.325 in.: 
 
3 3(5.92065 kips/in. )( 2.50 in.) (10.39584 kips/in. )( 2.325 in.)
9.3687 k 9.37 ksi (T)si
x
 Ans. 
 
(d) Orientation of neutral axis 
Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis 
from Eq. (8.23) to help identify points of maximum stress. 
 
4
4
tan
( 4.75 kip-ft)(7.3906 in. )
( 4.75 kip-ft)(4.2091 in. )
1.7
60.34
559
(i.e., 60.34 CW from axis)
y z z yz
z y y yz
M I M I
M I M I
z 
 
 
 
 
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(c) Maximum tension and compression bending stresses 
Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are 
farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To 
compute bending stresses at the upper point, use (y, z) coordinates of y = 3 in. and z = −0.175 in.: 
 
3 3(5.92065 kips/in. )(3 in.) (10.39584 kips/in. )( 0.175 in
19.58 ksi (T
.)
19.5812 ksi Maximum tension bending) stress
x
 Ans. 
 
To compute bending stresses at the lower point, use (y, z) coordinates of y = −3 in. and z = 0.175 in.: 
 
3 3(5.92065 kips/in. )( 3 in.) (10.39584 kips/in. )(0.175 in.)
19.5812 ksi Maximum compre19.58 ssion bending strksi e(C) ss
x
 Ans. 
 
 
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8.75 For the cross section shown in Fig. P8.75, 
determine the maximum magnitude of the bending 
moment M so that the bending stress in the zee 
shape does not exceed 24 ksi. 
 
 Fig. P8.75 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi – y Area Ai d²A IC + d²A 
 (in.
4
) (in.) (in.
2
) (in.
4
) (in.
4
) 
top flange 0.0260 2.75 1.25 9.4531 9.4792 
web 3.6458 0 1.75 0 3.6458 
bottom flange 0.0260 −2.75 1.25 9.4531 9.4792 
Moment of inertia about the z axis (in.
4
) = 22.6042 
 
Moment of inertia about the y axis: 
Shape IC d = zi – z Area Ai d²A IC + d²A 
 (in.
4
) (in.) (in.
2
) (in.
4
) (in.
4
) 
top flange 0.6510 1.075 1.25 1.4445 2.0956 
web 68,266.7 0 1.75 0 0.0179 
bottom flange 0.6510 −1.075 1.25 1.4445 2.0956 
Moment of inertia about the y axis (in.
4
) = 4.2091 
 
Product of inertia about the centroidal axes: 
Shape yc zc Area Ai yc zc Ai Iyz 
 (in.) (in.) (in.
2
) (in.
4
) (in.
4
) 
top flange 2.75 1.075 1.25 3.6953 3.6953 
web 0 0 1.75 0 0 
bottom flange −2.75 −1.075 1.25 3.6953 3.6953 
Product of inertia (in.
4
) = 7.3906 
 
Bending stresses in the section 
Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the 
bending stresses. Equation (8.21) will be used here. For this problem, My = 0 and from the sketch, Mz is 
observed to be negative. The bending stress in the zee cross section is described by: 
 
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2 2
4 4
4 4 4 2 4 4 4 2
4
(4.2091 in. ) (7.3906 in. )
(4.2091 in. )(22.6042 in. ) (7.3906 in. ) (4.2091 in. )(22.6042 in. ) (7.3906 in. )
(0.103871 in. ) (
z y y yz y z z yz
x
y z yz y z yz
z z
z
M I M I y M I M I z
I I I I I I
M M
y z
M y 4
4 4
0.182383 in. )
(0.103871 in. ) (0.182383 in. )
z
z
M z
M y z 
 
Orientation of neutral axis 
Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral 
axis from Eq. (8.23) before beginning the stress calculations: 
 
4
4
(7.3906 in. )
tan 1.7559
(4.2091 i
60.
n. )
(i.e., 60.34 CW from axi3 s)4
y z z yz z
z y y yz z
M I M I M
M I M I Mz 
 
 
 
 
Allowable moments based on maximum tension and compression bending stresses 
Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are 
farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To 
compute bending stresses at the upper point, coordinates of y = 3 in. and z = −0.175 in. are used. Set the 
bending stress at this point to the 24-ksi allowable bending stress and solve for the moment magnitude: 
 
4 4
3
(0.103871 in. )(3 in.) (0.182383 in. )( 0.175 in.) 24 ksi
24 ksi
69.86287 kip-in.
0
5.82 kip-ft
.343530 in.
x z
z
M
M Ans. 
 
 
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8.76 A stainless-steel spring (shown in Fig. 
P8.76) has a thickness of ¾ in. and a change in 
depth at section B from D = 1.50 in. to d = 1.25 
in. The radius of the fillet between the two 
sections is r = 0.125 in. If the bending moment 
applied to the spring is M = 2,000 lb-in., 
determine the maximum normal stress in the 
spring. 
 Fig. P8.76 
Solution 
From Figure 8.18 
 
0.125 in. 1.50 in.
0.10 1.20 1.69
1.25 in. 1.25 in.
r D
K
d d
 
 
Moment of inertia at minimum depth section: 
 3
4(0.75 in.)(1.25 in.) 0.122070 in.
12
I
 
 
Nominal bending stress at minimum depth section: 
 
nom 4
(2,000 lb-in.)(1.25 in./2)
10.2400 ksi
0.122070 in.
My
I
 
 
Maximum bending stress: 
 
max nom 1.69(10.2400 ksi) 17.3056 ksi 17.31 ksiK
 Ans. 
 
 
 
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8.77 An alloy-steel spring (shown in Fig. P8.77) 
has a thickness of 25 mm and a change in depth at 
section B from D = 75 mm to d = 50 mm. If the 
radius of the fillet between the two sections is r = 
8 mm, determine the maximum moment that the 
spring can resist if the maximum bending stress in 
the spring must not exceed 120 MPa. 
 
 Fig. P8.77 
Solution 
From Figure 8.18 
 
8 mm 75 mm
0.16 1.50 1.57
50 mm 50 mm
r D
K
d d
 
 
Determine maximum nominal bending stress: 
 
max
nom
120 MPa
76.4331 MPa
1.57K
 
 
Moment of inertia at minimum depth section: 
 3
4(25 mm)(50 mm) 260,416.67 mm
12
I
 
 
Maximum bending moment: 
 2 4
nom
max
(76.4331 N/mm )(260,416.67 mm )
796,178.3 N-mm
50 mm/2
796 N-m
I
M
y
 Ans. 
 
 
 
 
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8.78 The notched bar shown in Fig. P8.78 is 
subjected to a bending moment of M = 300 N-m. 
The major bar width is D = 75 mm, the minor bar 
width at the notches is d = 50 mm, and the radius 
of each notch is r = 10 mm. If the maximum 
bending stress in the bar must not exceed 90 MPa, 
determine the minimum required bar thickness b. 
 
 Fig. P8.78 
Solution 
From Figure 8.17 
 
10 mm 75 mm
0.20 1.50 1.76
50 mm 50 mm
r D
K
d d
 
 
Determine maximum nominal bending stress: 
 
max
nom
90 MPa
51.1364 MPa
1.76K
 
 
Minimum bar thickness b: 
 
nom 3 2
2 2 2
nom
( /2) 6
/12
6 6(300 N-m)(1,000 mm/m)
(51.1364 N/mm
1
)(50 mm)
4.08 mm
M y M d M
I bd bd
M
b
d
 Ans. 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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8.79 The machine part shown in Fig. P8.79 is made of 
cold-rolled 18-8 stainless steel (see Appendix D for 
properties). The major bar width is D = 1.50 in., the 
minor bar width at the notches is d = 1.00 in., the radius 
of each notch is r = 0.125 in., and the bar thickness is b 
= 0.25 in. Determine the maximum safe moment M 
that may be applied to the bar if a factor of safety of 2.5 
with respect to failure by yield is specified. 
 
 Fig. P8.79 
Solution 
From Figure 8.17 
 
0.125 in. 1.50 in.
0.125 1.50 2.05
1.00 in. 1.00 in.
r D
K
d d
 
 
Moment of inertia at minimum depth section: 
 3
4(0.25 in.)(1.00 in.) 0.020833 in.
12
I
 
 
Maximum allowable bending moment: 
From the specified factor of safety and the yield stress of the material, the allowable bending stress is: 
allow
165 ksi
66 ksi
FS 2.5
Y
 
 
Thus, the maximum allowable bending moment can be determined from: 
 
allow
4
allow
max
(66 ksi)(0.020833 in. )
1.3415 kip-in.
(2.05)(1.00 in./
11
2)
1.8 lb-ft
My
K
I
I
M
K y
 Ans. 
 
 
 
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8.80 The shaft shown in Fig. P8.80 is supported 
at each end by self-aligning bearings. The major 
shaft diameter is D = 2.00 in., the minor shaft 
diameter is d = 1.50 in., and the radius of the 
fillet between the major and minor diameter 
sections is r = 0.125 in. The shaft length is L = 
24 in. and the fillets are located at x = 8 in. and x 
= 16 in. Determine the maximum load P that 
may be applied to the shaft if the maximum 
normal stress must be limited to 24,000 psi. 
 
 Fig. P8.80 
Solution 
From Figure 8.20 
 
0.125 in. 2.00 in.
0.083 1.33 1.78
1.50 in. 1.50 in.
r D
K
d d
 
 
Moment of inertia at minimum diameter section: 
 
4 4(1.50 in.) 0.248505 in.
64
I
 
 
Maximum allowable bending moment: 
 
allow
4
allow
max
(24,000 psi)(0.248505 in. )
4,467.50 lb-in.
(1.78)(1.50 in./2)
My
K
I
I
M
K y
 
 
Bending moment at x = 8 in.: 
 
(8 in.) (4 in.)
2 2
P P
M x P
 
 
Maximum load P: 
 
(4 in.) 4,467.50 lb-in.
1,116.88 lb 1,117 lb
P
P
 Ans. 
 
Check stress at midspan: 
 
midspan
4 4
midspan 4
(1,116.88 lb)(24 in.)
6,701.28 lb-in.
4 4
(2.00 in.) 0.785398 in.
64
(6,701.28 lb-in.)(2.00 in./2)
8,532 psi 24,000 psi OK
0.785398 in.
PL
M
I
M y
I
 
 
 
 
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8.81 The C86100 bronze (see Appendix D for 
properties) shaft shown in Fig. P8.81 is supported 
at each end by self-aligning bearings. The major 
shaft diameter is D = 40 mm, the minor shaft 
diameter is d = 25 mm, and the radius of the fillet 
between the major and minor diameter sections is r 
= 5 mm. The shaft length is L = 500 mm and the 
fillets are located at x = 150 mm and x = 350 mm. 
Determine the maximum load P that may be 
applied to the shaft if a factor of safety of 3.0 with 
respect to failure by yield is specified. 
 
 Fig. P8.81 
Solution 
From Figure 8.20 
 
5 mm 40 mm
0.20 1.60 1.48
25 mm 25 mm
r D
K
d d
 
 
Moment of inertia at minimum diameter section: 
 
4 4(25 mm) 19,174.76 mm
64
I
 
 
Maximum allowable bending moment: 
 
yield
allow
331 MPa
110.33 MPa
FS 3.0
 
 
allow
2 4
allow
max
(110.33 N/mm )(19,174.76 mm )
114,357.58 N-mm
(1.48)(25 mm/2)
My
K
I
I
M
K y
 
 
Bending moment at x = 150 mm: 
 
(150 mm) (75 mm)
2 2
P P
M x P
 
 
Maximum load P: 
 
(75 mm) 114,357.58 N-mm
1,524.77 N 1,525 N
P
P
 Ans. 
 
Check stress at midspan: 
 
midspan
4 4
midspan 4
(1,524.77 N)(500 mm)
190,596.25 N-mm
4 4
(40 mm) 125,663.71 mm
64
(190,596.25 N-mm)(40 mm/2)
30.33 MPa 110.33 MPa OK
125,663.71 mm
PL
M
I
M y
I
 
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8.82 The machine shaft shown in Fig. P8.82 is 
made of 1020 cold-rolled steel (see Appendix D 
for properties). The major shaft diameter is D = 
1.000 in., the minor shaft diameter is d = 0.625 in., 
and the radius of the fillet between the major and 
minor diameter sections is r = 0.0625 in. The fillet 
is located at x = 4 in. from C. If a load of P = 125 
lb is applied at C, determine the factor of safety 
with respect to failure by yield in the fillet at B. 
 Fig. P8.82 
Solution 
For 1020 cold-rolled steel: 
 
62,000 psiY
 
 
From Figure 8.20 
 
0.0625 in. 1.000 in.
0.10 1.6 1.74
0.625 in. 0.625 in.
r D
K
d d
 
 
Moment of inertia at minimum diameter section: 
 
4 4(0.625 in.) 0.0074901 in.
64
I
 
 
Bending moment at x = 4 in.: 
 
(125 lb)(4 in.) 500 lb-in.M Px
 
 
Maximum bending stress: 
 
max 4
(500 lb-in.)(0.625 in./2)
(1.74) 36,297.7 psi
0.0074901 in.
My
K
I
 
 
Factor of safety: 
max
62,000 psi
FS
36,297.7 psi
1.708Y
 Ans. 
 
 
 
 
 
 
 
 
 
 
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8.83 The machine shaft shown in Fig. P8.83 is 
made of 1020 cold-rolled steel (see Appendix D 
for properties). The major shaft diameter is D = 30 
mm, the minor shaft diameter is d = 20 mm, and 
the radius of the fillet between the major and minor 
diameter sections is r = 3 mm. The fillet is located 
at x = 90 mm from C. Determine the maximum 
load P that can be applied to the shaft at C if a 
factor of safety of 1.5 with respect to failure by 
yield is specified for the fillet at B. 
 
 Fig. P8.83 
Solution 
From Figure 8.20 
 
3 mm 30 mm
0.10 1.5 1.58
20 mm 20 mm
r D
K
d d
 
 
Moment of inertia at minimum diameter section: 
 
4 4(20 mm) 7,853.98 mm
64
I
 
 
Maximum allowable bending moment: 
 
allow
427 MPa
284.6667 MPa
FS 1.5
Y
 
 
allow
2 4
allow
max
(284.6667 N/mm )(7,853.98 mm )
141,504.2261 N-mm
(1.58)(20 mm/2)
My
K
I
I
M
K y
 
 
Bending moment at x = 90 mm: 
 
(90 mm)M Px P
 
 
Maximum load P: 
 
(90 mm) 141,504.2261 N-mm
1,572.3 N 1,572 N
P
P
 Ans. 
 
 
 
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8.84 The grooved shaft shown in Fig. P8.84 is made 
of C86100 bronze (see Appendix D for properties). 
The major shaft diameter is D = 50 mm, the minor 
shaft diameter at the groove is d = 34 mm, and the 
radius of the groove is r = 4 mm. Determine the 
maximum allowable moment M that may be applied 
to the shaft if a factor of safety of 1.5 with respect to 
failure by yield is specified. 
 
 Fig. P8.84 
Solution 
From Figure 8.19 
 
4 mm 50 mm
0.20 1.471 1.96
34 mm 34 mm
r D
K
d d
 
 
Moment of inertia at minimum diameter section: 
 
4 4(34 mm) 65,597.24 mm
64
I
 
 
Maximum allowable bending moment: 
 
allow
331 MPa
220.6667 MPa
FS 1.5
Y
 
 
allow
2 4
allow
max
(220.6667 N/mm )(65,597.24 mm )
(1.96)(34 mm/2)
434,427. 435 N-m N mm 4 -
My
K
I
I
M
K y
 Ans. 
 
 
 
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9.1 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (1) of the 20-in.-long beam segment, which is subjected to internal bending moments of 
MA = 24 kip-ft and MB = 28 kip-ft. 
 
 
Fig. P9.1a Beam segment Fig. P9.1b Cross-sectional dimensions 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left web 864.000 0.000 0.000 864.000 
top flange 12.505 10.250 1,287.016 1,299.521 
bottom flange 12.505 –10.250 1,287.016 1,299.521 
right web 864.0000.000 0.000 864.000 
Moment of inertia about the z axis (in.
4
) = 4,327.042 
 
 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(798.699 psi 565.745 psi)(3.5 in.)(3.5 in.) 8,357. 8.36 kips (C)227 lb
2
AF
 Ans. 
and on section B, the horizontal resultant force on area (1) is 
 
1
(931.816 psi 660.036 psi)(3.5 in.)(3.5 in.) 9,750. 9.75 kips (C)098 lb
2
BF
 Ans. 
 
(c) Equilibrium of area (1) 
 
8,357.227 lb 9,750.098 lb 1,392.87
1.393 kip
1 b
s
 l 0x
H
F
F
 Ans. 
The horizontal shear force is directed from section A toward section B at the interface between area (1) 
and the web elements. 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.2 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (1) of the 12-in.-long beam segment, which is subjected to internal bending moments of 
MA = 700 lb-ft and MB = 400 lb-ft. 
 
Fig. P9.2a Beam segment Fig. P9.2b Cross-sectional dimensions 
Solution 
Centroid location in y direction: (reference axis at bottom of tee shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
top flange 4.5 1.0 4.50 6.50 29.25 
stem 1.0 6.0 6.00 3.00 18.00 
 
 
10.50 47.25 
 
 3
2
47.25 in.
4.50 in.
10.50 in.
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 0.375 2.000 18.000 18.375 
stem 18.000 –1.500 13.500 31.500 
Moment of inertia about the z axis (in.
4
) = 49.875 
 
 
 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(421.053 psi 252.632 psi)(4.5 in.)(1 in.) 1,515.792 lb 1,516 lb (
2
C)AF
 Ans. 
and on section B, the horizontal resultant force on area (1) is 
 
1
(240.602 psi 144.361 psi)(4.5 in.)(1 in.) 866.167 lb 866 lb (
2
C)BF
 Ans. 
 
(c) Equilibrium of area (1) 
 
1,515.792 lb 866.167 lb 649.625 lb 0
650 lb
x
H
F
F Ans. 
The horizontal shear force is directed from section B toward section A at the interface between area (1) 
and the stem. 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.3 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (1) of the 500-mm-long beam segment, which is subjected to internal bending moments of 
MA = −5.8 kN-m and MB = −3.2 kN-m. 
 
Fig. P9.3a Beam segment Fig. P9.3b Cross-sectional dimensions 
Solution 
Centroid location in y direction: (reference axis at bottom of shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
top flange 160 30 4,800 285 1,368,000 
left stem 20 270 5,400 135 729,000 
right stem 20 270 5,400 135 729,000 
 
 
15,600 2,826,000 
 
 3
2
2,826,000 mm
181.154 mm
15,600 mm
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 360,000 103.846 51,763,160.3 52,123,160.3 
left stem 32,805,000 –46.154 11,503,035.3 44,308,035.3 
right stem 32,805,000 –46.154 11,503,035.3 44,308,035.3 
Moment of inertia about the z axis (mm
4
) = 140,739,231 
 
 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(4.898 MPa 3.661 MPa)(160 mm)(30 mm) 20,542 20.5 kN (T)N
2
AF
 Ans. 
and on section B, the horizontal resultant force on area (1) is 
 
1
(2.702 MPa 2.020 MPa)(160 mm)(30 mm) 11,3 11.33 kN (T)34 N
2
BF
 Ans. 
 
(c) Equilibrium of area (1) 
 
20,542 N 11,334 N 9,209 N
9.21 kN
0x
H
F
F Ans. 
The horizontal shear force is directed from section A toward section B at the interface between area (1) 
and the stems. 
 
 
 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.4 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (1) of the 16-in.-long beam segment, which is subjected to internal bending moments of 
MA = −3,300 lb-ft and MB = −4,700 lb-ft. 
 
Fig. P9.4a Beam segment Fig. P9.4b Cross-sectional dimensions 
Solution 
Centroid location in y direction: (reference axis at bottom of shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
left flange (1) 1.50 3.50 5.25 10.25 53.8125 
right flange (2) 1.50 3.50 5.25 10.25 53.8125 
central stem 1.50 12.00 18.00 6.00 108.0000 
 
 
28.50 215.6250 
 
 3
2
215.625 in.
7.5658 in.
28.50 in.
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left flange (1) 5.3594 2.6842 37.8262 43.1856 
right flange (2) 5.3594 2.6842 37.8262 43.1856 
central stem 216.0000 –1.5658 44.1305 260.1305 
Moment of inertia about the z axis (in.
4
) = 346.5016 
 
 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(506.765 psi 106.767 psi)(1.5 in.)(3.5 in.) 1,610.522 lb 1,611 lb 
2
(T)AF
 Ans. 
and on section B, the horizontal resultant force on area (1) is 
 
1
(721.756 psi 152.061 psi)(1.5 in.)(3.5 in.) 2,293.773 lb 2,290 lb 
2
(T)BF
 Ans. 
 
(c) Equilibrium of area (1) 
 
1,610.522 lb 2,293.773 lb 683.252 lb 0
683 lb
x
H
F
F Ans. 
The horizontal shear force is directed from section B toward section A at the interface between area (1) 
and the stem. 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.5 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (1) of the 18-in.-long beam segment, which is subjected to internal bending moments of 
MA = −42 kip-in. and MB = −36 kip-in. 
 
Fig. P9.5a Beam segment Fig. P9.5b Cross-sectional dimensions 
Solution 
Centroid location in y direction: (reference axis at bottom of shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
top flange (1) 6 2 12 11 132 
bottom flange (2) 10 2 20 1 20 
web 2 8 16 6 96 
 
 
48 248 
 
 3
2
248 in.
5.1667 in.
48 in.
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange (1) 4.0000 5.8333 408.3333 412.3333 
bottom flange (2) 6.6667 –4.1667 347.2222 353.8889 
web 85.3333 0.8333 11.1111 96.4444 
Moment of inertia about the z axis (in.
4
) = 862.6667 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(332.690 psi 235.317 psi)(6 in.)(2 in.) 3,408.0 3.41 kips (43 ) lb T
2
AF
 Ans. 
and on section B, the horizontal resultant force on area (1) is 
 
1
(251.546 psi 154.173 psi)(6 in.)(2 in.) 2,921.1 2.92 kips (80 ) lb T
2
BF
 Ans. 
 
(c) Equilibrium of area (1) 
 
3,408.043 lb 2,921.180 lb 486.863 lb
0.487 ki s
0
p
x
H
F
F
 Ans. 
The horizontal shear force is directed from section A toward section B at the interface between area (1) 
and the web. 
 
 
 
 
 
 
 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.6 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (2) of thebeam segment shown in Problem 9.5. 
 
Fig. P9.6a Beam segment Fig. P9.6b Cross-sectional dimensions 
Solution 
Centroid location in y direction: (reference axis at bottom of shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
top flange (1) 6 2 12 11 132 
bottom flange (2) 10 2 20 1 20 
web 2 8 16 6 96 
 
 
48 248 
 
 3
2
248 in.
5.1667 in.
48 in.
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange (1) 4.0000 5.8333 408.3333 412.3333 
bottom flange (2) 6.6667 –4.1667 347.2222 353.8889 
web 85.3333 0.8333 11.1111 96.4444 
Moment of inertia about the z axis (in.
4
) = 862.6667 
 
 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (2) 
On section A, the resultant force on area (2) in the x direction is 
 
1
(251.546 psi 154.173 psi)(10 in.)(2 in.) 4,057.1 4.06 kips (95 ) lb C
2
AF
 Ans. 
and on section B, the horizontal resultant force on area (2) is 
 
1
(215.611 psi 132.149 psi)(10 in.)(2 in.) 3,477.5 3.48 kips (95 ) lb C
2
BF
 Ans. 
 
(c) Equilibrium of area (2) 
 
4,057.195 lb 3,477.595 lb 579.599 l
0.580 p
b
 
0
ki s
x
H
F
F
 Ans. 
The horizontal shear force is directed from section B toward section A at the interface between area (2) 
and the web. 
 
 
 
 
 
 
 
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9.7 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Consider area (1) of the 300-mm-long beam segment, which is subjected to internal bending moments of 
MA = 7.5 kN-m and MB = 8.0 kN-m. 
 
Fig. P9.7a Beam segment Fig. P9.7b Cross-sectional dimensions 
Solution 
Centroid location in y direction: (reference axis at bottom of shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
left stiff (1) 40 90 3,600 275 990,000 
flange (2) 150 40 6,000 300 1,800,000 
right stiff (3) 40 90 3,600 275 990,000 
stem 40 280 11,200 140 1,568,000 
 
 
24,400 5,348,000 
 
 3
2
5,348,000 mm
219.180 mm
24,400 mm
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
 
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Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
left stiff (1) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87 
flange (2) 800,000.00 80.8197 39,190,916.42 39,990,916.42 
right stiff (3) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87 
stem 73,173,333.33 –79.1803 70,218,672.40 143,392,005.73 
Moment of inertia about the z axis (mm
4
) = 210,676,939.89 
 
(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N 7.15 kN (C
2
)AF
 Ans. 
and on section B, the horizontal resultant force on area (1) is 
 
1
(3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N 7.63 kN (C
2
)BF
 Ans. 
 
(c) Equilibrium of area (1) 
 
7,153.755 N 7,630.672 N 476.917 N
0.477 kN
0x
H
F
F Ans. 
The horizontal shear force is directed from section A toward section B at the interface between area (1) 
and area (2). 
 
 
 
 
 
 
 
 
 
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9.8 For the following problems, a beam segment subjected to internal bending moments at sections A 
and B is shown along with a sketch of the cross-sectional dimensions. For each problem: 
(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at 
sections A and B. Indicate the magnitude of key bending stresses on the sketch. 
(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and 
show these resultant forces on the sketch. 
(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine 
the horizontal force required to satisfy equilibrium for the specified area and show the location and 
direction of this force on the sketch. 
 
Combine areas (1), (2), and (3) of the beam segment shown in Problem 9.7. 
 
Fig. P9.8a Beam segment Fig. P9.8b Cross-sectional dimensions 
Solution 
(a) Centroid location in y direction: (reference axis at bottom of shape) 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
left stiff (1) 40 90 3,600 275 990,000 
flange (2) 150 40 6,000 300 1,800,000 
right stiff (3) 40 90 3,600 275 990,000 
stem 40 280 11,200 140 1,568,000 
 
 
24,400 5,348,000 
 
 3
2
5,348,000 mm
219.180 mm
24,400 mm
i i
i
y A
y
A
 (measured upward from bottom edge of shape) 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
left stiff (1) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87 
flange (2) 800,000.00 80.8197 39,190,916.42 39,990,916.42 
right stiff (3) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87 
stem 73,173,333.33 -79.1803 70,218,672.40 143,392,005.73 
Moment of inertia about the z axis (mm
4
) = 210,676,939.89 
 
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(a) Bending stress distribution 
 
 
(b) Resultant forces acting on area (1) 
On section A, the resultant force on area (1) in the x direction is 
 
1
(3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N
2
AF
 
and on section B, the horizontal resultant force on area (1) is 
 
1
(3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N
2
BF
 
 
Resultant forces acting on area (3) 
The forces acting on area (3) are identical to those acting on area (1). 
 
Resultant forces acting on area (2) 
On section A, the resultant force on area (2) in the x direction is 
 
1
(3.589 MPa 2.165 MPa)(150 mm)(40 mm) 17,262.855 N
2
AF
 
and on section B, the horizontal resultant force on area (2) is 
 
1
(3.828 MPa 2.309 MPa)(150 mm)(40 mm) 18,413.697 N
2
BF
 
 
Resultant forces acting on combined areas (1), (2), and (3) 
On section A, the resultant force on combined areas (1), (2), and (3) is 
 
2(7,153.755 N) 17,262.855 N 31,570.36 31.6 kN (C3 N )AF
 Ans. 
and on section B, the horizontal resultant force on area (2) is 
 
2(7,630.672 N) 18,413.697 N 33,675.05 33.7 kN (C4 N )BF
 Ans. 
 
(c) Equilibrium of combined areas (1), (2), and (3) 
 
31,570.363 N 33,675.054 N 2,104.691 N 0
2.10 kN
x
H
F
F Ans. 
The horizontal shear force is directed from section A toward section B at the interface between area (2) 
and the stem of the tee. 
 
 
 
 
 
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9.9 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beam is 
made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate the maximum 
horizontal shear stresses at points located 35 mm, 70 mm, 105 mm, and 140 mm below the top surface 
of the beam. From these results, plot a graph showing the distribution of shear stresses from top to 
bottom of the beam. 
 
Fig. P9.9a Cantilever beam Fig. P9.9b Cross-sectional dimensions 
Solution 
Shear force in cantilever beam: 
 V = 7.2 kN = 7,200 N 
 
Shear stress formula: 
 
V Q
I t
 
 
 
Section properties: 
 3
6 4(120 mm)(280 mm) 219.52 10 mm
12
I   
 
 
 t = 120 mm 
 
 
Distance below top 
surface of beam 
y Q 
35 mm 105 mm 514,500 mm
3
 140.6 kPa 
70 mm 70 mm 882,000 mm
3
 241 kPa 
105 mm 35 mm 1,102,500 mm
3
 301 kPa 
140 mm 0 mm 1,176,000 mm
3
 321 kPa 
 
 
 
 
 
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9.10 A 14-ft long simply supported timber beam carries a 6-kip concentrated load at midspan, as shown 
in Fig. P9.10a. The cross-sectional dimensions of the timber are shown in Fig. P9.10b. 
(a) At section a–a, determine the magnitude of the shear stress in the beam at point H. 
(b) At section a–a, determine the magnitude of the shear stress in the beam at point K. 
(c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 
14-ft span length. 
(d) Determine the maximum tension bending stress that occurs in the beam at any location within the 
14-ft span length. 
 
 
Fig. P9.10a Simply supported timber beam 
Fig. P9.10b Cross-sectional 
dimensions 
Solution 
Section properties: 
 3
4(6 in.)(15 in.) 1,687.5 in. 6 in.
12
I t  
 
 
 
(a) Shear stress at H: 
 
3
3
4
(6 in.)(3 in.)(6 in.) 108 in.
(3,000 lb)(108 in. )
(1,687.500 in. )(6
32.0 ps
.
i
 in )
Q
V Q
I t

 

 
 Ans. 
 
(b) Shear stress at K: 
 
3
3
4
(6 in.)(1 in.)(7 in.) 42 in.
(3,000 lb)(42 in. )
(1,687.500 in. )(6 in.)
12.44 psi
Q
V Q
I t

 

 
 Ans. 
 
 
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(c) Maximum shear stress at any location: 
 
3
max
3
4
(6 in.)(7.5 in.)(3.75 in.) 168.75 in.
(3,000 lb)(168.75 in. )
(1,687.500 in. )(6 in.)
50.0 psi
Q
V Q
I t

 
   Ans. 
 
(d) Maximum bending stress at any location: 
 
max
4
21 kip-ft 21,000 lb-ft
(21,000 lb-ft)(7.5 in.)(12 in./ft)
1,687.5
1,120 psi (T) and (C
00 in.
)x
M
M c
I

 
   Ans. 
 
 
 
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9.11 A 5-m long simply supported timber beam carries a uniformly distributed load of 12 kN/m, 
as shown in Fig. P9.11a. The cross-sectional dimensions of the beam are shown in Fig. P9.11b. 
(a) At section a–a, determine the magnitude of the shear stress in the beam at point H. 
(b) At section a–a, determine the magnitude of the shear stress in the beam at point K. 
(c) Determine the maximum horizontal shear stress that occurs in the beam at any location 
within the 5-m span length. 
(d) Determine the maximum compression bending stress that occurs in the beam at any location 
within the 5-m span length. 
 
 
Fig. P9.11a Simply supported timber beam Fig. P9.11b Cross-sectional dimensions 
Solution 
Section properties: 
 3
6 4(100 mm)(300 mm) 225 10 mm 100 mm
12
I t   
 
 
 
(a) Shear stress magnitude at H: 
 
3
3
6 4
(100 mm)(90 mm)(105 mm)
945,000 mm
(18,000 N)(945,000 mm )
(225 10 mm )(100 m
756 kP
m)
a
Q
V Q
I t






 Ans. 
 
(b) Shear stress magnitude at K: 
 
3
3
6 4
(100 mm)(40 mm)(130 mm)
520,000 mm
(18,000 N)(520,000 mm )
(225 10 mm )(100 m
416 kP
m)
a
Q
V Q
I t






 Ans. 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.(c) Maximum shear stress at any location: 
 
3
max
3
6 4
(100 mm)(150 mm)(75 mm) 1,125,000 mm
(30,000 N)(1,125,000 mm )
(225 10 mm )(100 mm)
1,500 kPa
Q
V Q
I t

 
  

 Ans. 
 
(d) Maximum compression bending stress at any location: 
 
max
6 4
37.5 kN-m
(37.5 kN-m)(150 mm)(1,000 N/kN)(1,000 mm/m)
225 10 mm
25.0 MPa 25.0 MPa (C)
x
M
M y
I


   

   Ans. 
 
 
 
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9.12 A 5-m long simply supported timber beam carries two concentrated loads, as shown in Fig. P9.12a. 
The cross-sectional dimensions of the beam are shown in Fig. P9.12b. 
(a) At section a–a, determine the magnitude of the shear stress in the beam at point H. 
(b) At section a–a, determine the magnitude of the shear stress in the beam at point K. 
(c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 5-
m span length. 
(d) Determine the maximum compression bending stress that occurs in the beam at any location within 
the 5-m span length. 
 
 
Fig. P9.12a Simply supported timber beam 
Fig. P9.12b Cross-sectional 
dimensions 
Solution 
Section properties: 
 3
6 4(150 mm)(450 mm) 1,139.1 10 mm 150 mm
12
I t   
 
 
 
(a) Shear stress magnitude at H: 
 
3
3
6 4
(150 mm)(150 mm)(150 mm)
3,375,000 mm
(39,200 N)(3,375,000 mm )
(1,139.1 10 mm )(150 m
7
m)
74 kPa
Q
V Q
I t






 Ans. 
 
(b) Shear stress magnitude at K: 
 
3
3
6 4
(150 mm)(100 mm)(175 mm)
2,625,000 mm
(39,200 N)(2,625,000 mm )
(1,139.1 10 mm )(150 m
6
m)
02 kPa
Q
V Q
I t






 Ans. 
 
 
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(c) Maximum shear stress at any location: 
 
3
max
3
6 4
(150 mm)(225 mm)(112.5 mm) 3,796,875 mm
(39,200 N)(3,796,875 mm )
(1,139.1 10 mm )(150 m
871 kP
m)
a
Q
V Q
I t

 
  

 Ans. 
 
(d) Maximum bending stress at any location: 
 
max
6 4
39.2 kN-m
(39.2 kN-m)(225 mm)(1,000 N/kN)(1,000 mm/m)
1,139.1 10 mm
7.74296 MPa 7,740 kPa (C)
x
M
M y
I


   

   Ans. 
 
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9.13 A laminated wood beam consists of eight 2 in. × 6-in. planks glued together to form a section 6 in. 
wide by 16 in. deep, as shown in Fig. P9.13a. If the allowable strength of the glue in shear is 160 psi, 
determine: 
(a) the maximum uniformly distributed load w that can be applied over the full length of the beam if the 
beam is simply supported and has a span of 20 ft. 
(b) the shear stress in the glue joint at H, which is located 4 in. above the bottom of the beam and 3 ft 
from the left support. Assume the beam is subjected to the load w determined in part (a). 
(c) the maximum tension bending stress in the beam when the load of part (a) is applied. 
 
 
Fig. P9.13a Simply supported timber beam 
Fig. P9.13b Cross-sectional 
dimensions 
Solution 
Section properties: 
 3
4(6 in.)(16 in.) 2,048 in. 6 in.
12
I t  
 
 
(a) Maximum Q: 
 
3(6 in.)(8 in.)(4 in.) 192 in.Q  
 
 
Maximum shear force V: 
 
4
3
160 psi
(160 psi)(2,048 in. )(6 in.)
10,240 lb
192 in.
V Q
I t
V
  
   
 
Maximum distributed load w: 
 
max
max
10,240 lb
2
2(10,240 lb)
20 ft
1,024 lb/ft
wL
V
w
 
   Ans. 
 
 
 
 
 
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(b) Shear force at x = 3 ft: 
 
10,240 lb (1,024 lb/ft)(3 ft) 7,168 lbV   
 
 
3
3
4
(6 in.)(4 in.)(6 in.) 144 in.
(7,168 lb)(144 in. )
(2,048 in. )(6 in.)
84.0 psi
Q
V Q
I t

 
   Ans. 
 
(c) Maximum tension bending stress at any location: 
 
2 2
max
4
(1,024 lb/ft)(20 ft)
51,200 lb-ft
8 8
(51,200 lb-ft)( 8 in.)(12 in./ft)
2,048 in.
2,400 psi (T)x
wL
M
M y
I

  

     Ans. 
 
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9.14 A 5-ft long simply supported wood beam carries a concentrated load P at midspan, as shown in Fig. 
P9.14a. The cross-sectional dimensions of the beam are shown in Fig. P9.14b. If the allowable shear 
strength of the wood is 80 psi, determine the maximum load P that may be applied at midspan. Neglect 
the effects of the beam’s self weight. 
 
 
Fig. P9.14a Simply supported timber beam Fig. P9.14b Cross-sectional dimensions 
Solution 
Section properties: 
 3
4(6 in.)(10 in.) 500 in. 6 in.
12
I t  
 
 
Maximum Q: 
 
3(6 in.)(5 in.)(2.5 in.) 75 in.Q  
 
 
Maximum shear force V: 
 
4
3
80 psi
(80 psi)(500 in. )(6 in.)
3,200 lb
75 in.
V Q
I t
V
  
   
 
Maximum concentrated load P: 
 
max
max
3,200 lb
2
2(3,200 lb) 6,400 lb
P
V
P
 
   Ans. 
 
 
 
 
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9.15 A wood beam supports the loads shown in Fig. P9.15a. The cross-sectional dimensions of the beam 
are shown in Fig. P9.15b. Determine the magnitude and location of: 
(a) the maximum horizontal shear stress in the beam. 
(b) the maximum tension bending stress in the beam. 
 
 
Fig. P9.15a Simply supported timber beam Fig. P9.15b Cross-sectional dimensions 
Solution 
Section properties: 
 3 3
4(75 mm)(240 mm) (20 mm)(100 mm)2 89,733,333 mm
12 12
I   (a) Maximum shear force: 
 Vmax = 9.54 kN = 9,540 N @ support A 
 
Check shear stress at neutral axis: 
 3
(75 mm)(120 mm)(60 mm)
2(20 mm)(50 mm)(25 mm) 590,000 mm
Q 
  
 
 3
4
(9,540 N)(590,000 mm )
545 kPa
(89,733,333 mm )(115 mm)
VQ
I t
   
 
 
Check shear stress at top edge of cover plates: 
 
3(75 mm)(70 mm)(85 mm) 446,250 mmQ  
 
 
 3
4
(9,540 N)(446,250 mm )
633 kPa
(89,733,333 mm )(75 mm)
VQ
I t
   
 
 
Maximum shear stress in beam: 
 
,max 633 kPaH 
 Ans. 
 
(b) Maximum bending moment: 
 Mmax = 6.49 kN-m (between support A and point B) 
 
Maximum tension bending stress: 
 
4
(6.49 kN-m)( 120 mm)(1,000 N/kN)(1,000 mm/m)
89,733,333 mm
8.67905 MPa 8,680 kPa (T)
x
M y
I
    
  Ans. 
 
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9.16 A 50-mm-diameter solid steel shaft 
supports loads PA = 1.5 kN and PC = 3.0 kN, 
as shown in Fig. P9.16. Assume L1 = 150 
mm, L2 = 300 mm, and L3 = 225 mm. The 
bearing at B can be idealized as a roller 
support and the bearing at D can be idealized 
as a pin support. Determine the magnitude 
and location of: 
(a) the maximum horizontal shear stress in 
the shaft. 
(b) the maximum tension bending stress in 
the shaft. 
 
Fig. P9.16 
Solution 
 
Section properties: 
 
4 4
4
3 3
3
(50 mm)
64 64
306,796.158 mm
(50 mm)
12 12
10,416.667 mm
I D
D
Q
 
 

 

 
 
 
Maximum shear force magnitude: 
 Vmax = 1.71 kN (between B and C) 
 
 
Maximum bending moment magnitude: 
 Mmax = 289.29 kN-mm (at C) 
 
 
 
 
 
(a) Maximum horizontal shear stress: 
 
3
4
(1,710 N)(10,416.667 mm )
(306,796.158 mm )(50 mm)
(at neutral axis 1.161 MPa between and )
V Q
I t
B C
  
 Ans. 
 
(b) Maximum tension bending stress: 
 
4
(289.29 kN-mm)( 50 mm/2)(1,000 N/kN)
306,796.15
23.6 MPa (T)
8 mm
23.574 MPa (on bottom of shaft at )
x
M y
I
C
    
  Ans. 
 
 
 
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9.17 A 1.25-in.-diameter solid steel shaft 
supports loads PA = 400 lb and PC = 900 lb, 
as shown in Fig. P9.17. Assume L1 = 6 in., 
L2 = 12 in., and L3 = 8 in. The bearing at B 
can be idealized as a roller support and the 
bearing at D can be idealized as a pin 
support. Determine the magnitude and 
location of: 
(a) the maximum horizontal shear stress in 
the shaft. 
(b) the maximum tension bending stress in 
the shaft. 
 
Fig. P9.17 
Solution 
 
Section properties: 
 
4 4
4
3 3
3
(1.25 in.)
64 64
0.119842 in.
(1.25 in.)
12 12
0.162760 in.
I D
D
Q
 
 

 

 
 
 
Maximum shear force magnitude: 
 Vmax = 480 lb (between B and C) 
 
 
Maximum bending moment magnitude: 
 Mmax = 3,360 lb-in. (at C) 
 
 
 
 
 
(a) Maximum horizontal shear stress: 
 
3
4
(480 lb)(0.162760 in. )
(0.119842 in. )(1.25 in.)
521.519 psi (at neutral axis b522 etween and )psi
V Q
I t
B C
  
  Ans. 
 
(b) Maximum tension bending stress: 
 
4
(3,360 lb-in.)( 1.25 in./2)
0.119842 in.
17, 17,520 psi (T523.022 psi (on bottom of) shaft at )
x
M y
I
C
    
  Ans.
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.18 A 1.00-in.-diameter solid steel shaft supports loads PA = 200 
lb and PD = 240 lb, as shown in Fig. P9.18. Assume L1 = 2 in., 
L2 = 5 in., and L3 = 4 in. The bearing at B can be idealized as a 
pin support and the bearing at C can be idealized as a roller 
support. Determine the magnitude and location of: 
(a) the maximum horizontal shear stress in the shaft. 
(b) the maximum tension bending stress in the shaft. 
 
Fig. P9.18 
Solution 
 
Section properties: 
 
4 4
4
3 3
3
(1.00 in.)
64 64
0.049087 in.
(1.00 in.)
12 12
0.083333 in.
I D
D
Q
 
 

 

 
 
 
Maximum shear force magnitude: 
 Vmax = 272 lb (between B and C) 
 
 
Maximum bending moment magnitude: 
 Mmax = 960 lb-in. (at C) 
 
 
 
 
 
(a) Maximum horizontal shear stress: 
 
3
4
(272 lb)(0.083333 in. )
(0.049087 in. )(1.00 in.)
461.762 psi (at neutral axis b462 etween and )psi
V Q
I t
B C
  
  Ans. 
 
(b) Maximum tension bending stress: 
 
4
(960 lb-in.)( 1.00 in./2)
0.049087 in.
9,7 9,780 psi (T)78.480 psi (on bottom of shaft at )
x
M y
I
C
    
  Ans.
 
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9.19 A 20-mm-diameter solid steel shaft supports loads PA = 900 
N and PD = 1,200 N, as shown in Fig. P9.19. Assume L1 = 50 
mm, L2 = 120 mm, and L3 = 90 mm. The bearing at B can be 
idealized as a pin support and the bearing at C can be idealized as 
a roller support. Determine the magnitude and location of: 
(a) the maximum horizontal shear stress in the shaft. 
(b) the maximum compression bending stress in the shaft. 
 
Fig. P9.19 
Solution 
 
Section properties: 
 
4 4
4
3 3
3
(20 mm)
64 64
7,853.982 mm
(20 mm)
12 12
666.667 mm
I D
D
Q
 
 

 

 
 
 
Maximum shear force magnitude: 
 Vmax = 1,275 N (between B and C) 
 
 
Maximum bending moment magnitude: 
 Mmax = 108,000 N-mm (at C) 
 
 
 
 
 
(a) Maximum horizontal shear stress: 
 
3
4
(1,275 N)(666.667 mm )
(7,853.982 mm )(20 mm)
5.411 MPa (at neutral axi5.41 M s between and )Pa
V Q
I t
B C
  
  Ans. 
 
(b) Maximum compression bending stress: 
 
4
(108,000 N-mm)(20 mm/2)
7,853.982 mm
1 137.5 MPa (C)37.510 MPa (on top of shaft at )
x
M y
I
C
    
   Ans. 
 
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9.20 A 1.25-in.-diameter solid steel shaft 
supports loads PA = 600 lb, PC = 1,600 lb, and 
PE = 400 lb, as shown in Fig. P9.20. Assume 
L1 = 6 in., L2 = 15 in., L3 = 8 in., and L4 = 10 
in. The bearing at B can be idealized as a roller 
support and the bearing at D can be idealized 
as a pin support. Determine the magnitude and 
location of: 
(a) the maximum horizontal shear stress in the 
shaft. 
(b) the maximum tension bending stress in the 
shaft. 
 
Fig. P9.20 
Solution 
 
Section properties: 
 
4 4
4
3 3
3
(1.25 in.)
64 64
0.119842 in.
(1.25 in.)
12 12
0.162760 in.
I D
D
Q
 
 

 

 
 
 
Maximum shear force magnitude: 
 Vmax = 1,060.9 lb (between C and D) 
 
 
Maximum bending moment magnitude: 
 Mmax = 4,487 lb-in. (at C) 
 
 
 
 
 
(a) Maximum horizontal shear stress: 
 
3
4
(1,060.9 lb)(0.162760 in. )
(0.119842 in. )(1.25 in.)
1,152.632 psi (at neutral axis1,153 p between ansi d )
V Q
I t
C D
  
  Ans. 
 
(b) Maximum tension bending stress: 
 
4
(4,487 lb-in.)( 1.25 in./2)
0.119842 in.
23, 23,400 psi (T400.309 psi (on bottom of) shaft at )
x
M y
I
C
    
  Ans.
 
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9.21 A 25-mm-diameter solid steel shaft 
supports loads PA = 1,000 N, PC = 3,200 N, 
and PE = 800 N, as shown in Fig. P9.21. 
Assume L1 = 80 mm, L2 = 200 mm, L3 = 100 
mm, and L4 = 125 mm. The bearing at B can be 
idealized as a roller support and the bearing at 
D can be idealized as a pin support. Determine 
the magnitude and location of: 
(a) the maximum horizontal shear stress in the 
shaft. 
(b) the maximum tension bending stress in the 
shaft. 
 
 
Fig. P9.21 
Solution 
 
Section properties: 
 
4 4
4
3 3
3
(25 mm)
64 64
19,174.760 mm
(25 mm)
12 12
1,302.083 mm
I D
D
Q
 
 

 

 
 
 
Maximum shear force magnitude: 
 Vmax = 2,200 N (between C and D) 
 
 
Maximum bending moment magnitude: 
 Mmax = 120,000 N-mm (at C) 
 
 
 
 
 
(a) Maximum horizontal shear stress: 
 
3
4
(2,200 N)(1,302.083 mm )
(19,174.760 mm )(25 mm)
(at neutral axis b5.98 M etween and )Pa
V Q
I t
C D
  
 Ans. 
 
(b) Maximum tension bending stress: 
 
4
(120,000 N-mm)( 25 mm/2)
19,174.760 mm
7 78.2 MPa 8.228 MPa (on bottom of shaft at )( T)
x
M y
I
C
    
  Ans. 
 
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9.22 A 3-in. standard steel pipe (D = 3.500 in.; d = 3.068 in.) supports a concentrated load of P = 900 lb, 
as shown in Fig. P9.22a. The span length of the cantilever beam is L = 3 ft. Determine the magnitude of: 
(a) the maximum horizontal shear stress in the pipe. 
(b) the maximum tension bending stress in the pipe. 
 
 
Fig. P9.22a Cantilever beam Fig. P9.22b Pipe cross section 
Solution 
Section properties: 
 
4 4 4 4 4
3 3 3 3 3
[ ] [(3.500 in.) (3.068 in.) ] 3.017157 in.
64 64
1 1
[ ] [(3.500 in.) (3.068 in.) ] 1.166422 in.
12 12
I D d
Q D d
 
    
    
 
 
Maximum shear force magnitude: 
 Vmax = 900 lb 
 
Maximum bending moment magnitude: 
 Mmax = (900 lb)(3 ft)(12 in./ft) = 32,400 lb-in. 
 
(a) Maximum horizontal shear stress: 
 3
4
(900 lb)(1.166422 in. )
805.410 psi
(3.017157 in. )(3.500 
805 psi
in. 3.068 in.)
VQ
I t
    

 Ans. 
 
(b) Maximum tension bending stress: 
 
4
( 32,400 lb-in.)(3.500 in./2)
18,792.529 psi
3.0
18,790 psi (T
1715 in
)
7 .
x
M y
I
       Ans. 
 
 
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9.23 A steel pipe (D = 170 mm; d = 150 mm) supports a concentrated load of P as shown in Fig. P9.23a. 
The span length of the cantilever beam is L = 1.2 m. 
(a) Compute the value of Q for the pipe. 
(b) If the allowable shear stress for the pipe shape is 75 MPa, determine the maximum load P than can 
be applied to the cantilever beam. 
 
 
Fig. P9.23a Cantilever beam Fig. P9.23b Pipe cross section 
Solution 
(a) Section properties: 
 
4 4 4 4 4
3 3 3 3 3 3
[ ] [(170 mm) (150 mm) ] 16,147,786.239 mm
64 64
1 1
[ ] [(170 mm) (150 mm) ] 128,166.667 mm
12
128,170
12
 mm
I D d
Q D d
 
    
      Ans. 
 
(b) Maximum load P: 
 
2 4
max 3
75 MPa
(75 N/mm )(16,147,786.239 mm )(170 mm 150 mm)
188,986 N
128,166.667 mm
V Q
I t
V
  

   
 
For the cantilever beam shown here, V = P; therefore, 
 
max max 189.0 kNP V 
 Ans. 
 
 
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9.24 A concentrated load P is applied to the upper end of a 1-m long 
pipe, as shown in Fig. P9.24. The outside diameter of the pipe is D = 
114 mm and the inside diameter is d = 102 mm. 
(a) Compute the value of Q for the pipe. 
(b) If the allowable shear stress for the pipe shape is 75 MPa, determine 
the maximum load P than can be applied to the cantilever beam. 
 
Fig. P9.24 
 
Solution 
(a) Section properties: 
 
4 4 4 4 4
3 3 3 3 3
[ ] [(114 mm) (102 mm) ] 2,977,287 mm
64 64
1 1
[ ] [(114 mm) 35,028 (102 m mm) ]
12 1
m
2
I D d
Q D d
 
    
     Ans. 
 
(b) Maximum load P: 
 
2 4
max 3
75 MPa
(75 N/mm )(2,977,287 mm )(114 mm 102 mm)
76,498 N
35,028 mm
V Q
I t
V
  

   
 
For the cantilever beam shown here, V = P; therefore, 
 
max max 76.5 kNP V 
 Ans. 
 
 
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9.25 A concentrated load of P = 6 kips is applied to the upper end of a 
4-ft long pipe, as shown in Fig. P9.25. The pipe is an 8 in. standardsteel pipe, which has an outside diameter of D = 8.625 in. and an 
inside diameter of d = 7.981 in. Determine the magnitude of: 
(a) the maximum vertical shear stress in the pipe. 
(b) the maximum tension bending stress in the pipe. 
 
Fig. P9.25 
 
Solution 
Section properties: 
 
4 4 4 4 4
3 3 3 3 3
[ ] [(8.625 in.) (7.981 in.) ] 72.489241 in.
64 64
1 1
[ ] [(8.625 in.) (7.981 in.) ] 11.104874 in.
12 12
I D d
Q D d
 
    
    
 
 
Maximum shear force magnitude: 
 Vmax = 6 kips = 6,000 lb 
 
Maximum bending moment magnitude: 
 Mmax = (6,000 lb)(4 ft)(12 in./ft) = 288,000 lb-in. 
 
(a) Maximum vertical shear stress: 
 3
4
(6,000 lb)(11.104874 in. )
1,427.268 psi
(72.489241 in. )(8.625 in. 7.981
1,427 p
 in.)
si
VQ
I t
    

 Ans. 
 
(b) Maximum tension bending stress: 
 
4
(288,000 lb-in.)(8.625 in./2)
17,133.578 psi
72.48
17,130 psi 
924
T
.
(
1 in
)x
M c
I
     Ans. 
 
 
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9.26 The cantilever beam shown in Fig. P9.26a is subjected to a concentrated load of P = 38 kips. The 
cross-sectional dimensions of the wide-flange shape are shown in Fig. P9.26b. Determine: 
(a) the shear stress at point H, which is located 4 in. below the centroid of the wide-flange shape. 
(b) the maximum horizontal shear stress in the wide-flange shape. 
 Fig. P9.26a Fig. P9.26b 
Solution 
Moment of inertia about the z axis: 
Shape Width b Height h IC d = yi – y d²A IC + d²A 
 (in.) (in.) (in.
4
) (in.) (in.
4
) (in.
4
) 
flange 6.75 0.455 0.0530 6.7725 140.8683 140.9213 
web 0.285 13.090 53.2700 0.0000 0.0000 53.2700 
flange 6.75 0.455 0.0530 –6.7725 140.8683 140.9213 
Moment of inertia about the z axis (in.
4
) = 335.1125 
 
 
(a) Shear stress at H: 
 
3
3
4
0.455 in.
(6.75 in.)(0.455 in.) 7 in.
2
7 in. 0.455 in. 4 in.
(0.285 in.)(7 in. 0.455 in. 4 in.) 4 in. 24.6243 in.
2
(38 kips)(24.6243 in. )
9.7974 ksi
(335.1125 in. )(0.285 in.)
9.80 ksi
H
H
Q

 
   
  
      
  
 Ans. 
 
(b) Maximum horizontal shear stress: 
 
max
3
3
max 4
0.455 in.
(6.75 in.)(0.455 in.) 7 in.
2
7 in. 0.455 in.
(0.285 in.)(7 in. 0.455 in.) 26.9043 in.
2
(38 kips)(26.9043 in. )
10.7046 ksi
(335.1125 in. )(0.28
10.70 k
5 i
s
)
i
n.
Q

 
   
 
    
  
 Ans. 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.27 The cantilever beam shown in Fig. P9.27a is subjected to a concentrated load of P. The cross-
sectional dimensions of the wide-flange shape are shown in Fig. P9.27b. 
(a) Compute the value of Q that is associated with point K, which is located 2 in. above the centroid of 
the wide-flange shape. 
(b) If the allowable shear stress for the wide-flange shape is 14 ksi, determine the maximum 
concentrated load P than can be applied to the cantilever beam. 
 Fig. P9.27a Fig. P9.27b 
Solution 
Moment of inertia about the z axis: 
Shape Width b Height h IC d = yi – y d²A IC + d²A 
 (in.) (in.) (in.
4
) (in.) (in.
4
) (in.
4
) 
flange 6.75 0.455 0.0530 6.7725 140.8683 140.9213 
web 0.285 13.090 53.2700 0.0000 0.0000 53.2700 
flange 6.75 0.455 0.0530 –6.7725 140.8683 140.9213 
Moment of inertia about the z axis (in.
4
) = 335.1125 
 
 
(a) Q associated with point K: 
 3
0.455 in.
(6.75 in.)(0.455 in.) 7 in.
2
7 in. 0.455 in. 2 in.
(0.285 in.)(7 in. 0.455 26.33in. 2 in.) 2 i 43 inn
2
..
KQ
 
   
  
      
 Ans. 
 
(b) Maximum load P: 
 
max
3
0.455 in.
(6.75 in.)(0.455 in.) 7 in.
2
7 in. 0.455 in.
(0.285 in.)(7 in. 0.455 in.) 26.9043 in.
2
Q
 
   
 
    
 
 
max
max
4
max 3
14 ksi
(14 ksi)(335.1125 in. )(0.285 in.)
49.6983 kips
26.9043 in.
VQ
It
V
  
  
 
 
For the cantilever beam shown here, V = P; therefore, 
 
max max 49.7 kipsP V 
 Ans. 
 
 
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9.28 The cantilever beam shown in Fig. P9.28a is subjected to a concentrated load of P. The cross-
sectional dimensions of the rectangular tube shape are shown in Fig. P9.28b. 
(a) Compute the value of Q that is associated with point H, which is located 90 mm above the centroid 
of the rectangular tube shape. 
(b) If the allowable shear stress for the rectangular tube shape is 125 MPa, determine the maximum 
concentrated load P than can be applied to the cantilever beam. 
 Fig. P9.28a 
Fig. P9.28b 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi –
y
 d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
outer rectangle 195,312,500 0.000 0.000 195,312,500 
inner rectangle −143,077,428 0.000 0.000 −143,077,428 
Moment of inertia about the z axis (mm
4
) = 52,235,072 
(a) Q associated with point H: 
 
3
250 mm 8 mm
(150 mm)(8 mm)
2 2
250 mm
8 mm 90 mm
250 mm 22(8 mm) 8 mm 90 mm 90 mm
189,912 mm
2 2
HQ
 
   
 
 
   
       
  
 Ans. 
(b) Maximum load P: 
 
max
3
250 mm 8 mm
(150 mm)(8 mm)
2 2
250 mm
8 mm
250 mm 22(8 mm) 8 mm 254,712 mm
2 2
Q
 
   
 

   
      
  
 
 
max
max
2 4
max 3
125 MPa
(125 N/mm )(52,235,072 mm )(2 8 mm)
410,150 N 410.15 kN
254,712 mm
VQ
It
V
  

   
 
 
For the cantilever beam shown here, V = P; therefore, 
 
max max 410 kNP V 
 Ans. 
 
 
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9.29 The cantilever beam shown in Fig. P9.29a is subjected to a concentrated load of P = 175 kN. The 
cross-sectional dimensions of the rectangular tube shape are shown in Fig. P9.29b. Determine: 
(a) the shear stress at point K, which is located 50 mm below the centroid of the rectangular tube shape. 
(b) the maximum horizontal shear stress in the rectangular tube shape. 
 Fig. P9.29a 
Fig. P9.29b 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi –
y
 d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
)outer rectangle 195,312,500 0.000 0.000 195,312,500 
inner rectangle −143,077,428 0.000 0.000 −143,077,428 
Moment of inertia about the z axis (mm
4
) = 52,235,072 
 
(a) Shear stress at K: 
 
3
250 mm 8 mm
(150 mm)(8 mm)
2 2
250 mm
8 mm 50 mm
250 mm 22(8 mm) 8 mm 50 mm 50 mm
2 2
234,712 mm
KQ
 
   
 
 
   
       
  
 
 
 3
4
(175,000 N)(234,712 mm )
49.1463 MPa
(52,235,072 mm )(2 8
49.1 
 mm
Pa
)
MK   

 Ans. 
 
(b) Maximum horizontal shear stress: 
 
max
3
250 mm 8 mm
(150 mm)(8 mm)
2 2
250 mm
8 mm
250 mm 22(8 mm) 8 mm 254,712 mm
2 2
Q
 
   
 

   
      
  
 
 3
max 4
(175,000 N)(254,712 mm )
53.3341 MPa
(52,235,072 mm )(2 8 
53.3 MP
)
a
mm
   

 Ans. 
 
 
 
 
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9.30 The internal shear force V at a certain section of an 
aluminum beam is 8 kN. If the beam has a cross section 
shown in Fig. P9.30, determine: 
(a) the shear stress at point H, which is located 30 mm 
above the bottom surface of the tee shape. 
(b) the maximum horizontal shear stress in the tee shape. 
 
 Fig. P9.30 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (mm
2
) (mm) (mm
3
) 
top flange 375.0 72.5 27,187.5 
stem 350.0 35.0 12,250.0 
 725.0 mm
2
 
 
39,437.5 mm
3 
 
 
3
2
39,437.5 mm
54.397 mm (from bottom of shape to centroid)
725.0 mm
20.603 mm (from top of shape to centroid)
i i
i
y A
y
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 781.250 18.103 122,900.565 123,681.815 
stem 142,916.667 −19.397 131,679.177 274,595.843 
Moment of inertia about the z axis (mm
4
) = 398,277.658 
 
(a) Shear stress at H: 
 
3
3
4
(5 mm)(30 mm)(39.397 mm) 5,909.550 mm
(8,000 N)(5,909.550 mm )
(398,277.658 mm )(5 mm
23.7 MP
)
a
H
H
Q

 
  Ans. 
 
(b) Maximum horizontal shear stress: 
At neutral axis: 
 
3
max
3
max 4
(5 mm)(54.397 mm)(27.199 mm) 7,397.720 mm
(8,000 N)(7,397.720 mm )
(398,277.658 mm )(5 mm)
29.7 MPa
Q

 
  Ans. 
 
 
 
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9.31 The internal shear force V at a certain section of a 
steel beam is 80 kN. If the beam has a cross section 
shown in Fig. P9.31, determine: 
(a) the shear stress at point H, which is located 30 mm 
below the centroid of the wide-flange shape. 
(b) the maximum horizontal shear stress in the wide-
flange shape. 
 
 Fig. P9.31 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 59,062.5 97.5 29,944,687.5 30,003,750.0 
web 4,860,000.0 0.0 0.0 4,860,000.0 
bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750.0 
Moment of inertia about the z axis (mm
4
) = 64,867,500.0 
 
(a) Shear stress at H: 
 
3
3
4
(210 mm)(15 mm)(97.5 mm) (10 mm)(60 mm)(60 mm) 343,125 mm
(80,000 N)(343,125 mm )
(64,867,500 mm )(10 mm)
42.3 MPa
H
H
Q

  
  Ans. 
 
(b) Maximum horizontal shear stress: 
At neutral axis: 
 
3
max
3
max 4
(210 mm)(15 mm)(97.5 mm) (10 mm)(90 mm)(45 mm) 347,625 mm
(80,000 N)(347,625 mm )
(64,867,500 mm )(10 mm
4
)
2.9 MPa
Q

  
  Ans. 
 
 
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9.32 The internal shear force V at a certain section 
of a steel beam is 110 kips. If the beam has a cross 
section shown in Fig. P9.32, determine: 
(a) the value of Q associated with point H, which is 
located 2 in. below the top surface of the flanged 
shape. 
(b) the maximum horizontal shear stress in the 
flanged shape. 
Fig. P9.32 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (in.
2
) (in.) (in.
3
) 
top flange 5.0 11.5 57.5 
web 10.0 6.0 60.0 
bottom flange 8.0 0.5 4.0 
 23.0 in.
2
 
 
121.5 in.
3 
 
 
3
2
121.5 in.
5.2826 in. (from bottom of shape to centroid)
23.0 in.
6.7174 in. (from top of shape to centroid)
i i
i
y A
y
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 0.4167 6.2174 193.2798 193.6964 
web 83.3333 0.7174 5.1465 88.4798 
bottom flange 0.6667 −4.7826 182.9868 183.6534 
Moment of inertia about the z axis (in.
4
) = 465.8297 
 
(a) Q at point H: 
 
3(5 in.)(1 in.)(6.2174 in.) (1 in.)(1 in.)(5.2174 36.304in.) 4 in.Q   
 Ans. 
 
(b) Maximum horizontal shear stress: 
At neutral axis: 
 
3
max
3
max 4
(5 in.)(1 in.)(6.2174 in.) (1 in.)(5.7174 in.)(2.8587 in.) 47.4313 in.
(110 kips)(47.4313 in. )
(465.8297 in. )(1 in.)
11.20 ksi
Q

  
  Ans. 
 
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9.33 The internal shear force V at a certain section of a 
steel beam is 75 kips. If the beam has a cross section 
shown in Fig. P9.33, determine: 
(a) the shear stress at point H, which is located 2 in. 
above the bottom surface of the flanged shape. 
(b) the shear stress at point K, which is located 4.5 in. 
below the top surface of the flanged shape. 
 
 Fig. P9.33 
Solution 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
left flange 62.5000 0.0000 0.0000 62.5000 
web 0.0885 0.0000 0.0000 0.0885 
right flange 62.5000 0.0000 0.0000 62.5000 
Moment of inertia about the z axis (in.
4
) = 125.0885 
 
(a) Shear stress at H: 
 
3
3
4
2(0.75 in.)(2 in.)(4 in.) 12 in.
(75 kips)(12 in. )
(125.0885 in. )(2 0.75 in.)
4.80 ksi
H
H
Q

 
 

 Ans. 
 
(b) Shear stress at K: 
 
3
3
4
2(0.75 in.)(4.5 in.)(2.75 in.) 18.5625 in.
(75 kips)(18.5625 in. )
(125.0885 in. )(2 0.75 in.)
7.42 ksi
K
K
Q

 
 

 Ans.Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
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9.34 Consider a 100-mm-long segment of a simply 
supported beam (Fig. P9.34a). The internal bending 
moments on the left and right sides of the segment are 
75 kN-m and 80 kN-m, respectively. The cross-
sectional dimensions of the flanged shape are shown in 
Fig. P9.34b. Determine the maximum horizontal shear 
stress in the beam at this location. 
 
 
Fig. P9.34b Cross-sectional dimensions Fig. P9.34a Beam segment (side view) 
Solution 
Centroid location in y direction: 
Shape Area Ai 
yi 
(from bottom) yi Ai 
 (mm
2
) (mm) (mm
3
) 
top flange 9,000 300 2,700,000 
web 8,400 165 1,386,000 
bottom flange 15,000 30 450,000 
 32,400 mm
2
 
 
4,536,000 mm
3 
 
3
2
4,536,000 mm
140 mm (from bottom of shape to centroid)
32,400 mm
190 mm (from top of shape to centroid)
i i
i
y A
y
A

  


 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 2,700,000 160 230,400,000 233,100,000 
web 30,870,000 25 5,250,000 36,120,000 
bottom flange 4,500,000 −110 181,500,000 186,000,000 
Moment of inertia about the z axis (mm
4
) = 455,220,000 
 
Shear force in beam: 
 
80 kN-m 75 kN-m 5 kN-m
50 kN
100 mm 0.1 m
M
V
x
 
   

 
 
Maximum horizontal shear stress: 
At neutral axis: 
 
3
max
3
max 4
(250 mm)(60 mm)(110 mm) (40 mm)(80 mm)(40 mm) 1,778,000 m
4.88 MP
m
(50,000 N)(1,778,000 mm )
(455,220,000 mm )(40 mm)
a
Q

  
  Ans. 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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9.35 A simply supported beam supports the loads shown in Fig. P9.35a. The cross-sectional dimensions 
of the wide-flange shape are shown in Fig. P9.35b. 
(a) Determine the maximum shear force in the beam. 
(b) At the section of maximum shear force, determine the shear stress in the cross section at point H, 
which is located 100 mm below the neutral axis of the wide-flange shape. 
(c) At the section of maximum shear force, determine the maximum horizontal shear stress in the cross 
section. 
(d) Determine the magnitude of the maximum bending stress in the beam. 
 
 
Fig. P9.35a Fig. P9.35b 
Solution 
 
(a) Maximum shear force magnitude: 
 Vmax = 175 kN (just to the right of B) 
 
 
Maximum bending moment magnitude: 
 Mmax = 156.25 kN-m (between B and C) 
 
 
 
 
 
 
 
 
 
 
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Section properties: 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
top flange 56,250 142.5 60,918,750 60,975,000 
web 16,402,500 0 0 16,402,500 
bottom flange 56,250 −142.5 60,918,750 60,975,000 
Moment of inertia about the z axis (mm
4
) = 138,352,500 
 
(b) Shear stress at H: 
 
3
3
4
(200 mm)(15 mm)(142.5 mm) (10 mm)(35 mm)(117.5 mm) 468,625 mm
(175,000 N)(468,625 mm )
(138,352,500 mm )(10 mm)
59.3 MPa
H
H
Q

  
  Ans. 
 
(c) Maximum horizontal shear stress: 
At neutral axis: 
 
3
max
3
max 4
(200 mm)(15 mm)(142.5 mm) (10 mm)(135 mm)(67.5 mm) 518,625 mm
(175,000 N)(518,625 mm )
(138,352,500 mm )(10 mm
65.6 MP
)
a
Q

  
  Ans. 
 
(d) Maximum tension bending stress: 
 
6
4
(156.25 10 N-mm)(300 mm/2)
138,352,500
169.4 MP
 mm
169.40 a a4 MP
x
M c
I
   
  Ans. 
 
 
 
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9.36 A simply supported beam supports the loads shown in Fig. P9.36a. The cross-sectional dimensions 
of the structural tube shape are shown in Fig. P9.36b. 
(a) At section a–a, which is located 4 ft to the right of pin support B, determine the bending stress and 
the shear stress at point H, which is located 3 in. below the top surface of the tube shape. 
(b) Determine the magnitude and the location of the maximum horizontal shear stress in the tube shape 
at section a–a. 
 
Fig. P9.36a Fig. P9.36b 
Solution 
 
Shear force magnitude at a–a: 
 V = 27.60 kips 
 
Bending moment at a–a: 
 M = 60.90 kip-ft 
 
Section properties: 
 
3 3
4
(12 in.)(16 in.) (11.25 in.)(15.25 in.)
12 12
771.0830 in.
I  
 
 
(a) Bending stress at H: 
 
4
(60,900 lb-ft)(5 in.)(12 in./ft)
771.0830 in.
4,738.79 psi 4,740 psi (C)
H
M y
I
  
 
   Ans. 
 
Shear stress at H: 
 
3
3
4
(12 in.)(0.375 in.)(7.8125 in.) 2(0.375 in.)(2.625 in.)(6.3125 in.) 47.5840 in.
(27,600 lb)(47.5840 in. )
(771.0830 in. )(2 0.375 in.)
2,270 psi
H
H
Q

  
 

 Ans. 
 
(b) Maximum shear force magnitude: 
 V = 39.60 kips (at pin B) 
 
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Maximum horizontal shear stress: 
At neutral axis: 
 
3
max
3
max 4
(12 in.)(0.375 in.)(7.8125 in.) 2(0.375 in.)(7.625 in.)(3.8125 in.) 56.9590 in.
(39,600 lb)(56.9590 in. )
(771.0830 in. )(2 0.375 in.)
3,900 psi
Q

  
 

 Ans. 
 
 
 
 
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9.37 A cantilever beam supports the loads shown in Fig. P9.37a. The cross-sectional dimensions of the 
shape are shown in Fig. P9.37b. Determine: 
(a) the maximum horizontal shear stress. 
(b) the maximum compression bending stress. 
(c) the maximum tension bending stress. 
 
Fig. P9.37a Fig. P9.37b 
SolutionCentroid location in y direction: 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (in.) (in.) (in.
2
) (in.) (in.
3
) 
top flange 12.0 0.5 6.00 5.75 34.5000 
left stem 0.5 5.5 2.75 2.75 7.5625 
right stem 0.5 5.5 2.75 2.75 7.5625 
 11.50 
 
49.6250
 
 
 
3
2
49.6250 in.
4.3152 in. (from bottom of shape to centroid)
11.50 in.
1.6848 in. (from top of shape to centroid)
i i
i
y A
y
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (in.
4
) (in.) (in.
4
) (in.
4
) 
top flange 0.1250 1.4348 12.3519 12.4769 
left stem 6.9323 −1.5652 6.7371 13.6694 
right stem 6.9323 −1.5652 6.7371 13.6694 
Moment of inertia about the z axis (in.
4
) = 39.8157 
 
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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Maximum shear force magnitude: 
 V = 5,800 lb 
 
Maximum positive bending moment: 
 Mpos = 8,850 lb-ft 
 
Maximum negative bending moment: 
 Mneg = −9,839 lb-ft 
 
(a) Maximum shear stress: 
 
max
3
3
max 4
2(0.5 in.)(4.3152 in.)(4.3152 in./2)
9.3105 in.
(5,800 lb)(9.3105 in. )
(39.8157 in. )(2 0.5 in.)
1,356 psi
Q





 Ans. 
 
 
(b) Maximum compression bending stress: 
Check two possibilities. First, check the bending stress created by the largest positive moment at the top 
of the cross section: 
 
pos top
4
(8,850 lb-ft)(1.6848 in.)(12 in./ft)
4,494 psi
39.8157 in.
x
z
M y
I
       
Next, for the largest negative moment, compute the bending stress at the bottom of the cross section: 
 
neg bot
4
( 9,839 lb-ft)( 4.3152 in.)(12 in./ft)
12,796 psi
39.8157 in.
x
z
M y
I
        
Therefore, the maximum compression bending stress is: 
 
comp 12,800 psi (C) 
 Ans. 
 
(c) Maximum tension bending stress: 
Check two possibilities. First, check the bending stress created by the largest positive moment at the 
bottom of the cross section: 
 
pos bot
4
(8,850 lb-ft)( 4.3152 in.)(12 in./ft)
11,510 psi
39.8157 in.
x
z
M y
I
      
Next, for the largest negative moment, compute the bending stress at the top of the cross section: 
 
neg top
4
( 9,839 lb-ft)(1.6848 in.)(12 in./ft)
4,996 psi
39.8157 in.
x
z
M y
I
      
Therefore, the maximum tension bending stress is: 
 
tens 11,510 psi (T) 
 Ans. 
 
 
 
 
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9.38 A cantilever beam supports the loads shown in Fig. 
P9.38a. The cross-sectional dimensions of the shape are 
shown in Fig. P9.38b. Determine: 
(a) the maximum vertical shear stress. 
(b) the maximum compression bending stress. 
(c) the maximum tension bending stress. 
Fig. P9.38a Fig. P9.38b 
Solution 
 
Maximum shear force magnitude: 
 Vmax = 5 kN 
 
Maximum positive bending moment: 
 Mpos = 2.00 kN-m 
 
Maximum negative bending moment: 
 Mneg = −1.50 kN-m 
 
 
 
 
 
 
 
 
 
 
Centroid location in y direction: 
Shape Width b Height h Area Ai 
yi 
(from bottom) yi Ai 
 (mm) (mm) (mm
2
) (mm) (mm
3
) 
flange 100 8 800 96 76,800 
stem 6 92 552 46 25,392 
 1,352 102,192 
 
 
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3
2
102,192 mm
75.5858 mm (from bottom of shape to centroid)
1,352 mm
24.4142 mm (from top of shape to centroid)
i i
i
y A
y
A

  


 
 
Moment of inertia about the z axis: 
Shape IC d = yi – y d²A IC + d²A 
 (mm
4
) (mm) (mm
4
) (mm
4
) 
flange 4,266.67 20.4142 333,391.69 337,658.35 
stem 389,344.00 -29.5858 483,176.36 872,520.36 
Moment of inertia about the z axis (mm
4
) = 1,210,178.71 
 
(a) Maximum vertical shear stress: 
At neutral axis: 
 
3
max
3
max 4
(6 mm)(75.5858 mm)(75.5858 mm/2) 17,139.640 mm
(5,000 N)(17,139.640 mm )
(1,210,178.71 mm )(6 mm)
11.80 MPa
Q

 
  Ans. 
 
(b) Maximum compression bending stress: 
Check two possibilities. First, check the bending stress created by the largest positive moment at the top 
of the cross section: 
 6
pos top
4
(2.00 10 N-mm)(24.4142 mm)
40.348 MPa
1,210,178.71 mm
x
z
M y
I
       
Next, for the largest negative moment, compute the bending stress at the bottom of the cross section: 
 6
neg bot
4
( 1.50 10 N-mm)( 75.5858 mm)
93.688 MPa
1,210,178.71 mm
x
z
M y
I
         
Therefore, the maximum compression bending stress is: 
 
comp 93.7 MPa (C) 
 Ans. 
 
(c) Maximum tension bending stress: 
Check two possibilities. First, check the bending stress created by the largest positive moment at the 
bottom of the cross section: 
 6
pos bot
4
(2.00 10 N-mm)( 75.5858 mm)
124.917 MPa
1,210,178.71 mm
x
z
M y
I
       
Next, for the largest negative moment, compute the bending stress at the top of the cross section: 
 6
neg top
4
( 1.50 10 N-mm)(24.4142 mm)
30.261 MPa
1,210,178.71 mm
x
z
M y
I
       
Therefore, the maximum tension bending stress is: 
 
tens 124.9 MPa (T) 
 Ans. 
 
 
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9.39 A simply supported beam fabricated from pultruded reinforced plastic supports the loads shown in 
Fig. P9.39a. The cross-sectional dimensions of the plastic wide-flange shape are shown in Fig. P9.39b. 
(a) Determine the magnitude of the maximum shear force in the beam. 
(b) At the section of maximum shear force, determine the shear stress magnitude in the cross section at 
point H, which is located 2 in. above the bottom surface of the wide-flange shape. 
(c) At the section of maximum shear force, determine the magnitude of the maximum horizontal shear 
stress in the cross section. 
(d) Determine the magnitude of the maximum compression bending stress in the beam. Where along the 
span does this stress occur? 
 
 
Fig. P9.39a Fig. P9.39b 
Solution 
 
Section properties: 
 
3 3
4
(4 in.)(8 in.) (3.625 in.)(7.25 in.)
12 12
55.5493 in.
zI  
 
 
(a) Maximum shear force magnitude: 
 V = 3,664 lb