philpot mom 2nd ch07-11 ism
TRANSCRIPT
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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
02
2
y y
y
A A
A
F A w L
A w L
LM M w L
w LM
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
02 2
2 2
a a
w L xM w Lx w x M
w L w xM
wL 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 PbC A
a b a b
Section a-a:
For the interval 0 ≤ x < a:
0
0
y y
a a y
PbV
a b
PbM x
a b
PbF A V V
a b
PbM A x M x M
a b
Section b-b:
For the interval a ≤ x < b:
0y y
PbF A P V P V
a b
PaV
a b
(
( )
) ( ) 0b b y
PbM A x P x a M x P x a M
a
Pb
b
M x P x aa 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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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
02 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 bM w a b w b M
a bM w a b w b
Section a-a:
For the interval 0 ≤ x < a:
2
2
0
02
y a
a a
a
aa
V w xF w x V
xM w x M
w xM
Section b-b:
For the interval a ≤ x < b:
2
2 2
0
02 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 aaM w a x
x aM w a x w x a 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
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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
( ) 02 2
( 2 )
2( )
( ) 02 2
(2 )
2( )
C a b y
a by
A a b y
a by
a bM w a b w b A a b
w a a b w bA
a b
a bM w a w b a C a b
w a w b a bC
a b
Section a-a:
For the interval 0 ≤ x < a:
2
2 22
0
02
2
( 2 )
2( )
( 2 )
2 2( )
y y a a y
a a y a
a ba
ya ba a
F A w x V V w x A
xM A x w x M
w xM
w a a b w bw x
a b
w x w a a b w bA
a bx x
Section b-b:
For the interval a ≤ x < b:
2( 2 )( )
2( ) 2( )
( ) 0
( )
a ba
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 bw a w x a
a b a b
a
22 2
( 2 )
2
02 2
2 22( ) 2(2 )
b b y a b
ay a ba b
b
a x aM A x w a x w x a M
x aaM A x w
x aaw a a b w bx w a x w
a b aa x w
b
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(b) Shear-force and bending-moment diagrams
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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
02 2
02 3 6
y y y
B B B
w L w LF B B
w L L w LM M M
Section a-a:
0
2
0
3
0
0 02
02 3
2
6
y
a a
w xw x xF V
L
w x x xM M
L
VL
w xM
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
02 3 6
20
2 3 3
B y y
A y y
w L L w LM A L A
w L L w LM B L B
Section a-a:
2
0 0 0
3
0 0
2
0 0
3
0 0
0
6 2
6
02 6 2
02
6
3 6 6
y y
a a y
w x x w L w xF A V V
L L
w x x x w Lx w xM A x M M
L
w L w xV
L
w x w LxM
L
L
(b) Shear-force and bending-moment diagrams
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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
06
0.577350
2
2 6 3 3
w L w xV
L
w x w L L
LL
Lx 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
60
350 )
6
w x w LxM
L
w L w L LM
Lw 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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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 positive bending 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 )
20
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) 02
(7 kips/ft)(150 kips) 1,350 kips
2
b b y
xM x B
M x
x M
xx 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/ftt
2)( ) 0
2
y
a a
F x V
xM 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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(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
t2
-
)
ft
0
b b y
xM 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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(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 ft120 kip-ft 5 kips/
5 40
ft 8 ft =02
kips
2.5 40 280 kip-f t
y
b b
F x V
xM 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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(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:
16 kips/ft 9 ft 17 kips 0
2
44 kips
16 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 1 644 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 xF A x V x V
x xM M A x x M
x xx x
xV
M
3
44 400 kip-ft9
x
M x
16 kips/ft 9 ft
2
1 44 kips 6 kips/ft 9 ft
17 ki
02
ps
y yF A V
V
V
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-
16 kips/ft 9 ft 6 ft
2
1 400 kip-ft 6 kips/ft 9 ft 6 ft 044
1
kips2
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/ft2
250 kip-ft 061.03 kips 7
7 61.03 kips
3.5 61.
kips/ft2
3 0
y y
b b y
F A x V x V
xM A x x M
xx 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/ft2
061.03 kips 7
7 61.03 kips
3.5 61.03
kips/ft2
kip- t f
y y
a a y
F A x V x V
xM A x x
V x
M
M
xx
x
M
x
x
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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/ft2
250 kip-ft
(17 ft)175 kips 113.97 kips 7 kips/ft2
7 175 kips
y y y
c c y y
F A C x V
x V
xM A x C x x
M
xx
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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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 0
32.00 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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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
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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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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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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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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
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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
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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 of this work beyond that
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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
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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
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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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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
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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
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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
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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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7.31 Draw the shear-force and bending-moment
diagram for the beam shown in Fig. P7.31. 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. P7.31
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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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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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
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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
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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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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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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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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
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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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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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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
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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
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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 2.5 m 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
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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
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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 ft6 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:
12
12
12
(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
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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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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 m2 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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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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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/ft9 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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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/ft12 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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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/m3 m 12 kN 5 m
2
M x x x x
x x Ans.
Shear-force and bending-moment diagrams:
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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:
12
12
12
(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/ft22 ft 30 ft 9 kips/ft 30 ft
8 ft 8 ft
w x x x x x
x x x Ans.
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/ft22 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/ft22 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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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/m3 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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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/m1 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/m1 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/m1 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/ft9 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/ft6 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/ft6 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/ft6 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 textbook has 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/m6.5 m 45 kN/m 6.5 m 114.38 kN 8 m
4.0 m
w x x x x x
x x x Ans.
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/m6.5 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/m6.5 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
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 = 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/m234.24 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/m234.24 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/m234.24 kN 1.5 m 7 m 7 m
2 6(7.0 m)
50 kN/m 50 kN/m215.76 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/m6 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/m6 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/m6 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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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 = 197.2 kN-m at x = 5.01 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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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 kt
si)
x
Ey
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.
8.2 A high-strength steel [E = 200 GPa] tube having an outside 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 MPm)
ax
Ey
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 mm2 2
From Eq. (8.3):
200,000 MPa
( 0.5 mm) 443.459 MPa225.5 mm
443 MPax
Ey
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 mm2
t
From Eq. (8.3):
12,000 MPa
7 MPa2
10,000 mm2
x
E ty
t
Solve for t:
12,000 MPa 7 MPa 10,000 mm2 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
(mm2) (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 mm2
687,500.0 mm
3
3
2
687,500.0 mm
6,250.0 mm110.0 mm
i i
i
y Ay
A
(measured upward from bottom edge of stem) Ans.
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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 mm286,858.974 mm
(175 mm 110 mm)
18,645,833.33 mm169,507.576 mm
110 mm
169,500 mm
z
z
IS
c
IS
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 i4.70
n.
20.0 i i
nn
..
i i
i
y Ay
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:
43
top
top
43
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
IS
c
IS
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
(mm2) (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 mm2
21,088.0 mm
3
3
2
21,088.0 mm
1,072.0 mm19.67 mm
i i
i
y Ay
A
(measured upward from bottom edge of section) Ans.
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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 mm8,494.814 mm
(50 mm 19.672 mm)
257,633.75 mm13,096.708 mm
19.672
8,495 mm
mm
z
z
IS
c
IS
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.
52.06.08 in.
in.
i i
i
y Ay
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
43
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
zbot
bot
ztop
top
c
c
IS
c
IS
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 i968 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 in666 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
1,285.021,295 psi
6 n.(T)
ix
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
(mm2) (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 mm2
27,937.5 mm
3
3
2
27,937.5 mm33.864 mm 33.9 mm
825.0 mm
i i
i
y Ay
A
(measured upward from bottom of section)
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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
IM
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
2130.8 MPa (
04,559.659 mmT)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 Ay
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
IM
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
01,114 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 12zI
Maximum internal bending moment Mz:
4(15,000 psi)(1.791667 in. )
26,875 lb-in.1 in.
2,240 lb-ft
zx
z
x z
M c
I
IM
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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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 Ay
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
10.09694.26 ksi
6 in. (C)x
z
M y
I 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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8.13 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
(mm4) (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 (mm4) = 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 MPa90
6
8.1 MPa (T)mm
H K
H K
HH 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-mm175 m
3 Nm
79 k -m
zx
z
x z
z
M c
I
IM
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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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
(mm4) (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 (mm4) = 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 MPa60 m
MPa (Cm
)
H K
H K
HH 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-mm90 mm
35.7 kN-m
zx
z
x z
M c
I
IM
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 Ay
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 i703 p
nsi 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 Ay
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 cross section 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
(mm2) (mm) (mm
3)
top flange 3,000.0 167.5 502,500.0
stem 1,440.0 80.0 115,200.0
4,440 mm2
617,700 mm
3
3
2
617,700 mm139.1216 mm
4,440 mm
i i
i
y Ay
A
(measured upward from bottom edge of stem)
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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 m157.8 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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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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8.18 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 Ay
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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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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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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8.19 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 Ay
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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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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8.22 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 Ay
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
49.625 in.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 Ay
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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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
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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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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8.24 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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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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8.25 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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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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8.26 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 mm64 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 mm64 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 m65.8 MPa
mx 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
MS
Section modulus for solid circular section
3
32
dS
Minimum shaft diameter
330.375 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-mm5,440 mm
25 N/mm
x
x
M
S
MS
Section modulus for solid circular section
3
32
dS
Minimum shaft diameter
335,440 m
38. m
m3
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.
wLM
Minimum required section modulus
3480,000 lb-in.342.8571 in.
1, 400 psi
x
x
M
S
MS
Section modulus for solid rectangular section
3 2/12
/ 2 6
I bh bhS
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 ) 2.250.3750
6 6 6
bh b b bS 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
bhI
Maximum allowable moment
4(900 psi)(500 in. )
90,000 lb-in. 7,500 lb-ft5 in.
xx
IMcM
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-ft8
wLM
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
wLM
Minimum required section modulus
6
6 3
2
12.5 10 N-mm1.3889 10 mm
9 N/mm
x
x
M
S
MS
Section modulus for solid rectangular section
3 2/12
/ 2 6
I bh bhS
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 ) 0.250.0416667
6 6 6
bh b b bS 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 bhS
c h
Maximum allowable bending moment:
2 6 3 6
allow (6 N/mm )(2 10 mm ) 12 10 N-mmx x
MM S
S
Maximum bending moment in cantilever span:
The maximum bending moment magnitude in the cantilever beam occurs at support A:
2
max2
wLM
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 mm2.67 kN/
/ )m
m ]
wLM
Mw
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
MS
(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-m
Minimum required section modulus
2
3 3
2
(206.63 kN-m)(1,000)1,252 10 mm
165 N/mm
x
x
M
S
MS
(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
MS
(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
MS
(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
MS
(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 ksi16.6667
1,800 ksi
En
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
ii
Myn
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
11 max
(1.200 ksi)(2,064 in. )412.80 kip-in.
6 in.
IMyM
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
22 max
(24.00 ksi)(2,064 in. )495.36 kip-in.
(16.667)(6 in.)
IMyn 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
MP 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 GPa3
10 GPa
En
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
(mm4) (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 mm4
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 mm11.12 MPa
Myn
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 m7.94
m MPa
Myn
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 GPa1.3333
75 GPa
En
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
(mm4) (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 mm4
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 mm40.0 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 m88.8 MPa
m
Myn
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 m53.3 MPa
m
Myn
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 ksi3
10,000 ksi
En
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.
5.500.7614 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 in4,970 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 i9,570 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 ksi3
10,000 ksi
En
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.
5.500.7614 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 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
11
(20 ksi)(0.5514 in. )14.484 kip-in.
0.7614 in.
My IM
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
22
(30 ksi)(0.5514 in. )11.285 kip-in.
(3)(1.25 in. 0.7614 in.)
My In 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 ksi16.6667
1,800 ksi
En
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
11
(1.200 ksi)(7,814.72 in. )1,172.208 kip-in.
8 in.
IMyM
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
22
(24 ksi)(7,814.72 in. )1,364.021 kip-in.
(16.6667)( 8.25 in.)
IMyn 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 GPa16.6667
12 GPa
En
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) (mm2) (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 mm137.294 mm
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
(mm4) (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 mm4
= 1.5106 ×109 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
wLM
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 mm5.83 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 m56.8 MPa (
mT)
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
611 max
(9 N/mm )(1.5106 10 mm )57.925 10 N-mm
(372 mm 137.294 mm)
IMyM
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
622 max
(165 N/mm )(1.5106 10 mm )108.926 10 N-mm
(16.6667)(137.294 mm)
IMyn 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 57.925×106 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 m18.54 kN/m
)
wLM
Mw
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 GPa9.3333
12 GPa
En
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) (mm2) (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 mm122.00 mm
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
(mm4) (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 mm4
= 134.261 ×106 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
PLM
(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 15.85 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)
134.261 10 m50.9 MPa (
mT)
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
611 max
(9 N/mm )(134.261 10 mm )9.224 10 N-mm
(253 mm 122.00 mm)
IMyM
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
622 max
(1,500 N/mm )(134.261 10 mm )176.867 10 N-mm
(9.3333)(122.00 mm)
IMyn 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 9.224×106 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
PLM
MP
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 ksi15
2,000 ksi
En
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.)
11.190 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 cn
I 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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8.51 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 ksi14
1,700 ksi
En
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 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)
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
11 max
(2.40 ksi)(970.747 in. )361.393 kip-in.
(12.125 in. 5.6783 in.)
IMyM
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
22 max
(175 ksi)(970.747 in. )2,137 kip-in.
(14)(5.6783 in.)
IMyn 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 ft5.02 kips
M P
MP 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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8.52 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
2 142 mm 2(6.5 mm) 129 mm
(142 mm) (129 mm) 2,766.958 mm4 4
(142 mm) (129 mm) 6,364,867 mm64 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 N7.951 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
34
(0.375 in.)(1.250 in.) 0.468750 in.
(0.375 in.)(1.250 in.)0.061035 in.
12z
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 lb746.667 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
39 4
(250 mm)(500 mm) 125,000 mm
(250 mm)(500 mm)2.60417 10 mm
12z
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 N5.20 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 lb134.36 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
34
(4 in.)(8 in.) 32 in.
(4 in.)(8 in.)170.6667 in.
12z
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 lb125 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 in88.3 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 PP
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/mm168,629 N
8.8953 10 mm168.6 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
34
(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 N2.00 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 Ax
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 lb104.545 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.
zH
z
zK
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 magnitude of 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 Ax
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.
zH
z
zK
z
F PP
A
M x PP
I
M x PP
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) (mm2) (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 mm101.0 mm (from left side to centroid)
5,000 mm
49.0 mm (from right side to centroid)
i i
i
x Ax
A
Moment of inertia about the z axis:
Shape IC d = xi – x d²A IC + d²A
(mm4) (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 (mm4) = 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 N5 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
zH
z
zK
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) (mm2) (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 mm101.0 mm (from left side to centroid)
5,000 mm
49.0 mm (from right side to centroid)
i i
i
x Ax
A
Moment of inertia about the z axis:
Shape IC d = xi – x d²A IC + d²A
(mm4) (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 (mm4) = 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
zH
z
zK
z
F PP
A
M x PP
I
M x PP
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) (mm2) (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 mm101.0 mm (from left side to centroid)
5,000 mm
49.0 mm (from right side to centroid)
i i
i
x Ax
A
Moment of inertia about the z axis:
Shape IC d = xi – x d²A IC + d²A
(mm4) (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 (mm4) = 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 N4.0958 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
zH
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
zH
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
zK
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
zK
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 34
3 34
(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 zx
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 zx
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 Ay
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 12yI
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 zx
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 zx
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 34
3 34
(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 zx
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 zx
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 zx
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 zx
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
(mm4) (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 (mm4) = 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 12yI
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 zx
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 zx
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
(mm4) (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 (mm4) = 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 12yI
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 zx
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 Ay
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 Az
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 zM
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 zM
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 zM
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 bending
moment 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 Ay
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 Az
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
53
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 zM 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
53
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 zM 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
(mm4) (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 (mm4) = 74,208,333.3
Moment of inertia about the y axis:
Shape IC d = zi – z Area Ai d²A IC + d²A
(mm4) (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 (mm4) = 13,054,933.3
Product of inertia about the centroidal axes:
Shape yc zc Area Ai yc zc Ai Iyz
(mm) (mm) (mm2) (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 (mm4) = −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 compute the 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) (mm2) (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 mm135.82 mm (from bottom of shape to centroid)
6,289 mm
64.18 mm (from top of shape to centroid)
i i
i
y Ay
A
Centroid location in z direction:
Shape Area Ai
zi
(from right edge) zi Ai
(mm2) (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 mm39.18 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 Az
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 25,059,086.23
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Moment of inertia about the y axis:
Shape IC d = zi – z d²A IC + d²A
(mm4) (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 (mm4) = 12,133,386.23
Product of inertia about the centroidal axes:
Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz
(mm4) (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 (mm4) = 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
y4 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 My 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 M
z
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 ksi69.86287 kip-in.
05.82 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 1.691.25 in. 1.25 in.
r DK
d d
Moment of inertia at minimum depth section:
3
4(0.75 in.)(1.25 in.)0.122070 in.
12I
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 1.5750 mm 50 mm
r DK
d d
Determine maximum nominal bending stress:
maxnom
120 MPa76.4331 MPa
1.57K
Moment of inertia at minimum depth section:
3
4(25 mm)(50 mm)260,416.67 mm
12I
Maximum bending moment:
2 4
nommax
(76.4331 N/mm )(260,416.67 mm )796,178.3 N-mm
50 mm/2796 N-m
IM
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 1.7650 mm 50 mm
r DK
d d
Determine maximum nominal bending stress:
maxnom
90 MPa51.1364 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/mm1
)(50 mm)4.08 mm
M y M d M
I bd bd
Mb
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 2.051.00 in. 1.00 in.
r DK
d d
Moment of inertia at minimum depth section:
3
4(0.25 in.)(1.00 in.)0.020833 in.
12I
Maximum allowable bending moment:
From the specified factor of safety and the yield stress of the material, the allowable bending stress is:
allow
165 ksi66 ksi
FS 2.5
Y
Thus, the maximum allowable bending moment can be determined from:
allow
4
allowmax
(66 ksi)(0.020833 in. )1.3415 kip-in.
(2.05)(1.00 in./11
2)1.8 lb-ft
MyK
I
IM
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 1.781.50 in. 1.50 in.
r DK
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
allowmax
(24,000 psi)(0.248505 in. )4,467.50 lb-in.
(1.78)(1.50 in./2)
MyK
I
IM
K y
Bending moment at x = 8 in.:
(8 in.) (4 in.)2 2
P PM 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.
PLM
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 1.4825 mm 25 mm
r DK
d d
Moment of inertia at minimum diameter section:
4 4(25 mm) 19,174.76 mm64
I
Maximum allowable bending moment:
yield
allow
331 MPa110.33 MPa
FS 3.0
allow
2 4
allowmax
(110.33 N/mm )(19,174.76 mm )114,357.58 N-mm
(1.48)(25 mm/2)
MyK
I
IM
K y
Bending moment at x = 150 mm:
(150 mm) (75 mm)2 2
P PM 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 mm64
(190,596.25 N-mm)(40 mm/2)30.33 MPa 110.33 MPa OK
125,663.71 mm
PLM
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 1.740.625 in. 0.625 in.
r DK
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.
MyK
I
Factor of safety:
max
62,000 psiFS
36,297.7 psi1.708Y 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 1.5820 mm 20 mm
r DK
d d
Moment of inertia at minimum diameter section:
4 4(20 mm) 7,853.98 mm64
I
Maximum allowable bending moment:
allow
427 MPa284.6667 MPa
FS 1.5
Y
allow
2 4
allowmax
(284.6667 N/mm )(7,853.98 mm )141,504.2261 N-mm
(1.58)(20 mm/2)
MyK
I
IM
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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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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8.84 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 1.9634 mm 34 mm
r DK
d d
Moment of inertia at minimum diameter section:
4 4(34 mm) 65,597.24 mm64
I
Maximum allowable bending moment:
allow
331 MPa220.6667 MPa
FS 1.5
Y
allow
2 4
allowmax
(220.6667 N/mm )(65,597.24 mm )
(1.96)(34 mm/2)
434,427. 435 N-m N mm 4 -
MyK
I
IM
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 864.000 0.000 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 lb2
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 lb2
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 Ay
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) (mm2) (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 mm181.154 mm
15,600 mm
i i
i
y Ay
A (measured upward from bottom edge of shape)
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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)N2
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 N2
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 Ay
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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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
(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 Ay
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 T2
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 T2
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 the beam 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 Ay
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 (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 C2
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 C2
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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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.
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) (mm2) (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 mm219.180 mm
24,400 mm
i i
i
y Ay
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
(mm4) (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 (mm4) = 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 (C2
)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 (C2
)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) (mm2) (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 mm219.180 mm
24,400 mm
i i
i
y Ay
A (measured upward from bottom edge of shape)
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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 N2
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 N2
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 N2
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 N2
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
12I
t = 120 mm
Distance below top
surface of beam y Q
35 mm 105 mm 514,500 mm3 140.6 kPa
70 mm 70 mm 882,000 mm3 241 kPa
105 mm 35 mm 1,102,500 mm3 301 kPa
140 mm 0 mm 1,176,000 mm3 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.
12I 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. )(632.0 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)
1,687.51,120 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
12I 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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(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
12I 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 m871 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.
12I 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 lb2
2(10,240 lb)
20 ft1,024 lb/ft
wLV
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
wLM
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.
12I 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 lb2
2(3,200 lb) 6,400 lb
PV
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 12I
(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
DQ
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
DQ
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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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
DQ
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
DQ
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
DQ
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
DQ
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
3.018,790 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 mm64 64
1 1[ ] [(170 mm) (150 mm) ] 128,166.667 mm
12128,170
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 mm64 64
1 1[ ] [(114 mm) 35,028 (102 m mm) ]
12 1m
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. standard
steel 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. 7.9811,427 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
72.4817,130 psi
924T
.(
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. )(0.2810.70 k
5 is
)i
n.
Q
Ans.
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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
maxmax
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
(mm4) (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 (mm4) = 52,235,072
(a) Q associated with point H:
3
250 mm 8 mm(150 mm)(8 mm)
2 2
250 mm8 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 mm8 mm
250 mm 22(8 mm) 8 mm 254,712 mm2 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
(mm4) (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 (mm4) = 52,235,072
(a) Shear stress at K:
3
250 mm 8 mm(150 mm)(8 mm)
2 2
250 mm8 mm 50 mm
250 mm 22(8 mm) 8 mm 50 mm 50 mm2 2
234,712 mm
KQ
3
4
(175,000 N)(234,712 mm )49.1463 MPa
(52,235,072 mm )(2 849.1
mmPa
)MK
Ans.
(b) Maximum horizontal shear stress:
max
3
250 mm 8 mm(150 mm)(8 mm)
2 2
250 mm8 mm
250 mm 22(8 mm) 8 mm 254,712 mm2 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
(mm2) (mm) (mm
3)
top flange 375.0 72.5 27,187.5
stem 350.0 35.0 12,250.0
725.0 mm2
39,437.5 mm
3
3
2
39,437.5 mm54.397 mm (from bottom of shape to centroid)
725.0 mm
20.603 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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 mm23.7 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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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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9.31 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
(mm4) (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 (mm4) = 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 mm4
)2.9 MPa
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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9.32 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 Ay
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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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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9.33 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.
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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
(mm2) (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 mm2
4,536,000 mm
3
3
2
4,536,000 mm140 mm (from bottom of shape to centroid)
32,400 mm
190 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 455,220,000
Shear force in beam:
80 kN-m 75 kN-m 5 kN-m
50 kN100 mm 0.1 m
MV
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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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
(mm4) (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 (mm4) = 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 mm65.6 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
Solution
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 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 Ay
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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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) (mm2) (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 mm75.5858 mm (from bottom of shape to centroid)
1,352 mm
24.4142 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (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 (mm4) = 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 mmx
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 mmx
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 mmx
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 mmx
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 Ans.
(b) Shear stress magnitude at H:
3
3
4
(4 in.)(0.375 in.)(3.8125 in.)
(0.375 in.)(1.625 in.)(2.8125 in.)
7.4326 in.
(3,664 lb)(7.4326 in. )
(55.5493 in. )(0.375 in.)
1,307 psi
H
H
Q
Ans.
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(c) Maximum horizontal shear stress magnitude:
At neutral axis:
3
max
3
max 4
(4 in.)(0.375 in.)(3.8125 in.) (0.375 in.)(3.625 in.)(1.8125 in.) 8.1826 in.
(3,664 lb)(8.1826 in. )
(55.5493 in. )(0.375 in1,439 psi
.)
Q
Ans.
(d) Maximum compression bending stress:
4
(7,719 lb-ft)(4 in.)(12 in./ft)6,669.965 psi
56,670 psi (
5.5493C)
in.H
M y
I Ans.
This stress occurs at 5.86 ft to the right of A.
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9.40 A wooden beam is fabricated from one 2 × 10 and two 2 × 4 pieces
of dimension lumber to form the I-beam cross section shown in Fig.
P9.40. The flanges of the beam are fastened to the web with nails that
can safely transmit a force of 120 lb in direct shear. If the beam is
simply supported and carries a 1,000-lb load at the center of a 12-ft
span, determine:
(a) the horizontal force transferred from each flange to the web in a 12-
in. long segment of the beam.
(b) the maximum spacing s (along the length of the beam) required for
the nails.
(c) the maximum horizontal shear stress in the I-beam.
Fig. P9.40
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)
top flange 4 2 2.667 6.000 288.000 290.667
web 2 10 166.667 0.000 0.000 166.667
bottom flange 4 2 2.667 –6.000 288.000 290.667
Moment of inertia about the z axis (in.4) = 748.000
Maximum shear force
For P = 1,000 lb, V = P/2 = 500 lb
(a) Horizontal force transferred from each flange
(in a 12-in. length):
3
3
4
(4 in.)(2 in.)(6 in.) 48 in.
(500 lb)(48 in. )32.086 lb/in.
748 in.
(32.086 lb/in.)(12 38 5 li bn.)H
Q
VQq
I
F Ans.
(b) Maximum nail spacing:
(1 nail)(120 lb/nail)
32.086 3.74 in
lb/ n.
i .
f f
f f
q s n V
n Vs
q Ans.
(c) Maximum horizontal shear stress:
3
max
3
max 4
(4 in.)(2 in.)(6 in.) (2 in.)(5 in.)(2.5 in.) 73 in.
(500 lb)(73 in. )
(748 in. )(2 in.)24.4 psi
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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9.41 A wooden beam is fabricated from one 2 × 10 and two 2 × 4 pieces
of dimension lumber to form the I-beam cross section shown in Fig.
P9.41. The I-beam will be used as a simply supported beam to carry a
concentrated load P at the center of a 20-ft span. The wood has an
allowable bending stress of 1,200 psi and an allowable shear stress of 90
psi. The flanges of the beam are fastened to the web with nails that can
safely transmit a force of 120 lb in direct shear.
(a) If the nails are uniformly spaced at an interval of s = 4.5 in. along the
span, what is the maximum concentrated load P that can be supported by
the beam? Demonstrate that the maximum bending and shear stresses
produced by P are acceptable.
(b) Determine the magnitude of load P that produces the allowable
bending stress in the span (i.e., b = 1,200 psi). What nail spacing s is
required to support this load magnitude? Demonstrate that the
maximum horizontal shear stresses produced by P are acceptable.
Fig. P9.41
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)
top flange 4 2 2.667 6.000 288.000 290.667
web 2 10 166.667 0.000 0.000 166.667
bottom flange 4 2 2.667 –6.000 288.000 290.667
Moment of inertia about the z axis (in.4) = 748.000
(a) Maximum concentrated load P:
3(4 in.)(2 in.)(6 in.) 48 in.
(1 nail)(120 lb/nail)26.667 lb/in.
4.5 in.
f f
f f
Q
q s n V
n Vq
s
4
3
max
(26.667 lb/in.)(748 in. )415.556 lb
831
48 in
lb
.
2
VQq
I
q IV
Q
PV P 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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Check maximum bending and shear stresses:
max 4
3
max
3
maxmax 4
(831.1 lb)(240 in.)/4 (7 in.)( / 4)466.7 psi 1,200 psi OK
748 in.
(2 in.)(5 in.)(2.5 in.) (4 in.)(2 in.)(6 in.) 73 in.
( / 2) (831.1 lb / 2)(73 in. )2
(748 in. )(2 in.)
z z
z z
Mc PL c
I I
Q
VQ P Q
I t I t0.3 psi 90 psi OK
(b) Magnitude of load P that produces the allowable bending stress in the span:
4
max
maxmax 2,1
1,200
40
psi
(1,200 psi)(748 in. )128,228.566 lb-in.
7 in.
4
4 4(128,228.566 lb-in.)2,137.143 lb
(20 ft)(12 in./ft)lb
x
M c
I
M
P LM
MP
L Ans.
Required nail spacing s:
maxmax
3
4
2,137.143 lb1,068.571 lb
2 2
(1,068.571 lb)(48 in. )68.571 lb/in.
748 in.
(1 nail)(1201.7
lb)
68.571 50 in.
lb/in.
f f
f f
PV
VQq
I
q s n V
n Vs
q Ans.
Check maximum shear stresses:
3
maxmax 4
( / 2) (2,137.143 lb / 2)(73 in. )52.143 psi 90 psi OK
(748 in. )(2 in.)z z
VQ P Q
I t I t
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9.42 A wooden box beam is fabricated from four boards, which are fastened together with nails, as
shown in Fig. P9.42b. The nails are installed at a spacing of s = 125 mm (Fig. P9.42a), and each nail can
provide a resistance of Vf = 500 N. In service, the box beam will be installed so that bending occurs
about the z axis. Determine the maximum shear force V that can be supported by the box beam based on
the shear capacity of the nailed connections.
Fig. P9.42a Fig. P9.42b
Solution
Moment of inertia Iz:
3 3
4(200 mm)(300 mm) (120 mm)(250 mm)293,750,000 mm
12 12zI
First moment of area Q:
3(200 mm)(25 mm)(137.5 mm) 687,500 mmQ
Shear flow q based on nail shear force:
(2 nails)(500 N/nail)
8 N/mm125 mm
f f
f f
q s n V
n Vq
s
Maximum shear force V:
4
3
(8 N/mm)(293,750,000 mm )3,418 N
687,500 mm3.42 kNz
z
VQ q Iq V
I 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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9.43 A wooden box beam is fabricated from four boards, which are fastened together with screws, as
shown in Fig. P9.43b. Each screw can provide a resistance of 800 N. In service, the box beam will be
installed so that bending occurs about the z axis, and the maximum shear force in the beam will be 9 kN.
Determine the maximum permissible spacing interval s for the screws (see Fig. P9.43a).
Fig. P9.43a Fig. P9.43b
Solution
Moment of inertia Iz:
3 3
4(190 mm)(250 mm) (140 mm)(150 mm)208,020,833 mm
12 12zI
First moment of area Q:
3(140 mm)(50 mm)(100 mm) 700,000 mmQ
Shear flow q based on beam shear force V:
3
4
(9,000 N)(750,000 mm )30.285 N/mm
208,020,833 mmz
VQq
I
Maximum spacing interval s:
(2 screws)(800 N/screw)
30.2852.8
5 N/m
m m
m
f f
f f
q s n V
n Vs
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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9.44 A wooden beam is fabricated by nailing together three pieces of dimension lumber, as shown in
Fig. P9.44a. The cross-sectional dimensions of the beam are shown in Fig. P9.44b. The beam must
support an internal shear force of V = 750 lb.
(a) Determine the maximum horizontal shear stress in the cross section for V = 750 lb.
(b) If each nail can provide 100 lb of horizontal resistance, determine the maximum spacing s for the
nails.
(c) If the three boards were connected by glue instead of nails, what minimum shear strength would be
necessary for the glued joints?
Fig. P9.44a Fig. P9.44b
Solution
Centroid location in y direction:
Shape Width b Height h Area Ai
yi
(from bottom) yi Ai
(in.) (in.) (in.2) (in.) (in.
3)
left board 2 8 16 4 64
flange board 4 2 8 7 56
right board 2 9 16 4 64
40 184
3
2
184 in.4.6 in. (from bottom of shape to centroid)
40 in.
3.4 in. (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(in.4) (in.) (in.
4) (in.
4)
left board 85.3333 –0.60 5.7600 91.0933
flange board 2.6667 2.40 46.0800 48.7467
right board 85.3333 –0.60 5.7600 91.0933
Moment of inertia about the z axis (in.4) = 230.9333
(a) Maximum horizontal shear stress:
At neutral axis:
3
max
3
max 4
2(2 in.)(4.6 in.)(4.6 in./2) 42.32 in.
(750 lb)(42.32 in. )
(230.9333 in. )(4 in.)34.4 psi
Q
Ans.
(b) Shear flow q based on beam shear force V:
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3
3
4
(4 in.)(2 in.)(3.4 in. 1 in.) 19.20 in.
(750 lb)(19.20 in. )62.356 lb/in.
230.9333 in.z
Q
VQq
I
Maximum spacing interval s:
(2 nails)(100 lb/nail)
62.356 3.21 in
lb/in..
f f
f f
q s n V
n Vs
q Ans.
(c) Glue joint shear stress:
62.356 lb/in.
2(15.59 psi
2 in.) Ans.
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9.45 A wooden beam is fabricated by gluing four dimension lumber
boards, each 40-mm wide and 90-mm deep, to a 32 × 400 plywood
web, as shown in Fig. P9.45. Determine the maximum allowable
shear force and the maximum allowable bending moment that this
section can carry if the allowable bending stress is 6 MPa, the
allowable shear stress in the plywood is 640 kPa, and the allowable
shear stress in the glued joints is 250 kPa.
Fig. P9.45
Solution
Moment of inertia Iz:
3 3
4(112 mm)(400 mm) (80 mm)(220 mm)526,346,667 mm
12 12zI
Maximum allowable bending moment:
2 4
max
(6 N/mm )(526,346,667 mm )15,790,4 15.79 kN-m00 N-mm
200 mm
M c
I
IM
c Ans.
Maximum allowable shear force:
Consider maximum shear stress, which occurs at the neutral axis:
3
4
3
(32 mm)(200 mm)(100 mm) 2(40 mm)(90 mm)(200 mm 90 mm/2) 1,756,000 mm
(0.640 N/mm)(526,346,667 mm )(32 mm)6,138.7 N 6.14 kN
1,756,000 mm
Q
VQ I tV
I t Q (a)
Consider shear stress in glue joints:
3(40 mm)(90 mm)(200 mm 90 mm/2) 558,000 mmQ
The shear stress in the glue joints can be found from the shear flow across the glue joint divided by the
width of the glue joint; thus,
glue
glue glue
4glue glue
3
/
(0.250 N/mm)(526,346,667 mm )(90 mm)21,224 N 21.2 kN
558,000 mm
q VQ I
t t
I tV
Q (b)
Compare results (a) and (b) to find that the maximum allowable shear force for the section is:
max 6.14 kNV Ans.
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9.46 A wooden beam is fabricated from one 2 × 12 and two 2 × 10
dimension lumber boards to form the double-tee cross section
shown in Fig. P9.46. The beam flange is fastened to the stem with
nails. Each nail can safely transmit a force of 175 lb in direct shear.
The allowable shear stress of the wood is 70 psi.
(a) If the nails are uniformly spaced at an interval of s = 4 in. along
the span, what is the maximum internal shear force V that can be
supported by the double-tee cross section?
(b) What nail spacing s would be necessary to develop the full
strength of the double-tee shape in shear? (Full strength means that
the maximum horizontal shear stress in the double-tee shape equals
the allowable shear stress of the wood.)
Fig. P9.46
Solution
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 12 2 24 11 264
left stem 2 10 20 5 100
right stem 2 10 20 5 100
64 464
3
2
464 in.7.25 in. (from bottom of shape to centroid)
64 in.
4.75 in. (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(in.4) (in.) (in.
4) (in.
4)
left board 8.000 3.750 337.500 345.500
flange board 166.667 –2.250 101.250 267.917
right board 166.667 –2.250 101.250 267.917
Moment of inertia about the z axis (in.4) = 881.333
(a) Maximum shear force based on capacity of nails at s = 4 in.:
3
4
max 3
(12 in.)(2 in.)(4.75 in. 2 in./2) 90.000 in.
(2 nails)(175 lb/nail)87.5 lb/in.
4 in.
(87.5 lb/in.)(881.333 in. )856.852 lb
90.000 in.
Q
q
V Q q Iq V
I Q
Maximum shear force based on full shear strength of double tee shape:
At neutral axis:
3
max
4
maxmax max 3
2(2 in.)(7.25 in.)(7.25 in./2) 105.125 in.
(70 psi)(881.333 in. )(2 2 in.)2,347.428 lb
90.000 in.
Q
V Q I tV
I t Q
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Comparison of two Vmax values:
The smaller of the two values computed above for Vmax gives the limiting shear force:
max 857 lbV Ans.
(b) Nail spacing necessary to develop the full shear strength of the section:
Consider the nailed portion of the beam (i.e., only the top flange) to establish the minimum required nail
spacing for Vmax = 2,347.428 lb, which is the shear force that produces a maximum horizontal shear
stress of 70 psi:
3
4
max
(2,347.428 lb)(90.000 in. )239.715 l
1.46
b/in.881.333 in.
(2 nails)(175 lb/nail)
239.715 lb/i0 n
ni .
.
f f
f f
V Qq
I
q s n V
n Vs
q Ans.
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9.47 A box beam is fabricated from two plywood webs that are
secured to dimension lumber boards at its top and bottom flanges
(Fig. P9.47b). The beam supports a concentrated load of P = 5,000 lb
at the center of a 15-ft span (Fig. P9.47a). Bolts (⅜-in. diameter)
connect the plywood webs and the lumber flanges at a spacing of s =
12 in. along the span. Supports A and C can be idealized as a pin and
a roller, respectively. Determine:
(a) the maximum horizontal shear stress in the plywood webs.
(b) the average shear stress in the bolts.
(c) the maximum bending stress in the lumber flanges.
Fig. P9.47a Fig. P9.47b
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 576 0 0 576
top flange 16 10 1,200 1,216
bottom flange 16 –10 1,200 1,216
right web 576 0 0 576
Moment of inertia about the z axis (in.4) = 3,584
Maximum shear force:
For P = 5,000 lb, V = P/2 = 2,500 lb
Maximum bending moment:
For P = 5,000 lb, M = PL/4 = 18,750 lb-ft = 225,000 lb-in.
(a) Maximum horizontal shear stress (in plywood webs):
max
3
3
max 4
(3 in.)(4 in.)(10 in.)
2(0.5 in.)(12 in.)(6 in.) 192 in.
(2,500 lb)(192 in. )
(3,584 in. )(2133.
0.5 in.)9 psi
Q
VQ
I t Ans.
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(b) Bolt shear stress:
Consider the dimension lumber boards that comprise the top flange.
3
flange
3flange
4
(3 in.)(4 in.)(10 in.) 120 in.
(2,500 lb)(120 in. )83.7054 lb/in.
3,584 in.
Q
VQq
I
Determine the force carried by one bolt:
(83.7054 lb/in.)(12 in.)
1,004.4643 lb/bolt1 bolt
f f
f
f
q s n V
q sV
n
The bolt cross-sectional area is:
2 2
bolt (0.375 in.) 0.110447 in.4
A
Each bolt acts in double shear; therefore, the shear stress in each bolt is:
bolt 2
1,004.4643 lb/bolt4,547.284 psi
2(0.110444,550 ps
)i
7 in. Ans.
(c) Maximum bending stress in lumber flanges:
4
(225,000 lb-in.)(12 in.)
3,584 in.753 psi
M c
I Ans.
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9.48 A box beam is fabricated from two plywood webs that are secured to
dimension lumber boards at its top and bottom flanges (Fig. P9.48b). The
lumber has an allowable bending stress of 1,500 psi. The plywood has an
allowable shear stress of 300 psi. The ⅜-in. diameter bolts have an
allowable shear stress of 6,000 psi, and they are spaced at intervals of s = 9
in. The beam span is L = 15 ft (Fig. P9.48a). Support A can be assumed to
be pinned and support C can be idealized as a roller.
(a) Determine the maximum load P that can be applied to the beam at
midspan.
(b) Report the bending stress in the lumber, the shear stress in the
plywood, and the average shear stress in the bolts at the load P determined
in part (a).
Fig. P9.48a Fig. P9.48b
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 576 0 0 576
top flange 16 10 1,200 1,216
bottom flange 16 –10 1,200 1,216
right web 576 0 0 576
Moment of inertia about the z axis (in.4) = 3,584
Maximum shear force:
V = P/2
Maximum bending moment:
M = PL/4
(a) Determine maximum load P:
Consider maximum bending stress:
4
1,500 psi
(1,500 psi)(3,584 in. )448,000 lb-in.
12 in.
M c
I
M
max
448,000 lb-in.4
4(448,000 lb-in.)9,956 lb
(15 ft)(12 in./ft)
PLM
P (a)
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Consider maximum horizontal shear stress (in plywood webs):
3
4
max 3
max
max max
(3 in.)(4 in.)(10 in.) 2(0.5 in.)(12 in.)(6 in.) 192 in.
300 psi
(300 psi)(3,584 in. )(2 0.5 in.)5,600 lb
192 in.
2
2 2(5,600 lb) 11,200 lb
Q
VQ
I t
V
PV
P V (b)
Consider bolt shear stress:
The bolt cross-sectional area is:
2 2
bolt (0.375 in.) 0.110447 in.4
A
Each bolt acts in double shear; therefore, the maximum shear force that can be carried by one bolt is:
2
bolt 2(0.110447 in. )(6,000 psi) 1,325.364 lbV
Determine the shear flow that can be allowed based on the bolt shear stress:
(1 bolt)(1,325.364 lb/bolt)
147.263 lb/in.9 in.
f f
f f
q s n V
n Vq
s
Consider the dimension lumber boards that comprise the top flange.
3
flange
flange
4
max 3
flange
max
max max
(3 in.)(4 in.)(10 in.) 120 in.
(147.263 lb/in.)(3,584 in. )4,398.245 lb
120 in.
2
2 2(4,398.245 lb) 8,796 lb
Q
VQq
I
q IV
Q
PV
P V (c)
Compare the three values obtained for Pmax in Eqs. (a), (b), and (c) to find
max 8,796 l 8.80 ipb k sP Ans.
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(b) Bending stress in lumber flanges for Pmax:
4
(8,796 lb)(15 ft)32,985 lb-ft
4 4
(32,985 lb-ft)(12 in.)(12 in./ft)
3,584 in.1,325 psi
PLM
M c
I Ans.
Maximum shear stress in plywood webs:
3
4
8,796 lb4,398 lb
2 2
(4,398 lb)(192 in. )
(3,584 in. )(2 0.5 23
in.)6 psi
PV
VQ
I t Ans.
Bolt shear stress:
bolt 6,000 psi Ans.
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9.49 A wooden beam is fabricated from three boards, which are fastened together with screws, as shown
in Fig. P9.49b. The screws are uniformly spaced along the span of the beam at intervals of 150 mm (see
Fig. P9.49a). In service, the beam will be positioned so that bending occurs about the z axis. The
maximum bending moment in the beam is Mz = −4.50 kN-m, and the maximum shear force in the beam
is Vy = −2.25 kN. Determine:
(a) the magnitude of the maximum horizontal shear stress in the beam.
(b) the shear force in each screw.
(c) the magnitude of the maximum bending stress in the beam.
Fig. P9.49a Fig. P9.49b
Solution
Centroid location in y direction:
Shape Width b Height h Area Ai
yi
(from bottom) yi Ai
(mm) (mm) (mm2) (mm) (mm
3)
left board 40 180 7,200 90 648,000
bottom board 140 40 5,600 20 112,000
right board 40 180 7,200 90 648,000
20,000 1,408,000
3
2
1,408,000 mm70.4 mm (from bottom of shape to centroid)
20,000 mm
109.6 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
left board 19,440,000.00 19.60 2,765,952.00 22,205,952.00
bottom board 746,666.67 –50.40 14,224,896.00 14,971,562.67
right board 19,440,000.00 19.60 2,765,952.00 22,205,952.00
Moment of inertia about the z axis (mm4) = 59,383,466.67
(a) Maximum horizontal shear stress:
At neutral axis:
3
max
3
max 4
2(40 mm)(109.6 mm)(109.6 mm/2) 480,486.4 mm
(2,250 N)(480,486.4 mm )0.2276 MPa
(59,383,466.228 kP
67 mm )(2 0 m )a
4 m
Q
V Q
I t Ans.
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(b) Shear force in each screw
Consider bottom board, which is held in place by two screws:
3
3
4
(140 mm)(40 mm)(70.4 mm 40 mm/2) 282,240 mm
(2,250 N)(282,240 mm )10.6939 N/mm
59,383,466.67 mm
(10.6939 N/mm)(150 mm)
2 screw802 N per e
s scr w
f f
f
f
Q
V Qq
I
q s n V
q sV
n Ans.
(c) Maximum bending stress:
6
4
( 4.50 10 N-mm)(109.6 mm)
59,383,466.67 8.31 MPa (T)
mm
zx
z
M y
I Ans.
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9.50 A wooden beam is fabricated by bolting together three
members, as shown in Fig. P9.50a. The cross-sectional
dimensions are shown in Fig. P9.50b. The 8-mm-diameter
bolts are spaced at intervals of s = 200 mm along the x axis
of the beam. If the internal shear force in the beam is V = 7
kN, determine the shear stress in each bolt.
Fig. P9.50a Fig. P9.50b
Solution
Centroid location in y direction:
Shape Width b Height h Area Ai
yi
(from bottom) yi Ai
(mm) (mm) (mm2) (mm) (mm
3)
left board 40 90 3,600 255 918,000
center board 40 300 12,000 150 1,800,000
right board 40 90 3,600 255 918,000
19,200 3,636,000
3
2
3,636,000 mm189.375 mm (from bottom of shape to centroid)
19,200 mm
110.625 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
left board 2,430,000 65.625 15,503,906.25 17,933,906.25
center board 90,000,000 -39.375 18,604,687.50 108,604,687.50
right board 2,430,000 65.625 15,503,906.25 17,933,906.25
Moment of inertia about the z axis (mm4) = 144,472,500.00
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Shear force in each bolt
Consider left board, which is held in place by the bolt:
3
3
4
(40 mm)(90 mm)(110.625 mm 90 mm/2) 236,250 mm
(7,000 N)(236,250 mm )11.4468 N/mm
144,472,500 mm
Q
V Qq
I
Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross
section. The cross-sectional area of the bolt is:
2 2
bolt (8 mm) 50.2655 mm4
A
Relate the shear flow and the bolt shear stress with Eq. (9.14):
2
(11.4468 N/mm)(200 mm)
(1 bolt surface)(50.45.5 M
2655 mmPa
)
f f f
f
f f
q s n A
q s
n A Ans.
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9.51 A wooden beam is fabricated by bolting together three
members, as shown in Fig. P9.51a. The cross-sectional
dimensions are shown in Fig. P9.51b. The allowable shear stress
of the wood is 850 kPa, and the allowable shear stress of the 10-
mm-diameter bolts is 40 MPa. Determine:
(a) the maximum internal shear force V that the cross section can
withstand based on the allowable shear stress in the wood.
(b) the maximum bolt spacing s required to develop the internal
shear force computed in part (a).
Fig. P9.51a Fig. P9.51b
Solution
Centroid location in y direction:
Shape Width b Height h Area Ai
yi
(from bottom) yi Ai
(mm) (mm) (mm2) (mm) (mm
3)
left board 40 90 3,600 255 918,000
center board 40 300 12,000 150 1,800,000
right board 40 90 3,600 255 918,000
19,200 3,636,000
3
2
3,636,000 mm189.375 mm (from bottom of shape to centroid)
19,200 mm
110.625 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
left board 2,430,000 65.625 15,503,906.25 17,933,906.25
center board 90,000,000 -39.375 18,604,687.50 108,604,687.50
right board 2,430,000 65.625 15,503,906.25 17,933,906.25
Moment of inertia about the z axis (mm4) = 144,472,500.00
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Consider maximum horizontal shear stress:
3
2 4
max 3
(40 mm)(189.375 mm)(189.375 mm/2) 717,257.813 mm
850 kPa 0.850 MPa
(0.850 N/mm )(144,472,500 mm )(40 mm)6,848.4 N
717,26.85 k
57.8 3 mmN
1
Q
VQ
I t
V Ans.
Maximum bolt spacing
Consider left board, which is held in place by the bolt:
3
3
4
(40 mm)(90 mm)(110.625 mm 90 mm/2) 236,250 mm
(6,848.4 N)(236,250 mm )11.1989 N/mm
144,472,500 mm
Q
V Qq
I
Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross
section. The cross-sectional area of the bolt is:
2 2
bolt (10 mm) 78.5398 mm4
A
Relate the shear flow and the bolt shear stress with Eq. (9.14):
2 2(1 bolt surface)(40 N/mm )(78.5398 mm )
11.1989 N/28
m1 mm
m
f f f
f f f
q s n A
n As
q Ans.
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9.52 A cantilever flexural member is fabricated by bolting two identical cold-
rolled steel channels back-to-back, as shown in Fig. P9.52a. The cantilever
beam has a span of L = 1,600 mm and supports a concentrated load of P = 600
N. The cross-sectional dimensions of the built-up shape are shown in Fig.
P9.52b. The effect of the rounded corners can be neglected in determining the
section properties for the built-up shape.
(a) If 4-mm-diameter bolts are installed at intervals of s = 75 mm, determine the
shear stress produced in the bolts.
(b) If the allowable average shear stress in the bolts is 96 MPa, determine the
minimum bolt diameter required if a spacing of s = 400 mm is used.
Fig. P9.52a
Fig. P9.52b
Solution
Centroid location in y direction for the upper channel shown in Figure 9.52b:
Shape Width b Height h Area Ai
yi
(from z axis) yi Ai
(mm) (mm) (mm2) (mm) (mm
3)
left element 3 40 120 20 2400
center element 59 3 177 1.5 265.5
right element 3 40 120 20 2400
417 5065.5
3
2
5,065.5 mm12.1475 mm
417 mm
i i
i
y Ay
A
Note: y is measured from the z axis to the centroid of the upper channel shown in Figure 9.52b.
Moment of inertia (both channels):
3 3
4(3 mm)(40 mm) (65 mm 2(3 mm))(3 mm)2 2 257,062 mm
3 3I
Shear flow:
2 3
3
4
(12.1475 mm)(417 mm ) 5,065.51 mm
(600 N)(5,065.51 mm )11.8232 N/mm
257,062 mm
Q
VQq
I
Bolt area:
2 2
bolt (4 mm) 12.5664 mm4
A
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(a) Bolt shear stress:
2
(11.8232 N/mm)(75 mm)
(1 bolt surface)(12.70.6 M
5664 mmPa
)
f f f
f
f f
q s n A
q s
n A Ans.
(b) Minimum bolt diameter for s = 400 mm:
2
2
2 2
bolt
2
bolt
(11.8232 N/mm)(400 mm)49.2633 mm
(1 bolt surface)(96 N/mm )
49.2633 mm4
4(49.2633 m7.92 m
mm
)
f f f
f
f f
q s n A
q sA
n
A D
D Ans.
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9.53 A W310 × 60 steel beam (see Appendix B) in an existing
structure is to be strengthened by adding a 250 mm wide by 16
mm thick cover plate to its lower flange, as shown in Fig.
P9.53. The cover plate is attached to the lower flange by pairs
of 24-mm-diameter bolts spaced at intervals of s along the
beam span. Bending occurs about the z centroidal axis.
(a) If the allowable bolt shear stress is 96 MPa, determine the
maximum bolt spacing interval s required to support an internal
shear force in the beam of V = 50 kN.
(b) If the allowable bending stress is 150 MPa, determine the
allowable bending moment for the existing W310 × 60 shape,
the allowable bending moment for the W310 × 60 with the
added cover plate, and the percentage increase in moment
capacity that is gained by adding the cover plate.
Fig. P9.53
Solution
Centroid location in y direction:
Shape Width b Height h Area Ai
yi
(from bottom) yi Ai
(mm) (mm) (mm2) (mm) (mm
3)
W310 × 60 - - 7,550 167 1,260,850
cover plate 250 16 4,000 8 32,000
11,550 1,292,850
3
2
1,292,850 mm111.935 mm (from bottom of shape to centroid)
11,550 mm
206.065 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
W310 × 60 128,000,000 55.065 22,892,764 150,892,764
cover plate 85,333.33 –103.935 43,209,937 43,295,270
Moment of inertia about the z axis (mm4) = 194,188,035
(a) Maximum bolt spacing
Consider the cover plate, which is connected to the W310 × 60 shape with two bolts:
3
3
4
(250 mm)(16 mm)(111.935 mm 16 mm/2) 415,740 mm
(50,000 N)(415,740 mm )107.0457 N/mm
194,188,035 mm
Q
V Qq
I
The cross-sectional area of a 24-mm-diameter bolt is:
2 2
bolt (24 mm) 452.389 mm4
A
Relate the shear flow and the bolt shear stress with Eq. (9.14):
2 2(2 bolts)(96 N/mm )(452.389 mm )
107.0457 N/mm811 mm
f f f
f f f
q s n A
n As
q Ans.
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(b) Allowable bending moment for W310 × 60 shape (without cover plate):
2 4
allow
150 MPa
(150 N/mm )(128,000,000 mm )127,152, 127318 N-mm
(3.2 kN-m
02 mm/2)
M c
I
M Ans.
Allowable bending moment for W310 × 60 shape (with cover plate):
2 4
allow
150 MPa
(150 N/mm )(194,188,035 mm )141,35 141.4 4,452 N-mm
(206.065 m-
m)kN m
M c
I
M Ans.
Percentage increase in moment capacity:
141,354,452 N-mm 127,152,318 N-mm
% increase (100%)127,152,318 N-mm
11.17% Ans.
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9.54 A W410 × 60 steel beam (see Appendix B) is simply supported at its ends and carries a
concentrated load P at the center of a 7-m span. The W410 × 60 shape will be strengthened by adding
two 250 mm wide by 16 mm thick cover plate to its flanges, as shown in Fig. P9.54. Each cover plate is
attached to its flange by pairs of 20-mm-diameter bolts spaced at intervals of s along the beam span. The
allowable bending stress is 150 MPa, the allowable average shear stress in the bolts is 96 MPa, and
bending occurs about the z centroidal axis.
(a) Based on the 150 MPa allowable bending stress, determine the
maximum concentrated load P that may be applied at the center of
a 7-m span for a W410 × 60 steel beam with two cover plates.
(b) For the internal shear force V associated with the concentrated
load P determined in part (a), compute the maximum spacing
interval s required for the bolts that attach the cover plates to the
flanges.
Fig. P9.54
Solution
Moment of inertia about the z axis (with cover plates):
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
top cover plate 85,333.33 211 178,084,000 178,169,333
W410 × 60 216,000,000 0 0 216,000,000
bottom cover plate 85,333.33 –211 178,084,000 178,169,333
Moment of inertia about the z axis (mm4) = 572,338,666
Maximum shear force:
V = P/2
Maximum bending moment:
M = PL/4
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Consider W410 × 60 with two cover plates:
150 MPaM c
I
2 3
allow
(150 N/mm )(572,338,666 mm )392,012,785 N-mm 392.013 kN-m
(438 mm/2)M
max
392.013 kN-m4
4(392.013 kN-m)224.007 kN
7 m224 kN
PLM
P Ans.
(b) Maximum bolt spacing
Consider top cover plate, which is held in place by two bolts:
3
3
4
(250 mm)(16 mm)(406 mm/2 16 mm/2) 844,000 mm
(223,418 N/2)(844,000 mm )165.166 N/mm
572,338,666 mm
Q
V Qq
I
Note that this value of q is the shear flow that must be transmitted across two bolt surfaces. The cross-
sectional area of the bolt is:
2 2
bolt (20 mm) 314.159 mm4
A
Relate the shear flow and the bolt shear stress with Eq. (9.14):
2 2(2 bolt surfaces)(96 N/mm )(314.159 mm )
165.166 N365 m
m m
/m
f f f
f f f
q s n A
n As
q Ans.
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9.55 A W410 × 60 steel beam (see Appendix B) is simply
supported at its ends and carries a concentrated load of P = 420 kN
at the center of a 7-m span. The W410 × 60 shape will be
strengthened by adding two 250 mm wide by 16 mm thick cover
plate to its flanges, as shown in Fig. P9.55. Each cover plate is
attached to its flange by pairs of bolts spaced at intervals of s = 250
mm along the beam span. The allowable average shear stress in the
bolts is 96 MPa, and bending occurs about the z centroidal axis.
Determine the minimum required diameter for the bolts.
Fig. P9.55
Solution
Moment of inertia about the z axis (with cover plates):
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
top cover plate 85,333.33 211 178,084,000 178,169,333
W410 × 60 216,000,000 0 0 216,000,000
bottom cover plate 85,333.33 –211 178,084,000 178,169,333
Moment of inertia about the z axis (mm4) = 565,618,666
Maximum shear force:
V = P/2 = 420 kN/2 = 210 kN
Minimum bolt diameter
Consider top cover plate, which is held in place by two bolts:
3
3
4
(250 mm)(16 mm)(406 mm/2 16 mm/2) 844,000 mm
(210,000 N)(844,000 mm )309.677 N/mm
572,338,666 mm
Q
V Qq
I
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Relate the shear flow and the required bolt area with Eq. (9.14). Note that the shear flow will be
transmitted by means of two fasteners.
2
2
(309.677 N/mm)(250 mm)403.225 mm
2(96 N/mm )
f f f
f
f f
q s n A
q sA
n
Use the minimum required cross-sectional area of the bolt to calculate the minimum bolt diameter:
2 2
bolt
min
403.225 m
22.7 mm
m4
22.658 mm
A D
D Ans.
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9.56 A W310 × 60 steel beam (see Appendix B) has a C250 × 45
channel bolted to the top flange, as shown in Fig. P9.56. The beam is
simply supported at its ends and carries a concentrated load of 100 kN at
the center of a 6-m span. Pairs of 24-mm-diameter bolts are spaced at
intervals of s along the beam. If the allowable average shear stress in the
bolts must be limited to 125 MPa, determine the maximum spacing
interval s for the bolts.
Fig. P9.56
Solution
Centroid location in y direction:
Shape Area Ai
yi
(from bottom) yi Ai
(mm2) (mm) (mm
3)
W310 × 60 7,550 151 1,140,050
C250 × 45 5,680 302 + 17.1 – 16.5 = 302.6 1,718,768
13,230 2,858,818
3
2
2,858,818 mm216.086 mm (from bottom of shape to centroid)
13,230 mm
103.014 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
W310 × 60 128,000,000 –65.086 31,983,215 159,983,315
C250 × 45 1,640,000 86.5140 42,512,938 44,152,938
Moment of inertia about the z axis (mm4) = 204,136,153
Maximum shear force:
V = P/2 = 100 kN/2 = 50 kN
Shear flow through the bolts
Consider the C250 × 45 shape, which is connected to the
W310 × 60 shape with two bolts:
2
3
3
4
(5,680 mm )(302.6 mm 216.086 mm)
491,399.5 mm
(50,000 N)(491,399.5 mm )
204,136,153 mm
120.361 N/mm
Q
V Qq
I
The cross-sectional area of a single 24-mm-diameter bolt is:
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2 2
bolt (24 mm) 452.389 mm4
A
Use Eq. (9.13) to determine the maximum spacing s:
2 2(2 bolts)(125 N/mm )(452.389 mm )
939.653 mm120.361 N/m
940 mmm
f f f
f f f
q s n A
n As
q Ans.
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9.57 A W310 × 60 steel beam (see Appendix B) has a C250 × 45
channel bolted to the top flange, as shown in Fig. P9.57. The beam is
simply supported at its ends and carries a concentrated load of 90 kN at
the center of an 8-m span. If pairs of bolts are spaced at 600-mm
intervals along the beam, determine:
(a) the shear force carried by each of the bolts.
(b) the bolt diameter required if the average shear stress in the bolts must
be limited to 75 MPa.
Fig. P9.57
Solution
Centroid location in y direction:
Shape Area Ai
yi
(from bottom) yi Ai
(mm2) (mm) (mm
3)
W310 × 60 7,550 151 1,140,050
C250 × 45 5,680 302 + 17.1 – 16.5 = 302.6 1,718,768
13,230 2,858,818
3
2
2,858,818 mm216.086 mm (from bottom of shape to centroid)
13,230 mm
103.014 mm (from top of shape to centroid)
i i
i
y Ay
A
Moment of inertia about the z axis:
Shape IC d = yi – y d²A IC + d²A
(mm4) (mm) (mm
4) (mm
4)
W310 × 60 128,000,000 –65.086 31,983,215 159,983,315
C250 × 45 1,640,000 86.5140 42,512,938 44,152,938
Moment of inertia about the z axis (mm4) = 204,136,153
Maximum shear force:
V = P/2 = 90 kN/2 = 45 kN
(a) Shear force in each bolt
Consider the C250 × 45 shape, which is connected to the
W310 × 60 shape with two bolts:
2
3
3
4
(5,680 mm )(302.6 mm 216.086 mm)
491,399.5 mm
(45,000 N)(491,399.5 mm )
204,136,153 mm
108.325 N/mm
Q
V Qq
I
Relate the shear flow and the bolt shear force with Eq. (9.13):
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(108.325 N/mm)(600 mm)
32,497 N2 bolts
32.5 kN per bolt
f f
f
f
q s n V
q sV
n Ans.
(b) Required bolt diameter
2bolt boltbolt 2
bolt allow
2 2
bolt bolt
2
bolt
32,497 N433.299 mm
75 N/mm
433.299
23.5
mm4
4(433.299m
m )m
m
V VA
A
A D
D Ans.
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10.1 For the loading shown, use the double-
integration method to determine (a) the
equation of the elastic curve for the
cantilever beam, (b) the deflection at the free
end, and (c) the slope at the free end.
Assume that EI is constant for each beam.
Fig. P10.1
Solution
Integration of moment equation:
2
02( )
d vEI M x M
dx
0 1
dvEI M x C
dx (a)
2
01 2
2
M xEI v C x C (b)
Boundary conditions:
0 at 0
0 at 0
dvx
dx
v x
Evaluate constants:
From Eq. (a), C1 = 0. From Eq. (b), C2 = 0
(a) Elastic curve equation:
0
2
0
2
2 2
M xEI v v
M x
EI Ans.
(b) Deflection at the free end:
2
0
2
0( )
22B
M Lv
E
M L
EI I Ans.
(c) Slope at the free end:
0 0( )B
B
dv M L
dx EI
M L
EI Ans.
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10.2 For the loading shown, use the double-
integration method to determine (a) the
equation of the elastic curve for the
cantilever beam, (b) the deflection at the free
end, and (c) the slope at the free end.
Assume that EI is constant for each beam.
Fig. P10.2
Solution
Integration of moment equation:
2 2
2( )
2
d v wxEI M x
dx
3
16
dv wxEI C
dx (a)
4
1 224
wxEI v C x C (b)
Boundary conditions:
0 at
0 at
dvx L
dx
v x L
Evaluate constants:
Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1:
3 3
1 1
( )(0)
6 6
w L wLEI C C
Substitute x = L and v = 0 into Eq. (b) to determine C2:
4 4 4 4
1 2 2 2
( )(0) ( )
24 24 6 8
w L wL wL wLEI C L C C C
(a) Elastic curve equation:
4 3
4 3 44
24 3
24 6 8 4
wx wL x wLEI v v
wx L x L
EI Ans.
(b) Deflection at the free end:
4
4 3 443
(0) 4 (0) 324 24 8
A
w wLv
wL L
EI E
L
EII Ans.
(c) Slope at the free end:
3 3 3(0
6 6
)
6A
A
dv w wL
dx EI EI
wL
EI Ans.
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10.3 For the loading shown, use the double-
integration method to determine (a) the
equation of the elastic curve for the cantilever
beam, (b) the deflection at the free end, and
(c) the slope at the free end. Assume that EI is
constant for each beam.
Fig. P10.3
Solution
Integration of moment equation:
2 3
0
2( )
6
d v w xEI M x
dx L
4
01
24
dv w xEI C
dx L (a)
5
01 2
120
w xEI v C x C
L (b)
Boundary conditions:
0 at
0 at
dvx L
dx
v x L
Evaluate constants:
Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1:
4 3
0 01 1
( )(0)
24 24
w L w LEI C C
L
Substitute x = L and v = 0 into Eq. (b) to determine C2:
5 5 3
0 0 01 2 2
4 4 4
0 0 02
( )(0) ( ) ( )
120 120 24
120 24 30
w L w L w LEI C L C L C
L L
w L w L w LC
(a) Elastic curve equation:
5 3
0 50
4
0 0 4 5
120 24 305 4
120
wx L x L
L E
w x w L w LEI v
L Ix v Ans.
(b) Deflection at the free end:
5 4 50
4
0(0) 5 (0) 412 30 0
A
w L
E
wv L L
L E II Ans.
(c) Slope at the free end:
34 3
0 0 0(0)
24 24 24A
A
dv w w L
dx L E
w L
EI II E Ans.
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10.4 For the beam and loading shown in
Fig. P10.4, use the double-integration
method to determine (a) the equation of
the elastic curve for segment AB of the
beam, (b) the deflection at B, and (c) the
slope at A. Assume that EI is constant for
the beam.
Fig. P10.4
Solution
Integration of moment equation:
2
2( )
2
d v PEI M x x
dx
2
14
dv PxEI C
dx (a)
3
1 212
PxEI v C x C (b)
Boundary conditions:
0 at 0
0 at2
v x
dv Lx
dx
Evaluate constants:
Substitute x = L/2 and dv/dx = 0 into Eq. (a) to determine C1:
2 2
1 1
( / 2)(0)
4 16
P L PLEI C C
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
3 2
2 2
(0) (0)(0) 0
12 16
P PLEI C C
(a) Elastic curve equation:
3
2
2
23 44
(0 )2812 16
P P xL x
x PL x L
EIEI v v x Ans.
(b) Deflection at B:
2
23( / 2)
3 448 2 48
B
P L L
I
PL
Lv
E EI Ans.
(c) Slope at A:
22 2(0)
4 16 61A
A
dv P PL
dx E
PL
EI II E Ans.
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10.5 For the beam and loading shown in
Fig. P10.5, use the double-integration
method to determine (a) the equation of
the elastic curve for the beam, (b) the
slope at A, (c) the slope at B, and (d) the
deflection at midspan. Assume that EI is
constant for the beam. Fig. P10.5
Solution
Beam FBD:
0
0 0
0
0
and
y y y
y y
A y
y y
F A B
A B
M B L M
M MB A
L L
Moment equation:
00 0
00
( ) ( ) 0
( )
a a y
MM M x A x M M x x M
L
M xM x M
L
Integration of moment equation:
2
002
( )d v M x
EI M x Mdx L
2
00 1
2
dv M xEI M x C
dx L (a)
2 3
0 01 2
2 6
M x M xEI v C x C
L (b)
Boundary conditions:
0 at 0
0 at
v x
v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
2 3
0 01 2 2
(0) (0)(0) (0) 0
2 6
M MEI C C C
L
Substitute x = L and v = 0 into Eq. (b) to determine C1:
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2 3
0 01
0 0 01
( ) ( )(0) ( )
2 6
6 2 3
M L M LEI C L
L
M L M L M LC
(a) Elastic curve equation:
0 2
3
20
2
0 0
23 2
66 3
M x M x M LxEI v v
L
M xx Lx L
L EI Ans.
(b) Slope at A:
2
0 0 00
(0)(0)
2 33A
A
dv M M LM
dx L EI E
M L
EI I Ans.
(c) Slope at B:
2
0 0 0 0 0( ) ( )6 3 2
2 3 66B
B
dv M L M L M L ML L L
dx EI L EI
M L
EIEI EI Ans.
(d) Deflection at midspan:
20 0
2
/ 2
2( / 2
1
)3 2
6 2 62x L
M L L Lv L
M L
EL
L I IE 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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10.6 For the beam and loading shown in Fig.
P10.6, use the double-integration method to
determine (a) the equation of the elastic curve for
the beam, (b) the maximum deflection, and (c) the
slope at A. Assume that EI is constant for the
beam. Fig. P10.6
Solution
Moment equation:
2
2
( ) 02 2
( )2 2
a a
wLx wxM M x
wx wLxM x
Integration of moment equation:
2 2
2( )
2 2
d v wx wLxEI M x
dx
3 2
16 4
dv wx wLxEI C
dx (a)
4 3
1 224 12
wx wLxEI v C x C (b)
Boundary conditions:
0 at 0
0 at
v x
v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
4 3
1 2 2
(0) (0)(0) (0) 0
24 12
w wLEI C C C
Substitute x = L and v = 0 into Eq. (b) to determine C1:
4 3 4 4 3
1 1
( ) ( ) ( ) ( )(0) ( )
24 12 24 12 24
w L wL L w L w L wLEI C L C
L L
(a) Elastic curve equation:
4 3
3 23
3224 2 2 42 1 4
wx wLx wL wxxEI v v x Lx L
EI Ans.
(b) Maximum deflection: At x = L/2:
3 2 3 32 2
4
max
( / 2)2
5
3824 2 2 48 2 48
w L L L wL L Lv L L L
EI EI
wL
EI Ans.
(c) Slope at A:
3 32 3(0) (0)
6 4 2 24 4A
A
dv w wL wL
dx EI EI
wL
EEI I Ans.
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10.7 For the beam and loading shown in
Fig. P10.7, use the double-integration
method to determine (a) the equation of
the elastic curve for segment AB of the
beam, (b) the deflection midway between
the two supports, (c) the slope at A, and
(d) the slope at B. Assume that EI is
constant for the beam. Fig. P10.7
Solution
Beam FBD:
0
30
2
3and
2 2
y y y
A y
y y
F A B P
LM B L P
P PB A
Moment equation:
( ) 0 ( )2 2
a a
P PxM M x x M x
Integration of moment equation:
2
2( )
2
d v PxEI M x
dx
2
14
dv PxEI C
dx (a)
3
1 212
PxEI v C x C (b)
Boundary conditions:
0 at 0
0 at
v x
v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
3
1 2 2
(0)(0) (0) 0
12
PEI C C C
Substitute x = L and v = 0 into Eq. (b) to determine C1:
3 2
1 1
( )(0) ( )
12 12
P L PLEI C L C
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(a) Elastic curve equation for segment AB of the beam:
3
22
2
12 1 22 1
Px PL x Px
II v L xE v
E Ans.
(b) Deflection at midspan:
2
2
2
3
/3
( / 2) 3
12 2 24 4 2x L
P PL
E
L L PL Lv L
I EI IE Ans.
(c) Slope at A:
22 2
12
(0)
4 12A
A
dv P PL
dx EI
PL
EI EI Ans.
(d) Slope at B:
22 2(
1 6
)
4 2B
B
dv P L PL
dx EI
P
EIEI
L Ans.
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10.8 For the beam and loading shown in Fig.
P10.8, use the double-integration method to
determine (a) the equation of the elastic curve
for segment BC of the beam, (b) the
deflection midway between B and C, and (c)
the slope at C. Assume that EI is constant for
the beam.
Fig. P10.8
Solution
Beam FBD:
(4 ) (5 ) 0
2 0
B y
y
y y y
y
M PL C L P L
C P
F B C P
B P
Moment equation:
( ) ( ) ( ) ( ) 0
( )
a a yM M x B x P L x M x Px P L x
M x PL
Integration of moment equation:
2
2( )
d vEI M x PL
dx
1
dvEI PLx C
dx (a)
2
1 22
PLxEI v C x C (b)
Boundary conditions:
0 at 0
0 at 4
v x
v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
2
1 2 2
(0)(0) (0) 0
2
PLEI C C C
Substitute x = 4L and v = 0 into Eq. (b) to determine C1:
2
2
1 1
(4 )(0) (4 ) 2
2
PL LEI C L C PL
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(a) Elastic curve equation for segment BC of the beam:
2
22 422
PLxEI v PL x
Pxv
LxL
EI Ans.
(b) Deflection at midspan:
3
2
(2 )4 ( )
22
2x L
PL Lv L L
EI
PL
EI Ans.
(c) Slope at C:
2 2(4 ) 2 2
C
C
dv PL L PL
dx E
P
I EIEI
L Ans.
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10.9 For the beam and loading shown in Fig.
P10.9, use the double-integration method to
determine (a) the equation of the elastic curve for
segment AB of the beam, (b) the deflection
midway between A and B, and (c) the slope at B.
Assume that EI is constant for the beam.
Fig. P10.9
Solution
Beam FBD:
2 50
2 4
5
2 4
0
2 4
A y
y
y y y
y
wL LM P B L
wL PB
F A B wL P
wL PA
Moment equation:
2 2
2
( ) ( ) 02 2 2 4
( )2 2 4
a a y
wx wx wLx PxM M x A x M x
wx wLx PxM x
Integration of moment equation:
2 2
2( )
2 2 4
d v wx wLx PxEI M x
dx
3 2 2
16 4 8
dv wx wLx PxEI C
dx (a)
4 3 3
1 224 12 24
wx wLx PxEI v C x C (b)
Boundary conditions:
0 at 0
0 at
v x
v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
4 3 3
1 2 2
(0) (0) (0)(0) (0) 0
24 12 24
w wL PEI C C C
Substitute x = L and v = 0 into Eq. (b) to determine C1:
4 3 3 3 2
1 1
( ) ( ) ( )(0) ( )
24 12 24 24 24
w L wL L P L wL PLEI C L C
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(a) Elastic curve equation for segment AB of the beam:
4 3 3
3 3 2
2
2
3
2
24 12 24 24
224 24
24
wx wLx Px wL x PL xEI v
vwx Px
x Lx L x LEI EI
Ans.
(b) Deflection at midspan:
4
3 2 2
3 2
/ 2
3
( / 2) ( / 2)2
24 2 2
5
38 6
24 2
4 4
x L
w L L L P L Lv L L L
EI E
wL PL
EI E
I
I Ans.
(c) Slope at B:
3 2 2 3 2 3 2( ) ( ) ( )
6 4 8 24 24 1224B
B
dv w L wL L P L wL PL
dx EI EI EI E
wL PL
EI EEI II Ans.
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10.10 For the beam and loading shown in Fig.
P10.10, use the double-integration method to
determine (a) the equation of the elastic curve for
segment AC of the beam, (b) the deflection at B,
and (c) the slope at A. Assume that EI is constant
for the beam. Fig. P10.10
Solution
Beam FBD:
3(3 ) (2 ) 0
2
9
4
(3 ) 0
9 33
4 4
A y
y
y y y
y
LM w L C L
wLC
F A C w L
wL wLA wL
Moment equation:
2
3( ) ( ) 0
2 4 2
3( )
2 4
a a y
x wL xM M x A x wx M x x wx
wx wLxM x
Integration of moment equation:
2 2
2
3( )
2 4
d v wx wLxEI M x
dx
3 2
1
3
6 8
dv wx wLxEI C
dx (a)
4 3
1 2
3
24 24
wx wLxEI v C x C (b)
Boundary conditions:
0 at 0
0 at 2
v x
v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2:
4 3
1 2 2
(0) 3 (0)(0) (0) 0
24 24
w wLEI C C C
Substitute x = 2L and v = 0 into Eq. (b) to determine C1:
4 3
1
3 3 3
1
(2 ) 3 (2 )(0) (2 )
24 24
8 12
24 24 6
w L wL LEI C L
wL wL wLC
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(a) Elastic curve equation for segment AC of the beam:
4 3 3
3 2 3
3 2 3
33 4
24 24 6 24
3 424
wx wLx wL x wxEI v x Lx
wxx Lx L
EI
L
v Ans.
(b) Deflection at B:
3 2 3
4( )( ) ) 4
4 123 (
2B
w Lv L L L
wL
EI
L
EI Ans.
(c) Slope at A:
3 32 3(0) 3 (0)
6 8 6 6A
A
dv w wL wL
dx EI EI
wL
EIEI 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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10.11 For the simply supported steel beam [E =
200 GPa; I = 129 × 106 mm
4] shown in Fig.
P10.11, use the double-integration method to
determine the deflection at B. Assume L = 4 m, P
= 60 kN, and w = 40 kN/m.
Fig. P10.11
Solution
Beam FBD:
( ) 02 2
2 2
( ) 0
2 2
A y
y
y y y
y
L LM wL P C L
wL PC
F A C w L P
wL PA
Moment equation: 2 2
2
( ) ( ) 02 2 2 2
( )2 2 2
a a y
wx wx wLx PxM M x A x M x
wx wLx PxM x
Integration of moment equation:
2 2
2( )
2 2 2
d v wx wLx PxEI M x
dx
3 2 2
16 4 4
dv wx wLx PxEI C
dx (a)
4 3 3
1 224 12 12
wx wLx PxEI v C x C (b)
Boundary conditions:
0 at 0
0 at2
v x
dv Lx
dx
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L/2 and dv/dx = 0 into
Eq. (b) to determine C1:
3 2 2
1
3 3 2 3 2
1
( / 2) ( / 2) ( / 2)(0)
6 4 4
48 16 16 24 16
w L wL L P LEI C
wL wL PL wL PLC
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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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Elastic curve equation:
4 3 3 3 2
3 2 3 2 2
24 12 12 24 16
2 3 424 48
wx wLx Px wL x PL xEI v
wx Pxv x Lx L L x
EI EI
Deflection at B: At x = L/2:
4 35
384 48B
wL PLv
EI EI
Let E = 200 GPa, I = 129 × 106 mm
4, w = 40 kN/m, P = 60 kN, and L = 4 m.
4 3
2 6 4 2 6 4
5(40 N/mm)(4,000 mm) (60,000 N)(4,000 mm)
384(200,000 N/mm )(129 10 mm ) 48(200,000 N/mm )(129 10 mm )
5.1680 mm 3.1008 mm
8.27 mm
Bv
Ans.
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10.12 For the cantilever steel beam [E = 200 GPa;
I = 129 × 106 mm
4] shown in Fig. P10.12, use the
double-integration method to determine the
deflection at A. Assume L = 2.5 m, P = 50 kN, and
w = 30 kN/m.
Fig. P10.12
Solution
Moment equation:
2 2
( ) 0 ( )2 2
a a
wx wxM M x Px M x Px
Integration of moment equation:
2 2
2( )
2
d v wxEI M x Px
dx
3 2
16 2
dv wx PxEI C
dx (a)
4 3
1 224 6
wx PxEI v C x C (b)
Boundary conditions:
0 at
0 at
v x L
dvx L
dx
Evaluate constants:
Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1:
3 2 3 2
1 1
( ) ( )(0)
6 2 6 2
w L P L wL PLEI C C
Substitute x = L and v = 0 into Eq. (b) to determine C2:
4 3 3 2 4 4 3 3
2 2
4 3
2
( ) ( )(0) ( ) ( )
24 6 6 2 24 6 6 2
8 3
w L P L wL PL wL wL PL PLEI L L C C
wL PLC
Elastic curve equation:
4 3 4 3 2 3
4 3 4 3 2 3
24 6 8 6 2 3
4 3 3 224 6
wx wL x wL Px PL x PLEI v
w Pv x L x L x L x L
EI EI
Deflection at A:
4 3
4 3 4 3 2 3 3(0) 4 (0) 3 (0) 3 (0) 2
24 6 24 3A
w P wL PLv L L L L
EI EI EI EI
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Let E = 200 GPa, I = 129 × 106 mm
4, w = 30 kN/m, P = 50 kN, and L = 2.5 m.
4 3
2 6 4 2 6 4
3(30 N/mm)(2,500 mm) (50,000 N)(2,500 mm)
24(200,000 N/mm )(129 10 mm ) 3(200,000 N/mm )(129 10 mm )
5.6777 mm 10.0937 mm
= 15.77 mm
Av
Ans.
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10.13 For the cantilever steel beam [E = 200 GPa;
I = 129 × 106 mm
4] shown in Fig. P10.13, use the
double-integration method to determine the
deflection at B. Assume L = 3 m, M0 = 70 kN-m,
and w = 15 kN/m.
Fig. P10.13
Solution
Moment equation:
2
0
2
0
( )( ) 0
2
( )( )
2
a a
w L xM M x M
w L xM x M
Integration of moment equation:
2 2 2 2
2 2
0 0 02
( )( ) 2
2 2 2 2
d v w L x w wL wxEI M x M L Lx x M wLx M
dx
2 2 3
0 12 2 6
dv wL x wLx wxEI M x C
dx (a)
2 2 3 4 2
01 2
4 6 24 2
wL x wLx wx M xEI v C x C (b)
Boundary conditions:
0 at 0
0 at 0
v x
dvx
dx
Evaluate constants:
Substitute x = 0 and dv/dx = 0 into Eq. (a) to determine C1 = 0. Next, substitute x = 0 and v = 0 into Eq.
(b) to determine C2 = 0.
Elastic curve equation:
2 2 3 4 2
0
24 3 2 2 0
4 6 24 2
4 624 2
wL x wLx wx M xEI v
w M xv x Lx L x
EI EI
Deflection at B:
2 4 2
4 3 2 2 0 0( )( ) 4 ( ) 6 ( )
24 2 8 2B
w M L wL M Lv L L L L L
EI EI EI EI
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Let E = 200 GPa, I = 129 × 106 mm
4, w = 15 kN/m, M0 = 70 kN-m, and L = 3 m.
4 2
2 6 4 2 6 4
(15 N/mm)(3,000 mm) (70 kN-m)(1,000 N/kN)(1,000 mm/m)(3,000 mm)
8(200,000 N/mm )(129 10 mm ) 2(200,000 N/mm )(129 10 mm )
5.8866 mm 12.2093 mm
= 18.10 mm
Bv
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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10.14 For the cantilever steel beam [E = 200 GPa;
I = 129 × 106 mm
4] shown in Fig. P10.14, use the
double-integration method to determine the
deflection at A. Assume L = 2.5 m, P = 50 kN-m,
and w0 = 90 kN/m.
Fig. P10.14
Solution
Moment equation:
0
3
0
( ) ( ) 02 3
( )6
a a
w x xM M x x Px
L
w xM x Px
L
Integration of moment equation:
2 3
0
2( )
6
d v w xEI M x Px
dx L
4 2
01
24 2
dv w x PxEI C
dx L (a)
5 3
01 2
120 6
w x PxEI v C x C
L (b)
Boundary conditions:
0 at
0 at
v x L
dvx L
dx
Evaluate constants:
Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1:
4 2 3 2
0 01 1
( ) ( )(0)
24 2 24 2
w L P L w L PLEI C C
L
Substitute x = L and v = 0 into Eq. (b) to determine C2:
5 3 3 2 4 4 3 3
0 0 0 02 2
4 3
02
( ) ( )(0) ( ) ( )
120 6 24 2 120 24 6 2
30 3
w L P L w L PL w L w L PL PLEI L L C C
L
w L PLC
Elastic curve equation:
5 3 3 2 4 3
0 0 0
5 4 5 3 2 30
120 6 24 2 30 3
5 4 3 2120 6
w x Px w L PL w L PLEI v x x
L
w Pv x L x L x L x L
L EI EI
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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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Deflection at A: Let E = 200 GPa, I = 129 × 10
6 mm
4, w0 = 90 kN/m, P = 50 kN, and L = 2.5 m.
5 4 5 3 2 30
4 3
0
4 3
2 6 4 2 6 4
(0) 5 (0) 4 (0) 3 (0) 2120 6
30 3
(90 N/mm)(2,500 mm) (50,000 N)(2,500 mm)
30(200,000 N/mm )(129 10 mm ) 3(200,000 N/mm )(129 10 mm )
4.5422 mm 10.0937
1
mm
=
A
w Pv L L L L
L EI EI
w L PL
EI EI
4.64 mm 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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10.15 For the beam and loading shown in Fig.
P10.15, use the double-integration method to
determine (a) the equation of the elastic curve for
the cantilever beam AB, (b) the deflection at the
free end, and (c) the slope at the free end. Assume
that EI is constant for each beam.
Fig. P10.15
Solution
Beam FBD:
0
2
0
0
0
20
2 3
3
02
2
A A
A
y y
y
w L LM M
w LM
w LF A
w LA
Moment equation:
0
2
0 0 0
3 2
0 0 0
( ) ( )2 3
( ) ( ) ( ) 03 2 3 2
( )6 2 3
a a A y
w x xM M x M x A x
L
w L w x x w LM x x x
L
w x w Lx w LM x
L
Integration of moment equation:
2 3 2
0 0 0
2( )
6 2 3
d v w x w Lx w LEI M x
dx L
4 2 2
0 0 01
24 4 3
dv w x w Lx w L xEI C
dx L (a)
5 3 2 2
0 0 01 2
120 12 6
w x w Lx w L xEI v C x C
L (b)
Boundary conditions:
0 at 0
0 at 0
v x
dvx
dx
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq.
(b) to determine C1 = 0.
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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) Elastic curve equation:
5 3 2 2
30
0
3
0
22
0
120 1
10 2012
2 6
0
w x w Lx w L xEI v
L
vw x
x L x LL EI
Ans.
(b) Deflection at the free end:
5 2 34
03 20 ( ) 10 ( ) 20 (11
12)
120 0B
wv L L L L L
L EI
w L
EI Ans.
(c) Slope at the free end:
4 2 2 3 3 3
0 0 0 0 0 0
3
0( ) ( ) ( ) 6 8
24 4 3 2 4 84 2 24B
B
dv w L w L w L
EI
L w L L w L w L w L
dx 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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10.16 For the beam and loading shown in Fig.
P10.16, use the double-integration method to
determine (a) the equation of the elastic curve for
the cantilever beam AB, (b) the deflection at the
free end, and (c) the slope at the free end. Assume
that EI is constant for each beam.
Fig. P10.16
Solution
Beam FBD:
0
2
0
0
0
02 3
6
02
2
A A
A
y y
y
w L LM M
w LM
w LF A
w LA
Moment equation:
3
0 ( )( ) 0
2 3a a
w L xM M x
L
30
3 2 2 30
2 2 3
0 0 0 0
( ) ( )6
( 3 3 )6
6 2 2 6
wM x L x
L
wL L x Lx x
L
w L w Lx w x w x
L
Integration of moment equation:
2 3 2 2
0 0 0 0
2( )
6 2 2 6
d v w x w x w Lx w LEI M x
dx L
4 3 2 2
0 0 0 01
24 6 4 6
dv w x w x w Lx w L xEI C
dx L (a)
5 4 3 2 2
0 0 0 01 2
120 24 12 12
w x w x w Lx w L xEI v C x C
L (b)
Boundary conditions:
0 at 0
0 at 0
v x
dvx
dx
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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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Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq.
(b) to determine C1 = 0.
(a) Elastic curve equation:
5 4 3 2 2
23 2 2 3
0 0 0
0
0
120 24 12
5 10 101
2
2
1
0
w
w xx Lx L x L
L
x w x w Lx
E
w LI v
I
xE
L
v Ans.
(b) Deflection at the free end:
5
5 4 2 3 3 20 0
4
04( ) 5 ( ) 10 ( ) 10 ( )
120 3120 0B
w w Lv L L L L L L L
L E
w
IEI EI
L
L Ans.
(c) Slope at the free end:
4 3 2 2
0 0 0 0
3
0( ) ( ) ( ) ( )
24 6 244 6B
B
dv w L w L w L L w L L
dx L EI EI EI EI
w L
EI Ans.
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10.17 For the beam and loading shown in Fig.
P10.17, use the double-integration method to
determine (a) the equation of the elastic curve for
the cantilever beam, (b) the deflection at B, (c) the
deflection at the free end, and (d) the slope at the
free end. Assume that EI is constant for the beam.
Fig. P10.17
Solution
Beam FBD:
2
02 2 4
3
8
02
2
A A
A
y y
y
wL L LM M
wLM
wLF A
wLA
Consider beam segment AB (0 ≤ x ≤ L/2)
Moment equation:
2
2
3( ) ( ) 0
8 2
3( )
8 2
a a A y
wL wLM M x M A x M x x
wL wLxM x
Integration of moment equation:
2 2
2
3( )
8 2
d v wL wLxEI M x
dx
2 2
1
3
8 4
dv wL x wLxEI C
dx (a)
2 2 3
1 2
3
16 12
wL x wLxEI v C x C (b)
Boundary conditions:
0 at 0
0 at 0
v x
dvx
dx
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq.
(b) to determine C1 = 0.
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Elastic curve equation for beam segment AB:
2 2 3
2
3
16 12
9 4 (0 / 2)48
wL x wLxEI v
wLxv L x x L
EI
Slope at B: Let x = L/2
2 2 33 ( / 2) ( / 2)
8 4 8B
B
dv wL L wL L wL
dx EI EI EI
Deflection at B: Let x = L/2
2 4( / 2) 7
9 448 2 192
B
wL L L wLv L
EI EI
Consider beam segment BC (L/2 ≤ x ≤ L)
Moment equation:
2
22
( )2 2
3( ) 0
8 2 2 2
b b A y
w LM M x M A x x
wL wL w LM x x x
2 2
2 2
3( )
2 2 2 8
2 2
w L wL wLM x x x
wx wLwLx
Integration of moment equation:
2 2 2
2( )
2 2
d v wx wLEI M x wLx
dx
3 2 2
36 2 2
dv wx wLx wL xEI C
dx (c)
4 3 2 2
3 424 6 4
wx wLx wL xEI v C x C (d)
Continuity conditions:
4
3
7at
192 2
at8 2
wL Lv x
EI
dv wL Lx
dx EI
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Evaluate constants:
Substitute the slope continuity condition into Eq. (c) for x = L/2 and solve for C3:
3 2 2 3
3
3
3
( / 2) ( / 2) ( / 2)
6 2 2 8
48
dv w L wL L wL L wLEI C
dx
wLC
Next, substitute the deflection continuity condition into Eq. (d) for x = L/2 and solve for C4
4 3 2 2 3 4
4
4
4
( / 2) ( / 2) ( / 2) 7( / 2)
24 6 4 48 192
384
w L wL L wL L wL wLEI v L C
wLC
Elastic curve equation for beam segment BC:
4 3 2 2 3 4
4 3 2 2 3 4
24 6 4 48 384
16 64 96 8 ( / 2 )384
wx wLx wL x wL x wLEI v
wv x Lx L x L x L L x L
EI
(a) Elastic curve equations for entire beam:
2
9 4 (0 / 2)48
wLxL x xv L
EI Ans.
4 3 2 2 3 416 64 96 8 ( / 2 )384
wx Lx L x L x L L x L
EIv Ans.
(b) Deflection at B:
47
192B
wLv
EI Ans.
(c) Deflection at free end of cantilever:
4 3 2 2 3 4
4
16( ) 64 ( ) 96 ( ) 8 ( )3
4
384
1
84C
wv L L L L L L L L
E
L
I
w
EI Ans.
(d) Slope at free end of cantilever:
2
3
3 2 3 38 ( ) 24 ( ) 24 ( ) 7
48 48 48 48 48
7
48C
C
dv w L wL L wL L wL wLEI
dx
dv
EIdx
wL Ans.
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10.18 For the beam and loading shown in Fig.
P10.18, use the double-integration method to
determine (a) the equation of the elastic curve for
the beam, and (b) the deflection at B. Assume that
EI is constant for the beam.
Fig. P10.18
Solution
Beam FBD:
( ) 02 4
8
02
3
8
A y
y
y y y
y
wL LM C L
wLC
wLF A C
wLA
Consider beam segment AB (0 ≤ x ≤ L/2)
Moment equation:
2 2
2
3( ) ( ) 0
2 2 8
3( )
2 8
a a y
wx wx wLM M x A x M x x
wx wLxM x
Integration of moment equation:
2 2
2
3( )
2 8
d v wx wLxEI M x
dx
3 2
1
3
6 16
dv wx wLxEI C
dx (a)
4 3
1 224 16
wx wLxEI v C x C (b)
Boundary conditions:
0 at 0v x
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0.
Slope at B: Let x = L/2 in Eq. (a).
3 2 3 3 3
1 1 1
( / 2) 3 ( / 2) 3 5
6 16 48 64 192B
B
dv w L wL L wL wL wLEI EI C C C
dx (c)
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Deflection at B: Let x = L/2 in Eq. (b).
4 3 4 4 4
1 11
( / 2) ( / 2)( / 2)
24 16 384 128 2 192 2B
w L wL L wL wL C L wL C LEI v C L (d)
Consider beam segment BC (L/2 ≤ x ≤ L)
Moment equation:
2
( ) ( ) ( ) ( ) 08
( ) ( )8 8 8
b b y
wLM M x C L x M x L x
wL wL wLxM x L x
Integration of moment equation:
2 2
2( )
8 8
d v wLx wLEI M x
dx
2 2
316 8
dv wLx wL xEI C
dx (e)
3 2 2
3 448 16
wLx wL xEI v C x C (f)
Boundary conditions:
0 atv x L
Evaluate constants:
Substitute x = L and v = 0 into Eq. (f) to find
3 2 2
3 4
4
3 4
( ) ( )(0) ( )
48 16
24
wL L wL LEI C L C
wLC L C (g)
Slope at B: Let x = L/2 in Eq. (e).
2 2 3 3 3
3 3 3
( / 2) ( / 2) 3
16 8 64 16 64B
B
dv wL L wL L wL wL wLEI EI C C C
dx (h)
Deflection at B: Let x = L/2 in Eq. (f).
3 2 2 4
33 4 4
( / 2) ( / 2) 5( / 2)
48 16 384 2B
wL L wL L wL C LEI v C L C C (i)
Continuity conditions:
Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h):
3 3
1 3
5 3
192 64
wL wLC C (j)
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Further, the deflection at B must be the same for both segments; therefore, equate Eqs. (d) and (i):
4 4
1 34
5
192 2 384 2
wL C L wL C LC (k)
Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C1,
C3, and C4:
3 3 4
1 3 4
9 17
384 384 384
wL wL wLC C C
(a) Elastic curve equation for beam segment AB:
3
3
4 3
3 216 24 9 (0 / 2)3
9
24 16 3
8
84
4
wx wLx wL xEI v
vwx
x Lx L x LEI
Ans.
(a) Elastic curve equation for beam segment BC:
3 2 2 3
3 2 2 3 417
48 16 38
8 24 17 ( /
4 384
2 )384
wLx wL x wL x wL
wLx Lx L x L L x L
I
E
E
I v
v Ans.
(b) Deflection at B:
4 44 49 5
192 76
5
7688 768B B
wL wL wLEI v
L
EIv
w Ans.
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10.19 For the beam and loading shown in Fig.
P10.19, use the double-integration method to
determine (a) the equation of the elastic curve for
the entire beam, (b) the deflection at C, and (c) the
slope at B. Assume that EI is constant for the
beam. Fig. P10.19
Solution
Beam FBD:
(3 ) 3 02
7
6
0
6
A y
y
y y y
y
LM B L wL L
wLB
F A B wL
wLA
Consider beam segment AB (0 ≤ x ≤ 3L)
Moment equation:
( ) ( ) 06
( )6
a a y
wLM M x A x M x x
wLxM x
Integration of moment equation:
2
2( )
6
d v wLxEI M x
dx
2
112
dv wLxEI C
dx (a)
3
1 236
wLxEI v C x C (b)
Boundary conditions:
0 at 0 and 0 at 3v x v x L
Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 3L and v = 0 into Eq. (b)
and solve for C1:
3 3 3
1 1
(3 ) 9(0) (3 )
36 36 4
wL L wL wLEI C L C
Slope at B: Let x = 3L in Eq. (a).
2 3 3(3 )
12 4 2B
B
dv wL L wL wLEI EI
dx (c)
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Consider beam segment BC (3L ≤ x ≤ 4L)
Moment equation:
2( ) (4 ) 02
b b
wM M x L x
2(4 )
( )2
w L xM x
Integration of moment equation:
2 2
2
(4 )( )
2
d v w L xEI M x
dx
3
3
(4 )
6
dv w L xEI C
dx (e)
4
3 4
(4 )
24
w L xEI v C x C (f)
Boundary conditions:
0 at 3v x L
Substitute x = 3L and v = 0 into Eq. (f) to find
4 4
3 4 3 4
4
4 3
(4 3 )(0) (3 ) (3 )
24 24
(3 )24
w L L wLEI C L C C L C
wLC L C (g)
Slope at B: Let x = 3L in Eq. (e).
3 3
3 3
(4 3 )
6 6B
B
dv w L L wLEI EI C C
dx (h)
Continuity conditions:
Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h):
3 3 3
3 3
2
2 6 3
wL wL wLC C (i)
Backsubstitute this result into Eq. (g) to determine C4:
4 4 3 4
4 3
2 49(3 ) (3 )
24 24 3 24
wL wL wL wLC L C L
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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) Elastic curve equation for beam segment AB:
3
2
3
2
9
3
9 (0 3 )36
6 36
wL
wL
xx L x L
x wL xEI v
EIv Ans.
Elastic curve equation for beam segment BC:
4 3
4 3 4
4(4 ) 16 49 (3 4 )2
(4 ) 2 49
2
4
4 3 24
w L x
wL x L x
w
L L x LEI
L x wLEI v
v Ans.
(b) Deflection at C:
44 3 4 4 4
4
15(4 4 ) 16 (4 ) 49 64 49
24 24 2
5
8
4C
C
w w wLv L L L L L L L
EI EI EI
wv
L
EI Ans.
(c) Slope at B: Let x = 3L in Eq. (a).
3 3
2 2B B
B B
dv wL dvEI EI
dx dx
wL
EI Ans.
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10.20 For the beam and loading shown in Fig.
P10.20, use the double-integration method to
determine (a) the equation of the elastic curve for
the beam, (b) the location of the maximum
deflection, and (c) the maximum beam deflection.
Assume that EI is constant for the beam. Fig. P10.20
Solution
Beam FBD:
0
0
0
0
20
2 3
3
02
6
A y
y
y y y
y
w L LM B L
w LB
w LF A B
w LA
Moment equation:
2
0
2
0 0
( )2 3
( ) 02 3 6
a a y
w x xM M x A x
L
w x x w LxM x
L
3
0 0( )6 6
w x w LxM x
L
Integration of moment equation:
2 3
0 0
2( )
6 6
d v w x w LxEI M x
dx L
4 2
0 01
24 12
dv w x w LxEI C
dx L (a)
5 3
0 01 2
120 36
w x w LxEI v C x C
L (b)
Boundary conditions:
0 at 0
0 at
v x
v x L
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Evaluate constants:
Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L and v = 0 into Eq. (b)
and solve for C1:
5 3 3
0 0 01 1
( ) ( ) 7(0) ( )
120 36 360
w L w L L w LEI C L C
L
(a) Elastic curve equation:
5 3 3
4 2 2 400 0 07
120 36 33 10 7
6 3600
ww x w Lx w xx L x L
L EI
L xEI v v
L Ans.
(b) Location of maximum deflection: The maximum deflection occurs where the beam slope is zero.
Therefore, set the beam slope equation [Eq. (a)] equal to zero:
4 2 3
0 0 070
24 12 360
dv w x w Lx w LEI
dx L
Multiply by −360L/w0 to obtain:
4 2 2 415 30 7 0x L x L
Solve this equation numerically to obtain:
= 0.51932962 0.523 1936 3x L L Ans.
(c) Maximum beam deflection:
4 2 2 4
4
0max
440 0 0
(0.51933 )3(0.51933 ) 10 (0.51933 ) 7
360
(0.51933) (0.0065222)4.52118
360
0.00652
w Lv L L L L
L EI
w w LL
EI EI
w L
EI 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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10.21 For the beam and loading shown in
Fig. P10.21, integrate the load distribution to
determine (a) the equation of the elastic
curve for the beam, and (b) the maximum
deflection for the beam. Assume that EI is
constant for the beam.
Fig. P10.21
Solution
Integrate the load distribution:
4
0
4
d v w xEI
dx L
3 2
013 2
d v w xEI C
dx L
2 3
01 22 6
d v w xEI C x C
dx L
4 2
0 12 3
24 2
dv w x C xEI C x C
dx L
5 3 2
0 1 23 4
120 6 2
w x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
3
13at 0, 0 0
d vx V EI C
dx
2
22at 0, 0 0
d vx M EI C
dx
4 3
0 03 3
( )at , 0 0
24 24
dv w L w Lx L C C
dx L
5 3 4
0 0 04 4
( ) ( )at , 0 0
120 24 30
w L w L L w Lx L v C C
L
(a) Elastic curve equation:
5 3
0 50
4
0 0 4 5
120 24 305 4
120
wx L x L
LE
w x w L x w LE v
L II v Ans.
(b) Maximum deflection:
5 4 50max
4
0(0) 5 (1 30
0) 420
wv L L
LEI
w L
EI 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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10.22 For the beam and loading shown in
Fig. P10.22, integrate the load distribution
to determine (a) the equation of the elastic
curve for the beam, and (b) the deflection
midway between the supports. Assume
that EI is constant for the beam. Fig. P10.22
Solution
Integrate the load distribution:
4
0
4
d v w xEI
dx L
3 2
013 2
d v w xEI C
dx L
2 3
01 22 6
d v w xEI C x C
dx L
4 2
0 12 3
24 2
dv w x C xEI C x C
dx L
5 3 2
0 1 23 4
120 6 2
w x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
2
22at 0, 0 0
d vx M EI C
dx
2 3
0 01 12
( )at , 0 ( ) 0
6 6
d v w L w Lx L M EI C L C
dx L
4at 0, 0 0x v C
5 3 3
0 0 03 3
( ) 7at , 0 0
120 36 360
w L w Lx w Lx L v C x C
L
(a) Elastic curve equation:
5 3 3
5 2 3 400 0 07
120 36 33 10 7
36060
wx L x L x
L
w x w Lx w L xEI v v
L EI Ans.
(b) Deflection midway between the supports:
5 2 3 40/2
4
03( / 2) 10 ( / 2) 7 ( / 2)360
5
768x L
wv L L L L L
LEI
w L
EI 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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10.23 For the beam and loading shown in
Fig. P10.23, integrate the load distribution to
determine (a) the equation of the elastic
curve, (b) the deflection at the left end of the
beam, and (c) the support reactions By and
MB. Assume that EI is constant for the beam.
Fig. P10.23
Solution
Integrate the load distribution:
4 3
0
4 3
d v w xEI
dx L
3 4
013 34
d v w xEI C
dx L
2 5
01 22 320
d v w xEI C x C
dx L
6 2
0 12 33120 2
dv w x C xEI C x C
dx L
7 3 2
0 1 23 43840 6 2
w x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
3
13at 0, 0 0
d vx V EI C
dx
2
22at 0, 0 0
d vx M EI C
dx
6 3
0 03 33
( )at , 0 0
120 120
dv w L w Lx L C C
dx L
7 3 4
0 0 04 43
( ) ( ) 6at , 0 0
840 120 840
w L w L L w Lx L v C C
L
(a) Elastic curve equation:
7 3 4
7 6 70
3
0 0 0
3
7 6
840 840 87 6
84040
wx L x
w x w L x w LEI v
LLL
EIv Ans.
(b) Deflection at left end of beam:
7
7 6 70 0max 3
4
0
3
6(0) 7 (0) 6
840 840 140
ww w Lv L L
L
L
EIEI L EI Ans.
(c) Support reactions By and MB:
3
0
3 4
0 0
3
( )
4 44B y
x L
d v w L w LV EI B
w
dx L
L Ans.
2 5 2
0 0
2 3
2
0( )
20 2
0(cw)
20B B
x L
d v w L w LM EI M
dx 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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10.24 For the beam and loading shown in
Fig. P10.24, integrate the load distribution
to determine (a) the equation of the elastic
curve, (b) the deflection midway between
the supports, and (c) the support reactions
Ay and By. Assume that EI is constant for
the beam. Fig. P10.24
Solution
Integrate the load distribution:
4 3
0
4 3
d v w xEI
dx L
3 4
013 34
d v w xEI C
dx L
2 5
01 22 320
d v w xEI C x C
dx L
6 2
0 12 33120 2
dv w x C xEI C x C
dx L
7 3 2
0 1 23 43840 6 2
w x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
2
22at 0, 0 0
d vx M EI C
dx
2 5
0 01 12 3
( )at , 0 ( ) 0
20 20
d v w L w Lx L M EI C L C
dx L
4at 0, 0 0x v C
7 3 3
0 0 03 33
( ) ( ) 6at , 0 ( ) 0
840 120 840
w L w L L w Lx L v C L C
L
(a) Elastic curve equation:
7 4 3 60
7 3 3
0 0
3
0
3
6
840 120 847 6
8 00 4
wx L x L x
L
w
E
x w Lx w L xEI
Iv v
L Ans.
(b) Deflection midway between the supports:
7 3
4 60/2 3
4
07 6840 2 2 2
13
5120x L
w L
EI
w L L Lv L L
L EI Ans.
(c) Support reactions Ay and By:
3 4
0 0 0
3 3
0
0(0)
4 20 0 02 2A y
x
d v w w L w LV EI A
dx L
w L Ans.
3 4
0 0 0
3 3
0( ) 4
4 20 2 50B y
x L
d v w L w L w LV EI B
dx 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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10.25 For the beam and loading shown in
Fig. P10.25, integrate the load distribution
to determine (a) the equation of the elastic
curve, (b) the deflection at the left end of the
beam, and (c) the support reactions By and
MB. Assume that EI is constant for the beam.
Fig. P10.25
Solution
Integrate the load distribution:
4
04cos
2
d v xEI w
dx L
3
013
2sin
2
d v w L xEI C
dx L
2 2
01 22 2
4cos
2
d v w L xEI C x C
dx L
3 2
0 12 33
8sin
2 2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
16cos
2 6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
3
13at 0, 0 0
d vx V EI C
dx
2 2 2
0 02 22 2 2
4 (0) 4at 0, 0 cos 0
2
d v w L w Lx M EI C C
dx L
3 2 3
0 0 03 33 2 3
8 ( ) 4 ( ) 4at , 0 sin 0 (2 )
2
dv w L L w L L w Lx L C C
dx L
4 2 2 3
0 0 044 2 3
4
04 3
16 ( ) 4 ( ) 4 ( )at , 0 cos (2 ) 0
2 2
2(4 )
w L L w L L w L Lx L v C
L
w LC
(a) Elastic curve equation:
4 2 2 3 4
0 0 0
4 2 2 2 3 40
0
4 2
4
3 3
16 4 4 2cos (2 ) (4 )
2
32 cos 4 8 (2 ) 4 (4 )2 2
2
w xL L x L x L
w L x w L x w L w LEI v
L
vEI L
Ans.
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(b) Deflection at left end of beam:
4 2 2 2 3 40
4
4
4 4
0
0
4
0
4 4
(0)32 cos 4 (0) 8 (0)(2 ) 4 (4 )
2 2
32 4 (4 ) 32 4 (4
0.1089
)2 2
A
wv L L L L
EI L
w w LL
w L
EI
LEI EI
Ans.
(c) Support reactions By and MB:
3
00 0
3
2 ( ) 2sin
2
2B y
x L
d v w L L w LV EI B
dx L
w L Ans.
2
0
2
2 2 2 2
0 0 0
2 2 2 2
4 ( ) 4 (cw)
4 4cos
2B B
x L
d v w L L w wL w LM EI M
dx L
L Ans.
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10.26 For the beam and loading shown
in Fig. P10.26, integrate the load
distribution to determine (a) the equation
of the elastic curve, (b) the deflection
midway between the supports, (c) the
slope at the left end of the beam, and (d)
the support reactions Ay and By. Assume
that EI is constant for the beam.
Fig. P10.26
Solution
Integrate the load distribution:
4
04sin
d v xEI w
dx L
3
013
cosd v w L x
EI Cdx L
2 2
01 22 2
sind v w L x
EI C x Cdx L
3 2
0 12 33
cos2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
sin6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
2
22at 0, 0 0
d vx M EI C
dx
2 2
01 12 2
( )at , 0 sin ( ) 0 0
d v w L Lx L M EI C L C
dx L
4at 0, 0 0x v C
4
03 34
( )at , 0 sin ( ) 0 0
w L Lx L v C L C
L
(a) Elastic curve equation:
0
4
44
0
4
ssin inw L x
EI v vL
w L x
EI L Ans.
(b) Deflection midway between the supports:
4
0/2
0
44
4( / 2)sinx L
w L Lv
EI L
w L
EI Ans.
(c) Slope at the left end of the beam:
3 3
0 0
3 3
3
0
3
(0)cosA A
A
dv w L w LEI EI
dx L
w L
EI Ans.
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(d) Support reactions Ay and By:
3
00 0
3
0
(0)cosA y
x
d v w L w LV EI A
dx L
w L Ans.
3
00 0
3
( )cosB y
x L
d v w L L w LV EI B
dx L
w L Ans.
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10.27 For the beam and loading shown in
Fig. P10.27, integrate the load distribution
to determine (a) the equation of the elastic
curve, (b) the deflection midway between
the supports, (c) the slope at the left end of
the beam, and (d) the support reactions Ay
and By. Assume that EI is constant for the
beam.
Fig. P10.27
Solution
Integrate the load distribution:
4
04sin
2
d v xEI w
dx L
3
013
2cos
2
d v w L xEI C
dx L
2 2
01 22 2
4sin
2
d v w L xEI C x C
dx L
3 2
0 12 33
8cos
2 2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
16sin
2 6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
2
22at 0, 0 0
d vx M EI C
dx
2 2
0 01 12 2 2
4 ( ) 4at , 0 sin ( ) 0
2
d v w L L w Lx L M EI C L C
dx L
4at 0, 0 0x v C
4 3 3
20 0 03 34 2 4
16 ( ) 4 ( ) 2at , 0 sin ( ) 0 (24 )
2 6 3
w L L w L L w Lx L v C L C
L
(a) Elastic curve equation:
4
4 2 3 2 30
3 320 0 0
4 2
4
4
16 4 2sin (24 )
224
2 6 3
sin (24 )3 2
w L x w Lx w L xEI v
w xL Lx L
EI
L
v xL
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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(b) Deflection midway between the supports:
3
4 2 2 30/ 2 4
4 2 2
0
4
4
0
4
4
0
4
0
2 ( / 2)24 sin (24 )
3 2 2 2
2 (24 )24sin
3 4 8 2
2(1.2694611)
3
0.008688 0.002 869
x L
w L L Lv L L L
EI L
w L
EI
w L
EI
w L
EI
w L
EI Ans.
(c) Slope at the left end of the beam:
3 2 320 0 0
3 2 4
3 32 3 30 0
0 03 4 3 4
3
0
2
8 (0) 2 (0) 2cos (24 )
2 3
8 2 8 16 2(24 ) 0.0262
0.0262
093 3
A
A
A
dv w L w L w LEI EI
dx L
w L w Lw L
w L
E
w L
I Ans.
(d) Support reactions Ay and By:
3
0 0 0
3 2 2
0
0
2
2 (0) 4 2cos ( 2)
2
2
( 2)
A
x
y
d v w L w L w LV EI
dx
wA
L
L
Ans.
3
0 0 0
3 2
0
2
2
2 ( ) 4 4cos
2
4
B
x L
y
d v w L L w L w LV EI
dx
L
L
Bw
Ans.
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10.28 For the beam and loading shown in
Fig. P10.28, integrate the load distribution
to determine (a) the equation of the elastic
curve, (b) the deflection at the left end of the
beam, and (c) the support reactions By and
MB. Assume that EI is constant for the beam.
Fig. P10.28
Solution
Integrate the load distribution:
4
04sin
2
d v xEI w
dx L
3
013
2cos
2
d v w L xEI C
dx L
2 2
01 22 2
4sin
2
d v w L xEI C x C
dx L
3 2
0 12 33
8cos
2 2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
16sin
2 6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
3
0 01 13
2 (0) 2at 0, 0 cos 0
2
d v w L w Lx V EI C C
dx L
2 2
0 02 22 2
4 (0) 2 (0)at 0, 0 sin 0 0
2
d v w L w Lx M EI C C
dx L
3 2 3
0 0 03 33
8 2 ( )( )at , 0 cos 0
2 2
w L w L L w Ldv Lx L C C
dx L
4 3 3
0 0 044
430
4 4
16 2 ( ) ( )( )at , 0 sin 0
2 6
2(24 )
3
w L w L L w L LLx L v C
L
w LC
(a) Elastic curve equation:
4 3 3 430 0 0 0
4
4 3 3 3 3 3 40
4
4
16 2 2sin (
48 sin 3 2(24
24 )2 6
)3 2
3
w L w Lx w L x w LxEI v
L
vw x
L Lx L x LEI 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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(b) Deflection at left end of beam:
4 3 3 3 3 3 40
4
4
3 40 0
4
4
0
(0)48 sin (0) 3 (0) 2(24 )
3 2
2(24 ) 0
0.04
.047
7
95093
95
A
wv L L
w L
E
L LEI L
w w LL
E
I
I EI
Ans.
(c) Support reactions By and MB:
3
0
3
0 0 02 ( ) 2 2cos
2
2B y
x L
d v w L L w w LL w LV EI B
dx L Ans.
2
2 2 2 2
0 0 0 0
2
0
2
2 2
2
0
2
4 ( ) 2 ( )
2( 2)
4 2sin
2
2( (cw)2)
B
x L
B
d v w L L w L L w L w LM EI
dx L
LwM
wL Ans.
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10.29 For the beam and loading shown, use
discontinuity functions to compute the deflection
of the beam at D. Assume a constant value of EI =
1,750 kip-ft2 for the beam.
Fig. P10.29
Solution
Support reactions: A FBD of the beam is shown to the
right.
(5 kips)(4 ft) (3 kips)(13 ft) (10 ft) 0
5.90 kips
5 kips 3 kips 0
2.10 kips
A y
y
y y y
y
M C
C
F A C
A
Load function w(x):
1 1 1 1
( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ftw x x x x x
Shear-force function V(x) and bending-moment function M(x):
0 0 0 0
( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ftV x x x x x
1 1 1 1
( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ftM x x x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
2
1 1 1 1
2( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ft
d vEI M x x x x x
dx
Integrate the moment function to obtain an expression for the beam slope:
2 2
2 2
1
2.10 kips 5 kips0 ft 4 ft
2 2
5.90 kips 3 kips10 ft 13 ft
2 2
dvEI x x
dx
x x C (a)
Integrate again to obtain the beam deflection function:
3 3
3 3
1 2
2.10 kips 5 kips0 ft 4 ft
6 6
5.90 kips 3 kips10 ft 13 ft
6 6
EI v x x
x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 10 ft). Substitute the
boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain:
2 0C
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Next, substitute the boundary condition v = 0 at x = 10 ft into Eq. (b) to obtain:
3 3
1
2
1
2.10 kips 5 kips(10 ft) (6 ft) (10 ft) 0
6 6
17 kip-ft
EI v C
C
The beam slope and elastic curve equations are now complete:
2 2
2 2 2
2.10 kips 5 kips0 ft 4 ft
2 2
5.90 kips 3 kips10 ft 13 ft 17 kip-ft
2 2
dvEI x x
dx
x x (c)
3 3
3 3 2
2.10 kips 5 kips0 ft 4 ft
6 6
5.90 kips 3 kips10 ft 13 ft (17 kip-ft )
6 6
EI v x x
x x x (d)
Beam deflection at D: At the tip of the overhang where x = 13 ft, the beam deflection is:
3 3 3 2
3
3
2
2.10 kips 5 kips 5.90 kips(13 ft) (9 ft) (3 ft) (17 kip-ft )(13 ft)
6 6 6
33.000 kip-ft
33.000 kip-ft0.018857 ft
1,750 kip-f0.226 in.
t
D
D
EI v
v Ans.
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10.30 The solid 30-mm-diameter steel [E = 200
GPa] shaft shown in Fig. P10.30 supports two
pulleys. For the loading shown, use discontinuity
functions to compute:
(a) the shaft deflection at pulley B.
(b) the shaft deflection at pulley C.
Fig. P10.30
Solution
Support reactions: A FBD of the beam is shown to the
right.
800 N 500 N 0
1,300 N
(800 N)(250 mm) (500 N)(600 mm) 0
500,000 N-mm
y y
y
A A
A
F A
A
M M
M
Load function w(x):
2 1
1 1
( ) 500,000 N-mm 0 mm 1,300 N 0 mm
800 N 250 mm 500 N 600 mm
w x x x
x x
Shear-force function V(x) and bending-moment function M(x):
1 0
0 0
( ) 500,000 N-mm 0 mm 1,300 N 0 mm
800 N 250 mm 500 N 600 mm
V x x x
x x
0 1
1 1
( ) 500,000 N-mm 0 mm 1,300 N 0 mm
800 N 250 mm 500 N 600 mm
M x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
20 1
2
1 1
( ) 500,000 N-mm 0 mm 1,300 N 0 mm
800 N 250 mm 500 N 600 mm
d vEI M x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
1 2
2 2
1
1,300 N500,000 N-mm 0 mm 0 mm
2
800 N 500 N250 mm 600 mm
2 2
dvEI x x
dx
x x C (a)
Integrate again to obtain the beam deflection function:
2 3
3 3
1 2
500,000 N-mm 1,300 N0 mm 0 mm
2 6
800 N 500 N250 mm 600 mm
6 6
EI v x x
x x C x C (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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Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 0 mm). Substitute the boundary
condition dv/dx = 0 at x = 0 mm into Eq. (a) to obtain:
1 0C
Next, substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain:
2 0C
The beam slope and elastic curve equations are now complete:
1 2
2 2
1,300 N500,000 N-mm 0 mm 0 mm
2
800 N 500 N250 mm 600 mm
2 2
dvEI x x
dx
x x
2 3
3 3
500,000 N-mm 1,300 N0 mm 0 mm
2 6
800 N 500 N250 mm 600 mm
6 6
EI v x x
x x
Section properties:
4 4 2
9 2
(30 mm) 39,750.782 mm 200 GPa 200,000 N/mm64
7.9522 10 N-mm
I E
EI
(a) Beam deflection at B: The beam deflection at B where x = 250 mm is:
2 3
9 3
9 2
500,000 N-mm 1,300 N(250 mm) (250 mm)
2 6
12.2396 10 N-mm1.5392 mm
7.9522 10 N-m1.539 m
mm
B
B
EI v
v Ans.
(b) Beam deflection at C: The beam deflection at C where x = 600 mm is:
2 3 3
9 3
9 2
500,000 N-mm 1,300 N 800 N(600 mm) (600 mm) (350 mm)
2 6 6
48.9167 10 N-mm6.1514 mm
7.9522 10 N-mm6.15 mm
C
C
EI v
v Ans.
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10.31 For the beam and loading shown, use
discontinuity functions to compute (a) the slope of
the beam at C and (b) the deflection of the beam at
C. Assume a constant value of EI = 560×106 N-
mm2 for the beam.
Fig. P10.31
Solution
Support reactions: A FBD of the beam is shown to the right.
(210 N-m)(1,000 mm/m) (1,400 N)(450 mm)
(700 mm) 0
600 N
1,400 N 0
800 N
A
y
y
y y y
y
M
E
E
F A E
A
Load function w(x):
1 2
1 1
( ) 800 N 0 mm 210,000 N-mm 200 mm
1,400 N 450 mm 600 N 700 mm
w x x x
x x
Shear-force function V(x) and bending-moment function M(x):
0 1
0 0
( ) 800 N 0 mm 210,000 N-mm 200 mm
1,400 N 450 mm 600 N 700 mm
V x x x
x x
1 0
1 1
( ) 800 N 0 mm 210,000 N-mm 200 mm
1,400 N 450 mm 600 N 700 mm
M x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 0
2
1 1
( ) 800 N 0 mm 210,000 N-mm 200 mm
1,400 N 450 mm 600 N 700 mm
d vEI M x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
2 1
2 2
1
800 N0 mm 210,000 N-mm 200 mm
2
1,400 N 600 N450 mm 700 mm
2 2
dvEI x x
dx
x x C (a)
Integrate again to obtain the beam deflection function:
3 2
3 3
1 2
800 N 210,000 N-mm0 mm 200 mm
6 2
1,400 N 600 N450 mm 700 mm
6 6
EI v x x
x x C x C (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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Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 700 mm). Substitute
the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain:
2 0C
Next, substitute the boundary condition v = 0 at x = 700 mm into Eq. (b) to obtain:
3 2 3
1
2
1
800 N 210,000 N-mm 1,400 N0 (700 mm) (500 mm) (250 mm) (700 mm)
6 2 6
22,625,000 N-mm
C
C
The beam slope and elastic curve equations are now complete:
2 1
2 2 2
800 N0 mm 210,000 N-mm 200 mm
2
1,400 N 600 N450 mm 700 mm 22,625,000 N-mm
2 2
dvEI x x
dx
x x
3 2
3 3 2
800 N 210,000 N-mm0 mm 200 mm
6 2
1,400 N 600 N450 mm 700 mm (22,625,000 N-mm )
6 6
EI v x x
x x x
(a) Beam slope at C: The beam slope at C is:
2 2
6 2
6 2
800 N(350 mm) (210,000 N-mm)(150 mm) 22,625,000 N-mm
2
5.125 10 N-mm0.009152 rad
50.00915 rad
60 10 N-mm
C
C
dvEI
dx
dv
dx Ans.
(b) Beam deflection at C: The beam deflection at C is:
3 2 2
9 3
9 3
6 2
800 N 210,000 N-mm(350 mm) (150 mm) (22,625,000 N-mm )(350 mm)
6 2
4.564583 10 N-mm
4.564583 10 N-mm8.1510 mm
560 10 N8.15 mm
-mm
C
C
EI v
v 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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10.32 The solid 30-mm-diameter steel [E = 200
GPa] shaft shown in Fig. P10.32 supports two
belt pulleys. Assume that the bearing at A can be
idealized as a pin support and that the bearing at
E can be idealized as a roller support. For the
loading shown, use discontinuity functions to
compute:
(a) the shaft deflection at pulley B.
(b) the shaft deflection at point C.
Fig. P10.32
Solution
Support reactions: A FBD of the beam is shown to the
right.
(600 N)(300 mm) (800 N)(800 mm)
(1,000 mm) 0
820 N
600 N 800 N 0
580 N
A
y
y
y y y
y
M
E
E
F A E
A
Load function w(x): 1 1 1 1
( ) 580 N 0 mm 600 N 300 mm 800 N 800 mm 820 N 1,000 mmw x x x x x
Shear-force function V(x) and bending-moment function M(x): 0 0 0 0
( ) 580 N 0 mm 600 N 300 mm 800 N 800 mm 820 N 1,000 mmV x x x x x
1 1 1 1( ) 580 N 0 mm 600 N 300 mm 800 N 800 mm 820 N 1,000 mmM x x x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 1
2
1 1
( ) 580 N 0 mm 600 N 300 mm
800 N 800 mm 820 N 1,000 mm
d vEI M x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
2 2 2
2
1
580 N 600 N 800 N0 mm 300 mm 800 mm
2 2 2
820 N1,000 mm
2
dvEI x x x
dx
x C (a)
Integrate again to obtain the beam deflection function:
3 3 3
3
1 2
580 N 600 N 800 N0 mm 300 mm 800 mm
6 6 6
820 N1,000 mm
6
EI v x x x
x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
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deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 1,000 mm). Substitute
the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain:
2 0C
Next, substitute the boundary condition v = 0 at x = 1,000 mm into Eq. (b) to obtain:
3 3 3
1
6 2
1
580 N 600 N 800 N0 (1,000 mm) (700 mm) (200 mm) (1,000 mm)
6 6 6
61.3 10 N-mm
C
C
The beam slope and elastic curve equations are now complete:
2 2 2
2 6 2
580 N 600 N 800 N0 mm 300 mm 800 mm
2 2 2
820 N1,000 mm 61.3 10 N-mm
2
dvEI x x x
dx
x
3 3 3
3 6 2
580 N 600 N 800 N0 mm 300 mm 800 mm
6 6 6
820 N1,000 mm (61.3 10 N-mm )
6
EI v x x x
x x
Section properties:
4 4 2
9 2
(30 mm) 39,750.782 mm 200 GPa 200,000 N/mm64
7.9522 10 N-mm
I E
EI
(a) Beam deflection at B: The beam deflection at B where x = 300 mm is:
3 6 2
9 3
9 2
580 N(300 mm) (61.3 10 N-mm )(300 mm)
6
15.7800 10 N-mm1.9844 mm
7.9522 10 N-m1.984 m
mm
B
B
EI v
v Ans.
(b) Beam deflection at C: The beam deflection at C where x = 500 mm is:
3 3 6 2
9 3
9 2
580 N 600 N(500 mm) (200 mm) (61.3 10 N-mm )(500 mm)
6 6
19.3667 10 N-mm2.4354 mm
7.9522 10 N-mm2.44 mm
C
C
EI v
v Ans.
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10.33 The cantilever beam shown in Fig. P10.33
consists of a W530 × 74 structural steel wide-
flange shape [E = 200 GPa; I = 410 × 106 mm
4].
Use discontinuity functions to compute the
deflection of the beam at C for the loading shown.
Fig. P10.33
Solution
Support reactions: A FBD of the beam is shown to the
right.
(30 kN/m)(3 m) 40 kN 0
130 kN
(30 kN/m)(3 m)(1.5 m) (40 kN)(5 m) 0
335 kN-m
y y
y
A A
A
F A
A
M M
M
Load function w(x):
1 2
0 0 1
( ) 130 kN 0 m 335 kN-m 0 m
30 kN/m 0 m 30 kN/m 3 m 40 kN 5 m
w x x x
x x x
Shear-force function V(x) and bending-moment function M(x):
0 1
1 1 0
( ) 130 kN 0 m 335 kN-m 0 m
30 kN/m 0 m 30 kN/m 3 m 40 kN 5 m
V x x x
x x x
1 0
2 2 1
( ) 130 kN 0 m 335 kN-m 0 m
30 kN/m 30 kN/m0 m 3 m 40 kN 5 m
2 2
M x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 0
2
2 2 1
( ) 130 kN 0 m 335 kN-m 0 m
30 kN/m 30 kN/m0 m 3 m 40 kN 5 m
2 2
d vEI M x x x
dx
x x x
Integrate the moment function to obtain an expression for the beam slope:
2 1
3 3 2
1
130 kN0 m 335 kN-m 0 m
2
30 kN/m 30 kN/m 40 kN0 m 3 m 5 m
6 6 2
dvEI x x
dx
x x x C (a)
Integrate again to obtain the beam deflection function:
3 2
4 4 3
1 2
130 kN 335 kN-m0 m 0 m
6 2
30 kN/m 30 kN/m 40 kN0 m 3 m 5 m
24 24 6
EI v x x
x x x C x C (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.
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary
condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain:
1 0C
Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain:
2 0C
The beam slope and elastic curve equations are now complete:
2 1
3 3 2
130 kN0 m 335 kN-m 0 m
2
30 kN/m 30 kN/m 40 kN0 m 3 m 5 m
6 6 2
dvEI x x
dx
x x x
3 2
4 4 3
130 kN 335 kN-m0 m 0 m
6 2
30 kN/m 30 kN/m 40 kN0 m 3 m 5 m
24 24 6
EI v x x
x x x
Beam deflection at C: For the W530 × 74 structural steel wide-flange shape, EI = 82,000 kN-m2. At
the tip of the overhang where x = 5 m, the beam deflection is:
3 2 4 4
3
3
2
130 kN 335 kN-m 30 kN/m 30 kN/m(5 m) (5 m) (5 m) (2 m)
6 2 24 24
2,240.416667 kN-m
2,240.416667 kN-m0.027322 m
82,000 kN-m27.3 mm
C
C
EI v
v 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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10.34 The cantilever beam shown in Fig. P10.34
consists of a W21 × 50 structural steel wide-flange
shape [E = 29,000 ksi; I = 984 in.4]. Use
discontinuity functions to compute the deflection
of the beam at D for the loading shown.
Fig. P10.34
Solution
Support reactions: A FBD of the beam is shown to
the right.
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
Load function w(x):
2 1 1 1
1 0
4 kips/ft( ) 270 kip-ft 0 ft 27 kips 0 ft 9 kips 4 ft 7 ft
9 ft
4 kips/ft16 ft 4 kips/ft 16 ft
9 ft
w x x x x x
x x
Shear-force function V(x) and bending-moment function M(x):
1 0 0 2
2 1
4 kips/ft( ) 270 kip-ft 0 ft 27 kips 0 ft 9 kips 4 ft 7 ft
2(9 ft)
4 kips/ft16 ft 4 kips/ft 16 ft
2(9 ft)
V x x x x x
x x
0 1 1 3
3 2
4 kips/ft( ) 270 kip-ft 0 ft 27 kips 0 ft 9 kips 4 ft 7 ft
6(9 ft)
4 kips/ft 4 kips/ft16 ft 16 ft
6(9 ft) 2
M x x x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
20 1 1
2
3 3 2
( ) 270 kip-ft 0 ft 27 kips 0 ft 9 kips 4 ft
4 kips/ft 4 kips/ft 4 kips/ft7 ft 16 ft 16 ft
6(9 ft) 6(9 ft) 2
d vEI M x x x x
dx
x x x
Integrate the moment function to obtain an expression for the beam slope:
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1 2 2
4 4 3
1
27 kips 9 kips270 kip-ft 0 ft 0 ft 4 ft
2 2
4 kips/ft 4 kips/ft 4 kips/ft7 ft 16 ft 16 ft
24(9 ft) 24(9 ft) 6
dvEI x x x
dx
x x x C
(a)
Integrate again to obtain the beam deflection function:
2 3 3
5 5 4
1 2
270 kip-ft 27 kips 9 kips0 ft 0 ft 4 ft
2 6 6
4 kips/ft 4 kips/ft 4 kips/ft7 ft 16 ft 16 ft
120(9 ft) 120(9 ft) 24
EI v x x x
x x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 0 ft). Substitute the boundary
condition dv/dx = 0 at x = 0 ft into Eq. (a) to obtain:
1 0C
Next, substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain:
2 0C
The beam slope and elastic curve equations are now complete:
1 2 2 4
4 3
27 kips 9 kips 4 kips/ft270 kip-ft 0 ft 0 ft 4 ft 7 ft
2 2 24(9 ft)
4 kips/ft 4 kips/ft16 ft 16 ft
24(9 ft) 6
dvEI x x x x
dx
x x
2 3 3 5
5 4
270 kip-ft 27 kips 9 kips 4 kips/ft0 ft 0 ft 4 ft 7 ft
2 6 6 120(9 ft)
4 kips/ft 4 kips/ft16 ft 16 ft
120(9 ft) 24
EI v x x x x
x x
Beam deflection at D: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2.
At the tip of the overhang where x = 16 ft, the beam deflection is:
2 3 3 5
3
3
2
270 kip-ft 27 kips 9 kips 4 kips/ft(16 ft) (16 ft) (12 ft) (9 ft)
2 6 6 120(9 ft)
18,938.7 kip-ft
18,938.7 kip-ft0.095570 ft
198,166.658 kip-f1.147
t in.
D
D
EI v
v 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.
10.35 The simply supported beam shown in Fig.
P10.35 consists of a W410 × 85 structural steel
wide-flange shape [E = 200 GPa; I = 316 × 106
mm4]. For the loading shown, use discontinuity
functions to compute (a) the slope of the beam at A
and (b) the deflection of the beam at midspan. Fig. P10.35
Solution
Support reactions: A FBD of the beam is shown to the
right.
(75 kN/m)(2.5 m)(1.25 m)
(75 kN/m)(2.5 m)(6.75 m) (8 m) 0
187.5 kN
(75 kN/m)(2.5 m) (75 kN/m)(2.5 m) 0
187.5 kN
A
y
y
y y y
y
M
D
D
F A D
A
Load function w(x):
1 0 0
0 0 1
( ) 187.5 kN 0 m 75 kN/m 0 m 75 kN/m 2.5 m
75 kN/m 5.5 m 75 kN/m 8 m 187.5 kN 8 m
w x x x x
x x x
Shear-force function V(x) and bending-moment function M(x):
0 1 1
1 1 0
( ) 187.5 kN 0 m 75 kN/m 0 m 75 kN/m 2.5 m
75 kN/m 5.5 m 75 kN/m 8 m 187.5 kN 8 m
V x x x x
x x x
1 2 2
2 2 1
75 kN/m 75 kN/m( ) 187.5 kN 0 m 0 m 2.5 m
2 2
75 kN/m 75 kN/m5.5 m 8 m 187.5 kN 8 m
2 2
M x x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 2 2
2
2 2 1
75 kN/m 75 kN/m( ) 187.5 kN 0 m 0 m 2.5 m
2 2
75 kN/m 75 kN/m5.5 m 8 m 187.5 kN 8 m
2 2
d vEI M x x x x
dx
x x x
Integrate the moment function to obtain an expression for the beam slope:
2 3 3
3 3 2
1
187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m
2 6 6
75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m
6 6 2
dvEI x x x
dx
x x x C (a)
Integrate again to obtain the beam deflection function:
3 4 4
4 4 3
1 2
187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m
6 24 24
75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m
24 24 6
EI v x x x
x x x C x C (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.
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the
boundary condition v = 0 at x = 0 m into Eq. (b) to obtain:
2 0C
Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain:
3 4 4 4
1
2
1
187.5 kN 75 kN/m 75 kN/m 75 kN/m0 (8 m) (8 m) (5.5 m) (2.5 m) (8 m)
6 24 24 24
742.1875 kN-m
C
C
The beam slope and elastic curve equations are now complete:
2 3 3
3 3 2 2
187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m
2 6 6
75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m 742.1875 kN-m
6 6 2
dvEI x x x
dx
x x x
3 4 4
4 4 3 2
187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m
6 24 24
75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m (742.1875 kN-m )
24 24 6
EI v x x x
x x x x
(a) Beam slope at A: For the W410 × 85 structural steel wide-flange shape, EI = 63,200 kN-m2. The
beam slope at A is:
2
2
2
742.1875 kN-m
742.1875 kN-m0.011743 rad
63,200 kN-0.01174 rad
m
A
A
dvEI
dx
dv
dx Ans.
(b) Beam deflection at midspan: At midspan where x = 4 m, the beam deflection is:
3 4 4 2
midspan
3
3
midspan 2
187.5 kN 75 kN/m 75 kN/m(4 m) (4 m) (1.5 m) (742.1875 kN-m )(4 m)
6 24 24
1,752.929687 kN-m
1,752.929687 kN-m0.027736 m
63,200 kN-27
m.7 mm
EI v
v 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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10.36 The simply supported beam shown in Fig.
P10.36 consists of a W14 × 30 structural steel
wide-flange shape [E = 29,000 ksi; I = 291 in.4].
For the loading shown, use discontinuity functions
to compute (a) the slope of the beam at A and (b)
the deflection of the beam at midspan.
Fig. P10.36
Solution
Support reactions: A FBD of the beam is shown to the
right.
(2.5 kips/ft)(12 ft)(12 ft) (24 ft) 0
15 kips
(2.5 kips/ft)(12 ft) 0
15 kips
A y
y
y y y
y
M D
D
F A D
A
Load function w(x):
1 0 0 1
( ) 15 kips 0 ft 2.5 kips/ft 6 ft 2.5 kips/ft 18 ft 15 kips 24 ftw x x x x x
Shear-force function V(x) and bending-moment function M(x):
0 1 1 0
( ) 15 kips 0 ft 2.5 kips/ft 6 ft 2.5 kips/ft 18 ft 15 kips 24 ftV x x x x x
1 2 2 12.5 kips/ft 2.5 kips/ft
( ) 15 kips 0 ft 6 ft 18 ft 15 kips 24 ft2 2
M x x x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 2
2
2 1
2.5 kips/ft( ) 15 kips 0 ft 6 ft
2
2.5 kips/ft18 ft 15 kips 24 ft
2
d vEI M x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
2 3
3 2
1
15 kips 2.5 kips/ft0 ft 6 ft
2 6
2.5 kips/ft 15 kips18 ft 24 ft
6 2
dvEI x x
dx
x x C (a)
Integrate again to obtain the beam deflection function:
3 4
4 3
1 2
15 kips 2.5 kips/ft0 ft 6 ft
6 24
2.5 kips/ft 15 kips18 ft 24 ft
24 6
EI v x x
x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 24 ft). Substitute the
boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain:
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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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2 0C
Next, substitute the boundary condition v = 0 at x = 24 ft into Eq. (b) to obtain:
3 4 4
1
2
1
15 kips 2.5 kips/ft 2.5 kips/ft0 (24 ft) (18 ft) (6 ft) (24 ft)
6 24 24
990 kip-ft
C
C
The beam slope and elastic curve equations are now complete:
2 3
3 2 2
15 kips 2.5 kips/ft0 ft 6 ft
2 6
2.5 kips/ft 15 kips18 ft 24 ft 990 kip-ft
6 2
dvEI x x
dx
x x
3 4
4 3 2
15 kips 2.5 kips/ft0 ft 6 ft
6 24
2.5 kips/ft 15 kips18 ft 24 ft (990 kip-ft )
24 6
EI v x x
x x x
(a) Beam slope at A: For the W14 × 30 structural steel wide-flange shape, EI = 58,604.164 kip-ft2.
The beam slope at A is:
2
2
2
990 kip-ft
990 kip-ft0.016893 rad
58,604.164 kip-ft0.01689 rad
A
A
dvEI
dx
dv
dx Ans.
(b) Beam deflection at midspan: At midspan where x = 12 ft, the beam deflection is:
3 4 2 3
midspan
3
midspan 2
15 kips 2.5 kips/ft(12 ft) (6 ft) (990 kip-ft )(12 ft) 7,695 kip-ft
6 24
7,695 kip-ft0.131305 ft
58,604.164 kip-f1.5 6 n
t7 i .
EI v
v 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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10.37 The simply supported beam shown in Fig.
P10.37 consists of a W21 × 50 structural steel
wide-flange shape [E = 29,000 ksi; I = 984 in.4].
For the loading shown, use discontinuity functions
to compute (a) the slope of the beam at A and (b)
the deflection of the beam at B.
Fig. P10.37
Solution
Support reactions: A FBD of the beam is shown to the right.
(7 kips/ft)(11 ft)(5.5 ft) (4 kips/ft)(9 ft)(15.5 ft)
(20 ft) 0
49.075 kips
(7 kips/ft)(11 ft) (4 kips/ft)(9 ft) 0
63.925 kips
A
y
y
y y y
y
M
C
C
F A C
A
Load function w(x):
1 0 0
0 0 1
( ) 63.925 kips 0 ft 7 kips/ft 0 ft 7 kips/ft 11 ft
4 kips/ft 11 ft 4 kips/ft 20 ft 49.075 kips 20 ft
w x x x x
x x x
Shear-force function V(x) and bending-moment function M(x):
0 1 1
1 1 0
( ) 63.925 kips 0 ft 7 kips/ft 0 ft 7 kips/ft 11 ft
4 kips/ft 11 ft 4 kips/ft 20 ft 49.075 kips 20 ft
V x x x x
x x x
1 2 2
2 2 1
7 kips/ft 7 kips/ft( ) 63.925 kips 0 ft 0 ft 11 ft
2 2
4 kips/ft 4 kips/ft11 ft 20 ft 49.075 kips 20 ft
2 2
M x x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 2 2
2
2 2 1
7 kips/ft 7 kips/ft( ) 63.925 kips 0 ft 0 ft 11 ft
2 2
4 kips/ft 4 kips/ft11 ft 20 ft 49.075 kips 20 ft
2 2
d vEI M x x x x
dx
x x x
Integrate the moment function to obtain an expression for the beam slope:
2 3 3
3 3 2
1
63.925 kips 7 kips/ft 7 kips/ft0 ft 0 ft 11 ft
2 6 6
4 kips/ft 4 kips/ft 49.075 kips11 ft 20 ft 20 ft
6 6 2
dvEI x x x
dx
x x x C (a)
Integrate again to obtain the beam deflection function:
3 4 4
4 4 3
1 2
63.925 kips 7 kips/ft 7 kips/ft0 ft 0 ft 11 ft
6 24 24
4 kips/ft 4 kips/ft 49.075 kips11 ft 20 ft 20 ft
24 24 6
EI v x x x
x x x C x C (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.
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 20 ft). Substitute the
boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain:
2 0C
Next, substitute the boundary condition v = 0 at x = 20 ft into Eq. (b) to obtain:
3 4 4 4
1
2
1
63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft0 (20 ft) (20 ft) (9 ft) (9 ft) (20 ft)
6 24 24 24
1,969.3396 kip-ft
C
C
The beam slope and elastic curve equations are now complete:
2 3 3 3
3 2 2
63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft0 ft 0 ft 11 ft 11 ft
2 6 6 6
4 kips/ft 49.075 kips20 ft 20 ft 1,969.3396 kip-ft
6 2
dvEI x x x x
dx
x x
3 4 4 4
4 3 2
63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft0 ft 0 ft 11 ft 11 ft
6 24 24 24
4 kips/ft 49.075 kips20 ft 20 ft (1,969.3396 kip-ft )
24 6
EI v x x x x
x x x
(a) Beam slope at A: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2.
The beam slope at A is:
2
2
2
1,969.3396 kip-ft
1,969.3396 kip-ft0.009938 rad
198,166.658 kip0.00994 ra
td
-f
A
A
dvEI
dx
dv
dx Ans.
(b) Beam deflection at B: At midspan where x = 11 ft, the beam deflection is:
3 4 2
3
2
63.925 kips 7 kips/ft(11 ft) (11 ft) (1,969.3396 kip-ft )(11 ft)
6 24
11,752.33123 kip-ft0.059305 ft
198,10.712 in
66.658 ki f.
p- t
B
B
EI v
v Ans.
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10.38 The simply supported beam shown in Fig.
P10.38 consists of a W200 × 59 structural steel
wide-flange shape [E = 200 GPa; I = 60.8 × 106
mm4]. For the loading shown, use discontinuity
functions to compute (a) the deflection of the beam
at C and (b) the deflection of the beam at F.
Fig. P10.38
Solution
Support reactions: A FBD of the beam is shown to the
right.
(20 kN)(2 m) (8 kN/m)(6 m)(7 m)
(10 kN)(12 m) (8 m) 0
62 kN
20 kN (8 kN/m)(6 m) 10 kN 0
16 kN
A
y
y
y y y
y
M
D
D
F A D
A
Load function w(x):
1 1 0 1
0 1
( ) 16 kN 0 m 20 kN 2 m 8 kN/m 4 m 62 kN 8 m
8 kN/m 10 m 10 kN 12 m
w x x x x x
x x
Shear-force function V(x) and bending-moment function M(x):
0 0 1 0
1 0
( ) 16 kN 0 m 20 kN 2 m 8 kN/m 4 m 62 kN 8 m
8 kN/m 10 m 10 kN 12 m
V x x x x x
x x
1 1 2 1
2 1
8 kN/m( ) 16 kN 0 m 20 kN 2 m 4 m 62 kN 8 m
2
8 kN/m10 m 10 kN 12 m
2
M x x x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 1 2 1
2
2 1
8 kN/m( ) 16 kN 0 m 20 kN 2 m 4 m 62 kN 8 m
2
8 kN/m10 m 10 kN 12 m
2
d vEI M x x x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
2 2 3 2
3 2
1
16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m
2 2 6 2
8 kN/m 10 kN10 m 12 m
6 2
dvEI x x x x
dx
x x C (a)
Integrate again to obtain the beam deflection function:
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3 3 4 3
4 3
1 2
16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m
6 6 24 6
8 kN/m 10 kN10 m 12 m
24 6
EI v x x x x
x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the
boundary condition v = 0 at x = 0 m into Eq. (b) to obtain:
2 0C
Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain:
3 3 4
1
2
1
16 kN 20 kN 8 kN/m0 (8 m) (6 m) (4 m) (8 m)
6 6 24
70 kN-m
C
C
The beam slope and elastic curve equations are now complete:
2 2 3 2
3 2 2
16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m
2 2 6 2
8 kN/m 10 kN10 m 12 m 70 kN-m
6 2
dvEI x x x x
dx
x x
3 3 4 3
4 3 2
16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m
6 6 24 6
8 kN/m 10 kN10 m 12 m (70 kN-m )
24 6
EI v x x x x
x x x
(a) Beam deflection at C: For the W200 × 59 structural steel wide-flange shape, EI = 12,160 kN-m2.
At C where x = 4 m, the beam deflection is:
3 3 2
C
3
3
2
16 kN 20 kN(4 m) (2 m) (70 kN-m )(4 m)
6 6
136 kN-m
136 kN-m0.011184 m
12,16011.18
mm
kN-mC
EI v
v Ans.
(b) Beam deflection at F: At F where x = 12 m, the beam deflection is:
3 3 4 3
4 2
3
3
2
16 kN 20 kN 8 kN/m 62 kN(12 m) (10 m) (8 m) (4 m)
6 6 24 6
8 kN/m(2 m) (70 kN-m )(12 m)
24
264 kN-m
264 kN-21.7 m
m0.021711 m
12,160 kN-mm
F
F
EI v
v Ans.
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10.39 The solid 0.50-in.-diameter steel [E =
30,000 ksi] shaft shown in Fig. P10.39 supports
two belt pulleys. Assume that the bearing at B
can be idealized as a pin support and that the
bearing at D can be idealized as a roller support.
For the loading shown, use discontinuity
functions to compute:
(a) the shaft deflection at pulley A.
(b) the shaft deflection at pulley C.
Fig. P10.39
Solution
Support reactions: A FBD of the beam is shown to the
right.
(90 lb)(5 in.) (120 lb)(10 in.) (20 in.) 0
37.5 lb
90 lb 120 lb 0
172.5 lb
B y
y
y y y
y
M D
D
F B D
B
Load function w(x):
1 1 1 1
( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.w x x x x x
Shear-force function V(x) and bending-moment function M(x):
0 0 0 0
( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.V x x x x x
1 1 1 1
( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.M x x x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
2
1 1 1 1
2( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.
d vEI M x x x x x
dx
Integrate the moment function to obtain an expression for the beam slope:
2 2 2 2
1
90 lb 172.5 lb 120 lb 37.5 lb0 in. 5 in. 15 in. 25 in.
2 2 2 2
dvEI x x x x C
dx (a)
Integrate again to obtain the beam deflection function:
3 3 3
3
1 2
90 lb 172.5 lb 120 lb0 in. 5 in. 15 in.
6 6 6
37.5 lb25 in.
6
EI v x x x
x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 5 in.) and at the roller support (x = 25 in.). Substitute the
boundary condition v = 0 at x = 5 in. into Eq. (b) to obtain:
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3
1 2
3
1 2
90 lb0 (5 in.) (5 in.)
6
(5 in.) 1,875.0 lb-in.
C C
C C (c)
Next, substitute the boundary condition v = 0 at x = 25 in. into Eq. (b) to obtain:
3 3 3
1 2
3
1 2
90 lb 172.5 lb 120 lb0 (25 in.) (20 in.) (10 in.) (25 in.)
6 6 6
(25 in.) 24,375.0 lb-in.
C C
C C (d)
Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2:
2 3
1 21,125 lb-in. and 3,750 lb-in.C C
The beam slope and elastic curve equations are now complete:
2 2 2
2 2
90 lb 172.5 lb 120 lb0 in. 5 in. 15 in.
2 2 2
37.5 lb25 in. 1,125 lb-in.
2
dvEI x x x
dx
x
3 3 3
3 2 3
90 lb 172.5 lb 120 lb0 in. 5 in. 15 in.
6 6 6
37.5 lb25 in. (1,125 lb-in. ) 3,750 lb-in.
6
EI v x x x
x x
Section properties:
4 3 4 6
3 2
(0.5 in.) 3.06796 10 in. 30,000 ksi 30 10 psi64
92.0388 10 lb-in.
I E
EI
(a) Beam deflection at A: The beam deflection at A where x = 0 in. is:
3
3
3 2
3,750 lb-in.
3,750 lb-in.0.040744 in.
92.0380.0407 i
8 10 lb- .n
i.
n
A
A
EI v
v Ans.
(b) Beam deflection at C: The beam deflection at C where x = 15 in. is:
3 3 2 3
3
3 2
90 lb 172.5 lb(15 in.) (10 in.) (1,125 lb-in. )(15 in.) 3,750 lb-in.
6 6
8,750 lb-in.0.095069 in.
92.03880.0951 in.
10 lb-in.
C
C
EI v
v 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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10.40 The cantilever beam shown in Fig. P10.40
consists of a W8 × 31 structural steel wide-flange
shape [E = 29,000 ksi; I = 110 in.4]. For the
loading shown, use discontinuity functions to
compute (a) the slope of the beam at A and (b) the
deflection of the beam at A.
Fig. P10.40
Solution
Support reactions: A FBD of the beam is shown to
the right.
(3.5 kips/ft)(10 ft) 0
35 kips
75 kip-ft (3.5 kips/ft)(10 ft)(5 ft) 0
100 kip-ft
y y
y
C C
C
F C
C
M M
M
Load function w(x):
2 0 0
1 2
( ) 75 kip-ft 0 ft 3.5 kips/ft 5 ft 3.5 kips/ft 15 ft
35 kips 15 ft 100 kip-ft 15 ft
w x x x x
x x
Shear-force function V(x) and bending-moment function M(x):
1 1 1
0 1
( ) 75 kip-ft 0 ft 3.5 kips/ft 5 ft 3.5 kips/ft 15 ft
35 kips 15 ft 100 kip-ft 15 ft
V x x x x
x x
0 2 2
1 0
3.5 kips/ft 3.5 kips/ft( ) 75 kip-ft 0 ft 5 ft 15 ft
2 2
35 kips 15 ft 100 kip-ft 15 ft
M x x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
20 2 2
2
1 0
3.5 kips/ft 3.5 kips/ft( ) 75 kip-ft 0 ft 5 ft 15 ft
2 2
35 kips 15 ft 100 kip-ft 15 ft
d vEI M x x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
1 3 3
2 1
1
3.5 kips/ft 3.5 kips/ft75 kip-ft 0 ft 5 ft 15 ft
6 6
35 kips15 ft 100 kip-ft 15 ft
2
dvEI x x x
dx
x x C
(a)
Integrate again to obtain the beam deflection function:
2 4 4
3 2
1 2
75 kip-ft 3.5 kips/ft 3.5 kips/ft0 ft 5 ft 15 ft
2 24 24
35 kips 100 kip-ft15 ft 15 ft
6 2
EI v x x x
x x C x C
(b)
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Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 15 ft). Substitute the boundary
condition dv/dx = 0 at x = 15 ft into Eq. (a) to obtain:
1 3
1
2
1
3.5 kips/ft0 (75 kip-ft)(15 ft) (10 ft)
6
541.666667 kip-ft
C
C
Next, substitute the boundary condition v = 0 at x = 15 ft into Eq. (b) to obtain:
2 4 2
2
3
2
75 kip-ft 3.5 kips/ft0 (15 ft) (10 ft) ( 541.666667 kip-ft )(15 ft)
2 24
1,145.833334 kip-ft
C
C
The beam slope and elastic curve equations are now complete:
1 3 3
2 1 2
3.5 kips/ft 3.5 kips/ft75 kip-ft 0 ft 5 ft 15 ft
6 6
35 kips15 ft 100 kip-ft 15 ft 541.666667 kip-ft
2
dvEI x x x
dx
x x
2 4 4
3 2 2 3
75 kip-ft 3.5 kips/ft 3.5 kips/ft0 ft 5 ft 15 ft
2 24 24
35 kips 100 kip-ft15 ft 15 ft (541.666667 kip-ft ) 1,145.833334 kip-ft
6 2
EI v x x x
x x x
(a) Beam slope at A: For the W8 × 31 structural steel wide-flange shape, EI = 22,152.777 kip-ft2. At
the tip of the overhang where x = 0 ft, the beam slope is:
2
2
2
541.666667 kip-ft
541.666667 kip-ft0.024451 rad
22,152.777 ki0.0245 r
p-a
ftd
A
A
dvEI
dx
dv
dx Ans.
(b) Beam deflection at A: At the tip of the overhang where x = 0 ft, the beam deflection is:
3
3
2
1,145.833334 kip-ft
1,145.833334 kip-ft0.051724 ft
22,152.777 kip-f0.621 i .
tn
A
A
EI v
v Ans.
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10.41 The simply supported beam shown in Fig.
P10.41 consists of a W14 × 34 structural steel
wide-flange shape [E = 29,000 ksi; I = 340 in.4].
For the loading shown, use discontinuity functions
to compute (a) the slope of the beam at E and (b)
the deflection of the beam at C. Fig. P10.41
Solution
Support reactions: A FBD of the beam is shown to the right.
1
2
1
2
8 ft(6 kips/ft)(8 ft) (4 kips/ft)(10 ft)(13 ft)
3
(22 ft) 0
20.727 kips
(6 kips/ft)(8 ft) (4 kips/ft)(10 ft) 0
43.273 kips
B
y
y
y y y
y
M
E
E
F B E
B
Load function w(x):
1 1 0 1
0 0 1
6 kips/ft 6 kips/ft( ) 0 ft 8 ft 6 kips/ft 8 ft 43.273 kips 8 ft
8 ft 8 ft
4 kips/ft 16 ft 4 kips/ft 26 ft 20.727 kips 30 ft
w x x x x x
x x x
Shear-force function V(x) and bending-moment function M(x):
2 2 1 0
1 1 0
6 kips/ft 6 kips/ft( ) 0 ft 8 ft 6 kips/ft 8 ft 43.273 kips 8 ft
2(8 ft) 2(8 ft)
4 kips/ft 16 ft 4 kips/ft 26 ft 20.727 kips 30 ft
V x x x x x
x x x
3 3 2 1
2 2 1
6 kips/ft 6 kips/ft 6 kips/ft( ) 0 ft 8 ft 8 ft 43.273 kips 8 ft
6(8 ft) 6(8 ft) 2
4 kips/ft 4 kips/ft16 ft 26 ft 20.727 kips 30 ft
2 2
M x x x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
23 3 2
2
1 2 2
1
6 kips/ft 6 kips/ft 6 kips/ft( ) 0 ft 8 ft 8 ft
6(8 ft) 6(8 ft) 2
4 kips/ft 4 kips/ft43.273 kips 8 ft 16 ft 26 ft
2 2
20.727 kips 30 ft
d vEI M x x x x
dx
x x x
x
Integrate the moment function to obtain an expression for the beam slope:
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4 4 3 2
3 3 2
1
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 ft 8 ft 8 ft 8 ft
24(8 ft) 24(8 ft) 6 2
4 kips/ft 4 kips/ft 20.727 kips16 ft 26 ft 30 ft
6 6 2
dvEI x x x x
dx
x x x C (a)
Integrate again to obtain the beam deflection function:
5 5 4 3
4 4 3
1 2
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 ft 8 ft 8 ft 8 ft
120(8 ft) 120(8 ft) 24 6
4 kips/ft 4 kips/ft 20.727 kips16 ft 26 ft 30 ft
24 24 6
EI v x x x x
x x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 8 ft) and at the roller support (x = 30 ft). Substitute the
boundary condition v = 0 at x = 8 ft into Eq. (b) to obtain:
5
1 2
3
1 2
6 kips/ft0 (8 ft) (8 ft)
120(8 ft)
(8 ft) 204.80 kips-ft
C C
C C (c)
Next, substitute the boundary condition v = 0 at x = 30 ft into Eq. (b) to obtain:
5 5 4 3
4 4
1 2
3
1 2
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 (30 ft) (22 ft) (22 ft) (22 ft)
120(8 ft) 120(8 ft) 24 6
4 kips/ft 4 kips/ft(14 ft) (4 ft) (30 ft)
24 24
(30 ft) 9,334.351 kip-ft
C C
C C (d)
Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2:
2 3
1 2433.598 kip-ft and 3,673.582 kip-ftC C
The beam slope and elastic curve equations are now complete:
4 4 3 2
3 3 2 2
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 ft 8 ft 8 ft 8 ft
24(8 ft) 24(8 ft) 6 2
4 kips/ft 4 kips/ft 20.727 kips16 ft 26 ft 30 ft 433.598 kip-ft
6 6 2
dvEI x x x x
dx
x x x
5 5 4 3
4 4 3
2 3
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 ft 8 ft 8 ft 8 ft
120(8 ft) 120(8 ft) 24 6
4 kips/ft 4 kips/ft 20.727 kips16 ft 26 ft 30 ft
24 24 6
(433.598 kip-ft ) 3,673.582 kip-ft
EI v x x x x
x x 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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(a) Beam slope at E: For the W14 × 34 structural steel wide-flange shape, EI = 68,472.219 kip-ft2.
The beam slope at E is:
4 4 3 2
3 3 2
2
2
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips(30 ft) (22 ft) (22 ft) (22 ft)
24(8 ft) 24(8 ft) 6 2
4 kips/ft 4 kips/ft(14 ft) (4 ft) 433.598 kip-ft
6 6
907.801 kip-ft0.0132
68,472.219 kip-ft
E
E
dvEI
dx
dv
dx0.01358 ra dd 26 ra Ans.
(b) Beam deflection at C: At C where x = 16 ft, the beam deflection is:
5 5 4 3
2 3
3
2
6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips(16 ft) (8 ft) (8 ft) (8 ft)
120(8 ft) 120(8 ft) 24 6
(433.598 kip-ft )(16 ft) 3,673.582 kip-ft
4,896.157 kip-ft0.071506 ft
68,472.219 kip-ft0.858
C
C
EI v
v in. 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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10.42 For the beam and loading shown, use
discontinuity functions to compute (a) the
deflection of the beam at A and (b) the deflection
of the beam at midspan (i.e., x = 2.5 m). Assume a
constant value of EI = 1,500 kN-m2 for the beam.
Fig. P10.42
Solution
Support reactions: A FBD of the beam is shown to the
right.
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/m1 m 4 m 6 kN 4 m
3 m 3 m
w x x x x
x x x
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/m1 m 4 m 6 kN 4 m
2(3 m) 2(3 m)
V x x x x
x x x
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/m1 m 4 m 6 kN 4 m
6(3 m) 6(3 m)
M x x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
20 1 2
2
3 3 1
18 kN/m( ) 9 kN-m 0 m 21 kN 1 m 1 m
2
18 kN/m 18 kN/m1 m 4 m 6 kN 4 m
6(3 m) 6(3 m)
d vEI M x x x x
dx
x x x
Integrate the moment function to obtain an expression for the beam slope:
1 2 3
4 4 2
1
21 kN 18 kN/m9 kN-m 0 m 1 m 1 m
2 6
18 kN/m 18 kN/m 6 kN1 m 4 m 4 m
24(3 m) 24(3 m) 2
dvEI x x x
dx
x x x C (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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Integrate again to obtain the beam deflection function:
2 3 4
5 5 3
1 2
9 kN-m 21 kN 18 kN/m0 m 1 m 1 m
2 6 24
18 kN/m 18 kN/m 6 kN1 m 4 m 4 m
120(3 m) 120(3 m) 6
EI v x x x
x x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 1 m) and at the roller support (x = 4 m). Substitute the
boundary condition v = 0 at x = 1 m into Eq. (b) to obtain:
2
1 2
3
1 2
9 kN-m0 (1 m) (1 m)
2
(1 m) 4.5 kN-m
C C
C C (c)
Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain:
2 3 4 5
1 2
3
1 2
9 kN-m 21 kN 18 kN/m 18 kN/m0 (4 m) (3 m) (3 m) (3 m) (4 m)
2 6 24 120(3 m)
(4 m) 26.10 kN-m
C C
C C (d)
Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2:
2 3
1 27.2 kN-m and 2.7 kN-mC C
The beam slope and elastic curve equations are now complete:
1 2 3
4 4 2 2
21 kN 18 kN/m9 kN-m 0 m 1 m 1 m
2 6
18 kN/m 18 kN/m 6 kN1 m 4 m 4 m 7.2 kN-m
24(3 m) 24(3 m) 2
dvEI x x x
dx
x x x
2 3 4
5 5 3 2 3
9 kN-m 21 kN 18 kN/m0 m 1 m 1 m
2 6 24
18 kN/m 18 kN/m 6 kN1 m 4 m 4 m (7.2 kN-m ) 2.7 kN-m
120(3 m) 120(3 m) 6
EI v x x x
x x x x
(a) Beam deflection at A: The beam deflection at A is:
3
3
2
2.7 kN-m
2.7 kN-m0.001800 m
1,51.800
00 kN mm
-m
A
A
EI v
v Ans.
(b) Beam deflection at midspan: At x = 2.5 m, the beam deflection is:
2 3 4
midspan
5 2 3
3
3
midspan 2
9 kN-m 21 kN 18 kN/m(2.5 m) (1.5 m) (1.5 m)
2 6 24
18 kN/m(1.5 m) (7.2 kN-m
2.95
)(2.5 m) 2.7 kN-m120(3 m)
4.429688 kN-m
4.429688 kN-m0.002953 m
1,500 kN-m
m m
EI v
v Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
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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10.43 For the beam and loading shown, use
discontinuity functions to compute (a) the slope of
the beam at B and (b) the deflection of the beam at
A. Assume a constant value of EI = 133,000 kip-
ft2 for the beam.
Fig. P10.43
Solution
Support reactions: A FBD of the beam is shown to the
right.
1
2
1
2
(4 kips/ft)(9 ft) 0
18 kips
(4 kips/ft)(9 ft)(11 ft) 0
198 kip-ft
y y
y
C C
C
F C
C
M M
M
Load function w(x):
0 1 1
1 2
4 kips/ft 4 kips/ft( ) 4 kips/ft 0 ft 0 ft 9 ft
9 ft 9 ft
18 kips 14 ft 198 kip-ft 14 ft
w x x x x
x x
Shear-force function V(x) and bending-moment function M(x):
1 2 2
0 1
4 kips/ft 4 kips/ft( ) 4 kips/ft 0 ft 0 ft 9 ft
2(9 ft) 2(9 ft)
18 kips 14 ft 198 kip-ft 14 ft
V x x x x
x x
2 3 3
1 0
4 kips/ft 4 kips/ft 4 kips/ft( ) 0 ft 0 ft 9 ft
2 6(9 ft) 6(9 ft)
18 kips 14 ft 198 kip-ft 14 ft
M x x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
22 3 3
2
1 0
4 kips/ft 4 kips/ft 4 kips/ft( ) 0 ft 0 ft 9 ft
2 6(9 ft) 6(9 ft)
18 kips 14 ft 198 kip-ft 14 ft
d vEI M x x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
3 4 4
2 1
1
4 kips/ft 4 kips/ft 4 kips/ft0 ft 0 ft 9 ft
6 24(9 ft) 24(9 ft)
18 kips14 ft 198 kip-ft 14 ft
2
dvEI x x x
dx
x x C (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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Integrate again to obtain the beam deflection function:
4 5 5
3 2
1 2
4 kips/ft 4 kips/ft 4 kips/ft0 ft 0 ft 9 ft
24 120(9 ft) 120(9 ft)
18 kips 198 kip-ft14 ft 14 ft
6 2
EI v x x x
x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 14 ft). Substitute the boundary
condition dv/dx = 0 at x = 14 ft into Eq. (a) to obtain:
3 4 4
1
2
1
4 kips/ft 4 kips/ft 4 kips/ft0 (14 ft) (14 ft) (5 ft)
6 24(9 ft) 24(9 ft)
1,129.5 kip-ft
C
C
Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain:
4 5 5 2
2
3
2
4 kips/ft 4 kips/ft 4 kips/ft0 (14 ft) (14 ft) (5 ft) (1,129.5 kip-ft )(14 ft)
24 120(9 ft) 120(9 ft)
11,390.7 kip-ft
C
C
The beam slope and elastic curve equations are now complete:
3 4 4
2 1 2
4 kips/ft 4 kips/ft 4 kips/ft0 ft 0 ft 9 ft
6 24(9 ft) 24(9 ft)
18 kips14 ft 198 kip-ft 14 ft 1,129.5 kip-ft
2
dvEI x x x
dx
x x
4 5 5
3 2 2 3
4 kips/ft 4 kips/ft 4 kips/ft0 ft 0 ft 9 ft
24 120(9 ft) 120(9 ft)
18 kips 198 kip-ft14 ft 14 ft (1,129.5 kip-ft ) 11,390.7 kip-ft
6 2
EI v x x x
x x x
(a) Beam slope at B: The beam slope at B (i.e., x = 9 ft) is:
3 4 2 2
2
2
4 kips/ft 4 kips/ft(9 ft) (9 ft) 1,129.5 kip-ft 765 kip-ft
6 24(9 ft)
765 kip-ft0.005752 rad
133,000 kip-ft0.00575 rad
B
B
dvEI
dx
dv
dx Ans.
(b) Beam deflection at A: The beam deflection at A is:
3
3
2
11,390.7 kip-ft
11,390.7 kip-ft0.085644 ft
133,000 k1.028 in.
ip-ft
A
A
EI v
v 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.
10.44 For the beam and loading shown, use
discontinuity functions to compute (a) the slope of
the beam at B and (b) the deflection of the beam at
C. Assume a constant value of EI = 34×106 lb-ft
2
for the beam.
Fig. P10.44
Solution
Support reactions: A FBD of the beam is shown to the
right.
1
2
1
2
(7,000 lb/ft)(9 ft)(3 ft) (14 ft) 0
6,750 lbs
(7,000 lb/ft)(9 ft) 0
24,750 lbs
B y
y
y y y
y
M D
D
F B D
B
Load function w(x):
1 0 1
1 1
7,000 lb/ft( ) 24,750 lbs 4 ft 7,000 lb/ft 4 ft 4 ft
9 ft
7,000 lb/ft13 ft 6,750 lbs 18 ft
9 ft
w x x x x
x x
Shear-force function V(x) and bending-moment function M(x):
0 1 2
2 0
7,000 lb/ft( ) 24,750 lbs 4 ft 7,000 lb/ft 4 ft 4 ft
2(9 ft)
7,000 lb/ft13 ft 6,750 lbs 18 ft
2(9 ft)
V x x x x
x x
1 2 3
3 1
7,000 lb/ft 7,000 lb/ft( ) 24,750 lbs 4 ft 4 ft 4 ft
2 6(9 ft)
7,000 lb/ft13 ft 6,750 lbs 18 ft
6(9 ft)
M x x x x
x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 2 3
2
3 1
7,000 lb/ft 7,000 lb/ft( ) 24,750 lbs 4 ft 4 ft 4 ft
2 6(9 ft)
7,000 lb/ft13 ft 6,750 lbs 18 ft
6(9 ft)
d vEI M x x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
2 3 4
4 2
1
24,750 lbs 7,000 lb/ft 7,000 lb/ft4 ft 4 ft 4 ft
2 6 24(9 ft)
7,000 lb/ft 6,750 lbs13 ft 18 ft
24(9 ft) 2
dvEI x x x
dx
x x C (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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Integrate again to obtain the beam deflection function:
3 4 5
5 3
1 2
24,750 lbs 7,000 lb/ft 7,000 lb/ft4 ft 4 ft 4 ft
6 24 120(9 ft)
7,000 lb/ft 6,750 lbs13 ft 18 ft
120(9 ft) 6
EI v x x x
x x C x C
(b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 4 ft) and at the roller support (x = 18 ft). Substitute the
boundary condition v = 0 at x = 4 ft into Eq. (b) to obtain:
1 2(4 ft) 0C C (c)
Next, substitute the boundary condition v = 0 at x = 18 ft into Eq. (b) to obtain:
3 4 5 5
1 2
3
1 2
24,750 lbs 7,000 lb/ft 7,000 lb/ft 7,000 lb/ft0 (14 ft) (14 ft) (14 ft) (5 ft)
6 24 120(9 ft) 120(9 ft)
(18 ft)
(18 ft) 3,579,975 lb-ft
C C
C C (d)
Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2:
2 3
1 2255,712.5 lb-ft and 1,022,850 lb-ftC C
The beam slope and elastic curve equations are now complete:
2 3 4
4 2 2
24,750 lbs 7,000 lb/ft 7,000 lb/ft4 ft 4 ft 4 ft
2 6 24(9 ft)
7,000 lb/ft 6,750 lbs13 ft 18 ft 255,712.5 lb-ft
24(9 ft) 2
dvEI x x x
dx
x x
3 4 5
5 3 2 3
24,750 lbs 7,000 lb/ft 7,000 lb/ft4 ft 4 ft 4 ft
6 24 120(9 ft)
7,000 lb/ft 6,750 lbs13 ft 18 ft (255,712.5 lb-ft ) 1,022,850 lb-ft
120(9 ft) 6
EI v x x x
x x x
(a) Beam slope at B: The beam slope at B is:
2
2
6 2
255,712.5 lb-ft
255,712.5 lb-ft0.0075210 rad
34 10.0
0 lb-ft0752 rad
B
B
dvEI
dx
dv
dx Ans.
(b) Beam deflection at C: At C where x = 13 ft, the beam deflection is:
3 4 5
2 3
3
6 2
24,750 lbs 7,000 lb/ft 7,000 lb/ft(9 ft) (9 ft) (9 ft)
6 24 120(9 ft)
(255,712.5 lb-ft )(13 ft) 1,022,850 lb-ft
825,187.5 kip-ft0.0242702 ft
34 10 lb-f0.291 .
tin
C
C
EI v
v 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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10.45 For the beam and loading shown, use
discontinuity functions to compute (a) the slope of
the beam at A and (b) the deflection of the beam at
B. Assume a constant value of EI = 370,000 kip-
ft2 for the beam.
Fig. P10.45
Solution
Support reactions: A FBD of the beam is shown to the
right.
1
2
1
2
1
2
(8 kips/ft)(12 ft)(8 ft)
(8 kips/ft)(12 ft)(16 ft) (24 ft) 0
48 kips
(8 kips/ft)(24 ft) 0
48 kips
A
y
y
y y y
y
M
C
C
F A C
A
Load function w(x):
1 1 1
0 0 1
1 1
1
8 kips/ft 8 kips/ft( ) 48 kips 0 ft 0 ft 12 ft
12 ft 12 ft
8 kips/ft8 kips/ft 12 ft 8 kips/ft 12 ft 12 ft
12 ft
8 kips/ft24 ft 48 kips 24 ft
12 ft
8 kips/ft48 kips 0 ft 0 f
12 ft
w x x x x
x x x
x x
x x1 1
1 1
2(8 kips/ft)t 12 ft
12 ft
8 kips/ft24 ft 48 kips 24 ft
12 ft
x
x x
Shear-force function V(x) and bending-moment function M(x):
0 2 2
2 0
8 kips/ft 2(8 kips/ft)( ) 48 kips 0 ft 0 ft 12 ft
2(12 ft) 2(12 ft)
8 kips/ft24 ft 48 kips 24 ft
2(12 ft)
V x x x x
x x
1 3 3
3 1
8 kips/ft 2(8 kips/ft)( ) 48 kips 0 ft 0 ft 12 ft
6(12 ft) 6(12 ft)
8 kips/ft24 ft 48 kips 24 ft
6(12 ft)
M x x x x
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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Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 3 3
2
3 1
8 kips/ft 2(8 kips/ft)( ) 48 kips 0 ft 0 ft 12 ft
6(12 ft) 6(12 ft)
8 kips/ft24 ft 48 kips 24 ft
6(12 ft)
d vEI M x x x x
dx
x x
Integrate the moment function to obtain an expression for the beam slope:
2 4 4
4 2
1
48 kips 8 kips/ft 2(8 kips/ft)0 ft 0 ft 12 ft
2 24(12 ft) 24(12 ft)
8 kips/ft 48 kips24 ft 24 ft
24(12 ft) 2
dvEI x x x
dx
x x C (a)
Integrate again to obtain the beam deflection function:
3 5 5
5 3
1 2
48 kips 8 kips/ft 2(8 kips/ft)0 ft 0 ft 12 ft
6 120(12 ft) 120(12 ft)
8 kips/ft 48 kips24 ft 24 ft
120(12 ft) 6
EI v x x x
x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 24 ft). Substitute the
boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain:
2 0C
Next, substitute the boundary condition v = 0 at x = 24 ft into Eq. (b) to obtain:
3 5 5
1
2
1
48 kips 8 kips/ft 2(8 kips/ft)0 (24 ft) (24 ft) (12 ft) (24 ft)
6 120(12 ft) 120(12 ft)
2,880 kip-ft
C
C
The beam slope and elastic curve equations are now complete:
2 4 4
4 2 2
48 kips 8 kips/ft 2(8 kips/ft)0 ft 0 ft 12 ft
2 24(12 ft) 24(12 ft)
8 kips/ft 48 kips24 ft 24 ft 2,880 kip-ft
24(12 ft) 2
dvEI x x x
dx
x x
3 5 5
5 3 2
48 kips 8 kips/ft 2(8 kips/ft)0 ft 0 ft 12 ft
6 120(12 ft) 120(12 ft)
8 kips/ft 48 kips24 ft 24 ft (2,880 kip-ft )
120(12 ft) 6
EI v x x x
x x x
(a) Beam slope at A: The beam slope at A is:
2
2
2
2,880 kip-ft
2,880 kip-ft0.0077838 rad
370,000 kip0.00778 rad
-ft
A
A
dvEI
dx
dv
dx 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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(b) Beam deflection at B: At B where x = 12 ft, the beam deflection is:
3 5 2 3
3 3
2
48 kips 8 kips/ft(12 ft) (12 ft) (2,880 kip-ft )(12 ft) 22,118.4 kip-ft
6 120(12 ft)
22,118.4 kip-ft kip-ft0.0597795 ft
370,000 kip0.71 n.
-f7 i
t
B
B
EI v
v 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.
10.46 For the beam and loading shown, use
discontinuity functions to compute (a) the slope of
the beam at B and (b) the deflection of the beam at
B. Assume a constant value of EI = 110,000 kN-
m2 for the beam.
Fig. P10.46
Solution
Support reactions: A FBD of the beam is shown to the
right.
1
2
1
2
(15 kN/m)(4 m) (25 kN/m)(4 m) 0
110 kN
(15 kN/m)(4 m)(2 m)
2(4 m)(25 kN/m)(4 m) 0
3
253.33 kN-m
y y
y
A
A
A
F A
C
M
M
M
Load function w(x):
2 1 0 1
0 1 0
25 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 15 kN/m 0 m 0 m
4 m
25 kN/m15 kN/m 4 m 4 m 25 kN/m 4 m
4 m
w x x x x x
x x x
Shear-force function V(x) and bending-moment function M(x):
1 0 1 2
1 2 1
25 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 15 kN/m 0 m 0 m
2(4 m)
25 kN/m15 kN/m 4 m 4 m 25 kN/m 4 m
2(4 m)
V x x x x x
x x x
0 1 2 3
2 3 2
15 kN/m 25 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 0 m 0 m
2 6(4 m)
15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m
2 6(4 m) 2
M x x x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
20 1 2
2
3 2 3 2
15 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 0 m
2
25 kN/m 15 kN/m 25 kN/m 25 kN/m0 m 4 m 4 m 4 m
6(4 m) 2 6(4 m) 2
d vEI M x x x x
dx
x x x x
Integrate the moment function to obtain an expression for the beam slope:
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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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1 2 3 4
3 4 3
1
110 kN 15 kN/m 25 kN/m253.33 kN-m 0 m 0 m 0 m 0 m
2 6 24(4 m)
15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m
6 24(4 m) 6
dvEI x x x x
dx
x x x C (a)
Integrate again to obtain the beam deflection function:
2 3 4 5
4 5 4
1 2
253.33 kN-m 110 kN 15 kN/m 25 kN/m0 m 0 m 0 m 0 m
2 6 24 120(4 m)
15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m
24 120(4 m) 24
EI v x x x x
x x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary
condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain:
1 0C
Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain:
2 0C
The beam slope and elastic curve equations are now complete:
1 2 3 4
3 4 3
110 kN 15 kN/m 25 kN/m253.33 kN-m 0 m 0 m 0 m 0 m
2 6 24(4 m)
15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m
6 24(4 m) 6
dvEI x x x x
dx
x x x
2 3 4 5
4 5 4
253.33 kN-m 110 kN 15 kN/m 25 kN/m0 m 0 m 0 m 0 m
2 6 24 120(4 m)
15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m
24 120(4 m) 24
EI v x x x x
x x x
(a) Beam slope at B: The beam slope at B is:
1 2 3 4
2
2
2
110 kN 15 kN/m 25 kN/m( 253.33 kN-m)(4 m) (4 m) (4 m) (4 m)
2 6 24(4 m)
360 kN-m
2,120 kN-m0.003273 rad
110,000 kN-m0.00327 rad
B
B
dvEI
dx
dv
dx Ans.
(b) Beam deflection at B: The beam deflection at B is:
2 3 4 5
3
3
2
253.33 kN-m 110 kN 15 kN/m 25 kN/m(4 m) (4 m) (4 m) (4 m)
2 6 24 120(4 m)
1,066.67 kN-m
1,066.67 kN-m0.009697 m
110,009.70 m
0 kN-mm
B
B
EI v
v 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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10.47 For the beam and loading shown, use
discontinuity functions to compute (a) the
deflection of the beam at A and (b) the deflection
of the beam at C. Assume a constant value of EI =
24,000 kN-m2 for the beam.
Fig. P10.47
Solution
Support reactions: A FBD of the beam is shown to the
right.
1
2
1
2
(35 kN)(2.5 m) (25 kN/m)(4.0 m)(2.0 m)
2(4.0 m)(45 kN/m)(4.0 m) (5.5 m) 0
3
64.09 kN
35 kN (25 kN/m)(4.0 m)
(45 kN/m)(4.0 m) 0
160.91 kN
B
y
y
y y y
y
M
D
D
F B D
B
Load function w(x):
1 1 0
0 1 1
0 1
( ) 35 kN 0 m 160.91 kN 2.5 m 25 kN/m 2.5 m
45 kN/m 45 kN/m25 kN/m 6.5 m 2.5 m 6.5 m
4.0 m 4.0 m
45 kN/m 6.5 m 64.09 kN 8 m
w x x x x
x x x
x x
Shear-force function V(x) and bending-moment function M(x):
0 0 1
1 2 2
1 0
( ) 35 kN 0 m 160.91 kN 2.5 m 25 kN/m 2.5 m
45 kN/m 45 kN/m25 kN/m 6.5 m 2.5 m 6.5 m
2(4.0 m) 2(4.0 m)
45 kN/m 6.5 m 64.09 kN 8 m
V x x x x
x x x
x x
1 1 2
2 3 3
2 1
1 1 2
25 kN/m( ) 35 kN 0 m 160.91 kN 2.5 m 2.5 m
2
25 kN/m 45 kN/m 45 kN/m6.5 m 2.5 m 6.5 m
2 6(4.0 m) 6(4.0 m)
45 kN/m6.5 m 64.09 kN 8 m
2
25 kN/m 4535 kN 0 m 160.91 kN 2.5 m 2.5 m
2
M x x x x
x x x
x x
x x x3
2 3 1
kN/m2.5 m
6(4.0 m)
70 kN/m 45 kN/m6.5 m 6.5 m 64.09 kN 8 m
2 6(4.0 m)
x
x 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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Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
21 1 2
2
3 2 3
1
25 kN/m( ) 35 kN 0 m 160.91 kN 2.5 m 2.5 m
2
45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m
6(4.0 m) 2 6(4.0 m)
64.09 kN 8 m
d vEI M x x x x
dx
x x x
x
Integrate the moment function to obtain an expression for the beam slope:
2 2 3
4 3 4
2
1
35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m
2 2 6
45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m
24(4.0 m) 6 24(4.0 m)
64.09 kN8 m
2
dvEI x x x
dx
x x x
x C (a)
Integrate again to obtain the beam deflection function:
3 3 4
5 4 5
3
1 2
35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m
6 6 24
45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m
120(4.0 m) 24 120(4.0 m)
64.09 kN8 m
6
EI v x x x
x x x
x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, the
deflection v is known at the pin support (x = 2.5 m) and at the roller support (x = 8 m). Substitute the
boundary condition v = 0 at x = 2.5 m into Eq. (b) to obtain:
3
1 2
3
1 2
35 kN0 (2.5 m) (2.5 m)
6
(2.5 m) 91.145833 kN-m
C C
C C (c)
Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain:
3 3 4 5
4 5
1 2
3
1 2
35 kN 160.91 kN 25 kN/m 45 kN/m0 (8 m) (5.5 m) (5.5 m) (5.5 m)
6 6 24 120(4.0 m)
70 kN/m 45 kN/m(1.5 m) (1.5 m) (8 m)
24 120(4.0 m)
(8 m) 65.666667 kN-m
C C
C C (d)
Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2:
2 3
1 228.511 kN-m and 162.424 kN-mC C
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The beam slope and elastic curve equations are now complete:
2 2 3
4 3 4
2 2
35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m
2 2 6
45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m
24(4.0 m) 6 24(4.0 m)
64.09 kN8 m 28.511 kN-m
2
dvEI x x x
dx
x x x
x
3 3 4
5 4 5
3 2 3
35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m
6 6 24
45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m
120(4.0 m) 24 120(4.0 m)
64.09 kN8 m (28.511 kN-m ) 162.424 kN-m
6
EI v x x x
x x x
x x
(a) Beam deflection at A: The beam deflection at A is:
3
3
2
162.424 kN-m
162.424 kN-m0.006768 m
24,000 kN-m6.77 mm
A
A
EI v
v Ans.
(b) Beam deflection at C: At x = 6.5 m, the beam deflection is:
3 3 4
5 2 3
3
3
2
35 kN 160.91 kN 25 kN/m(6.5 m) (4.0 m) (4.0 m)
6 6 24
45 kN/m(4.0 m) (28.511 kN-m )(6.5 m) 162.424 kN-m
120(4.0 m)
271.1797 kN-m
271.1797 kN-m0.0011299 m
24,000 k11.3 m
N-m0 m
C
C
EI v
v Ans.
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10.48 For the beam and loading shown, use
discontinuity functions to compute (a) the slope of
the beam at B and (b) the deflection of the beam at
A. Assume a constant value of EI = 54,000 kN-m2
for the beam.
Fig. P10.48
Solution
Support reactions: A FBD of the beam is shown to
the right.
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
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
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/m3 m 105 kN 4 m 240 kN-m 4 m
2
M x x x x x
x x x
Equations for beam slope and beam deflection:
From Eq. (10.1), we can write:
22 3 2
2
3 2 1 0
20 kN/m 30 kN/m 20 kN/m( ) 0 m 0 m 3 m
2 6(3 m) 2
30 kN/m 30 kN/m3 m 3 m 105 kN 4 m 240 kN-m 4 m
6(3 m) 2
d vEI M x x x x
dx
x x x x
Integrate the moment function to obtain an expression for the beam slope:
3 4 3 4
3 2 1
1
20 kN/m 30 kN/m 20 kN/m 30 kN/m0 m 0 m 3 m 3 m
6 24(3 m) 6 24(3 m)
30 kN/m 105 kN3 m 4 m 240 kN-m 4 m
6 2
dvEI x x x x
dx
x x x C (a)
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Integrate again to obtain the beam deflection function:
4 5 4 5
4 3 2
1 2
20 kN/m 30 kN/m 20 kN/m 30 kN/m0 m 0 m 3 m 3 m
24 120(3 m) 24 120(3 m)
30 kN/m 105 kN 240 kN-m3 m 4 m 4 m
24 6 2
EI v x x x x
x x x C x C (b)
Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection
v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the
slope dv/dx and the deflection v are known at the fixed support (x = 4 m). Substitute the boundary
condition dv/dx = 0 at x = 4 m into Eq. (a) to obtain:
3 4 3 4 3
1
2
1
20 kN/m 30 kN/m 20 kN/m 30 kN/m 30 kN/m0 (4 m) (4 m) (1 m) (1 m) (1 m)
6 24(3 m) 6 24(3 m) 6
311.25 kN-m
C
C
Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain:
4 5 4 5
4 2
2
3
2
20 kN/m 30 kN/m 20 kN/m 30 kN/m0 (4 m) (4 m) (1 m) (1 m)
24 120(3 m) 24 120(3 m)
30 kN/m(1 m) (311.25 kN-m )(4 m)
24
948.50 kN-m
C
C
The beam slope and elastic curve equations are now complete:
3 4 3 4
3 2 1 2
20 kN/m 30 kN/m 20 kN/m 30 kN/m0 m 0 m 3 m 3 m
6 24(3 m) 6 24(3 m)
30 kN/m 105 kN3 m 4 m 240 kN-m 4 m 311.25 kN-m
6 2
dvEI x x x x
dx
x x x
4 5 4 5
4 3 2
3
20 kN/m 30 kN/m 20 kN/m 30 kN/m0 m 0 m 3 m 3 m
24 120(3 m) 24 120(3 m)
30 kN/m 105 kN 240 kN-m3 m 4 m 4 m
24 6 2
(311.25 kN-m) 948.50 kN-m
EI v x x x x
x x x
x
(a) Beam slope at B: The beam slope at B is:
3 4 2
2
2
2
20 kN/m 30 kN/m(3 m) (3 m) 311.25 kN-m
6 24(3 m)
187.5 kN-m
187.5 kN-m0.003472 rad
54,000 kN0.00347 d
-mra
B
B
dvEI
dx
dv
dx Ans.
(b) Beam deflection at A: The beam deflection at A is:
3
3
2
948.50 kN-m
948.50 kN-m0.017565 m
54,000 kN-m17.56 mm
A
A
EI v
v Ans.
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10.49a For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 8 × 104 kN-m
2 is
constant for each beam.
Fig. P10.49a
Solution
Determine beam slope at A.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
6
A
ML
EI (slope magnitude)
Values:
M = 150 kN-m, L = 8 m, EI = 8 × 104 kN-m
2
Computation:
4 2
(150 kN-m)(8 m)0.00250 rad
6 6(8 10 kN-m )A
ML
EI
Determine beam deflection at H. [Skill 1]
(3 m)(0.00250 rad) 0.00750 7.50m mmHv Ans.
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10.49b For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 8 × 104 kN-m
2 is constant
for each beam.
Fig. P10.49b
Solution
Determine beam deflection at A. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m
2
Computation:
4 4
4 2
(6 kN/m)(4 m)0.00240 m
8 8(8 10 kN-m )A
wLv
EI
Determine beam slope at A. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
6A
wL
EI (slope magnitude)
Values:
w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m
2
Computation:
3 3
4 2
(6 kN/m)(4 m)0.00080 rad
6 6(8 10 kN-m )A
wL
EI
Determine beam deflection at H. [Skill 2]
0.00240 m (2 m)(0.00080 rad) 0.00400 m 4.00 mmHv 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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10.49c For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 8 × 104 kN-m
2 is
constant for each beam.
Fig. P10.49c
Solution
Determine beam deflection at H. [Skill 3]
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
H
Pbxv L b x
LEI (elastic curve)
Values:
P = 30 kN-m, L = 12 m, b = 4 m, x = 4 m,
EI = 8 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(30 kN)(4 m)(4 m)(12 m) (4 m) (4 m)
6(12 m)(8 10 kN-m )
0.0 9.33 mm0933 m
H
Pbxv L b x
LEI
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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10.49d For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 8 × 104 kN-m
2 is
constant for each beam.
Fig. P10.49d
Solution
Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
3
,cant3
H
PLv
EI (assuming fixed support at B)
Values:
P = 15 kN, L = 4 m, EI = 8 × 104 kN-m
2
Computation:
3 3
,cant 4 2
(15 kN)(4 m)0.004000 m
3 3(8 10 kN-m )H
PLv
EI
Determine beam slope at B. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (15 kN)(4 m) = 60 kN-m, L = 8 m,
EI = 8 × 104 kN-m
2
Computation:
4 2
(60 kN-m)(8 m)0.002000 rad
3 3(8 10 kN-m )B
ML
EI
Determine beam deflection at H. [Skill 4]
0.00400 m (4 m)(0.00200 rad) 0.01200 m 12.00 mmHv 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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10.50a For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 1.2 × 107 kip-in.
2
is constant for each beam.
Fig. P10.50a
Solution
Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated moment.]
Relevant equation from Appendix C:
2
2B
MLv
EI
Values:
M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.
2
Computation:
2 2 3
7 2
(40 kip-ft)(9 ft) (12 in./ft)0.23328 in.
2 2(1.2 10 kip-in. )B
MLv
EI
Determine beam slope at B. [Appendix C, Cantilever beam with concentrated moment.]
Relevant equation from Appendix C:
B
ML
EI (slope magnitude)
Values:
M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.
2
Computation:
2
7 2
(40 kip-ft)(9 ft)(12 in./ft)0.004320 rad
(1.2 10 kip-in. )B
ML
EI
Determine beam deflection at H. [Skill 2]
0.23328 in. (6 ft)(12 in./ft)(0.004320 rad) 0.54432 in 0.. 544 in.Hv 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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10.50b For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 1.2 × 107 kip-
in.2 is constant for each beam.
Fig. P10.50b
Solution
Determine beam slope at C. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6C
Pa L a
LEI
(slope magnitude)
Values:
P = 25 kips, L = 18 ft, a = 12 ft,
EI = 1.2 × 107 kip-in.
2
Computation:
2 2
2 2 2
7 2
( )
6
(25 kips)(12 ft)(18 ft) (12 ft) (12 in./ft) 0.00600 rad
6(18 ft)(1.2 10 kip-in. )
C
Pa L a
LEI
Determine beam deflection at H. [Skill 1]
(7 ft)(12 in./ft)(0.00600 rad) 0.5040 0.504 in.in.Hv 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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10.50c For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 1.2 × 107 kip-in.
2 is
constant for each beam.
Fig. P10.50c
Solution
Determine beam deflection at H. [Skill 3]
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 )24
H
wxv L Lx x
EI (elastic curve)
Values:
w = 2.5 kips/ft, L = 15 ft, x = 9 ft,
EI = 1.2 × 107 kip-in.
2
Computation:
2
2 2
2
2 2 3
7 2
(6 4 )24
(2.5 kips/ft)(9 ft)6(15 ft) 4(15 ft)(9 ft) (9 ft) (12 in./ft)
24(1.2 10 ki
1.0
p-in. )
1.082565 83 in.in.
H
wxv L Lx x
EI
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.
10.50d For the beams and loadings
shown below, determine the beam
deflection at point H. Assume that EI =
1.2 × 107 kip-in.
2 is constant for each
beam.
Fig. P10.50d
Solution
Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.]
Relevant equation from Appendix C:
4
,cant8
H
wLv
EI (assuming fixed support at A)
Values:
w = 5 kips/ft, L = 8 ft, EI = 1.2 × 107 kip-in.
2
Computation:
4 4 3
,cant 7 2
(5 kips/ft)(8 ft) (12 in./ft)0.36864 in.
8 8(1.2 10 kip-in. )H
wLv
EI
Determine beam slope at A. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
A
ML
EI (slope magnitude)
Values:
M = (5 kips/ft)(8 ft)(4 ft) = 160 kip-ft, L = 22 ft,
EI = 1.2 × 107 kip-in.
2
Computation:
2
7 2
(160 kip-ft)(22 ft)(12 in./ft)0.014080 rad
3 3(1.2 10 kip-in. )A
ML
EI
Determine beam deflection at H. [Skill 4]
0.36864 in. (8 ft)(12 in./ft)(0.014080 rad) 1.72032 in 1.. 720 in.Hv 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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10.51a For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 6 × 104 kN-m
2
is constant for each beam.
Fig. P10.51a
Solution
Determine beam deflection at H. [Skill 3]
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )6
H
M xv L Lx x
LEI (elastic curve)
Values:
M = −60 kN-m, L = 12 m, x = 6 m,
EI = 6 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 60 kN-m)(6 m)2(12 m) 3(12 m)(6 m) (6 m)
6(12 m)(6 10 kN-
9.
m )
0.009 0000 mmm 0
H
M xv L Lx x
LEI
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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10.51b For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 6 × 104 kN-m
2 is constant
for each beam.
Fig. P10.51b
Solution
Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.]
Relevant equation from Appendix C:
4
,cant8
H
wLv
EI (assuming fixed support at A)
Values:
w = 7.5 kN/m, L = 3 m, EI = 6 × 104 kN-m
2
Computation:
4 4
,cant 4 2
(7.5 kN/m)(3 m)0.00126563 m
8 8(6 10 kN-m )H
wLv
EI
Determine beam slope at A. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
A
ML
EI (slope magnitude)
Values:
M = (7.5 kN/m)(3 m)(1.5 m) = 33.75 kN-m,
L = 6 m, EI = 6 × 104 kN-m
2
Computation:
4 2
(33.75 kN-m)(6 m)0.001125 rad
3 3(6 10 kN-m )A
ML
EI
Determine beam deflection at H. [Skill 4]
0.00126563 m (3 m)(0.001125 rad) 0.00464063 m 4.64 mmHv 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.
10.51c For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 6 × 104 kN-m
2 is
constant for each beam.
Fig. P10.51c
Solution
Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = 30 kN, L = 3 m, EI = 6 × 104 kN-m
2
Computation:
3 3
4 2
(30 kN)(3 m)0.004500 m
3 3(6 10 kN-m )B
PLv
EI
Determine beam slope at B. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
2
2B
PL
EI (slope magnitude)
Values:
P = 30 kN, L = 3 m, EI = 6 × 104 kN-m
2
Computation:
2 2
4 2
(30 kN)(3 m)0.002250 rad
2 2(6 10 kN-m )B
PL
EI
Determine beam deflection at H. [Skill 2]
0.004500 m (3 m)(0.002250 rad) 0.01125 m 11.25 mmHv Ans.
Alternative solution for beam deflection at B. [Appendix C, Cantilever beam with concentrated load at midspan.]
Relevant equation from Appendix C: 35
48H
PLv
EI
Values: P = 30 kN, L = 6 m, EI = 6 × 104 kN-m
2
Computation: 3 3
4 2
5 5(30 kN)(6 m)0.011250 m 11.25 mm
48 48(6 10 kN-m )H
PLv
EI
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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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10.51d For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 6 × 104 kN-m
2 is
constant for each beam.
Fig. P10.51d
Solution
Determine beam slope at C. [Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
2(2 )24
C
waL a
LEI (slope magnitude)
Values:
w = 5 kN/m, L = 9 m, a = 6 m,
EI = 6 × 104 kN-m
2
Computation:
2 2
22
4 2
(5 kN/m)(6 m)(2 ) 2(9 m) (6 m) 0.00200 rad
24 24(9 m)(6 10 kN-m )C
waL a
LEI
Determine beam deflection at H. [Skill 1]
(3 m)(0.00200 rad) 0.00600 6.00m mmHv 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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10.52a For the beams and loadings shown
below, determine the beam deflection at
point H. Assume that EI = 3.0 × 106 kip-
in.2 is constant for each beam.
Fig. P10.52a
Solution
Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated moment.]
Relevant equation from Appendix C:
2
,cant2
H
MLv
EI (assuming fixed support at A)
Values:
M = 50 kip-ft, L = 6 ft, EI = 3.0 × 106 kip-in.
2
Computation:
2 2 3
,cant 6 2
(50 kip-ft)(6 ft) (12 in./ft)0.51840 in.
2 2(3.0 10 kip-in. )H
MLv
EI
Determine beam slope at A. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
A
ML
EI (slope magnitude)
Values:
M = 50 kip-ft, L = 18 ft, EI = 3.0 × 106 kip-in.
2
Computation:
2
6 2
(50 kip-ft)(18 ft)(12 in./ft)0.01440 rad
3 3(3.0 10 kip-in. )A
ML
EI
Determine beam deflection at H. [Skill 4]
0.51840 in. (6 ft)(12 in./ft)(0.01440 rad) 1.5552 in 1.. 555 in.Hv Ans.
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10.52b For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 3.0 × 106 kip-in.
2 is
constant for each beam.
Fig. P10.52b
Solution
Determine beam deflection at H. [Skill 3]
[Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
2
(3 )6
H
Pxv L x
EI (elastic curve)
Values:
P = 10 kips, L = 10 ft, x = 7 ft,
EI = 3.0 × 106 kip-in.
2
Computation:
2
2 3
6 2
(3 )6
(10 kips)(7 ft) (12 in./ft)[3(10 ft) (7 ft)] 1.081920 in.
6(3.0 101.082 in.
kip-in. )
H
Pxv L x
EI
Ans.
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10.52c For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 3.0 × 106 kip-in.
2 is
constant for each beam.
Fig. P10.52c
Solution
Determine beam slope at A. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
6
A
ML
EI (slope magnitude)
Values:
M = (2 kips/ft)(8 ft)(4 ft) = 64 kip-ft,
L = 18 ft, EI = 3.0 × 106 kip-in.
2
Computation:
2
6 2
(64 kip-ft)(18 ft)(12 in./ft)0.009216 rad
6 6(3.0 10 kip-in. )A
ML
EI
Determine beam deflection at H. [Skill 1]
(6 ft)(12 in./ft)(0.009216 rad) 0.663552 0.664 inn. . iHv Ans.
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10.52d For the beams and loadings shown
below, determine the beam deflection at point
H. Assume that EI = 3.0 × 106 kip-in.
2 is
constant for each beam.
Fig. P10.52d
Solution
Determine beam deflection at B. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.
2
Computation:
4 4 3
6 2
(1.5 kips/ft)(10 ft) (12 in./ft)1.0800 in.
8 8(3.0 10 kip-in. )B
wLv
EI
Determine beam slope at B. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
6B
wL
EI (slope magnitude)
Values:
w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.
2
Computation:
3 3 2
6 2
(1.5 kips/ft)(10 ft) (12 in./ft)0.01200 rad
6 6(3.0 10 kip-in. )B
wL
EI
Determine beam deflection at H. [Skill 2]
1.0800 in. (4 ft)(12 in./ft)(0.0120 rad) 1.6560 in 1.. 656 in.Hv Ans.
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10.53 The simply supported beam shown
in Fig. P10.53 consists of a W24 × 94
structural steel wide-flange shape [E =
29,000 ksi; I = 2,700 in.4]. For the loading
shown, determine the beam deflection at
point C.
Fig. P10.53
Solution
Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 3.2 kips/ft, L = 28 ft, a = 21 ft,
EI = 7.830 × 107 kip-in.
2
Computation:
32 2
3 32 2
7 2
(4 7 3 )24
(3.2 kips/ft)(21 ft) (12 in./ft)4(28 ft) 7(21 ft)(28 ft) 3(21 ft) 0.333822 in.
24(28 ft)(7.830 10 kip-in. )
C
wav L aL a
LEI
Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
2 2(3 4 )48
C
Pxv L x
EI (elastic curve)
Values:
P = 36 kips, L = 28 ft, x = 7 ft,
EI = 7.830 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(3 4 )48
(36 kips)(7 ft)(12 in./ft)3(28 ft) 4(7 ft) 0.249799 in.
48(7.830 10 kip-in. )
C
Pxv L x
EI
Beam deflection at C
0.333822 in. 0.249799 in. 0.583620 in. 0.584 in.Cv Ans.
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10.54 The simply supported beam shown
in Fig. P10.54 consists of a W460 × 82
structural steel wide-flange shape [E = 200
GPa; I = 370 × 106 mm
4]. For the loading
shown, determine the beam deflection at
point C.
Fig. P10.54
Solution
Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 26 kN/m, L = 8 m, a = 6 m,
EI = 7.4 × 104 kN-m
2
Computation:
32 2
32 2
4 2
(4 7 3 )24
(26 kN/m)(6 m)4(8 m) 7(6 m)(8 m) 3(6 m) 0.011068 m
24(8 m)(7.40 10 kN-m )
C
wav L aL a
LEI
Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6C
Pbxv L b x
LEI (elastic curve)
Values:
P = 60 kN, L = 8 m, b = 3 m, x = 2 m,
EI = 7.4 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(60 kN)(3 m)(2 m)(8 m) (3 m) (2 m) 0.005169 m
6(8 m)(7.40 10 kN-m )
C
Pbxv L b x
LEI
Beam deflection at C
0.011068 m 0.005169 m 0.016237 m 16.24 mmCv Ans.
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10.55 The simply supported beam shown
in Fig. P10.55 consists of a W410 × 60
structural steel wide-flange shape [E = 200
GPa; I = 216 × 106 mm
4]. For the loading
shown, determine the beam deflection at
point B.
Fig. P10.55
Solution
Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
B
Pabv L a b
LEI
Values:
P = 60 kN, L = 9 m, a = 3 m, b = 6 m,
EI = 4.32 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(60 kN)(3 m)(6 m)(9 m) (3 m) (6 m) 0.016667 m
6(9 m)(4.32 10 kN-m )
B
Pabv L a b
LEI
Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI (elastic curve)
Values:
M = −45 kN-m, L = 9 m, x = 6 m,
EI = 4.32 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 45 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.004167 m
6(9 m)(4.32 10 kN-m )
B
M xv L Lx x
LEI
Beam deflection at B
0.016667 m 0.004167 m 0.012500 m 12.50 mmBv Ans.
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10.56 The simply supported beam shown
in Fig. P10.56 consists of a W21 × 44
structural steel wide-flange shape [E =
29,000 ksi; I = 843 in.4]. For the loading
shown, determine the beam deflection at
point B.
Fig. P10.56
Solution
Consider uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
Values:
w = 5 kips/ft, L = 24 ft, a = 16 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
32 2
3 32 2
7 2
(4 7 3 )24
(5 kips/ft)(16 ft) (12 in./ft)4(24 ft) 7(16 ft)(24 ft) 3(16 ft) 0.965066 in.
24(24 ft)(2.4447 10 kip-in. )
B
wav L aL a
LEI
Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI (elastic curve)
Values:
M = −200 kip-ft, L = 24 ft, x = 8 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(2 3 )6
( 200 kip-ft)(8 ft)(12 in./ft)2(24 ft) 3(24 ft)(8 ft) (8 ft) 0.502638 in.
6(24 ft)(2.4447 10 kip-in. )
B
M xv L Lx x
LEI
Beam deflection at B
0.965066 in. 0.502638 in. 0.462428 in. 0.462 in.Bv Ans.
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10.57 The cantilever beam shown in Fig.
P10.57 consists of a rectangular
structural steel tube shape [E = 29,000
ksi; I = 476 in.4]. For the loading shown,
determine:
(a) the beam deflection at point B.
(b) the beam deflection at point C.
Fig. P10.57
Solution
(a) Beam deflection at point B
Consider uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.
2
Computation:
4 4 3
7 2
(2 kips/ft)(6 ft) (12 in./ft)0.040559 in.
8 8(1.3804 10 kip-in. )B
wLv
EI
Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
(3 )6
B
Pxv L x
EI (elastic curve)
Values:
P = 12 kips, L = 10 ft, x = 6 ft,
EI = 1.3804 × 107 kip-in.
2
Computation:
2 2 3
7 2
(12 kips)(6 ft) (12 in./ft)(3 ) 3(10 ft) (6 ft) 0.216313 in.
6 6(1.3804 10 kip-in. )B
Pxv L x
EI
Beam deflection at B
0.040559 in. 0.216313 in. 0.256872 in. 0.257 in.Bv Ans.
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(b) Beam deflection at point C
Consider uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
6B
wL
EI
Values:
w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.
2
Computation:
3 3 26
7 2
6
(2 kips/ft)(6 ft) (12 in./ft)751.0866 10 rad
6 6(1.3804 10 kip-in. )
0.040559 in. (4 ft)(12 in./ft)(751.0866 10 rad) 0.076611 in.
B
C
wL
EI
v
Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3C
PLv
EI
Values:
P = 12 kips, L = 10 ft, EI = 1.3804 × 107 kip-in.
2
Computation:
3 3 3
7 2
(12 kips)(10 ft) (12 in./ft)0.500724 in.
3 3(1.3804 10 kip-in. )C
PLv
EI
Beam deflection at C
0.076611 in. 0.500724 in. 0.577336 in. 0.577 in.Cv Ans.
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10.58 The cantilever beam shown in Fig.
P10.58 consists of a rectangular structural
steel tube shape [E = 200 GPa; I = 400 ×
106 mm
4]. For the loading shown,
determine:
(a) the beam deflection at point A.
(b) the beam deflection at point B.
Fig. P10.58
Solution
(a) Beam deflection at point A
Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = 25 kN/m, L = 4 m, EI = 8.0 × 104 kN-m
2
Computation:
4 4
4 2
(25 kN/m)(4 m)0.010000 m
8 8(8.0 10 kN-m )A
wLv
EI
Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI
Values:
P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m
2
Computation:
3 3
4 2
2 2
4 2
(55 kN)(2.5 m)0.003581 m
3 3(8.0 10 kN-m )
(55 kN)(2.5 m)0.002148 rad
2 2(8.0 10 kN-m )
0.003581 m (1.5 m)(0.002148 rad) 0.006803 m
B
B
A
PLv
EI
PL
EI
v
Beam deflection at A
0.010000 m 0.006803 m 0.016803 m 16.80 mmAv Ans.
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(b) Beam deflection at point B
Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 )24
B
wxv L Lx x
EI (elastic curve)
Values:
w = 25 kN/m, L = 4 m, x = 2.5 m,
EI = 8.0 × 104 kN-m
2
Computation:
22 2
22 2
4 2
(6 4 )24
(25 kN/m)(2.5 m)6(4.0 m) 4(4.0 m)(2.5 m) (2.5 m) 0.005066 m
24(8.0 10 kN-m )
B
wxv L Lx x
EI
Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3
3B
PLv
EI
Values:
P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m
2
Computation:
3 3
4 2
(55 kN)(2.5 m)0.003581 m
3 3(8.0 10 kN-m )B
PLv
EI
Beam deflection at B
0.005066 m 0.003581 m 0.008647 m 8.65 mmBv Ans.
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10.59 The solid 1.25-in.-diameter steel [E =
29,000 ksi] shaft shown in Fig. P10.59
supports two pulleys. For the loading
shown, determine:
(a) the shaft deflection at point B.
(b) the shaft deflection at point C.
Fig. P10.59
Solution
Section properties:
4 4(1.25 in.) 0.119842 in.64
I
(a) Shaft deflection at point B
Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.
2
Computation:
3 3
6 2
(200 lb)(10 in.)0.019182 in.
3 3(3.47543 10 lb-in. )B
PLv
EI
Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
(3 )6
B
Pxv L x
EI (elastic curve)
Values:
P = 120 lb, L = 25 in., x = 10 in.,
EI = 3.47543 × 106 lb-in.
2
Computation:
2 2
6 2
(120 lb)(10 in.)(3 ) 3(25 in.) (10 in.) 0.037405 in.
6 6(3.47543 10 lb-in. )B
Pxv L x
EI
Shaft deflection at B
0.019182 in. 0.037405 in. 0.056588 0.0566 in.in.Bv Ans.
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(b) Shaft deflection at point C
Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI (magnitude)
Values:
P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.
2
Computation:
3 3
6 2
(200 lb)(10 in.)0.019182 in.
3 3(3.47543 10 lb-in. )B
PLv
EI
2 2
6 2
(200 lb)(10 in.)0.0028773 rad
2 2(3.47543 10 lb-in. )
0.019182 in. (15 in.)(0.0028773 rad) 0.062342 in.
B
C
PL
EI
v
Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3C
PLv
EI
Values:
P = 120 lb, L = 25 in.,
EI = 3.47543 × 106 lb-in.
2
Computation:
3 3
6 2
(120 lb)(25 in.)0.179834 in.
3 3(3.47543 10 lb-in. )C
PLv
EI
Shaft deflection at C
0.062342 in. 0.179834 in. 0.242176 in. 0.242 in.Cv Ans.
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10.60 The cantilever beam shown in Fig.
P10.60 consists of a rectangular structural
steel tube shape [E = 29,000 ksi; I = 1,710
in.4]. For the loading shown, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point B.
Fig. P10.60
Solution
(a) Beam deflection at point A
Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.]
Relevant equation from Appendix C:
2
2A
MLv
EI
Values:
M = −200 kip-ft, L = 15 ft,
EI = 4.959 × 107 kip-in.
2
Computation:
2 2 3
7 2
( 200 kip-ft)(15 ft) (12 in./ft)0.784029 in.
2 2(4.959 10 kip-in. )A
MLv
EI
Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI (slope magnitude)
Values:
P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.
2
Computation:
3 3 3
7 2
2 2 2
7 2
( 18 kips)(9 ft) (12 in./ft)0.152415 in.
3 3(4.959 10 kip-in. )
(18 kips)(9 ft) (12 in./ft)0.0021169 rad
2 2(4.959 10 kip-in. )
0.152415 in. (6 ft)(12 in./ft)(0.0021169 rad) 0.3
B
B
A
PLv
EI
PL
EI
v
04830 in.
Beam deflection at A
0.784029 in. 0.304830 in. 1.088860 1.in. 089 in.Av 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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(b) Beam deflection at point B
Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.]
Relevant equation from Appendix C:
2
2B
Mxv
EI (elastic curve)
Values:
M = −200 kip-ft, L = 15 ft, x = 9 ft,
EI = 4.959 × 107 kip-in.
2
Computation:
2 2 3
7 2
( 200 kip-ft)(9 ft) (12 in./ft)0.282250 in.
2 2(4.959 10 kip-in. )B
Mxv
EI
Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −18 kips, L = 9 ft,
EI = 4.959 × 107 kip-in.
2
Computation:
3 3 3
7 2
( 18 kips)(9 ft) (12 in./ft)0.152415 in.
3 3(4.959 10 kip-in. )B
PLv
EI
Beam deflection at B
0.282250 in. 0.152415 in. 0.434665 0.in. 435 in.Bv Ans.
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10.61 The simply supported beam shown in
Fig. P10.61 consists of a W21 × 44
structural steel wide-flange shape [E =
29,000 ksi; I = 843 in.4]. For the loading
shown, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point C.
Fig. P10.61
Solution
(a) Beam deflection at point A
Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.]
Relevant equation from Appendix C:
4
8A
wLv
EI (assuming fixed support at B)
Values:
w = 4 kips/ft, L = 8 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 3
7 2
(4 kips/ft)(8 ft) (12 in./ft)0.144760 in.
8 8(2.4447 10 kip-in. )A
wLv
EI
Consider deflection at A resulting from rotation at B caused by distributed load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (4 kips/ft)(8 ft)(4 ft) = 128 kip-ft,
L = 22 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2
7 2
(128 kip-ft)(22 ft)(12 in./ft)0.0055290 rad
3 3(2.4447 10 kip-in. )
(8 ft)(12 in./ft)(0.0055290 rad) 0.530786 in.
B
A
ML
EI
v
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Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
2
16B
PL
EI (slope magnitude)
Values:
P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2 2
7 2
(45 kips)(22 ft) (12 in./ft)0.0080182 rad
16 16(2.4447 10 kip-in. )
(8 ft)(12 in./ft)(0.0080182 rad) 0.769744 in.
B
A
PL
EI
v
Beam deflection at A
0.144760 in. 0.530786 in. 0.769744 in. 0.094198 in. 0.0942 in.Av Ans.
(b) Beam deflection at point C
Consider distributed load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = (4 kips/ft)(8 ft)(4 ft) = −128 kip-ft,
L = 22 ft, x = 11 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(2 3 )6
( 128 kip-ft)(11 ft)(12 in./ft)2(22 ft) 3(22 ft)(11 ft) (11 ft) 0.273687 in.
6(22 ft)(2.4447 10 kip-in. )
C
M xv L Lx x
LEI
Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48C
PLv
EI
Values:
P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
3 3 3
7 2
(45 kips)(22 ft) (12 in./ft)0.705598 in.
48 48(2.4447 10 kip-in. )C
PLv
EI
Beam deflection at C
0.273687 in. 0.705598 in. 0.431912 0.432 in.in.Cv Ans.
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10.62 The simply supported beam shown
in Fig. P10.62 consists of a W530 × 66
structural steel wide-flange shape [E =
200 GPa; I = 351 × 106 mm
4]. For the
loading shown, determine:
(a) the beam deflection at point B.
(b) the beam deflection at point D. Fig. P10.62
Solution
(a) Beam deflection at point B
Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
45
384B
wLv
EI
Values:
w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m
2
Computation:
4 4
4 2
5 5(55 kN/m)(7.2 m)0.027415 m
384 384(7.02 10 kN-m )B
wLv
EI
Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI (elastic curve)
Values:
M = (55 kN/m)(2.8 m)(1.4 m) = −215.6 kN-m,
L = 7.2 m, x = 3.6 m, EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
7 2
(2 3 )6
( 215.6 kN-m)(3.6 m)2(7.2 m) 3(7.2 m)(3.6 m) (3.6 m) 0.009951 m
6(7.2 m)(7.02 10 kN-m )
B
M xv L Lx x
LEI
Beam deflection at B
0.027415 m 0.009951 m 0.017464 m 17.46 mmBv Ans.
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(b) Beam deflection at point D
Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
24C
wL
EI (slope magnitude)
Values:
w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m
2
Computation:
3 3
4 2
(55 kN/m)(7.2 m)0.0121846 rad
24 24(7.02 10 kN-m )
(2.8 m)(0.0121846 rad) 0.034117 m
C
D
wL
EI
v
Consider deflection at D resulting from rotation at C caused by distributed load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
C
ML
EI (slope magnitude)
Values:
M = (55 kN/m)(2.8 m)(1.4 m) = 215.6 kN-m,
L = 7.2 m, EI = 7.02 × 104 kN-m
2
Computation:
7 2
(215.6 kN-m)(7.2 m)0.0073709 rad
3 3(7.02 10 kN-m )
(2.8 m)(0.0073709 rad) 0.020639 m
C
D
ML
EI
v
Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.]
Relevant equation from Appendix C:
4
8D
wLv
EI (assuming fixed support at C)
Values:
w = 55 kN/m, L = 2.8 m, EI = 7.02 × 104 kN-m
2
Computation:
4 4
4 2
(55 kN/m)(2.8 m)0.006020 m
8 8(7.02 10 kN-m )D
wLv
EI
Beam deflection at D
0.034117 m 0.020639 m 0.006020 m 0.007459 7m .46 mmDv Ans.
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10.63 The simply supported beam shown in Fig.
P10.63 consists of a W21 × 44 structural steel
wide-flange shape [E = 29,000 ksi; I = 843 in.4].
For a loading of w = 6 kips/ft, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point C. Fig. P10.63
Solution
(a) Beam deflection at point A
Determine cantilever deflection due to linearly distributed load on overhang. [Appendix C, Cantilever beam with linear load.]
Relevant equation from Appendix C:
4
0
30A
w Lv
EI (assuming fixed support at B)
Values:
w0 = −6 kips/ft, L = 12 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 3
0
7 2
( 6 kips/ft)(12 ft) (12 in./ft)0.293139 in.
30 30(2.4447 10 kip-in. )A
w Lv
EI
Consider deflection at A resulting from rotation at B caused by linear load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft,
L = 18 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2
7 2
(144 kip-ft)(18 ft)(12 in./ft)0.0050892 rad
3 3(2.4447 10 kip-in. )
(12 ft)(12 in./ft)(0.0050892 rad) 0.732847 in.
B
A
ML
EI
v
Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
24B
wL
EI (slope magnitude)
Values:
w = 6 kips/ft, L = 18 ft,
EI = 2.4447 × 107 kip-in.
2
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Computation:
3 3 2
7 2
(6 kips/ft)(18 ft) (12 in./ft)0.0085880 rad
24 24(2.4447 10 kip-in. )
(12 ft)(12 in./ft)(0.0085880 rad) 1.236679 in.
B
A
wL
EI
v
Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
6
B
ML
EI (slope magnitude)
Values:
M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft,
L = 18 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2
7 2
(108 kip-ft)(18 ft)(12 in./ft)0.0019085 rad
6 6(2.4447 10 kip-in. )
(12 ft)(12 in./ft)(0.0019085 rad) 0.274818 in.
B
A
ML
EI
v
Beam deflection at A
0.293139 in. 0.732847 in. 1.236679 in. 0.274818 in. 0.064124 in 0.0641 in. .Av Ans.
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(b) Beam deflection at point C
Consider deflection at C from moment caused by linear load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft,
L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(2 3 )6
(144 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.206112 in.
6(18 ft)(2.4447 10 kip-in. )
C
M xv L Lx x
LEI
Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
45
384C
wLv
EI
Values:
w = −6 kips/ft, L = 18 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 2
7 2
5 5( 6 kips/ft)(18 ft) (12 in./ft)0.579693 in.
384 384(2.4447 10 kip-in. )C
wLv
EI
Consider deflection at C resulting from moment caused by uniform load on overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = (6 kips/ft)(6 ft)(3 ft) = 108 kip-ft,
L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(2 3 )6
(108 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.154585 in.
6(18 ft)(2.4447 10 kip-in. )
C
M xv L Lx x
LEI
Beam deflection at C
0.206112 in. 0.579693 in. 0.154585 in. 0.218995 in 0.219 in. .Cv Ans.
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10.64 The simply supported beam shown in Fig.
P10.64 consists of a W21 × 44 structural steel
wide-flange shape [E = 29,000 ksi; I = 843 in.4].
For a loading of w = 8 kips/ft, determine:
(a) the beam deflection at point C.
(b) the beam deflection at point E. Fig. P10.64
Solution
(a) Beam deflection at point C
Consider deflection at C from moment caused by linear load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = ½(8 kips/ft)(12 ft)(4 ft) = 192 kip-ft,
L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2
3
2 2
7 2
(2 3 )6
(192 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.274816 in.
6(18 ft)(2.4447 10 kip-in. )
C
M xv L Lx x
LEI
Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
45
384C
wLv
EI
Values:
w = −8 kips/ft, L = 18 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 2
7 2
5 5( 8 kips/ft)(18 ft) (12 in./ft)0.772924 in.
384 384(2.4447 10 kip-in. )C
wLv
EI
Consider deflection at C resulting from moment caused by uniform load on overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = (8 kips/ft)(6 ft)(3 ft) = 144 kip-ft,
L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
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2 2
3
2 2
7 2
(2 3 )6
(144 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.206113 in.
6(18 ft)(2.4447 10 kip-in. )
C
M xv L Lx x
LEI
Beam deflection at C
0.274816 in. 0.772924 in. 0.206113 in. 0.291995 in 0.292 in. .Cv Ans.
(b) Beam deflection at point E
Consider deflection at E resulting from rotation at D caused by linear load on overhang.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
6
D
ML
EI (slope magnitude)
Values:
M = ½(8 kips/ft)(12 ft)(4 ft) = 192 kip-ft,
L = 18 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2
7 2
(192 kip-ft)(18 ft)(12 in./ft)0.0033928 rad
6 6(2.4447 10 kip-in. )
(6 ft)(12 in./ft)(0.0033928 rad) 0.244282 in.
D
E
ML
EI
v
Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
24D
wL
EI (slope magnitude)
Values:
w = 8 kips/ft, L = 18 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
3 3 2
7 2
(8 kips/ft)(18 ft) (12 in./ft)0.0114507 rad
24 24(2.4447 10 kip-in. )
(6 ft)(12 in./ft)(0.0114507 rad) 0.824448 in.
D
E
wL
EI
v
Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
D
ML
EI (slope magnitude)
Values:
M = (8 kips/ft)(6 ft)(3 ft) = 144 kip-ft,
L = 18 ft, EI = 2.4447 × 107 kip-in.
2
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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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Computation:
2
7 2
(144 kip-ft)(18 ft)(12 in./ft)0.0050892 rad
3 3(2.4447 10 kip-in. )
(6 ft)(12 in./ft)(0.0050892 rad) 0.366422 in.
D
E
ML
EI
v
Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8E
wLv
EI (assuming fixed support at D)
Values:
w = −8 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 3
7 2
( 8 kips/ft)(6 ft) (12 in./ft)0.091605 in.
8 8(2.4447 10 kip-in. )E
wLv
EI
Beam deflection at E
0.244282 in. 0.824448 in. 0.366422 in. 0.091605 in.
0.122139 i 0n. .1221 in.
Ev
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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10.65 The solid 30-mm-diameter steel [E = 200
GPa] shaft shown in Fig. P10.65 supports two
belt pulleys. Assume that the bearing at B can
be idealized as a roller support and that the
bearing at D can be idealized as a pin support.
For the loading shown, determine:
(a) the shaft deflection at pulley A.
(b) the shaft deflection at pulley C.
Fig. P10.65
Solution
Section properties:
4 4(30 mm) 39,760.78 mm64
I
(a) Shaft deflection at pulley A
Determine cantilever deflection due to pulley A load. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
3
3A
PLv
EI (assuming fixed support at B)
Values:
P = 700 N, L = 500 mm,
EI = 7.95216 × 109 N-mm
2
Computation:
3 3
9 2
(700 N)(500 mm)3.6678 mm
3 3(7.95216 10 N-mm )A
PLv
EI
Consider deflection at A resulting from rotation at B caused by pulley A load.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (700 N)(500 mm) = 350,000 N-mm,
L = 1,800 mm, EI = 7.95216 × 109 N-mm
2
Computation:
9 2
(350,000 N-mm)(1,800 mm)0.0264079 rad
3 3(7.95216 10 N-mm )
(500 mm)(0.0264079 rad) 13.2040 mm
B
A
ML
EI
v
Consider deflection at A resulting from rotation at B caused by pulley C load.
[Appendix C, SS beam with concentrated load.]
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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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Relevant equation from Appendix C:
2
16B
PL
EI (slope magnitude)
Values:
P = 1,000 N, L = 1,800 mm,
EI = 7.95216 × 109 N-mm
2
Computation:
2 2
9 2
(1,000 N)(1,800 mm)0.0254648 rad
16 16(7.95216 10 N-mm )
(500 mm)(0.0254648 rad) 12.7324 mm
B
A
PL
EI
v
Shaft deflection at A
3.6678 mm 13.2040 mm 12.7324 mm 4.1393 mm 4.14 mmAv Ans.
(b) Shaft deflection at pulley C
Consider pulley A load. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = (700 N)(500 mm) = −350,000 N-mm,
L = 1,800 mm, x = 900 mm,
EI = 7.95216 × 109 N-mm
2
Computation:
2 2
2 2
9 2
(2 3 )6
( 350,000 N-mm)(900 mm)2(1,800 mm) 3(1,800 mm)(900 mm) (900 mm)
6(1,800 mm)(7.95216 10 N-mm )
8.9127 mm
C
M xv L Lx x
LEI
Consider pulley C load. [Appendix C, SS beam with concentrated load.]
Relevant equation from Appendix C:
3
48C
PLv
EI
Values:
P = 1,000 N, L = 1,800 mm,
EI = 7.95216 × 109 N-mm
2
Computation:
3 3
9 2
(1,000 N)(1,800 mm)15.2789 mm
48 48(7.95216 10 N-mm )C
PLv
EI
Shaft deflection at C
8.9127 mm 15.2789 mm 6.3662 mm 6.37 mmCv 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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10.66 The cantilever beam shown in Fig. P10.66
consists of a W530 × 92 structural steel wide-
flange shape [E = 200 GPa; I = 552 × 106 mm
4].
For the loading shown, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point B.
Fig. P10.66
Solution
(a) Beam deflection at point A
Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = −85 kN/m, L = 4 m, EI = 1.104 × 105 kN-m
2
Computation:
4 4
5 2
( 85 kN/m)(4 m)0.024638 m
8 8(1.104 10 kN-m )A
wLv
EI
Consider a downward 85 kN/m uniformly distributed load acting over span BC.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4 3
and8 6
B B
wL wLv
EI EI (magnitude)
Values:
w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m
2
Computation:
4 4
5 2
3 3
5 2
(85 kN)(2.5 m)0.003759 m
8 8(1.104 10 kN-m )
(85 kN)(2.5 m)0.0020050 rad
6 6(1.104 10 kN-m )
0.003759 m (1.5 m)(0.0020050 rad) 0.006767 m
B
B
A
wLv
EI
wL
EI
v
Beam deflection at A
0.024638 m 0.006767 m 0.017871 17 m .87 mmAv 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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(b) Beam deflection at point B
Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 )24
B
wxv L Lx x
EI (elastic curve)
Values:
w = −85 kN/m, L = 4 m, x = 2.5 m,
EI = 1.104 × 105 kN-m
2
Computation:
22 2
22 2
5 2
(6 4 )24
( 85 kN/m)(2.5 m)6(4 m) 4(4 m)(2.5 m) (2.5 m) 0.012481 m
24(1.104 10 kN-m )
B
wxv L Lx x
EI
Consider a downward 85 kN/m uniformly distributed load acting over span BC.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m
2
Computation:
4 4
5 2
(85 kN)(2.5 m)0.003759 m
8 8(1.104 10 kN-m )B
wLv
EI
Beam deflection at B
0.012481 m 0.003759 m 0.008722 m 8.72 mmBv Ans.
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10.67 The solid 30-mm-diameter steel [E
= 200 GPa] shaft shown in Fig. P10.67
supports two belt pulleys. Assume that
the bearing at A can be idealized as a pin
support and that the bearing at E can be
idealized as a roller support. For the
loading shown, determine the shaft
deflection at pulley B.
Fig. P10.67
Solution
Section properties:
4 4(30 mm) 39,760.78 mm64
I
Shaft deflection at pulley B
Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6B
Pabv L a b
LEI
Values:
P = 750 N, L = 1,000 mm, a = 300 mm,
b = 700 mm, EI = 7.95216 × 109 N-mm
2
Computation:
2 2 2
2 2 2
9 2
( )6
(750 N)(300 mm)(700 mm)(1,000 mm) (300 mm) (700 mm)
6(1,000 mm)(7.95216 10 N-mm )
1.38642 mm
B
Pabv L a b
LEI
Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6B
Pbxv L b x
LEI (elastic curve)
Values:
P = 500 N, L = 1,000 mm, x = 300 mm,
b = 200 mm, EI = 7.95216 × 109 N-mm
2
Computation:
2 2 2
2 2 2
9 2
( )6
(500 N)(200 mm)(300 mm)(1,000 mm) (200 mm) (300 mm)
6(1,000 mm)(7.95216 10 N-mm )
0.54702 mm
B
Pbxv L b x
LEI
Shaft deflection at B
1.38642 mm 0.54702 mm 1.93344 mm 1.933 mmBv Ans.
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10.68 The solid 30-mm-diameter steel [E
= 200 GPa] shaft shown in Fig. P10.68
supports two belt pulleys. Assume that
the bearing at A can be idealized as a pin
support and that the bearing at E can be
idealized as a roller support. For the
loading shown, determine the shaft
deflection at pulley D.
Fig. P10.68
Solution
Section properties:
4 4(30 mm) 39,760.78 mm64
I
Shaft deflection at pulley D
Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
D
Pbxv L b x
LEI (elastic curve)
Values:
P = 750 N, L = 1,000 mm, x = 200 mm,
b = 300 mm, EI = 7.95216 × 109 N-mm
2
Computation:
2 2 2
2 2 2
9 2
( )6
(750 N)(300 mm)(200 mm)(1,000 mm) (300 mm) (200 mm)
6(1,000 mm)(7.95216 10 N-mm )
0.82053 mm
D
Pbxv L b x
LEI
Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6D
Pabv L a b
LEI
Values:
P = 500 N, L = 1,000 mm, a = 800 mm,
b = 200 mm, EI = 7.95216 × 109 N-mm
2
Computation:
2 2 2
2 2 2
9 2
( )6
(500 N)(800 mm)(200 mm)(1,000 mm) (800 mm) (200 mm)
6(1,000 mm)(7.95216 10 N-mm )
0.53654 mm
D
Pabv L a b
LEI
Shaft deflection at D
0.82053 mm 0.53654 mm 1.35707 mm 1.357 mmDv Ans.
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10.69 The simply supported beam shown in Fig.
P10.69 consists of a W410 × 60 structural steel
wide-flange shape [E = 200 GPa; I = 216 × 106
mm4]. For the loading shown, determine the
beam deflection at point B.
Fig. P10.69
Solution
Beam deflection at point B Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
B
M xv L Lx x
LEI (elastic curve)
Values:
M = −180 kN-m, L = 6 m, x = 1.5 m,
EI = 4.32 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 180 kN-m)(1.5 m)2(6 m) 3(6 m)(1.5 m) (1.5 m) 0.008203 m
6(6 m)(4.32 10 kN-m )
B
M xv L Lx x
LEI
Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6B
Pabv L a b
LEI
Values:
P = 70 kN, L = 6 m, a = 1.5 m, b = 4.5 m,
EI = 4.32 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(70 kN)(1.5 m)(4.5 m)(6 m) (1.5 m) (4.5 m) 0.004102 m
6(6 m)(4.32 10 kN-m )
B
Pabv L a b
LEI
Consider uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
3 2 2 2 2(2 6 4 )24
B
wav x Lx a x L x a L
LEI
Values:
w = 80 kN/m, L = 6 m, a = 3 m, x = 4.5 m,
EI = 4.32 × 104 kN-m
2
Computation:
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23 2 2 2 2
23 2 2 2 2
4 2
(2 6 4 )24
(80 kN/m)(3 m)2(4.5) 6(6)(4.5) (3) (4.5) 4(6) (4.5) (3) (6)
24(6.0 m)(4.32 10 kN-m )
0.010156 m
B
wav x Lx a x L x a L
LEI
Beam deflection at B
0.008203 m 0.004102 m 0.010156 m 0.006055 m 6.06 mmBv Ans.
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10.70 The simply supported beam shown in Fig.
P10.70 consists of a W410 × 60 structural steel
wide-flange shape [E = 200 GPa; I = 216 × 106
mm4]. For the loading shown, determine the
beam deflection at point C.
Fig. P10.70
Solution
Beam deflection at point C Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = −180 kN-m, L = 6 m, x = 3.0 m,
EI = 4.32 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 180 kN-m)(3.0 m)2(6 m) 3(6 m)(3.0 m) (3.0 m) 0.009375 m
6(6 m)(4.32 10 kN-m )
C
M xv L Lx x
LEI
Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6C
Pbxv L b x
LEI (elastic curve)
Values:
P = 70 kN, L = 6 m, x = 3.0 m, b = 1.5 m,
EI = 4.32 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(70 kN)(1.5 m)(3.0 m)(6 m) (1.5 m) (3.0 m) 0.005013 m
6(6 m)(4.32 10 kN-m )
C
Pbxv L b x
LEI
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Consider uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 80 kN/m, L = 6 m, a = 3 m,
EI = 4.32 × 104 kN-m
2
Computation:
32 2
32 2
4 2
(4 7 3 )24
(80 kN/m)(3 m)4(6 m) 7(3 m)(6 m) 3(3 m) 0.015625 m
24(6.0 m)(4.32 10 kN-m )
C
wav L aL a
LEI
Beam deflection at C
0.009375 m 0.005013 m 0.015625 m 0.011263 11.26 mm mCv 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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10.71 The simply supported beam shown in Fig.
P10.71 consists of a W530 × 66 structural steel
wide-flange shape [E = 200 GPa; I = 351 × 106
mm4]. If w = 80 kN/m, determine
(a) the beam deflection at point A.
(b) the beam deflection at point C.
Fig. P10.71
Solution
(a) Beam deflection at point A
Determine cantilever deflection due to concentrated load on overhang AB. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
3
3A
PLv
EI (assuming fixed support at B)
Values:
P = 35 kN, L = 4 m, EI = 7.02 × 104 kN-m
2
Computation:
3 3
4 2
(35 kN)(4 m)0.0106363 m
3 3(7.02 10 kN-m )A
PLv
EI
Consider deflection at A resulting from rotation at B caused by concentrated load on overhang
AB. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (35 kN)(4 m) = 140 kN-m, L = 8 m,
EI = 7.02 × 104 kN-m
2
Computation:
4 2
(140 kN-m)(8 m)0.0053181 rad
3 3(7.02 10 kN-m )
(4 m)(0.0053181 rad) 0.0212726 m
B
A
ML
EI
v
Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equations from Appendix C:
2
2 2(2 )24
B
waL a
LEI (slope magnitude)
Values:
w = 80 kN/m, L = 8 m, a = 4 m,
EI = 7.02 × 104 kN-m
2
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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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Computation:
2 2
2 2 2 2
4 2
(80 kN/m)(4 m)(2 ) 2(8 m) (4 m) 0.0106363 rad
24 24(8 m)(7.02 10 kN-m )
(4 m)(0.0106363 rad) 0.0425451 m
B
A
waL a
LEI
v
Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
6
B
ML
EI (slope magnitude)
Values:
M = (80 kN/m)(2 m)(1 m) = 160 kN-m,
L = 8 m, EI = 7.02 × 104 kN-m
2
Computation:
4 2
(160 kN-m)(8 m)0.0030389 rad
6 6(7.02 10 kN-m )
(4 m)(0.0030389 rad) 0.0121557 m
B
A
ML
EI
v
Beam deflection at A
0.0106363 m 0.0212726 m 0.0425451 m 0.0121557 m
0.0015195 m 1.520 mm
Av
Ans.
(b) Beam deflection at point C
Consider concentrated moment from overhang AB.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI(elastic curve)
Values:
M = (35 kN)(4 m) = −140 kN-m, L = 8 m,
x = 4 m, EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 140 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0079772 m
6(8 m)(7.02 10 kN-m )
C
M xv L Lx x
LEI
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Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equations from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 80 kN/m, L = 8 m, a = 4 m,
EI = 7.02 × 104 kN-m
2
Computation:
32 2
32 2
4 2
(4 7 3 )24
(80 kN/m)(4 m)4(8 m) 7(4 m)(8 m) 3(4 m) 0.0303894 m
24(8 m)(7.02 10 kN-m )
C
wav L aL a
LEI
Consider concentrated moment from overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = (80 kN/m)(2 m)(1 m) = 160 kN-m,
L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 160 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0091168 m
6(8 m)(7.02 10 kN-m )
C
M xv L Lx x
LEI
Beam deflection at C
0.0079772 m 0.0303894 m 0.0091168 m 0.0132954 1m 3.30 mmCv Ans.
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10.72 The simply supported beam shown in Fig.
P10.72 consists of a W530 × 66 structural steel
wide-flange shape [E = 200 GPa; I = 351 × 106
mm4]. If w = 90 kN/m, determine:
(a) the beam deflection at point C.
(b) the beam deflection at point E.
Fig. P10.72
Solution
(a) Beam deflection at point C
Consider concentrated moment from overhang AB.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI(elastic curve)
Values:
M = (35 kN)(4 m) = −140 kN-m, L = 8 m,
x = 4 m, EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 140 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0079772 m
6(8 m)(7.02 10 kN-m )
C
M xv L Lx x
LEI
Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equations from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 90 kN/m, L = 8 m, a = 4 m,
EI = 7.02 × 104 kN-m
2
Computation:
32 2
3
2 2
4 2
(4 7 3 )24
(90 kN/m)(4 m)4(8 m) 7(4 m)(8 m) 3(4 m) 0.0341881 m
24(8 m)(7.02 10 kN-m )
C
wav L aL a
LEI
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Consider concentrated moment from overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = (90 kN/m)(2 m)(1 m) = 180 kN-m,
L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 180 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0102564 m
6(8 m)(7.02 10 kN-m )
C
M xv L Lx x
LEI
Beam deflection at C
0.0079772 m 0.0341881 m 0.0102564 m 0.0159545 1m 5.95 mmCv Ans.
(b) Beam deflection at point E
Consider deflection at E resulting from rotation at D caused by concentrated load on overhang
AB. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
6
D
ML
EI (slope magnitude)
Values:
M = (35 kN)(4 m) = 140 kN-m, L = 8 m,
EI = 7.02 × 104 kN-m
2
Computation:
4 2
(140 kN-m)(8 m)0.0026591 rad
6 6(7.02 10 kN-m )
(2 m)(0.0026591 rad) 0.0053181 m
D
E
ML
EI
v
Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equations from Appendix C:
2
2(2 )24
D
waL a
LEI (slope magnitude)
Values:
w = 90 kN/m, L = 8 m, a = 4 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2
22
4 2
(90 kN/m)(4 m)(2 ) 2(8 m) (4 m) 0.0153846 rad
24 24(8 m)(7.02 10 kN-m )
(2 m)(0.0153846 rad) 0.0307692 m
D
E
waL a
LEI
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
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Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
D
ML
EI (slope magnitude)
Values:
M = (90 kN/m)(2 m)(1 m) = 180 kN-m,
L = 8 m, EI = 7.02 × 104 kN-m
2
Computation:
4 2
(180 kN-m)(8 m)0.0068376 rad
3 3(7.02 10 kN-m )
(2 m)(0.0068376 rad) 0.0136753 m
D
E
ML
EI
v
Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with distributed load.]
Relevant equation from Appendix C:
4
8E
wLv
EI (assuming fixed support at D)
Values:
w = 90 kN/m, L = 2 m, EI = 7.02 × 104 kN-m
2
Computation: 4 4
4 2
(90 kN/m)(2 m)0.0025641 m
8 8(7.02 10 kN-m )E
wLv
EI
Beam deflection at E
0.0053181 m 0.0307692 m 0.0136753 m 0.0025641 m
0.0092117 9.21 mmm
Ev
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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10.73 The simply supported beam shown in
Fig. P10.73 consists of a W16 × 40
structural steel wide-flange shape [E =
29,000 ksi; I = 518 in.4]. For the loading
shown, determine:
(a) the beam deflection at point C.
(b) the beam deflection at point F.
Fig. P10.73
Solution
(a) Beam deflection at point C
Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
C
Pbxv L b x
LEI (elastic curve)
Values:
P = 40 kips, L = 18 ft, b = 4 ft, x = 10 ft,
EI = 1.5022 × 107 kip-in.
2
Computation:
2 2 2
32 2 2
7 2
( )6
(40 kips)(4 ft)(10 ft)(12 in./ft)(18 ft) (4 ft) (10 ft) 0.354467 in.
6(18 ft)(1.5022 10 kip-in. )
C
Pbxv L b x
LEI
Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6C
Pbxv L b x
LEI (elastic curve)
Values:
P = 30 kips, L = 18 ft, b = 6 ft, x = 8 ft,
EI = 1.5022 × 107 kip-in.
2
Computation:
2 2 2
32 2 2
7 2
( )6
(30 kips)(6 ft)(8 ft)(12 in./ft)(18 ft) (6 ft) (8 ft) 0.343560 in.
6(18 ft)(1.5022 10 kip-in. )
C
Pbxv L b x
LEI
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Consider 20-kip concentrated load at F. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft,
x = 10 ft, EI = 1.5022 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(2 3 )6
( 120 kip-ft)(10 ft)(12 in./ft)2(18 ft) 3(18 ft)(10 ft) (10 ft) 0.265850 in.
6(18 ft)(1.5022 10 kip-in. )
C
M xv L Lx x
LEI
Beam deflection at C
0.354467 in. 0.343560 in. 0.265850 in. 0.432177 in. 0.432 in.Cv Ans.
(b) Beam deflection at point F
Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6E
Pa L a
LEI (slope magnitude)
Values:
P = 40 kips, L = 18 ft, a = 4 ft,
EI = 1.5022 × 107 kip-in.
2
Computation:
2 2 22 2
7 2
( ) (40 kips)(4 ft)(12 in./ft)(18 ft) (4 ft) 0.0043740 rad
6 6(18 ft)(1.5022 10 kip-in. )
(6 ft)(12 in./ft)(0.0043740 rad) 0.314930 in.
E
F
Pa L a
LEI
v
Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6E
Pa L a
LEI (slope magnitude)
Values:
P = 30 kips, L = 18 ft, x = 8 ft, a = 12 ft,
EI = 1.5022 × 107 kip-in.
2
Computation:
2 2 22 2
7 2
( ) (30 kips)(12 ft)(12 in./ft)(18 ft) (12 ft) 0.0057516 rad
6 6(18 ft)(1.5022 10 kip-in. )
(6 ft)(12 in./ft)(0.0057516 rad) 0.414113 in.
E
F
Pa L a
LEI
v
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permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Consider deflection at F resulting from rotation at E caused by 20-kip load on overhang EF.
[Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
E
ML
EI (slope magnitude)
Values:
M = (20 kips)(6 ft) = 120 kip-ft, L = 18 ft,
EI = 1.5022 × 107 kip-in.
2
Computation:
2
7 2
(120 kip-ft)(18 ft)(12 in./ft)0.0069019 rad
3 3(1.5022 10 kip-in. )
(6 ft)(12 in./ft)(0.0069019 rad) 0.496935 in.
E
F
ML
EI
v
Determine cantilever deflection due to concentrated load on overhang EF. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
3
3F
PLv
EI (assuming fixed support at E)
Values:
P = 20 kips, L = 6 ft, EI = 1.5022 × 107 kip-in.
2
Computation: 3 3 3
7 2
(20 kips)(6 ft) (12 in./ft)0.165645 in.
3 3(1.5022 10 kip-in. )F
PLv
EI
Beam deflection at F
0.314930 in. 0.414113 in. 0.496935 in. 0.165645 in.
0.066463 in 0.0665 in. .
Fv
Ans.
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10.74 The cantilever beam shown in Fig. P10.74
consists of a rectangular structural steel tube
shape [E = 200 GPa; I = 170 × 106 mm
4]. For
the loading shown, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point B.
Fig. P10.74
Solution
(a) Beam deflection at point A
Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = −65 kN/m, L = 6 m, EI = 3.4 × 104 kN-m
2
Computation:
4 4
4 2
( 65 kN/m)(6 m)0.309706 m
8 8(3.4 10 kN-m )A
wLv
EI
Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3
3A
PLv
EI
Values:
P = 90 kN, L = 6 m, EI = 3.4 × 104 kN-m
2
Computation:
3 3
4 2
(90 kN)(6 m)0.190588 m
3 3(3.4 10 kN-m )A
PLv
EI
Consider 30-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI (magnitude)
Values:
P = 30 kN, L = 3.5 m, EI = 3.4 × 104 kN-m
2
Computation:
3 3
4 2
(30 kN)(3.5 m)0.012610 m
3 3(3.4 10 kN-m )B
PLv
EI (a)
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2 2
4 2
(30 kN)(3.5 m)0.0054044 rad
2 2(3.4 10 kN-m )
0.012610 m (2.5 m)(0.0054044 rad) 0.026121 m
B
A
PL
EI
v
Consider 225 kN-m concentrated moment at B. [Appendix C, Cantilever beam with concentrated moment at tip.]
Relevant equations from Appendix C:
2
and2
B B
ML MLv
EI EI (slope magnitude)
Values:
M = 225 kN-m, L = 3.5 m, EI = 3.4 × 104 kN-m
2
Computation:
2 2
4 2
(225 kN-m)(3.5 m)0.040533 m
2 2(3.4 10 kN-m )B
MLv
EI (b)
4 2
(225 kN-m)(3.5 m)0.0231618 rad
(3.4 10 kN-m )
0.040533 m (2.5 m)(0.0231618 rad) 0.098438 m
B
A
ML
EI
v
Beam deflection at A
0.309706 m 0.190588 m 0.026121 m 0.098438 m 0.005441 m 5.44 mmAv Ans.
(b) Beam deflection at point B
Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 )24
B
wxv L Lx x
EI (elastic curve)
Values:
w = −65 kN/m, L = 6 m, x = 3.5 m,
EI = 3.4 × 104 kN-m
2
Computation:
22 2
22 2
4 2
(6 4 )24
( 65 kN/m)(3.5 m)6(6 m) 4(6 m)(3.5 m) (3.5 m) 0.140759 m
24(3.4 10 kN-m )
B
wxv L Lx x
EI
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Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
2
(3 )6
B
Pxv L x
EI (elastic curve)
Values:
P = 90 kN, L = 6 m, x = 3.5 m,
EI = 3.4 × 104 kN-m
2
Computation:
2 2
4 2
(90 kN)(3.5 m)(3 ) 3(6 m) (3.5 m) 0.078364 m
6 6(3.4 10 kN-m )B
Pxv L x
EI
Consider 30-kN concentrated load at B. Previously calculated in Eq. (a).
Consider 225 kN-m concentrated moment at B. Previously calculated in Eq. (b).
Beam deflection at B
0.140759 m 0.078364 m 0.012610 m 0.040533 m 0.009252 m 9.25 mmBv Ans.
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10.75 The simply supported beam shown in Fig.
P10.75 consists of a rectangular structural steel
tube shape [E = 200 GPa; I = 350 × 106 mm
4].
For the loading shown, determine:
(a) the beam deflection at point C.
(b) the beam deflection at point E.
Fig. P10.75
Solution
(a) Beam deflection at point C
Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = −315 kN-m, L = 9 m, x = 6 m,
EI = 7.0 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 315 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.018000 m
6(9 m)(7.0 10 kN-m )
C
M xv L Lx x
LEI
Consider 120 kN/m uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
3 2 2 2 2(2 6 4 )24
C
wav x Lx a x L x a L
LEI
Values:
w = 120 kN/m, L = 9 m, a = 3 m, x = 6 m,
EI = 7.0 × 104 kN-m
2
Computation: 2
3 2 2 2 2
23 2 2 2 2
4 2
(2 6 4 )24
(120 kN/m)(3 m)2(6 m) 6(9 m)(6 m) (3 m) (6 m) 4(9 m) (6 m) (3 m) (9 m)
24(9 m)(7.0 10 kN-m )
0.028929 m
C
wav x Lx a x L x a L
LEI
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Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
C
Pabv L a b
LEI
Values:
P = 100 kN, L = 9 m, a = 6 m, b = 3 m,
EI = 7.0 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(100 kN)(6 m)(3 m)(9 m) (6 m) (3 m) 0.017143 m
6(9 m)(7.0 10 kN-m )
C
Pabv L a b
LEI
Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m,
L = 9 m, x = 3 m, EI = 7.0 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 270 kN-m)(3 m)2(9 m) 3(9 m)(3 m) (3 m) 0.019286 m
6(9 m)(7.0 10 kN-m )
C
M xv L Lx x
LEI
Beam deflection at C
0.018000 m 0.028929 m 0.017143 m 0.019286 m 0.008786 m 8.79 mmCv Ans.
(b) Beam deflection at point E
Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
6
D
ML
EI (slope magnitude)
Values:
M = −315 kN-m, L = 9 m,
EI = 7.0 × 104 kN-m
2
Computation:
4 2
( 315 kN-m)(9 m)0.0067500 rad
6 6(7.0 10 kN-m )
(3 m)( 0.0067500 rad) 0.020250 m
D
E
ML
EI
v
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Consider 120 kN/m uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
2 2(2 )24
D
waL a
LEI (slope magnitude)
Values:
w = 120 kN/m, L = 9 m, a = 3 m,
EI = 7.0 × 104 kN-m
2
Computation:
2 22 2 2 2
4 2
(120 kN/m)(3 m)(2 ) 2(9 m) (3 m) 0.0109286 rad
24 24(9 m)(7.0 10 kN-m )
(3 m)(0.0109286 rad) 0.032786 m
D
E
waL a
LEI
v
Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6D
Pa L a
LEI (slope magnitude)
Values:
P = 100 kN, L = 9 m, a = 6 m,
EI = 7.0 × 104 kN-m
2
Computation:
2 22 2
4 2
( ) (100 kN)(6 m)(9 m) (6 m) 0.0071429 rad
6 6(9 m)(7.0 10 kN-m )
(3 m)(0.0071429 rad) 0.021429 m
D
E
Pa L a
LEI
v
Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.]
Relevant equation from Appendix C:
3
D
ML
EI (slope magnitude)
Values:
M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m,
L = 9 m, EI = 7.0 × 104 kN-m
2
Computation:
4 2
(270 kN-m)(9 m)0.0115714 rad
3 3(7.0 10 kN-m )
(3 m)(0.0115714 rad) 0.034714 m
D
E
ML
EI
v
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Determine cantilever deflection due to 60 kN/m uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with concentrated load.]
Relevant equation from Appendix C:
4
8E
wLv
EI (assuming fixed support at D)
Values:
w = 60 kN/m, L = 3 m, EI = 7.0 × 104 kN-m
2
Computation: 4 4
4 2
(60 kN-m)(3 m)0.008679 m
8 8(7.0 10 kN-m )E
wLv
EI
Beam deflection at E
0.020250 m 0.032786 m 0.021429 m 0.034714 m 0.008679 m
0.009429 m 9.43 mm
Ev
Ans.
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10.76 The cantilever beam shown in Fig. P10.76
consists of a rectangular structural steel tube
shape [E = 200 GPa; I = 95 × 106 mm
4]. For the
loading shown, determine the beam deflection at
point B.
Fig. P10.76
Solution
Consider the downward 50 kN/m uniformly distributed load acting over span AB.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
4 4
4 2
(50 kN/m)(2 m)0.0052632 m
8 8(1.9 10 kN-m )B
wLv
EI
Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 )24
B
wxv L Lx x
EI (elastic curve)
Values:
w = −25 kN/m, L = 5 m, x = 2 m,
EI = 1.9 × 104 kN-m
2
Computation:
22 2
22 2
4 2
(6 4 )24
( 25 kN/m)(2 m)6(5 m) 4(5 m)(2 m) (2 m) 0.0250000 m
24(1.9 10 kN-m )
B
wxv L Lx x
EI
Consider a downward 25 kN/m uniformly distributed load acting over span AB.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
4 4
4 2
(25 kN/m)(2 m)0.0026316 m
8 8(1.9 10 kN-m )B
wLv
EI
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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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Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3
3B
PLv
EI
Values:
P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
3 3
4 2
( 20 kN)(2 m)0.0028070 m
3 3(1.9 10 kN-m )B
PLv
EI
Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
2
(3 )6
B
Pxv L x
EI (elastic curve)
Values:
P = 50 kN, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
2 2
4 2
(50 kN)(2 m)(3 ) 3(5 m) (2 m) 0.0228070 m
6 6(1.9 10 kN-m )B
Pxv L x
EI
Beam deflection at B
0.0052632 m 0.0250000 m 0.0026316 m 0.0028070 m 0.0228070 m
0.0028947 m 2.89 mm
Bv
Ans.
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10.77 The cantilever beam shown in Fig. P10.77
consists of a rectangular structural steel tube
shape [E = 200 GPa; I = 95 × 106 mm
4]. For the
loading shown, determine the beam deflection at
point C.
Fig. P10.77
Solution
Consider the downward 50 kN/m uniformly distributed load acting over span AB.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI and
3
6B
wL
EI (slope magnitude)
Values:
w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
4 4
4 2
(50 kN/m)(2 m)0.0052632 m
8 8(1.9 10 kN-m )B
wLv
EI
3 3
4 2
(50 kN/m)(2 m)0.0035088 rad
6 6(1.9 10 kN-m )
0.0052632 m (3 m)(0.0035088 rad) 0.0157895 m
B
C
wL
EI
v
Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8C
wLv
EI
Values:
w = −25 kN/m, L = 5 m, EI = 1.9 × 104 kN-m
2
Computation:
4 4
4 2
( 25 kN/m)(5 m)0.1027961 m
8 8(1.9 10 kN-m )C
wLv
EI
Consider a downward 25 kN/m uniformly distributed load acting over span AB.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI and
3
6B
wL
EI (slope magnitude)
Values:
w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
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4 4
4 2
(25 kN/m)(2 m)0.0026316 m
8 8(1.9 10 kN-m )B
wLv
EI
3 3
4 2
(25 kN)(2 m)0.0017544 rad
6 6(1.9 10 kN-m )
0.0026316 m (3 m)(0.0017544 rad) 0.0078948 m
B
C
wL
EI
v
Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3
3B
PLv
EI and
2
2B
PL
EI (slope magnitude)
Values:
P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m
2
Computation:
3 3
4 2
( 20 kN)(2 m)0.0028070 m
3 3(1.9 10 kN-m )B
PLv
EI
2 2
4 2
(20 kN)(2 m)0.0021053 rad
2 2(1.9 10 kN-m )
0.0028070 m (3 m)(0.0021053 rad) 0.0091228 m
B
C
PL
EI
v
Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3
3C
PLv
EI
Values:
P = 50 kN, L = 5 m, EI = 1.9 × 104 kN-m
2
Computation:
3 3
4 2
(50 kN)(5 m)0.1096491 m
3 3(1.9 10 kN-m )C
PLv
EI
Beam deflection at C
0.0157895 m 0.1027961 m 0.0078948 m 0.0091228 m 0.1096491 m
0.0214145 m 21.4 mm
Cv
Ans.
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10.78 The simply supported beam shown in Fig.
P10.78 consists of a W10 × 30 structural steel
wide-flange shape [E = 29,000 ksi; I = 170 in.4].
If w = 5 kips/ft, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point C.
Fig. P10.78
Solution
(a) Beam deflection at point A
Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2
2A
MLv
EI
Values:
M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2 2 3
6 2
(85 kip-ft)(3 ft) (12 in./ft)0.134069 in.
2 2(4.93 10 kip-in. )A
MLv
EI
Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = 85 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2
6 2
(85 kip-ft)(15 ft)(12 in./ft)0.0124138 rad
3 3(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0124138 rad) 0.446897 in.
B
A
ML
EI
v
Consider cantilever beam deflection of 5 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = 5 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.
2
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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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Computation:
4 4 3
6 2
(5 kips/ft)(3 ft) (12 in./ft)0.017744 in.
8 8(4.93 10 kip-in. )A
wLv
EI
Consider rotation at B caused by 5 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft,
L = 15 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2
6 2
(22.5 kip-ft)(15 ft)(12 in./ft)0.0032860 rad
3 3(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0032860 rad) 0.118296 in.
B
A
ML
EI
v
Consider 5 kips/ft uniformly distributed load on segment BC.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
2(2 )24
B
waL a
LEI (slope magnitude)
Values:
w = 5 kips/ft, L = 15 ft, a = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2 222
6 2
(5 kips/ft)(5 ft) (12 in./ft)(2 ) 2(15 ft) (5 ft) 0.0063387 rad
24 24(15 ft)(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0063387 rad) 0.228195 in.
B
A
waL a
LEI
v
Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6B
Pb L b
LEI (slope magnitude)
Values:
P = 25 kips, L = 15 ft, b = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2 22 2
6 2
( ) (25 kips)(5 ft)(12 in./ft)(15 ft) (5 ft) 0.0081136 rad
6 6(15 ft)(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0081136 rad) 0.292089 in.
B
A
Pb L b
LEI
v
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permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Beam deflection at A
0.134069 in. 0.446897 in. 0.017744 in. 0.118296 in. 0.228195 in. 0.292089 in.
0.196722 in. 0.1967 in.
Av
Ans.
(b) Beam deflection at point C
Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = −85 kip-ft, L = 15 ft, x = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2
32 2
6 2
(2 3 )6
( 85 kip-ft)(5 ft)(12 in./ft)2(15 ft) 3(15 ft)(5 ft) (5 ft) 0.413793 in.
6(15 ft)(4.93 10 kip-in. )
C
M xv L Lx x
LEI
Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft,
L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2 2
32 2
6 2
(2 3 )6
( 22.5 kip-ft)(5 ft)(12 in./ft)2(15 ft) 3(15 ft)(5 ft) (5 ft) 0.109533 in.
6(15 ft)(4.93 10 kip-in. )
C
M xv L Lx x
LEI
Consider 5 kips/ft uniformly distributed load on segment BC.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 5 kips/ft, L = 15 ft, a = 5 ft,
EI = 4.93 × 106 kip-in.
2
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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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Computation:
32 2
3 32 2
6 2
(4 7 3 )24
(5 kips/ft)(5 ft) (12 in./ft)4(15 ft) 7(5 ft)(15 ft) 3(5 ft) 0.273834 in.
24(15 ft)(4.93 10 kip-in. )
C
wav L aL a
LEI
Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
C
Pbxv L b x
LEI (elastic curve)
Values:
P = 25 kips, L = 15 ft, b = 5 ft, x = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2 2
32 2 2
6 2
( )6
(25 kips)(5 ft)(5 ft)(12 in./ft)(15 ft) (5 ft) (5 ft) 0.425963 in.
6(15 ft)(4.93 10 kip-in. )
C
Pbxv L b x
LEI
Beam deflection at C
0.413793 in. 0.109533 in. 0.273834 in. 0.425963 in.
0.176471 i 0n. .1765 in.
Cv
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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10.79 The simply supported beam shown in Fig.
P10.79 consists of a W10 × 30 structural steel
wide-flange shape [E = 29,000 ksi; I = 170 in.4].
If w = 9 kips/ft, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point D.
Fig. P10.79
Solution
(a) Beam deflection at point A
Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2
2A
MLv
EI
Values:
M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2 2 3
6 2
(85 kip-ft)(3 ft) (12 in./ft)0.134069 in.
2 2(4.93 10 kip-in. )A
MLv
EI
Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = 85 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2
6 2
(85 kip-ft)(15 ft)(12 in./ft)0.0124138 rad
3 3(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0124138 rad) 0.446897 in.
B
A
ML
EI
v
Consider cantilever beam deflection of 9 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = 9 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.
2
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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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Computation:
4 4 3
6 2
(9 kips/ft)(3 ft) (12 in./ft)0.031939 in.
8 8(4.93 10 kip-in. )A
wLv
EI
Consider rotation at B caused by 9 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (9 kips/ft)(3 ft)(1.5 ft) = 40.5 kip-ft,
L = 15 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2
6 2
(40.5 kip-ft)(15 ft)(12 in./ft)0.0059148 rad
3 3(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0059148 rad) 0.212933 in.
B
A
ML
EI
v
Consider 9 kips/ft uniformly distributed load on segment BC.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
2(2 )24
B
waL a
LEI (slope magnitude)
Values:
w = 9 kips/ft, L = 15 ft, a = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2 222
6 2
(9 kips/ft)(5 ft) (12 in./ft)(2 ) 2(15 ft) (5 ft) 0.0114097 rad
24 24(15 ft)(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0114097 rad) 0.410748 in.
B
A
waL a
LEI
v
Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6B
Pb L b
LEI (slope magnitude)
Values:
P = 25 kips, L = 15 ft, b = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2 22 2
6 2
( ) (25 kips)(5 ft)(12 in./ft)(15 ft) (5 ft) 0.0081136 rad
6 6(15 ft)(4.93 10 kip-in. )
(3 ft)(12 in./ft)(0.0081136 rad) 0.292089 in.
B
A
Pb L b
LEI
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
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Beam deflection at A
0.134069 in. 0.446897 in. 0.031939 in. 0.212933 in. 0.410748 in. 0.292089 in.
0.123001 in. 0.1230 in.
Av
Ans.
(b) Beam deflection at point D
Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
D
M xv L Lx x
LEI (elastic curve)
Values:
M = −85 kip-ft, L = 15 ft, x = 10 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2
32 2
6 2
(2 3 )6
( 85 kip-ft)(10 ft)(12 in./ft)2(15 ft) 3(15 ft)(10 ft) (10 ft) 0.331034 in.
6(15 ft)(4.93 10 kip-in. )
D
M xv L Lx x
LEI
Consider moment at B caused by 9 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6D
M xv L Lx x
LEI (elastic curve)
Values:
M = −(9 kips/ft)(3 ft)(1.5 ft) = −40.5 kip-ft,
L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2 2
3
2 2
6 2
(2 3 )6
( 40.5 kip-ft)(10 ft)(12 in./ft)2(15 ft) 3(15 ft)(10 ft) (10 ft) 0.157729 in.
6(15 ft)(4.93 10 kip-in. )
D
M xv L Lx x
LEI
Consider 9 kips/ft uniformly distributed load on segment BC.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
3 2 2 2 2(2 6 4 )24
D
wav x Lx a x L x a L
LEI
Values:
w = 9 kips/ft, L = 15 ft, a = 5 ft, x = 10 ft,
EI = 4.93 × 106 kip-in.
2
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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.
Computation: 2
3 2 2 2 2
2 3
3 2 2 2 2
6 2
(2 6 4 )24
(9 kips/ft)(5 ft) (12 in./ft)2(10 ft) 6(15 ft)(10 ft) (5 ft) (10 ft) 4(15 ft) (10 ft) (5 ft) (15 ft)
24(15 ft)(4.93 10 kip-in. )
0.410751 in.
D
wav x Lx a x L x a L
LEI
Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
D
Pabv L a b
LEI
Values:
P = 25 kips, L = 15 ft, a = 10 ft, b = 5 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
2 2 2
32 2 2
6 2
( )6
(25 kips)(10 ft)(5 ft)(12 in./ft)(15 ft) (10 ft) (5 ft) 0.486815 in.
6(15 ft)(4.93 10 kip-in. )
D
Pabv L a b
LEI
Beam deflection at D
0.331034 in. 0.157729 in. 0.410751 in. 0.486815 in.
0.408803 in. 0.409 in.
Dv
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.
10.80 The simply supported beam shown in
Fig. P10.80 consists of a W10 × 30 structural
steel wide-flange shape [E = 29,000 ksi; I =
170 in.4]. For the loading shown, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point C.
Fig. P10.80
Solution
(a) Beam deflection at point A
Consider cantilever beam deflection of linearly distributed load on overhang AB. [Appendix C, Cantilever beam with linearly distributed load.]
Relevant equation from Appendix C:
4
0
30A
w Lv
EI
Values:
w0 = 8 kips/ft, L = 9 ft, EI = 4.93 × 106 kip-in.
2
Computation:
4 4 3
0
6 2
(8 kips/ft)(9 ft) (12 in./ft)0.613247 in.
30 30(4.93 10 kip-in. )A
w Lv
EI
Consider rotation at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = ½(8 kips/ft)(9 ft)(3 ft) = 108 kip-ft,
L = 18 ft, EI = 4.93 × 106 kip-in.
2
Computation:
2
6 2
(108 kip-ft)(18 ft)(12 in./ft)0.0189274 rad
3 3(4.93 10 kip-in. )
(9 ft)(12 in./ft)(0.0189274 rad) 2.044157 in.
B
A
ML
EI
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
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Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.]
Relevant equation from Appendix C:
3
0
45B
w L
EI (slope magnitude)
Values:
w0 = 8 kips/ft, L = 18 ft, EI = 4.93 × 106 kip-in.
2
Computation:
3 3 2
0
6 2
(8 kips/ft)(18 ft) (12 in./ft)0.0302838 rad
45 45(4.93 10 kip-in. )
(9 ft)(12 in./ft)(0.0302838 rad) 3.270652 in.
B
A
w L
EI
v
Consider 4 kips/ft uniformly distributed load on segment CD.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
2 2(2 )24
B
waL a
LEI (slope magnitude)
Values:
w = 4 kips/ft, L = 18 ft, a = 9 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
22 2
2 22 2
6 2
(2 )24
(4 kips/ft)(9 ft) (12 in./ft)2(18 ft) (9 ft) 0.0124211 rad
24(18 ft)(4.93 10 kip-in. )
(9 ft)(12 in./ft)(0.0124211 rad) 1.341478 in.
B
A
waL a
LEI
v
Beam deflection at A
0.613247 in. 2.044157 in. 3.270652 in. 1.341478 in. 1.954726 in. 1.955 in.Av Ans.
(b) Beam deflection at point C
Consider moment at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI
Values:
M = −½(8 kips/ft)(9 ft)(3 ft) = −108 kip-ft,
L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.
2
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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.
Computation:
2 2
32 2
6 2
(2 3 )6
( 108 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.766559 in.
6(18 ft)(4.93 10 kip-in. )
C
M xv L Lx x
LEI
Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.]
Relevant equation from Appendix C:
4 2 2 40 (7 10 3 )360
C
w xv L L x x
LEI
Values:
w0 = 8 kips/ft, L = 18 ft, x = 9 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
4 2 2 40
34 2 2 4
6 2
(7 10 3 )360
(8 kips/ft)(9 ft)(12 in./ft)7(18 ft) 10(18 ft) (9 ft) 3(9 ft) 1.916398 in.
360(18 ft)(4.93 10 kip-in. )
C
w xv L L x x
LEI
Consider 4 kips/ft uniformly distributed load on segment CD.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
C
wav L aL a
LEI
Values:
w = 4 kips/ft, L = 18 ft, a = 9 ft,
EI = 4.93 × 106 kip-in.
2
Computation:
32 2
3 32 2
6 2
(4 7 3 )24
(4 kips/ft)(9 ft) (12 in./ft)4(18 ft) 7(9 ft)(18 ft) 3(9 ft) 0.958199 in.
24(18 ft)(4.93 10 kip-in. )
C
wav L aL a
LEI
Beam deflection at C
0.766559 in. 1.916398 in. 0.958199 in. 2.108037 in. 2.11 in.Cv 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.
10.81 The simply supported beam shown
in Fig. P10.81 consists of a W21 × 44
structural steel wide-flange shape [E =
29,000 ksi; I = 843 in.4]. For the loading
shown, determine:
(a) the beam deflection at point A.
(b) the beam deflection at point C.
Fig. P10.81
Solution
(a) Beam deflection at point A
Consider cantilever beam deflection of downward 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = 4 kips/ft, L = 12 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 3
7 2
(4 kips/ft)(12 ft) (12 in./ft)0.732847 in.
8 8(2.4447 10 kip-in. )A
wLv
EI
Consider cantilever beam deflection of upward 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4 3
and8 6
wL wLv
EI EI(slope magnitude)
Values:
w = −4 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
4 4 3
7 2
3 3 2
7 2
( 4 kips/ft)(6 ft) (12 in./ft)0.045803 in.
8 8(2.4447 10 kip-in. )
(4 kips/ft)(6 ft) (12 in./ft)0.0008482 rad
6 6(2.4447 10 kip-in. )
0.045803 in. (6 ft)(12 in./ft)(0.0008482 rad)A
wLv
EI
wL
EI
v 0.106873 in.
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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.
Consider rotation at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (4 kips/ft)(6 ft)(9 ft) = 216 kip-ft,
L = 24 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2
7 2
(216 kip-ft)(24 ft)(12 in./ft)0.0101784 rad
3 3(2.4447 10 kip-in. )
(12 ft)(12 in./ft)(0.0101784 rad) 1.465693 in.
B
A
ML
EI
v
Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2( )
6B
Pb L b
LEI (slope magnitude)
Values:
P = 42 kips, L = 24 ft, b = 18 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
2 2 22 2
7 2
( ) (42 kips)(18 ft)(12 in./ft)(24 ft) (18 ft) 0.0077929 rad
6 6(24 ft)(2.4447 10 kip-in. )
(12 ft)(12 in./ft)(0.0077929 rad) 1.122172 in.
B
A
Pb L b
LEI
v
Consider 4 kips/ft uniformly distributed load on 6-ft segment near D.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
2 2(2 )24
B
waL a
LEI (slope magnitude)
Values:
w = 4 kips/ft, L = 24 ft, a = 6 ft,
EI = 2.4447 × 107 kip-in.
2
Computation:
2 2 22 2 2 2
7 2
(4 kips/ft)(6 ft) (12 in./ft)(2 ) 2(24 ft) (6 ft) 0.0016434 rad
24 24(24 ft)(2.4447 10 kip-in. )
(12 ft)(12 in./ft)(0.0016434 rad) 0.236648 in.
B
A
waL a
LEI
v
Beam deflection at A
0.732847 in. 0.106873 in. 1.465693 in. 1.122172 in. 0.236648 in.
0.732847 in. 0.733 in.
Av
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.
(b) Beam deflection at point C
Consider moment at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
C
M xv L Lx x
LEI (elastic curve)
Values:
M = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft,
L = 24 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2
32 2
7 2
(2 3 )6
( 216 kip-ft)(12 ft)(12 in./ft)2(24 ft) 3(24 ft)(12 ft) (12 ft) 0.549635 in.
6(24 ft)(2.4447 10 kip-in. )
C
M xv L Lx x
LEI
Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6C
Pbxv L b x
LEI (elastic curve)
Values:
P = 42 kips, L = 24 ft, b = 6 ft,
x = 12 ft, EI = 2.4447 × 107 kip-in.
2
Computation:
2 2 2
32 2 2
7 2
( )6
(42 kips)(6 ft)(12 ft)(12 in./ft)(24 ft) (6 ft) (12 ft) 0.587804 in.
6(24 ft)(2.4447 10 kip-in. )
C
Pbxv L b x
LEI
Consider 4 kips/ft uniformly distributed load on 6-ft segment near D.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C: 2
3 2 2 2 2(2 6 4 )24
C
wav x Lx a x L x a L
LEI
(elastic curve)
Values:
w = 4 kips/ft, L = 24 ft, a = 6 ft, x = 12 ft,
EI = 2.4447 × 107 kip-in.
2
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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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Computation: 2
3 2 2 2 2
3 2 22 3
7 2 2 2
(2 6 4 )24
2(12 ft) 6(24 ft)(12 ft) (6 ft) (12 ft)(4 kips/ft)(6 ft) (12 in./ft)
24(24 ft)(2.4447 10 kip-in. ) 4(24 ft) (12 ft) (6 ft) (24 ft)
0.175578 in.
C
wav x Lx a x L x a L
LEI
Beam deflection at C
0.549635 in. 0.587804 in. 0.175578 in. 0.213747 in 0.. 214 in.Cv 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.
10.82 The simply supported beam shown in Fig.
P10.82 consists of a W530 × 66 structural steel
wide-flange shape [E = 200 GPa; I = 351 × 106
mm4]. If w = 85 kN/m, determine the beam
deflection at point B.
Fig. P10.82
Solution
Beam deflection at point B
Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI (elastic curve)
Values:
M = −300 kN-m, L = 9 m, x = 4 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 300 kN-m)(4 m)2(9 m) 3(9 m)(4 m) (4 m) 0.022159 m
6(9 m)(7.02 10 kN-m )
B
M xv L Lx x
LEI
Consider 85 kN/m uniformly distributed load on segment AB.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
Values:
w = 85 kN/m, L = 9 m, a = 4 m,
EI = 7.02 × 104 kN-m
2
Computation:
32 2
32 2
4 2
(4 7 3 )24
(85 kN/m)(4 m)4(9 m) 7(4 m)(9 m) 3(4 m) 0.043052 m
24(9 m)(7.02 10 kN-m )
B
wav L aL a
LEI
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Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
B
Pbxv L b x
LEI (elastic curve)
Values:
P = 140 kN, L = 9 m, b = 3 m, x = 4 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(140 kN)(3 m)(4 m)(9 m) (3 m) (4 m) 0.024818 m
6(9 m)(7.02 10 kN-m )
B
Pbxv L b x
LEI
Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI (elastic curve)
Values:
M = −175 kN-m, L = 9 m, x = 5 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 175 kN-m)(5 m)2(9 m) 3(9 m)(5 m) (5 m) 0.012003 m
6(9 m)(7.02 10 kN-m )
B
M xv L Lx x
LEI
Beam deflection at B
0.022159 m 0.043052 m 0.024818 m 0.012003 m 0.033708 m 33.7 mmBv Ans.
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10.83 The simply supported beam shown in Fig.
P10.83 consists of a W530 × 66 structural steel
wide-flange shape [E = 200 GPa; I = 351 × 106
mm4]. If w = 115 kN/m, determine the beam
deflection at point C.
Fig. P10.83
Solution
(b) Beam deflection at point C
Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = −300 kN-m, L = 9 m, x = 6 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 300 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.017094 m
6(9 m)(7.02 10 kN-m )
C
M xv L Lx x
LEI
Consider 115 kN/m uniformly distributed load on segment AB.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
2
3 2 2 2 2(2 6 4 )24
C
wav x Lx a x L x a L
LEI
Values:
w = 115 kN/m, L = 9 m, a = 4 m, x = 6 m,
EI = 7.02 × 104 kN-m
2
Computation: 2
3 2 2 2 2
2
3 2 2 2 2
4 2
(2 6 4 )24
(115 kN/m)(4 m)2(6 m) 6(9 m)(6 m) (4 m) (6 m) 4(9 m) (6 m) (4 m) (9 m)
24(9 m)(7.02 10 kN-m )
0.046597 m
C
wav x Lx a x L x a L
LEI
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Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
C
Pabv L a b
LEI
Values:
P = 140 kN, L = 9 m, a = 6 m, b = 3 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2 2
2 2 2
4 2
( )6
(140 kN)(6 m)(3 m)(9 m) (6 m) (3 m) 0.023932 m
6(9 m)(7.02 10 kN-m )
C
Pabv L a b
LEI
Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6C
M xv L Lx x
LEI (elastic curve)
Values:
M = −175 kN-m, L = 9 m, x = 3 m,
EI = 7.02 × 104 kN-m
2
Computation:
2 2
2 2
4 2
(2 3 )6
( 175 kN-m)(3 m)2(9 m) 3(9 m)(3 m) (3 m) 0.012464 m
6(9 m)(7.02 10 kN-m )
C
M xv L Lx x
LEI
Beam deflection at C
0.017094 m 0.046597 m 0.023932 m 0.012464 m 0.040971 m 41.0 mmCv Ans.
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10.84 A 25-ft-long soldier beam is used as a key
component of an earth retention system at an
excavation site. The soldier beam is subjected to
a soil loading that is linearly distributed from
520 lb/ft to 260 lb/ft, as shown in Fig. P10.84.
The soldier beam can be idealized as a
cantilever with a fixed support at A. Added
support is supplied by a tieback anchor at B,
which exerts a force of 5,000 lb on the soldier
beam. Determine the horizontal deflection of the
soldier beam at point C. Assume EI = 5 × 108
lb-
in.2.
Fig. P10.84
Solution
Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8C
wLv
EI
Values:
w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.
2
Computation:
4 4 3
8 2
(260 lb/ft)(25 ft) (12 in./ft)43.875 in.
8 8(5.0 10 lb-in. )C
wLv
EI
Consider a linearly distributed load that varies from 260 lb/ft at A to 0 lb/ft at C. [Appendix C, Cantilever beam with linearly distributed load.]
Relevant equation from Appendix C:
4
0
30C
w Lv
EI
Values:
w0 = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.
2
Computation:
4 4 3
0
8 2
(260 lb/ft)(25 ft) (12 in./ft)11.700 in.
30 30(5.0 10 lb-in. )C
w Lv
EI
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Consider 5,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI (slope magnitude)
Values:
P = 5,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.
2
Computation:
3 3 3
8 2
2 2 2
8 2
(5,000 lb)(18 ft) (12 in./ft)33.592320 in.
3 3(5.0 10 lb-in. )
(5,000 lb)(18 ft) (12 in./ft)0.2332800 rad
2 2(5.0 10 lb-in. )
33.592320 in. (7 ft)(12 in./ft)(0.2332800 rad) 53.187
B
B
C
PLv
EI
PL
EI
v 840 in.
Beam deflection at C
43.875 in. 11.700 in. 53.187840 in. 2.387160 in. 2.39 in.Cv Ans.
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10.85 A 25-ft-long soldier beam is used as a key
component of an earth retention system at an
excavation site. The soldier beam is subjected to a
uniformly distributed soil loading of 260 lb/ft, as
shown in Fig. P10.85. The soldier beam can be
idealized as a cantilever with a fixed support at A.
Added support is supplied by a tieback anchor at B,
which exerts a force of 4,000 lb on the soldier beam.
Determine the horizontal deflection of the soldier
beam at point C. Assume EI = 5 × 108
lb-in.2.
Fig. P10.85
Solution
Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8C
wLv
EI
Values:
w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.
2
Computation:
4 4 3
8 2
(260 lb/ft)(25 ft) (12 in./ft)43.875 in.
8 8(5.0 10 lb-in. )C
wLv
EI
Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI (slope magnitude)
Values:
P = 4,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.
2
Computation:
3 3 3
8 2
2 2 2
8 2
(4,000 lb)(18 ft) (12 in./ft)26.873856 in.
3 3(5.0 10 lb-in. )
(4,000 lb)(18 ft) (12 in./ft)0.1866240 rad
2 2(5.0 10 lb-in. )
26.873856 in. (7 ft)(12 in./ft)(0.1866240 rad) 42.550
B
B
C
PLv
EI
PL
EI
v 272 in.
Beam deflection at C
43.875 in. 42.550272 in. 1.324728 in. 1.325 in.Cv Ans.
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11.1 A beam is loaded and supported as
shown in Fig. P11.1. Use the double-
integration method to determine the
magnitude of the moment M0 required to
make the slope at the left end of the beam
zero.
Fig. P11.1
Solution
Moment equation:
0
2
0
( ) 02
( )2
a a
xM M x wx M
wxM x M
Integration:
2 2
02( )
2
d v wxEI M x M
dx
3
0 16
dv wxEI M x C
dx
2 4
01 2
2 24
M x wxEI v C x C
Boundary conditions and evaluate constants:
3
0 1
3
1 0
( )at , 0 ( ) 0
6
6
dv w Lx L M L C
dx
wLC M L
Beam slope equation:
3 3
0 06 6
dv wx wLEI M x M L
dx
Constraint:
At x = 0, the slope of the beam is to be zero; therefore,
3
2
3
0 0
0
(0)(0) 0
6 6
6
A
dv w wLEI M M L
dx
MwL
Ans.
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11.2 When moment M0 is applied to the left
end of the cantilever beam shown in Fig.
P11.2, the slope of the beam is zero. Use the
double-integration method to determine the
magnitude of the moment M0.
Fig. P11.2
Solution
Moment equation:
0
0
( ) 0
( )
a aM M x Px M
M x Px M
Integration:
2
02( )
d vEI M x Px M
dx
2
0 12
dv PxEI M x C
dx
3 2
01 2
6 2
Px M xEI v C x C
Boundary conditions and evaluate constants:
2
0 1
2
1 0
( )at , 0 ( ) 0
2
2
dv P Lx L M L C
dx
PLC M L
Beam slope equation:
2 2
0 02 2
dv Px PLEI M x M L
dx
Constraint:
At x = 0, the slope of the beam is to be zero; therefore,
2 2
0 0
0
(0)(0) 0
2 2
2
A
dv P PLEI M M
MP
Ldx
L Ans.
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11.3 When the load P is applied to the right
end of the cantilever beam shown in Fig.
P11.3, the deflection at the right end of the
beam is zero. Use the double-integration
method to determine the magnitude of the
load P.
Fig. P11.3
Solution
Moment equation:
( ) ( ) ( ) 02
a a
L xM w L x P L x M x
2( ) ( ) ( )2
wM x L x P L x
Integration:
2
2
2( ) ( ) ( )
2
d v wEI M x L x P L x
dx
3 2
1( ) ( )6 2
dv w PEI L x L x C
dx
4 3
1 2( ) ( )24 6
w PEI v L x L x C x C
Boundary conditions and evaluate constants:
3 2
1
3 2
1
at 0, 0 ( 0) ( 0) 06 2
6 2
dv w Px L L C
dx
wL PLC
4 3
1 2
4 3
2
at 0, 0 ( 0) ( 0) (0) 024 6
24 6
w Px v L L C C
wL PLC
Beam elastic curve equation:
3 2 4 34 3
3 4 2 34 3
( ) ( )24 6 6 2 24 6
( ) ( )24 6 24 6 2 6
w P wLx PL x wL PLEI v L x L x
w wLx wL P PL x PLL x L x
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Constraint:
At x = L, the deflection of the beam is to be zero; therefore,
3 4 2 3
4 3( ) ( )( ) ( ) 0
24 6 24 6 2 6B
w wL L wL P PL L PLEI v L L L L
which simplifies to
4 4 3 3 4 3
06 24 2 6 8 3
B
wL wL PL PL wL PLEI v
Therefore, the magnitude of P is
3
8P
wL Ans.
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11.4 A beam is loaded and supported as
shown in Fig. P11.4. Use the double-
integration method to determine the
reactions at supports A and B.
Fig. P11.4
Solution
Beam FBD:
0
0
0
y y y
A A y
F A B
M M B L M
Moment equation:
0 0( ) ( ) 0 ( ) ( )a a y yM M x M B L x M x B L x M
Integration:
2
02( ) ( )y
d vEI M x B L x M
dx
2
0 1( )2
yBdvEI L x M x C
dx
2
3 01 2( )
6 2
yB M xEI v L x C x C
Boundary conditions and evaluate constants:
2
2
0 1 1at 0, 0 ( 0) (0) 02 2
y yB B Ldvx L M C C
dx
32
3 0
1 2 2
(0)at 0, 0 ( 0) (0) 0
6 2 6
y yB B LMx v L C C C
2 32
3 0
3 2
0 00
( )at , 0 ( ) ( ) 0
6 2 2 6
3
3 2 2
3
2
y y y
y
y
B B L B LM Lx L v L L L
B L M M M
L
LB
L Ans.
Backsubstitute into equilibrium equations:
0 030
2
3
2y y y y y y
MMF A B A B A
L L Ans.
0
0 0 0
00
30 0
2
(cw)22
A A y A y
A
MM M B L M M B L
M
M L ML
MM Ans.
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11.5 A beam is loaded and supported as
shown in Fig. P11.5.
(a) Use the double-integration method to
determine the reactions at supports A and B.
(b) Draw the shear-force and bending-
moment diagrams for the beam.
Fig. P11.5
Solution
Beam FBD:
0
02
y y y
A B y
F A B wL
LM M B L wL
Moment equation:
2
( ) 0 ( )2 2
a a y y
x wxM M x wx A x M x A x
Integration:
2 2
2( )
2y
d v wxEI M x A x
dx
23
16 2
yA xdv wxEI C
dx
34
1 224 6
yA xwxEI v C x C
Boundary conditions and evaluate constants:
34
1 2 2
(0)(0)at 0, 0 (0) 0 0
24 6
yAwx v C C C
2 23 3
1 1
( )( )at , 0 0
6 2 6 2
y yA L A Ldv w L wLx L C C
dx (a)
3 24 3
1 1
( )( )at , 0 ( ) 0
24 6 24 6
y yA L A Lw L wLx L v C L C (b)
Solve Eqs. (a) and (b) simultaneously to find:
3
1
3and
48 8
3
8y
wL wLC A
wL Ans.
Backsubstitute into equilibrium equations:
3 5
05
88 8y y y y y y
wL wLF A B wL B wL
wLA wL B Ans.
2
2 2 2 2
2
50
2
(cw)
2 2 8 8
8 8
A B y B y
B
L wL wL wL wL
wL
M M B L wL M B L
wLM 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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11.6 A beam is loaded and supported as
shown in Fig. P11.6. Use the double-
integration method to determine the
reactions at supports A and B.
Fig. P11.6
Solution
Beam FBD:
0
0
02
20
2 3
y y y
A B y
w LF A B
w L LM M B L
Moment equation:
2
0
3
0
( ) 02 3
( )6
a a y
y
w x xM M x A x
L
w xM x A x
L
Integration:
2 3
0
2( )
6y
d v w xEI M x A x
dx L
24
01
24 2
yA xdv w xEI C
dx L
35
01 2
120 6
yA xw xEI v C x C
L
Boundary conditions and evaluate constants:
35
01 2 2
(0)(0)at 0, 0 (0) 0 0
120 6
yAwx v C C C
L
2 24 3
0 01 1
( )( )at , 0 0
24 2 24 2
y yA L A Ldv w L w Lx L C C
dx L (a)
3 25 3
0 0
1 1
( )( )at , 0 ( ) 0
120 6 120 6
y yA L A Lw L w Lx L v C L C
L (b)
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Solve Eqs. (a) and (b) simultaneously to find:
3
0 0 01 and
120 110 0y
w L w LC A
w L Ans.
Backsubstitute into equilibrium equations:
0 0 0 00 040
2 2 2 10 1 50
2y y y y y y
w L w L w L w L w LF A B B A B
w L Ans.
2 2 2 2
0 0 0 0 0
2 2
0 0
2 20
2 3 3 3 5 15
(cw15
)1 5
A B y B y
B
w L L w L w L w L w LM M B L M B L
w Lw LM 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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11.7 A beam is loaded and supported as
shown in Fig. P11.7. Use the fourth-order
integration method to determine the reaction
at roller support B.
Fig. P11.7
Solution
Integrate the load distribution:
4 2
0
4 2
d v w xEI
dx L
3 3
013 23
d v w xEI C
dx L
2 4
01 22 212
d v w xEI C x C
dx L
5 2
0 12 3260 2
dv w x C xEI C x C
dx L
6 3 2
0 1 23 42360 6 2
w x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
6 3 2
0 1 23 4 42
(0) (0) (0)at 0, 0 (0) 0 0
360 6 2
w C Cx v C C C
L
5 2
0 12 3 32
(0) (0)at 0, 0 (0) 0 0
60 2
dv w Cx C C C
dx L
6 3 2 2
0 1 2 01 22
( ) ( ) ( )at , 0 0 3
360 6 2 60
w L C L C L w Lx L v C L C
L (a)
2 4 2
0 01 2 1 22 2
( )at , 0 ( ) 0
12 12
d v w L w Lx L M EI C L C C L C
dx L (b)
Solve Eqs. (a) and (b) simultaneously to obtain:
2 2 2 2
0 0 0 02 2
42
60 12 60 30
w L w L w L w LC C
2 2 2
0 0 0 01 1
7 7
12 30 60 60
w L w L w L w LC L C
Roller reaction at B:
3 3
0 00 0 0 0
3 2
( ) 7 20 7 13
3 60 60 60 60
13
60B y
x L
d v w L w L w L w L w LV EI B
dx 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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11.8 A beam is loaded and supported as
shown in Fig. P11.8. Use the fourth-order
integration method to determine the reaction
at roller support A.
Fig. P11.8
Solution
Integrate the load distribution:
4 2
0
4 2
d v w xEI
dx L
3 3
013 23
d v w xEI C
dx L
2 4
01 22 212
d v w xEI C x C
dx L
5 2
0 12 3260 2
dv w x C xEI C x C
dx L
6 3 2
0 1 23 42360 6 2
w x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
6 3 2
0 1 23 4 42
(0) (0) (0)at 0, 0 (0) 0 0
360 6 2
w C Cx v C C C
L
2 4
01 2 22 2
(0)at 0, 0 (0) 0 0
12
d v wx M EI C C C
dx L
5 2 3
20 1 03 1 32
( ) ( )at , 0 0 2
60 2 30
dv w L C L w Lx L C C L C
dx L (a)
6 3 3
20 1 03 1 32
( ) ( )at , 0 ( ) 0 6
360 6 60
w L C L w Lx L v C L C L C
L (b)
Solve Eqs. (a) and (b) simultaneously to obtain:
3 3 3 3
0 0 0 03 34
30 60 60 240
w L w L w L w LC C
3 3 3
2 0 0 0 0 01 1
5 5
30 120 120 120 24
w L w L w L w L w LC L C
Roller reaction at A:
3 3
0 0 0
3 2
0
0(0)
3 24 4 42 2A y
x
d v w w L w LV EI A
dx L
w L Ans.
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11.9 A beam is loaded and supported as
shown in Fig. P11.9. Use the fourth-order
integration method to determine the
reaction at roller support A.
Fig. P11.9
Solution
Integrate the load distribution:
4
04sin
2
d v xEI w
dx L
3
013
2cos
2
d v w L xEI C
dx L
2 2
01 22 2
4sin
2
d v w L xEI C x C
dx L
3 2
0 12 33
8cos
2 2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
16sin
2 6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
4 3 2
0 1 23 4 44
16 (0) (0) (0)at 0, 0 sin (0) 0 0
2 6 2
w L C Cx v C C C
L
2 2
01 2 22 2
4 (0)at 0, 0 sin (0) 0 0
2
d v w Lx M EI C C C
dx L
3 2
20 13 1 33
8 ( ) ( )at , 0 cos 0 2 0
2 2
dv w L L C Lx L C C L C
dx L (a)
4 3 3
20 1 03 1 34 4
16 ( ) ( ) 96at , 0 sin ( ) 0 6
2 6
w L L C L w Lx L v C L C L C
L (b)
Solve Eqs. (a) and (b) simultaneously to obtain:
3 3
0 03 34 4
96 244
w L w LC C
3
2 0 01 14 4
24 482
w L w LC L C
Roller reaction at A:
3
0 0 0
3 4
0
4
0
2 (0) 48cos
2
2 48A y
x
d v w L w LV EI
w L
x
LA
w
d L Ans.
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11.10 A beam is loaded and supported as
shown in Fig. P11.10. Use the fourth-order
integration method to determine the
reactions at supports A and B.
Fig. P11.10
Solution
Integrate the load distribution:
4
04cos
2
d v xEI w
dx L
3
013
2sin
2
d v w L xEI C
dx L
2 2
01 22 2
4cos
2
d v w L xEI C x C
dx L
3 2
0 12 33
8sin
2 2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
16cos
2 6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
3 2
0 12 3 33
8 (0) (0)at 0, 0 sin (0) 0 0
2 2
dv w L Cx C C C
dx L
4 3 2 4
0 1 2 04 44 4
16 (0) (0) (0) 16at 0, 0 cos 0
2 6 2
w L C C w Lx v C C
L
3 2 2
0 1 02 1 23 3
8 ( ) ( ) 16at , 0 sin ( ) 0 2
2 2
dv w L L C L w Lx L C L C L C
dx L(a)
4 3 2 4 2
0 1 2 0 01 24 4 4
16 ( ) ( ) ( ) 16 96at , 0 cos 0 3
2 6 2
w L L C L C L w L w Lx L v C L C
L(b)
Solve Eqs. (a) and (b) simultaneously to obtain:
2 2 2
0 0 02 23 4 4
16 96 166
w L w L w LC C
2 2
0 0 01 13 4 4
48 192 484
w L w L w LC L C
Reactions at supports A and B
3
0 0 0
3 4 4
0
0
4
2 (0) 48 48sin 4 4
4
2
84
A
x
y
d v w L w L
w
w
L
LV EI
dx L
A Ans.
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330 0
4
4
30
0
3 4
2 ( ) 48 2si
296
n 4 24 9
4
62
2
B
x L
y
d v w L L w L w LV EI
dx L
w LB Ans.
2 2 2
0 0 0
2 2 4 4
22
0
2 2
0 0
2 4
0
4
4 (0) 48 (0) 16cos 4
424 4 (c
62
4 1
)
66
w
A
x
A
d v w L w L w LM EI
dx L
w L w L
Mw L
Ans.
2 2 2
0 0 0
2 2 4 4
2 2 2 2
0 0 0 0
4 4 4
2
0
4
4
4 ( ) 48 (
323 (c
) 16cos 4 6
2
48 16
cw
16 164 6 3 12 6 2 6
)
B
x L
B
d v w L L w L L w LM EI
dx L
w L w L
L
w L w L
wM 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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11.11 A beam is loaded and supported as
shown in Fig. P11.11. Use the fourth-order
integration method to determine the
reactions at supports A and B.
Fig. P11.11
Solution
Integrate the load distribution:
4
04sin
d v xEI w
dx L
3
013
cosd v w L x
EI Cdx L
2 2
01 22 2
sind v w L x
EI C x Cdx L
3 2
0 12 33
cos2
dv w L x C xEI C x C
dx L
4 3 2
0 1 23 44
sin6 2
w L x C x C xEI v C x C
L
Boundary conditions and evaluate constants:
3 2 3
0 1 02 3 33 3
(0) (0)at 0, 0 cos (0) 0
2
dv w L C w Lx C C C
dx L
4 3 2
0 1 23 4 44
(0) (0) (0)at 0, 0 sin (0) 0 0
6 2
w L C Cx v C C C
L
3 2 3 2
0 1 0 02 1 23 3 3
( ) ( ) 4at , 0 cos ( ) 0 2
2
dv w L L C L w L w Lx L C L C L C
dx L (a)
4 3 2 3 2
0 1 2 0 01 24 3 3
( ) ( ) ( ) 6at , 0 sin ( ) 0 3
6 2
w L L C L C L w L w Lx L v L C L C
L (b)
Solve Eqs. (a) and (b) simultaneously to obtain:
2 2 2
0 0 02 23 3 3
6 4 2w L w L w LC C
2 2
0 01 13 3
4 22 0
w L w LC L C
Reactions at supports A and B
3
0
0 0
3
0
(0)cosA
x
y
d v w L w LV EI
dx L
Aw L
Ans.
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3
0
3
0 0( )cosB
x L
y
d v w L L w LV EI
dx L
Bw L
Ans.
2 2 2 2
0 0 0
2 2 3 3
0
2
0
3
(0) 2
2 )
sin
(cw
2A
x
A
d v w L w L w LM
w
Idx L
L
E
M Ans.
2 2 2 2
0 0 0
2 2 3 3
2
0
3
( ) 2
2 (ccw)
2sinB
x L
B
d v w L L w L w
w
LM E
L
Idx L
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
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11.12 A beam is loaded and supported as
shown in Fig. P11.12.
(a) Use the double-integration method to
determine the reactions at supports A and C.
(b) Draw the shear-force and bending-
moment diagrams for the beam.
(c) Determine the deflection in the middle
of the span. Fig. P11.12
Solution
Beam FBD:
from symmetry,
2
and
y y
A C
PA C
M M
Moment equation:
( ) 0 ( )2 2
a a A A
P PxM M x M x M x M
Integration:
2
2( )
2A
d v PxEI M x M
dx
2
14
A
dv PxEI M x C
dx
3 2
1 212 2
APx M xEI v C x C
Boundary conditions and evaluate constants:
2
1 1
(0)at 0, 0 (0) 0 0
4A
dv Px M C C
dx
3 2
2 2
(0) (0)at 0, 0 0 0
12 2
AP Mx v C C
(a) Beam reaction forces:
2
y yAP
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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(a) Beam reaction moments:
2
(ccw) (cw)8
( / 2)at , 0 0
2 4 2
8 8
A
A C
L dv P L Lx M
dx
PL PL PM M
L Ans.
Elastic curve equation:
3 2 3 2 2
2
3 412 2 12 16 48
3 448
APx M x Px PLx PxEI v L x
Pxv L x
EI
(c) Midspan deflection:
2 3( / 2)
3 4(192
/ 2)48
B
P Lv L L
E
PL
EII 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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11.13 A beam is loaded and supported as
shown in Fig. P11.13.
(a) Use the double-integration method to
determine the reactions at supports A and B.
(b) Draw the shear-force and bending-
moment diagrams for the beam.
(c) Determine the deflection in the middle
of the span.
Fig. P11.13
Solution
Beam FBD:
from symmetry,
2
and
y y
A B
wLA B
M M
Moment equation:
2
( ) 02 2
( )2 2
a a A
A
x wLM M x M wx x
wx wLxM x M
Integration:
2 2
2( )
2 2A
d v wx wLxEI M x M
dx
3 2
16 4
A
dv wx wLxEI M x C
dx
4 3 2
1 224 12 2
Awx wLx M xEI v C x C
Boundary conditions and evaluate constants:
3 2
1 1
(0) (0)at 0, 0 (0) 0 0
6 4A
dv w wLx M C C
dx
4 3 2
2 2
(0) (0) (0)at 0, 0 0 0
24 12 2
Aw wL Mx v C C
(a) Beam reaction forces:
2
y yA BwL
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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(a) Beam reaction moments:
3
2
2
2 22
(ccw) (cw)
( / 2) ( / 2)at , 0 0
2
12 12
6 4 2
12 12
A
A B
L dv w L wL L Lx M
dx
wL wLM M
wL wL Ans.
Elastic curve equation:
4 3 2 4 3 2 2 2 22 2 2
22
2 ( )24 12 2 24 12 24 24 24
( )24
Awx wLx M x wx wLx wL x wx wxEI v x Lx L x L
wxv x L
EI
(c) Midspan deflection:
22
2
4
/
( / 2)
4 2 3)
842x L
w L Lv L
EI
wL
EI 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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11.14 A beam is loaded and supported as
shown in Fig. P11.14.
(a) Use the double-integration method to
determine the reactions at supports A and C.
(b) Determine the deflection in the middle
of the span.
Fig. P11.14
Solution
Beam FBD:
0
from symmetry,
2
and
y y
A C
w LA C
M M
Moment equation:
2
0 0
3
0 0
( ) 02 3 2
( )6 2
a a A
A
w x x w LM M x M x
L
w x w LxM x M
L
Integration:
2 3
0 0
2( )
6 2A
d v w x w LxEI M x M
dx L
4 2
0 01
24 4A
dv w x w LxEI M x C
dx L
5 3 2
0 01 2
120 12 2
Aw x w Lx M xEI v C x C
L
Boundary conditions and evaluate constants:
4 2
0 01 1
(0) (0)at 0, 0 (0) 0 0
24 4A
dv w w Lx M C C
dx L
5 3 2
0 02 2
(0) (0) (0)at 0, 0 0 0
120 12 2
Aw w L Mx v C C
L
(a) Beam reaction forces:
0
2y yA C
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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(a) Beam reaction moments:
2
4 2
0 0
2 2
0 0
2
0 0
( ) ( )at , 0 ( ) 0
24 4
5 5 5 (ccw) (cw)
24
5
24 24 24
A
A C
d
w
v w L w L Lx L M L
dx L
w L w LL w LM M Ans.
Elastic curve equation:
5 3 2 5 3 2 2
0 0 0 0 0
5 2 3 3 2 23 2 30 0 0 0
23 2 30
5
120 12 2 120 12 48
2 20 252 20 25
240 240 240 240
2 20 25240
Aw x w Lx M x w x w Lx w L xEI v
L L
w x w L x w L x w xx L x L
L L L L
w xv x L x L
L EI
(c) Midspan deflection:
2
3 2 34
0 0( )2( ) 20 ( ) 25
2
7
0 24 40B
w Lv L L L L
w
L EI
L
EI 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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11.15 A beam is loaded and supported as
shown in Fig. P11.15.
(a) Use the double-integration method to
determine the reactions at supports A and C.
(b) Draw the shear-force and bending-
moment diagrams for the beam.
(c) Determine the deflection in the middle
of the span. Fig. P11.15
Solution
Beam FBD:
0
02
y y y
A C y
F A C P
LM M C L P
Moment equation:
( ) 0
( ) 02
a a y
y
M M x A x
LM x A x x
( ) 02
( )2 2
b b y
y
LM M x A x P x
PL LM x A x Px x L
Integration:
For beam segment AB:
2
2( ) y
d vEI M x A x
dx
2
12
yA xdvEI C
dx
3
1 26
yA xEI v C x C
For beam segment BC:
2
2( )
2y
d v PLEI M x A x Px
dx
2 2
32 2 2
yA xdv Px PLxEI C
dx
3 3 2
3 46 6 4
yA x Px PLxEI v C x C
Boundary conditions and evaluate constants:
3
1 2 2
(0)at 0, 0 (0) 0 0
6
yAx v C C C
2 22
3 3
( ) ( ) ( )at , 0 0
2 2 2 2
y yA L A Ldv P L PL Lx L C C
dx
3 2 33 2 3
4 4
( ) ( ) ( )at , 0 ( ) 0
6 6 4 2 3 12
y y yA L A L A LP L PL L PLx L v L C C
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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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Slope continuity condition at x = L/2:
2 2 22
1
22
1
at ,2
( / 2) ( / 2) ( / 2) ( / 2)
2 2 2 2 2
8 2
AB BC
y y y
y
L dv dvx
dx dx
A L A L A LP L PL LC
A LPLC
Deflection continuity condition at x = L/2:
3 2 3 2 32 3 2 3
at ,2
6 8 2 6 6 4 2 3 12
B BAB BC
y y y y y
Lx v v
A x A L x A x A L x A LPL x Px PLx PL
eliminate terms and rearrange:
3 3 2 2 3
3 6 4 8 12
yA L Px PLx PL x PL
Substitute x = L/2 to obtain:
3 3 2 2 3 3( / 2) ( / 2) ( / 2) 5
3 6 4 8 12 48
5
16
y
y
A L P L PL L PL L PL PL
PA
(a) Beam reaction forces:
5 11
16 16y yA C
P P Ans.
(a) Beam reaction moment:
11 3
(cw)16
3 3
2 2 16 16 16C y C
L PL PL PL PLM P C L
PM
L Ans.
Elastic curve equation for beam segment AB:
3 22 3 2 2 3 2
2 2
5 5 5 3
6 8 2 96 8 32 96 96
5 396
y yA x A L xPL x Px PL x PL x Px PL xEI v
Pxv x L
EI
(c) Midspan deflection:
2 32( / 2)
5 396 2
7
768B
P L PL
EI
Lv L
EI 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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11.16 A beam is loaded and supported as
shown in Fig. P11.16.
(a) Use the double-integration method to
determine the reactions at supports A and C.
(b) Draw the shear-force and bending-
moment diagrams for the beam.
Fig. P11.16
Solution
Beam FBD:
0
0
0
y y y y y
A A y
F A C C A
M M C L M
Moment equation:
( ) 0
( ) 02
a a y A
y A
M M x A x M
LM x A x M x
0
0
( ) 0
( )2
b b y A
y A
M M x A x M M
LM x A x M M x L
Integration:
For beam segment AB:
2
2( ) y A
d vEI M x A x M
dx
2
12
y
A
A xdvEI M x C
dx
3 2
1 26 2
y AA x M x
EI v C x C
For beam segment BC:
2
02( ) y A
d vEI M x A x M M
dx
2
0 32
y
A
A xdvEI M x M x C
dx
3 2 2
03 4
6 2 2
y AA x M x M x
EI v C x C
Boundary conditions and evaluate constants for segment AB:
3 2
1 2 2
(0) (0)at 0, 0 (0) 0 0
6 2
y AA M
x v C C C
2
1 1
(0)at 0, 0 (0) 0 0
2
y
A
Advx M C C
dx
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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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Slope continuity condition at x = L/2:
2 2
0 3
03
at ,2
2 2
2
AB BC
y y
A A
L dv dvx
dx dx
A x A xM x M x M x C
M LC
Deflection continuity condition at x = L/2:
3 32 2 2
0 04
2
04
at ,2
6 2 6 2 2 2
8
B BAB BC
y yA A
Lx v v
A x A xM x M x M x M LxC
M LC
Boundary condition for segment BC:
3 2 2 2
0 0 0 0( ) ( ) ( ) 3
at , 0 ( ) 0 36 2 2 2 8 4
y Ay A
A L M L M L M L M L Mx L v L A L M
Also, the beam moment equilibrium equation can be written as:
0y AA L M M
(a) Beam Reactions: Solve these two equations simultaneously to obtain:
0 0 0 00 9 9 (cw)
8 88 8
9
8A y y
M MM A C
L
M M M
L 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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11.17 A beam is loaded and supported as shown
in Fig. P11.17.
(a) Use the double-integration method to
determine the reactions at supports A and C.
(b) Draw the shear-force and bending-moment
diagrams for the beam.
Fig. P11.17
Solution
Beam FBD:
02
02 4
y y y
A C y
wLF A C
wL LM M C L
Moment equation:
2
( ) 02
( ) 02 2
a a y
y
xM M x wx A x
wx LM x A x x
( ) 02 4
( )2 4 2
b b y
y
wL LM M x x A x
wL L LM x x A x x L
Integration:
For beam segment AB:
2 2
2( )
2y
d v wxEI M x A x
dx
23
16 2
yA xdv wxEI C
dx
34
1 224 6
yA xwxEI v C x C
For beam segment BC:
2
2( )
2 4y
d v wL LEI M x x A x
dx
2 2
34 4 2
yA xdv wL LEI x C
dx
3 3
3 412 4 6
yA xwL LEI v x C x C
Boundary conditions and evaluate constants for segment AB:
34
1 2 2
(0)(0)at 0, 0 (0) 0 0
24 6
yAwx v C C C
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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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Slope continuity condition at x = L/2:
Equate the slope expressions for the two beam segments:
22 23
1 36 2 4 4 2
y yA x A xwx wL LC x C
Set x = L/2 and solve for the constant C1:
2 23 3 3 3
1 3 3 3
3
1 3
( / 2)
6 4 4 6 4 2 4 48 64
192
wx wL L w L wL L L wL wLC C x C C
wLC C
Deflection continuity condition at x = L/2:
Equate the deflection expressions for the two beam segments:
33 34
1 3 424 6 12 4 6
y yA x A xwx wL LC x x C x C
Set x = L/2 and solve for the constant C4:
34 3
3 3 4
4 4 4
3 3 4
4
4
( / 2)
24 192 2 12 2 4 2
384 2 384 768 2
768
w L wL L wL L L LC C C
wL L wL wL LC C C
wLC
Boundary conditions and evaluate constants for segment BC:
2 2 23
3 3
( ) 9at , 0 0
4 4 2 64 2
y yA L A Ldv wL L wLx L L C C
dx
at x = L, v = 0
3 3 4
3
3 24 3 4
3 34 4 4
3 3 4 4
3 4
( )( ) 0
12 4 6 768
27 9( ) 0
768 6 64 2 768
27 90
768 6 64 2 768
3 26 108
6 6 768 768
82 41
3 768 128
y
y y
y y
y y
y
y
A LwL L wLL C L
A L A LwL wL wLL
A L A LwL wL wL
A L A L wL wL
A L wL wLA
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Solve for C3:
3 3 3
3
9 41 5
64 256 256
wL wL wLC
and for C1:
3 3 3
1
5 11
256 192 768
wL wL wLC
(a) Beam force reactions:
41
128y
wLA Ans.
41 23
2 2 128
23
12 281 8y y y
wL wL wL wLC
wA C
L Ans.
Beam moment reaction:
2 2 22 2 223 7 7
8 8 1
7 (cw)
12828 128 128C y C
wL wL wL wL wLM
wLC L 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
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11.18 A propped cantilever beam is loaded as shown
in Fig. P11.18. Assume EI = 200,000 kN-m2. Use
discontinuity functions to determine:
(a) the reactions at A and C.
(b) the beam deflection at B.
Fig. P11.18
Solution
Moment equation:
(a) Support reactions:
150 kN 0
150 kN
y y y
y y
F A C
A C (a)
(150 kN)(7 m) (12 m) 0
(12 m) 1,050 kN-m
A y A
A y
M C M
M C (b)
Discontinuity expressions:
2 1 1 1
1 0 0 0
0 1 1 1
20 1 1
2
0 m 0 m 150 kN 7 m 12 m
( ) 0 m 0 m 150 kN 7 m 12 m
( ) 0 m 0 m 150 kN 7 m 12 m
( ) 0 m 0 m 150 kN 7 m 1
A y y
A y y
A y y
A y y
w x M x A x x C x
V x w x dx M x A x x C x
M x V x dx M x A x x C x
d vEI M x M x A x x C x
dx
12 m
1 2 2 2
1
150 kN0 m 0 m 7 m 12 m
2 2 2
y y
A
A CdvEI M x x x x C
dx (c)
2 3 3 3
1 2
150 kN0 m 0 m 7 m 12 m
2 6 6 6
y yAA CM
EI v x x x x C x C (d)
Boundary conditions and evaluate constants:
1at 0 m, 0 0dv
x Cdx
2at 0 m, 0 0x v C
2 3 3
2 3 3
150 kNat 12 m, 0 0 (12 m) (12 m) (5 m)
2 6 6
(72 m ) (288 m ) 3,125 kN-m
yA
A y
AMx v
M A (e)
(a) Solve for Ay, Cy, and MA:
Solve equations (a), (b), and (c) simultaneously to obtain the results:
88.3 kN 61.7 kN
310 kN-
88.3247
m
kN 61.6753 kN
309.8958 kN-m (ccw)
y y
A
A C
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
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(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 7 m) is computed as follows:
2 3
3
3
2
309.8958 kN-m 88.3247 kN(7 m) (7 m)
2 6
2,543.2219 kN-m
2,543.2219 kN-m0.0127161 m
200,12.72
000 kmm
N-m
B
B
EI v
v 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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11.19 A propped cantilever beam is loaded as
shown in Fig. P11.19. Assume EI = 200,000 kN-
m2. Use discontinuity functions to determine:
(a) the reactions at A and B.
(b) the beam deflection at C.
Fig. P11.19
Solution
Moment equation:
(a) Support reactions:
0y y y
y y
F A B
A B (a)
750 kN-m (5 m) 0
(5 m) 750 kN-m
A y A
A y
M B M
M B (b)
Discontinuity expressions:
2 1 1 2
1 0 0 1
0 1 1 0
20 1 1
2
0 m 0 m 5 m 750 kN-m 7.5 m
( ) 0 m 0 m 5 m 750 kN-m 7.5 m
( ) 0 m 0 m 5 m 750 kN-m 7.5 m
( ) 0 m 0 m 5 m
A y y
A y y
A y y
A y y
w x M x A x B x x
V x w x dx M x A x B x x
M x V x dx M x A x B x x
d vEI M x M x A x B x
dx
0750 kN-m 7.5 mx
1 2 2 1
10 m 0 m 5 m 750 kN-m 7.5 m2 2
y y
A
A BdvEI M x x x x C
dx (c)
2 3 3 2
1 2
750 kN-m0 m 0 m 5 m 7.5 m
2 6 6 2
y yAA BM
EI v x x x x C x C (d)
Boundary conditions and evaluate constants:
1at 0 m, 0 0dv
x Cdx
2at 0 m, 0 0x v C
2 3
2 3
at 5 m, 0 0 (5 m) (5 m)2 6
(12.5 m ) (20.83333 m ) 0
yA
A y
AMx v
M A (e)
(a) Solve for Ay, By, and MA:
Solve equations (a), (b), and (c) simultaneously to obtain the results:
225 kN 225 kN225.000 kN 225.000 kN
375.000 kN-m 375 kN-m (cw)
y y
A
A B
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
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(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 7.5 m) is computed as follows:
2 3 3
3
3
2
(7.5 m) (7.5 m) (2.5 m)2 6 6
4,687.500 kN-m
4,687.500 kN-m0.023438 m
200,000 kN-m23.4 mm
y yAC
C
A BMEI v
v 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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11.20 A propped cantilever beam is loaded as shown
in Fig. P11.20. Assume EI = 100,000 kip-ft2. Use
discontinuity functions to determine:
(a) the reactions at A and E.
(b) the beam deflection at C.
Fig. P11.20
Solution
Moment equation:
(a) Support reactions:
20 kips 30 kips 20 kips 0
70 kips
y y y
y y
F A E
A E (a)
(28 ft) 20 kips(21 ft)
30 kips(14 ft) 20 kips(7 ft) 0
(28 ft) 980 kip-ft
E y
E
y E
M A
M
A M (b)
Discontinuity expressions:
1 1 1 1
2 1
0 0 0 0
1 0
1
0 ft 20 kips 7 ft 30 kips 14 ft 20 kips 21 ft
28 ft 28 ft
( ) 0 ft 20 kips 7 ft 30 kips 14 ft 20 kips 21 ft
28 ft 28 ft
( ) 0 ft 20
y
E y
y
E y
y
w x A x x x x
M x E x
V x w x dx A x x x x
M x E x
M x V x dx A x1 1 1
0 1
21 1 1 1
2
0 1
kips 7 ft 30 kips 14 ft 20 kips 21 ft
28 ft 28 ft
( ) 0 ft 20 kips 7 ft 30 kips 14 ft 20 kips 21 ft
28 ft 28 ft
E y
y
E y
x x x
M x E x
d vEI M x A x x x x
dx
M x E x
2 2 2 2
1 2
1
20 kips 30 kips 20 kips0 ft 7 ft 14 ft 21 ft
2 2 2 2
28 ft 28 ft2
y
y
E
AdvEI x x x x
dx
EM x x C (c)
3 3 3 3
2 3
1 2
20 kips 30 kips 20 kips0 ft 7 ft 14 ft 21 ft
6 6 6 6
28 ft 28 ft2 6
y
yE
AEI v x x x x
EMx x C x C (d)
Boundary conditions and evaluate constants:
2at 0 ft, 0 0x v C
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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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3 3 3 3
1
at 28 ft, 0
20 kips 30 kips 20 kips(28 ft) (21 ft) (14 ft) (7 ft) (28 ft) 0
6 6 6 6
y
x v
AC (e)
2 2 2 2
1
at 28 ft, 0
20 kips 30 kips 20 kips(28 ft) (21 ft) (14 ft) (7 ft) 0
2 2 2 2
y
dvx
dx
AC (f)
(a) Solve for Ay, Ey, and ME:
Solve equations (e) and (f) simultaneously to obtain:
2
1 1,470.000 kip-ft
23.7500 ki 23.8 kips psy
C
A Ans.
With the value of Ay, calculate Ey and ME from equations (a) and (b), respectively.
46.2500 kips
315.000 kip-f
46.3 kips
315 kip-ft ( wt c )
y
E
E
M Ans.
(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 14 ft) is computed as follows:
3 3 2
3
3
2
23.75 kips 20 kips(14 ft) (7 ft) (1,470 kip-ft )(14 ft)
6 6
10,861.6667 kip-ft
10,861.6667 kip-ft0.108617 ft 1.3034 in.
100,000 ki1.303 in.
p-ft
C
C
EI v
v 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.
11.21 A propped cantilever beam is loaded as
shown in Fig. P11.21. Assume EI = 100,000 kip-
ft2. Use discontinuity functions to determine:
(a) the reactions at A and B.
(b) the beam deflection at x = 7 ft.
Fig. P11.21
Solution
Moment equation:
(a) Support reactions:
1(12 kips/ft)(16 ft) 0
2
96 kips
y y y
y y
F A B
A B (a)
1 2(16 ft)(16 ft) (12 kips/ft)(16 ft)
2 3
0
(16 ft) 1,024 kip-ft
B y
B
y B
M A
M
A M (b)
Discontinuity expressions:
1 0 1
2 1
0 1 2
1 0
1
12 kips/ft0 ft 12 kips/ft 0 ft 0 ft
16 ft
16 ft 16 ft
12 kips/ft( ) 0 ft 12 kips/ft 0 ft 0 ft
2(16 ft)
16 ft 16 ft
12 kips/ft( ) 0 ft 0
2
y
B y
y
B y
y
w x A x x x
M x B x
V x w x dx A x x x
M x B x
M x V x dx A x x2 3
0 1
21 2 3
2
0 1
12 kips/ft ft 0 ft
6(16 ft)
16 ft 16 ft
12 kips/ft 12 kips/ft( ) 0 ft 0 ft 0 ft
2 6(16 ft)
16 ft 16 ft
B y
y
B y
x
M x B x
d vEI M x A x x x
dx
M x B x
2 3 4
1 2
1
12 kips/ft 12 kips/ft0 ft 0 ft 0 ft
2 6 24(16 ft)
16 ft 16 ft2
y
y
B
AdvEI x x x
dx
BM x x C
(c)
3 4 5
2 3
1 2
12 kips/ft 12 kips/ft0 ft 0 ft 0 ft
6 24 120(16 ft)
16 ft 16 ft2 6
y
yB
AEI v x x x
BMx x C x C (d)
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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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Boundary conditions and evaluate constants:
2at 0 ft, 0 0x v C
3 4 5
1
at 16 ft, 0
12 kips/ft 12 kips/ft(16 ft) (16 ft) (16 ft) (16 ft) 0
6 24 120(16 ft)
y
x v
AC (e)
2 3 4
1
at 16 ft, 0
12 kips/ft 12 kips/ft(16 ft) (16 ft) (16 ft) 0
2 6 24(16 ft)
y
dvx
dx
AC (f)
(a) Solve for Ay, By, and MB:
Solve equations (e) and (f) simultaneously to obtain:
2
1 614.4000 kip-ft
52.8000 ki 52ps .8 kipsy
C
A Ans.
With the value of Ay, calculate By and MB from equations (a) and (b), respectively.
43.200 kips
179.200 kip-ft
43.2 kips
179.2 kip-ft (cw)
y
B
B
M Ans.
(b) Beam deflection at x = 7 ft: From Eq. (d), the beam deflection at x = 7 ft is computed as follows:
3 4 5 2
3
3
2
52.8 kips 12 kips/ft 12 kips/ft(7 ft) (7 ft) (7 ft) (614.400 kip-ft )(7 ft)
6 24 120(16 ft)
2,377.85625 kip-ft
2,377.85625 kip-ft0.023779 ft 0.2853 in.
100,000 ki0.285
p- i
fn
t.
EI v
v 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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11.22 A propped cantilever beam is loaded as
shown in Fig. P11.22. Assume EI = 200,000 kN-
m2. Use discontinuity functions to determine:
(a) the reactions at A and B.
(b) the beam deflection at C.
Fig. P11.22
Solution
Moment equation:
(a) Support reactions:
1(120 kN/m)(8 m) 0
2
480 kN
y y y
y y
F A B
A B (a)
1 2(8 m)(120 kN/m)(8 m) (6 m)
2 3
0
(6 m) 2,560 kN-m
A y
A
y A
M B
M
B M (b)
Discontinuity expressions:
2 1 1 1
1 0 2 0
0 1 3 1
20 1
2
120 kN/m0 m 0 m 0 m 6 m
8 m
120 kN/m( ) 0 m 0 m 0 m 6 m
2(8 m)
120 kN/m( ) 0 m 0 m 0 m 6 m
6(8 m)
1( ) 0 m 0 m
A y y
A y y
A y y
A y
w x M x A x x B x
V x w x dx M x A x x B x
M x V x dx M x A x x B x
d vEI M x M x A x
dx
3 120 kN/m0 m 6 m
6(8 m)yx B x
1 2 4 2
1
120 kN/m0 m 0 m 0 m 6 m
2 24(8 m) 2
y y
A
A BdvEI M x x x x C
dx (c)
2 3 5 3
1 2
120 kN/m0 m 0 m 0 m 6 m
2 6 120(8 m) 6
y yAA BM
EI v x x x x C x C (d)
Boundary conditions and evaluate constants:
1at 0 m, 0 0dv
x Cdx
2at 0 m, 0 0x v C
2 3 5120 kN/m
at 6 m, 0 (6 m) (6 m) (6 m) 02 6 120(8 m)
yAAM
x v (e)
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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) Solve for Ay, By, and MA:
Solve equations (a), (b), and (e) simultaneously to obtain:
66.5 kN 414 kN66.5000 kN
79.0 kN-m
413.5000 kN
79.0000 kN-m (ccw)
y y
A
A B
M Ans.
(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 8 m) is computed as follows:
2 3 5 3
3
3
2
79.0 kN-m 66.5 kN 120 kN/m 413.5 kN(8 m) (8 m) (8 m) (2 m)
2 6 120(8 m) 6
398.0000 kN-m
398.0000 kN-m0.001990 m 1.990 mm
200,1.990
000 kN-mm
m
EI v
v 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.
11.23 For the beam shown in Fig. P11.23, assume
EI = 200,000 kN-m2 and use discontinuity
functions to determine:
(a) the reactions at A, C, and D.
(b) the beam deflection at B.
Fig. P11.23
Solution
Moment equation:
(a) Support reactions:
(120 kN/m)(6 m) 0
720 kN
y y y y
y y y
F A C D
A C D (a)
(120 kN/m)(6 m)(3 m) (6 m)
(10 m) 0
(6 m) (10 m) 2,160 kN-m
A y
y
y y
M C
D
C D (b)
Discontinuity expressions:
1 0 0
1 1
0 1 1
0 0
1 2 2
1
0 m 120 kN/m 0 m 120 kN/m 6 m
6 m 10 m
( ) 0 m 120 kN/m 0 m 120 kN/m 6 m
6 m 10 m
120 kN/m 120 kN/m( ) 0 m 0 m 6 m
2 2
6 m 10 m
y
y y
y
y y
y
y y
w x A x x x
C x D x
V x w x dx A x x x
C x D x
M x V x dx A x x x
C x D x1
21 2 2
2
1 1
120 kN/m 120 kN/m( ) 0 m 0 m 6 m
2 2
6 m 10 m
y
y y
d vEI M x A x x x
dx
C x D x
2 3 3
2 2
1
120 kN/m 120 kN/m0 m 0 m 6 m
2 6 6
6 m 10 m2 2
y
y y
AdvEI x x x
dx
C Dx x C (c)
3 4 4
3 3
1 2
120 kN/m 120 kN/m0 m 0 m 6 m
6 24 24
6 m 10 m6 6
y
y y
AEI v x x x
C Dx x C x C (d)
Boundary conditions and evaluate constants:
2at 0 m, 0 0x v C
3 4
1
120 kN/mat 6 m, 0 (6 m) (6 m) (6 m) 0
6 24
yAx v C (e)
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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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3 4 4 3
1
at 10 m, 0
120 kN/m 120 kN/m(10 m) (10 m) (4 m) (4 m) (10 m) 0
6 24 24 6
y y
x v
A CC (f)
(a) Solve for Ay, Cy, and Dy:
Solve equations (a), (b), (e), and (f) simultaneously to obtain:
2
1 756.000 kN-m
306.0000 kN
4
306 kN
495 kN95.0000 kN
81. 81.0 k0000 N N k
y
y
y
C
A
C
D Ans.
(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 3 m) is computed as follows:
3 4 2
3
3
2
306.00 kN 120 kN/m(3 m) (3 m) (756.000 kN-m )(3 m)
6 24
1,296.0000 kN-m
1,296.0000 kN-m0.006480 m 6.48 mm
200,006.
0 48
k
N-m
mm
B
B
EI v
v 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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11.24 For the beam shown in Fig. P11.24, assume
EI = 100,000 kip-ft2 and use discontinuity
functions to determine:
(a) the reactions at A, C, and D.
(b) the beam deflection at B.
Fig. P11.24
Solution
Moment equation:
(a) Support reactions:
75 kips (7 kips/ft)(16 ft) 0
187 kips
y y y y
y y y
F A C D
A C D (a)
(75 kips)(8 ft) (7 kips/ft)(16 ft)(24 ft)
(16 ft) (32 ft) 0
(16 ft) (32 ft) 3,288 kip-ft
A
y y
y y
M
C D
C D (b)
Discontinuity expressions:
1 1 1
0 0 1
0 0 0
1 1 0
0 ft 75 kips 8 ft 16 ft
7 kips/ft 16 ft 7 kips/ft 32 ft 32 ft
( ) 0 ft 75 kips 8 ft 16 ft
7 kips/ft 16 ft 7 kips/ft 32 ft 32 ft
( ) 0 f
y y
y
y y
y
y
w x A x x C x
x x D x
V x w x dx A x x C x
x x D x
M x V x dx A x1 1 1
2 2 1
21 1 1
2
2 2 1
t 75 kips 8 ft 16 ft
7 kips/ft 7 kips/ft16 ft 32 ft 32 ft
2 2
( ) 0 ft 75 kips 8 ft 16 ft
7 kips/ft 7 kips/ft16 ft 32 ft 32 ft
2 2
y
y
y y
y
x C x
x x D x
d vEI M x A x x C x
dx
x x D x
2 2 2
3 3 2
1
75 kips0 ft 8 ft 16 ft
2 2 2
7 kips/ft 7 kips/ft16 ft 32 ft 32 ft
6 6 2
y y
y
A CdvEI x x x
dx
Dx x x C (c)
3 3 3
4 4 3
1 2
75 kips0 ft 8 ft 16 ft
6 6 6
7 kips/ft 7 kips/ft16 ft 32 ft 32 ft
24 24 6
y y
y
A CEI v x x x
Dx x x C x C (d)
Boundary conditions and evaluate constants:
2at 0 ft, 0 0x v C
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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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3 3
1
75 kipsat 16 ft, 0 (16 ft) (8 ft) (16 ft) 0
6 6
yAx v C (e)
3 3 3 4
1
at 32 ft, 0
75 kips 7 kips/ft(32 ft) (24 ft) (16 ft) (16 ft) (32 ft) 0
6 6 6 24
y y
x v
A CC (f)
(a) Solve for Ay, Cy, and Dy:
Solve equations (a), (b), (e), and (f) simultaneously to obtain:
2
1 601.3333 kip-ft
23.4688 kips
12
23.5 kips
121.6 kip1.5625 kips
41.9688 kips
s
42.0 kips
y
y
y
C
A
C
D Ans.
(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 8 ft) is computed as follows:
3 2
3
3
2
23.4688 kips(8 ft) (601.3333 kip-ft )(8 ft)
6
2,808.0000 k
0.337 in
ip-ft
2,808.0000 kip-ft0.028080 ft 0.3370 in.
100,000 ki.
p-ft
B
B
EI v
v Ans.
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11.25 For the propped cantilever beam shown in
Fig. P11.25, assume EI = 100,000 kip-ft2 and use
discontinuity functions to determine:
(a) the reactions at B and D.
(b) the beam deflection at C.
Fig. P11.25
Solution
Moment equation:
(a) Support reactions:
1(5 kips/ft)(10 ft)
2
(5 kips/ft)(12 ft) 0
85 kips
y y y
y y
F B D
B D (a)
1 10 ft(5 kips/ft)(10 ft) 20 ft
2 3
(5 kips/ft)(12 ft)(14 ft)
(20 ft) 0
(20 ft) 1,423.3333 kip-ft
D
y D
y D
M
B M
B M (b)
Discontinuity expressions:
1 1 0
1 0 0
2 1
1 1
5 kips/ft 5 kips/ft0 ft 10 ft 5 kips/ft 10 ft
10 ft 10 ft
10 ft 5 kips/ft 10 ft 5 kips/ft 22 ft
30 ft 30 ft
5 kips/ft 5 kips/ft0 ft 10 ft 5 kips/ft 22 f
10 ft 10 ft
y
D y
w x x x x
B x x x
M x D x
x x x0
2 1
2 2 0
1 1 0
3
t
30 ft 30 ft
5 kips/ft 5 kips/ft( ) 0 ft 10 ft 10 ft
2(10 ft) 2(10 ft)
5 kips/ft 22 ft 30 ft 30 ft
5 kips/ft 5 kips/ft( ) 0 ft 10 f
6(10 ft) 6(10 ft)
D y
y
D y
M x D x
V x w x dx x x B x
x M x D x
M x V x dx x x3 1
2 0 1
23 3 1
2
2 0 1
t 10 ft
5 kips/ft22 ft 30 ft 30 ft
2
5 kips/ft 5 kips/ft( ) 0 ft 10 ft 10 ft
6(10 ft) 6(10 ft)
5 kips/ft22 ft 30 ft 30 ft
2
y
D y
y
D y
B x
x M x D x
d vEI M x x x B x
dx
x M x D x
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4 4 2 3
1 2
1
5 kips/ft 5 kips/ft 5 kips/ft0 ft 10 ft 10 ft 22 ft
24(10 ft) 24(10 ft) 2 6
30 ft 30 ft2
y
y
D
BdvEI x x x x
dx
DM x x C (c)
5 5 3 4
2 3
1 2
5 kips/ft 5 kips/ft 5 kips/ft0 ft 10 ft 10 ft 22 ft
120(10 ft) 120(10 ft) 6 24
30 ft 30 ft2 6
y
yD
BEI v x x x x
DMx x C x C (d)
Boundary conditions and evaluate constants:
5
1 2
at 10 ft, 0
5 kips/ft(10 ft) (10 ft) 0
120(10 ft)
x v
C C (e)
5 5 3
4
1 2
at 30 ft, 0
5 kips/ft 5 kips/ft(30 ft) (20 ft) (20 ft)
120(10 ft) 120(10 ft) 6
5 kips/ft(8 ft) (30 ft) 0
24
y
x v
B
C C (f)
4 4 2 3
1
at 30 ft, 0
5 kips/ft 5 kips/ft 5 kips/ft(30 ft) (20 ft) (20 ft) (8 ft) 0
24(10 ft) 24(10 ft) 2 6
y
dvx
dx
BC (g)
(a) Solve for By, Dy, and MD:
Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain:
2
1
3
2
65.9 kip
59.0000 k
s
19.13 kips
105
ip-ft
1,006.6667 kip-ft
65.8700 kips
19.13
.9 kip-
00 kips
105.93 ft (cw33 )kip-ft
y
y
D
C
C
B
D
M Ans.
(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 22 ft) is computed as follows:
5 5 3
2 3
3
3
2
5 kips/ft 5 kips/ft 65.8700 kips(22 ft) (12 ft) (12 ft)
120(10 ft) 120(10 ft) 6
(59.0000 kip-ft )(22 ft) 1,006.6667 kip-ft
1,757.4400 kip-ft
1,757.4400 kip-ft0.017574 ft
100,000 kip-ft
C
C
EI v
v 0.2109 in. 0.211 in. Ans.
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11.26 For the beam shown in Fig. P11.26, assume
EI = 200,000 kN-m2 and use discontinuity
functions to determine:
(a) the reactions at B, C, and D.
(b) the beam deflection at A.
Fig. P11.26
Solution
Moment equation:
(a) Support reactions:
120 kN (60 kN/m)(12 m) 0
840 kN
y y y y
y y y
F B C D
B C D (a)
(120 kN)(3 m) (60 kN/m)(12 m)(6 m)
(6 m) (12 m) 0
(6 m) (12 m) 3,960 kN-m
B
y y
y y
M
C D
C D (b)
Discontinuity expressions:
1 1 0
1 1
0 0 1
0 0
1 1 2
1 1
2
120 kN 0 m 3 m 60 kN/m 3 m
9 m 15 m
( ) 120 kN 0 m 3 m 60 kN/m 3 m
9 m 15 m
60 kN/m( ) 120 kN 0 m 3 m 3 m
2
9 m 15 m
y
y y
y
y y
y
y y
w x x B x x
C x D x
V x w x dx x B x x
C x D x
M x V x dx x B x x
C x D x
d vEI
1 1 2
2
1 1
60 kN/m( ) 120 kN 0 m 3 m 3 m
2
9 m 15 m
y
y y
M x x B x xdx
C x D x
2 2 3
2 2
1
120 kN 60 kN/m0 m 3 m 3 m
2 2 6
9 m 15 m2 2
y
y y
BdvEI x x x
dx
C Dx x C (c)
3 3 4
3 3
1 2
120 kN 60 kN/m0 m 3 m 3 m
6 6 24
9 m 15 m6 6
y
y y
BEI v x x x
C Dx x C x C (d)
Boundary conditions and evaluate constants:
3
1 2
at 3 m, 0
120 kN(3 m) (3 m) 0
6
x v
C C (e)
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3 3 4
1 2
at 9 m, 0
120 kN 60 kN/m(9 m) (6 m) (6 m) (9 m) 0
6 6 24
y
x v
BC C (f)
3 3 4 3
1 2
at 15 m, 0
120 kN 60 kN/m(15 m) (12 m) (12 m) (6 m) (15 m) 0
6 6 24 6
y y
x v
B CC C (g)
(a) Solve for By, Cy, and Dy:
Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain:
2
1
3
2
330 kN
900.0000
360 kN
150.0 kN
kN-m
2,160.0000 kN-m
330.0000 kN
360.0000 kN
150.0000 kN
y
y
y
C
C
B
C
D Ans.
(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows:
3
3
2
2,160.0000 kN-m
2,160.0000 kN-m0.010800 m 10.80 mm
200,000 kN-m10.80 mm
A
A
EI v
v 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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11.27 For the beam shown in Fig. P11.27, assume
EI = 200,000 kN-m2 and use discontinuity
functions to determine:
(a) the reactions at B, C, and D.
(b) the beam deflection at A.
Fig. P11.27
Solution
Moment equation:
(a) Support reactions:
(60 kN/m)(6 m) 0
360 kN
y y y y
y y y
F B C D
B C D (a)
420 kN-m (60 kN/m)(6 m)(3 m)
(6 m) (12 m) 0
(6 m) (12 m) 660 kN-m
B
y y
y y
M
C D
C D (b)
Discontinuity expressions:
2 1 0
1 0 1
1 0 1
0 1 0
0 1
420 kN-m 0 m 3 m 60 kN/m 3 m
9 m 60 kN/m 9 m 15 m
( ) 420 kN-m 0 m 3 m 60 kN/m 3 m
9 m 60 kN/m 9 m 15 m
60 ( ) 420 kN-m 0 m 3 m
y
y y
y
y y
y
w x x B x x
C x x D x
V x w x dx x B x x
C x x D x
M x V x dx x B x2
1 2 1
20 1 2
2
1 2 1
kN/m3 m
2
60 kN/m9 m 9 m 15 m
2
60 kN/m( ) 420 kN-m 0 m 3 m 3 m
2
60 kN/m9 m 9 m 15 m
2
y y
y
y y
x
C x x D x
d vEI M x x B x x
dx
C x x D x
1 2 3
2 3 2
1
60 kN/m420 kN-m 0 m 3 m 3 m
2 6
60 kN/m9 m 9 m 15 m
2 6 2
y
y y
BdvEI x x x
dx
C Dx x x C (c)
2 3 4
3 4 3
1 2
420 kN-m 60 kN/m0 m 3 m 3 m
2 6 24
60 kN/m9 m 9 m 15 m
6 24 6
y
y y
BEI v x x x
C Dx x x C x C (d)
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Boundary conditions and evaluate constants:
2
1 2
at 3 m, 0
420 kN-m(3 m) (3 m) 0
2
x v
C C (e)
2 3 4
1 2
at 9 m, 0
420 kN-m 60 kN/m(9 m) (6 m) (6 m) (9 m) 0
2 6 24
y
x v
BC C (f)
2 3 4
3 4
1 2
at 15 m, 0
420 kN-m 60 kN/m(15 m) (12 m) (12 m)
2 6 24
60 kN/m(6 m) (6 m) (15 m) 0
6 24
y
y
x v
B
CC C (g)
(a) Solve for By, Cy, and Dy:
Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain:
2
1
3
2
1,590.0000 kN-m
2,880.0000 kN-m
245.0000 kN
120.0000 kN
245 kN
120 kN
5.00 kN5.0000 kN
y
y
y
C
C
B
C
D Ans.
(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows:
3
3
2
2,880.0000 kN-m
2,880.0000 kN-m0.014400 m 14.40 mm
200,000 kN-m14.40 mm
A
A
EI v
v 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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11.28a For the beams and loadings shown
below, assume that EI = 3.0 × 104 kN-m
2 is
constant for each beam.
(a) For the beam in Fig. P11.28a, determine
the concentrated upward force P required to
make the total beam deflection at B equal to
zero (i.e., vB = 0).
Fig. P11.28a
Solution
Downward deflection at B due to 15 kN/m uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
Values:
w = 15 kN/m, L = 7 m, a = 3.5 m,
EI = 3.0 × 104 kN-m
2
Computation:
32 2
3 32 2
(4 7 3 )24
(15 kN/m)(3.5 m) 234.472656 kN-m4(7 m) 7(3.5 m)(7 m) 3(3.5 m)
24(7 m)
B
wav L aL a
LEI
EI EI
Upward deflection at B due to concentrated load P.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48B
PLv
EI
Values:
L = 7 m, EI = 3.0 × 104 kN-m
2
Computation:
3 3 3(7 m) (7.145833 m )
48 48B
PL P Pv
EI EI EI
Compatibility equation at B:
3 3
3
3
234.472656 kN-m (7.145833 m )0
234.472656 kN-m32.8125 kN
7.14532.8 kN
833 m
P
EI EI
P Ans.
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11.28b For the beams and loadings shown
below, assume that EI = 3.0 × 104 kN-m
2 is
constant for each beam.
(b) For the beam in Fig. P11.28b, determine
the concentrated moment M required to
make the total beam slope at A equal to zero
(i.e., A = 0).
Fig. P11.28b
Solution
Slope at A due to 32 kN concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
2A
PL
EI (slope magnitude)
Values:
P = 32 kN, L = 4 m, EI = 3.0 × 104 kN-m
2
Computation:
2 2 2(32 kN)(4 m) 256 kN-m
(negative slope by inspection)2 2
A
PL
EI EI EI
Slope at A due to concentrated moment M.
[Appendix C, Cantilever beam with concentrated moment at tip.]
Relevant equation from Appendix C:
A
ML
EI (slope magnitude)
Values:
L = 4 m, EI = 3.0 × 104 kN-m
2
Computation:
(4 m) (4 m)
(positive slope by inspection)A
ML M M
EI EI EI
Compatibility equation at A:
2
2
64.
256 kN-m (4 m)0
256 kN-m0 kN-m
4 m
M
EI EI
M Ans.
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11.29a For the beams and loadings shown
below, assume that EI = 5.0 × 106 kip-in.
2 is
constant for each beam.
(a) For the beam in Fig. P11.29a, determine
the concentrated upward force P required to
make the total beam deflection at B equal to
zero (i.e., vB = 0).
Fig. P11.29a
Solution
Downward deflection at B due to 4 kips/ft uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 4 kips/ft, L = 13 ft, EI = 5.0 × 106 kip-in.
2
Computation:
4 4 3(4 kips/ft)(13 ft) 14,280.5 kip-ft
8 8B
wLv
EI EI EI
Upward deflection at B due to concentrated load P.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
L = 13 ft, EI = 5.0 × 106 kip-in.
2
Computation:
3 3 3(13 ft) (732.333333 ft )
3 3B
PL P Pv
EI EI EI
Compatibility equation at B:
3 3
3
319.50 kip
14,280.5 kip-ft (732.333333 ft )0
14,280.5 kip-ft
732.333333 s
ft
P
EI EI
P 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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11.29b For the beams and loadings shown
below, assume that EI = 5.0 × 106 kip-in.
2 is
constant for each beam.
(b) For the beam in Fig. P11.29b, determine
the concentrated moment M required to
make the total beam slope at C equal to zero
(i.e., C = 0).
Fig. P11.29b
Solution
Slope at C due to 40-kip concentrated load.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
2
16C
PL
EI (slope magnitude)
Values:
P = 40 kips, L = 18 ft, EI = 5.0 × 106 kip-in.
2
Computation:
2 2 2(40 kips)(18 ft) 810 kip-ft
(negative slope by inspection)16 16
C
PL
EI EI EI
Slope at C due to concentrated moment M.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
C
ML
EI (slope magnitude)
Values:
L = 18 ft, EI = 5.0 × 106 kip-in.
2
Computation:
(18 ft) (6 ft)
(positive slope by inspection)3 3
C
ML M M
EI EI EI
Compatibility equation at C:
2
2
810 kip-ft (6 ft)0
810 kip-f135.0
t
6 kip-f
ftt
M
EI EI
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
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11.30a For the beams and loadings shown
below, assume that EI = 5.0 × 104 kN-m
2
is constant for each beam.
(a) For the beam in Fig. P11.30a,
determine the concentrated downward
force P required to make the total beam
deflection at B equal to zero (i.e., vB = 0).
Fig. P11.30a
Solution
Upward deflection at B due to 105 kN-m concentrated moment.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
B
M xv L Lx x
LEI (elastic curve)
Values:
M = −105 kN-m, L = 8 m, x = 4 m,
EI = 5.0 × 104 kN-m
2
Computation:
2 2
32 2
(2 3 )6
( 105 kN-m)(4 m) 420 kN-m2(8 m) 3(8 m)(4 m) (4 m)
6(8 m)
B
M xv L Lx x
LEI
EI EI
Downward deflection at B due to concentrated load P.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48B
PLv
EI
Values:
L = 8 m, EI = 5.0 × 104 kN-m
2
Computation:
3 3 3(8 m) (10.666667 m )
48 48B
PL P Pv
EI EI EI
Compatibility equation at B:
3 3
3
3
420 kN-m (10.666667 m )0
420 kN-m39.375 kN
10.66666739.4 kN
m
P
EI EI
P 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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11.30b For the beams and loadings shown
below, assume that EI = 5.0 × 104 kN-m
2 is
constant for each beam.
(b) For the beam in Fig. P11.30b, determine
the concentrated moment M required to
make the total beam slope at A equal to zero
(i.e., A = 0).
Fig. P11.30b
Solution
Slope at A due to 6 kN/m uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
6A
wL
EI (slope magnitude)
Values:
w = 6 kN/m, L = 5 m, EI = 5.0 × 104 kN-m
2
Computation:
3 3 2(6 kN/m)(5 m) 125 kN-m
(positive slope by inspection)6 6
A
wL
EI EI EI
Slope at A due to concentrated moment M.
[Appendix C, Cantilever beam with concentrated moment at tip.]
Relevant equation from Appendix C:
A
ML
EI (slope magnitude)
Values:
L = 5 m, EI = 5.0 × 104 kN-m
2
Computation:
(5 m)
(negative slope by inspection)A
ML M
EI EI
Compatibility equation at A:
2
2
25.0 kN-
125 kN-m (5 m)0
125 kN-m
mm
5
M
EI EI
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
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11.31a For the beams and loadings shown
below, assume that EI = 8.0 × 106 kip-in.
2
is constant for each beam.
(a) For the beam in Fig. P11.31a,
determine the concentrated downward
force P required to make the total beam
deflection at B equal to zero (i.e., vB = 0).
Fig. P11.31a
Solution
Upward deflection at B due to 125 kip-ft concentrated moment.
[Appendix C, Cantilever beam with concentrated moment at tip.]
Relevant equation from Appendix C:
2
2B
MLv
EI
Values:
M = −125 kip-ft, L = 15 ft, EI = 8.0 × 106 kip-in.
2
Computation:
2 2 3( 125 kip-ft)(15 ft) 14,062.5 kip-ft
2 2B
MLv
EI EI EI
Downward deflection at B due to concentrated load P.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
L = 15 ft, EI = 8.0 × 106 kip-in.
2
Computation:
3 3 3(15 ft) (1,125 ft )
3 3B
PL P Pv
EI EI EI
Compatibility equation at B:
3 3
3
3
14,062.5 kip-ft (1,125 ft )0
14,062.5 kip-ft
1,125 12.50 kips
ft
P
EI EI
P 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.
11.31b For the beams and loadings shown
below, assume that EI = 8.0 × 106 kip-in.
2
is constant for each beam.
(b) For the beam in Fig. P11.31b,
determine the concentrated moment M
required to make the total beam slope at A
equal to zero (i.e., A = 0). Fig. P11.31b
Solution
Slope at A due to 7 kips/ft uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over portion of span.]
Relevant equation from Appendix C:
2
2 2(2 )24
A
waL a
LEI (slope magnitude)
Values:
w = 7 kips/ft, L = 23 ft, a = 15 ft,
EI = 8.0 × 106 kip-in.
2
Computation:
22 2
22 2
2
(2 )24
(7 kips/ft)(15 ft)2(23 ft) (15 ft)
24(23 ft)
2,376.766304 kip-ft(negative slope by inspection)
A
waL a
LEI
EI
EI
Slope at A due to concentrated moment M.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
A
ML
EI (slope magnitude)
Values:
L = 23 ft, EI = 8.0 × 106 kip-in.
2
Computation:
(23 ft) (7.666667 ft)
(positive slope by inspection)3 3
A
ML M M
EI EI EI
Compatibility equation at A:
2
2
310 kip-f
2,376.766304 kip-ft (7.666667 ft)0
2,376.766304 kip-ft
7.666667 ftt
M
EI EI
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
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11.32 For the beam and loading shown
below, derive an expression for the reactions
at supports A and B. Assume that EI is
constant for the beam.
Fig. P11.32
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at B due to concentrated moment M0.
[Appendix C, Cantilever beam with concentrated moment at tip.]
Relevant equation from Appendix C:
2 2
0
2 2B
ML M Lv
EI EI
Consider upward deflection of cantilever beam at B due to concentrated load By.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
33
3 3
y
B
B LPLv
EI EI
Compatibility equation for deflection at B:
32
0 002
3
3 2
y
y
B LM LB
EI E
M
I L Ans.
Equilibrium equations
for entire beam:
0 03
2
3
20y y y y y
MF A B A B
L
M
L Ans.
0 0A A yM M M B L
0 00 0 0
03 3( )
2 (c )
22wA y
M MM B L M L M
L
MM 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.
11.33 For the beam and loading shown
below, derive an expression for the reactions
at supports A and B. Assume that EI is
constant for the beam.
Fig. P11.33
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at B due to linearly distributed load.
[Appendix C, Cantilever beam with linearly distributed load.]
Relevant equation from Appendix C:
4
0
30B
w Lv
EI
Consider upward deflection of cantilever beam at B due to concentrated load By.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
33
3 3
y
B
B LPLv
EI EI
Compatibility equation for deflection at B:
3
0
4
0 030 3 10
y
y
B Lw LB
EI E
w L
I Ans.
Equilibrium equations for entire beam:
0 0 0 00 40
2 2 10
2
510y y y y
w L w L w L w LF A B A
w L Ans.
0 02 3
A A y
w L LM M B L
2 2 2
0 0 0 0
2
0( )6 1
(ccw)150 6 15
A y
w L w L w L w LM B
LL L
w 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.
11.34 For the beam and loading shown below,
derive an expression for the reactions at
supports A and B. Assume that EI is constant
for the beam.
Fig. P11.34
Solution
Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at A due to concentrated load P.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
(3 )6
A
Pxv L x
EI (elastic curve)
2 3
3Let ,
2
( ) 3 73
6 2 12A
Lx L L
P L L PLv L
EI EI
Consider upward deflection of cantilever beam at A due to concentrated load Ay.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
33
3 3
y
A
A LPLv
EI EI
Compatibility equation for deflection at A:
3370
12 3
7
4
y
y
A LPLA
EI EI
P Ans.
Equilibrium equations for entire beam:
0
7 3
4 4
3
4
y y y
y
F A B P
P PB P
P Ans.
3
02
B B y
LM M A L P
3 7 3
( (cc42
w)4 4
)2
B y
L P PL PLM A L P L
PL 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.
11.35 For the beam and loading shown below,
derive an expression for the reactions at
supports A and B. Assume that EI is constant
for the beam.
Fig. P11.35
Solution
Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at A due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 )24
A
wxv L Lx x
EI (elastic curve)
22 42
3Let ,
2
( ) 3 3 176 4 ( ) ( )
24 2 2 48A
Lx L L
w L L L wLv L L
EI EI
Consider upward deflection of cantilever beam at A due to concentrated load Ay.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
33
3 3
y
A
A LPLv
EI EI
Compatibility equation for deflection at A:
34170
48 3
17
16
y
y
A LwL
EI
LA
E
w
I 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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Equilibrium equations for entire beam:
3 3 17 7 7
11 60
2 2 16 6y y y y
wL wL wL wLF A B B
wL Ans.
3 30
2 4B B y
wL LM M A L
2 22 29 17 9
( )8 16 8
6
(cw)161
B y
wL wL wL w wLLM A L 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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11.36 For the beam and loading shown
below, derive an expression for the reactions
at supports A and B. Assume that EI is
constant for the beam.
Fig. P11.36
Solution
Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at A due to concentrated load P.
[Appendix C, Cantilever beam with concentrated load at midspan.]
Relevant equation from Appendix C:
35
48A
PLv
EI
3 3
Let 2
5 (2 ) 5
48 6A
L L
P L PLv
EI EI
Consider upward deflection of cantilever beam at A due to concentrated load Ay.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3A
PLv
EI
3 3
Let 2 ,
( )(2 ) 8
3 3
y
y y
A
L L P A
A L A Lv
EI EI
Compatibility equation for deflection at A:
33 850
6 3
5
16
y
y
A LPLA
EI EI
P Ans.
Equilibrium equations for entire beam:
0
5 11 11
11 16 66
y y y
y
F A B P
P PB
PP Ans.
(2 ) ( ) 0B B yM M A L P L
3
(c5 3
(2 ) ( )8 8
w)8
B y
PL PLM A L P
PLL PL 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.
11.37 For the beam and loading shown
below, derive an expression for the
reactions at supports A and C. Assume that
EI is constant for the beam.
Fig. P11.37
Solution
Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at C due to concentrated moment M0.
[Appendix C, Cantilever beam with concentrated moment.]
Relevant equations from Appendix C:
2
and2
B B
ML MLv
EI EI
2 2
0 0 03( )
2 2C
M L M L M Lv L
EI EI EI
Consider upward deflection of cantilever beam at C due to concentrated load Cy.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3C
PLv
EI
3 3
Let 2 ,
( )(2 ) 8
3 3
y
y y
C
L L P C
C L C Lv
EI EI
Compatibility equation for deflection at C:
32
0083
02 3
9
16
y
y
C LM L MC
E EI LI Ans.
Equilibrium equations for entire beam:
0 0
0
9
16
9
16
y y y
y
M
F A
L
C
MA
L Ans.
0 (2 ) 0A A yM M M C L
0 00 0
09(2 ) (2 ) (c )
6w
8 81A y
M MM C L M M
ML
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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
11.38 For the beam and loading shown
below, derive an expression for the
reactions at supports A and C. Assume that
EI is constant for the beam.
Fig. P11.38
Solution
Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at C due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equations from Appendix C:
4 3
and8 6
B B
wL wLv
EI EI
4 3 47
( )8 6 24
C
wL wL wLv L
EI EI EI
Consider upward deflection of cantilever beam at C due to concentrated load Cy.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3C
PLv
EI
3 3
Let 2 ,
( )(2 ) 8
3 3
y
y y
C
L L P C
C L C Lv
EI EI
Compatibility equation for deflection at C:
34 870
24 3
7
64
y
y
C LwL
EI I
LC
w
E Ans.
Equilibrium equations for entire beam:
57
64
0
7
64
y y y
y
F A C wL
wLA wL
wL Ans.
(2 ) 02
A A y
LM M wL C L
2 22 2 27 18 9
(2 ) (2 )2 64 2 64
9 (ccw)
3 322A y
wL wL wL wLwL wLM C L 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
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
11.39 For the beam and loading shown
below, derive an expression for the reaction
forces at A, C, and D. Assume that EI is
constant for the beam. (Reminder: The
roller symbol implies that both upward and
downward displacement is restrained.)
Fig. P11.39
Solution
Choose the reaction force at C as the redundant; therefore, the released beam is simply supported.
Consider downward deflection of simply supported beam at C due to P.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6C
Pbxv L b x
LEI
2 2 2
32
Let 3 , ,
( )( )(3 ) ( ) ( )
6(3 )
77
18 18
C
L L b L x L
P L Lv L L L
L EI
PL PLL
EI EI
Consider upward deflection of simply supported beam at C due to concentrated load Cy.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6C
Pabv L a b
LEI
2 2 2
3
2
Let 3 , 2 , ,
( )(2 )( )(3 ) (2 ) ( )
6(3 )
2 84
18 18
y
y
C
y y
L L a L b L P C
C L Lv L L L
L EI
C L C LL
EI EI
Compatibility equation for deflection at C:
33 870
18 1
7
8 8
y
y
C LPLC
EI EI
P Ans.
Equilibrium equations for entire beam:
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(2 ) (3 ) 0A y yM PL C L D L
7 3(3 ) (2 ) (2 )
8 4
4 4
y y
y
P PLD L PL C L PL L
PD
P Ans.
0
7 5 3
8 4 8 8
3
8
y y y y
y y y
F A C D P
P PP P PA P C D P P 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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11.40 For the beam and loading shown
below, derive an expression for the reaction
force at B. Assume that EI is constant for
the beam. (Reminder: The roller symbol
implies that both upward and downward
displacement is restrained.)
Fig. P11.40
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is simply supported.
Consider upward deflection of simply supported beam at B due to M0.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 ) (elastic curve)
6B
M xv L Lx x
LEI
0
22 2 20 0 0
Let 2 , ,
( )( )2(2 ) 3(2 )( ) ( ) 3
6(2 ) 12 4B
L L x L M M
M L M M Lv L L L L L
L EI EI EI
Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48B
PLv
EI
3 3
Let 2 ,
( )(2 )
48 6
y
y y
B
L L P B
B L B Lv
EI EI
Compatibility equation for deflection at B:
32
0 0 030
4 6 2
3
2
y
y
B LM L MB
EI EI
M
LL 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.
11.41 For the beam and loading shown
below, derive an expression for the reaction
force at B. Assume that EI is constant for
the beam.
Fig. P11.41
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is simply supported.
Consider downward deflection of simply supported beam at B due to uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
3 2 42 2 2
Let 3 , 2
(2 ) 24(3 ) 7(2 )(3 ) 3(2 ) 6
24(3 ) 9 3B
L L a L
w L wL wLv L L L L L
L EI EI EI
Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6B
Pabv L a b
LEI
3
2 2 2 2
Let 3 , , 2 ,
( )( )(2 ) 4(3 ) ( ) (2 ) 4
6(3 ) 9 9
y
y y y
B
L L a L b L P B
B L L B L B Lv L L L L
L EI EI EI
Compatibility equation for deflection at B:
34 42 30
3 9
3
22
y
y
B LwL wLB
EI EI
wL 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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11.42 For the beam and loading shown
below, derive an expression for the reaction
force at B. Assume that EI is constant for
the beam. (Reminder: The roller symbol
implies that both upward and downward
displacement is restrained.)
Fig. P11.42
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is simply supported.
Consider upward deflection of simply supported beam at B due to uniformly distributed load.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 ) (elastic curve)
6B
M xv L Lx x
LEI
2
2
2 42 2 2
Let 2 , , 2 4 8
( )8
2(2 ) 3(2 )( ) ( ) 36(2 ) 96 32
B
L L wLL L x L M w
wLL
wL wLv L L L L L
L EI EI EI
Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48B
PLv
EI
3 3
Let 2 ,
( )(2 )
48 6
y
y y
B
L L P B
B L B Lv
EI EI
Compatibility equation for deflection at B:
34 30
32 6 16
3
16
y
y
B LwL wLB
EI EI
wL 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.
11.43 For the beam and loading shown
below, derive an expression for the reaction
force at B. Assume that EI is constant for
the beam.
Fig. P11.43
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is simply supported.
Consider downward deflection of simply supported beam at B due to one concentrated load P.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6B
Pbxv L b x
LEI
32 2 2 2
Let 4 , , 2
( )(2 ) 11(4 ) ( ) (2 ) 11
6(4 ) 12 12B
L L b L x L
P L L PL PLv L L L L
L EI EI EI
The second concentrated load will cause an additional deflection at B of the same magnitude.
Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48B
PLv
EI
3 3 3
Let 4 ,
( )(4 ) 64 16
48 48 12
y
y y y
B
L L P B
B L B L B Lv
EI EI EI
Compatibility equation for deflection at B:
33 3 1611 110
12 1 82
11
2 1
y
y
B LPL PLB
I EI
P
E EI 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.
11.44 For the beam and loading shown
below, derive an expression for the reaction
force at B. Assume that EI is constant for
the beam.
Fig. P11.44
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is simply supported.
Consider downward deflection of simply supported beam at B due to uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
3 2 42 2 2
Let 5 , 3
(3 ) 27 994(5 ) 7(3 )(5 ) 3(3 ) 22
24(5 ) 120 20B
L L a L
w L wL wLv L L L L L
L EI EI EI
Consider downward deflection of simply supported beam at B due to concentrated load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6B
Pbxv L b x
LEI
2 42 2 2 2
Let 5 , , 3 , 3
( )(3 )3
(5 ) ( ) (3 ) 156(5 ) 30 2
B
wLL L b L x L P
wLL L
wL wLv L L L L
L EI EI EI
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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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Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
B
Pabv L a b
LEI
3
2 2 2 2
Let 5 , 3 , 2 ,
( )(3 )(2 ) 12(5 ) (3 ) (2 ) 12
6(5 ) 5 5
y
y y y
B
L L a L b L P B
B L L B L B Lv L L L L
L EI EI EI
Compatibility equation for deflection at B:
34 4
2.21299 109
020 2 5 48
70833y
y
B LwL wL wLB
EI EI EIwL 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.
11.45 The beam shown in Fig. P11.45 consists of a
W360 × 79 structural steel wide-flange shape [E =
200 GPa; I = 225 × 106 mm
4]. For the loading
shown, determine:
(a) the reactions at A, B, and C.
(b) the magnitude of the maximum bending stress
in the beam.
Fig. P11.45
Solution
(a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released
beam is simply supported between A and C.
Consider downward deflection of simply supported beam at B due to uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
Values:
w = 90 kN/m, L = 9 m, a = 6 m
Calculation:
32 2
3 32 2
(4 7 3 )24
(90 kN/m)(6 m) 4,860 kN-m4(9 m) 7(6 m)(9 m) 3(6 m)
24(9 m)
B
wav L aL a
LEI
EI EI
Consider downward deflection of simply supported beam at B due to concentrated moment.
[Appendix C, SS beam with concentrated moment at one end of span.]
Relevant equation from Appendix C:
2 2(2 3 ) (elastic curve)
6B
M xv L Lx x
LEI
Values:
M = 180 kN-m, L = 9 m, x = 3 m
Calculation:
2 2
32 2
(2 3 )6
(180 kN-m)(3 m) 900 kN-m2(9 m) 3(9 m)(3 m) (3 m)
6(9 m)
B
M xv L Lx x
LEI
EI EI
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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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Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
B
Pabv L a b
LEI
Values:
P = −By, L = 9 m, a = 3 m, b = 6 m
Calculation:
2 2 2
3
2 2 2
( )6
( )(3 m)(6 m) (12 m )(9 m) (3 m) (6 m)
6(9 m)
B
y y
Pabv L a b
LEI
B B
EI EI
Compatibility equation for deflection at B:
33 3
3
3
(12 m )4,860 kN-m 900 kN-m0
5,760 kN-m480 480 kkN
12 N
m
y
y
B
EI EI EI
B Ans.
Equilibrium equations for entire beam: (3 m) (9 m) 180 kN-m (90 kN/m)(6 m)(6 m) 0A y yM B C
180 kN-m (90 kN/m)(6 m)(6 m) (480 kN)(3 m)
9 m
220.0 kN 220 kN
yC
Ans.
(90 kN/m)(6 m) 0y y y yF A B C
(90 kN/m)(6 m) 480 kN 220 kN 160.0 kN 160.0 kNyA Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
6 4
3 3
225 10 mm 353 mm
1,270 10 mm
I d
S
Maximum bending moment magnitude
Mmax = 300 kN-m
Bending stresses at maximum moment
2
6 4
(300 kN-m)(353 mm/2)(1,000)
225 10
235 MPa
mmx
Ans.
or using the tabulated section modulus value:
2
3 3
(300 kN-m)(1,000)
1,270 10 m
23 M
m
6 Pa
x
Ans.
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11.46 The beam shown in Fig. P11.46 consists of a
W610 × 140 structural steel wide-flange shape [E
= 200 GPa; I = 1,120 × 106 mm
4]. For the loading
shown, determine:
(a) the reactions at A, B, and D.
(b) the magnitude of the maximum bending stress
in the beam.
Fig. P11.46
Solution
(a) Reactions at A, B, and D. Choose the reaction force at B as the redundant; therefore, the released
beam is simply supported between A and D.
Consider downward deflection of simply supported beam at B due to uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3 2 3( 2 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 60 kN/m, L = 7.5 m, x = 1.5 m
Calculation:
3 2 3
3
3 2 3
( 2 )24
(60 kN/m)(1.5 m) 1,468.125 kN-m(7.5 m) 2(7.5 m)(1.5 m) (1.5 m)
24
B
wxv L Lx x
EI
EI EI
Consider downward deflection of simply supported beam at B due to concentrated load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6B
Pbxv L b x
LEI
Values:
P = 125 kN, L = 7.5 m, b = 2.5 m, x = 1.5 m
Calculation:
2 2 2
32 2 2
( )6
(125 kN)(2.5 m)(1.5 m) 497.396 kN-m(7.5 m) (2.5 m) (1.5 m)
6(7.5 m)
B
Pbxv L b x
LEI
EI EI
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Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
B
Pabv L a b
LEI
Values:
P = −By, L = 7.5 m, a = 1.5 m, b = 6 m
Calculation:
2 2 2
3
2 2 2
( )6
( )(1.5 m)(6 m) (3.6 m )(7.5 m) (1.5 m) (6 m)
6(7.5 m)
B
y y
Pabv L a b
LEI
B B
EI EI
Compatibility equation for deflection at B:
33 3
3
3
(3.6 m )1,468.125 kN-m 497.396 kN-m0
1,965.521 kN-m545.978 kN
3.6 546 k
mN
y
y
B
EI EI EI
B Ans.
Equilibrium equations for entire beam: (1.5 m) (7.5 m) (60 kN/m)(7.5 m)(3.75 m) (125 kN)(5 m) 0A y yM B D
(60 kN/m)(7.5 m)(3.75 m) (125 kN)(5 m) (545.978 kN)(1.5 m)
7.5 m
199.138 kN 199.1 kN
yD
Ans.
(60 kN/m)(7.5 m) 125 kN 0y y y yF A B D
(60 kN/m)(7.5 m) 125 kN 545.978 kN 199.138 kN
170.116 170.1 kNkN
yA
Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
6 4
3 3
1,120 10 mm 617 mm
3,640 10 mm
I d
S
Maximum bending moment magnitude
Mmax = 322.67 kN-m
Bending stresses at maximum moment
2
6 4
(322.67 kN-m)(617 mm/2)(1,000)
1,120 10 mm
88.9 MPa
x
Ans.
or using the tabulated section modulus value:
2
3 3
(322.67 kN-m)(1,000
88.6
)
3,640 10 m
M a
m
P
x
Ans.
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11.47 A propped cantilever beam is loaded as
shown in Fig. P11.47. Determine the reactions at A
and D for the beam. Assume EI = 12.8 × 106 lb-in.
2.
Fig. P11.47
Solution
Choose the reaction force at D as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at D due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equations from Appendix C:
4 3
and8 6
C C
wL wLv
EI EI
Values:
w = 25 lb/in., L = 72 in.
Calculation:
4 4 3
3 3 2
3 2 3
(25 lb/in.)(48 in.) 16,588,800 lb-in.
8 8
(25 lb/in.)(48 in.) 460,800 lb-in.
6 6
16,588,800 lb-in. 460,800 lb-in. 27,648,000 lb-in.(24 in.)
C
C
D
wLv
EI EI EI
wL
EI EI EI
vEI EI EI
Consider downward deflection of cantilever beam at D due to the 360-lb concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI
Values:
P = 360 lb, L = 24 in.
Calculation:
3 3 3
2 2 2
3 2 3
(360 lb)(24 in.) 1,658,880 lb-in.
3 3
(360 lb)(24 in.) 103,680 lb-in.
2 2
1,658,880 lb-in. 103,680 lb-in. 6,635,520 lb-in.(48 in.)
B
B
D
PLv
EI EI EI
PL
EI EI EI
vEI EI EI
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Consider upward deflection of cantilever beam at D due to concentrated load Dy.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3D
PLv
EI
Values:
P = −Dy, L = 72 in.
Calculation:
3 33 ( )(72 in.) (124,416 in. )
3 3
y y
D
D DPLv
EI EI EI
Compatibility equation for deflection at D:
33 3
3
3
(124,416 in. )27,648,000 lb-in. 6,635,520 lb-in.0
34,283,520 lb-in.275.556 lb
124,416 276 l
in.b
y
y
D
EI EI EI
D Ans.
Equilibrium equations for entire beam:
Shear-force and bending-moment diagrams
(25 lb/in.)(48 in.) 360 lb 0
(25 lb/in.)(48 in.) 360 lb 275.556 lb 1,28 1,4 284 l.4 b44 lb
y y y
y
F A D
A Ans.
(25 lb/in.)(48 in.)(24 in.) (360 lb)(24 in.) (72 in.) 0A A yM M D
(275.556 lb)(72 in.) (25 lb/in.)(48 in.)(24 in.) (360 lb)(24 in.)
17,600 17,600 lb-in. (clb-in. cw)
AM
Ans.
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11.48 A propped cantilever beam is loaded as
shown in Fig. P11.48. Assume EI = 24 × 106
kip-in.2. Determine:
(a) the reactions at B and C for the beam.
(b) the beam deflection at A.
Fig. P11.48
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at B due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8B
wLv
EI
Values:
w = 8 kips/ft, L = 24 ft
Calculation:
4 4 3(8 kips/ft)(24 ft) 331,776 kip-ft
8 8B
wLv
EI EI EI (a)
Consider downward deflection of cantilever beam at B due to the 40-kip concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
(3 ) (elastic curve)6
B
Pxv L x
EI
Values:
P = 40 kips, L = 36 ft, x = 24 ft
Calculation:
2 2 3(40 kips)(24 ft) 322,560 kip-ft
(3 ) 3(36 ft) (24 ft)6 6
B
Pxv L x
EI EI EI
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Consider upward deflection of cantilever beam at B due to concentrated load By.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −By, L = 24 ft
Calculation:
3 33 ( )(24 ft) (4,608 ft )
3 3
y y
B
B BPLv
EI EI EI (b)
Compatibility equation for deflection at B:
33 3
3
3
(4,608
142.0
ft )331,776 kip-ft 322,560 kip-ft0
654,336 kip-ft142.0 kips
4,608 ftkips
y
y
B
EI EI EI
B Ans.
Equilibrium equations for entire beam:
40 kips (8 kips/ft)(24 ft) 0
40 kips (8 kips/ft)(24 ft) 142. 90.0 k0 kip ss ip
y y y
y
F B C
C Ans.
(40 kips)(36 ft) (8 kips/ft)(24 ft)(12 ft) (24 ft) 0C C yM M B
(142.0 kips)(24 ft) (40 kips)(36 ft) (8 kips/ft)(24 ft)(12 ft)
336. 30 kip-f 36 kip-ft t (cw)
CM
Ans.
(b) Beam deflection at A:
Consider downward deflection of cantilever beam at A due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equations from Appendix C:
4 3
and (slope magnitude)8 6
B B
wL wLv
EI EI
Values:
w = 8 kips/ft, L = 24 ft, EI = 24 × 106 kip-in.
2
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Calculation:
3
3 3 2
3 2 3
331,776 kip-ftcalculated previously in Eq. (a)
(8 kips/ft)(24 ft) 18,432 kip-ft
6 6
331,776 kip-ft 18,432 kip-ft 552,960 kip-ft(12 ft)
B
B
A
vEI
wL
EI EI EI
vEI EI EI
Consider downward deflection of cantilever beam at A due to the 40-kip concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3A
PLv
EI
Values:
P = 40 kips, L = 36 ft, EI = 24 × 106 kip-in.
2
Calculation:
3 3 3(40 kips)(36 ft) 622,080 kip-ft
3 3A
PLv
EI EI EI
Consider upward deflection of cantilever beam at B due to concentrated load By.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and (slope magnitude)3 2
B B
PL PLv
EI EI
Values:
P = −By = 142 kips, L = 24 ft, EI = 24 × 106 kip-in.
2
Calculation:
3 3
2 2 2
3 2 3
(4,608 ft )(142.0 kips) 654,336 kip-ftusing the results from Eq. (b)
(142.0 kips)(24 ft) 40,896 kip-ft
2 2
654,336 kip-ft 40,896 kip-ft 1,145,088 kip-ft(12 ft)
B
B
A
vEI EI
PL
EI EI EI
vEI EI EI
Beam deflection at A.
3 3 3
3 3 3
6 2
552,960 kip-ft 622,080 kip-ft 1,145,088 kip-ft
29,952 kip-ft (29,952 kip-ft )(12 in./ft)2.156544 in.
24 10 kip-i2.16 i
n.n.
AvEI EI EI
EI 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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11.49 A propped cantilever beam is loaded
as shown in Fig. P11.49. Assume EI = 86.4
× 106 N-mm
2. Determine:
(a) the reactions at A and C for the beam.
(b) the beam deflection at B.
Fig. P11.49
Solution
Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at C due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 ) (elastic curve)24
C
wxv L Lx x
EI
Values:
w = 25 N/mm, L = 400 mm, x = 300 mm,
EI = 86.4 × 106 N-mm
2
Calculation:
22 2
22 2
9 3
(6 4 )24
(25 N/mm)(300 mm)6(400 mm) 4(400 mm)(300 mm) (300 mm)
24
53.43750 10 N-mm
C
wxv L Lx x
EI
EI
EI
Consider downward deflection of cantilever beam at C due to the 4,000-N concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and (slope magnitude)3 2
B B
PL PLv
EI EI
Values:
P = 4,000 N, L = 120 mm, EI = 86.4 × 106 N-mm
2
Calculation:
3 3 9 3
2 2 7 2
9 3 7 2 9 3
(4,000 N)(120 mm) 2.304 10 N-mm
3 3
(4,000 N)(120 mm) 2.88 10 N-mm
2 2
2.304 10 N-mm 2.88 10 N-mm 7.488 10 N-mm(180 mm)
B
B
C
PLv
EI EI EI
PL
EI EI EI
vEI EI EI
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Consider upward deflection of cantilever beam at C due to concentrated load Cy.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3C
PLv
EI
Values:
P = −Cy, L = 300 mm, EI = 86.4 × 106 N-mm
2
Calculation:
3 6 33 ( )(300 mm) (9 10 mm )
3 3
y y
C
C CPLv
EI EI EI
Compatibility equation for deflection at C:
6 39 3 9 3
9 3
6 3
(9 10 mm )53.43
6,
750 10 N-mm 7.488 10 N-mm0
60.9255 10770 N
N-mm6,769.5 N
9 10 mm
y
y
C
EI EI EI
C Ans.
Equilibrium equations for entire beam:
4,000 N (25 N/mm)(400 mm) 0
4,000 N (25 N/mm)(400 mm) 6,769.5 7,230 NN 7,230.5 N
y y y
y
F A C
A Ans.
(4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm) (300 mm) 0A A yM M C
(6,769.5 N)(300 mm) (4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm)
449,15 449,000 N-mm (c0 N-mm cw)
AM
Ans.
(b) Beam deflection at B:
Consider downward deflection of cantilever beam at B due to uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 25 N/mm, L = 400 mm, x = 120 mm,
EI = 86.4 × 106 N-mm
2
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Calculation:
22 2
22 2
9 3
(6 4 )24
(25 N/mm)(120 mm)6(400 mm) 4(400 mm)(120 mm) (120 mm)
24
11.736 10 N-mm
B
wxv L Lx x
EI
EI
EI
Consider downward deflection of cantilever beam at B due to the 4,000-N concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = 4,000 N, L = 120 mm, EI = 86.4 × 106 N-mm
2
Calculation:
3 3 9 3(4,000 N)(120 mm) 2.304 10 N-mm
3 3B
PLv
EI EI EI
Consider upward deflection of cantilever beam at B due to concentrated load Cy.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
(3 )6
B
Pxv L x
EI
Values:
P = −Cy = −6,769.5 N, L = 300 mm, x = 120 mm,
EI = 86.4 × 106 N-mm
2
Calculation:
2 2
9 3
( 6,769.5 N)(120 mm)(3 ) 3(300 mm) (120 mm)
6 6
12.6725 10 N-mm
B
Pxv L x
EI EI
EI
Beam deflection at B.
9 3 9 3 9 3
9 3 9 3
6 2
11.736 10 N-mm 2.304 10 N-mm 12.6725 10 N-mm
1.367496 10 N-mm 1.367496 10 N-mm15.8275 mm
86.4 1015.83 m
mm
N-m
BvEI EI EI
EI 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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11.50 The beam shown in Fig. P11.50 consists of a
W610 × 82 structural steel wide-flange shape [E =
200 GPa; I = 562 × 106 mm
4]. For the loading shown,
determine:
(a) the reaction force at C.
(b) the beam deflection at A. Fig. P11.50
Solution
(a) Reaction force at C. Choose the reaction force at C as the redundant; therefore, the released beam
is simply supported between B and D.
Consider upward deflection of simply supported beam at C due to uniformly distributed load on
overhang AB. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 ) (elastic curve)
6C
M xv L Lx x
LEI
Values:
M = −(105 kN/m)(3 m)(1.5 m) = −472.5 kN-m,
L = 14 m, x = 7 m
Calculation:
2 2
32 2
(2 3 )6
( 472.5 kN-m)(7 m) 5,788.125 kN-m2(14 m) 3(14 m)(7 m) (7 m)
6(14 m)
C
M xv L Lx x
LEI
EI EI
Consider downward deflection of simply supported beam at C due to uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
45
384C
wLv
EI
Values:
w = 105 kN/m, L = 14 m
Calculation:
4 4 35 5(105 kN/m)(14 m) 52,521.875 kN-m
384 384C
wLv
EI EI EI
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Consider upward deflection of simply supported beam at C due to concentrated load Cy.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48C
PLv
EI
Values:
P = −Cy, L = 14 m
Calculation:
3 33 ( )(14 m) (57.1667 m )
48 48
y y
C
C CPLv
EI EI EI
Compatibility equation for deflection at C:
33 3
3
3
(57.1667
818
m )5,788.125 kN-m 52,521.875 kN-m0
46,733.750 kN-m817.5 kN
57.1667 mkN
y
y
C
EI EI EI
C Ans.
(b) Beam deflection at A.
Consider downward cantilever beam deflection caused by uniformly distributed load on overhang
AB. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
4
8A
wLv
EI
Values:
w = 105 kN/m, L = 3 m
Calculation:
4 4 3(105 kN/m)(3 m) 1,063.125 kN-m
8 8A
wLv
EI EI EI
Consider downward deflection at A resulting from rotation at B caused by concentrated load on
overhang AB. [Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
3
B
ML
EI (slope magnitude)
Values:
M = (105 kN/m)(3 m)(1.5 m) = 472.5 kN-m,
L = 14 m
Computation:
2
2 3
(472.5 kN-m)(14 m) 2,205 kN-m
3 3
2,205 kN-m 6,615 kN-m(3 m)
B
A
ML
EI EI EI
vEI EI
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Consider upward deflection at A due to uniformly distributed load between B and D.
[Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
3
(slope magnitude)24
B
wL
EI
Values:
w = 105 kN/m, L = 14 m
Calculation:
3 3 2
2 3
(105 kN/m)(14 m) 12,005 kN-m
24 24
12,005 kN-m 36,015 kN-m(3 m)
B
A
wL
EI EI EI
vEI EI
Consider downward deflection at A due to concentrated load Cy.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
2
(slope magnitude)16
B
PL
EI
Values:
P = −Cy = −817.5 kN, L = 14 m
Calculation:
2 2 2
2 3
(817.5 kN)(14 m) 10,014.375 kN-m
16 16
10,014.375 kN-m 30,043.125 kN-m(3 m)
B
A
PL
EI EI EI
vEI EI
Beam deflection at A.
3 3 3 3
3 3
2
1,063.125 kN-m 6,615 kN-m 36,015 kN-m 30,043.125 kN-m
1,706.25 kN-m 1,706.25 kN-m0.015180 m
112,400 kN15.18
-mm
m
AvEI EI EI EI
EI Ans.
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11.51 The beam shown in Fig. P11.51
consists of a W8 × 15 structural steel wide-
flange shape [E = 29,000 ksi; I = 48 in.4].
For the loading shown, determine:
(a) the reactions at A and B.
(b) the magnitude of the maximum bending
stress in the beam.
(Reminder: The roller symbol implies that
both upward and downward displacement is
restrained.)
Fig. P11.51
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever.
Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam
span. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = −80 lb/in., L = 150 in., x = 100 in.
Calculation:
22 2
22 2
9 3
(6 4 )24
( 80 lb/in.)(100 in.)6(150 in.) 4(150 in.)(100 in.) (100 in.)
24
2.8333 10 lb-in.
B
wxv L Lx x
EI
EI
EI
Consider deflection of cantilever beam at B due to the force caused by the linear portion of the
distributed load. [Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −½(50 in.)(60 lb/in.) = −1,500 lb, L = 100 in.
Calculation:
3 3 6 3( 1,500 lb)(100 in.) 500 10 lb-in.
3 3B
PLv
EI EI EI
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Consider deflection of cantilever beam at B due to the moment caused by the linear portion of the
distributed load. [Appendix C, Cantilever beam with concentrated moment at tip.]
Relevant equation from Appendix C:
2
2B
MLv
EI
Values:
M = −½(50 in.)(60 lb/in.)[⅔(50 in.)]
= −50,000 lb-in., L = 100 in.
Calculation:
2 2 6 3( 50,000 lb-in.)(100 in.) 250 10 lb-in.
2 2B
MLv
EI EI EI
Consider deflection of cantilever beam at B due to concentrated load By.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −By, L = 100 in.
Calculation:
3 33 ( )(100 in.) (333.3333 in. )
3 3
y y
B
B BPLv
EI EI EI
Compatibility equation for deflection at B:
3 39 3 6 3 6 3
9 3
3 3
(333.3333 10 in. )2.8333 10 lb-in. 500 10 lb-in. 250 10 lb-in.0
3.5833 10 lb-in.10,750 lb
333.3333 10 in.10,750 lb
y
y
B
EI EI EI EI
B Ans.
Equilibrium equations for entire beam:
1(80 lb/in.)(150 in.) (140 lb/in. 80 lb/in.)(50 in.) 0
2
(60 lb/in.)(50 in.)(80
2,
lb/in.)(150 in.) ( 10,750 lb)2
2,750 l 75 bb 0 l
y y y
y
F A B
A
Ans.
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(80 lb/in.)(150 in.)(75 in.)
(140 lb/in. 80 lb/in.)(50 in.) 2100 in. (50 in.) (100 in.) 0
2 3
A A
y
M M
B
(60 lb/in.)(50 in.)(80 lb/in.)(150 in.)(75 in.) (133.3333 in.) ( 10,750 in.)(100 in.)
2
25,000 lb-in 25,000 l. b-in. (cw)
AM
Ans.
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
4 348 in. 8.11 in. 11.8 in.I d S
Maximum bending moment magnitude
Mmax = 150,000 lb-in. (at B)
Bending stresses at maximum moment
4
(150,000 lb-in.)(8.11 in./2)12,671.875 psi
4812
in.,670 psix Ans.
or, using the tabulated value for the section modulus:
3
12,710 ps150,000 lb-in.
12,711.864 psi11.8 in.
ix Ans.
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11.52 The beam shown in Fig. P11.52
consists of a W24 × 94 structural steel wide-
flange shape [E = 29,000 ksi; I = 2,700 in.4].
For the loading shown, determine:
(a) the reactions at A and D.
(b) the magnitude of the maximum bending
stress in the beam.
Fig. P11.52
Solution
Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever.
Consider downward deflection of cantilever beam at A due to the 50-kip concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and (slope magnitude)3 2
B B
PL PLv
EI EI
Values:
P = 50 kips, L = 20 ft
Calculation:
3 3 3
2 2 3
3 3 3
(50 kips)(20 ft) 133,333.333 kip-ft
3 3
(50 kips)(20 ft) 10,000 kip-ft
2 2
133,333.333 kip-ft 10,000 kip-ft 193,333.333 kip-ft(6 ft)
B
B
A
PLv
EI EI EI
PL
EI EI EI
vEI EI EI
Consider downward deflection of cantilever beam at A due to the uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equations from Appendix C:
4 3
and (slope magnitude)8 6
C C
wL wLv
EI EI
Values:
w = 4 kips/ft, L = 14 ft
Calculation:
4 4 3
3 3 3
3 3 3
(4 kips/ft)(14 ft) 19,208 kip-ft
8 8
(4 kips/ft)(14 ft) 1,829.333 kip-ft
6 6
19,208 kip-ft 1,829.333 kip-ft 41,160 kip-ft(12 ft)
C
C
A
wLv
EI EI EI
wL
EI EI EI
vEI EI EI
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Consider upward deflection of cantilever beam at A due to concentrated load Ay.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3A
PLv
EI
Values:
P = −Ay, L = 26 ft
Calculation:
3 33 ( )(26 ft) (5,858.667 ft )
3 3
y y
A
A APLv
EI EI EI (b)
Compatibility equation for deflection at A:
33 3
3
3
(5,858.667 ft )193,333.333 kip-ft 41,160 kip-ft0
234,493.333 kip-ft40.025 kips
5,858.667 40.0 kips
ft
y
y
A
EI EI EI
A Ans.
Equilibrium equations for entire beam:
50 kips (4 kips/ft)(14 ft) 0
50 kips (4 kips/ft)(14 ft) 40.025 kips 65.975 kips 66.0 kips
y y y
y
F A D
D Ans.
(50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft) (26 ft) 0D D yM M A
(40.025 kips)(26 ft) (50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft)
351.350 kip- 351 kip-ft (cw)ft
DM
Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
4
3
2,700 in.
24.3 in.
222 in.
I
d
S
Maximum bending moment magnitude
Mmax = 351.350 kip-ft
Bending stresses at maximum moment
4
(351.350 kip-ft)(24.3 in./2)(12 in./ft)
2,700
18.97 ksi
in.x
Ans.
or, using the tabulated section modulus:
3
(351.350 kip-ft)(12 in./ft)
222 in
18.99 ksi
.x
Ans.
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11.53 The solid 20-mm-diameter steel [E = 200
GPa] shaft shown in Fig. P11.53 supports two
belt pulleys. Assume that the bearing at A can be
idealized as a pin support and that the bearings at
C and E can be idealized as roller supports. For
the loading shown, determine:
(a) the reaction forces at bearings A, C, and E.
(b) the magnitude of the maximum bending
stress in the shaft.
Fig. P11.53
Solution
(a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the
released beam is simply supported between A and E.
Consider downward deflection of simply supported beam at C due to pulley B load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6C
Pbxv L b x
LEI
Values:
P = 850 N, L = 2,000 mm, b = 600 mm,
x = 1,000 mm
Calculation:
2 2 2
2 2 2
9 3
( )6
(850 N)(600 mm)(1,000 mm)(2,000 mm) (600 mm) (1,000 mm)
6(2,000 mm)
112.2 10 N-mm
C
Pbxv L b x
LEI
EI
EI
Consider downward deflection of simply supported beam at C due to pulley D load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6C
Pbxv L b x
LEI
Values:
P = 600 N, L = 2,000 mm, b = 400 mm,
x = 1,000 mm
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Calculation:
2 2 2
2 2 2
9 3
( )6
(600 N)(400 mm)(1,000 mm)(2,000 mm) (400 mm) (1,000 mm)
6(2,000 mm)
56.8 10 N-mm
C
Pbxv L b x
LEI
EI
EI
Consider upward deflection of simply supported beam at C due to concentrated load Cy.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48C
PLv
EI
Values:
P = −Cy, L = 2,000 mm
Calculation:
3 6 33 ( )(2,000 mm) (166.6667 10 mm )
48 48
y y
C
C CPLv
EI EI EI
Compatibility equation for deflection at C:
6 39 3 9 3
9 3
6 3
(166.6667 10 mm )112.2 10 N-mm 56.8 10 N-mm0
169.0 10 N-mm1,014 N
166.661,014 N
67 10 mm
y
y
C
EI EI EI
C Ans.
Equilibrium equations for entire beam: (850 N)(600 mm) (600 N)(1,600 mm) (1,000 mm) (2,000 mm) 0A y yM C E
(850 N)(600 mm) (600 N)(1,600 mm) (1,014 N)(1,000 mm)
2,000 mm228 NyE Ans.
850 N 600 N 0
850 N 600 N 1,014 N 228 N 208 N
y y y y
y
F A C E
A Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties:
4 4(20 mm) 7,853.9816 mm64
I
Maximum bending moment magnitude
Mmax = 132,000 N-mm
Bending stresses at maximum moment
4
(132,000 N-mm)(20 mm/2)
7,853.9816 m
168.1 MPa
mx
Ans.
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11.54 The solid 1.00-in.-diameter steel [E
= 29,000 ksi] shaft shown in Fig. P11.54
supports three belt pulleys. Assume that
the bearing at A can be idealized as a pin
support and that the bearings at C and E
can be idealized as roller supports. For the
loading shown, determine:
(a) the reaction forces at bearings A, C,
and E.
(b) the magnitude of the maximum
bending stress in the shaft.
Fig. P11.54
Solution
(a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the
released beam is simply supported between A and E.
Consider downward deflection of simply supported beam at C due to pulley B load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6C
Pbxv L b x
LEI
Values:
P = 200 lb, L = 60 in., b = 15 in., x = 30 in.
Calculation:
2 2 2
2 2 2
3
( )6
(200 lb)(15 in.)(30 in.)(60 in.) (15 in.) (30 in.)
6(60 in.)
618,750 lb-in.
C
Pbxv L b x
LEI
EI
EI
Consider downward deflection of simply supported beam at C due to pulley D load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6C
Pbxv L b x
LEI
Values:
P = 280 lb, L = 60 in., b = 15 in., x = 30 in.
Calculation:
2 2 2
2 2 2
3
( )6
(280 lb)(15 in.)(30 in.)(60 in.) (15 in.) (30 in.)
6(60 in.)
866,250 lb-in.
C
Pbxv L b x
LEI
EI
EI
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Consider upward deflection of simply supported beam at C due to pulley F load.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 ) (elastic curve)6
C
M xv L Lx x
LEI
Values:
M = −(120 lb)(10 in.) = −1,200 lb-in.,
L = 60 in., x = 30 in.
Calculation:
2 2
2 2
3
(2 3 )6
( 1,200 lb-in.)(30 in.)2(60 in.) 3(60 in.)(30 in.) (30 in.)
6(60 in.)
270,000 lb-in.
C
M xv L Lx x
LEI
EI
EI
Consider upward deflection of simply supported beam at C due to concentrated load Cy.
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48C
PLv
EI
Values:
P = −Cy, L = 60 in.
Calculation:
3 33 ( )(60 in.) (4,500 in. )
48 48
y y
C
C CPLv
EI EI EI
Compatibility equation for deflection at C:
33 3 3
3
3
(4,500 in. )618,750 lb-in. 866,250 lb-in. 270,000 lb-in.0
1,215,000 lb-in.270 lb
4,500 in270 lb
.
y
y
C
EI EI EI EI
C Ans.
Equilibrium equations for entire beam: (200 lb)(15 in.) (280 lb)(45 in.) (120 lb)(70 in.) (30 in.) (60 in.) 0A y yM C E
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(200 lb)(15 in.) (280 lb)(45 in.) (120 lb)(70 in.) (215 lb)(30 in.)
60 in.
265.0 lb 265 lb
yE
Ans.
200 lb 280 lb 120 lb 0
200 lb 280 lb 120 lb 270 lb 2 65.0 65 l lbb
y y y y
y
F A C E
A Ans.
Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties:
4 4(1.00 in.) 0.0490874 in.64
I
Maximum bending moment magnitude
Mmax = 1,200 lb-in.
Bending stresses at maximum moment
4
(1,200 lb-in.)(1.00 in./2)
0.0490874 in.
12,223.1 psi
12,220 psi
x
Ans.
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11.55 The solid 1.00-in.-diameter steel [E =
29,000 ksi] shaft shown in Fig. P11.55
supports two belt pulleys. Assume that the
bearing at E can be idealized as a pin
support and that the bearings at B and C can
be idealized as roller supports. For the
loading shown, determine:
(a) the reaction forces at bearings B, C, and
E.
(b) the magnitude of the maximum bending
stress in the shaft.
Fig. P11.55
Solution
(a) Reaction forces at B, C, and E. Choose the reaction force at C as the redundant; therefore, the
released beam is simply supported between B and E.
Consider upward deflection of simply supported beam at C due to pulley A load.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 ) (elastic curve)
6C
M xv L Lx x
LEI
Values:
M = −(90 lb)(7 in.) = −630 lb-in.,
L = 45 in., x = 15 in.
Calculation:
2 2
2 2
3
(2 3 )6
( 630 lb-in.)(15 in.)2(45 in.) 3(45 in.)(15 in.) (15 in.)
6(45 in.)
78,750 lb-in.
C
M xv L Lx x
LEI
EI
EI
Consider downward deflection of simply supported beam at C due to pulley D load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( ) (elastic curve)
6C
Pbxv L b x
LEI
Values:
P = 240 lb, L = 45 in., b = 15 in., x = 15 in.
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Calculation:
2 2 2
2 2 2
3
( )6
(240 lb)(15 in.)(15 in.)(45 in.) (15 in.) (15 in.)
6(45 in.)
315,000 lb-in.
C
Pbxv L b x
LEI
EI
EI
Consider upward deflection of simply supported beam at C due to concentrated load Cy.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
C
Pabv L a b
LEI
Values:
P = −Cy, L = 45 in., a = 15 in., b = 30 in.
Calculation:
2 2 2
3
2 2 2
( )6
( )(15 in.)(30 in.) (1,500 in. )(45 in.) (15 in.) (30 in.)
6(45 in.)
C
y y
Pabv L a b
LEI
C C
EI EI
Compatibility equation for deflection at C:
33 3
3
3
(1,500 i
157.
n. )78,750 lb-in. 315,000 lb-in.0
236,250 lb-in.157.50 lb
1,500 inl
.5 b
y
y
C
EI EI EI
C Ans.
Equilibrium equations for entire beam: (240 lb)(15 in.) (90 lb)(52 in.) (45 in.) (30 in.) 0E y yM B C
(240 lb)(15 in.) (90 lb)(52 in.) (157.5 lb)(30 in.)
45 in.
79.0 lb 79.0 lb
yB
Ans.
90 lb 240 lb 0
90 lb 240 lb 157.5 lb 79.0 lb 93.5 lb 93.5 lb
y y y y
y
F B C E
E Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties:
4 4(1.00 in.) 0.0490874 in.64
I
Maximum bending moment magnitude
Mmax = 1,402.5 lb-in.
Bending stresses at maximum moment
4
(1,402.5 lb-in.)(1.00 in./2)
0.0490874 in.
14,285.74 psi
14,290 psi
x
Ans.
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11.56 The beam shown in Fig. P11.56 consists
of a W360 × 101 structural steel wide-flange
shape [E = 200 GPa; I = 301 × 106 mm
4]. For
the loading shown, determine:
(a) the reactions at A and B.
(b) the magnitude of the maximum bending
stress in the beam.
Fig. P11.56
Solution
Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever.
Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam
span. [Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equations from Appendix C:
2
2 2(6 4 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 30 kN/m, L = 8 m, x = 5.5 m
Calculation:
22 2
22 2
3
(6 4 )24
(30 kN/m)(5.5 m)6(8 m) 4(8 m)(5.5 m) (5.5 m)
24
9,008.828125 kN-m
B
wxv L Lx x
EI
EI
EI
Consider deflection of cantilever beam at B due a linearly distributed load.
[Appendix C, Cantilever beam with linearly distributed load.]
Relevant equation from Appendix C: 2
3 2 2 30 (10 10 5 )120
B
w xv L L x Lx x
LEI (elastic curve)
Values:
w0 = 60 kN/m, L = 8 m, x = 5.5 m
Calculation:
23 2 2 30
23 2 2 3
3
(10 10 5 )120
(60 kN/m)(5.5 m)10(8 m) 10(8 m) (5.5 m) 5(8 m)(5.5 m) (5.5 m)
120(8 m)
4,998.103516 kN-m
B
w xv L L x Lx x
LEI
EI
EI
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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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Consider deflection of cantilever beam at B due to concentrated load By.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −By, L = 5.5 m
Calculation:
3 33 ( )(5.5 m) (55.458333 m )
3 3
y y
B
B BPLv
EI EI EI
Compatibility equation for deflection at B:
33 3
3
3
(55.458333
25
m )9,008.828125 kN-m 4,998.103516 kN-m0
14,006.93164 kN-m252.56676 kN
55.458333 N
m3 k
y
y
B
EI EI EI
B Ans.
Equilibrium equations for entire beam:
1(30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 0
2
1(30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 252.56676 kN
227
2
227.43324 kN kN
y y y
y
F A B
A
Ans.
(90 kN/m 30 kN/m)(8 m) 8 m
(30 kN/m)(8 m)(4 m) (5.5 m) 02 3
A A yM M B
960 kN-m 640 kN-m (252.56676 kN)(5.5 m)
210.88281 211 kN-m (ccw) kN-m
AM
Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
6 4
3 3
301 10 mm
356 mm
1,690 10 mm
I
d
S
Maximum bending moment magnitude
Mmax = 210.88281 kN-m
Bending stresses at maximum moment
2
6 4
(210.88281 kN-m)(356 mm/2)(1,000)
301 10
124.7 MPa
mmx
or, using the tabulated value for the section
modulus:
2
3 3
(210.88281 kN-m)(1,000)
1,690 10 mm
124.8 MPa
x
Ans.
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11.57 A W530 × 92 structural steel wide-
flange shape [E = 200 GPa; I = 554 × 106
mm4] is loaded and supported as shown in
Fig. P11.57. Determine:
(a) the force and moment reactions at
supports A and C.
(b) the maximum bending stress in the
beam.
(c) the deflection of the beam at B.
Fig. P11.57
Solution
(a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a
simply supported beam between A and C as the released beam.
Determine the slopes at A and C caused by the 150-kN concentrated load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equations from Appendix C: 2 2 2 2( ) ( )
and6 6
A C
Pb L b Pa L a
LEI LEI
Values:
P = 150 kN, L = 10 m, a = 6 m, b = 4 m
Calculation:
2 2 22 2
2 2 22 2
( ) (150 kN)(4 m) 840 kN-m(10 m) (4 m)
6 6(10 m)
( ) (150 kN)(6 m) 960 kN-m(10 m) (6 m)
6 6(10 m)
A
C
Pb L b
LEI EI EI
Pa L a
LEI EI EI
Determine the slopes at A and C caused by moment reaction MA.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equations from Appendix C:
and3 6
A C
ML ML
EI EI
Values:
M = MA, L = 10 m
Calculation:
(10 m) (3.333333 m)
3 3
(10 m) (1.666667 m)
6 6
A AA
A AC
ML M M
EI EI EI
ML M M
EI EI EI
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Determine the slopes at A and C caused by moment reaction MC.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equations from Appendix C:
and6 3
A C
ML ML
EI EI
Values:
M = MC, L = 10 m
Calculation:
(10 m) (1.666667 m)
6 6
(10 m) (3.333333 m)
3 3
C CA
C CC
ML M M
EI EI EI
ML M M
EI EI EI
Compatibility equation for slope at A:
2840 kN-m (3.333333 m) (1.666667 m)
0A CM M
EI EI EI (a)
Compatibility equation for slope at C:
2960 kN-m (1.666667 m) (3.333333 m)
0A CM M
EI EI EI (b)
Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as:
2
2
(3.333333 m) (1.666667 m) 840 kN-m
(1.666667 m) (3.333333 m) 960 kN-m
A C
A C
M M
M M
and solved simultaneously for MA and MC:
144 kN-m (ccw144 kN-m )AM Ans.
2216 kN-m 16 kN-m (cw)CM Ans.
Equilibrium equations for entire beam: (150 kN)(6 m) (10 m) 0A A C yM M M C
(150 kN)(6 m) ( 144 kN-m) ( 216 kN-m)
10 m
97 97.2 kN.2 kN
yC
Ans.
150 kN 0y y yF A C
150 kN 97.2 kN 52.8 k 52.8 kNNyA Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
6 4
3 3
554 10 mm
533 mm
2,080 10 mm
I
d
S
Maximum bending moment magnitude
Mmax = 216 kN-m (at C)
Bending stresses at maximum moment
2
6 4
(216 kN-m)(533 mm/2)(1,000)
554 10 m
103.
m
9 MPa
x
Ans.
or using the tabulated value for the section modulus:
2
3 3
(216 kN-m)(1,000)
2,080 10
103.8 M a
mm
P
x
Ans.
(c) Beam deflection at B:
Determine the deflection at B caused by the 150-kN concentrated load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6B
Pabv L a b
LEI
Values:
P = 150 kN, L = 10 m, a = 6 m,
b = 4 m
Calculation:
3
2 2 2 2 2 2(150 kN)(6 m)(4 m) 2,880 kN-m( ) (10 m) (6 m) (4 m)
6 6(10 m)B
Pabv L a b
LEI EI EI
Determine the deflection at B caused by MA.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI
Values:
M = −144 kN-m, L = 10 m, x = 6 m
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Calculation:
2 2
32 2
(2 3 )6
( 144 kN-m)(6 m) 806.4 kN-m2(10 m) 3(10 m)(6 m) (6 m)
6(10 m)
B
M xv L Lx x
LEI
EI EI
Determine the deflection at B caused by MC.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
B
M xv L Lx x
LEI
Values:
M = −216 kN-m, L = 10 m, x = 4 m
Calculation:
2 2
32 2
(2 3 )6
( 216 kN-m)(4 m) 1,382.4 kN-m2(10 m) 3(10 m)(4 m) (4 m)
6(10 m)
B
M xv L Lx x
LEI
EI EI
Beam deflection vB:
3 3 3
3 3
2
2,880 kN-m 806.4 kN-m 1,382.4 kN-m
691.2 kN-m 691.2 kN-m0.006238 m
110,806.
0 24 mm
kN-m
BvEI EI EI
EI
Ans.
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11.58 A W530 × 92 structural steel wide-
flange shape [E = 200 GPa; I = 554 × 106
mm4] is loaded and supported as shown in
Fig. P11.58. Determine:
(a) the force and moment reactions at
supports A and C.
(b) the maximum bending stress in the beam.
(c) the deflection of the beam at B.
Fig. P11.58
Solution
(a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a
simply supported beam between A and C as the released beam.
Determine the slopes at A and C caused by the 80 kN/m uniformly distributed load.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equations from Appendix C:
22
22 2
(2 ) and24
(2 )24
A
C
waL a
LEI
waL a
LEI
Values:
w = 80 kN/m, L = 9 m, a = 4.5 m
Calculation:
2 2 2
22
2 2 22 2 2 2
(80 kN/m)(4.5 m) 1,366.875 kN-m(2 ) 2(9 m) (4.5 m)
24 24(9 m)
(80 kN/m)(4.5 m) 1,063.125 kN-m(2 ) 2(9 m) (4.5 m)
24 24(9 m)
A
C
waL a
LEI EI EI
waL a
LEI EI EI
Determine the slopes at A and C caused by moment reaction MA.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equations from Appendix C:
and3 6
A C
ML ML
EI EI
Values:
M = MA, L = 9 m
Calculation:
(9 m) (3 m)
3 3
(9 m) (1.5 m)
6 6
A AA
A AC
ML M M
EI EI EI
ML M M
EI EI EI
Determine the slopes at A and C caused by moment reaction MC.
[Appendix C, SS beam with concentrated moment at one end.]
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Relevant equations from Appendix C:
and6 3
A C
ML ML
EI EI
Values:
M = MC, L = 9 m
Calculation:
(9 m) (1.5 m)
6 6
(9 m) (3 m)
3 3
C CA
C CC
ML M M
EI EI EI
ML M M
EI EI EI
Compatibility equation for slope at A:
21,366.875 kN-m (3 m) (1.5 m)
0A CM M
EI EI EI (a)
Compatibility equation for slope at C:
21,063.125 kN-m (1.5 m) (3 m)
0A CM M
EI EI EI (b)
Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as:
2
2
(3 m) (1.5 m) 1,366.875 kN-m
(1.5 m) (3 m) 1,063.125 kN-m
A C
A C
M M
M M
and solved simultaneously for MA and MC:
371.25 kN-m (ccw371.25 kN-m )AM Ans.
168.75 kN- 168.8 kN-m (cw)mCM Ans.
Equilibrium equations for entire beam: (80 kN/m)(4.5 m)(2.25 m) (9 m) 0A A C yM M M C
(80 kN/m)(4.5 m)(2.25 m) ( 371.25 kN-m) ( 168.75 kN-m)
9 m
67.5 kN 67.5 kN
yC
Ans.
(80 kN/m)(4.5 m) 0y y yF A C
(80 kN/m)(4.5 m) 67.5 kN 292.5 kN 293 kNyA Ans.
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Shear-force and bending-moment diagrams
(b) Magnitude of maximum bending stress:
Section properties (from Appendix B):
6 4
3 3
554 10 mm
533 mm
2,080 10 mm
I
d
S
Maximum bending moment magnitude
Mmax = 371.25 kN-m (at A)
Bending stresses at maximum moment
2
6 4
(371.25 kN-m)(533 mm/2)(1,000)
554 10
178.6 MPa
mmx
Ans.
or using the tabulated value for the section modulus:
2
3 3
(371.25 kN-m)(1,000)
2,080 10 mm
178.5 MPa
x
Ans.
(c) Beam deflection at B:
Determine the deflection at B caused by the 80 kN/m uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
Values:
w = 80 kN/m, L = 9 m, a = 4.5 m
Calculation:
32 2
3 32 2
(4 7 3 )24
(80 kN/m)(4.5 m) 3,417.1875 kN-m4(9 m) 7(4.5 m)(9 m) 3(4.5 m)
24(9 m)
B
wav L aL a
LEI
EI EI
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Determine the deflection at B caused by MA.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )6
B
M xv L Lx x
LEI
Values:
M = −371.25 kN-m, L = 9 m,
x = 4.5 m
Calculation:
2 2
32 2
(2 3 )6
( 371.25 kN-m)(4.5 m) 1,879.4531 kN-m2(9 m) 3(9 m)(4.5 m) (4.5 m)
6(9 m)
B
M xv L Lx x
LEI
EI EI
Determine the deflection at B caused by MC.
[Appendix C, SS beam with concentrated moment at one end.]
Relevant equation from Appendix C:
2 2(2 3 )
6B
M xv L Lx x
LEI
Values:
M = −168.75 kN-m, L = 9 m,
x = 4.5 m
Calculation:
2 2
32 2
(2 3 )6
( 168.75 kN-m)(4.5 m) 854.2969 kN-m2(9 m) 3(9 m)(4.5 m) (4.5 m)
6(9 m)
B
M xv L Lx x
LEI
EI EI
Beam deflection vB:
3 3 3
3 3
2
3,417.1875 kN-m 1,879.4531 kN-m 854.2969 kN-m
683.4375 kN-m 683.4375 kN-m0.006168 m
110,800 k6.
N-17 mm
m
BvEI EI EI
EI
Ans.
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11.59 A timber [E = 1,800 ksi] beam is
loaded and supported as shown in Fig.
P11.59. The cross section of the timber beam
is 4-in. wide and 8-in. deep. The beam is
supported at B by a ½-in.-diameter steel [E =
30,000 ksi] rod, which has no load before the
distributed load is applied to the beam. After
a distributed load of 900 lb/ft is applied to
the beam, determine:
(a) the force carried by the steel rod.
(b) the maximum bending stress in the
timber beam.
(c) the deflection of the beam at B.
Fig. P11.59
Solution
Section properties:
34
beam beam
2 2
1 1
(4 in.)(8 in.)Beam: 170.6667 in. 1,800,000 psi
12
Rod (1): (0.50 in.) 0.1963495 in. 30,000,000 psi4
I E
A E
(a) Force carried by the steel rod.
The reaction force from rod (1) will be taken as the
redundant, leaving a simply supported beam
between A and C as the released beam.
For this analysis, a tension force is assumed to exist
in axial member (1).
Beam free-body diagram
Downward deflection of wood beam at B due to 900 lb/ft uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
45
384B
wLv
EI
Values:
w = 900 lb/ft, L = 14 ft, EI = 307.2 ×106 lb-in.
2
Calculation:
4 4 3
6 2
5 5(900 lb/ft)(14 ft) (12 in./ft)2.532305 in.
384 384(307.2 10 lb-in. )B
wLv
EI
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Upward deflection of wood beam at B due to force F1 in rod (1).
[Appendix C, SS beam with concentrated load at midspan.]
Relevant equation from Appendix C:
3
48B
PLv
EI
Values:
P = −F1, L = 14 ft, EI = 307.2 ×106 lb-in.
2
Calculation:
3 3 3
6116 2
( )(14 ft) (12 in./ft)(321.5625 10 in./lb)
48 48(307.2 10 lb-in. )B
PL Fv F
EI
Elongation of steel rod (1) due to force F1.
61 1
1 1 12 6
1 1
(16 ft)(12 in./ft)(32.59493 10 in./lb)
(0.1963495 in. )(30 10 psi)
F LF F
A E
The deflection of the wood beam at B will not equal zero in this instance because the rod that supports
the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore,
6
1 1(32.59493 10 in./lb)Bv F
Compatibility equation at B:
The sum of the downward deflection caused by the uniformly distributed load and the upward deflection
caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will
produce a downward (i.e., negative) deflection of the wood beam at B.
6 6
1 1
1 6
2.532305 in. (321.5625 10 in./lb) (32.59493 10 in./lb)
2.532305 in.7,150.22 lb
354.1574 10 in.7,150 lb (T)
/lb
F F
F
Ans.
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(b) Determine maximum bending stress in wood
beam:
Maximum bending moment magnitude
Mmax = 4,125 lb-ft = 49,500 lb-in.
Bending stress at maximum moment
4
(49,500 lb-in.)(8 in./2)
170.6667 in1,160 psi
.x Ans.
(c) Beam deflection at B.
The beam deflection at B is equal to the elongation of the steel rod:
1 11 2 6
1 1
(7,150.22 lb)(16 ft)(12 in./ft)0.233 in.
(0.1963495 in. )(30 10 p0.233
s )i
in.B
F Lv
A E
Ans.
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11.60 A W360 × 72 structural steel [E = 200
GPa] wide-flange shape is loaded and
supported as shown in Fig. P11.60. The beam
is supported at B by 20-mm-diameter solid
aluminum [E = 70 GPa] rod. After a
concentrated load of 40 kN is applied to the tip
of the cantilever, determine:
(a) the force produced in the aluminum rod.
(b) the maximum bending stress in the beam.
(c) the deflection of the beam at B.
Fig. P11.60
Solution
Section properties:
6 4
beam beam
3 3
beam
2 2
1 1
W360 × 72: 201 10 mm 351 mm 200,000 MPa
1,150 10 mm
Rod (1): (20 mm) 314.159266 mm 70,000 MPa4
I d E
S
A E
(a) Force in the aluminum rod.
The reaction force from rod (1) will be taken as the
redundant, leaving a cantilever beam as the released
beam.
For this analysis, a tension force is assumed to exist
in axial member (1).
Downward deflection of W360 × 72 beam at B due to 40-kN concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
2
(3 ) (elastic curve)6
B
Pxv L x
EI
Values:
P = 40 kN, L = 5 m, x = 3.6 m,
EI = 40,200 kN-m2
Calculation:
2 2
3
2
(40 kN)(3.6 m)(3 ) 3(5 m) (3.6 m) 24.50149 10 m
6 6(40,200 kN-m )B
Pxv L x
EI
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Upward deflection of W360 × 72 beam at B due to force F1 in rod (1).
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −F1, L = 3.6 m, EI = 40,200 kN-m2
Calculation:
3 3
6112
( )(3.6 m)(386.8657 10 m/kN)
3 3(40,200 kN-m )B
PL Fv F
EI
Elongation of aluminum rod (1) due to force F1.
61 1
1 1 12 2
1 1
(3 m)(136.4185 10 m/kN)
(314.159266 mm )(70,000 N/mm )(1 kN/1,000 N)
F LF F
A E
The deflection of the W360 × 72 beam at B will not equal zero in this instance because the rod that
supports the beam at B will elongate, thus permitting the W360 × 72 beam to deflect downward.
Therefore,
6
1 1(136.4185 10 m/kN)Bv F
Compatibility equation at B:
The sum of the downward deflection caused by the 40-kN concentrated load and the upward deflection
caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the
aluminum rod will produce a downward (i.e., negative) deflection of the W360 × 72 beam at B.
3 6 6
1 1
3
1 6
24.50149 10 m (386.8657 10 m/kN) (136.4185 10 m/kN)
24.50149 10 m46.82254 46.8 kN (T) kN
523.2842 10 m/kN
F F
F
Ans.
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(b) Determine maximum bending stress in the
W360 × 72 beam:
Maximum bending moment magnitude
Mmax = 56 kN-m (at B)
Bending stress at maximum moment
2
6 4
(56 kN-m)(351 mm/2)(1,000)
201 10 m
48.9 MPa
mx
Ans.
or using the tabulated value for the section modulus
of the W360 × 72 beam:
2
3 3
(56 kN-m)(1,000)
1,150 10 mm48.7 MPax
Ans.
(c) Beam deflection at B.
The beam deflection at B is equal to the elongation of the aluminum rod:
1 11 2 2
1 1
(46,822.54 N)(3,000 mm)6.38746 mm
(314.159266 mm )(70,000 N/6.39 mm
mm )B
F Lv
A E Ans.
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11.61 A W18 × 55 structural steel [E = 29,000
ksi] wide-flange shape is loaded and supported
as shown in Fig. P11.61. The beam is supported
at C by a ¾-in.-diameter aluminum [E = 10,000
ksi] rod, which has no load before the
distributed load is applied to the beam. After a
distributed load of 4 kips/ft is applied to the
beam, determine:
(a) the force carried by the aluminum rod.
(b) the maximum bending stress in the steel
beam.
(c) the deflection of the beam at C. Fig. P11.61
Solution
Section properties:
4
beam beam
3
beam
2 2
1 1
Beam: 890 in. 18.1 in. 29,000 ksi
98.3 in.
Rod (1): (0.75 in.) 0.4417865 in. 10,000 ksi4
I d E
S
A E
(a) Force in the aluminum rod.
The reaction force from rod (1) will be taken as
the redundant, leaving a cantilever beam as the
released beam.
For this analysis, a tension force is assumed to
exist in axial member (1).
Downward deflection of W18 × 55 beam at C due to 4 kips/ft uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equations from Appendix C:
4 3
and8 6
B B
wL wLv
EI EI
Values:
w = 4 kips/ft, L = 11 ft, EI = 25.81×106 kip-in.
2
Calculation:
4 4 3
6 2
3 3 23
6 2
(4 kips/ft)(11 ft) (12 in./ft)0.490113 in.
8 8(25.81 10 kip-in. )
(4 kips/ft)(11 ft) (12 in./ft)4.950639 10 rad
6 6(25.81 10 kip-in. )
0.490113 in. (5 ft)(12 in./ft)( 4.9
B
B
C
wLv
EI
wL
EI
v
350639 10 rad) 0.787152 in.
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Upward deflection of W18 × 55 beam at C due to force F1 in rod (1).
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3C
PLv
EI
Values:
P = −F1, L = 16 ft, EI = 25.81×106 kip-in.
2
Calculation:
3 3 3
3116 2
( )(16 ft) (12 in./ft)(91.41015 10 in./kip)
3 3(25.81 10 kip-in. )C
PL Fv F
EI
Elongation of aluminum rod (1) due to force F1.
31 1
1 1 12
1 1
(14 ft)(12 in./ft)(38.02742 10 in./kip)
(0.4417865 in. )(10,000 ksi)
F LF F
A E
The deflection of the W18 × 55 beam at C will not equal zero in this instance because the rod that
supports the beam at C elongates, thus permitting the W18 × 55 beam to deflect downward. Therefore,
3
1 1(38.02742 10 in./kip)Cv F
Compatibility equation at C:
The sum of the downward deflection caused by the uniformly distributed load and the upward deflection
caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the
aluminum rod will produce a downward (i.e., negative) deflection of the W18 × 55 beam at C.
3 3
1 1
1 3
0.787152 in. (91.41015 10 in./kip) (38.02742 10 in./kip)
0.787152 in.6.081323 kips
129.4376 106.08 kips (T
in./ p)
ki
F F
F
Ans.
(b) Determine maximum bending stress in W18
× 55 beam:
Maximum bending moment magnitude
Mmax = 144.70 kip-ft (at A)
Bending stress at maximum moment
4
(144.70 kip-ft)(18.1 in./2)(12 in./ft)
890
17.66 ksi
in.x
Ans.
or using the tabulated value for the section
modulus of the W18 × 55 beam:
3
(144.70 kip-ft)(12 in./ft)
98.3 in
17.66 ksi
.x
Ans.
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(c) Beam deflection at C.
The beam deflection at C is equal to the elongation of the aluminum rod:
1 11 2
1 1
(6.081323 kips)(14 ft)(12 in./ft)0.231257 in.
(0.4417865 in. )(10,000 ksi)0.231 in.C
F Lv
A E Ans.
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11.62 A W250 × 32.7 structural steel [E =
200 GPa] wide-flange shape is loaded and
supported as shown in Fig. P11.62. A
uniformly distributed load of 16 kN/m is
applied to the beam, causing the roller
support at B to settle downward (i.e.,
displace downward) by 15 mm. Determine:
(a) the reactions at supports A, B, and C.
(b) the maximum bending stress in the
beam.
Fig. P11.62
Solution
Section properties:
6 4 3 3250 32.7 : 49.1 10 mm 259 mm 380 10 mmW I d S
(a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released
beam is simply supported between A and C.
Consider downward deflection of simply supported beam at B due to uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3 2 3( 2 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 16 kN/m, L = 10 m, x = 4 m Calculation:
3 2 3
33 2 3
( 2 )24
(16 kN/m)(4 m) 1,984 kN-m(10 m) 2(10 m)(4 m) (4 m)
24
B
wxv L Lx x
EI
EI EI
Consider upward deflection of simply supported beam at B due to concentrated load By.
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )
6B
Pabv L a b
LEI
Values:
P = −By, L = 10 m, a = 4 m, b = 6 m
Calculation:
2 2 2
3
2 2 2
( )6
( )(4 m)(6 m) (19.2 m )(10 m) (4 m) (6 m)
6(10 m)
B
y y
Pabv L a b
LEI
B B
EI EI
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Compatibility equation for deflection at B:
2 6 4 12 2 3 2
33
3 2 3
3
(200,000 N/mm )(49.1 10 mm ) 9.82 10 N-mm 9.82 10 kN-m
(19.2 m )1,984 kN-m0.015 m
( 0.015 m)(9.82 10 kN-m ) 1,984 kN-m95.6615 kN
1995.7 kN
.2 m
y
y
EI
B
EI EI
B
Ans.
Equilibrium equations for entire beam: (4 m) (10 m) (16 kN/m)(10 m)(5 m) 0A y yM B C
(16 kN/m)(10 m)(5 m) (95.6615 kN)(4 m)
10 m
41.7354 kN 41.7 kN
yC
Ans.
(16 kN/m)(10 m) 0y y y yF A B C
(16 kN/m)(10 m) 95.6615 kN 41.7354 kN
22.6031 k 22.6 kNN
yA
Ans.
Shear-force and bending-moment diagrams
(b) Determine maximum bending stress in W250 ×
32.7 beam:
Maximum bending moment magnitude
Mmax = 54.4327 kN-m
Bending stress at maximum moment
2
6 4
(54.4327 kN-m)(259 mm/2)(1,000)
49.1 10
143.6 MPa
mmx
Ans.
or using the tabulated value for the section modulus
of the W250 × 32.7 beam:
2
3 3
(54.4327 kN-m)(1,000)
380 10
143.2 M a
mm
P
x
Ans.
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11.63 A W10 × 22 structural steel [E =
29,000 ksi] wide-flange shape is loaded
and supported as shown in Fig. P11.63.
The beam is supported at C by a timber
[E = 1,800 ksi] post having a cross-
sectional area of 16 in.2. After a
concentrated load of 10 kips is applied to
the beam, determine:
(a) the reactions at supports A and C.
(b) the maximum bending stress in the
beam.
(c) the deflection of the beam at C. Fig. P11.63
Solution
Section properties:
4
beam beam
3
beam
1 1
W10 22: 118 in. 10.2 in. 29,000 ksi
23.2 in.
Post (1): 16 in. 1,800 ksi
I d E
S
A E
(a) Reactions at supports A and C.
The reaction force from post (1) will be taken as the
redundant, leaving a cantilever beam as the released
beam. To be consistent with earlier sign
conventions (e.g., Chapter 5), we will assume that
the force in the axial member is tension (even
though intuitively we recognize that the post must be
in compression).
Beam free-body diagram
Downward deflection of W10 × 22 beam at C due to 10-kip concentrated load.
[Appendix C, Cantilever beam with concentrated load at tip.]
Relevant equations from Appendix C:
3 2
and3 2
B B
PL PLv
EI EI
Values:
P = 10 kips, L = 14 ft, EI = 3,422,000 kip-in.2
Calculation:
3 3 3
2
2 2 2
2
(10 kips)(14 ft) (12 in./ft)4.618773 in.
3 3(3,422,000 kip-in. )
(10 kips)(14 ft) (12 in./ft)0.0412390 rad
2 2(3,422,000 kip-in. )
4.618773 in. (0.0412390 rad)(6 ft)(12 in./f
B
B
C
PLv
EI
PL
EI
v
t) 7.587984 in.
Since we are assuming that a tension force exists in post (1), the tension force from the post will
cause a downward deflection of W10 × 22 beam at C.
[Appendix C, Cantilever beam with concentrated load at tip.]
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Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = F1, L = 20 ft, EI = 3,422,000 kip-in.2
Calculation:
3 3 3
112
( )(20 ft) (12 in./ft)(1.346581 in./kip)
3 3(3,422,000 kip-in. )B
PL Fv F
EI
Axial deformation of post (1) due to force F1.
1 11 1 12
1 1
(12 ft)(12 in./ft)(0.005 in./kip)
(16 in. )(1,800 ksi)
F LF F
A E
The deflection of the steel beam will not equal zero at C in this instance because the post deforms. If we
are consistent and assume that there is tension in the post, then the steel beam must deflect upward at C.
Therefore,
1 1(0.005 in./kip)Cv F
Compatibility equation at C:
The sum of the downward deflection caused by the 10-kip concentrated load and the upward deflection
caused by tension in post (1) must equal the elongation of the post.
1
1
7.587984 in. (1.346581 in./kip) (0.005 in./kip)
7.587984 in.5.614154 kips
1.351582 in./k5.61 kips (C)
ipy
F F
F C
Ans.
Equilibrium equations for entire beam: 1(20 ft) (10 kips)(14 ft) 0A AM M F
( 5.614154 kips)(20 ft) (10 kips)(14 ft) 27.7169 kip 27.7 kip-ft (-ft ccw)AM Ans.
1 (10 kips) 0y yF A F
10 kips ( 5.614154 kips) 4.3858 kips 4.39 kipsyA Ans.
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(b) Determine maximum bending stress in the
beam:
Maximum bending moment magnitude
Mmax = 33.685 kip-ft = 404.218 kip-in. (at B)
Bending stress at maximum moment
4
(404.218 kip-in.)(10.2 in./2)
118 in
17.47 ksi
.x
Ans.
or using the tabulated value for the section
modulus of the W10 × 22 beam:
3
404.218 kip-in.
23.2 in.17.42 ksix Ans.
(c) Beam deflection at C.
The beam deflection at C is equal to the deformation of the post:
1 11 2
1 1
( 5.614154 kips)(12 ft)(12 in./ft)0.0281 in.
(16 in. )(1,800 k0.0281
)i
in.
sC
F Lv
A E
Ans.
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11.64 A timber [E = 12 GPa] beam is loaded
and supported as shown in Fig. P11.64. The
cross section of the timber beam is 100-mm.
wide and 300-mm deep. The beam is
supported at B by a 12-mm-diameter steel [E
= 200 GPa] rod, which has no load before the
distributed load is applied to the beam. After
a distributed load of 7 kN/m is applied to the
beam, determine:
(a) the force carried by the steel rod.
(b) the maximum bending stress in the timber
beam.
(c) the deflection of the beam at B.
Fig. P11.64
Solution
Section properties:
36 4
beam beam
2 2
1 1
(100 mm)(300 mm)Beam: 225 10 mm 12,000 MPa
12
Rod (1): (12 mm) 113.097336 mm 200,000 MPa4
I E
A E
(a) Force carried by the steel rod.
The reaction force from rod (1) will be taken as the
redundant, leaving a simply supported beam between
A and C as the released beam.
For this analysis, a tension force is assumed to exist in
axial member (1).
Downward deflection of wood beam at B due to 7 kN/m uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load.]
Relevant equation from Appendix C:
3 2 3( 2 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 7 kN/m, L = 6 m, x = 4 m, EI = 2,700 kN-m2
Calculation:
3 2 3
3 2 3 3
2
( 2 )24
(7 kN/m)(4 m)(6 m) 2(6 m)(4 m) (4 m) 38.02469 10 m
24(2,700 kN-m )
B
wxv L Lx x
EI
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Upward deflection of wood beam at B due to force F1 in rod (1).
[Appendix C, SS beam with concentrated load not at midspan.]
Relevant equation from Appendix C:
2 2 2( )6
B
Pabv L a b
LEI
Values:
P = −F1, L = 6 m, a = 4 m, b = 2 m,
EI = 2,700 kN-m2
Calculation:
2 2 2
2 2 2 3112
( )6
( )(4 m)(2 m)(6 m) (4 m) (2 m) (1.316872 10 m/kN)
6(6 m)(2,700 kN-m )
B
Pabv L a b
LEI
FF
Elongation of steel rod (1) due to force F1.
61 1
1 1 12 2
1 1
(5 m)(221.0485 10 m/kN)
(113.097336 mm )(200,000 N/mm )(1 kN/1,000 N)
F LF F
A E
The deflection of the wood beam at B will not equal zero in this instance because the rod that supports
the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore,
6
1 1(221.0485 10 m/kN)Bv F
Compatibility equation at B:
The sum of the downward deflection caused by the uniformly distributed load and the upward deflection
caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will
produce a downward (i.e., negative) deflection of the wood beam at B.
3 3 6
1 1
3
1 3
38.02469 10 m (1.316872 10 m/kN) (221.0485 10 m/kN)
38.02469 10 m24.72474 24.7 kN (T) kN
1.537921 10 m/kN
F F
F
Ans.
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(b) Determine maximum bending stress in wood
beam:
Maximum bending moment magnitude
Mmax = 11.6270 kN-m
Bending stress at maximum moment
2
6 4
(11.6270 kN-m)(300 mm/2)(1,000)
225 10 mm
7.75 MPa
x
Ans.
(c) Beam deflection at B.
The beam deflection at B is equal to the elongation of the steel rod:
1 1
1
1 1
2 2
(24.72474 kN)(5 m)(1,000 N/kN)(1,000 mm/m)
(113.097336 mm )(200,000 N/mm )
5.4653 5.47 mm67 mm
B
F Lv
A E
Ans.
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11.65 A W360 × 72 structural steel [E = 200
GPa] wide-flange shape is loaded and
supported as shown in Fig. P11.65. The
beam is supported at B by a timber [E = 12
GPa] post having a cross-sectional area of
20,000 mm2. After a uniformly distributed
load of 50 kN/m is applied to the beam,
determine:
(a) the reactions at supports A, B, and C.
(b) the maximum bending stress in the
beam.
(c) the deflection of the beam at B.
Fig. P11.65
Solution
Section properties:
6 4
beam beam
3 3
beam
2
1 1
W360 × 72: 201 10 mm 351 mm 200,000 MPa
1,150 10 mm
Post (1): 20,000 mm 12,000 MPa
I d E
S
A E
(a) Reactions at supports A, B, and C.
The reaction force from post (1) will be taken as the
redundant, leaving a simply supported beam as the
released beam. To be consistent with earlier sign
conventions (e.g., Chapter 5), we will assume that the
force in the axial member is tension (even though
intuitively we recognize that the post must be in
compression).
Downward deflection of W360 × 72 beam at B due to 50 kN/m uniformly distributed load.
[Appendix C, SS beam with uniformly distributed load over a portion of the span.]
Relevant equation from Appendix C:
3
2 2(4 7 3 )24
B
wav L aL a
LEI
Values:
P = 50 kN/m, L = 13 m, a = 6 m,
EI = 40,200 kN-m2
Calculation:
32 2
32 2 3
2
(4 7 3 )24
(50 kN/m)(6 m)4(13 m) 7(6 m)(13 m) 3(6 m) 204.937 10 m
24(13 m)(40,200 kN-m )
B
wav L aL a
LEI
Since we are assuming that a tension force exists in post (1), the tension force from the post will
cause a downward deflection of W360 × 72 beam at B.
[Appendix C, SS beam with concentrated load not at midspan.]
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Relevant equation from Appendix C:
2 2 2( )6
B
Pabv L a b
LEI
Values:
P = F1, L = 13 m, a = 6 m, b = 7 m,
EI = 40,200 kN-m2
Calculation:
2 2 2
2 2 2 3112
( )6
( )(6 m)(7 m)(13 m) (6 m) (7 m) (1.12514 10 m/kN)
6(13 m)(40,200 kN-m )
B
Pabv L a b
LEI
FF
Deformation of wood post (1) due to force F1.
61 1
1 1 12 2
1 1
(5 m)(20.8333 10 m/kN)
(20,000 mm )(12,000 N/mm )(1 kN/1,000 N)
F LF F
A E
The deflection of the W360 × 72 beam at B will not equal zero in this instance because the post deforms.
If we are consistent and assume that there is tension in the post, then the steel beam must deflect upward
at B. Therefore,
6
1 1(20.8333 10 m/kN)Bv F
Compatibility equation at B:
The sum of the downward deflection caused by the 50 kN/m uniformly distributed load and the
downward deflection caused by the force in post (1) must equal the deformation of the wood post.
3 3 6
1 1
3
1 3178.8 kN
204.937 10 m (1.12514 10 m/kN) (20.8333 10 m/kN)
204.937 10 m178.832 kN
1.145(C)
98 10 m/kNy
F F
F B
Ans.
Equilibrium equations for entire beam: 1(6 m) (13 m) (50 kN/m)(6 m)(3 m) 0A yM F C
(50 kN/m)(6 m)(3 m) ( 178.832 kN)(6 m)
13.3071 kN13 m
13.31 kNyC
Ans.
1 (50 kN/m)(6 m) 0y y yF A F C
(50 kN/m)(6 m) ( 178.832 kN) ( 13.3071 kN)
134.475 kN 134.5 kN
yA
Ans.
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(b) Determine maximum bending stress in the
W360 × 72 beam:
Maximum bending moment magnitude
Mmax = 180.836 kN-m
Bending stress at maximum moment
2
6 4
(180.836 kN-m)(351 mm/2)(1,000)
201 10
157.9 MPa
mmx
Ans.
or using the tabulated value for the section modulus
of the W360 × 72 beam:
2
3 3
(180.836 kN-m)(1,000
157.2
)
1,150 10 mm
MPa
x
Ans.
(c) Beam deflection at B.
The beam deflection at B is equal to the deformation of the wood post:
1 11 2 2
1 1
( 178.832 kN)(5 m)(1,000 N/kN)(1,000 mm/m)
(20,000 mm )(12,000 N/
3
mm )
3.7257 mm .73 mm
B
F Lv
A E
Ans.
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11.66 A timber [E = 1,800 ksi] beam is loaded
and supported as shown in Fig. P11.66. The
cross section of the timber beam is 4-in. wide
and 8-in. deep. The beam is supported at B by
a ¾-in.-diameter aluminum [E = 10,000 ksi]
rod, which has no load before the distributed
load is applied to the beam. After a distributed
load of 800 lb/ft is applied to the beam,
determine:
(a) the force carried by the aluminum rod.
(b) the maximum bending stress in the timber
beam.
(c) the deflection of the beam at B.
Fig. P11.66
Solution
Section properties:
34
beam beam
2 2
1 1
(4 in.)(8 in.)Beam: 170.6667 in. 1,800,000 psi
12
Rod (1): (0.75 in.) 0.441786 in. 10,000,000 psi4
I E
A E
(a) Force carried by the aluminum rod.
The reaction force from rod (1) will be taken as the
redundant, leaving a cantilever beam between A and C
as the released beam.
For this analysis, a tension force is assumed to exist in
axial member (1).
Downward deflection of wood beam at B due to 800 lb/ft uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 800 lb/ft, L = 16 ft, x = 12 ft,
EI = 307.2×106 lb-in.
2
Calculation:
22 2
2 32 2
6 2
(6 4 )24
(800 lb/ft)(12 ft) (12 in./ft)6(16 ft) 4(16 ft)(12 ft) (12 ft) 24.624 in.
24(307.2 10 lb-in. )
B
wxv L Lx x
EI
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Upward deflection of wood beam at B due to force F1 in rod (1).
[Appendix C, SS beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −F1, L = 12 ft, EI = 307.2×106 lb-in.
2
Calculation:
3
3 331
16 2
3
( )(12 ft) (12 in./ft)(3.24 10 in./lb)
3(307.2 10 lb-in. )
B
PLv
EI
FF
Elongation of aluminum rod (1) due to force F1.
61 1
1 1 12 6
1 1
(14 ft)(12 in./ft)(38.02746 10 in./lb)
(0.441786 in. )(10 10 psi)
F LF F
A E
The deflection of the wood beam at B will not equal zero in this instance because the rod that supports
the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore,
6
1 1(38.02746 10 in./lb)Bv F
Compatibility equation at B:
The sum of the downward deflection caused by the uniformly distributed load and the upward deflection
caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the
aluminum rod will produce a downward (i.e., negative) deflection of the wood beam at B.
3 6
1 1
1 3
24.624 in. (3.24 10 in./lb) (38.02746 10 in./lb)
24.624 in.7,511.835 lb
3.278027 10 in./l7,510 lb (
bT)
F F
F
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.
(b) Determine maximum bending stress in wood
beam:
Maximum bending moment magnitude
Mmax = 12,258 lb-ft
Bending stress at maximum moment
4
(12,258 lb-ft)(8 in./2)(12 in./ft)
170.6667 in.
3,447.56 psi 3,450 psi
x
Ans.
(c) Beam deflection at B.
The beam deflection at B is equal to the elongation of the aluminum rod:
1 1
1
1 1
2 6
(7,511.835 lb)(14 ft)(12 in./ft)
(0.441786 in. )(10 10 psi)
0.28566 0.286 in. in.
B
F Lv
A E
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.
11.67 A W530 × 66 structural steel [E = 200
GPa] wide-flange shape is loaded and
supported as shown in Fig. P11.67. A
uniformly distributed load of 70 kN/m is
applied to the beam, causing the roller support
at B to settle downward (i.e., displace
downward) by 10 mm. Determine:
(a) the reactions at supports A and B.
(b) the maximum bending stress in the beam.
Fig. P11.67
Solution
Section properties:
6 4 3 3W530 66: 351 10 mm 526 mm 1,340 10 mmI d S
(a) Reactions at supports A and B.
The reaction force at B will be taken as the redundant, leaving a cantilever beam between A and C as the
released beam.
Downward deflection of beam at B due to 70 kN/m uniformly distributed load.
[Appendix C, Cantilever beam with uniformly distributed load.]
Relevant equation from Appendix C:
2
2 2(6 4 ) (elastic curve)24
B
wxv L Lx x
EI
Values:
w = 70 kN/m, L = 6 m, x = 4.5 m,
EI = 70,200 kN-m2
Calculation:
22 2
22 2
2
(6 4 )24
(70 kN/m)(4.5 m)6(6 m) 4(6 m)(4.5 m) (4.5 m) 0.107903 m
24(70,200 kN-m )
B
wxv L Lx x
EI
Upward deflection of beam at B due to reaction force By.
[Appendix C, SS beam with concentrated load at tip.]
Relevant equation from Appendix C:
3
3B
PLv
EI
Values:
P = −By, L = 4.5 m, EI = 70,200 kN-m2
Calculation:
3
3
6
2
3
( )(4.5 m)(432.6923 10 m/kN)
3(70,200 kN-m )
B
y
y
PLv
EI
BB
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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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Compatibility equation at B:
The sum of the downward deflection caused by the uniformly distributed load and the upward deflection
caused by the reaction force By must equal the support settlement:
6
3
6
0.107903 m (432.6923 10 m/kN) 0.010 m
97.90264 10 m226.2639 kN
432.6923 10 m/k226 k
NN
y
y
B
B
Ans.
Equilibrium equations for entire beam: (4.5 m) (70 kN/m)(6 m)(3 m) 0A A yM M B
(226.2639 kN)(4.5 m) (70 kN/m)(6 m)(3 m)
241.8125 242 kN-m (cc-m )kN w
AM
Ans.
(70 kN/m)(6 m) 0y y yF A B
(70 kN/m)(6 m) 226.2639 kN 193.7361 kN 193.7 kNyA Ans.
(b) Determine maximum bending stress in
beam:
Maximum bending moment magnitude
Mmax = 241.8125 kN-m (at A)
Bending stress at maximum moment
2
6 4
(241.8125 kN-m)(526 mm/2)(1,000)
351 10 mm
181.187 MPa 181.2 MPa
x
Ans.
or using the tabulated value for the section
modulus:
2
3 3
(241.8125 kN-m)(1,000)
1,340 10 mm
180.457 MPa 180.5 MPa
x
Ans.