PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond t
he limited
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without permission.
PROBLEM 5.1
For the beam and loading shown, (
a
) draw the shear and bendingmoment
diagrams, (
b
) determine the equations of the shear and bendingmoment
curves.
SOLUTION
Reactions:
0:0
C
P
b
M LA bP A
L
!
=
"
= =
0:0
A
P
a
M LC aP C
L
!
=
"
= =
From
A
to
B
:
0
x
a
< <
0:0
y
Pb
F V
L
!
=
"
=
P
b
V
L
=
W
0:0
J
Pb
M M x
L
!
=
"
=
P
bx
M
L
=
W
From
B
to
C
:
a x L
< <
0:0
y
Pa
F V
L
!
= + =
P
a
V
L
=
"
W
0:( ) 0
K
Pa
M M L x
L
!
=
"
+
"
=
( )
P
a L x
M
L
"
=
W
At section
B
:
2
P
ab
M
L
=
W
PROPRIETARY MATERIAL.
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PROBLEM 5.3
For the beam and loading shown, (
a
) draw the shear and bending
moment diagrams, (
b
) determine the equations of the shear and
bendingmoment curves.
SOLUTION
From
A
to
B
(0 )
x
a
< <
:
0:0
y
F wx V
¦
=
""
=
V wx
=
"
W
0:( ) 0
2
J
x
M wx M
¦
= + =
2
2
wx
M
=
"
W
From
B
to
C
( ):
a x L
< <
0:0
y
F wa V
¦
=
""
=
V wa
=
"
W
0:( ) 0
2
J
a
M wa x M
§ ·
¦
=
"
+ =
¨ ¸
© ¹
2
a
M wa x
§ ·
=
""
¨ ¸
© ¹
W
PROPRIETARY MATERIAL.
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PROBLEM 5.4
For the beam and loading shown, (
a
) draw the shear and bendingmoment
diagrams, (
b
) determine the equations of the shear and bendingmoment
curves.
SOLUTION
0
1
0:0
2
y
w x
F x V
L
!
=
"#"
=
2
0
2
w x
V
L
=
"
W
0
1
0:0
2 3
J
w x x
M x M
L
!
=
##
+ =
3
0
6
w x
M
L
=
"
W
At
,
x
L
=
0
2
w L
V
=
"
0
max
 
2
w L
V
=
W
2
0
6
w L
M
=
"
2
0
max
 
6
w L
M
=
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.7
Draw the shear and bendingmoment diagrams for the beam and
loading shown, and determine the maximum absolute value (
a
) of the
shear, (
b
) of the bending moment.
SOLUTION
Reactions:
0:(300)(4) (240)(3) (360)(7) 12 0 170 lb
C
M B
¦
=
""
+ = =
$
B
0:300 240 360 170 0 730 lb
y
F C
¦
=
"
+
""
+ = =
$
C
From
A
to
C
:
0:300 0 300 lb
y
F V V
¦
=
""
= =
"
1
0:(300)( ) 0 300
M
x M M x
¦
= + = =
"
From
C
to
D
:
0:300 730 0 430 lb
y
F V V
¦
=
"
+
"
= = +
2
0:(300) (730)( 4) 0
2920 430
M x x M
M
x
¦
=
""
+ =
=
"
+
From
D
to
E
:
0:360 170 0 190 lb
y
F V V
¦
=
"
+ = = +
3
0:(170)(16 ) (360)(11 ) 0
1240 190
M x x M
M
x
¦
=
""""
=
=
"
+
From
E
to
B
:
0:170 0 170lb
y
F V V
¦
= + = =
"
4
0:(170)(16 ) 0
2720 170
M x M
M
x
¦
=
""
=
=
"
(
a
)
max
430 lb
V
=
W
(
b
)
max
1200 lb in
M
=
#
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.11
Draw the shear and bendingmoment diagrams for the beam and
loading shown, and determine the maximum absolute value (
a
) of
the shear, (
b
) of the bending moment.
SOLUTION
Reactions:
0:3 (8)(60) (24)(60) 0
A EF
M F
¦
=
""
=
640 kips
EF
F
=
0:640 0 640 kips
x x x
F A A
¦
=
"
= =
%
0:60 60 0 120 kips
y y y
F A A
¦
=
""
= =
$
From
A
to
C
:
(0 8 in.)
x
< <
0:
y
F
¦
=
120 0
120 kips
V
V
"
=
=
0:120 0 120 kip in
J
M M x M x
¦
=
"
= =
#
From
C
to
D
:
(8 in.16 in.)
x
< <
0:120 60 0 60 kips
Y
F V V
¦
=
""
= =
0:120 60( 8) 0
(60 480) kips in
J
M M x x
M x
¦
=
"
+
"
=
= +
#
From
D
to
B
:
(16 in.24 in.)
x
< <
0:60 0 60 kips
y
F V V
¦
=
"
= =
0:60(24 ) 0
(60 1440) kip in
J
M M x
M x
¦
=
"""
=
=
"#
(
a
)
max
120.0 kips
V
=
W
(
b
)
max
1440 kip in 120.0 kip ft
M
=
#
=
#
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.13
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bendingmoment diagrams for the beam
AB
and determine
the maximum absolute value (
a
) of the shear, (
b
) of the bending
moment.
SOLUTION
Over the whole beam,
0:12 (3)(2) 24 (3)(2) 0
y
F w
!
=
"""
=
3 kips/ft
w
=
A
to
C
:
(0 3 ft)
x
&
<
0:3 2 0
y
F x x V
!
=
""
=
( ) kips
V x
=
0:(3 ) (2 ) 0
2 2
J
x x
M x x M
+
!
=
"
+ + =
2
(0.5 ) kip ft
M x
=
#
At
C
,
3 ft
x
=
3 kips,4.5 kip ft
V M
= =
#
C
to
D
:
(3 ft 6 ft)
x
&
<
0:3 (2)(3) 0
y
F x V
!
=
""
=
(3 6) kips
V x
=
"
3
0:(3 ) (2)(3) 0
2 2
x
MK x x M
§ · § ·
!
=
"
+
"
+ =
¨ ¸ ¨ ¸
© ¹ © ¹
2
(1.5 6 9) kip ft
M x x
=
"
+
#
At
,
D
"
6 ft
x
=
12 kips,27 kip ft
V M
= =
#
D
to
B
: Use symmetry to evaluate.
(
a
)
max
  12.00 kips
V
=
W
(
b
)
max
  27.0 kip ft
M
=
#
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.15
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at
C
.
SOLUTION
Reaction at
A
:
0:4.5 (3.0)(3) (1.5)(3) (1.8)(4.5)(2.25) 0
B
M A
=
"
+ + + =
7.05 kN
=
$
A
Use
AC
as free body.
3
0:(7.05)(1.5) (1.8)(1.5)(0.75) 0
8.55 kN m 8.55 10 N m
C C
C
M M
M
!
=
"
+ =
=
#
=
'#
3 3 6 4
6 4
1 1
(80)(300) 180 10 mm
12 12
180 10 m
1
(300) 150 mm 0.150 m
2
I bh
c
"
= = =
'
=
'
= = =
3
6
6
(8.55 10 )(0.150)
7.125 10 Pa
180 10
Mc
I
(
"
'
= = =
'
'
7.13 MPa
(
=
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.19
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at
C
.
SOLUTION
Use entire beam as free body.
0:
B
M
¦
=
90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0
A
"
+ + + + + =
9.5 kips
A
=
Use portion AC as free body.
0:(15)(9.5) 0
142.5 kip in
C
M M
M
¦
=
"
=
=
#
For
3
8 18.4,14.4 in
S S
'
=
Normal stress:
142.5
14.4
M
S
(
= =
9.90 ksi
(
=
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.44
Draw the shear and bendingmoment diagrams for the beam and loading
shown, and determine the maximum absolute value (
a
) of the shear, (
b
) of
the bending moment.
SOLUTION
Reaction at
A
:
0: 3.0 (1.5)(3.0)(3.5) (1.5)(3) 0
6.75 kN
B
M A
!
=
"
+ + =
=
#
A
Reaction at
B
:
6.75 kN
=
#
B
Beam
ACB
and loading: (See sketch.)
Areas of load diagram:
A
to
C
:
(2.4)(3.5) 8.4 kN
=
C
to
B
:
(0.6)(3.5) 2.1 kN
=
Shear diagram:
6.75 kN
6.75 8.4 1.65 kN
1.65 3 4.65 kN
4.65 2.1 6.75 kN
A
C
C
B
V
V
V
V
"
+
=
=
"
=
"
=
""
=
"
=
""
=
"
Over
A
to
C
,
6.75 3.5
V x
=
"
At
G
,
6.75 3.5 0 1.9286 m
G G
V x x
=
"
= =
Areas of shear diagram:
A
to
G
:
1
(1.9286)(6.75) 6.5089 kN m
2
=
$
G
to
C
:
1
(0.4714)( 1.65) 0.3889 kN m
2
"
=
"$
C
to
B
:
1
(0.6)( 4.65 6.75) 3.42 kN m
2
""
=
"$
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.44
(Continued)
Bending moments:
0
0 6.5089 6.5089 kN m
6.5089 0.3889 6.12 kN m
6.12 2.7 3.42 kN m
3.42 3.42 0
A
G
C
C
B
M
M
M
M
M
"
+
=
= + =
#
=
"
=
#
=
"
=
#
=
"
=
(
a
)
max
  6.75 kN
V
=
W
(
b
)
max
  6.51 kN m
M
=
#
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.45
Draw the shear and bendingmoment diagrams for the beam and loading
shown, and determine the maximum absolute value (
a
) of the shear, (
b
) of
the bending moment.
SOLUTION
0:
3 (1)(4) (0.5)(4) 0
2 kN
B
M
A
¦
=
"
+ + =
=
$
A
0:3 (2)(4) (2.5)(4) 0
6 kN
A
M B
¦
=
""
=
=
$
B
Shear diagram:
A
to
C
:
2 kN
V
=
C
to
D
:
2 4 2 kN
V
=
"
=
"
D
to
B
:
2 4 6 kN
V
=
""
=
"
Areas of shear diagram:
A
to
C
:
(1)(2) 2 kN m
Vdx
= =
#
³
C
to
D
:
(1)( 2) 2 kN m
Vdx
=
"
=
"#
³
D
to
E
:
(1)( 6) 6 kN m
Vdx
=
"
=
"#
³
Bending moments:
0
A
M
=
0 2 2 kN m
2 4 6 kN m
6 2 4 kN m
4 2 6kN m
6 6 0
C
C
D
D
B
M
M
M
M
M
"
+
"
+
= + =
#
= + =
#
=
"
=
#
= + =
#
=
"
=
(
a
)
max
6.00 kN
V
=
W
(
b
)
max
6.00 kN m
M
=
#
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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he limited
distribution to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this m
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PROBLEM 5.50
For the beam and loading shown, determine the equations of the shear
and bendingmoment curves, and the maximum absolute value of the
bending moment in the beam, knowing that (
a
)
k
=
1, (
b
)
k
=
0.5.
SOLUTION
0 0 0
0
0
2
0
0 1
1
2
0
0
( )
(1 ).
(1 )
(1 )
2
0 at 0 0
(1 )
2
w x kw L x w x
w k kw
L L L
w x
dV
w kw k
dx L
w x
V kw x k C
L
V x C
w x
dM
V kw x k
d x L
"
=
"
= +
"
=
"
=
"
+
=
"
+ +
= = =
= =
"
+
2 3
0 0
2
2
2 3
0 0
(1 )
2 6
0 at 0 0
(1 )
2 6
kw x w x
M
k C
L
M x C
kw x k w x
M
L
=
"
+ +
= = =
+
=
"
(
a
)
1.
k
=
2
0
0
w x
V w x
L
=
"
W
2 3
0 0
2 3
w x w x
M
L
=
"
W
Maximum
M
occurs at
.
x
L
=
2
0
max
6
w L
M
=
W
(
b
)
1
.
2
k
=
2
0 0
3
2 4
w x w x
V
L
=
"
W
2 3
0 0
4 4
w x w x
M
L
=
"
W
2
0 at
3
V x L
= =
At
( )
( )
2 3
2 2
2
0 0
3 3
2
0
0
2
,0.03704
3 4 4 27
w L w L
w L
x
L M w L
L
= =
"
= =
At
,0
x L M
= =
2
0
max
 
27
w L
M
=
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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PROBLEM 5.55
Draw the shear and bendingmoment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
0:(2)(1) (3)(4)(2) 4 0
5.5 kN
C
M B
B
¦
=
"
+ =
=
0:(5)(2) (3)(4)(2) 4 0
8.5 kN
B
M C
C
¦
= +
"
=
=
Shear:
A
to
C
:
2 kN
V
=
"
:
C
+
2 8.5 6.5 kN
V
=
"
+ =
B
:
6.5 (3)(4) 5.5 kN
V
=
"
=
"
Locate point
D
where
0.
V
=
4
12 26
6.5 5.5
d d
d
"
= =
2.1667m 4 3.8333 m
d d
=
"
=
Areas of the shear diagram:
A
to
C:
( 2.0)(1) 2.0 kN m
Vdx
³
=
"
=
"#
C
to
D
:
1
(2.16667)(6.5) 7.0417 kN m
2
Vdx
³
= =
#
D
to
B
:
1
(3.83333)( 5.5) 5.0417 kN m
2
Vdx
³
=
"
=
"#
Bending moments:
0
A
M
=
0 2.0 2.0 kN m
C
M
=
"
=
"#
2.0 7.0417 5.0417 kN m
D
M
=
"
+ =
#
5.0417 5.0417 0
B
M
=
"
=
Maximum
3
5.0417 kN m 5.0417 10 N m
M
=
#
=
'#
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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he limited
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PROBLEM 5.55
(Continued)
For pipe:
1 1 1 1
(160) 80 mm,(140) 70 mm
2 2 2 2
o o i i
c d c d
= = = = = =
( )
4 4 4 4 6 4
(80) (70) 13.3125 10 mm
4 4
o i
I c c
) )
ª º
=
"
=
"
=
'
¬ ¼
6
3 3 6 3
13.3125 10
166.406 10 mm 166.406 10 m
80
o
I
S
c
"
'
= = =
'
=
'
Normal stress:
3
6
6
5.0417 10
30.3 10 Pa
166.406 10
M
S
(
"
'
= = =
'
'
30.3 MPa
(
=
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
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he limited
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PROBLEM 5.67
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 1750 psi.
SOLUTION
Reactions: By symmetry,
A
=
D.
1 1
0:(3)(1.5) (6)(1.5) (3)(1.5) 0
2 2
6.75 kips
y
F A D
A D
¦
=
""""
=
= =
$
Shear diagram:
6.75 kips
A
V
=
1
6.75 (3)(1.5) 4.5 kips
2
B
V
=
"
=
4.5 (6)(1.5) 4.5 kips
C
V
=
"
=
"
1
4.5 (3)(1.5) 6.75 kips
2
D
V
=
""
=
"
Locate point
E
where
0
V
=
:
By symmetry,
E
is the midpoint of
BC
.
Areas of the shear diagram:
2
to:(3)(4.5) (3)(2.25) 18 kip ft
3
A B
+ =
#
1
to:(3)(4.5) 6.75 kip ft
2
B E
=
#
1
to:(3)( 4.5) 6.75 kip ft
2
E C
"
=
"#
to:Byantisymmetry,18 kip ft
C D
"#
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond t
he limited
distribution to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this m
anual is using it
without permission.
PROBLEM 5.67
(Continued)
Bending moments:
0
A
M
=
0 18 18 kip ft
B
M
= + =
#
18 6.75 24.75 kip ft
E
M
= + =
#
24.75 6.75 18 kip ft
C
M
=
"
=
#
18 18 0
D
M
=
"
=
3
max max
max
max
(24.75 kip ft)(12 in/ft)
169.714 in
1.750 ksi
M M
S
S
(
(
#
= = = =
For a rectangular section,
2
1
6
S bh
=
6 6(169.714)
14.27 in.
5
S
h
b
= = =
14.27 in.
h
=
W
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond t
he limited
distribution to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this m
anual is using it
without permission.
PROBLEM 5.73
Knowing that the allowable stress for the steel used is 160 MPa, select the
most economical wideflange beam to support the loading shown.
SOLUTION
2
1
1
2
2 3
2
2
2 3
18 6
6 (6 2 ) kN/m
6
6 2
6
0 at 0,0
6
1
3
3
0 at 0,0
1
3
3
w x x
dV
w x
dx
V x x C
V x C
dM
V x x
dx
M x x C
M x C
M x x
"
§ ·
= + = +
¨ ¸
© ¹
=
"
=
""
=
""
+
= = =
= =
""
=
""
+
= = =
=
""
max
occurs at 6 m.
M x
=
2 3 3
max
1
= (3)(6) (6) 80 kN m 180 10 N m
3
M
§ ·
""
=
#
=
'#
¨ ¸
© ¹
6
all
160 MPa 160 10 Pa
(
= =
'
3
3 3 3 3
min
6
all
180 10
1.125 10 m 1125 10 mm
160 10
M
S
(
"
'
= = =
'
=
'
'
Shape
S
, (
3 3
10 mm
)
W530 66
'
1340
*
W460 74
'
1460
Lightest acceptable wide flange beam:
W530 66
'
W
W410 85
'
1510
W360 79
'
1270
W310 107
'
1600
W250 101
'
1240
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond t
he limited
distribution to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this m
anual is using it
without permission.
PROBLEM 5.74
Knowing that the allowable stress for the steel used is 160 MPa, select the most
economical wideflange beam to support the loading shown.
SOLUTION
all
Sectionmodulus
160 Mpa
!
=
6 3
max
min
all
3 3
286 kN m
1787 10 m
160 MPa
1787 10 mm
M
S
!
"
#
= = =
$
=
$
Use
W530 92
$
W
Shape
S
, (
3 3
10 mm
)
W610 101
$
2520
W530 92
$
2080
%
W
W460 113
$
2390
W410 114
$
2200
W360 122
$
2020
W310 143
$
2150
PROPRIETARY MATERIAL.
© 2012 The McGrawHill Companies, Inc.
All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond t
he limited
distribution to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this m
anual is using it
without permission.
PROBLEM 5.89
A 54kip is load is to be supported at the center of the 16ft span shown.
Knowing that the allowable normal stress for the steel used is 24 ksi,
determine (
a
) the smallest allowable length
l
of beam
CD
if the
W12 50
!
beam
AB
is not to be overstressed, (
b
) the most economical
W shape that can be used for beam
CD
. Neglect the weight of both
beams.
SOLUTION
(
a
)
8ft 16ft 2
2
l
d l d
=
"
=
"
(1)
Beam
AB
(Portion
AC
):
For
W12 50,
!
3
64.2 in
x
S
=
all
24 ksi
#
=
all all
(24)(64.2) 1540.8 kip in 128.4 kip ft
27 128.4 kip ft 4.7556 ft
x
C
M S
M d d
#
= = =
$
=
$
= =
$
=
Using (1),
16 2 16 2(4.7556) 6.4888 ft
l d
=
"
=
"
=
6.49 ft
l
=
W
(
b
)
Beam
CD
:
all
6.4888 ft 24 ksi
l
#
= =
max
min
all
3
(87.599 12) kip in
24 ksi
43.800 in
M
S
#
!$
= =
=
Shape
3
(in )
S
W18 35
!
57.6
W16 31
!
47.2
ĸ
W14 38
!
54.6
W12 35
!
45.6
W10 45
!
49.1
W16 31.
!
W
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