PROBLEM 5.1
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions:
From A to B:
ΣM C = 0: LA − bP = 0
A=
Pb
L
ΣM A = 0: LC − aP = 0
C =
Pa
L
0< x<a
ΣFy = 0:
Pb
−V = 0
L
Pb
L
V =
ΣM J = 0: M −
Pb
x=0
L
M =
From B to C:
Pbx
L
a< x< L
ΣFy = 0: V +
Pa
=0
L
V =−
ΣM K = 0: − M +
Pa
( L − x) = 0
L
M =
At section B:
Pa
L
Pa( L − x)
L
M =
Pab
L2
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.3
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and
bending-moment curves.
From A to B (0 < x < a) :
SOLUTION
Fy = 0 : − wx − V = 0
V = −wx
M J = 0 : (wx)
x
+M =0
2
M =−
wx 2
2
From B to C (a < x < L) :
Fy = 0 : − wa − V = 0
M J = 0 : (wa) x −
a
+M =0
2
V = − wa
M = −wa x −
a
2
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.4
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
ΣFy = 0:
ΣM J = 0:
−
1 w0 x
⋅ x −V = 0
2 L
V =−
w0 x 2
2L
M =−
w0 x3
6L
|V |max =
w0 L
2
1 w0 x
x
⋅x⋅ +M = 0
2 L
3
At x = L,
V =−
w0 L
2
M =−
w0 L2
6
|M |max =
w0 L2
6
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.7
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the
shear, (b) of the bending moment.
SOLUTION
Reactions:
M C = 0 : (300)(4) − (240)(3) − (360)(7) + 12 B = 0
Fy = 0 :
− 300 + C − 240 − 360 + 170 = 0
B = 170 lb ↑
C = 730 lb ↑
From A to C:
Fy = 0 :
M1 = 0 :
− 300 − V = 0
V = −300 lb
(300)( x) + M = 0
M = −300 x
From C to D:
Fy = 0 :
M2 = 0 :
− 300 + 730 − V = 0
V = +430 lb
(300) x − (730)( x − 4) + M = 0
M = −2920 + 430 x
From D to E:
Fy = 0 : V − 360 + 170 = 0
M3 = 0 :
V = +190 lb
(170)(16 − x) − (360)(11 − x) − M = 0
M = −1240 + 190 x
From E to B:
Fy = 0 : V + 170 = 0
M4 = 0 :
V = −170lb
(170)(16 − x) − M = 0
M = 2720 − 170 x
(a)
(b)
V
M
max
max
= 430 lb
= 1200 lb ⋅ in
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.11
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
Reactions:
M A = 0 : 3FEF − (8)(60) − (24)(60) = 0
FEF = 640 kips
Fx = 0 : Ax − 640 = 0
Ax = 640 kips →
Fy = 0 : Ay − 60 − 60 = 0
Ay = 120 kips ↑
(0 < x < 8 in.)
From A to C:
Fy = 0 :
120 − V = 0
V = 120 kips
M J = 0 : M − 120 x = 0
M = 120 x kip ⋅ in
(8 in. < x < 16 in.)
From C to D:
FY = 0 : 120 − 60 − V = 0 V = 60 kips
MJ = 0 :
M − 120 x + 60( x − 8) = 0
M = (60 x + 480) kips ⋅ in
(16 in. < x < 24 in.)
From D to B:
Fy = 0 : V − 60 = 0 V = 60 kips
M J = 0 : − M − 60(24 − x) = 0
M = (60 x − 1440) kip ⋅ in
(a) V
(b)
M
max
max
= 120.0 kips
= 1440 kip ⋅ in = 120.0 kip ⋅ ft
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.13
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment 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,
ΣFy = 0: 12w − (3)(2) − 24 − (3)(2) = 0
A to C:
w = 3 kips/ft
(0 ≤ x < 3 ft)
ΣFy = 0: 3x − 2 x − V = 0
+ΣM J = 0: − (3x)
At C,
V = ( x) kips
x
x
+ (2 x) + M = 0
2
2
M = (0.5x 2 ) kip ⋅ ft
x = 3 ft
V = 3 kips, M = 4.5 kip ⋅ ft
C to D:
(3 ft ≤ x < 6 ft)
ΣFy = 0: 3x − (2)(3) − V = 0
ΣMK = 0: −(3x)
V = (3x − 6) kips
x
3
+ (2)(3) x −
+M =0
2
2
M = (1.5 x 2 − 6 x + 9) kip ⋅ ft
At D −,
x = 6 ft
V = 12 kips,
D to B:
M = 27 kip ⋅ ft
Use symmetry to evaluate.
(a)
|V |max = 12.00 kips
(b)
|M |max = 27.0 kip ⋅ ft
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
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:
M B = 0:
− 4.5 A + (3.0)(3) + (1.5)(3) + (1.8)(4.5)(2.25) = 0
A = 7.05 kN ↑
Use AC as free body.
ΣM C = 0: M C − (7.05)(1.5) + (1.8)(1.5)(0.75) = 0
M C = 8.55 kN ⋅ m = 8.55 × 103 N ⋅ m
I =
1 3
1
(80)(300)3 = 180 × 106 mm 4
bh =
12
12
= 180 × 10−6 m 4
c=
1
(300) = 150 mm = 0.150 m
2
σ =
Mc (8.55 × 103 )(0.150)
=
= 7.125 × 106 Pa
−6
I
180 × 10
σ = 7.13 MPa
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
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.
MB = 0 :
−90 A + (75)(5) + (60)(5) + (45)(2) + (30)(2) + (15)(2) = 0
A = 9.5 kips
Use portion AC as free body.
M C = 0 : M − (15)(9.5) = 0
M = 142.5 kip ⋅ in
For S 8 × 18.4, S = 14.4 in 3
Normal stress:
σ =
M
142.5
=
S
14.4
σ = 9.90 ksi
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.44
Draw the shear and bending-moment 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:
ΣM B = 0: −3.0 A + (1.5)(3.0)(3.5) + (1.5)(3) = 0
A = 6.75 kN ↑
B = 6.75 kN ↑
Reaction at 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:
VA = 6.75 kN
VC − = 6.75 − 8.4 = −1.65 kN
VC + = −1.65 − 3 = −4.65 kN
VB = −4.65 − 2.1 = −6.75 kN
Over A to C,
V = 6.75 − 3.5x
At G,
V = 6.75 − 3.5xG = 0 xG = 1.9286 m
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 McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.44 (Continued)
Bending moments:
MA = 0
M G = 0 + 6.5089 = 6.5089 kN ⋅ m
M C − = 6.5089 − 0.3889 = 6.12 kN ⋅ m
M C + = 6.12 − 2.7 = 3.42 kN ⋅ m
M B = 3.42 − 3.42 = 0
(a)
|V |max = 6.75 kN
(b)
|M |max = 6.51 kN ⋅ m
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.45
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
SOLUTION
M B = 0:
− 3 A + (1)(4) + (0.5)(4) = 0
A = 2 kN ↑
M A = 0: 3B − (2)(4) − (2.5)(4) = 0
B = 6 kN ↑
Shear diagram:
A to C:
V = 2 kN
C to D:
V = 2 − 4 = −2 kN
D to B:
V = −2 − 4 = −6 kN
Areas of shear diagram:
A to C:
Vdx = (1)(2) = 2 kN ⋅ m
C to D:
Vdx = (1)(−2) = −2 kN ⋅ m
D to E:
Vdx = (1)(−6) = −6 kN ⋅ m
Bending moments:
MA = 0
M C − = 0 + 2 = 2 kN ⋅ m
M C + = 2 + 4 = 6 kN ⋅ m
M D − = 6 − 2 = 4 kN ⋅ m
M D + = 4 + 2 = 6 kN ⋅ m
MB = 6 − 6 = 0
(a)
(b)
V
M
max
max
= 6.00 kN
= 6.00 kN ⋅ m
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.50
For the beam and loading shown, determine the equations of the shear
and bending-moment curves, and the maximum absolute value of the
bending moment in the beam, knowing that (a) k = 1, (b) k = 0.5.
SOLUTION
w0 x kw0 ( L − x)
wx
−
= (1 + k ) 0 − kw.
L
L
L
w0 x
= − w = kw0 − (1 + k )
L
2
wx
= kw0 x − (1 + k ) 0 + C1
2L
= 0 at x = 0 C1 = 0
w=
dV
dx
V
V
w x2
dM
= V = kw0 x − (1 + k ) 0
dx
2L
kw0 x 2
w x3
− (1 + k ) 0 + C2
2
6L
M = 0 at x = 0 C2 = 0
M=
M=
(a)
w0 x 2
L
2
wx
w x3
M= 0 − 0
2
3L
k = 1.
V = w0 x −
x = L.
Maximum M occurs at
(b)
kw0 x 2 (1 + k ) w0 x3
−
2
6L
k=
M
1
.
2
V = 0 at
At
2
x = L,
3
At
x = L,
M=
x=
w0 ( 23 L )
4
2
−
max
=
w0 L2
6
V=
w0 x 3w0 x 2
−
2
4L
M=
w0 x 2 w0 x3
−
4
4L
2
L
3
w0 ( 23 L )
4L
3
=
w0 L2
= 0.03704 w0 L2
27
M =0
|M |max =
w0 L2
27
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.55
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
M C = 0 : (2)(1) − (3)(4)(2) + 4B = 0
B = 5.5 kN
M B = 0 : (5)(2) + (3)(4)(2) − 4C = 0
C = 8.5 kN
Shear:
A to C:
V = −2 kN
C+ :
V = −2 + 8.5 = 6.5 kN
B:
V = 6.5 − (3)(4) = −5.5 kN
Locate point D where V = 0.
d
4−d
=
6.5
5.5
d = 2.1667 m
12d = 26
4 − d = 3.8333 m
Areas of the shear diagram:
A to C:
Vdx = (−2.0)(1) = −2.0 kN ⋅ m
C to D:
Vdx =
1
(2.16667)(6.5) = 7.0417 kN ⋅ m
2
D to B:
Vdx =
1
(3.83333)(−5.5) = −5.0417 kN ⋅ m
2
Bending moments:
MA = 0
M C = 0 − 2.0 = −2.0 kN ⋅ m
M D = −2.0 + 7.0417 = 5.0417 kN ⋅ m
M B = 5.0417 − 5.0417 = 0
Maximum M = 5.0417 kN ⋅ m = 5.0417 × 103 N ⋅ m
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.55 (Continued)
For pipe:
I =
Normal stress:
1
1
1
1
do = (160) = 80 mm, ci = di = (140) = 70 mm
2
2
2
2
co =
π
(c
4
4
o
)
− ci4 =
π
4
(80) 4 − (70)4 = 13.3125 × 106 mm 4
S =
I
13.3125 × 106
=
= 166.406 × 103 mm3 = 166.406 × 10−6 m3
co
80
σ =
M
5.0417 × 103
=
= 30.3 × 106 Pa
S
166.406 × 10−6
σ = 30.3 MPa
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
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
By symmetry, A = D.
Reactions:
1
1
(3)(1.5) − (6)(1.5) − (3)(1.5) − D = 0
2
2
A = D = 6.75 kips ↑
Fy = 0 : A −
Shear diagram:
VB = 6.75 −
VA = 6.75 kips
1
(3)(1.5) = 4.5 kips
2
VC = 4.5 − (6)(1.5) = −4.5 kips
VD = −4.5 −
1
(3)(1.5) = −6.75 kips
2
Locate point E where V = 0 :
By symmetry, E is the midpoint of BC.
Areas of the shear diagram:
A to B : (3)(4.5) +
2
(3)(2.25) = 18 kip ⋅ ft
3
B to E :
1
(3)(4.5) = 6.75 kip ⋅ ft
2
E to C :
1
(3)(−4.5) = −6.75 kip ⋅ ft
2
C to D : By antisymmetry, − 18 kip ⋅ ft
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.67 (Continued)
Bending moments: M A = 0
M B = 0 + 18 = 18 kip ⋅ ft
M E = 18 + 6.75 = 24.75 kip ⋅ ft
M C = 24.75 − 6.75 = 18 kip ⋅ ft
M D = 18 − 18 = 0
σ max =
M
max
S
S =
M
max
σ max
For a rectangular section,
=
(24.75 kip ⋅ ft)(12 in/ft)
= 169.714 in 3
1.750 ksi
S =
1 2
bh
6
h=
6S
=
b
6(169.714)
= 14.27 in.
5
h = 14.27 in.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.73
Knowing that the allowable stress for the steel used is 160 MPa, select the
most economical wide-flange beam to support the loading shown.
SOLUTION
w= 6+
18 − 6
x = (6 + 2 x) kN/m
6
dV
= −w = −6 − 2 x
dx
V = −6 x − x 2 + C1
V = 0 at
x = 0, C1 = 0
dM
= V = −6 x − x 2
dx
1
M = −3x 2 − x3 + C2
3
M = 0 at x = 0, C2 = 0
M = −3x 2 −
M
M
1 3
x
3
max
occurs at x = 6 m.
max
= −(3)(6)2 −
1
(6)3 = 80 kN ⋅ m = 180 × 103 N ⋅ m
3
σ all = 160 MPa = 160 × 106 Pa
Smin =
Shape
W530 × 66
M
σ all
=
180 × 103
= 1.125 × 10−3 m3 = 1125 × 103 mm3
160 × 106
S, ( 103 mm3 )
W460 × 74
1340 ←
1460
W410 × 85
1510
W360 × 79
1270
W310 × 107
1600
W250 × 101
1240
Lightest acceptable wide flange beam: W530 × 66
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.74
Knowing that the allowable stress for the steel used is 160 MPa, select the most
economical wide-flange beam to support the loading shown.
SOLUTION
Shape
Section modulus
σ all = 160 Mpa
Smin =
M max
σ all
=
286 kN ⋅ m
= 1787 × 10−6 m3
160 MPa
= 1787 × 103 mm3
S, ( 103 mm3 )
W610 × 101
2520
W530 × 92
2080 ←
W460 × 113
2390
W410 × 114
2200
W360 × 122
2020
W310 × 143
2150
Use W530 × 92
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 5.89
A 54-kip is load is to be supported at the center of the 16-ft 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)
d = 8ft −
l
l = 16 ft − 2d
2
(1)
Beam AB (Portion AC):
For W12 × 50,
S x = 64.2 in 3 σ all = 24 ksi
M all = σ all S x = (24)(64.2) = 1540.8 kip ⋅ in = 128.4 kip ⋅ ft
M C = 27d = 128.4 kip ⋅ ft
Using (1),
d = 4.7556 ft
l = 16 − 2d = 16 − 2(4.7556) = 6.4888 ft
l = 6.49 ft
(b)
l = 6.4888 ft σ all = 24 ksi
Beam CD:
M
(87.599 × 12) kip ⋅ in
Smin = max =
24 ksi
σ all
= 43.800 in 3
Shape
W18 × 35
W16 × 31
W14 × 38
W12 × 35
W10 × 45
S (in 3 )
57.6
47.2
54.6
45.6
49.1
W16 × 31.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.