PROBLEM 5.1
w
B
A
L
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:
M B  0:  AL  wL 
L
0
2
A
wL
2
M A  0:
L
0
2
B
wL
2
BL  wL 
Free body diagram for determining reactions:
Over whole beam,
0 x L
Place section at x.
Replace distributed load by equivalent concentrated load.
Fy  0:
wL
 wx  V  0
2
L

V  w  x  
2

M J  0: 
M 
wL
x
x  wx  M  0
2
2
w
( Lx  x 2 )
2
M 
Maximum bending moment occurs at x 
w
x ( L  x) 
2
L
.
2
M max 
wL2

8
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681
PROBLEM 5.2
P
A
B
C
a
b
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.
L
SOLUTION
Reactions:

M C  0: LA  bP  0
A
Pb
L
M A  0: LC  aP  0
C 
Pa
L
0 xa
From A to B,
Fy  0:
Pb
V  0
L

Pb

L
V 
M J  0: M 
Pb
x0
L
M 
Pbx

L
a x L
From B to C,
Fy  0: V 
Pa
0
L
V 
M K  0:  M 
Pa
( L  x)  0
L
M 
Pa( L  x)

L
M 
At section B,
Pa

L
Pab

L2
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682
PROBLEM 5.3
w0
A
B
L
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bendingmoment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy  0: RA 
w0 L
0
2
RA 
w0 L
2
 w L  2L 
M A  0: M A   0 
0
 2  3 
MA  
w0 L2
w L2
 0
3
3
Use portion to left of the section as the free body.
Replace distributed load with equivalent concentrated load.
Fy  0:
w0 L 1 w0 x

 x V  0
2
2 L
V 
w0 L w0 x 2


2
2L
M J  0:
w0 L2  w0 L 
 1 w0 x  x 

 x    M  0
 ( x)  
3
 2 
2 L
 3 
M 
w0 L2 w0 Lx w0 x3



3
2
6L
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683
PROBLEM 5.4
w
B
A
L
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bendingmoment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy  0: RA  wL  0
RA  wL
L
M A  0: M A  (wL)    0
2
MA 
w0 L2
2
Use portion to the right of the section as the free body.
Replace distributed load by equivalent concentrated load.
Fy  0: V  w( L  x)  0
V  w( L  x) 
L  x
M J  0:  M  w( L  x) 
0
 2 
M 
w
( L  x) 2 
2
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684
P
PROBLEM 5.5
P
B
C
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.
A
a
a
SOLUTION
0 xa
From A to B:
Fy  0 :
 P V  0
V   P 

M J  0 :
Px  M  0
M   Px 
a  x  2a
From B to C:
Fy  0 :
 P  P V  0
V   2 P 
M J  0 :
Px  P( x  a)  M  0
M  2 Px  Pa 

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685
w
B
A
PROBLEM 5.6
w
C
a
D
a
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.
L
SOLUTION
Reactions:
A  D  wa
From A to B,
0 xa
Fy  0:
wa  wx  V  0
V  w(a  x) 
M J  0: wax  (wx)
x
M 0
2

x2 
M  w  ax 
 
2 

a x La
From B to C,
Fy  0:
wa  wa  V  0
V 0 
a

wax  wa  x    M  0
2

M J  0:
From C to D,
Fy  0:
M 
1 2
wa 
2
La x L
V  w( L  x)  wa  0
V  w( L  x  a) 
L  x
 M  w( L  x) 
  wa( L  x)  0
 2 
M J  0:
1


M  w  a ( L  x)  ( L  x ) 2  
2


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686
3 kN
A
C
0.3 m
5 kN
2 kN
PROBLEM 5.7
E
B
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.
2 kN
D
0.3 m
0.3 m
0.4 m
SOLUTION
Origin at A:

Reaction at A:
Fy  0: RA  3  2  5  2  0
RA  2 kN
M A  0: M A  (3 kN)(0.3 m)  (2 kN)(0.6 m)  (5 kN)(0.9 m)  (2 kN)(1.3 m)  0
M A  0.2 kN  m


From A to C:
Fy  0:
V  2 kN
M1  0:
0.2 kN  m  (2 kN)x  M  0
M  0.2  2 x
From C to D:
Fy  0: 2  3  V  0
V  1 kN
M 2  0:
 0.2 kN  m  (2 kN)x  (3 kN)( x  0.3)  M  0
M  0.7  x
From D to E:
Fy  0: V  5  2  0
M 3  0:
V  3 kN
 M  5(0.9  x)  (2)(1.3  x)  0
M  1.9  3x
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687
PROBLEM 5.7 (Continued)
From E to B:
Fy  0: V  2 kN
M 4  0:
 M  2(1.3  x)  0
M  2.6  2 x


(a)
(b) M
V
max
max
 3.00 kN 
 0.800 kN  m 
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688
100 lb
250 lb
C
100 lb
D
E
B
A
15 in.
20 in.
25 in.
PROBLEM 5.8
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.
10 in.
SOLUTION
Reactions:
M C  0:
RE (45 in.)  100 lb(15 in.)  250 lb(20 in.)  100 lb(55 in.)  0
RE  200 lb
Fy  0:
RC  200 lb  100 lb  250 lb  100 lb  0
RC  250 lb
At any point, V is the sum of the loads and reactions to the left (assuming + ) and M the sum of their moments
about that point (assuming ).
(a) Vmax  150.0 lb 
(b) M max  1500 lb  in. 

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689
PROBLEM 5.8 (Continued)
Detailed computations of moments:
MA  0
M C  (100 lb)(15 in.)  1500 lb  in.
M D  (100 lb)(35 in.)  (250 lb)(20 in.)  1500 lb  in.
M E  (100 lb)(60 in.)  (250 lb)(45 in.)  (250 lb)(25 in.)  1000 lb  in.
M B  (100 lb)(70 in.)  (250 lb)(55 in.)  (250 lb)(35 in.)  (200 lb)(10 in.)  0
(Checks)
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690
PROBLEM 5.9
25 kN/m
C
D
B
A
40 kN
0.6 m
40 kN
1.8 m
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.
0.6 m
SOLUTION
The distributed load is replaced with an equivalent concentrated load of 45 kN to compute the reactions.
(25 kN/m)(1.8 m)  45 kN
M A  0:  (40 kN)(0.6 m)  45 kN(1.5 m)  40 kN(2.4 m)  RB (3.0 m)  0
RB  62.5 kN
Fy  0:
RA  62.5 kN  40 kN  45 kN  40 kN  0
RA  62.5 kN
At C:
Fy  0: V  62.5 kN
M1  0:
M  (62.5 kN)(0.6 m)  37.5kN  m
At centerline of the beam:
Fy  0:
62.5 kN  40 kN  (25 kN/m)(0.9 m)  V  0
V 0
M 2  0:
M  (62.5 kN)(1.5 m)  (40 kN)(0.9 m)  (25 kN/m)(0.9 m)(0.45 m)  0
M  47.625 kN  m
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691
PROBLEM 5.9 (Continued)
Shear and bending-moment diagrams:
(a)
(b)
M
V
max
max
 62.5 kN 
 47.6 kN  m 
From A to C and D to B, V is uniform; therefore M is linear.
From C to D, V is linear; therefore M is parabolic.
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692
2.5 kips/ft
PROBLEM 5.10
15 kips
C
D
B
A
6 ft
3 ft
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.
6 ft
SOLUTION
M B  0: 15RA  (12)(6)(2.5)  (6)(15)  0
RA  18 kips
M A  0: 15RB  (3)(6)(2.5)  (9)(15)  0
RB  12 kips
Shear:
VA  18 kips
VC  18  (6)(2.5)  3 kips
C to D :
V  3 kips
D to B :
V  3  15  12 kips
Areas under shear diagram:
A to C :
 
 V dx   2  (6)(18  3)  63 kip  ft
 
C to D :
 V dx  (3)(3)  9 kip  ft
D to B :
 V dx  (6)(12)  72 kip  ft
1
MA  0
Bending moments:
M C  0  63  63 kip  ft
M D  63  9  72 kip  ft
M B  72  72  0
V


M
max
max
 18.00 kips 
 72.0 kip  ft 
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693
3 kN
3 kN
PROBLEM 5.11
E
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.
450 N ? m
A
C
D
300 mm
B
300 mm
200 mm
SOLUTION
M B  0: (700)(3)  450  (300)(3)  1000 A  0
A  2.55 kN
M A  0:  (300)(3)  450  (700)(3)  1000B  0
B  3.45 kN
At A:
V  2.55 kN
A to C:
V  2.55 kN
M 0
M C  0:
At C:
(300)(2.55)  M  0
M  765 N  m
C to E:
V  0.45 N  m
M D  0:
At D:
(500)(2.55)  (200)(3)  M  0
M  675 N  m
M D  0:
At D:
(500)(2.55)  (200)(3)  450  M  0
M  1125 N  m
E to B:
V  3.45 kN
M E  0:
At E:
M  (300)(3.45)  0
M  1035 N  m
At B:
V  3.45 kN,
M 0
(a)
(b)
M
V
max
max
 3.45 kN 
 1125 N  m 
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694
400 lb
1600 lb
PROBLEM 5.12
400 lb
G
D
E
8 in.
F
A
B
8 in.
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.
C
12 in.
12 in.
12 in.
12 in.
SOLUTION
M G  0:  16C  (36)(400)  (12)(1600)
 (12)(400)  0
C  1800 lb
Fx  0:  C  Gx  0
Gx  1800 lb
Fy  0: 400  1600  G y  400  0
G y  2400 lb
A to E:
V  400 lb
E to F:
V  2000 lb
F to B:
V  400 lb
At A and B,
M 0
At D ,
M D  0: (12)(400)  M  0
At D +,
M D  0: (12)(400)  (8)(1800)  M  0
M  9600 lb  in.
At E,
M E  0: (24)(400)  (8)(1800)  M  0
M  4800 lb  in.

M  4800 lb  in.
At F,
M F  0: M  (8)(1800)  (12)(400)  0 M  19, 200 lb  in.
At F ,+
M F  0: M  (12)(400)  0
(a)
(b)
M  4800 lb  in.
Maximum |V |  2000 lb 
Maximum |M |  19, 200 lb  in. 
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695
1.5 kN
1.5 kN
C
D
A
PROBLEM 5.13
B
0.3 m
0.9 m
0.3 m
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,
Fy  0: 1.5w  1.5  1.5  0
A to C:
w  2 kN/m
0  x  0.3 m
Fy  0: 2 x  V  0
V  (2 x) kN
x
M J  0: (2 x)    M  0
2
At C ,
M  ( x 2 ) kN  m
x  0.3 m
V  0.6 kN, M  0.090 kN  m
 90 N  m
C to D:
0.3 m  x  1.2 m
Fy  0: 2 x  1.5  V  0
V  (2 x  1.5) kN
 x
M J  0:  (2 x)    (1.5)( x  0.3)  M  0
2
M  ( x 2  1.5x  0.45) kN  m
At the center of the beam, x  0.75 m
V 0
M  0.1125 kN  m
 112.5 N  m
At C +,
x  0.3 m,
V  0.9 kN
(a) Maximum |V |  0.9 kN  900 N 
(b)
Maximum |M |  112.5 N  m 
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696
24 kips
2 kips/ft
C
A
3 ft
D
3 ft
PROBLEM 5.14
2 kips/ft
E
3 ft
B
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.
3 ft
SOLUTION
Over the whole beam,
Fy  0: 12w  (3)(2)  24  (3)(2)  0
A to C:
w  3 kips/ft
(0  x  3 ft)
Fy  0: 3x  2 x  V  0
M 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)
Fy  0: 3x  (2)(3)  V  0
V  (3x  6) kips
3
x

MK  0: (3x)    (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 
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697
10 kN
PROBLEM 5.15
100 mm
3 kN/m
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
C
A
B
1.5 m
1.5 m
200 mm
2.2 m
SOLUTION
Using CB as a free body,
M C  0:  M  (2.2)(3  103 )(1.1)  0
M  7.26  103 N  m
Section modulus for rectangle:
S 

1 2
bh
6
1
(100)(200)2  666.7  103 mm3
6
 666.7  106 m3
Normal stress:
 
M
7.26  103

 10.8895  106 Pa
S
666.7  106
  10.89 MPa 
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698
PROBLEM 5.16
750 lb
750 lb
150 lb/ft
A
C
4 ft
B
D
4 ft
3 in.
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
12 in.
4 ft
SOLUTION
C  A by symmetry.
Reactions:
Fy  0:
A  C  (2)(750)  (12)(150)  0
A  C  1650 lb
Use left half of beam as free body.
M E  0:
(1650)(6)  (750)(2)  (150)(6)(3)  M  0
M  5700 lb  ft  68.4  103 lb  in.
Section modulus:
S 
1 2 1
bh    (3)(12)2  72 in 3
6
6
Normal stress:
 
M
68.4  103

 950 psi
S
72
  950 psi 
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699
PROBLEM 5.17
150 kN 150 kN
90 kN/m
C
D
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
E
A
B
W460 ⫻ 113
2.4 m
0.8 m
0.8 m
0.8 m
SOLUTION
Use entire beam as free body.
M B  0:
4.8 A  (3.6)(216)  (1.6)(150)  (0.8)(150)  0
A  237 kN
Use portion AC as free body.
M C  0:
M  (2.4)(237)  (1.2)(216)  0
M  309.6 kN  m
For W460  113, S  2390  106 mm3
Normal stress:
 
M
309.6  103 N  m

S
2390  106 m3
 129.5  106 Pa
  129.5 MPa 
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700
PROBLEM 5.18
30 kN 50 kN 50 kN 30 kN
W310 3 52
a
B
A
a
For the beam and loading shown, determine the maximum normal
stress due to bending on section a-a.
2m
5 @ 0.8 m 5 4 m
SOLUTION
Reactions:
A B
By symmetry,
Fy  0 :
A  B  80 kN
Using left half of beam as free body,
M J  0:
(80)(2)  (30)(1.2)  (50)(0.4)  M  0
M  104 kN  m  104  103 N  m
For
W310  52, S  747  103 mm3
 747  106 m3
Normal stress:
 
M
104  103

 139.2  106 Pa
S
747  106
  139.2 MPa 
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701
PROBLEM 5.19
8 kN
3 kN/m
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
C
A
B
W310 ⫻ 60
1.5 m
2.1 m
SOLUTION
Use portion CB as free body.
M C  0:
 M  (3)(2.1)(1.05)  (8)(2.1)  0
M  23.415 kN  m  23.415  103 N  m
For W310  60, S  844  103 mm3
 844  106 m3
Normal stress:  
M
23.415  103

 27.7  106 Pa
S
844  106
  27.7 MPa 
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702
PROBLEM 5.20
5 5 2
2 2
kips kips kips kips kips
C
D
E
F
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
G
B
A
S8 3 18.4
6 @ 15 in. 5 90 in.
SOLUTION
Use entire beam as free body.
 M B  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:
 
M
142.5

S
14.4
  9.90 ksi 
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703
25 kips
25 kips
25 kips
C
D
E
A
PROBLEM 5.21
B
S12 ⫻ 35
1 ft 2 ft
6 ft
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
2 ft
SOLUTION
 M B  0:
(11)(25)  10C  (8)(25)  (2)(25)  0
C  52.5 kips
 M C  0:
(1)(25)  (2)(25)  (8)(25)  10B  0 B  22.5 kips
Shear:
A to C  :
V  25 kips
C to D:
V  27.5 kips
D to E:
V  2.5 kips
E to B:
V  22.5 kips
Bending moments:
At C,
 M C  0: (1)(25)  M  0
M  25 kip  ft
At D,
 M D  0: (3)(25)  (2)(52.5)  M  0
M  30 kip  ft
At E,
 M E  0:
 M  (2)(22.5)  0 M  45 kip  ft
max M  45 kip  ft  540 kip  in.
For S12  35 rolled steel section,
Normal stress:
 
S  38.1 in 3
M
540

 14.17 ksi
S
38.1
  14.17 ksi 
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704
PROBLEM 5.22
160 kN
80 kN/m
B
C
D
A
E
W310 ⫻ 60
Hinge
2.4 m
1.5 m
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
1.5 m
0.6 m
SOLUTION
Statics: Consider portion AB and BE separately.
Portion BE:
 M E  0:
(96)(3.6)  (48)(3.3)  C (3)  (160)(1.5)  0
C  248kN 
E  56 kN 
MA  MB  ME  0
At midpoint of AB:
 Fy  0:
V 0
 M  0:
M  (96)(1.2)  (96)(0.6)  57.6 kN  m
Just to the left of C:
 Fy  0:
V  96  48  144 kN
 M C  0: M  (96)(0.6)  (48)(0.3)  72 kN
Just to the left of D:
 Fy  0:
 M D  0:
V  160  56  104 kN
M  (56)(1.5)  84 kN  m

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705
PROBLEM 5.22 (Continued)

From the diagram,
M

max
 84 kN  m  84  103 N  m 
For W310  60 rolled-steel shape,
S x  844  103 mm3
 844  106 m3
Stress:  m 
m 
M
max
S
84  103
 99.5  106 Pa
844  106
 m  99.5 MPa 
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706
300 N
B
300 N
C
D
40 N
E
300 N
F
G
30 mm
H
A
PROBLEM 5.23
20 mm
Hinge
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
7 @ 200 mm ⫽ 1400 mm
SOLUTION

Free body EFGH. Note that M E  0 due to hinge.

M E  0: 0.6 H  (0.2)(40)  (0.40)(300)  0

H  213.33 N

Fy  0: VE  40  300  213.33  0





VE  126.67 N
Shear:
E to F :
V  126.67 N  m
F to G :
V  86.67 N  m
G to H :
V  213.33 N  m
Bending moment at F:
M F  0: M F  (0.2)(126.67)  0
M F  25.33 N  m
Bending moment at G:
M G  0: M G  (0.2)(213.33)  0
M G  42.67 N  m
Free body ABCDE.
M B  0: 0.6 A  (0.4)(300)  (0.2)(300)
 (0.2)(126.63)  0
A  257.78 N
M A  0: (0.2)(300)  (0.4)(300)  (0.8)(126.67)  0.6D  0
D  468.89 N
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707
PROBLEM 5.23 (Continued)
Bending moment at B.
max M  51.56 N  m
 M B  0:  (0.2)(257.78)  M B  0
M B  51.56 N  m
S 

1 2 1
bh  (20)(30) 2
6
6
 3  103 mm3  3  106 m3
Bending moment at C.
Normal stress:
 M C  0:  (0.4)(257.78)  (0.2)(300)
 MC  0
 
51.56
 17.19  106 Pa
3  106
  17.19 MPa 
M C  43.11 N  m
V
Bending moment at D.
M
 M D  0:  M D  (0.2)(213.33)  0
max
max
 342 N 
 516 N  m 
M D  25.33 N  m
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708
64 kN ? m
C
PROBLEM 5.24
24 kN/m
D
A
B
S250 ⫻ 52
2m
2m
SOLUTION
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
2m
Reactions:
 M D  0: 4 A  64  (24)(2)(1)  0
A  28 kN
 Fy  0:  28  D  (24)(2)  0 D  76 kN
A to C:
0  x  2m
 Fy  0:  V  28  0
V  28 kN
 M J  0: M  28 x  0
M  (28 x) kN  m
C to D:
2m  x  4m
 Fy  0:  V  28  0
V  28 kN
 M J  0:
M  28 x  64  0
M  (28 x  64) kN  m
D to B:
4m  x  6m
 Fy  0:
V  24(6  x)  0
V  (24 x  144) kN
 M J  0:
6  x
M  24(6  x) 
0
 2 
M  12(6  x)2 kN  m
max M  56 kN  m  56  103 N  m
S  482  103 mm3
For S250  52 section,
Normal stress:  
M
S

56  103 N  m
482  10 6 m 3
 116.2  106 Pa
  116.2 MPa 
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709
5 kips
PROBLEM 5.25
10 kips
C
D
A
B
W14 ⫻ 22
5 ft
8 ft
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
5 ft
SOLUTION
Reaction at C:
 M B  0: (18)(5)  13C +(5)(10)  0
C  10.769 kips
Reaction at B:
M C  0: (5)(5)  (8)(10)  13B  0
B  4.231 kips
Shear diagram:
A to C :
V  5 kips
C  to D :
V  5  10.769  5.769 kips

D to B :
V  5.769  10  4.231 kips
At A and B,
M 0
At C,
M C  0: (5)(5)  M C  0
M C  25 kip  ft
At D,
M D  0: M D  (5)(4.231)
M D  21.155 kip  ft
V
max
 5.77 kips 
|M |max  25 kip  ft  300 kip  in. 
|M |max occurs at C.
For W14  22 rolled-steel section,
S  29.0 in 3
Normal stress:
 
M
300

S
29.0
  10.34 ksi 
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710
PROBLEM 5.26
Knowing that W  12 kN , draw the shear and bending-moment
diagrams for beam AB and determine the maximum normal
stress due to bending.
W
8 kN
C
8 kN
D
W310 ⫻ 23.8
E
B
A
1m
1m
1m
1m
SOLUTION
By symmetry, A  B
 Fy  0: A  8  12  8  B  0
A  B  2 kN
Shear:
A to C :
V  2 kN


C  to D :
V  6 kN 


D  to E :
V  6 kN 


E  to B :
V  2 kN 

V
Bending moment:
max
 6.00 kN 
 M C  0: M C  (1)(2)  0
At C,
M C  2 kN  m
At D,
M D  0: M D  (2)(2)  (8)(1)  0
By symmetry,
M  2 kN  m at D.
M D  4 kN  m 
M E  2 kN  m
max|M |  4.00 kN  m occurs at E.
For W310  23.8,
Normal stress:

S x  280  103 mm3  280  106 m3
 max 
|M |max
4  103

Sx
280  106
 14.29  106 Pa
 max  14.29 MPa 
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711
PROBLEM 5.27
W
8 kN
C
8 kN
D
W310 ⫻ 23.8
E
Determine (a) the magnitude of the counterweight W for which
the maximum absolute value of the bending moment in the beam
is as small as possible, (b) the corresponding maximum normal
stress due to bending. (Hint: Draw the bending-moment diagram
and equate the absolute values of the largest positive and
negative bending moments obtained.)
B
A
1m
1m
1m
1m
SOLUTION
By symmetry,
AB
 Fy  0: A  8  W  8  B  0
A  B  8  0.5W
Bending moment at C:
 M C  0: (8  0.5W )(1)  M C  0
M C  (8  0.5W ) kN  m
Bending moment at D:
M D  0:  (8  0.5W )(2)  (8)(1)  M D  0
M D  (8  W ) kN  m
M D  M C
Equate:
W  8  8  0.5W
W  10.67 kN 
(a)
W  10.6667 kN
M C  2.6667 kN  m
M D  2.6667 kN  m  2.6667.103 N  m
|M |max  2.6667 kN  m
For W310  23.8 rolled-steel shape,
S x  280  103 mm3  280  106 m3
(b)
 max 
|M |max
2.6667  103

 9.52  106 Pa
Sx
280  106
 max  9.52 MPa 
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712
5 kips
PROBLEM 5.28
10 kips
C
Determine (a) the distance a for which the maximum absolute
value of the bending moment in the beam is as small as
possible, (b) the corresponding maximum normal stress due to
bending. (See hint of Prob. 5.27.)
D
A
B
W14 ⫻ 22
a
8 ft
5 ft
SOLUTION
Reaction at B:
 M C  0: 5a  (8)(10)  13RB  0
RB 
1
(80  5a)
18
Bending moment at D:
 M D  0: M D  5RB  0
M D  5RB 
5
(80  5a)
13
Bending moment at C:
M C  0 5a  M C  0
M C  5a
Equate:
M C  M D
5a 
5
(80  5a)
13
(a) a  4.44 ft 
a  4.4444 ft
Then
M C  M D  (5)(4.4444)  22.222 kip  ft
|M |max  22.222 kip  ft  266.67 kip  in.
For W14  22 rolled-steel section, S  29.0 in 3
Normal stress:
 
M
266.67

 9.20 ksi
S
29.0
(b) 9.20 ksi 
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713
P
500 mm
Q
C
D
A
PROBLEM 5.29
12 mm
500 mm
18 mm
B
a
Knowing that P  Q  480 N, determine (a) the distance a for
which the absolute value of the bending moment in the beam is
as small as possible, (b) the corresponding maximum normal
stress due to bending. (See hint of Prob. 5.27.)
SOLUTION
P  480 N
Q  480 N
 M D  0:  Aa  480(a  0.5)
Reaction at A:
 480(1  a)  0
720 

A   960 
N
a 

Bending moment at C:
 M C  0:  0.5A  M C  0
360 

M C  0.5A   480 
Nm
a 

Bending moment at D:
 M D  0:  M D  480(1  a)  0
M D  480(1  a) N  m
(a)
M D  M C
Equate:
480(1  a)  480 
360
a
a  0.86603 m
A  128.62 N
(b)
For rectangular section, S 
S 
 max 
M C  64.31 N  m
a  866 mm 
M D  64.31 N  m
1 2
bh
6
1
(12)(13) 2  648 mm3  648  109 m3
6
|M |max
64.31

 99.2  106 Pa
S
6.48  109
 max  99.2 MPa 
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714
PROBLEM 5.30
P
500 mm
Q
12 mm
500 mm
C
D
A
Solve Prob. 5.29, assuming that P  480 N and Q  320 N.
18 mm
B
a
PROBLEM 5.29 Knowing that P  Q  480 N, determine
(a) the distance a for which the absolute value of the bending
moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See
hint of Prob. 5.27.)
SOLUTION
P  480 N
Reaction at A:
Q  320 N
 M D  0: Aa  480(a  0.5)  320(1  a)  0
560 

A   800 
N
a 

Bending moment at C:
 M C  0: 0.5A  M C  0
280 

M C  0.5A   400 
 Nm
a 

Bending moment at D:
 M D  0: M D  320 (1  a)  0
M D  (320  320a) N  m
(a)
M D  M C
Equate:
280
a
a  0.81873 m, 1.06873 m
320  320a  400 
320a 2  80a  280  0
a  819 mm 
Reject negative root.
A  116.014 N
(b)
M C  58.007 N  m
M D  58.006 N  m
1 2
bh
6
1
S  (12)(18) 2  648 mm3  648  109 m3
6
For rectangular section, S 
 max 
|M |max
58.0065

 89.5  106 Pa
9
S
648  10
 max  89.5 MPa 
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715
PROBLEM 5.31
4 kips/ft
B
C
A
a
W14 ⫻ 68
Hinge
18 ft
Determine (a) the distance a for which the absolute value
of the bending moment in the beam is as small as possible,
(b) the corresponding maximum normal stress due to bending.
(See hint of Prob. 5.27.)
SOLUTION
S x  103 in 3
For W14  68,
Let b  (18  a) ft
Segment BC:
By symmetry,
VB  C
 Fy  0: VB  C  4b  0
VB  2b
 x
 M J  0: VB x  (4 x)    M  0
2
M  VB x  2 x 2  2bx  2 x 2 lb  ft
dM
1
 2b  xm  0
xm  b
dx
2
1
1
M max  b 2  b 2  b 2
2
2
Segment AB:
(a  x )
2
VB (a  x)  M  0
 M K  0:  4(a  x)
M  2(a  x)2  2b (a  x)
|M max | occurs at x  0.
|M max |  2a 2  2ab  2a 2  2a(18  a)  36a
(a)
Equate the two values of |M max |:
36a 
1 2 1
1
b  (18  a) 2  162  18a  a 2
2
2
2
1 2
a  54a  162  0
a  54  (54) 2  (4)
2
a  54  50.9118  3.0883 ft
(b)
 12 (162)
a  3.09 ft 
|M |max  36a  111.179 kip  ft  1334.15 kip  in.
 
|M |max 1334.15

 12.95 kips/in 2
Sx
103
 m  12.95 ksi 
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716
PROBLEM 5.32
d
A
B
A solid steel rod of diameter d is supported as shown. Knowing that for
steel   490 lb/ft 3 , determine the smallest diameter d that can be
used if the normal stress due to bending is not to exceed 4 ksi.
L ⫽ 10 ft
SOLUTION
Let W  total weight.
W  AL 

4
d 2 L
Reaction at A:
A
1
W
2
Bending moment at center of beam:
 W  L   W  L 
 M C  0:          M  0
 2  2   2  4 
 2 2
WL

M 
d L
8
32

For circular cross section, c  1 d
2

I 

4
c4 ,
S 
I

 3
 c3 
d
c
4
32
Normal stress:
 
Solving for d,
Data:
d 
M

S

32
d 2 L2

32
d
3

L2
d
L2

L  10 ft  (12)(10)  120 in.
  490 lb/ft 3 
490
 0.28356 lb/in 3
123
  4 ksi  4000 lb/in 2
d 
(120)2 (0.28356)
4000
d  1.021 in. 
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717
PROBLEM 5.33
b
A
C
D
B
A solid steel bar has a square cross section of side b and is
supported as shown. Knowing that for steel   7860 kg / m3 ,
determine the dimension b for which the maximum normal stress
due to bending is (a) 10 MPa, (b) 50 MPa.
b
1.2 m
1.2 m
1.2 m
SOLUTION
Weight density:    g
Let L  total length of beam.
W  AL  g  b 2 L  g
Reactions at C and D:
C  D
W
2
Bending moment at C:
 L  W 
 M C  0:     M  0
 6  3 
WL
M 
18
Bending moment at center of beam:
 L  W   L  W 
 M E  0:         M  0
 4  2   6  2 
max|M | 
S 
For a square section,
Normal stress:
Solve for b:
Data:
 
M 
WL
24
WL b 2 L2  g

18
18
1 3
b
6
|M | b 2 L2  g /18 L2  g


S
3b
b3 /6
b
L  3.6 m   7860 kg/m3
L2  g
3
g  9.81 m/s 2
(a)   10  106 Pa
(b)   50  106 Pa
(a)
b
(3.6) 2 (7860)(9.81)
 33.3  103 m
(3)(10  106 )
b  33.3 mm 
(b)
b
(3.6) 2 (7860)(9.81)
 6.66  103 m
6
(3)(50  10 )
b  6.66 mm 
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718
PROBLEM 5.34
w
B
A
L
Using the method of Sec. 5.2, solve Prob. 5.1a.
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
 M B  0: AL  wL 
 M A  0: BL  wL 
L
0
2
L
0
2
A
wL
2
B
wL
2
dV
 w
dx
x
V  VA  0 w dx   wx
V  VA  wx  A  wx
V 
wL
 wx 
2
dM
V
dx
x
x  wL

M  M A  0 V dx  0 
 wx  dx
2



wLx wx 2

2
2
M  MA 
Maximum M occurs at x 
V 
wLx wx 2

2
2
M 
w
( Lx  x 2 ) 
2
1
, where
2
dM
0
dx
|M |max 
wL2

8
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719
PROBLEM 5.35
P
A
B
C
a
Using the method of Sec. 5.2, solve Prob. 5.2a.
PROBLEM 5.2 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.
b
L
SOLUTION
At A,
M C  0: LA  bP  0
A
Pb
L
M A  0: LC  aP  0
C 
Pa
L
V  A
A to B:
Pb
L
M 0
0 xa
x
0 w dx  0
w0
V  VA  0
a Pb
a
M B  M A   0 V dx  0
At B +,
V  AP 
B to C:
Pb

L
V 
L
dx 
Pba
L
MB 
Pba

L
Pb
Pa
P
L
L
a x L
w0
x
a w dx  0
VC  VB  0
MC  M B 
V 
Pa
Pa

L
Pab
L
a V dx   L ( L  a)   L
MC  M B 
Pab
Pba Pab


0
L
L
L
|M | max 
Pab

L
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720
PROBLEM 5.36
w0
Using the method of Sec. 5.2, solve Prob. 5.3a.
A
B
L
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 bending-moment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy  0 : VA 
w0 L
0
2
VA 
w0 L
2
 w L  2 L 
M A  0 :  M A   0 
0
 2  3 
MA  
w  w0
w0 L2
3
x
wL
w L2
, VA  0 , M A   0
2
3
L
dV
wx
 w   0
dx
L
x
V  VA    0
w0 x
w x2
dx   0
2L
L
V 
w0 L w0 x 2


2
2L
dM
w L w x2
V  o  o
2
2L
dx
w x2 
x
x w L
M  M A   0 V dx   0  0  0  dx
2L 
 2

w0 L
w x3
x 0
2
6L
M 
w0 L2 w0 L
w x3

x 0 
3
2
6L
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721
PROBLEM 5.37
w
Using the method of Sec. 5.2, solve Prob. 5.4a.
B
A
L
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
Fy  0: VA  wL  0
VA  wL
L
M A  0:  M  (wL)    0
2
MA  
wL2
2
dV
 w
dx
x
V  VA   0 w dx   wx
V  wL  wx 
dM
 V  wL  wx
dx
x
M  M A   0 (wL  wx)dx  wLx 
wx 2
2
M 
V
M
max
 wL
max

wL2
wx 2
 wLx 

2
2
wL2
2
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722
P
PROBLEM 5.38
P
B
C
A
a
a
Using the method of Sec. 5.2, solve Prob. 5.5a.
PROBLEM 5.5 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
At A+:
VA   P
Over AB:
dV
 w  0
dx
dM
 V  VA   P 
dx
M   Px  C
M  0 at x  0
C1  0
M   Px 
At point B:
xa
M   Pa
At point B+:
V   P  P  2P
Over BC:
dV
 w  0
dx
dM
 V  2 P 
dx
M  2 Px  C2
xa
At B:
M   Pa
 Pa  2 Pa  C2
C2  Pa
M  2 Px  Pa 
x  2a
At C:
M  3Pa 
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723
w
B
A
PROBLEM 5.39
w
C
a
D
a
L
Using the method of Sec. 5.2, solve Prob. 5.6a.
PROBLEM 5.6 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:
A  D  wa
A to B:
0 xa
ww
VA  A  wa,
MA  0
x
V  VA   0 w dx  wx
V  w(a  x) 
dM
 V  wa  wx
dx
M  MA 
x
x
 0 V dx   0 (wa  wx)dx
M  wax 
VB  0
B to C:
MB 
1 2
wx 
2
1 2
wa
2
a x La
V 0 
dM
V 0
dx
x
M  M B  a V dx  0
M  MB
M 
1 2
wa 
2






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724
PROBLEM 5.39 (Continued)
La x L
C to D:
x
V  VC  L  a w dx  w[ x  ( L  a)]
V  w[ L  x  a)] 
x
x
M  M C  L  a V dx  L  a w [x  ( L  a)]dx
 x2

  w   ( L  a) x 
2

x
La
 x2

( L  a) 2
  w   ( L  a) x 
 ( L  a) 2 
2
2

 x2
( L  a) 2 
  w   ( L  a) x 

2
2

M 
 x2
1 2
( L  a) 2 
wa  w   ( L  a) x 

2
2
2


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725
3 kN
A
2 kN
C
0.3 m
D
0.3 m
5 kN
2 kN
E
B
0.3 m
0.4 m
PROBLEM 5.40
Using the method of Sec. 5.2, solve Prob. 5.7.
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
Free body diagram for determining reactions.
Reactions:
Fy  0: VA  3 kN  2 kN  5 kN  2 kN  0
VA  2 kN
M A  0: M A  (3 kN)(0.3 m)  (2 kN)(0.6 m)  (5 kN)(0.9 m)  (2 kN)(1.3 m)  0
M A  0.2 kN  m
Between concentrated loads and the vertical reaction,
the scope of the shear diagram is  , i.e., the shear is
constant. Thus, the area under the shear diagram is equal
to the change in bending moment.
A to C:
V  2 kN M C  M A   0.6 M C   0.4 kN
C to D:
V   1 kN M D  M C   0.3 M D   0.1 kN  m
D to E:
V  3 kN M E  M D   0.9 M E   0.8 kN  m
E to B:
V  2 kN M B  M E   0.8 M B  0 (Checks)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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726
100 lb
250 lb
C
100 lb
D
PROBLEM 5.41
Using the method of Sec. 5.2, solve Prob. 5.8.
E
B
A
15 in.
20 in.
25 in.
10 in.
PROBLEM 5.8 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
Free body diagram for determining reactions.
Reactions:
FY  0 : VC  VE  100 lb  250 lb  100 lb  0
VC  VE  450 lb 
M C  0 : VE (45 in.)  (100 lb)(15 in.)  (250 lb)(20 in.)  (100 lb)(55 in.)  0
VE  200 lb
 VC  250 lb
Between concentrated loads and the vertical reaction, the
scope of the shear diagram is  , i.e., the shear is constant.
Thus, the area under the shear diagram is equal to he change
in bending moment.
A to C:
V  100 lb, M C  M A  1500, M C  1500 lb  in.
C to D:
V  150 lb M D  M C  3000, M D  1500 lb  in.
D to E:
V  100 lb, M E  M D  2500, M E  1000 lb  in.
E to B:
V  100 lb, M B  M E  1000, M B  0 (Checks) 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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727
PROBLEM 5.42
25 kN/m
C
D
B
A
40 kN
0.6 m
PROBLEM 5.9 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.
40 kN
1.8 m
Using the method of Sec. 5.2, solve Prob. 5.9.
0.6 m
SOLUTION
Free body diagram to determine reactions:
M A  0:
VB (3.0 m)  45 kN(1.5 m)  (40 kN)(0.6 m)  (40 kN)(2.4 m)  0
VB  62.5 kN 
Fy  0: VA  40 kN  45 kN  40 kN  62.5 kN  0
VA  62.5 kN
Change in bending moment is equal to area under shear diagram.
A to C:
(62.5 kN)(0.6 m)  37.5 kN  m
C to E:
1
(0.9 m)(22.5 kN)  10.125 kN  m
2
E to D:
1
(0.9 m)( 22.5 kN)  10.125 kN  m
2
D to B:
(62.5 kN)(0.6 m)  37.5 kN  m



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728
2.5 kips/ft
PROBLEM 5.43
15 kips
C
D
B
A
6 ft
3 ft
6 ft
Using the method of Sec. 5.2, solve Prob. 5.10.
PROBLEM 5.10 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 at supports A and B:
M B  0: 15( RA )  (12)(6)(2.5)  (6)(15)  0
RA  18 kips 
M A  0: 15RB  (3)(6)(2.5)  (9)(15)  0
RB  12 kips
Areas under shear diagram:
1
(6)(15)  63 kip  ft
2
A to C:
(6)(3) 
C to D:
(3)(3)  9 kip  ft
D to B:
(6)(12)  72 kip  ft
Bending moments:
MA  0
M C  0  63  63 kip  ft
M D  63  9  72 kip  ft
M B  72  72  0 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
729