CHAPTER 7
PROBLEM 7.1
Determine the internal forces (axial force, shearing force, and
bending moment) at Point J of the structure indicated.
Frame and loading of Problem 6.75.
SOLUTION
From Problem 6.75:
C x = 720 lb
C y = 140 lb
FBD of JC:
ΣFx = 0: F − 720 lb = 0
F = +720 lb
F = 720 lb

ΣFy = 0: V − 140 lb = 0
V = 140 lb
V = 140.0 lb 
Σ M J = 0: M − (140 lb)(8 in.) = 0
M = +1120 lb ⋅ in.
M = 1120 lb ⋅ in.

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1003
PROBLEM 7.2
Determine the internal forces (axial force, shearing force, and bending
moment) at Point J of the structure indicated.
Frame and loading of Problem 6.78.
SOLUTION
From Problem 6.78:
D x = 900 lb
D y = 750 lb
FBD of JD:
ΣM J = 0: − M + (750 lb)(4 ft) − (900 lb)(1.5 ft) = 0
M = +1650 lb ⋅ ft
M = 1650 lb ⋅ ft

ΣF = 0: −V + (750 lb) cos 20.56° − (900 lb)sin 20.56° = 0
V = +386.2 lb
V = 386 lb
69.4° 
ΣF = 0: F − (750 lb)sin 20.56° + (900 lb) cos 20.56° = 0
F = +1106.1 lb
F = 1106 lb
20.6° 
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1004
PROBLEM 7.3
Determine the internal forces at Point J when α = 90°.
SOLUTION
Reactions (α = 90°)
ΣM A = 0: B y = 0
 12 
ΣFy = 0: A   − 780 N = 0
 13 
A = 845 N
ΣFx = 0: (845 N)
A = 845 N
5
+ Bx = 0
13
Bx = −325 N
B x = 325 N
FBD BJ:
ΣF = 0: 125 N − F = 0
F = 125.0 N
67.4° 
ΣF = 0: V − 300 N = 0
V = 300 N
22.6° 
Σ M = 0: (325 N)(0.480 m) − M = 0
M = +156 N ⋅ m
M = 156.0 N ⋅ m

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1005
PROBLEM 7.4
Determine the internal forces at Point J when α = 0.
SOLUTION
Reactions (α = 0)
ΣM A = 0: (780 N)(0.720 m) + By (0.3 m) = 0
By = −1872 N
B y = 1872 N
 12 
ΣFy = 0: A   − 1872 N = 0
 13 
A = 2028 N
 5
ΣFx = 0: (2028 N)   + 780 N + Bx = 0
 13 
Bx = −1560 N
B x = 1560 N
FBD BJ:
ΣF = 0: 1728 N + 600 N − F = 0
F = +2328 N
F = 2330 N
67.4° 
V = 720 N
22.6° 
ΣF = 0: 720 N − 1440 N + V = 0
V = +720 N
BJ = 4802 + 2002 = 520 mm
ΣM J = 0: (1440 N)(0.520 m) − (720 N)(0.520 m) − M = 0
M = +374.4 N ⋅ m
Alternate
M = 374 N ⋅ m

Computation of M using B x + B y :
ΣM J = 0: (1560 N)(0.48 m) − (1872 N)(0.2 m) − M = 0
M = +374.4 N ⋅ m
M = 374 N ⋅ m

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1006
PROBLEM 7.5
Knowing that the turnbuckle has been tightened until the tension in wire AD
is 850 N, determine the internal forces at point indicated:
Point J.
SOLUTION
 160 
ΣFx = 0: − V + 
 (850 N) = 0
 340 
V = +400 N
V = 400 N

 300 
ΣFy = 0: F − 
 (850 N) = 0
 340 
F = +750 N
F = 750 N 
 300 
 160 
ΣM J = 0: M − 
 (850 N)(120 mm) −  340  (850 N)(100 mm) = 0
340




M = +130 N ⋅ m
M = 130 N ⋅ m

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1007
PROBLEM 7.6
Knowing that the turnbuckle has been tightened until the tension in wire AD
is 850 N, determine the internal forces at point indicated:
Point K.
SOLUTION
Free body AK:
AD = 1602 + 3002
= 340 mm
On portion KBA:
 160 
ΣFx = 0: − V + 
 (850 N) = 0
 340 
V = +400 N
V = 400 N

 300 
ΣFy = 0: F − 
 (850 N) = 0
 340 
F = +750 N
F = 750 N 
 300 
 160 
(850 N)(120 mm) − 
ΣM J = 0: M − 

 (850 N)(200 mm) = 0
 340 
 340 
M = +170 N ⋅ m
M = 170.0 N ⋅ m
Internal forces acting on KCD are equal and opposite
F = 750 N , V = 400 N
, M = 170.0 N ⋅ m

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1008
PROBLEM 7.7
Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and
support a 75-lb load at A. Determine the internal forces
at Point J.
SOLUTION
Free body: Entire frame
ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0
F = 100 lb ΣFx = 0: Cx = 0
ΣFy = 0: C y − 75 lb − 100 lb = 0
C = 175 lb C y = + 175 lb
Free body: Member BEDF
ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0
D = 125 lb ΣFx = 0: Bx = 0
ΣFy = 0: By + 125 lb − 100 lb = 0
B = 25 lb By = − 25 lb
Free body: BJ
ΣFx = 0: F − (25 lb)sin 30° = 0
F = 12.50 lb
30.0° 
ΣFy = 0: V − (25 lb) cos 30° = 0
V = 21.7 lb
60.0° 
ΣM J = 0: − M + (25 lb)(3 in.) = 0
M = 75.0 lb ⋅ in.

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1009
PROBLEM 7.8
Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and
support a 75-lb load at A. Determine the internal forces at
Point K.
SOLUTION
Free body: Entire frame
ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0
F = 100 lb ΣFx = 0: Cx = 0
ΣFy = 0: C y − 75 lb − 100 lb = 0
C = 175 lb C y = + 175 lb
Free body: Member BEDF
ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0
D = 125 lb ΣFx = 0: Bx = 0
ΣFy = 0: By + 125 lb − 100 lb = 0
By = − 25 lb
B = 25 lb Free body: DK
We found in Problem 7.11 that
D = 125 lb on BEDF.
D = 125 lb on DK. Thus
ΣFx = 0: F − (125 lb) cos 30° = 0
F = 108.3 lb
60.0° 
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1010
PROBLEM 7.8 (Continued)
ΣFy = 0: V − (125 lb)sin 30° = 0
V = 62.5 lb
30.0° 
ΣM K = 0: M − (125 lb)d = 0
M = (125 lb)d = (125 lb)(0.8038 in.)
= 100.5 lb ⋅ in.
M = 100.5 lb ⋅ in.

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1011
PROBLEM 7.9
A semicircular rod is loaded as shown. Determine the internal forces at Point J.
SOLUTION
FBD Rod:
ΣM B = 0: Ax (2r ) = 0
Ax = 0
ΣFx′ = 0: V − (120 N) cos 60° = 0
V = 60.0 N

F = 103.9 N


FBD AJ:
ΣFy′ = 0: F + (120 N)sin 60° = 0
F = −103.923 N
ΣM J = 0: M − [(0.180 m)sin 60°](120 N) = 0
M = 18.7061
M = 18.71 
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1012
PROBLEM 7.10
A semicircular rod is loaded as shown. Determine the internal forces at Point K.
SOLUTION
FBD Rod:
ΣFy = 0: By − 120 N = 0 B y = 120 N
ΣM A = 0: 2rBx = 0 B x = 0
ΣFx′ = 0: V − (120 N) cos 30° = 0
V = 103.923 N
V = 103.9 N

F = 60.0 N

M = 10.80 N ⋅ m

FBD BK:
ΣFy′ = 0: F + (120 N)sin 30° = 0
F = − 60 N
ΣM K = 0: M − [(0.180 m)sin 30°](120 N) = 0
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1013
PROBLEM 7.11
A semicircular rod is loaded as shown. Determine the internal
forces at Point J knowing that θ = 30°.
SOLUTION
FBD AB:
4 
3 
ΣM A = 0: r  C  + r  C  − 2r (280 N) = 0
5 
5 
C = 400 N
ΣFx = 0: − Ax +
4
(400 N) = 0
5
A x = 320 N
3
ΣFy = 0: Ay + (400 N) − 280 N = 0
5
A y = 40.0 N
FBD AJ:
ΣFx′ = 0: F − (320 N) sin 30° − (40.0 N) cos 30° = 0
F = 194.641 N
F = 194.6 N
60.0° 
ΣFy ′ = 0: V − (320 N) cos 30° + (40 N) sin 30° = 0
V = 257.13 N
V = 257 N
30.0° 
ΣM 0 = 0: (0.160 m)(194.641 N) − (0.160 m)(40.0 N) − M = 0
M = 24.743
M = 24.7 N ⋅ m

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1014
PROBLEM 7.12
A semicircular rod is loaded as shown. Determine the magnitude
and location of the maximum bending moment in the rod.
SOLUTION
Free body: Rod ACB
4

3

ΣM A = 0:  FCD  (0.16 m) +  FCD  (0.16 m)
5

5

− (280 N)(0.32 m) = 0
ΣFx = 0: Ax +
FCD = 400 N
A x = 320 N
4
(400 N) = 0
5
Ax = − 320 N
3
+ ΣFy = 0: Ay + (400 N) − 280 N = 0
5
Ay = + 40.0 N
A y = 40.0 N Free body: AJ (For θ < 90°)
ΣM J = 0: (320 N)(0.16 m) sin θ − (40.0 N)(0.16 m)(1 − cos θ ) − M = 0
M = 51.2sin θ + 6.4 cos θ − 6.4
(1)
For maximum value between A and C:
dM
= 0: 51.2 cos θ − 6.4sin θ = 0
dθ
tan θ =
51.2
=8
6.4
θ = 82.87° 
Carrying into (1):
M = 51.2sin 82.87° + 6.4cos82.87° − 6.4 = + 45.20 N ⋅ m 
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1015
PROBLEM 7.12 (Continued)
Free body: BJ (For θ > 90°)
ΣM J = 0: M − (280 N)(0.16 m)(1 − cos φ ) = 0
M = (44.8 N ⋅ m)(1 − cos φ )
Largest value occurs for φ = 90°, that is, at C, and is
M C = 44.8 N ⋅ m We conclude that
M max = 45.2 N ⋅ m
for
θ = 82.9° 
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1016
PROBLEM 7.13
The axis of the curved member AB is a parabola with vertex at A. If a vertical
load P of magnitude 450 lb is applied at A, determine the internal forces at J
when h = 12 in., L = 40 in., and a = 24 in.
SOLUTION
ΣFy = 0: − 450 lb + By = 0
Free body AB
B y = 450 lb
ΣM A = 0: Bx (12 in.) − (450 lb)(40 in.) = 0
B x = 1500 lb
ΣFx = 0: A = 1500 lb
y = k x2
Parabola:
12 in. = k (40 in.) 2
At B:
Equation of parabola:
y = 0.0075 x 2
slope =
At J:
k = 0.0075
dy
= 0.015 x
dx
xJ = 24 in.
y J = 0.0075(24) 2 = 4.32 in.
slope = 0.015(24) = 0.36, tan θ = 0.36, θ = 19.8°
Free body AJ
ΣM J = 0: (450 lb)(24 in.) − (1500 lb)(4.32 in.) − M = 0
M = 4320 lb ⋅ in.
M = 4320 lb ⋅ in.

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1017
PROBLEM 7.13 (Continued)
ΣF = 0: F −(450 lb) sin19.8° − (1500 lb) cos19.8° = 0
F = +1563.8 lb
F = 1564 lb
19.8° 
V = 84.7 lb
70.2° 
ΣF = 0: −V − (450 lb) cos19.8° + (1500 lb)sin19.8° = 0
V = +84.71 lb
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1018
PROBLEM 7.14
Knowing that the axis of the curved member AB is a parabola with vertex
at A, determine the magnitude and location of the maximum bending
moment.
SOLUTION
Parabola
y = kx 2
At B:
h = kL2
k = h /L2
Equation of parabola
y = hx 2/L2
Σ M B = 0: P( L) − A(h) = 0
Free body AJ
At J: xJ = a
A = PL /h
y J = ha 2/L2
Σ M J = 0: Pa − ( PL /h)(ha 2 /L2 ) + M = 0
 a2

M = P 
− a 
 L

For maximum:
dM
 2a 
= P
− 1 = 0
da
 L

M max occurs at: a =
 ( L /2)2 L 
PL
M max = P 
− =−
L
2
4


|M |max =
1
L 
2
1
PL 
4
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1019
PROBLEM 7.15
Knowing that the radius of each pulley is 200 mm and neglecting
friction, determine the internal forces at Point J of the frame shown.
SOLUTION
FBD Frame with pulley and cord:
ΣM A = 0: (1.8 m) Bx − (2.6 m)(360 N)
− (0.2 m)(360 N) = 0
B x = 560 N
FBD BE:
Note: Cord forces have been moved to pulley hub as per Problem 6.91.
ΣM E = 0: (1.4 m)(360 N) + (1.8 m)(560 N)
− (2.4 m) B y = 0
B y = 630 N
FBD BJ:
3
ΣFx′ = 0: F + 360 N − (630 N − 360 N)
5
4
− (560 N) = 0
5
F = 250 N
ΣFy′ = 0: V +
36.9° 
4
3
(630 N − 360 N) − (560 N) = 0
5
5
V = 120.0 N
53.1° 
ΣM J = 0: M + (0.6 m)(360 N) + (1.2 m)(560 N)
− (1.6 m)(630 N) = 0
M = 120.0 N ⋅ m

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1020
PROBLEM 7.16
Knowing that the radius of each pulley is 200 mm and neglecting
friction, determine the internal forces at Point K of the frame
shown.
SOLUTION
Free body: frame and pulleys
Σ M A = 0: − Bx (1.8 m) − (360 N)(0.2 m)
− (360 N)(2.6 m) = 0
Bx = −560 N
B x = 560 N
A x = +920 N
ΣFx = 0: Ax − 560 N − 360 N = 0
Ax = +920 N
ΣFy = 0: Ay + By − 360 N = 0
Ay + By = 360 N
(1)
Free body: member AE
We recall that the forces applied to a pulley may be applied
directly to the axle of the pulley.
Σ M E = 0: − Ay (2.4 m) − (360 N)(1.8 m) = 0
Ay = −270 N
From (1):
A y = 270 N By = 360 N + 270 N
By = 630 N
B y = 630 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1021
PROBLEM 7.16 (Continued)
Free body: AK
ΣFx = 0: 920 N − 360 N − F = 0
F = +560 N
F = 560 N

ΣFy = 0: 360 N − 270 N − V = 0
V = +90.0 N
V = 90.0 N 
Σ M K = 0: (270 N)(1.6 m) − (360 N)(1 m) − M = 0
M = +72.0 N ⋅ m
M = 72.0 N ⋅ m

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1022
PROBLEM 7.17
A 5-in.-diameter pipe is supported every 9 ft by a small frame
consisting of two members as shown. Knowing that the combined
weight of the pipe and its contents is 10 lb/ft and neglecting the
effect of friction, determine the magnitude and location of the
maximum bending moment in member AC.
SOLUTION
Free body: 10-ft section of pipe
4
ΣFx = 0: D − (90 lb) = 0
5
D = 72 lb
3
ΣFy = 0: E − (90 lb) = 0
5
E = 54 lb
Free body: Frame
ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.)
+ (54 lb)(8.75 in.) = 0
A y = 34.8 lb Ay = + 34.8 lb
4
3
ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0
5
5
B y = 55.2 lb By = + 55.2 lb
3
4
ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0
5
5
Ax + Bx = 0
(1)
Free body: Member AC
ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0
Ax = − 26.4 lb
A x = 26.4 lb
B x = 26.4 lb
Bx = − Ax = + 26.4 lb
From (1):
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1023
PROBLEM 7.17 (Continued)
Free body: Portion AJ
For x ≤ 12.5 in. ( AJ ≤ AD ) :
3
4
ΣM J = 0: (26.4 lb) x − (34.8 lb) x + M = 0
5
5
M = 12 x
M max = 150 lb ⋅ in. for x = 12.5 in.
M max = 150.0 lb ⋅ in. at D For x > 12.5 in.( AJ > AD ):
3
4
ΣM J = 0: (26.4 lb) x − (34.8 lb) x + (72 lb)( x − 12.5) + M = 0
5
5
M = 900 − 60 x
M max = 150 lb ⋅ in. for x = 12.5 in.
M max = 150.0 lb ⋅ in. at D 
Thus:
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1024
PROBLEM 7.18
For the frame of Problem 7.17, determine the magnitude and location
of the maximum bending moment in member BC.
PROBLEM 7.17 A 5-in.-diameter pipe is supported every 9 ft by a
small frame consisting of two members as shown. Knowing that the
combined weight of the pipe and its contents is 10 lb/ft and neglecting
the effect of friction, determine the magnitude and location of the
maximum bending moment in member AC.
SOLUTION
Free body: 10-ft section of pipe
4
ΣFx = 0: D − (90 lb) = 0
5
D = 72 lb
3
ΣFy = 0: E − (90 lb) = 0
5
E = 54 lb
Free body: Frame
ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.)
+ (54 lb)(8.75 in.) = 0
A y = 34.8 lb Ay = + 34.8 lb
4
3
ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0
5
5
B y = 55.2 lb By = + 55.2 lb
3
4
ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0
5
5
Ax + Bx = 0
(1)
Free body: Member AC
ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.)
− Ax (9 in.) = 0
Ax = − 26.4 lb
From (1):
A x = 26.4 lb
B x = 26.4 lb
Bx = − Ax = + 26.4 lb
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1025
PROBLEM 7.18 (Continued)
Free body: Portion BK
For x ≤ 8.75 in.( BK ≤ BE ):
3
4
ΣM K = 0: (55.2 lb) x − (26.4 lb) x − M = 0
5
5
M = 12 x
M max = 105.0 lb ⋅ in. for
x = 8.75 in.
M max = 105.0 lb ⋅ in. at E For x > 8.75 in.( BK > BE ):
3
4
ΣM K = 0: (55.2 lb) x − (26.4 lb) x − (54 lb)( x − 8.75 in.) − M = 0
5
5
M = 472.5 − 42 x
M max = 105.0 lb ⋅ in. for
x = 8.75 in.
M max = 105.0 lb ⋅ in. at E 
Thus
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1026
PROBLEM 7.19
Knowing that the radius of each pulley is 150 mm, that α = 20°,
and neglecting friction, determine the internal forces at (a) Point J,
(b) Point K.
SOLUTION
Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter
computations: (cf. Problem 6.90)
(a)
Free body: AJ
ΣFx = 0: 500 N − F = 0
F = 500 N
F = 500 N

ΣFy = 0: V − 500 N = 0
V = 500 N
V = 500 N 
ΣM J = 0: (500 N)(0.6 m) = 0
M = 300 N ⋅ m
(b)
M = 300 N ⋅ m

V = 171.0 N

Free body: ABK
ΣFx = 0: 500 N − 500 N + (500 N)sin 20° − V = 0
V = 171.01 N
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1027
PROBLEM 7.19 (Continued)
ΣFy = 0: − 500 N − (500 N) cos 20° + F = 0
F = 969.8 N
F = 970 N 
ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 20°(0.9 m) − M = 0
M = 446.1 N ⋅ m
M = 446 N ⋅ m

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1028
PROBLEM 7.20
Knowing that the radius of each pulley is 150 mm, that α = 30°,
and neglecting friction, determine the internal forces at (a) Point J,
(b) Point K.
SOLUTION
Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter
computations: (cf. Problem 6.90)
(a)
Free body: AJ:
ΣFx = 0: 500 N − F = 0
F = 500 N
F = 500 N

ΣFy = 0: V − 500 N = 0
V = 500 N
V = 500 N 
ΣM J = 0: (500 N)(0.6 m) = 0
M = 300 N ⋅ m
(b)
M = 300 N ⋅ m

V = 250 N

FBD: Portion ABK:
ΣFx = 0: 500 N − 500 N + (500 N)sin 30° − V
ΣFy = 0: − 500 N − (500 N) cos 30° + F = 0
ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 30°(0.9 m) − M = 0
F = 933 N 
M = 375 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1029
PROBLEM 7.21
A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.
SOLUTION
(a)
FBD Rod:
ΣFx = 0:
Ax = 0
ΣM D = 0: aP − 2aAy = 0
Ay =
P
2
ΣFx = 0: V = 0 
FBD AJ:
ΣFy = 0:
P
−F =0
2
F=
P

2
ΣM J = 0: M = 0 
(b)
FBD Rod:
4 
3 
ΣM A = 0: 2a  D  + 2a  D  − aP = 0
5 
5 
ΣFx = 0: Ax −
4 5
P=0
5 14
ΣFy = 0: Ay − P +
3 5
P=0
5 14
D=
5P
14
Ax =
2P
7
Ay =
11P
14
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1030
PROBLEM 7.21 (Continued)
FBD AJ:



(c)
ΣFx = 0:
2
P −V = 0
7
ΣFy = 0:
11P
−F =0
14
ΣM J = 0: a
2P
−M =0
7
ΣM A = 0:
a  4D 
− aP = 0
2  5 
V=
2P
7
F=
M=

11P

14
2
aP
7

FBD Rod:
D=
5P
2
Ax −
4 5P
=0
5 2
Ax = 2 P
ΣFy = 0: Ay − P −
3 5P
=0
5 2
Ay =
ΣFx = 0:
5P
2
FBD AJ:
ΣFx = 0:
2P − V = 0
ΣFy = 0:
5P
−F =0
2
ΣM J = 0: a(2 P) − M = 0
V = 2P
F=

5P

2
M = 2aP


PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1031
PROBLEM 7.22
A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.
SOLUTION
(a)
FBD Rod:
ΣM D = 0: aP − 2aA = 0
A=
ΣFx = 0: V −
FBD AJ:
P
=0
2
V=
ΣFy = 0:
P
2

F=0 
ΣM J = 0: M − a
(b)
P
2
P
=0
2
M=
aP
2

FBD Rod:
ΣM D = 0: aP −
a4 
A =0
2  5 
A=
5P
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1032
PROBLEM 7.22 (Continued)
FBD AJ:
(c)
ΣFx = 0:
3 5P
−V = 0
5 2
ΣFy = 0:
4 5P
−F =0
5 2
V=
3P
2

F = 2P 
M=
3
aP
2

V=
3P
14

FBD Rod:
3 
4 
ΣM D = 0: aP − 2a  A  − 2a  A  = 0
5 
5 
A=
 3 5P 
ΣFx = 0: V − 
=0
 5 14 
ΣFy = 0:
4 5P
−F =0
5 14
 3 5P 
ΣM J = 0: M − a 
=0
 5 14 
5P
14
F=
M=
2P
7
3
aP
14


PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1033
PROBLEM 7.23
A quarter-circular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when θ = 30°.
SOLUTION
FBD Rod:
ΣFx = 0: A x = 0
2r
ΣM B = 0:
FBD AJ:
π
W − rAy = 0
α = 15°, weight of segment = W
r=
r
α
ΣFy ′ = 0:
sin α =
r
π
Ay =
2W
π
30° W
=
90° 3
sin15° = 0.9886r
12
2W
π
cos 30° −
W
cos 30° − F = 0
3
F=
W 3  2 1
 − 
2 π 3
2W 
W

ΣM 0 = M + r  F −
 + r cos15° 3 = 0
π


M = 0.0557 Wr

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1034
PROBLEM 7.24
A quarter-circular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when θ = 30°.
SOLUTION
FBD Rod:
ΣM A = 0: rB −
2r
π
W =0
B=
α = 15° =
FBD BJ:
r=
r
π
2W
π
π
12
sin15° = 0.98862r
12
Weight of segment = W
30° W
=
90° 3
ΣFy′ = 0: F −
W
2W
cos 30° −
sin 30° = 0
3
π
 3 1
F=
+
W
 6 π 


ΣM 0 = 0: rF − ( r cos15°)
W
−M =0
3
 3 1 
cos15° 
M = rW 
+
− 0.98862
Wr
 6 π  
3 


M = 0.289Wr

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1035
PROBLEM 7.25
For the rod of Problem 7.23, determine the magnitude and location of the maximum
bending moment.
PROBLEM 7.23 A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at Point J when θ = 30°.
SOLUTION
ΣFx = 0: Ax = 0
FBD Rod:
ΣM B = 0:
2r
π
W − rAy = 0
α=
θ
2
Ay =
2W
r=
r
,
Weight of segment = W
2α
π
=
F=
FBD AJ:
2W

ΣM 0 = 0: M +  F −
π

M=
But,
so
sin α cos α =
M=
2W
π
4α
π
2W
π
W cos 2α +
α
4α
2
ΣFx′ = 0: − F −
π
π
2W
π
sin α
W
cos 2α = 0
(1 − 2α ) cos 2α =
2W
π
(1 − θ ) cos θ
4α

 r + (r cos α ) π W = 0

(1 + θ cos θ − cos θ ) r −
4αW r
π
α
sin α cos α
1
1
sin 2α = sin θ
2
2
2r
π
W (1 − cos θ + θ cos θ − sin θ )
dM 2rW
(sin θ − θ sin θ + cos θ − cos θ ) = 0
=
dθ
π
for
(1 − θ )sin θ = 0
dM
= 0 for θ = 0, 1, nπ ( n = 1, 2,)
dθ
Only 0 and 1 in valid range
At
θ = 0 M = 0, at θ = 1 rad
at
θ = 57.3°
M = M max = 0.1009Wr 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1036
PROBLEM 7.26
For the rod of Problem 7.24, determine the magnitude and location of the maximum
bending moment.
PROBLEM 7.24 A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at Point J when θ = 30°.
SOLUTION
FBD Bar:
ΣM A = 0: rB −
2r
π
W =0
B=
θ
so
α=
r=
2
r
α
2W
π
0 ≤α ≤
π
4
sin α
Weight of segment = W
2α
π
2
=
ΣFx′ = 0: F −
4α
π
π
W
W cos 2α −
F=
2W
=
2W
π
π
4α
2W
π
sin 2α = 0
(sin 2α + 2α cos 2α )
(sin θ + θ cos θ )
FBD BJ:
ΣM 0 = 0: rF − (r cos α )
M=
But,
sin α cos α =
4α
π
W −M =0
r
 4α
Wr (sin θ + θ cos θ ) −  sin α cos α 
W
π
α
 π
2
1
1
sin 2α = sin θ
2
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1037
PROBLEM 7.26 (Continued)
so
M=
or
M=
2Wr
π
2
π
(sin θ + θ cos θ − sin θ )
Wrθ cos θ
dM 2
= Wr (cos θ − θ sin θ ) = 0 at θ tan θ = 1
dθ π
Solving numerically
θ = 0.8603 rad
M = 0.357 Wr
and

at θ = 49.3° 
(Since M = 0 at both limits, this is the maximum)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1038
PROBLEM 7.27
A half section of pipe rests on a frictionless horizontal surface as shown. If the
half section of pipe has a mass of 9 kg and a diameter of 300 mm, determine the
bending moment at Point J when θ = 90°.
SOLUTION
For half section
m = 9 kg
W = mg = (9)(9.81) = 88.29 N
Portion JC:
1
Weight = W = 44.145 N
2
From Fig. 5.8B:
x=
2r
π
=
2(150)
π
x = 95.49 mm
ΣM J = 0: (44.145 N)(0.15 m) − (44.145 N)(0.09549 m) − M = 0
M = +2.406 N ⋅ m
M = 2.41 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1039
PROBLEM 7.28
A half section of pipe rests on a frictionless horizontal surface as shown. If the
half section of pipe has a mass of 9 kg and a diameter of 300 mm, determine
the bending moment at point J when θ = 90°.
SOLUTION
m = 9 kg
For half section
W = mg = (9 kg)(9.81 m/s 2 ) = 88.29 N
Free body JC
Weight of portion JC,
1
= W = 44.145 N
2
r = 150 mm
From Fig. 5.8B:
x=
2r
π
=
2(150)
π
= 95.49 mm
ΣM J = 0: M − (44.145 N)(0.09549 m) = 0
M = +4.2154 N ⋅ m
M = 4.22 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1040
PROBLEM 7.29
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
ΣFy = 0: − wx − V = 0
V = − wx
 x
ΣM1 = 0: wx   + M = 0
2
1
M = − wx 2
2
|V |max = wL 
| M |max =
1 2
wL 
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1041
PROBLEM 7.30
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
1
1
x
1
x2
wx =  w0  x = w0
2
2 L
2
L
ΣFy = 0: −
1
x2
w0
−V = 0
2
L
1
x2  x
ΣFy = 0:  w0
 +M =0
L  3
2
By similar D’s
1
x2
V = − w0
2
L
1
x3
M = − w0
6
L
|V |max =
1
w0 L 
2
| M |max =
1
w0 L2 
6
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1042
PROBLEM 7.31
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: entire beam
 2L 
L
ΣM D = 0: P 
− P   − AL = 0

 3 
3
A = P /3
ΣFx = 0: Dx = 0
ΣFy = 0:
P
− P + P + Dy = 0
3
Dy = − P /3
(a)
D = P /3
Shear and bending moment. Since the loading consists of concentrated loads,
the shear diagram is made of horizontal straight-line segments and the
B. M. diagram is made of oblique straight-line segments.
Just to the right of A:
ΣFy = 0: − V1 +
P
=0
3
ΣM1 = 0: M 1 −
P
(0) = 0
3
V1 = + P /3 M1 = 0 Just to the right of B:
ΣFy = 0: − V2 +
ΣM 2 = 0: M 2 −
P
− P = 0,
3
PL
+ P(0) = 0
3  3 
V2 = −2 P /3 M 2 = + PL /9 Just to the right of C:
ΣFy = 0:
P
− P + P − V3 = 0
3
ΣM 3 = 0: M 3 −
P  2L 
L
+ P − P(0) = 0


3 3 
3
V3 = + P /3 M 3 = − PL /9 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1043
PROBLEM 7.31 (Continued)
Just to the left of D:
ΣFy = 0: V4 −
P
=0
3
ΣM 4 = 0: − M 4 −
P
(0) = 0
3
V4 = +
P
3
M4 = 0 |V |max = 2 P /3; | M |max = PL /9 
(b)
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1044
PROBLEM 7.32
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
(a)
Shear and bending moment.
Just to the right of A:
V1 = + P;
M1 = 0 Just to the right of B:
ΣFy = 0: P − P − V2 = 0;
L
ΣM 2 = 0: M 2 − P   = 0;
2
V2 = 0 M 2 = + PL /2 Just to the left of C:
ΣFy = 0: P − P − V3 = 0,
L
ΣM 3 = 0: M 3 + P   − PL = 0,
2
V3 = 0 M 3 = + PL /2 |V |max = P 
(b)
| M |max = PL /2 
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1045
PROBLEM 7.33
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
(a)
FBD Beam:
ΣM C = 0: LAy − M 0 = 0
Ay =
M0
L
ΣFy = 0: − Ay + C = 0
C=
M0
L
Along AB:
ΣFy = 0: −
M0
−V = 0
L
V =−
ΣM J = 0: x
M0
L
M0
+M =0
L
Straight with
M =−
M0
x
L
M =−
M0
at B
2
Along BC:
ΣFy = 0: −
M0
−V = 0
L
ΣM K = 0: M + x
Straight with
(b)
M=
V =−
M0
− M0 = 0
L
M0
at B
2
M0
L
x

M = M 0 1 − 
L

M = 0 at C
|V |max = M 0 /L 
From diagrams:
| M |max =
M0
at B 
2
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1046
PROBLEM 7.34
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Portion AJ
ΣFy = 0: − P − V = 0
ΣM J = 0: M + Px − PL = 0
(a)
V = −P 
M = P( L − x) 
The V and M diagrams are obtained by plotting the functions V and M.
|V |max = P 
(b)
| M |max = PL 
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1047
PROBLEM 7.35
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
Just to the right of A:
ΣFy = 0 V1 = +15 kN M1 = 0
Just to the left of C:
V2 = +15 kN M 2 = +15 kN ⋅ m
Just to the right of C:
V3 = +15 kN M 3 = +5 kN ⋅ m
Just to the right of D:
V4 = −15 kN M 4 = +12.5 kN ⋅ m
Just to the right of E:
V5 = −35 kN M 5 = +5 kN ⋅ m
At B:
(b)
M B = −12.5 kN ⋅ m
|V |max = 35.0 kN
| M |max = 12.50 kN ⋅ m 
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1048
PROBLEM 7.36
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(3.2 m) − (40 kN)(0.6 m) − (32 kN)(1.5 m) − (16 kN)(3 m) = 0
B = +37.5 kN
B = 37.5 kN 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 37.5 kN − 40 kN − 32 kN − 16 kN = 0
Ay = +50.5 kN
(a)
A = 50.5 kN 
Shear and bending moment
Just to the right of A:
V1 = 50.5 kN
M1 = 0 
Just to the right of C:
ΣFy = 0: 50.5 kN − 40 kN − V2 = 0
ΣM 2 = 0: M 2 − (50.5 kN)(0.6 m) = 0
V2 = +10.5 kN 
M 2 = +30.3 kN ⋅ m 
Just to the right of D:
ΣFy = 0: 50.5 − 40 − 32 − V3 = 0
ΣM 3 = 0: M 3 − (50.5)(1.5) + (40)(0.9) = 0
V3 = −21.5 kN 
M 3 = +39.8 kN ⋅ m 
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1049
PROBLEM 7.36 (Continued)
Just to the right of E:
ΣFy = 0: V4 + 37.5 = 0
V4 = −37.5 kN 
ΣM 4 = 0: − M 4 + (37.5)(0.2) = 0
VB = M B = 0
At B:
M 4 = +7.50 kN ⋅ m 

|V |max = 50.5 kN 
(b)
| M |max = 39.8 kN ⋅ m 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1050
PROBLEM 7.37
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: E (6 ft) − (6 kips)(2 ft) − (12 kips)(4 ft) − (4.5 kips)(8 ft) = 0
E = +16 kips
E = 16 kips 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 16 kips − 6 kips − 12 kips − 4.5 kips = 0
Ay = +6.50 kips
(a)
A = 6.50 kips 
Shear and bending moment
Just to the right of A:
V1 = +6.50 kips M1 = 0

Just to the right of C:
ΣFy = 0: 6.50 kips − 6 kips − V2 = 0
ΣM 2 = 0: M 2 − (6.50 kips)(2 ft) = 0
V2 = +0.50 kips 
M 2 = +13 kip ⋅ ft 
Just to the right of D:
ΣFy = 0: 6.50 − 6 − 12 − V3 = 0
ΣM 3 = 0: M 3 − (6.50)(4) − (6)(2) = 0
V3 = +11.5 kips 
M 3 = +14 kip ⋅ ft 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1051
PROBLEM 7.37 (Continued)
Just to the right of E:
ΣFy = 0: V4 − 4.5 = 0
ΣM 4 = 0: − M 4 − (4.5)2 = 0
VB = M B = 0
At B:
V4 = +4.5 kips 
M 4 = −9 kip ⋅ ft 

|V |max = 11.50 kips 
(b)
| M |max = 14.00 kip ⋅ ft 
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1052
PROBLEM 7.38
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Entire beam
ΣM C = 0: (120 lb)(10 in.) − (300 lb)(25 in.) + E (45 in.) − (120 lb)(60 in.) = 0
E = +300 lb
E = 300 lb 
ΣFx = 0: Cx = 0
ΣFy = 0: C y + 300 lb − 120 lb − 300 lb − 120 lb = 0
C y = +240 lb
(a)
C = 240 lb 
Shear and bending moment
Just to the right of A:
ΣFy = 0: − 120 lb − V1 = 0
V1 = −120 lb, M 1 = 0 
Just to the right of C:
ΣFy = 0: 240 lb − 120 lb − V2 = 0
ΣM C = 0: M 2 + (120 lb)(10 in.) = 0
V2 = +120 lb 
M 2 = −1200 lb ⋅ in. 
Just to the right of D:
ΣFy = 0: 240 − 120 − 300 − V3 = 0
ΣM 3 = 0: M 3 + (120)(35) − (240)(25) = 0,
V3 = −180 lb 
M 3 = +1800 lb ⋅ in. 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1053
PROBLEM 7.38 (Continued)
Just to the right of E:
+ΣFy = 0: V4 − 120 lb = 0
ΣM 4 = 0: − M 4 − (120 lb)(15 in.) = 0
V4 = +120 lb 
M 4 = −1800 lb ⋅ in. 
VB = M B = 0 
At B:
|V |max = 180.0 lb 
(b)
| M |max = 1800 lb ⋅ in. 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1054
PROBLEM 7.39
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(5 m) − (60 kN)(2 m) − (50 kN)(4 m) = 0
B = +64.0 kN
B = 64.0 kN 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 64.0 kN − 6.0 kN − 50 kN = 0
Ay = +46.0 kN
(a)
A = 46.0 kN 
Shear and bending-moment diagrams.
From A to C:
ΣFy = 0: 46 − V = 0
ΣM y = 0: M − 46 x = 0
V = +46 kN 
M = (46 x)kN ⋅ m 
From C to D:
ΣFy = 0: 46 − 60 − V = 0
V = −14 kN 
ΣM j = 0: M − 46 x + 60( x − 2) = 0
M = (120 − 14 x)kN ⋅ m
For
x = 2 m: M C = +92.0 kN ⋅ m

For
x = 3 m: M D = +78.0 kN ⋅ m

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1055
PROBLEM 7.39 (Continued)
From D to B:
ΣFy = 0: V + 64 − 25μ = 0
V = (25μ − 64)kN
μ
ΣM j = 0: 64 μ − (25μ )   − M = 0
2
M = (64μ − 12.5μ 2 )kN ⋅ m
For
μ = 0: VB = −64 kN
MB = 0 
|V |max = 64.0 kN 
(b)
| M |max = 92.0 kN ⋅ m 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1056
PROBLEM 7.40
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Entire beam
ΣM B = 0: (24 kN)(9 m) − C (6 m) + (24 kN)(4.5 m) = 0
C = 54 kN 
ΣFy = 0: 54 − 24 − 24 + B = 0
B = −6 kN
B = 6 kN
From A to C:
ΣFy = 0: − 24 − V = 0
V = −24 kN
ΣM1 = 0: (24)( x) + M = 0
M = (−24 x)kN ⋅ m
From C to D:
ΣFy = 0: − 24 − 8( x − 3) − V + 54 = 0
V = (−8 x + 54)kN
 x −3
ΣM 2 = 0: (24)( x) + 8( x − 3) 
 − (54)( x − 3) + M = 0
 2 
M = ( −4 x 2 + 54 x − 198)kN ⋅ m
From D to B:
ΣFy = 0: V − 6 = 0
ΣM 3 = 0: − M − (6)(9 − x) = 0
V = +6 kN
M = (6 x − 54)kN ⋅ m
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1057
PROBLEM 7.40 (Continued)
|V |max = 30.0 kN 
(b)
| M |max = 72.0 kN ⋅ m 
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1058
PROBLEM 7.41
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and bending
moment.
SOLUTION
(a)
By symmetry:
Ay = B = 8 kips +
1
(4 kips)(5 ft) A y = B = 18 kips
2
Along AC:
ΣFy = 0 : 18 kips − V = 0 V = 18 kips
ΣM J = 0: M − x(18 kips) M = (18 kips) x
M = 36 kip ⋅ ft at C ( x = 2 ft)
Along CD:
ΣFy = 0: 18 kips − 8 kips − (4 kips/ft) x1 − V = 0
V = 10 kips − (4 kips/ft) x1
V = 0 at x1 = 2.5 ft (at center)
ΣM K = 0: M +
x1
(4 kips/ft) x1 + (8 kips) x1 − (2 ft + x1 )(18 kips) = 0
2
M = 36 kip ⋅ ft + (10 kips/ft) x1 − (2 kips/ft) x12
M = 48.5 kip ⋅ ft at x1 = 2.5 ft
Complete diagram by symmetry
(b)
|V |max = 18.00 kips 
From diagrams:
|M |max = 48.5 kip ⋅ ft 
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1059
PROBLEM 7.42
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and
bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(10 ft) − (15 kips)(3 ft) − (12 kips)(6 ft) = 0
B = +11.70 kips
B = 11.70 kips 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 15 − 12 + 11.70 = 0
Ay = +15.30 kips
(a)
A = 15.30 kips 
Shear and bending-moment diagrams
From A to C:
ΣFy = 0: 15.30 − 2.5 x − V = 0
V = (15.30 − 2.5 x) kips
 x
ΣM J = 0: M + (2.5 x)   − 15.30 x = 0
2
M = 15.30 x − 1.25 x 2
For x = 0:
VA = +15.30 kips
For x = 6 ft:
VC = +0.300 kip
MA = 0 
M C = +46.8 kip ⋅ ft 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
1060
PROBLEM 7.42 (Continued)
From C to B:
ΣFy = 0: V + 11.70 = 0
V = −11.70 kips 
ΣM J = 0: 11.70μ − M = 0
M = (11.70μ ) kip ⋅ ft
M C = +46.8 kip ⋅ ft 
For μ = 4 ft:
For μ = 0:
MB = 0 
(b)
|V |max = 15.30 kips 






|M |max = 46.8 kip ⋅ ft 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1061
PROBLEM 7.43
Assuming the upward reaction of the ground on beam AB to be uniformly
distributed and knowing that P = wa, (a) draw the shear and bending-moment
diagrams, (b) determine the maximum absolute values of the shear and bending
moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (4a) − 2wa − wa = 0
wg =
(a)
3
w 
4
Shear and bending-moment diagrams
From A to C:
ΣFy = 0:
3
wx − wx − V = 0
4
1
V = − wx
4
ΣM J = 0 : M + ( wx)
x 3 x
wx
−
=0
2  4  2
1
M = − wx 2
8
For x = 0:
VA = M A = 0 
1
VC = − wa
4
For x = a :
1
M C = − wa 2 
8
From C to D:
ΣFy = 0:
3
wx − wa − V = 0
4
3

V =  x − aw
4


a 3  x

ΣM J = 0: M + wa  x −  − wx   = 0
2 4 2

3
a

M = wx 2 − wa  x − 
8
2

(1)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1062
PROBLEM 7.43 (Continued)
For x = a :
1
VC = − wa
4
1
M C = − wa 2 
8
For x = 2a :
1
VD = + wa
2
MD = 0 
Because of the symmetry of the loading, we can deduce the values of V and
M for the right-hand half of the beam from the values obtained for its
left-hand half.
|V |max =
(b)
To find | M |max , we differentiate Eq. (1) and set
dM
dx
1
wa 
2
= 0:
dM 3
4
= wx − wa = 0, x = a
dx 4
3
2
3 4 
wa 2
4 1
M = w  a  − wa 2  −  = −
8 3 
6
3 2
|M |max =
1 2
wa 
6
Bending-moment diagram consists of four distinct arcs of parabola.
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1063
PROBLEM 7.44
Solve Problem 7.43 knowing that P = 3wa.
PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB
to be uniformly distributed and knowing that P = wa, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (4a ) − 2wa − 3wa = 0
wg =
(a)
5
w 
4
Shear and bending-moment diagrams
From A to C:
ΣFy = 0:
5
wx − wx − V = 0
4
1
V = + wx
4
ΣM J = 0 : M + ( wx)
x 5 x
−
wx
=0
2  4  2
1
M = + wx 2
8
For x = 0:
VA = M A = 0 
1
VC = + wa
4
For x = a :
1
M C = + wa 2 
8
From C to D:
ΣFy = 0:
5
wx − wa − V = 0
4
5

V =  x − aw
4


a 5  x

ΣM J = 0: M + wa  x −  − wx   = 0
2 4 2

M=
5 2
a

wx − wa  x − 
8
2

(1)
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1064
PROBLEM 7.44 (Continued)
1
1
VC = + wa, M C = + wa 2 
4
8
For x = a :
For x = 2a :
3
VD = + wa, M D = + wa 2 
2
Because of the symmetry of the loading, we can deduce the values
of V and M for the right-hand half of the beam from the values
obtained for its left-hand half.
|V |max =
(b)
To find |M |max , we differentiate Eq. (1) and set
dM
dx
3
wa 
2
= 0:
dM 5
= wx − wa = 0
dx 4
4
x = a < a (outside portion CD)
5
The maximum value of |M | occurs at D:
|M |max = wa 2 
Bending-moment diagram consists of four distinct arcs of parabola.
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1065
PROBLEM 7.45
Assuming the upward reaction of the ground on beam AB to be uniformly
distributed and knowing that a = 0.3 m, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear
and bending moment.
SOLUTION
(a)
FBD Beam:
ΣFy = 0: w(1.5 m) − 2(3.0 kN) = 0
w = 4.0 kN/m
Along AC:
ΣFy = 0: (4.0 kN/m) x − V = 0
V = (4.0 kN/m) x
x
(4.0 kN/m) x = 0
2
M = (2.0 kN/m) x 2
ΣM J = 0: M −
Along CD:
ΣFy = 0: (4.0 kN/m) x − 3.0 kN − V = 0
V = (4.0 kN/m) x − 3.0 kN
ΣM K = 0: M + ( x − 0.3 m)(3.0 kN) −
x
(4.0 kN/m) x = 0
2
M = 0.9 kN ⋅ m − (3.0 kN) x + (2.0 kN/m) x 2
Note: V = 0 at x = 0.75 m, where M = −0.225 kN ⋅ m
Complete diagrams using symmetry.
|V |max = 1.800 kN 
(b)
|M |max = 0.225 kN ⋅ m 
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1066
PROBLEM 7.46
Solve Problem 7.45 knowing that a = 0.5 m.
PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB
to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute values
of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (1.5 m) − 3 kN − 3 kN = 0
(a)
wg = 4 kN/m 
Shear and bending moment
From A to C:
ΣFy = 0: 4 x − V = 0 V = (4 x) kN
ΣM J = 0: M − (4 x)
x
= 0, M = (2 x 2 ) kN ⋅ m
2
For x = 0:
For x = 0.5 m:
From C to D:
VA = M A = 0 
VC = 2 kN,
M C = 0.500 kN ⋅ m 
ΣFy = 0: 4 x − 3 kN − V = 0
V = (4 x − 3) kN
ΣM J = 0: M + (3 kN)( x − 0.5) − (4 x)
x
=0
2
M = (2 x 2 − 3 x + 1.5) kN ⋅ m
For x = 0.5 m:
VC = −1.00 kN,
For x = 0.75 m:
VC = 0,
For x = 1.0 m:
VC = 1.00 kN,
M C = 0.500 kN ⋅ m 
M C = 0.375 kN ⋅ m 
M C = 0.500 kN ⋅ m 
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1067
PROBLEM 7.46 (Continued)
From D to B:
ΣFy = 0: V + 4μ = 0 V = −(4μ ) kN
ΣM J = 0: (4 μ )
μ
2
− M = 0, M = 2μ 2
For μ = 0:
For μ = 0.5 m:
VB = M B = 0 
VD = −2 kN,
M D = 0.500 kN ⋅ m 
(b)
|V |max = 2.00 kN 

|M |max = 0.500 kN ⋅ m 









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1068
PROBLEM 7.47
Assuming the upward reaction of the ground on beam AB to be uniformly
distributed, (a) draw the shear and bending-moment diagrams, (b) determine
the maximum absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0
wg = 1.5 kips/ft 
(a)
Shear and bending-moment diagrams from A to C:
ΣFy = 0: 1.5 x − V = 0
ΣM J = 0: M − (1.5 x)
V = (1.5 x)kips
x
2
M = (0.75 x 2 )kip ⋅ ft
For x = 0:
VA = M A = 0 
For x = 3 ft: VC = 4.5 kips,
M C = 6.75 kip ⋅ ft 
From C to D:
ΣFy = 0: 1.5 x − 3( x − 3) − V = 0
V = (9 − 1.5 x)kips
ΣM J = 0: M + 3( x − 3)
x−3
x
− (1.5 x) = 0
2
2
M = [0.75 x 2 − 1.5( x − 3)2 ]kip ⋅ ft
For x = 3 ft: VC = 4.5 kips,
M C = 6.75 kip ⋅ ft 
For x = 6 ft: VcL = 0,
McL= 13.50 kip ⋅ ft 
For x = 9 ft: VD = −4.5 kips,
M D = 6.75 kip ⋅ ft 
VB = M B = 0 
At B:
|V |max = 4.50 kips 
(b)
|M |max = 13.50 kip ⋅ ft 
Bending-moment diagram consists or three distinct arcs of parabola,
all located above the x axis.
Thus: M ≥ 0 everywhere 
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1069
PROBLEM 7.48
Assuming the upward reaction of the ground on beam AB to be uniformly
distributed, (a) draw the shear and bending-moment diagrams, (b) determine
the maximum absolute values of the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0
(a)
wg = 1.5 kips/ft 
Shear and bending-moment diagrams.
From A to C:
ΣFy = 0: 1.5 x − 3x − V = 0
V = (−1.5 x) kips
x
x
− (1.5 x) = 0
2
2
2
M = ( −0.75 x ) kip ⋅ ft
ΣM J = 0: M + (3x)
For x = 0:
VA = M A = 0 
For x = 3 ft:
VC = −4.5 kips M C = −6.75 kip ⋅ ft 
From C to D:
ΣFy = 0: 1.5 x = 9 − V = 0, V = (1.5 x − 9) kips
ΣM J = 0: M + 9( x − 1.5) − (1.5 x)
x
=0
2
M = 0.75 x 2 − 9 x + 13.5
For x = 3 ft:
VC = −4.5 kips,
For x = 6 ft:
VC = 0,
M C = −6.75 kip ⋅ ft 
M C = −13.50 kip ⋅ ft 
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1070
PROBLEM 7.48 (Continued)
For x = 9 ft:
VD = 4.5 kips, M D = −6.75 kip ⋅ ft 
VB = M B = 0 
At B:
(b)
|V |max = 4.50 kips 



Bending-moment diagram consists of three distinct arcs of parabola.
|M |max = 13.50 kip ⋅ ft 
M ≤ 0 everywhere 
Since entire diagram is below the x axis:
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1071
PROBLEM 7.49
Draw the shear and bending-moment diagrams for the beam AB, and determine
the maximum absolute values of the shear and bending moment.
SOLUTION
Reactions:
ΣM A = 0: By (0.4) − (120)(0.2) = 0
B y = 60 N
ΣFx = 0:
Bx = 0
ΣFy = 0:
A = 180 N
Equivalent loading on straight part of beam AB.
From A to C:
ΣFy = 0: V = +180 N
ΣM1 = 0: M = +180 x
From C to B:
ΣFy = 0: 180 − 120 − V = 0
V = 60 N
ΣM x = 0: − (180 N)( x) + 24 N ⋅ m + (120 N)( x − 0.2) + M = 0
M = +60 x
|V |max = 180.0 N 
|M |max = 36.0 N ⋅ m 
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1072
PROBLEM 7.50
Draw the shear and bending-moment diagrams for the beam AB, and
determine the maximum absolute values of the shear and bending
moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: B(0.9 m) − (400 N)(0.3 m) − (400 N)(0.6 m)
− (400 N)(0.9 m) = 0
B = +800 N
B = 800 N 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 800 N − 3(400 N) = 0
Ay = +400 N
A = 400 N 
We replace the loads by equivalent force-couple systems at C, D, and E.
We consider successively the following F-B diagrams.
V5 = −400 N
V1 = +400 N
M1 = 0
M 5 = +180 N ⋅ m
V2 = +400 N
V6 = −400 N
M 2 = +60 N ⋅ m
M 6 = +60 N ⋅ m
V3 = 0
V7 = −800 N
M 3 = +120 N ⋅ m
M 7 = +120 N ⋅ m
V8 = −800 N
V4 = 0
M 4 = +120 N ⋅ m
M8 = 0
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1073
PROBLEM 7.50 (Continued)
(b)
|V |max = 800 N 
|M |max = 180.0 N ⋅ m 



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1074
PROBLEM 7.51
Draw the shear and bending-moment diagrams for the beam AB, and
determine the maximum absolute values of the shear and bending
moment.
SOLUTION
Slope of cable CD is
∴
Dy
7
=
Dx
10
 7

ΣM H = 0: (50 lb)(26 in.) + Dx (4 in.) −  Dx  (20 in.) + (100 lb)(10 in.) − (50 lb)(6 in.) = 0
 10 
2000 + (4 − 14) Dx = 0
Dy =
D x = 200 lb
7
7
Dx = (200)
10
10
ΣFx = 0: 200 lb − H x = 0
D y = 140 lb
H x = 200 lb
ΣFy = 0: 140 − 50 − 100 − 50 + H y = 0
H y = 60 lb
Equivalent loading on straight part of beam AB.
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1075
PROBLEM 7.51 (Continued)
From A to E:
ΣFy = 0: V = −50 lb
ΣM1 = 0: M = −50 x
From E to F:
ΣFy = 0: −50 + 140 − V = 0
V = +90 lb
ΣM 2 = 0: (50 lb) x − (140 lb)( x − 6) − 800 lb ⋅ in. + M = 0
M = −40 + 90 x
From F to G:
ΣFy = 0: −50 + 140 − 100 − V = 0
V = −10 lb
ΣM 3 = 0: 50 x − 140( x − 6) + 100( x − 16) − 800 + M = 0
M = 1560 − 10 x
From G to B:
ΣFy = 0: V − 50 = 0
V = 50 lb
ΣM 4 = 0: − M − (50)(32 − x) = 0
M = −1600 + 50 x
|V |max = 90.0 lb; |M |max = 1400 lb ⋅ in. 
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1076
PROBLEM 7.52
Draw the shear and bending-moment diagrams for the beam
AB, and determine the maximum absolute values of the shear
and bending moment.
SOLUTION
ΣFG = 0: T (9 in.) − (45 lb)(9 in.) − (120 lb)(21 in.) = 0
T = 325 lb
ΣFx = 0: − 325 lb + Gx = 0
G x = 325 lb
ΣFy = 0: G y − 45 lb − 120 lb = 0
G y = 165 lb
Equivalent loading on straight part of beam AB
From A to E:
ΣFy = 0: V = +165 lb
ΣM1 = 0: + 1625 lb ⋅ in. − (165 lb) x + M = 0
M = −1625 + 165 x
From E to F:
ΣFy = 0: 165 − 45 − V = 0
V = +120 lb
ΣM 2 = 0 : + 1625 lb ⋅ in. − (165 lb) x + (45 lb)( x − 9) + M = 0
M = −1220 + 120 x
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1077
PROBLEM 7.52 (Continued)
From F to B:
ΣFy = 0 V = +120 lb
ΣM 3 = 0: − (120)(21 − x) − M = 0
M = −2520 + 120 x
|V |max = 165.0 lb 
|M |max = 1625lb ⋅ in. 
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1078
PROBLEM 7.53
Two small channel sections DF and EH have been welded to
the uniform beam AB of weight W = 3 kN to form the rigid
structural member shown. This member is being lifted by
two cables attached at D and E. Knowing that θ = 30° and
neglecting the weight of the channel sections, (a) draw the
shear and bending-moment diagrams for beam AB, (b)
determine the maximum absolute values of the shear and
bending moment in the beam.
SOLUTION
FBD Beam + channels:
(a)
T1 = T2 = T
By symmetry:
ΣFy = 0: 2T sin 60° − 3 kN = 0
T=
T1x =
3
3
3
kN
2 3
3
T1 y = kN
2
M = (0.5 m)
FBD Beam:
3
2 3
= 0.433 kN ⋅ m
kN
With cable force replaced by equivalent force-couple system at F and G
Shear Diagram:
V is piecewise linear
 dV

 dx = −0.6 kN/m  with 1.5 kN


discontinuities at F and H.
VF − = −(0.6 kN/m)(1.5 m) = 0.9 kN
+
V increases by 1.5 kN to +0.6 kN at F
VG = 0.6 kN − (0.6 kN/m)(1 m) = 0
Finish by invoking symmetry
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1079
PROBLEM 7.53 (Continued)
Moment diagram: M is piecewise parabolic
 dM

 dx decreasing with V 


with discontinuities of .433 kN at F and H.
1
M F − = − (0.9 kN)(1.5 m)
2
= −0.675 kN ⋅ m
M increases by 0.433 kN m to –0.242 kN ⋅ m at F+
M G = −0.242 kN ⋅ m +
1
(0.6 kN)(1 m)
2
= 0.058 kN ⋅ m
Finish by invoking symmetry
|V |max = 900 N 
(b)
−
+
at F and G
|M |max = 675 N ⋅ m 
at F and G
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1080
PROBLEM 7.54
Solve Problem 7.53 when θ = 60°.
PROBLEM 7.53 Two small channel sections DF and EH
have been welded to the uniform beam AB of weight
W = 3 kN to form the rigid structural member shown. This
member is being lifted by two cables attached at D and E.
Knowing that θ = 30° and neglecting the weight of the
channel sections, (a) draw the shear and bending-moment
diagrams for beam AB, (b) determine the maximum absolute
values of the shear and bending moment in the beam.
SOLUTION
Free body: Beam and channels
From symmetry:
E y = Dy
Ex = Dx = Dy tan θ
Thus:
(1)
ΣFy = 0: D y + E y − 3 kN = 0
D y = E y = 1.5 kN 
D x = (1.5 kN) tan θ
From (1):
E = (1.5 kN) tan θ

We replace the forces at D and E by equivalent force-couple systems at F and H, where
M 0 = (1.5 kN tan θ )(0.5 m) = (750 N ⋅ m) tan θ
(2)
We note that the weight of the beam per unit length is
w=
(a)
W 3 kN
=
= 0.6 kN/m = 600 N/m
L
5m
Shear and bending moment diagrams
From A to F:
ΣFy = 0: − V − 600 x = 0 V = (−600 x) N
ΣM J = 0: M + (600 x)
For x = 0:
x
= 0, M = (−300 x 2 ) N ⋅ m
2
VA = M A = 0 
For x = 1.5 m:
VF = −900 N, M F = −675 N ⋅ m 
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1081
PROBLEM 7.54 (Continued)
From F to H:
ΣFy = 0: 1500 − 600 x − V = 0 

V = (1500 − 600 x) N 

ΣM J = 0: M + (600 x)

x
− 1500( x − 1.5) − M 0 = 0
2
M = M 0 − 300 x 2 + 1500( x − 1.5) N ⋅ m 
For x = 1.5 m:
VF = +600 N, M F = ( M 0 − 675) N ⋅ m 
For x = 2.5 m:
VG = 0, M G = ( M 0 − 375) N ⋅ m 
From G to B, The V and M diagrams will be obtained by symmetry,
(b)
Making θ = 60° in Eq. (2):
|V |max = 900 N 
M 0 = 750 tan 60° = 1299 N ⋅ m
M = 1299 − 675 = 624 N ⋅ m 
Thus, just to the right of F:
M G = 1299 − 375 = 924 N ⋅ m 
and
(b)
|V |max = 900 N 


|M |max = 924 N ⋅ m 

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1082
PROBLEM 7.55
For the structural member of Problem 7.53, determine (a) the angle
θ for which the maximum absolute value of the bending moment in
beam AB is as small as possible, (b) the corresponding value of
| M |max. (Hint: Draw the bending-moment diagram and then equate
the absolute values of the largest positive and negative bending
moments obtained.)
PROBLEM 7.53 Two small channel sections DF and EH have
been welded to the uniform beam AB of weight W = 3 kN to form
the rigid structural member shown. This member is being lifted by
two cables attached at D and E. Knowing that θ = 30° and neglecting
the weight of the channel sections, (a) draw the shear and bendingmoment diagrams for beam AB, (b) determine the maximum absolute
values of the shear and bending moment in the beam.
SOLUTION
See solution of Problem 7.50 for reduction of loading or beam AB to the following:
M 0 = (750 N ⋅ m) tan θ 
where
[Equation (2)]
The largest negative bending moment occurs Just to the left of F:
 1.5 m 
ΣM1 = 0: M 1 + (900 N) 
=0
 2 
M1 = −675 N ⋅ m 
The largest positive bending moment occurs
At G:
ΣM 2 = 0: M 2 − M 0 + (1500 N)(1.25 m − 1 m) = 0
M 2 = M 0 − 375 N ⋅ m 
Equating M 2 and − M 1 :
M 0 − 375 = +675
M 0 = 1050 N ⋅ m
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1083
PROBLEM 7.55 (Continued)
(a)
From Equation (2):
tan θ =
1050
= 1.400
750
θ = 54.5° 
|M |max = 675 N ⋅ m 
(b)
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1084
PROBLEM 7.56
For the beam of Problem 7.43, determine (a) the ratio k = P/wa for which the
maximum absolute value of the bending moment in the beam is as small as
possible, (b) the corresponding value of |M |max. (See hint for Problem 7.55.)
PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB
to be uniformly distributed and knowing that P = wa, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wg (4a) − 2wa − kwa = 0
wg =
wg
Setting
w
w
(2 + k )
4
=α
(1)
k = 4α − 2
We have
(2)
Minimum value of B.M. For M to have negative values, we must have wg < w. We verify that M will then be
negative and keep decreasing in the portion AC of the beam. Therefore, M min will occur between C and D.
From C to D:
a

 x
ΣM J = 0: M + wa  x −  − α wx   = 0
2

2
M=
We differentiate and set
dM
= 0:
dx
1
w(α x 2 − 2ax + a 2 )
2
αx − a = 0
xmin =
a
α
(3)
(4)
Substituting in (3):
1 2 1 2

wa  − + 1
2
α
α


α
1
−
= − wa 2
2α
M min =
M min
(5)
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1085
PROBLEM 7.56 (Continued)
Maximum value of bending moment occurs at D
 3a 
ΣM D = 0: M D + wa   − (2α wa)a = 0
 2 
3

M max = M D = wa 2  2α − 
2

(6)
Equating − M min and M max :
wa 2
1−α
3

= wa 2  2α − 
2α
2

4α 2 − 2α − 1 = 0
α=
(a)
Substitute in (2):
(b)
Substitute for α in (5):
2 + 20
8
α=
1+ 5
= 0.809
4
k = 4(0.809) − 2
|M |max = − M min = − wa 2
Substitute for α in (4):
k = 1.236 
1 − 0.809
2(0.809)
|M |max = 0.1180wa 2 
xmin =
a
1.236a 
0.809
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1086
PROBLEM 7.57
For the beam of Problem 7.45, 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 value of |M |max. (See hint for Problem 7.55.)
PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB
to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear
and bending-moment diagrams, (b) determine the maximum absolute values
of the shear and bending moment.
SOLUTION
Force per unit length exerted by ground:
wg =
6 kN
= 4 kN/m
1.5 m
The largest positive bending moment occurs Just to the left of C:
ΣM1 = 0: M 1 = (4a)
a
2
M 1 = 2a 2 
The largest negative bending moment occurs
At the center line:
ΣM 2 = 0: M 2 + 3(0.75 − a ) − 3(0.375) = 0
Equating M1 and − M 2 :
M 2 = −(1.125 − 3a) 
2a 2 = 1.125 − 3a
a 2 + 1.5a − 0.5625 = 0
(a)
Solving the quadratic equation:
(b)
Substituting:
a = 0.31066,
|M |max = M 1 = 2(0.31066) 2
a = 0.311 m 
|M |max = 193.0 N ⋅ m 
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1087
PROBLEM 7.58
For the beam and loading shown, 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 value of |M |max. (See hint for Problem 7.55.)
SOLUTION
Free body: Entire beam
ΣFx = 0: Ax = 0
ΣM E = 0: − Ay (2.4) + (3)(1.8) + 3(1) − (2)a = 0
5
Ay = 3.5 kN − a
6
5
A = 3.5 kN − a 
6
Free body: AC
5 

ΣM C = 0: M C −  3.5 − a  (0.6 m) = 0,
6 

M C = +2.1 −
a

2
Free body: AD
5 

ΣM D = 0: M D −  3.5 − a  (1.4 m) + (3 kN)(0.8 m) = 0
6 

M D = +2.5 −
ΣM E = 0: − M E − (2 kN)a = 0
Free body: EB
7
a 
6
M E = −2a 
We shall assume that M C > M D and, thus, that M max = M C .
We set M max = | M min |
or
M C = | M E | = 2.1 −
a
= 2a
2
a = 0.840 m 
| M |max = M C = | M E | = 2a = 2(0.840)
|M |max = 1.680 N ⋅ m 
We must check our assumption.
M D = 2.5 −
7
(0.840) = 1.520 N ⋅ m
6
Thus, M C > M D , O.K.
The answers are
(a)
a = 0.840 m 
(b)
|M |max = 1.680 N ⋅ m 
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1088
PROBLEM 7.59
A uniform beam is to be picked up by crane cables attached at A and B.
Determine the distance a from the ends of the beam to the points where
the cables should be attached if the maximum absolute value of the
bending moment in the beam is to be as small as possible. (Hint: Draw
the bending-moment diagram in terms of a, L, and the weight w per
unit length, and then equate the absolute values of the largest positive
and negative bending moments obtained.)
SOLUTION
w = weight per unit length
To the left of A:
 x
ΣM1 = 0: M + wx   = 0
2
1
M = − wx 2
2
1 2
M A = − wa
2
Between A and B:
1

1 
ΣM 2 = 0: M −  wL  ( x − a) + ( wx)  x  = 0
2

2 
1
1
1
M = − wx 2 + wLx − wLa
2
2
2
At center C:
x=
L
2
2
1 L
1
L 1
M C = − w   + wL   − wLa
2 2
2
2 2
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1089
PROBLEM 7.59 (Continued)
We set
| M A| = | M C | :
1
1
1
1
1
1
− wa 2 = wL2 − wLa + wa 2 = wL2 − wLa
2
8
2
2
8
2
a 2 + La − 0.25L2 = 0
1
1
( L ± L2 + L2 ) = ( 2 − 1) L
2
2
1
2
= w(0.207 L) = 0.0214 wL2
2
a=
M max
a = 0.207 L 
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1090
PROBLEM 7.60
Knowing that P = Q = 150 lb, determine (a) the distance a for which the
maximum absolute value of the bending moment in beam AB is as small as
possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)
SOLUTION
Free body: Entire beam
ΣM A = 0: Da − (150)(30) − (150)(60) = 0
D=
13,500
a

Free body: CB
ΣM C = 0: − M C − (150)(30) +
13,500
(a − 30) = 0
a
 45 
M C = 9000 1 − 
a 


Free body: DB
ΣM D = 0: − M D − (150)(60 − a) = 0
M D = −150(60 − a)
(a)
We set
 45 
M C = − M D : 9000 1 −  = 150(60 − a)
a 

M max = | M min | or
60 −
2700
= 60 − a
a
a 2 = 2700 a = 51.96 in.
(b)

a = 52.0 in. 
|M |max = − M D = 150(60 − 51.96)
| M | max = 1206 lb ⋅ in. 
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1091
PROBLEM 7.61
Solve Problem 7.60 assuming that P = 300 lb and Q = 150 lb.
PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a
for which the maximum absolute value of the bending moment in beam AB is
as small as possible, (b) the corresponding value of |M |max . (See hint for
Problem 7.55.)
SOLUTION
Free body: Entire beam
ΣM A = 0: Da − (300)(30) − (150)(60) = 0
D=
18,000

a
Free body: CB
ΣM C = 0: − M C − (150)(30) +
18,000
(a − 30) = 0
a
 40 
M C = 13,500 1 −  
a 

Free body: DB
ΣM D = 0: − M D − (150)(60 − a) = 0
M D = −150(60 − a) 
(a)
We set
 40 
M max = | M min | or M C = − M D : 13,500 1 −  = 150(60 − a)
a 

90 −
3600
= 60 − a
a
a 2 + 30a − 3600 = 0
a=
−30 + 15.300
= 46.847
2
a = 46.8 in. 
(b)
|M |max = − M D = 150(60 − 46.847)
|M | max = 1973 lb ⋅ in. 
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1092
PROBLEM 7.62*
In order to reduce the bending moment in the cantilever beam
AB, a cable and counterweight are permanently attached at end
B. Determine the magnitude of the counterweight for which the
maximum absolute value of the bending moment in the beam is
as small as possible and the corresponding value of |M |max .
Consider (a) the case when the distributed load is permanently
applied to the beam, (b) the more general case when the
distributed load may either be applied or removed.
SOLUTION
M due to distributed load:
ΣM J = 0: − M −
x
wx = 0
2
1
M = − wx 2
2
M due to counter weight:
ΣM J = 0: − M + xw = 0
M = wx
(a)
Both applied:
M = Wx −
w 2
x
2
dM
W
= W − wx = 0 at x =
dx
w
And here M =
W2
> 0 so Mmax; Mmin must be at x = L
2w
So M min = WL −
so
1 2
wL . For minimum | M |max set M max = − M min ,
2
W2
1
= − WL + wL2
2w
2
or
W 2 + 2 wLW − w2 L2 = 0
W = − wL ± 2w2 L2 (need + )
W = ( 2 − 1) wL = 0.414 wL 
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1093
PROBLEM 7.62* (Continued)
(b)
w may be removed
M max =
Without w,
With w (see Part a)
W 2 ( 2 − 1) 2 2
=
wL
2w
2
M max = 0.0858 wL2 
M = Wx
M max = WL at A
M = Wx −
w 2
x
2
W2
W
at x =
2w
w
1 2
= WL − wL at x = L
2
M max =
M min
For minimum M max , set M max (no w) = − M min (with w)
WL = − WL +
1 2
1
wL → W = wL →
2
4
With
M max =
1 2
wL 
4
W=
1
wL 
4
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1094
PROBLEM 7.63
Using the method of Section 7.6, solve Problem 7.29.
PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: wL − B = 0
B = wL
Shear diagram
L
ΣM B = 0: ( wL)   − M B = 0
2
MB =
wL2
2
Moment diagram
(b)
|V |max = wL;
|M |max =
wL2

2
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1095
PROBLEM 7.64
Using the method of Section 7.6, solve Problem 7.30.
PROBLEM 7.30 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
1
ΣFy = 0: B − ( w0 )( L) = 0
2
B=
1
w0 L
2
Shear diagram
ΣM B = 0:
1
L
( w0 )( L)   − M B = 0
2
3
MB =
1
w0 L2
6
Moment diagram
|V |max = w0 L /2 
(b)
| M |max = w0 L2 /6 
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1096
PROBLEM 7.65
Using the method of Section 7.6, solve Problem 7.31.
PROBLEM 7.31 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
 2L 
L
ΣM D = 0: P 
− P   − AL = 0

 3 
3
A = P /3
Shear diagram
We note that
VA = A = + P /3
|V |max = 2 P /3 
Bending diagram
We note that M A = 0
|M |max = PL /9 
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1097
PROBLEM 7.66
Using the method of Section 7.6, solve Problem 7.32.
PROBLEM 7.32 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: C = 0
ΣM C = 0: M C =
1
PL
2
Shear diagram
At A:
VA = + P
|V |max = P 
Moment diagram
At A:
MA = 0
|M |max =
1
PL 
2
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1098
PROBLEM 7.67
Using the method of Section 7.6, solve Problem 7.33.
PROBLEM 7.33 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: A = C
ΣM C = 0: Al − M 0 = 0
A=C =
M0
L
VA = −
M0
L
Shear diagram
At A:
|V | max =
M0

L
Bending-moment diagram
MA = 0
At A:
At B, M increases by M0 on account of applied couple.
|M | max = M 0 /2 
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1099
PROBLEM 7.68
Using the method of Section 7.6, solve Problem 7.34.
PROBLEM 7.34 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
ΣFy = 0: B − P = 0
B=P
Σ M B = 0: M B − M 0 + PL = 0
MB = 0
Shear diagram
VA = − P
At A:
|V | max = P 
Bending-moment diagram
M A = M 0 = PL
At A:
|M | max = PL 
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1100
PROBLEM 7.69
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Reactions
ΣFx = 0
Ax = 0
ΣM D = 0: + 3 kN ⋅ m + (6 kN)(3.5 m) + (3 kN)(2 m) − 1.2 kN ⋅ m − Ay (4.5 m) = 0
Ay = + 6.4 kN
(b)
|V |max = 6.40 kN;
Ay = + 6.4 kN
|M |max = 4.00 kN ⋅ m 
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1101
PROBLEM 7.70
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Reactions
Σ M A = 0: 12 kN ⋅ m − 3(5 kN)(4 m) + E (8 m) = 0
E = 6 kN
Σ Fy = 0
A = 9 kN
(b)
|V |max = 9.00 kN;
|M |max = 14.00 kN ⋅ m 
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1102
PROBLEM 7.71
Using the method of Section 7.6, solve Problem 7.39.
PROBLEM 7.39 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM B = 0: (60 kN)(3 m) + (50 kN)(1 m) − Ay (5 m) = 0
Ay = + 46.0 kN 䉰
ΣFy = 0: B + 46.0 kN − 60 kN − 50 kN = 0
B = + 64.0 kN 䉰
Shear diagram
At A:
VA = Ay = + 46.0 kN
|V |max = 64.0 kN 
Bending-moment diagram
At A:
MA = 0
|M |max = 92.0 kN ⋅ m 
Parabola from D to B. Its slope at D is same as that of straight-line segment
CD since V has no discontinuity at D.
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1103
PROBLEM 7.72
Using the method of Section 7.6, solve Problem 7.40.
PROBLEM 7.40 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Entire beam
Σ M B = 0: (24 kN)(9 m) − C (6 m) + (24 kN)(4.5 m) = 0
C = 54 kN
Shear diagram
Σ Fy = 0: 54 − 24 − 24 − B = 0
B = 6 kN
(b)
|V |max = 30.0 kN;
|M |max = 72.0 kN ⋅ m 
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1104
PROBLEM 7.73
Using the method of Section 7.6, solve Problem 7.41.
PROBLEM 7.41 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Reactions at supports.
Because of the symmetry:
A= B=
1
(8 + 8 + 4 × 5) kips
2
A = B = 18 kips 䉰
Shear diagram
At A:
VA = + 18 kips
|V |max = 18.00 kips 
Bending-moment diagram
At A:
MA = 0
|M |max = 48.5 kip ⋅ ft 
Discontinuities in slope at C and D, due to the discontinuities of V.
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1105
PROBLEM 7.74
Using the method of Section 7.6, solve Problem 7.42.
PROBLEM 7.42 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Beam
Σ Fx = 0: Ax = 0
Σ M B = 0: (12 kips)(4 ft) + (15 kips)(7 ft) − Ay (10 ft) = 0
Ay = + 15.3 kips 䉰
Σ Fy = 0: B + 15.3 − 15 − 12 = 0
B = + 11.7 kips 䉰
Shear diagram
At A:
VA = Ay = 15.3 kips
|V |max = 15.30 kips 
Bending-moment diagram
At A:
MA = 0
|M |max = 46.8 kip ⋅ ft 
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1106
PROBLEM 7.75
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Reactions
Σ M D = 0: (16)(9) + (45)(4) − (8)(3) − A(12) = 0
A = +25 kips
A = 25 kips
ΣFy = 0: 25 − 16 − 45 − 8 + Dy = 0
Dy = +44,
Dy = 44 kips
Σ Fx = 0: Dx = 0
(b)
|V |max = 36.0 kips;
|M |max = 120.0 kip ⋅ ft 
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1107
PROBLEM 7.76
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
Reactions
Σ Fx = 0: Bx = 0
Σ M E = 0: (20 lb/in.)(9 in.)(40.5 in.) + (125 in.)(24 in.) + (125 in.)(12 in.) − B(36 in.) = 0
By = +327.5 lb
B y = 327.5 lb
Σ Fy = 0: −(20 lb/in.)(9 in.) − 125 lb − 125 lb + 327.5 lb + E = 0
E = +102.5 lb
(b)
|V |max = 180.0 lb;
E = 102.5 lb
|M |max = 1230 lb ⋅ in. 
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1108
PROBLEM 7.77
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum
absolute value of the bending moment.
SOLUTION
Reactions
Σ M A = 0: + 45 kN ⋅ m − (90 kN)(3 m) + C (7.5 m) = 0
C = +30 kN
C = 30 kN
Σ Fy = 0: A − 90 kN + 30 kN = 0
A = 60 kN
75.0 kN ⋅ m, 4.00 m from A 
(b)
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1109
PROBLEM 7.78
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum
absolute value of the bending moment.
SOLUTION
ΣM A = 0: (8.75)(1.75) − B(2.5) = 0
B = 6.125 kN
ΣFy = 0:
A = 2.625 kN
Similar D’s
x
2.5 − x 2.5
=
=
2.625 
3.625

 6.25
}
Add numerators
and denominators
x = 1.05 m
1.378 kN ⋅ m, 1.050 m from A 
(b)
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1110
PROBLEM 7.79
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the
maximum absolute value of the bending moment.
SOLUTION
Free body: Beam
ΣM A = 0: B(4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0
B = + 55 kN 䉰
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 55 − 100 = 0
Ay = + 45 kN 䉰
Shear diagram
VA = Ay = + 45 kN
At A:
To determine Point C where V = 0:
VC − VA = − wx
0 − 45 kN = − (25 kN ⋅ m) x
x = 1.8 m 䉰
We compute all areas bending-moment
Bending-moment diagram
At A:
MA = 0
At B:
M B = − 20 kN ⋅ m
| M |max = 40.5 kN ⋅ m 


1.800 m from A 
Single arc of parabola
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1111
PROBLEM 7.80
Solve Problem 7.79 assuming that the 20-kN ⋅ m couple applied at B is
counterclockwise.
PROBLEM 7.79 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the magnitude and
location of the maximum absolute value of the bending moment.
SOLUTION
Free body: Beam
ΣM A = 0: B(4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0
B = + 45 kN 䉰
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 45 − 100 = 0
Ay = + 55 kN 䉰
Shear diagram
VA = Ay = + 55 kN
At A:
To determine Point C where V = 0 :
VC − VA = − wx
0 − 55 kN = − (25 kN/m) x
x = 2.20 m 䉰
We compute all areas bending-moment
Bending-moment diagram
At A:
MA = 0
At B:
M B = + 20 kN ⋅ m
| M |max = 60.5 kN ⋅ m 


2.20 m from A 
Single arc of parabola
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1112
PROBLEM 7.81
For the beam shown, draw the shear and bending-moment diagrams, and
determine the magnitude and location of the maximum absolute value of
the bending moment, knowing that (a) M = 0, (b) M = 24 kip · ft.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM B = 0: (16 kips)(6 ft) − Ay (8 ft) − M = 0
1
Ay = 12 kips − M
8
(1) 
1
ΣFy = 0: B + 12 − M − 16 = 0
8
1
B = 4 kips + M
8
(a)
(2) 
M = 0:
Load diagram
Making M = 0 in. (1) and (2).
Ay = +12 kips
B = +4 kips
VA = Ay = +12 kips
Shear diagram
To determine Point D where V = 0:
VD − VA = − wx
0 − 12 kips = −(4 kips/ft)x
x = 3 ft 
We compute all areas
B. M. Diagram
At A:
MA = 0
| M |max = 18.00 kip ⋅ ft, 
3.00 ft from A 
Parabola from A to C
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1113
PROBLEM 7.81 (Continued)
(b)
M = 24 kip ⋅ ft
Load diagram
Making M = 24 kip ⋅ ft in (1) and (2)
1
A = 12 − (24) = +9 kips
8
1
B = 4 + (24) = +7 kips
8
Shear diagram
VA = Ay = +9 kips
To determine Point D where V = 0:
VD = VA = − wx
0 − 9 kips = −(4 kips/ft) x
x = 2.25 ft 
B. M. Diagram
At A:
M A = +24 kip ⋅ ft
|M |max = 34.1 kip ⋅ ft, 
2.25 ft from A 

Parabola from A to C.
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1114
PROBLEM 7.82
For the beam shown, draw the shear and bending-moment diagrams, and
determine the magnitude and location of the maximum absolute value of
the bending moment, knowing that (a) P = 6 kips, (b) P = 3 kips.
SOLUTION
Free body: Beam
ΣFx = 0: Ax = 0
ΣM A = 0: C (6 ft) − (12 kips)(3 ft) − P(8 ft) = 0
C = 6 kips +
4
P
3
(1) 
4 

ΣFy = 0: Ay +  6 + P  − 12 − P = 0
3 

1
Ay = 6 kips − P
3
(a)
(2) 
P = 6 kips.
Load diagram
Substituting for P in Eqs. (2) and (1):
1
Ay = 6 − (6) = 4 kips
3
4
C = 6 + (6) = 14 kips
3
VA = Ay = +4 kips
Shear diagram
To determine Point D where V = 0:
VD − VA = − wx
0 − 4 kips = (2 kips/ft)x
x = 2 ft 
We compute all areas
Bending-moment diagram
At A:
MA = 0
|M |max = 12.00 kip ⋅ ft, at C 
Parabola from A to C
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1115
PROBLEM 7.82 (Continued)
(b)
P = 3 kips
Load diagram
Substituting for P in Eqs. (2) and (1):
1
A = 6 − (3) = 5 kips
3
4
C = 6 + (3) = 10 kips
3
Shear diagram
VA = Ay = +5 kips
To determine D where V = 0:
VD − VA = − wx
0 − (5 kips) = −(2 kips/ft) x
x = 2.5 ft 
We compute all areas
Bending-moment diagram
At A:
MA = 0
|M |max = 6.25 kip ⋅ ft 
2.50 ft from A 

Parabola from A to C.
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1116
PROBLEM 7.83
(a) Draw the shear and bending-moment diagrams for beam AB, (b) determine
the magnitude and location of the maximum absolute value of the bending
moment.
SOLUTION
Reactions at supports
Because of symmetry of load
A= B=
1
(300 × 8 + 300)
2
A = B = 1350 lb 䉰
Load diagram for AB
The 300-lb force at D is replaced by an equivalent force-couple system at C.
Shear diagram
VA = A = 1350 lb
At A:
To determine Point E where V = 0:
VE − VA = − wx
0 − 1350 lb = − (300 lb/ft) x
x = 4.50 ft 䉰
We compute all areas
Bending-moment diagram
At A:
MA = 0
Note 600 − lb ⋅ ft drop at C due to couple
|M |max = 3040 lb ⋅ ft 


4.50 ft from A 
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1117
PROBLEM 7.84
Solve Problem 7.83 assuming that the 300-lb force applied at D is directed
upward.
PROBLEM 7.83 (a) Draw the shear and bending-moment diagrams for
beam AB, (b) determine the magnitude and location of the maximum absolute
value of the bending moment.
SOLUTION
Reactions at supports
Because of symmetry of load:
A= B=
1
(300 × 8 − 300)
2
A = B = 1050 lb 䉰
Load diagram
The 300-lb force at D is replaced by an equivalent force-couple
system at C.
Shear diagram
VA = A = 1050 lb
At A:
To determine Point E where V = 0:
VE − VA = − wx
0 − 1050 lb = − (300 lb/ft) x
x = 3.50 ft 䉰
We compute all areas
Bending-moment diagram
MA = 0
At A:
Note 600 − lb ⋅ ft increase at C due to couple
|M |max = 1838 lb ⋅ ft 


3.50 ft from A 
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1118
PROBLEM 7.85
For the beam and loading shown, (a) write the equations of the shear and
bending-moment curves, (b) determine the magnitude and location of the
maximum bending moment.
SOLUTION
π x
dv
= − w = w0 cos
2 L
dx
 2L  π x
V = − wdx = − w0 
 sin 2 L + C1
 π 

(1)
dM
 2L  π x
= V = − w0 
 sin 2 L + C1
dx
 π 
2
πx
 2L 
M = Vdx = + w0 
 cos 2 L + C1 x + C2
π



(2)
Boundary conditions
At x = 0:
V = C1 = 0 C1 = 0
At x = 0:
 2L 
M = + w0 
 cos(0) + C2 = 0
 π 
2
 2L 
C2 = − w0 

 π 
2
 2L  π x
V = − w0 
 sin 2 L 
 π 
Eq. (1)
2
πx
 2L  
−1 + cos

M = w0 


2 L 
 π  
2
4
 2L 
|M max | = w0 
| −1 + 0| = 2 w0 L2 

π
 π 
M max at x = L :
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1119
PROBLEM 7.86
For the beam and loading shown, (a) write the equations of the shear
and bending-moment curves, (b) determine the magnitude and location
of the maximum bending moment.
SOLUTION
dV
x
= − w = − w0
dx
L
Eq. (7.1):

V = − w0
(1)
dM
1
x2
= V = − w0
+ C1
2
dx
L
Eq. (7.3):
M=
(a)
x
1
x2
dx = − w0
+ C1
2
L
L

 1

x2
1
x3
+ C1  dx = − w0 =
+ C1 x + C2
 − w0
L
6
L
 2

(2)
Boundary conditions
At
x = 0: M = 0 = 0 + 0 + C2
C2 = 0
1
x = L : M = 0 = − w0 L2 + C1 L
6
C1 =
1
w0 L
6
Substituting C1 and C2 into (1) and (2):
1
x2 1
V = − w0
+ w0 L
L 6
2
1
x3 1
M = − w0
+ w0 Lx
6
L 6
(b)
V=
M=
 1 x2 
1
w0 L  − 2  
2
3 L 
 x x3 
1
w0 L2  − 3  
6
L L 
Max moment occurs when V = 0:
1− 3
x2
=0
L2
M max =
Note: At x = 0,
x
1
=
L
3
 1  1 3 
1
−
w0 L2 
 
6
 3  3  
A = VA =
1
1
w0 L  
2
3
x = 0.577 L 
M max = 0.0642w0 L2 
A=
1
w0 L 
6
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1120
PROBLEM 7.87
For the beam and loading shown, (a) write the equations of the
shear and bending-moment curves, (b) determine the magnitude and
location of the maximum bending moment.
SOLUTION
(a)
We note that at
Load:
B( x = L): VB = 0, M B = 0
(1)
x 1 x

 4x 
w( x) = w0 1 −  − w0   = w0 1 −

L
3
L


 
 3L 
Shear: We use Eq. (7.2) between C ( x = x) and B( x = L):
VB − VC = −
L

V ( x) = w0
x

w( x) dx 0 − V ( x) = −

L
x
w( x)dx
L
4x 
1 − 3L  dx


x
L


2 x2 
2L
2x2 
w
L
x
= w0  x −
=
−
−
+



0
3L  x
3
3L 


V ( x) =
w0
(2 x 2 − 3Lx + L2 )
3L
(2) 
Bending moment: We use to Eq. (7.4) between C ( x = x) and B( x = L) :
M B − MC =
=

L
x
w0
3L
V ( x)dx 0 − M ( x)

L
x
(2 x 2 − 3Lx + L2 )dx
L
w 2
3

M ( x) = − 0  x3 − Lx 2 + L2 x 
3L  3
2

w
= − 0 [4 x3 − 9 Lx 2 + 6 L2 x]Lx
18L
w
= − 0 [(4 L3 − 9 L3 + 6 L3 ) − (4 x3 − 9 Lx 2 + 6 L2 x)]
18L
M ( x) =
w0
(4 x3 − 9 Lx 2 + 6 L2 x − L3 )
18L
(3) 
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1121
PROBLEM 7.87 (Continued)
(b)
Maximum bending moment
dM
=V = 0
dx
Eq. (2):
2 x 2 − 3Lx + L2 = 0
x=
Carrying into (3):
Note:
M max =
|M |max =
3− 9−8
L
L=
4
2
w0 L2
, At
72
w0 L2
18
At
x=
L
2

x =0
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1122
PROBLEM 7.88
For the beam and loading shown, (a) write the equations of the
shear and bending-moment curves, (b) determine the maximum
bending moment.
SOLUTION
(a)
Reactions at supports:
1
W
A = B = W , where
= Total load
2
L
W=

L
wdx = w0
0

πx
1 − sin
 dx
L 

L
0
L
πx
L

= w0  x + cos 
x
L 0

2

= w0 L 1 − 
 π
1
1
2

VA = A = W = w0 L 1 − 
2
2
 π
Thus
MA = 0
(1)
πx

w( x) = w0 1 − sin
L 

Load:
Shear: From Eq. (7.2):
V ( x ) − VA = −

x
w( x) dx
0
= − w0

πx
1 − sin L  dx


x
0
Integrating and recalling first of Eqs. (1),
x
1
2
L
πx


V ( x) − w0 L 1 −  = − w0  x + cos 
π
2
L 0
 π

V ( x) =
πx
1
2
L
L


w0 L 1 −  − w0  2 + cos
 + w0
π
π
2
L 
 π

πx
L
L
V ( x) = w0  − x − cos
π
2
L 

(2) 
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1123
PROBLEM 7.88 (Continued)
Bending moment: From Eq. (7.4) and recalling that M A = 0.
M ( x) − M A =

x
V ( x) dx
0
x
2
L
1
πx
L
= w0  x − x 2 −   sin

2
L 
π 
 2
0
M ( x) =
(b)
1 
2 L2
πx
w0  Lx − x 2 − 2 sin

2 
L 
π
(3) 
Maximum bending moment
dM
= V = 0.
dx
This occurs at x =
L
2
as we may check from (2):
π
L
L L L
V   = w0  − − cos  = 0
π
2
2
2
2
 

From (3):
2
L2 2 L2
π
L 1 L
M   = w0  −
− 2 sin 
4 π
2
2 2  2
1
8 

= w0 L2 1 − 2 
8
 π 
= 0.0237 w0 L2
M max = 0.0237 w0 L2 , at
x=
L
2

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1124
PROBLEM 7.89
The beam AB is subjected to the uniformly distributed load shown and to two
unknown forces P and Q. Knowing that it has been experimentally
determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E,
(a) determine P and Q, (b) draw the shear and bending-moment diagrams for
the beam.
SOLUTION
(a)
Free body: Portion AD
ΣFx = 0: C x = 0
ΣM D = 0: −C y (0.3 m) + 0.800 kN ⋅ m + (6 kN)(0.45 m) = 0
C y = +11.667 kN
C = 11.667 kN 
Free body: Portion EB
ΣM E = 0: B(0.3 m) − 1.300 kN ⋅ m = 0
B = 4.333 kN 
Free body: Entire beam
ΣM D = 0: (6 kN)(0.45 m) − (11.667 kN)(0.3 m)
− Q (0.3 m) + (4.333 kN)(0.6 m) = 0
Q = 6.00 kN 
ΣM y = 0: 11.667 kN + 4.333 kN
−6 kN − P − 6 kN = 0
P = 4.00 kN 
Load diagram
(b)
Shear diagram
At A: VA = 0
|V |max = 6 kN 
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1125
PROBLEM 7.89 (Continued)
Bending-moment diagram
MA = 0
At A:
|M |max = 1300 N ⋅ m 
We check that
M D = +800 N ⋅ m and M E = +1300 N ⋅ m
As given:
M C = −900 N ⋅ m 
At C:
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1126
PROBLEM 7.90
Solve Problem 7.89 assuming that the bending moment was found to
be +650 N ⋅ m at D and +1450 N ⋅ m at E.
PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load
shown and to two unknown forces P and Q. Knowing that it has been
experimentally determined that the bending moment is +800 N ⋅ m at D and
+1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.
SOLUTION

(a)
Free body: Portion AD
ΣFx = 0: C x = 0
ΣM D = 0: −C (0.3 m) + 0.650 kN ⋅ m + (6 kN)(0.45 m) = 0
C y = +11.167 kN
C = 11.167 kN 
Free body: Portion EB
ΣM E = 0: B(0.3 m) − 1.450 kN ⋅ m = 0


B = 4.833 kN 

Free body: Entire beam
ΣM D = 0: (6 kN)(0.45 m) − (11.167 kN)(0.3 m)
−Q (0.3 m) + (4.833 kN)(0.6 m) = 0
Q = 7.50 kN 


ΣM y = 0: 11.167 kN + 4.833 kN
− 6 kN − P − 7.50 kN = 0
P = 2.50 kN 
Load diagram


(b)
Shear diagram
VA = 0
At A:
|V |max = 6 kN 


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1127

PROBLEM 7.90 (Continued)

Bending-moment diagram
MA = 0
At A:
|M |max = 1450 N ⋅ m 
We check that
M D = +650 N ⋅ m and M E = +1450 N ⋅ m
As given:

M C = −900 N ⋅ m 
At C:
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1128
PROBLEM 7.91*
The beam AB is subjected to the uniformly distributed load shown and to two
unknown forces P and Q. Knowing that it has been experimentally
determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft
at E, (a) determine P and Q, (b) draw the shear and bending-moment
diagrams for the beam.
SOLUTION
(a)
Free body: Portion DE
ΣM E = 0: 5.50 kip ⋅ ft − 6.10 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0
VD = +0.350 kip
ΣFy = 0: 0.350 kip − 1kip − VE = 0
VE = −0.650 kip
Free body: Portion AD
ΣM A = 0: 6.10 kip ⋅ ft − P(2ft) − (1 kip)(2 ft) − (0.350 kip)(4 ft) = 0
P = 1.350 kips 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 1 kip − 1.350 kip − 0.350 kip = 0
Ay = +2.70 kips
A = 2.70 kips 
Free body: Portion EB
ΣM B = 0: (0.650 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 5.50 kip ⋅ ft = 0
Q = 0.450 kip 
ΣFy = 0: B − 0.450 − 1 − 0.650 = 0
B = 2.10 kips 
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1129
PROBLEM 7.91* (Continued)
(b)
Load diagram
Shear diagram
VA = A = +2.70 kips
At A:
To determine Point G where V = 0, we write
VG − VC = − wμ
0 − 0.85 kips = −(0.25 kip/ft)μ
μ = 3.40 ft 
|V |max = 2.70 kips at A 
We next compute all areas
Bending-moment diagram
At A: M A = 0
Largest value occurs at G with
AG = 2 + 3.40 = 5.40 ft
|M |max = 6.345 kip ⋅ ft 
5.40 ft from A 
Bending-moment diagram consists of 3 distinct arcs of parabolas.
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1130
PROBLEM 7.92*
Solve Problem 7.91 assuming that the bending moment was found to
be +5.96 kip ⋅ ft at D and +6.84 kip ⋅ ft at E.
PROBLEM 7.91* The beam AB is subjected to the uniformly distributed
load shown and to two unknown forces P and Q. Knowing that it has been
experimentally determined that the bending moment is +6.10 kip ⋅ ft at D
and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.
SOLUTION
(a)
Free body: Portion DE
ΣM E = 0: 6.84 kip ⋅ ft − 5.96 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0
VD = +0.720 kip
ΣFy = 0: 0.720 kip − 1kip − VE = 0
VE = −0.280 kip
Free body: Portion AD
ΣM A = 0: 5.96 kip ⋅ ft − P (2 ft) − (1 kip)(2 ft) − (0.720 kip)(4 ft) = 0
P = 0.540 kip 
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 1 kip − 0.540 kip − 0.720 kip = 0
Ay = +2.26 kips
A = 2.26 kips 
Free body: Portion EB
ΣM B = 0: (0.280 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 6.84 kip ⋅ ft = 0
Q = 1.860 kips 
ΣFy = 0: B − 1.860 − 1 − 0.280 = 0
B = 3.14 kips 
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1131
PROBLEM 7.92* (Continued)
(b)
Load diagram
Shear diagram
VA = A = +2.26 kips
At A:
To determine Point G where V = 0, we write
VG − VC = − wμ
0 − (1.22 kips) = −(0.25 kip/ft)μ
μ = 4.88 ft 
|V |max = 3.14 kips at B 
We next compute all areas
Bending-moment diagram
At A: M A = 0
Largest value occurs at G with
AG = 2 + 4.88 = 6.88 ft
|M |max = 6.997 kip ⋅ ft 
6.88 ft from A 
Bending-moment diagram consists of 3 distinct arcs of parabolas.
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1132
PROBLEM 7.93
Three loads are suspended as shown from the cable ABCDE.
Knowing that dC = 3 m, determine (a) the components of the
reaction at E, (b) the maximum tension in the cable.
SOLUTION
(a)
ΣM A = 0: E y (16 m) − (2 kN)(4 m) − (3 kN)(8 m) − (4 kN)(12 m) = 0
E y = + 5 kN
E y = 5.00 kN 
Portion CDE:
ΣM C = 0: (5 kN)(8 m) − (4 kN)(4 m) − E x (3 m) = 0
E x = 8 kN
(b)
E x = 8.00 kN

Maximum tension occurs in DE:
Tm = Ex2 + E y2 = 82 + 52
Tm = 9.43 kN 
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1133
PROBLEM 7.94
Knowing that the maximum tension in cable ABCDE is 13 kN,
determine the distance dC.
SOLUTION
Maximum tension of 13 kN occurs in DE. See solution of Problem 7.93 for the determination of E y = 5.00 kN
From force triangle
Portion CDE:
Ex2 + 52 = 132
Ex = 12 kN
ΣM C = 0: (5 kN)(8 m) − (12 kN)hC − (4 kN)(4 m) = 0
hC = 2.00 m 
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1134
PROBLEM 7.95
If dC = 8 ft, determine (a) the reaction at A, (b) the reaction at E.
SOLUTION
Free body: Portion ABC
ΣM C = 0
2 Ax − 16 Ay + 300(8) = 0
Ax = 8 Ay − 1200
(1)
Free body: Entire cable
ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0
3 Ax + 16 Ay − 6400 = 0
Substitute from Eq. (1):
3(8 Ay − 1200) + 16 Ay − 6400 = 0
Eq. (1)
A y = 250 lb
Ax = 8(250) − 1200
A x = 800 lb
ΣFx = 0: − Ax + Ex = 0 − 800 lb + Ex = 0
E x = 800 lb
ΣFy = 0: 250 + E y − 300 − 200 − 300 = 0
E y = 550 lb
(a)
A = 838 lb
(b)
E = 971 lb
17.35° 
34.5° 
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1135
PROBLEM 7.96
If dC = 4.5 ft, determine (a) the reaction at A, (b) the reaction at E.
SOLUTION
Free body: Portion ABC
ΣM C = 0: − 1.5 Ax − 16 Ay + 300 × 8 = 0
Ax =
Free body: Entire cable
(2400 − 16 Ay )
(1)
1.5
ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0
3 Ax + 16 Ay − 6400 = 0
Substitute from Eq. (1):
3(2400 − 16 Ay )
1.5
+ 16 Ay − 6400 = 0
Ay = −100 lb
A y = 100 lb
Thus Ay acts downward
Eq. (1)
(2400 − 16( −100))
= 2667 lb
1.5
A x = 2667 lb
ΣFx = 0: − Ax + Ex = 0 − 2667 + Ex = 0
E x = 2667 lb
Ax =
ΣFy = 0: Ay + E y − 300 − 200 − 300 = 0
−100 lb + E y − 800 lb = 0
E y = 900 lb
A = 2670 lb
(a)
E = 2810 lb
(b)
2.10° 
18.65° 
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1136
PROBLEM 7.97
Knowing that dC = 3 m, determine (a) the distances dB and dD (b) the
reaction at E.
SOLUTION
Free body: Portion ABC
ΣM C = 0: 3 Ax − 4 Ay + (5 kN)(2 m) = 0
Ax =
4
10
Ay −
3
3
(1)
Free body: Entire cable
ΣM E = 0: 4 Ax − 10 Ay + (5 kN)(8 m) + (5 kN)(6 m) + (10 kN)(3 m) = 0
4 Ax − 10 Ay + 100 = 0
Substitute from Eq. (1):
10 
4
4  Ay −  − 10 Ay + 100 = 0
3
3
Ay = +18.571 kN
Eq. (1)
A y = 18.571 kN
4
10
(18.511) −
= +21.429 kN
3
3
A x = 21.429 kN
ΣFx = 0: − Ax + Ex = 0 − 21.429 + Ex = 0
E x = 21.429 kN
ΣFy = 0: 18.571 kN + E y + 5 kN + 5 kN + 10 kN = 0
E y = 1.429 kN
Ax =
E = 21.5 kN
(b)
3.81° 
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1137
PROBLEM 7.97 (Continued)
(a)
Portion AB
ΣM B = 0: (18.571 kN)(2 m) − (21.429 kN)d B = 0
d B = 1.733 m 
Portion DE
Geometry
h = (3 m) tan 3.8°
= 0.199 m
d D = 4 m + 0.199 m
d D = 4.20 m 

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1138
PROBLEM 7.98
Determine (a) distance dC for which portion DE of the cable is horizontal,
(b) the corresponding reactions at A and E.
SOLUTION
Free body: Entire cable
(b)
ΣFy = 0: Ay − 5 kN − 5 kN − 10 kN = 0
A y = 20 kN
ΣM A = 0: E (4 m) − (5 kN)(2 m) − (5 kN)(4 m) − (10 kN)(7 m) = 0
E = +25 kN
E = 25.0 kN
ΣFx = 0: − Ax + 25 kN = 0
A x = 25 kN
A = 32.0 kN
(a)

38.7° 
Free body: Portion ABC
ΣM C = 0: (25 kN)dC − (20 kN)(4 m) + (5 kN)(2 m) = 0
25dC − 70 = 0
dC = 2.80 m 
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1139
PROBLEM 7.99
An oil pipeline is supported at 6-ft intervals by vertical hangers
attached to the cable shown. Due to the combined weight of the
pipe and its contents the tension in each hanger is 400 lb.
Knowing that dC = 12 ft, determine (a) the maximum tension in
the cable, (b) the distance dD.
SOLUTION
FBD Cable: Hanger forces at A and F act on the supports, so A y and Fy act on the cable.
ΣM F = 0: (6 ft + 12 ft + 18 ft + 24 ft)(400 lb)
− (30 ft)Ay − (5 ft) Ax = 0
Ax + 6 Ay = 4800 lb
FBD ABC:
(1)
ΣM C = 0: (7 ft)Ax − (12 ft)Ay + (6 ft)(400 lb) = 0
(2)
Solving (1) and (2)
A x = 800 lb
Ay =
From FBD Cable:
2000
lb
3
ΣFx = 0: − 800 lb + Fx = 0
Fx = 800 lb
FBD DEF:
ΣFy = 0:
200
lb − 4 (400 lb) + Fy = 0
3
Fy =
Since Ax = Fx and Fy > Ay ,
 2800 
lb 
Tmax = TEF = (800 lb) 2 + 
 3

2800
lb
3
2
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1140
PROBLEM 7.99 (Continued)
(a)
Tmax = 1229.27 lb,
Tmax = 1229 lb 
 2800 
ΣM D = 0: (12 ft) 
lb  − d D (800 lb) − (6 ft)(400 lb) = 0
 3

d D = 11.00 ft 
(b)
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1141
PROBLEM 7.100
Solve Problem 7.99 assuming that dC = 9 ft.
PROBLEM 7.99 An oil pipeline is supported at 6-ft intervals by
vertical hangers attached to the cable shown. Due to the combined
weight of the pipe and its contents the tension in each hanger is
400 lb. Knowing that dC = 12 ft, determine (a) the maximum
tension in the cable, (b) the distance dD.
SOLUTION
FBD CDEF:
ΣM C = 0: (18 ft)Fy − (9 ft)Fy − (6 ft + 12 ft)(400 lb) = 0
Fx − 2 Fy = −800 lb
(1)
EM A = 0: (30 ft)Fy − (5 ft)Fx
FBD Cable:
− (6 ft)(1 + 2 + 3 + 4)(400 lb) = 0
Fx − 6 Fy = −4800 lb
(2)
Solving (1) and (2),
Fx = 1200 lb
, Fy = 1000 lb
ΣFx = 0: − Ax + 1200 lb = 0, A x = 1200 lb
Point F:
ΣFy = 0: Ay + 1000 lb − 4(400 lb) = 0,
A y = 600 lb
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1142
PROBLEM 7.100 (Continued)
Since
Ax = Ay
and Fy > Ay , Tmax = TEF
Tmax = (1 kip) 2 + (1.2 kips)2
Tmax = 1.562 kips 
(a)
FBD DEF:
ΣM D = 0: (12 ft)(1000 lb) − d D (1200 lb)
− (6 ft)(400 lb) = 0
d D = 8.00 ft 
(b)
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1143
PROBLEM 7.101
Knowing that mB = 70 kg and mC = 25 kg, determine the
magnitude of the force P required to maintain equilibrium.
SOLUTION
Free body: Portion CD
ΣM C = 0: D y (4 m) − Dx (3 m) = 0
Dy =
3
Dx
4
Free body: Entire cable
ΣM A = 0:
3
Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0
4
ΣM B = 0:
3
Dx (10 m) − Dx (5 m) − WC (6 m) = 0
4
(1)
Free body: Portion BCD
Dx = 2.4WC
For
mB = 70 kg mC = 25 kg
g = 9.81 m/s 2 :
WB = 70 g
Eq. (2):
Dx = 2.4WC = 2.4(25g ) = 60 g
Eq. (1):
(2)
WC = 25 g
3
60 g (14) − 70 g (4) − 25 g (10) − 5P = 0
4
100 g − 5 P = 0: P = 20 g
P = 20(9.81) = 196.2 N
P = 196.2 N 
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1144
PROBLEM 7.102
Knowing that mB = 18 kg and mC = 10 kg, determine the magnitude
of the force P required to maintain equilibrium.
SOLUTION
Free body: Portion CD
ΣM C = 0: D y (4 m) − Dx (3 m) = 0
Dy =
3
Dx
4
Free body: Entire cable
ΣM A = 0:
3
Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0
4
ΣM B = 0:
3
Dx (10 m) − Dx (5 m) − WC (6 m) = 0
4
(1)
Free body: Portion BCD
Dx = 2.4WC
For
mB = 18 kg mC = 10 kg
g = 9.81 m/s 2 :
WB = 18 g
Eq. (2):
Dx = 2.4WC = 2.4(10 g ) = 24 g
Eq. (1):
3
24 g (14) − (18 g )(4) − (10 g )(10) − 5P = 0
4
(2)
WC = 10 g
80 g − 5P : P = 16 g
P = 16(9.81) = 156.96 N
P = 157.0 N 
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1145
PROBLEM 7.103
Cable ABC supports two loads as shown. Knowing that b = 21 ft,
determine (a) the required magnitude of the horizontal force P,
(b) the corresponding distance a.
SOLUTION
Free body: ABC
ΣM A = 0: P(12 ft) − (140 lb) a − (180 lb)b = 0
(1)
ΣM B = 0: P(9 ft) − (180 lb)(b − a) = 0
(2)
Free body: BC
Data
b = 21 ft
Equate (3) and (4) through P :
Eq. (3):
P=
20
a + 180
3
Eq. (1): 21P − 140a − 180(21) = 0
P=
Eq. (2): 9 P − 180(21 − a) = 0
P = −20a + 420
(4)
(b)
a = 9.00 ft 
(a)
P = 240 lb 
20
a + 180 = −20a + 420
3
20
(9.00) + 180
3
(3)
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1146
PROBLEM 7.104
Cable ABC supports two loads as shown. Determine the distances a
and b when a horizontal force P of magnitude 200 lb is applied at A.
SOLUTION
Free body: ABC
ΣM A = 0: P(12 ft) − (140 lb) a − (180 lb)b = 0
(1)
ΣM B = 0: P(9 ft) − (180 lb)(b − a) = 0
(2)
Free body: BC
Data
P = 200 lb
Eq. (1): 200(21) − 140a − 180b = 0
(3)
Eq. (2): 200(9) + 180a − 180b = 0
(4)
(3) − (4) : 2400 − 320a = 0
a = 7.50 ft 
Eq. (2): (200 lb)(9 ft) − (180 lb)(b − 7.50 ft) = 0
b = 17.50 ft 
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1147
PROBLEM 7.105
If a = 3 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.
SOLUTION
Free body: Portion DE
ΣM D = 0: E y (4 m) − Ex (5 m) = 0
Ey =
5
Ex
4
Free body: Portion CDE
ΣM C = 0:
5
Ex (8 m) − E x (7 m) − P(2 m) = 0
4
Ex =
2
P
3
(1)
Free body: Portion BCDE
ΣM B = 0:
5
Ex (12 m) − E x (5 m) − (120 kN)(4 m) = 0
4
10 Ex − 480 = 0; Ex = 48 kN
Eq. (1):
48 kN =
2
P
3
P = 72.0 kN 
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1148
PROBLEM 7.105 (Continued)
Free body: Entire cable
ΣM A = 0:
5
Ex (16 m) − Ex (2 m) + P(3 m) − Q (4 m) − (120 kN)(8 m) = 0
4
(48 kN)(20 m − 2 m) + (72 kN)(3 m) − Q (4 m) − 960 kN ⋅ m = 0
4Q = 120
Q = 30.0 kN 
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1149
PROBLEM 7.106
If a = 4 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.
SOLUTION
Free body: Portion DE
ΣM D = 0: E y (4 m) − Ex (6 m) = 0
Ey =
3
Ex
2
Free body: Portion CDE
ΣM C = 0:
3
Ex (8 m) − E x (8 m) − P(2 m) = 0
2
Ex =
1
P
2
(1)
Free body: Portion BCDE
ΣM B = 0:
3
Ex (12 m) − E x (6 m) + (120 kN)(4 m) = 0
2
12 Ex = 480
Eq (1):
Ex =
1
P;
2
Ex = 40 kN
40 kN =
1
P
2
P = 80.0 kN 
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1150
PROBLEM 7.106 (Continued)
Free body: Entire cable
ΣM A = 0:
3
Ex (16 m) − Ex (2 m) + P (4 m) − Q (4 m) − (120 kN)(8 m) = 0
2
(40 kN)(24 m − 2 m) + (80 kN)(4 m) − Q (4 m) − 960 kN ⋅ m = 0
4Q = 240
Q = 60.0 kN 
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1151
PROBLEM 7.107
A transmission cable having a mass per unit length of 0.8 kg/m is strung between two insulators at the same
elevation that are 75 m apart. Knowing that the sag of the cable is 2 m, determine (a) the maximum tension in
the cable, (b) the length of the cable.
SOLUTION
w = (0.8 kg/m)(9.81 m/s 2 )
= 7.848 N/m
W = (7.848 N/m)(37.5 m)
W = 294.3 N
(a)
1

ΣM B = 0: T0 (2 m) − W  37.5 m  = 0
2

1
T0 (2 m) − (294.3 N) (37.5 m) = 0
2
T0 = 2759 N
Tm2 = (294.3 N) 2 + (2759 N)2
(b)

sB = xB 1 +


Tm = 2770 N 
2

2  yB 

 + 
3  xB 


 2  2 m 2

= 37.5 m 1+ 
+ 

 3  37.5 m 

= 37.57 m
Length = 2sB = 2(37.57 m)
Length = 75.14 m 
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1152
PROBLEM 7.108
The total mass of cable ACB is 20 kg. Assuming that the mass of the
cable is distributed uniformly along the horizontal, determine (a) the sag h,
(b) the slope of the cable at A.
SOLUTION
Free body: Entire frame
Σ M D = 0: Ax (4.5 m) − (196.2 N)(4 m) − (1471.5 N)(6 m) = 0
Ax = 2136.4 N
Free body: Entire cable
Σ M D = 0: Ay (8 m) − (196.2 N)(4 m) = 0
Ay = 98.1 N
(a)
Free body: Portion AC
Σ Fx = 0: T0 = Ax = 2136.4 N
Σ M A = 0: T0 h − (98.1 N)(2 m) = 0
(2136.4 N)h − 196.2 N ⋅ m = 0
h = 0.09183 m
(b)
tan θ A =
h = 91.8 mm 
Ax
98.1 N
=
Ay 2136.4 N
tan θ A = 0.045918
θ A = 2.629°
θ A = 2.63° 
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1153
PROBLEM 7.109
The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four
cables. The uniform load supported by each cable is w = 10.8 kips/ft along the horizontal. Knowing that the
span L is 4260 ft and that the sag h is 390 ft, determine (a) the maximum tension in each cable, (b) the length
of each cable.
SOLUTION
(a)
yB =
At B:
390 ft =
wxB2
2T0
(10.8 kips/ft)(2130 ft) 2
2T0
T0 = 62,819 kips
Tm = T02 + w2 xB2
= (62,819 kips) 2 + (10.8 kips/ft) 2 (2130 ft) 2
= 62,8192 + (23, 004) 2 = 66,898 kips
(b)
Tm = 66,900 kips 
Length of cable

sB = xB 1 +


2
4
2  yB 
2  yB  
−



 
3  xB 
5  xB  

2
4
 2  390 ft 
2  390 ft  

= (2130 ft) 1 + 
 − 
  = 2176.65 ft
5  2130 ft  
 3  2130 ft 


Total length:
2 S B = 2(2176.65 ft) = 4353.3 ft
L = 4353 ft 
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1154
PROBLEM 7.110
The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four
cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of the
center span to vary from hw = 386 ft in winter to hs = 394 ft in summer. Knowing that the span is L = 4260 ft,
determine the change in length of the cables due to extreme temperature changes.
SOLUTION
Eq. 7.10:
 2  y 2 2  y 4

sB = xB 1 +  B  −  B  + 
 3  xB  5  x B 



Winter:
yB = h = 386 ft, xB =
1
L = 2130 ft
2
 2  386 2 2  386 4

sB = (2130) 1 + 
− 
+  = 2175.715 ft


5  2130 
 3  2130 

Summer:
yB = h = 394 ft, xB =

sB = (2130) 1 +

1
L = 2130 ft
2
2
4

2  394 
2  394 
−
+  = 2177.59 ft




3  2130 
5  2130 

Δ = 2( Δ sB ) = 2(2177.59 ft − 2175.715 ft) = 2(1.875 ft)
Change in length = 3.75 ft 
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1155
PROBLEM 7.111
Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft along the horizontal. Knowing that the
span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length
of each cable.
SOLUTION
Eq. (7.8) Page 386:
At B:
(a)
yB =
wxB2
2T0
T0 =
wxB2 (11.1 kip/ft)(2075 ft)2
=
2 yB
2(464 ft)
T0 = 51.500 kips
W = wxB = (11.1 kips/ft)(2075 ft) = 23.033 kips
Tm = T02 + W 2 = (51.500 kips)2 + (23.033 kips) 2
Tm = 56, 400 kips 
(b)

sB = xB 1 +


2
4

2  yB 
2  yB 

−
+





3  xB  5  y B 


yB
464 ft
=
= 0.22361
xB 2075 ft
2
 2

sB = (2075 ft) 1 + (0.22361) 2 − (0.22361) 4 +  = 2142.1 ft
5
 3

Length = 2sB = 2(2142.1 ft)
Length = 4284 ft 
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1156
PROBLEM 7.112
Two cables of the same gauge are attached to a transmission tower
at B. Since the tower is slender, the horizontal component of the
resultant of the forces exerted by the cables at B is to be zero.
Knowing that the mass per unit length of the cables is 0.4 kg/m,
determine (a) the required sag h, (b) the maximum tension in each
cable.
SOLUTION
W = wxB
ΣM B = 0: T0 yB − ( wxB )
Horiz. comp. = T0 =
(a)
Cable AB:
w(45 m) 2
2h
xB = 30 m,
T0 =
Equate T0 = T0
wxB2
2 yB
xB = 45 m
T0 =
Cable BC:
yB
=0
2
yB = 3 m
w(30 m) 2
2(3 m)
w (45 m) 2 w (30 m) 2
=
2h
2(3 m)
h = 6.75 m 
Tm2 = T02 + W 2
(b)
Cable AB:
w = (0.4 kg/m)(9.81 m/s) = 3.924 N/m
xB = 45 m,
T0 =
yB = h = 6.75 m
wxB2 (3.924 N/m)(45 m)2
=
= 588.6 N
2 yB
2(6.75 m)
W = wxB = (3.924 N/m)(45 m) = 176.58 N
Tm2 = (588.6 N) 2 + (176.58 N)2
Tm = 615 N 
For AB:
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1157
PROBLEM 7.112 (Continued)
Cable BC:
xB = 30 m,
T0 =
yB = 3 m
wxB2 (3.924 N/m)(30 m)2
=
= 588.6 N (Checks)
2 yB
2(3 m)
W = wxB = (3.924 N/m)(30 m) = 117.72 N
Tm2 = (588.6 N) 2 + (117.72 N) 2
Tm = 600 N 
For BC:
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1158
PROBLEM 7.113
A 50.5-m length of wire having a mass per unit length of 0.75 kg/m is used to span a horizontal distance of
50 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use only
the first two terms of Eq. (7.10).]
SOLUTION
First two terms of Eq. 7.10
1
(50.5 m) = 25.25 m,
2
1
xB = (50 m) = 25 m
2
yB = h
sB =
(a)

sB = xB 1 +


2
2  yB  

 
3  xB  

 2  y 2 
25.25 m = 25 m 1 −  B  
 3  xB  


2
 yB 
3

 = 0.01  = 0.015
x
2
 B
yB
= 0.12247
xB
2
h
= 0.12247
25m
h = 3.0619 m
(b)
h = 3.06 m 
Free body: Portion CB
w = (0.75 kg/m) (9.81m) = 7.3575 N/m
W = sB w = (25.25 m) (7.3575 N/m)
W = 185.78 N
ΣM 0 = 0: T0 (3.0619 m) − (185.78 N) (12.5 m) = 0
T0 = 758.4 N
Bx = T0 = 758.4 N
+ ΣFy = 0: By − 185.78 N = 0 By = 185.78 N
Tm = Bx2 + By2 = (758.4 N) 2 + (185.78 N) 2
Tm = 781 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1159
PROBLEM 7.114
A cable of length L + Δ is suspended between two points that are at the same elevation and a distance L apart.
(a) Assuming that Δ is small compared to L and that the cable is parabolic, determine the approximate sag in
terms of L and Δ. (b) If L = 100 ft and Δ = 4 ft, determine the approximate sag. [Hint: Use only the first two
terms of Eq. (7.10).
SOLUTION
Eq. 7.10
(First two terms)
(a)
 2  y 2 
sB = xB 1 +  B  
 3  xB  


xB = L/2
1
( L + Δ)
2
yB = h
sB =

1
L
( L + Δ) = 1 +
2
2

2
2 h  
 
3  L2  

Δ 4 h2
3
=
; h 2 = LΔ;
2 3 L
8
(b)
L = 100 ft, h = 4 ft.
h=
h=
3
(100)(4);
8
3
LΔ 
8
h = 12.25 ft 
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1160
PROBLEM 7.115
The total mass of cable AC is 25 kg. Assuming that the mass of the cable
is distributed uniformly along the horizontal, determine the sag h and the
slope of the cable at A and C.
SOLUTION
Cable:
m = 25 kg
W = 25 (9.81)
= 245.25 N
Block:
m = 450 kg
W = 4414.5 N
ΣM B = 0: (245.25) (2.5) + (4414.5)(3) − C x (2.5) = 0
C x = 5543 N
ΣFx = 0: Ax = C x = 5543 N
ΣM A = 0: C y (5) − (5543) (2.5) − (245.25) (2.5) = 0
C y = 2894 N
+ ΣFy = 0: C y − Ay − 245.25 N = 0
2894 N − Ay − 245.25 N = 0
Point A:
tan θ A =
Ay
Point C:
tan θC =
Cy
Free body: Half cable
Ax
Cx
A y = 2649 N
=
2649
= 0.4779;
5543
θ A = 25.5° 
=
2894
= 0.5221;
5543
θ A = 27.6° 
W = (12.5 kg) g = 122.6
ΣM 0 = 0: (122.6 N) (1.25 M) + (2649 N) (2.5 m) − (5543 N) yd = 0
yd = 1.2224 m; sag = h = 1.25 m − 1.2224 m
h = 0.0276 m = 27.6 mm 
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1161
PROBLEM 7.116
Cable ACB supports a load uniformly distributed along the
horizontal as shown. The lowest Point C is located 9 m to the right
of A. Determine (a) the vertical distance a, (b) the length of the
cable, (c) the components of the reaction at A.
SOLUTION
Free body: Portion AC
ΣFy = 0: Ay − 9 w = 0
A y = 9w
ΣM A = 0: T0 a − (9 w)(4.5 m) = 0
(1)
Free body: Portion CB
ΣFy = 0: By − 6w = 0
B y = 6w
Free body: Entire cable
ΣM A = 0: 15w (7.5 m) − 6 w (15 m) − T0 (2.25 m) = 0
T0 = 10 w
(a)
Eq. (1):
T0 a − (9w)(4.5 m) = 0
10wa = (9w)(4.5) = 0
a = 4.05 m 
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1162
PROBLEM 7.116 (Continued)
(b)
Length = AC + CB
Portion AC:
x A = 9 m,
s AC
s AC
Portion CB:
y A = a = 4.05 m;
y A 4.05
=
= 0.45
9
xA
 2  y 2 2  y 4

= xB 1 +  A  −  B  + 
5  xA 
 3  xA 



2
 2

= 9 m 1+ 0.452 − 0.454 +   = 10.067 m
5
 3

xB = 6 m,
yB = 4.05 − 2.25 = 1.8 m;
yB
= 0.3
xB
2
2


sCB = 6 m 1 + 0.32 − 0.34 +   = 6.341 m
3
5


Total length = 10.067 m + 6.341 m
(c)
Total length = 16.41 m 
Components of reaction at A.
Ay = 9 w = 9(60 kg/m)(9.81 m/s 2 )
= 5297.4 N
Ax = T0 = 10 w = 10(60 kg/m)(9.81 m/s 2 )
= 5886 N
A x = 5890 N

A y = 5300 N 
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1163
PROBLEM 7.117
Each cable of the side spans of the Golden Gate Bridge supports a
load w = 10.2 kips/ft along the horizontal. Knowing that for the side
spans the maximum vertical distance h from each cable to the chord
AB is 30 ft and occurs at midspan, determine (a) the maximum
tension in each cable, (b) the slope at B.
SOLUTION
FBD AB:
ΣM A = 0: (1100 ft)TBy − (496 ft)TBx − (550 ft)W = 0 
11TBy − 4.96TBx = 5.5W
(1)
FBD CB:
ΣM C = 0: (550 ft)TBy − (278 ft)TBx − (275 ft)
W
=0
2
11TBy − 5.56TBx = 2.75W
Solving (1) and (2)
TBy = 28, 798 kips
Solving (1) and (2
TBx = 51, 425 kips 
Tmax = TB = TB2x + TB2y
So that
tan θ B =
TBy
TBx
(2)

(a)
(b)
Tmax = 58,900 kips 
θ B = 29.2° 
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1164
PROBLEM 7.118
A steam pipe weighting 45 lb/ft that passes between two
buildings 40 ft apart is supported by a system of cables as
shown. Assuming that the weight of the cable system is
equivalent to a uniformly distributed loading of 5 lb/ft,
determine (a) the location of the lowest Point C of the
cable, (b) the maximum tension in the cable.
SOLUTION
Note:
xB − x A = 40 ft
or
x A = xB − 40 ft
(a)
Use Eq. 7.8
Point A:
yA =
wx A2
w ( xB − 40) 2
; 9=
2T0
2T0
(1)
Point B:
yB =
wxB2
wx 2
; 4= B
2T0
2T0
(2)
Dividing (1) by (2):
(b)
9 ( xB − 40) 2
; xB = 16 ft
=
4
xB2
Point C is 16.00 ft to left of B 
Maximum slope and thus Tmax is at A
x A = xB − 40 = 16 − 40 = −24 ft
yA =
wx A2
(50 lb/ft)(−24 ft) 2
; 9 ft =
; T0 = 1600 lb
2T0
2T0
WAC = (50 lb/ft)(24 ft) = 1200 lb
Tmax = 2000 lb 
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1165
PROBLEM 7.119*
A cable AB of span L and a simple beam A′B′ of the same span are
subjected to identical vertical loadings as shown. Show that the
magnitude of the bending moment at a point C′ in the beam is equal to
the product T0h, where T0 is the magnitude of the horizontal component
of the tension force in the cable and h is the vertical distance between
Point C and the chord joining the points of support A and B.
SOLUTION
ΣM B = 0: LACy + aT0 − ΣM B loads = 0
(1)
FBD Cable:
(Where ΣM B loads includes all applied loads)
x

ΣM C = 0: xACy −  h − a  T0 − ΣM C left = 0
L


(2)
FBD AC:
(Where ΣM C left includes all loads left of C)
x
x
(1) − (2): hT0 − ΣM B loads + ΣM C left = 0
L
L
(3)
ΣM B = 0: LABy − ΣM B loads = 0
(4)
ΣM C = 0: xABy − ΣM C left − M C = 0
(5)
FBD Beam:
FBD AC:
Comparing (3) and (6)
x
x
(4) − (5): − ΣM B loads + ΣM C left + M C = 0
L
L
M C = hT0
(6)
Q.E.D.
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1166
PROBLEM 7.120
Making use of the property established in Problem 7.119,
solve the problem indicated by first solving the corresponding
beam problem.
PROBLEM 7.94 Knowing that the maximum tension in
cable ABCDE is 13 kN, determine the distance dC.
SOLUTION
Free body: beam AE
ΣM E = 0: − A(16) + 2(12) + 3(8) + 4(4) = 0
ΣFy = 0: 4 − 2 − 3 − 4 + E = 0
At E:
At C:
Tm2 = T02 + E 2
M C = T0 hC ;
132 = T02 + 52
A = 4 kN
E = 5 kN
T0 = 12 kN
24 kN ⋅ m = (12 kN)hC
hC = dC = 2.00 m
2.00 m 
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1167
PROBLEM 7.121
Making use of the property established in Problem 7.119, solve the problem
indicated by first solving the corresponding beam problem.
PROBLEM 7.97 (a) Knowing that dC = 3 m, determine the distances dB
and dD .
SOLUTION
ΣM B = 0: A(10 m) − (5 kN)(8 m) − (5 kN)(6 m) − (10 kN)(3 m) = 0
A = 10 kN
Geometry:
dC = 1.6 m + hC
3 m = 1.6 m + hC
hC = 1.4 m
Since M = T0h, h is proportional to M, thus
h
hB
h
= C = D ;
M B MC M D
hB
hD
1.4 m
=
=
20 kN ⋅ m 30 kN ⋅ m 30 kN ⋅ m
 20 
hB = 1.4   = 0.9333 m
 30 
 30 
hD = 1.4   = 1.4 m
 30 
d B = 0.8 m + 0.9333 m
d D = 2.8 m + 1.4 m
d B = 1.733 m 
d D = 4.20 m 
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1168
PROBLEM 7.122
Making use of the property established in Problem 7.119,
solve the problem indicated by first solving the corresponding
beam problem.
PROBLEM 7.99 An oil pipeline is supported at 6-ft
intervals by vertical hangers attached to the cable shown.
Due to the combined weight of the pipe and its contents the
tension in each hanger is 400 lb. Knowing that dC = 12 ft,
determine (a) the maximum tension in the cable.
SOLUTION
A= B=
1
(4 × 400) = 800 lb
2
Geometry
dC = hC + 3 ft
12 ft = hC + 3 ft
hC = 9 ft
d D = hD + 2 ft
At C:
M C = T0 hC
7200 lb ⋅ ft = T0 (9 ft)
At D:
T0 = 800 lb
M D = T0 hD
7200 lb ⋅ ft = (800 lb)hD
Eq. (1):
(1)
h0 = 9 ft
d D = 9 ft + 2 ft
d D = 11.00 ft 
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1169
PROBLEM 7.123
Making use of the property established in Problem 7.119,
solve the problem indicated by first solving the corresponding
beam problem.
PROBLEM 7.100 An oil pipeline is supported at 6-ft
intervals by vertical hangers attached to the cable shown.
Due to the combined weight of the pipe and its contents the
tension in each hanger is 400 lb. Knowing that dC = 9 ft,
determine (b) the distance dD.
SOLUTION
1
(4 × 400)
2
A = B = 800 lb
A= B=
At any point:
M = T0 h
We note that since M C = M D , we have hC = hD
Geometry
dC = hC + 3 ft
9 ft = hC + 3 ft
hC = 6 ft
and
hD = 6 ft
d D = hD + 2 ft = 6 ft + 2 ft
d D = 8.00 ft 
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1170
PROBLEM 7.124*
Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential
equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.
SOLUTION
FBD Elemental segment:
ΣFy = 0: Ty ( x + Δ x) − Ty ( x) − w( x)Δ x = 0
So
Ty ( x + Δ x )
T0
−
Ty ( x )
Ty
But
So
In lim :
Δ x →0
T0
T0
dy
dx
−
x +Δx
Δx
dy
dx
x
=
w( x)
Δx
T0
=
dy
dx
=
w( x)
T0
d 2 y w( x)
=
T0
dx 2
Q.E.D.
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1171
PROBLEM 7.125*
Using the property indicated in Problem 7.124, determine the curve assumed by a cable of span L and sag h
carrying a distributed load w = w0 cos (π x/L), where x is measured from mid-span. Also determine the
maximum and minimum values of the tension in the cable.
PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by
the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.
SOLUTION
w( x) = w0 cos
πx
L
From Problem 7.124
πx
d 2 y w( x) w0
=
=
cos
2
T0
T0
L
dx
So


dy
= 0
 using
dx 0


πx
dy W0 L
=
sin
dx T0π
L
y=
But
w L2
L
y  = h = 0 2
T0π
2
And
T0 = Tmin
w0 L2
π

1 − cos 2  so T0 = 2
π h


Tmin =
so
Tmax = TA = TB :
TBy =
w0 L2 
πx
1 − cos
[using y (0) = 0] 
2 
L 
T0π 
TBy
T0
=
w0 L
dy
dx
=
x = L/2
π 2h

w0 L
T0π
2
+ T02 =
TB = TBy
π
w0 L2
w0 L
π
 L 
1+ 

πh 
2

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1172
PROBLEM 7.126*
If the weight per unit length of the cable AB is w0/cos2 θ, prove that the curve
formed by the cable is a circular arc. (Hint: Use the property indicated in
Problem 7.124.)
PROBLEM 7.124 Show that the curve assumed by a cable that carries a
distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0,
where T0 is the tension at the lowest point.
SOLUTION
Elemental Segment:
Load on segment*
w ( x)dx =
w0
cos 2 θ
ds
dx = cos θ ds, so w( x) =
But
w0
cos3 θ
From Problem 7.119
w0
d 2 y w( x)
=
=
2
T0
dx
T0 cos3 θ
In general
d 2 y d  dy  d
dθ
=   = (tan θ ) = sec2 θ
2
dx
dx
dx
dx
dx
 
So
or
Giving r =
w0
w0
dθ
=
=
dx T0 cos3 θ sec 2 θ T0 cos θ
T0
cos θ dθ = dx = rdθ cos θ
w0
T0
= constant. So curve is circular arc
w0
Q.E.D.
*For large sag, it is not appropriate to approximate ds by dx.
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1173
PROBLEM 7.127
A 20-m chain of mass 12 kg is suspended between two points at the same elevation. Knowing that the sag is
8 m, determine (a) the distance between the supports, (b) the maximum tension in the chain.
SOLUTION
mass/meter = (12 kg)/(20 m) = 0.6 kg/m
w = (0.6 kg/m)(9.81 m/s 2 ) = 5.886 N/m
Eq. 7.17:
yB2 − sB2 = c 2 ; (8 + c)2 − 102 = c 2
64 + 16c + c 2 − 100 = c 2
16c = 36 c = 2.25 m
Eq. 7.18:
Tm = wyB = (5.886 N/m)(8 m + 2.25 m)
Tm = 60.33 N
Eq. 7.15:
Tm = 60.3 N 
xB
x
; 10 = (2.25 m) sinh B
c
c
xB
xB
sinh
= 4.444;
= 2.197
c
c
sB = c sinh
xB = 2.197(2.25 m) = 4.944 m; L = 2 xB = 2(4.944 m) = 9.888 m
L = 9.89 m 
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1174
PROBLEM 7.128
A 600-ft-long aerial tramway cable having a weight per unit length of 3.0 lb/ft is suspended between two
points at the same elevation. Knowing that the sag is 150 ft, find (a) the horizontal distance between the
supports, (b) the maximum tension in the cable.
SOLUTION
Given:
Length = 600 ft
Unit mass = 3.0 lb/ft
h = 150 ft
sB = 300 ft
Then,
yB = h + c = 150 ft + c
yB2
− sB2 = c 2 ; (150 + c) 2 − (300) 2 = c 2
1502 + 300c + c 2 − 3002 = c 2
c = 225 ft
sB = c sinh
xB
x
; 300 = 225sinh B
225
c
xB = 247.28 ft
span = L = 2 xB = 2(247.28 ft)
Tm = wyB = (3 lb/ft)(150 ft + 225 ft)
L = 495 ft 
Tm = 1125 lb 
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1175
PROBLEM 7.129
A 40-m cable is strung as shown between two buildings. The
maximum tension is found to be 350 N, and the lowest point of
the cable is observed to be 6 m above the ground. Determine
(a) the horizontal distance between the buildings, (b) the total
mass of the cable.
SOLUTION
sB = 20 m
Tm = 350 N
h = 14 m − 6 m = 8 m
yB = h + c = 8 m + c
Eq. 7.17:
yB2 − sB2 = c 2 ; (8 + c)2 − 202 = c 2
64 + 16c + c 2 − 400 = c 2
c = 21.0 m
xB
x
; 20 = (21.0)sinh B
21.0
c
Eq. 7.15:
sB = c sinh
(a)
xB = 17.7933 m; L = 2 xB
(b)
Eq. 7.18:
L = 35.6 m 
Tm = wyB ; 350 N = w(8 + 21.0)
w = 12.0690 N/m
W = 2sB w = (40 m)(12.0690 N/m)
= 482.76 N
W 482.76 N
m=
=
g 9.81 m/s 2
Total mass = 49.2 kg 
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1176
PROBLEM 7.130
A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched between two points at the same elevation and
pulled until the tension at each end is 16 lb, determine the horizontal distance between the ends of the tape.
Neglect the elongation of the tape due to the tension.
SOLUTION
sB = 100 ft
 4 lb 
w=
 = 0.02 lb/ft Tm = 16 m
 200 ft 
Eq. 7.18:
Eq. 7.17:
Eq. 7.15:
Tm = wyB ; 16 lb = (0.02 lb/ft) yB ;
yB = 800 ft
yB2 − sB2 = c 2 ; (800)2 − (100) 2 = c 2 ; c = 793.73 ft
sB = c sinh
xB
x
; 100 = 793.73 sinh B
c
c
xB
= 0.12566; xB = 99.737 ft
c
L = 2 xB = 2(99.737 ft)
L = 199.5 ft 
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1177
PROBLEM 7.131
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the force P for which h = 8 m, (b) the
corresponding span L.
SOLUTION
FBD Cable:
20 m


sT = 20 m  so sB =
= 10 m 
2


w = (0.2 kg/m)(9.81 m/s 2 )
= 1.96200 N/m
hB = 8 m
yB2 = (c + hB )2 = c 2 + sB2
So
c=
sB2 − hB2
2hB
c=
(10 m)2 − (8 m)2
2(8 m)
= 2.250 m
Now
xB
s
→ xB = c sinh −1 B
c
c


10 m
= (2.250 m)sinh −1 

2.250
m

sB = c sinh
xB = 4.9438 m
P = T0 = wc = (1.96200 N/m)(2.250 m)
L = 2 xB = 2(4.9438 m)
(a)
(b)
P = 4.41 N

L = 9.89 m 
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1178
PROBLEM 7.132
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Knowing that the
magnitude of the horizontal force applied to the collar is P = 20 N,
determine (a) the sag h, (b) the span L.
SOLUTION
FBD Cable:
sT = 20 m, w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m
P = T0 = wc c =
c=
P
w
20 N
= 10.1937 m
1.9620 N/m
yB2 = (hB + c) 2 = c 2 + sB2
h 2 + 2ch − sB2 = 0 sB =
20 m
= 10 m
2
h 2 + 2(10.1937 m)h − 100 m 2 = 0
h = 4.0861 m
sB = c sinh
(a)
h = 4.09 m 
xA
s
 10 m 
→ xB = c sinh −1 B = (10.1937 m)sinh −1 

c
c
 10.1937 m 
= 8.8468 m
L = 2 xB = 2(8.8468 m)
(b)
L = 17.69 m 
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1179
PROBLEM 7.133
A 20-m length of wire having a mass per unit length of 0.2 kg/m is
attached to a fixed support at A and to a collar at B. Neglecting the
effect of friction, determine (a) the sag h for which L = 15 m, (b) the
corresponding force P.
SOLUTION
FBD Cable:
20 m
= 10 m
2
w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m
L = 15 m
sT = 20 m → sB =
L
xB
= c sinh 2
c
c
7.5 m
10 m = c sinh
c
sB = c sinh
Solving numerically:
c = 5.5504 m
x
yB = c cosh  B
 c
yB = 11.4371 m

 7.5 
 = (5.5504) cosh  5.5504 



hB = yB − c = 11.4371 m − 5.5504 m
(a)
P = wc = (1.96200 N/m)(5.5504 m)
(b)
hB = 5.89 m 
P = 10.89 N

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1180
PROBLEM 7.134
Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.
SOLUTION
30 ft
= 15 ft
2
L
xB = = 10 ft
2
x
sB = c sinh B
c
10 ft
15 ft = c sinh
c
sB =
Solving numerically:
L = 20 ft
c = 6.1647 ft
yB = c cosh
xB
c
= (6.1647 ft)cosh
10 ft
6.1647 ft
= 16.2174 ft
hB = yB − c = 16.2174 ft − 6.1647 ft
hB = 10.05 ft 
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1181
PROBLEM 7.135
A 10-ft rope is attached to two supports A and B as shown. Determine (a) the
span of the rope for which the span is equal to the sag, (b) the corresponding
angle θB.
SOLUTION
Note: Since L = h,
xB =
Eq. 7.16:
L h
=
2 2
xB
c
h /2
h + c = c cosh
c
h
1 h
+ 1 = cosh 

c
2 c
yB = cosh
Solve for h/c:
h
= 4.933
c
Eq. 7.16:
y = c cosh
x
c
dy
x
= sinh
dx
c
At B:
Substitute
tan θ B =
dy
dx
= sinh
B
xB
c
h
1 h
1

xB = : tan θ B = sinh 
= sinh  × 4.933 

2
2 c
2

tan θ B = 5.848
θ B = 80.3° 
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1182
PROBLEM 7.135 (Continued)
Eq. 7.17:
Recall that
xB
1 h
= c sinh 

c
2 c
1

5 ft = c sinh  × 4.933 
2

5 ft = c(5.848)
c = 0.855
sB = c sinh
h
= 4.933
c
h = 4.933(0.855) = 4.218
h = 4.22 ft 
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1183
PROBLEM 7.136
A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the
maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.
SOLUTION
sB = 45 m
Eq. 7.17:
Eq. 7.16:
xB
c
30
45 = c sinh ; c = 18.494 m
c
sB = c sinh
yB = c cosh
xB
c
yB = (18.494) cosh
30
18.494
yB = 48.652 m
yB = h + c
48.652 = h + 18.494
h = 30.158 m
Eq. 7.18:
h = 30.2 m 
Tm = wyB
300 N = w(48.652 m)
w = 6.166 N/m
Total weight of cable
W = w(Length)
= (6.166 N/m)(90 m)
= 554.96 N
Total mass of cable
m=
W 554.96 N
=
= 56.57 kg
9.81 m/s
g
m = 56.6 kg 
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1184
PROBLEM 7.137
A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart.
Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.
SOLUTION
Eq. 7.18:
Eq. 7.16:
Tm = wyB ; 400 lb = (2 lb/ft) yB ;
yB = 200 ft
xB
c
80 ft
200 ft = c cosh
c
yB = c cosh
Solve for c: c = 182.148 ft and c = 31.592 ft
yB = h + c; h = yB − c
For
c = 182.148 ft;
h = 200 − 182.147 = 17.852 ft 
For
c = 31.592 ft;
h = 200 − 31.592 = 168.408 ft 
For
Tm ≤ 400 lb:
smallest h = 17.85 ft 
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1185
PROBLEM 7.138
A uniform cord 50 in. long passes over a pulley at B and is attached to a
pin support at A. Knowing that L = 20 in. and neglecting the effect of
friction, determine the smaller of the two values of h for which the cord
is in equilibrium.
SOLUTION
b = 50 in. − 2sB
Length of overhang:
Weight of overhang equals max. tension
Tm = TB = wb = w(50 in. − 2sB )
Eq. 7.15:
sB = c sinh
xB
c
Eq. 7.16:
yB = c cosh
xB
c
Eq. 7.18:
Tm = wyB
w(50 in. − 2sB ) = wyB
x 
x

w  50 in. − 2c sinh B  = wc cosh B
c 
c

xB = 10: 50 − 2c sinh
Solve by trial and error:
For c = 5.549 in.
c = 5.549 in.
and
10
10
= c cosh
c
c
c = 27.742 in.
10 in.
= 17.277 in.
5.549 in.
yB = h + c; 17.277 in. = h + 5.549 in.
yB = (5.549 in.) cosh
h = 11.728 in.
For c = 27.742 in.
h = 11.73 in. 
10 in.
= 29.564 in.
27.742 in.
yB = h + c; 29.564 in. = h + 27.742 in.
yB = (27.742 in.) cosh
h = 1.8219 in.
h = 1.822 in. 
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1186
PROBLEM 7.139
A motor M is used to slowly reel in the cable shown. Knowing that the
mass per unit length of the cable is 0.4 kg/m, determine the maximum
tension in the cable when h = 5 m.
SOLUTION
w = 0.4 kg/m L = 10 m hB = 5 m
xB
c
L
hB + c = c cosh
2c
5m 

5 m = c  cosh
− 1
c


yB = c cosh
Solving numerically:
c = 3.0938 m
yB = hB + c = 5 m + 3.0938 m
Tmax
= 8.0938 m
= TB = wyB
= (0.4 kg/m)(9.81 m/s 2 )(8.0938 m)
Tmax = 31.8 N 
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1187
PROBLEM 7.140
A motor M is used to slowly reel in the cable shown. Knowing that the
mass per unit length of the cable is 0.4 kg/m, determine the maximum
tension in the cable when h = 3 m.
SOLUTION
w = 0.4 kg/m, L = 10 m, hB = 3 m
xB
L
= c cosh
c
2c
5m 

− 1
3 m = c  c cosh
c


yB = hB + c = c cosh
Solving numerically:
c = 4.5945 m
yB = hB + c = 3 m + 4.5945 m
Tmax
= 7.5945 m
= TB = wyB
= (0.4 kg/m)(9.81 m/s 2 )(7.5945 m)
Tmax = 29.8 N 
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1188
PROBLEM 7.141
The cable ACB has a mass per unit length of 0.45 kg/m. Knowing
that the lowest point of the cable is located at a distance a = 0.6 m
below the support A, determine (a) the location of the lowest
Point C, (b) the maximum tension in the cable.
SOLUTION
xB − x A = 12 m
Note:
− x A = 12 m − xB
or,
Point A:
y A = c cosh
− xA
12 − xB
; c + 0.6 = c cosh
c
c
(1)
Point B:
yB = c cosh
xB
x
; c + 2.4 = c cosh B
c
c
(2)
From (1):
12 xB
 c + 0.6 
−
= cosh −1 

c
c
 c 
(3)
From (2):
xB
 c + 2.4 
= cosh −1 

c
 c 
(4)
Add (3) + (4):
12
 c + 0.6 
 c + 2.4 
= cosh −1 
+ cosh −1 


c
 c 
 c 
c = 13.6214 m
Solve by trial and error:
Eq. (2):
13.6214 + 2.4 = 13.6214 cosh
cosh
xB
= 1.1762;
c
xB
c
xB
= 0.58523
c
xB = 0.58523(13.6214 m) = 7.9717 m
Point C is 7.97 m to left of B 
yB = c + 2.4 = 13.6214 + 2.4 = 16.0214 m
Eq. 7.18:
Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(16.0214 m)
Tm = 70.726 N
Tm = 70.7 N 
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1189
PROBLEM 7.142
The cable ACB has a mass per unit length of 0.45 kg/m.
Knowing that the lowest point of the cable is located at a
distance a = 2 m below the support A, determine (a) the
location of the lowest Point C, (b) the maximum tension in
the cable.
SOLUTION
Note:
xB − x A = 12 m
or
− x A = 12 m − xB
Point A:
y A = c cosh
− xA
12 − xB
; c + 2 = c cosh
c
c
(1)
Point B:
yB = c cosh
xB
x
; c + 3.8 = c cosh B
c
c
(2)
12 xB
c+2
−
= cosh −1 

c
c
 c 
From (1):
(3)
From (2):
xB
 c + 3.8 
= cosh −1 

c
 c 
Add (3) + (4):
12
c+2
 c + 3.8 
= cosh −1 
+ cosh −1 


c
 c 
 c 
c = 6.8154 m
Solve by trial and error:
Eq. (2):
(4)
6.8154 m + 3.8 m = (6.8154 m) cosh
cosh
xB
c
xB
xB
= 1.5576
= 1.0122
c
c
xB = 1.0122(6.8154 m) = 6.899 m
Point C is 6.90 m to left of B 
yB = c + 3.8 = 6.8154 + 3.8 = 10.6154 m
Eq. (7.18):
Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(10.6154 m)
Tm = 46.86 N
Tm = 46.9 N 
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1190
PROBLEM 7.143
A uniform cable weighing 3 lb/ft is held in the position shown by a
horizontal force P applied at B. Knowing that P = 180 lb and θA = 60°,
determine (a) the location of Point B, (b) the length of the cable.
SOLUTION
T0 = P = cw
Eq. 7.18:
c=
c = 60 ft
P
cos 60°
cw
=
= 2cw
0.5
Tm =
At A:
(a)
P 180 lb
=
w 3 lb/ft
Tm = w(h + c)
Eq. 7.18:
2cw = w(h + c)
2c = h + c h = b = c
b = 60.0 ft 
xA
c
x
h + c = c cosh A
c
y A = c cosh
Eq. 7.16:
(60 ft + 60 ft) = (60 ft) cosh
cosh
xA
=2
60 m
xA
60
xA
= 1.3170
60 m
x A = 79.02 ft
(b)
Eq. 7.15:
a = 79.0 ft 
xB
79.02 ft
= (60 ft)sinh
60 ft
c
s A = 103.92 ft
s A = c sinh
length = s A
s A = 103.9 ft 
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1191
PROBLEM 7.144
A uniform cable weighing 3 lb/ft is held in the position shown by a
horizontal force P applied at B. Knowing that P = 150 lb and θA = 60°,
determine (a) the location of Point B, (b) the length of the cable.
SOLUTION
Eq. 7.18:
T0 = P = cw
c=
P
cos 60°
cw
=
= 2 cw
0.5
Tm =
At A:
(a)
P 150 lb
=
= 50 ft
w 3 lb/ft
Tm = w(h + c)
Eq. 7.18:
2cw = w(h + c)
2c = h + c h = c = b
Eq. 7.16:
xA
c
xA
h + c = c cosh
c
y A = c cosh
(50 ft + 50 ft) = (50 ft) cosh
cosh
xA
=2
c
xA
c
xA
= 1.3170
c
x A = 1.3170(50 ft) = 65.85 ft
(b)
Eq. 7.15:
b = 50.0 ft 
s A = c sinh
a = 65.8 ft 
xA
65.85 ft
= (50 ft)sinh
50 ft
c
s A = 86.6 ft
length = s A = 86.6 ft 
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1192
PROBLEM 7.145
To the left of Point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the mass per unit length of
the cable is 2 kg/m, determine the force F when a = 3.6 m.
SOLUTION
xD = a = 3.6 m h = 4 m
xD
c
a
h + c = c cosh
c
3.6 m 

− 1
4 m = c  cosh
c


yD = c cosh
Solving numerically:
Then
c = 2.0712 m
yB = h + c = 4 m + 2.0712 m = 6.0712 m
F = Tmax = wyB = (2 kg/m)(9.81 m/s 2 )(6.0712 m)
F = 119.1 N

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1193
PROBLEM 7.146
To the left of Point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the mass per unit length of the
cable is 2 kg/m, determine the force F when a = 6 m.
SOLUTION
xD = a = 6 m h = 4 m
xD
c
a
h + c = c cosh
c
6m 

− 1
4 m = c  cosh
c


y D = c cosh
Solving numerically:
c = 5.054 m
yB = h + c = 4 m + 5.054 m = 9.054 m
F = TD = wyD = (2 kg/m)(9.81 m/s 2 )(9.054 m)
F = 177.6 N

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1194
PROBLEM 7.147*
The 10-ft cable AB is attached to two collars as shown. The collar at
A can slide freely along the rod; a stop attached to the rod prevents
the collar at B from moving on the rod. Neglecting the effect of
friction and the weight of the collars, determine the distance a.
SOLUTION
Collar at A: Since μ = 0, cable ⬜ rod
Point A:
x dy
x
= sinh
y = c cosh ;
c
dx
c
xA
dy
tan θ =
= sinh
c
dx A
xA
= sinh(tan (90° − θ ))
c
x A = c sinh(tan (90° − θ ))
Length of cable = 10 ft
(1)
10 ft = AC + CB
xA
x
+ c sinh B
c
c
xB 10
xA
sinh
=
− sinh
c
c
c
10 = c sinh
x 
10
xB = c sinh −1  − sinh A 
c 
c
x
x
y A = c cosh A yB = c cosh B
c
c
In Δ ABD:
tan θ =
yB − y A
xB + x A
(2)
(3)
(4)
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1195
PROBLEM 7.147* (Continued)
Method of solution:
For given value of θ, choose trial value of c and calculate:
From Eq. (1): xA
Using value of xA and c, calculate:
From Eq. (2): xB
From Eq. (3): yA and yB
Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ
Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ.
For θ = 30°
Result of trial and error procedure:
c = 1.803 ft
x A = 2.3745 ft
xB = 3.6937 ft
y A = 3.606 ft
yB = 7.109 ft
a = yB − y A
= 7.109 ft − 3.606 ft
= 3.503 ft
a = 3.50 ft 
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1196
PROBLEM 7.148*
Solve Problem 7.147 assuming that the angle θ formed by the rod and
the horizontal is 45°.
PROBLEM 7.147 The 10-ft cable AB is attached to two collars as
shown. The collar at A can slide freely along the rod; a stop attached to
the rod prevents the collar at B from moving on the rod. Neglecting
the effect of friction and the weight of the collars, determine the
distance a.
SOLUTION
Collar at A: Since μ = 0, cable ⬜ rod
Point A:
x dy
x
= sinh
y = c cosh ;
c
dx
c
x
dy
tan θ =
= sinh A
c
dx A
xA
= sinh(tan (90° − θ ))
c
x A = c sinh(tan (90° − θ ))
Length of cable = 10 ft
(1)
10 ft = AC + CB
x
x
10 = c sinh A + c sinh B
c
c
xB 10
xA
sinh
=
− sinh
c
c
c
x 
10
xB = c sinh −1  − sinh A 
c 
c
y A = c cosh
xA
c
yB = c cosh
(2)
xB
c
(3)
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1197
PROBLEM 7.148* (Continued)
In Δ ABD:
tan θ =
yB − y A
xB + x A
(4)
Method of solution:
For given value of θ, choose trial value of c and calculate:
From Eq. (1): xA
Using value of xA and c, calculate:
From Eq. (2): xB
From Eq. (3): yA and yB
Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ
Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ.
For θ = 45°
Result of trial and error procedure:
c = 1.8652 ft
x A = 1.644 ft
xB = 4.064 ft
y A = 2.638 ft
yB = 8.346 ft
a = yB − y A
= 8.346 ft − 2.638 ft
= 5.708 ft
a = 5.71 ft 
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1198
PROBLEM 7.149
Denoting by θ the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan θ,
(b) y = c sec θ.
SOLUTION
dy
x
= sinh
dx
c
x
s = c sinh = c tan θ
c
tan θ =
(a)
(b)
Q.E.D.
Also
y 2 = s 2 + c 2 (cosh 2 x = sinh 2 x + 1)
so
y 2 = c 2 (tan 2 θ + 1)c 2 sec2 θ
and
y = c secθ
Q.E.D.
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1199
PROBLEM 7.150*
(a) Determine the maximum allowable horizontal span for a uniform cable of weight per unit length w if the
tension in the cable is not to exceed a given value Tm. (b) Using the result of part a, determine the maximum
span of a steel wire for which w = 0.25 lb/ft and Tm = 8000 lb.
SOLUTION
Tm = wyB
(a)
xB
c

x
1 
 cosh B
xB 
c

c 
= wc cosh


= wxB 


We shall find ratio
( ) for when T
xB
c
m
is minimum
2







d Tm
x
xB 
1
1
= wxB 
=0
sinh B − 
cosh

 xB
c  xB 
c 
 xB 
d 


 c

 c 
 c 


xB
sinh
c = 1
x
xB
cosh B
c
c
xB
c
=
tanh
c
xB
Solve by trial and error for:
sB = c sinh
Eq. 7.17:
xB
= c sinh(1.200):
c
xB
= 1.200
c
(1)
sB
= 1.509
c
yB2 − sB2 = c 2
  s 2 
yB2 = c 2 1 +  B   = c 2 (1 + 1.5092 )
  c  
yB = 1.810c
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1200
PROBLEM 7.150* (Continued)
Eq. 7.18:
Tm = wyB
= 1.810 wc
Tm
c=
1.810 w
Eq. (1):
Span:
(b)
xB = 1.200c = 1.200
Tm
T
= 0.6630 m
w
1.810 w
L = 2 xB = 2(0.6630)
Tm
w
L = 1.326
Tm

w
For w = 0.25lb/ft and Tm = 8000 lb
8000 lb
0.25lb/ft
= 42, 432ft
L = 1.326
L = 8.04 miles 
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1201
PROBLEM 7.151*
A cable has a mass per unit length of 3 kg/m and is supported
as shown. Knowing that the span L is 6 m, determine the two
values of the sag h for which the maximum tension is 350 N.
SOLUTION
L
=h+c
2c
w = (3 kg/m)(9.81 m/s 2 ) = 29.43 N/m
ymax = c cosh
Tmax = wymax
Tmax
w
350 N
=
= 11.893 m
29.43 N/m
ymax =
ymax
c cosh
Solving numerically:
3m
= 11.893 m
c
c1 = 0.9241 m
c2 = 11.499 m
h = ymax − c
h1 = 11.893 m − 0.9241 m
h1 = 10.97 m 
h2 = 11.893 m − 11.499 m
h2 = 0.394 m 
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1202
PROBLEM 7.152*
Determine the sag-to-span ratio for which the maximum
tension in the cable is equal to the total weight of the entire
cable AB.
SOLUTION
Tmax = wyB = 2wsB
y B = 2 sB
L
2c
L
tanh
2c
L
2c
hB
c
c cosh
= 2c sinh
=
L
2c
1
2
1
= 0.549306
2
y −c
L
= B
= cosh
− 1 = 0.154701
2c
c
= tanh −1
h
B
hB
0.5(0.154701)
= cL =
= 0.14081
0.549306
L 2 ( 2c )
hB
= 0.1408 
L
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1203
PROBLEM 7.153*
A cable of weight per unit length w is suspended between
two points at the same elevation that are a distance L apart.
Determine (a) the sag-to-span ratio for which the maximum
tension is as small as possible, (b) the corresponding values
of θB and Tm.
SOLUTION
L
2c
L
L
L

= w  cosh
− sinh 
2
2
2
c
c
c

Tmax = wyB = wc cosh
(a)
dTmax
dc
For
min Tmax ,
dTmax
=0
dc
tanh
L 2c
L
=
→
= 1.1997
2c L
2c
yB
L
= cosh
= 1.8102
2c
c
h yB
=
− 1 = 0.8102
c
c
0.8102
h  1 h  2c  
=
= 0.3375
=


L  2 c  L   2(1.1997)
T0 = wc Tmax = wc cosh
(b)
But
T0 = Tmax cos θ B
So
θ B = sec −1 
 yB
 c
Tmax = wyB = w
L
2c
h
= 0.338 
L
Tmax
L yB
= cosh
=
T0
c
2c
Tmax
= sec θ B
T0

−1
 = sec (1.8102) = 56.46°

yB  2c  L 
L
 = w(1.8102) 2(1.1997)
c  L 
2
 
θ B = 56.5° 
Tmax = 0.755wL 
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1204
PROBLEM 7.154
Determine the internal forces at Point J of the structure shown.
SOLUTION
FBD ABC:
ΣM D = 0: (0.375 m)(400 N) − (0.24 m)C y = 0
C y = 625 N
ΣM B = 0: − (0.45 m)Cx + (0.135 m)(400 N) = 0
C x = 120 N
FBD CJ:
ΣFy = 0: 625 N − F = 0
ΣFx = 0: 120 N − V = 0
F = 625 N 
V = 120.0 N

M = 27.0 N ⋅ m

ΣM J = 0: M − (0.225 m)(120 N) = 0
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1205
PROBLEM 7.155
Determine the internal forces at Point K of the structure shown.
SOLUTION
FBD AK:
ΣFx = 0: V = 0
V=0 
ΣFy = 0: F − 400 N = 0
F = 400 N 
ΣM K = 0: (0.135 m)(400 N) − M = 0
M = 54.0 N ⋅ m

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1206
PROBLEM 7.156
An archer aiming at a target is pulling with a 45-lb force on the bowstring.
Assuming that the shape of the bow can be approximated by a parabola,
determine the internal forces at Point J.
SOLUTION
FBD Point A:
By symmetry
T1 = T2
3 
ΣFx = 0: 2  T1  − 45 lb = 0 T1 = T2 = 37.5 lb
5 
Curve CJB is parabolic: x = ay 2
FBD BJ:
At B : x = 8 in.
y = 32 in.
8 in.
1
a=
=
2
128
in.
(32 in.)
x=
y2
128
Slope of parabola = tan θ =
At J:
So
dx 2 y
y
=
=
dy 128 64
 16 
 = 14.036°
 64 
θ J = tan −1 
α = tan −1
4
− 14.036° = 39.094°
3
ΣFx′ = 0: V − (37.5 lb) cos(39.094°) = 0
V = 29.1 lb
14.04° 
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1207
PROBLEM 7.156 (Continued)
ΣFy ′ = 0: F + (37.5 lb) sin (39.094°) = 0
F = − 23.647
F = 23.6 lb
76.0° 
3

4

ΣMJ = 0: M + (16 in.)  (37.5 lb)  + [(8 − 2) in.]  (37.5 lb)  = 0
5
5




M = 540 lb ⋅ in.

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1208
PROBLEM 7.157
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point J
of the frame shown.
SOLUTION
Free body: Frame and pulleys
ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0
Bx = − 520 N
B x = 520 N
䉰
A x = 520 N
䉰
ΣFx = 0: Ax − 520 N = 0
Ax = + 520 N
ΣFy = 0: Ay + By − 360 N = 0
Ay + By = 360 N
(1)
Free body: Member AE
ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0
Ay = − 240 N
From (1):
A y = 240 N 䉰
By = 360 N + 240 N
B y = 600 N 䉰
By = + 600 N
Free body: BJ
We recall that the forces applied to a pulley may be applied directly to its axle.
ΣFy = 0:
3
4
(600 N) + (520 N)
5
5
3
− 360 N − (360 N) − F = 0
5
F = + 200 N
F = 200 N
36.9° 
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1209
PROBLEM 7.157 (Continued)
ΣFx = 0:
4
3
4
(600 N) − (520 N) − (360 N) + V = 0
5
5
5
V = + 120.0 N
V = 120.0 N
53.1° 
ΣM J = 0: (520 N)(1.2 m) − (600 N)(1.6 m) + (360 N)(0.6 m) + M = 0
M = + 120.0 N ⋅ m
M = 120.0 N ⋅ m

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1210
PROBLEM 7.158
For the beam shown, determine (a) the magnitude P of the two upward
forces for which the maximum absolute value of the bending moment in the
beam is as small as possible, (b) the corresponding value of |M |max .
SOLUTION
By symmetry:
Ay = B = 60 kips − P
Along AC:
ΣM J = 0: M − x(60 kips − P) = 0
M = (60 kips − P) x
M = 120 kips ⋅ ft − (2 ft) P at
x = 2 ft
Along CD:
ΣM K = 0: M + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0
M = 120 kip ⋅ ft − Px
M = 120 kip ⋅ ft − (4 ft) P at
x = 4 ft
Along DE:
ΣM L = 0: M − ( x − 4 ft) P + ( x − 2 ft)(60 kips)
− x(60 kips − P) = 0
M = 120 kip ⋅ ft − (4 ft) P (const)
Complete diagram by symmetry
For minimum |M |max , set M max = − M min
120 kip ⋅ ft − (2 ft) P = − [120 kip ⋅ ft − (4 ft) P]
M min = 120 kip ⋅ ft − (4 ft) P
(a)
P = 40.0 kips 
(b)
| M |max = 40.0 kip ⋅ ft 
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1211
PROBLEM 7.159
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
ΣM A = 0: (8)(2) + (4)(3.2) − 4C = 0
C = 7.2 kN
ΣFy = 0: A = 4.8 kN
(a)
Shear diagram
Similar Triangles:
x
3.2 − x 3.2
=
=
; x = 2.4 m
4.8
1.6
6.4
↑
Add num. & den.
Bending-moment diagram
(b)
|V | max = 7.20 kN 

|M | max = 5.76 kN ⋅ m 
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1212
PROBLEM 7.160
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
Free body: Beam
ΣM B = 0: 6 kip ⋅ ft + 12 kip ⋅ ft + (2 kips)(18 ft)
+ (3 kips)(12 ft) + (4 kips)(6 ft) − Ay (24 ft) = 0
Ay = + 4.75 kips 䉰
ΣFx = 0: Ax = 0
Shear diagram
At A:
VA = Ay = + 4.75 kips
|V |max = 4.75 kips 
Bending-moment diagram
At A:
M A = − 6 kip ⋅ ft
| M |max = 39.0 kip ⋅ ft 
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1213
PROBLEM 7.161
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the maximum absolute
value of the bending moment.
SOLUTION
Free body: Entire beam
ΣM A = 0: (6 kips)(10 ft) − (9 kips)(7 ft) + 8(4 ft) = 0
B = + 0.75 kips
B = 0.75 kips
ΣFy = 0: A + 0.75 kips − 9 kips + 6 kips = 0
A = + 2.25 kips
A = 2.25 kips
M max = 12.00 kip ⋅ ft 
6.00 ft from A
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1214
PROBLEM 7.162
The beam AB, which lies on the ground, supports the parabolic
load shown. Assuming the upward reaction of the ground to be
uniformly distributed, (a) write the equations of the shear and
bending-moment curves, (b) determine the maximum bending
moment.
SOLUTION
ΣFy = 0: wg L −
(a)

L
0
4 w0
L2
( Lx − x 2 ) dx = 0
4 w0  1 2 1 3  2
 LL − 3 L  = 3 w0 L
L2  2

2w
wg = 0
3
wg L =
x
dx
so d ξ =
L
L
 x  x 2  2
net load w = 4 w0  −    − w0
 L  L   3
Define
ξ=
or
 1

w = 4 w0  − + ξ − ξ 2 
6


 1

4w0 L  − + ξ − ξ 2  d ξ
 6

1
1 
2
1
= 0 + 4w0 L  ξ + ξ 2 − ξ 3 
V = w0 L (ξ − 3ξ 2 + 2ξ 3 ) 
6
2
3
3


x
ξ
2
M = M 0 + Vdx = 0 + w0 L2 (ξ − 3ξ 2 + 2ξ 3 )d ξ
0
0
3
2
1  1
1
= w0 L2  ξ 2 − ξ 3 + ξ 4  = w0 L2 (ξ 2 − 2ξ 3 + ξ 4 )

3
2
2  3

V = V (0) −

ξ
0

(b)
Max M occurs where
V =0

1 − 3ξ + 2ξ 2 = 0
1 1

M  ξ =  = w0 L2
2 3

ξ=
1
2
2
 1 2 1  w0 L
−
+
=


48
 4 8 16 
M max =
w0 L2
at center of beam 
48
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1215
PROBLEM 7.163
Two loads are suspended as shown from the cable ABCD.
Knowing that d B = 1.8 m, determine (a) the distance dC , (b) the
components of the reaction at D, (c) the maximum tension in the
cable.
SOLUTION
ΣFx = 0: − Ax + Dx = 0
FBD Cable:
Ax = Dx
ΣM A = 0: (10 m)Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0
D y = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
FDB AB:
ΣM B = 0: (1.8 m) Ax − (3 m)(8.2 kN) = 0
Ax =
From above
Dx = Ax =
41
kN
3
41
kN
3
FBD CD:
 41

kN  = 0
ΣM C = 0: (4 m)(7.8 kN) − dC 
 3

dC = 2.283 m
dC = 2.28 m 
(a)
D x = 13.67 kN
(b)

D y = 7.80 kN 
Since Ax = Bx and Ay > By , max T is TAB
2
 41

kN  + (8.2 kN)2
TAB = Ax2 + Ay2 = 
 3

Tmax = 15.94 kN 
(c)
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1216
PROBLEM 7.164
A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation that
are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum tension in
the wire.
SOLUTION
Eq. 7.16:
Solve by trial and error:
Eq. 7.15:
xB
c
60
30m + c = c cosh
c
yB = c cosh
c = 64.459 m
sB = c sin h
xB
c
sB = (64.456 m) sinh
60 m
64.459 m
sB = 69.0478 m
Length = 2 sB = 2(69.0478 m) = 138.0956 m
Eq. 7.18:
L = 138.1 m 
Tm = wyB = w(h + c)
= (0.65 kg/m)(9.81 m/s 2 )(30 m + 64.459 m)
Tm = 602.32 N
Tm = 602 N 
PROPRIETARY MATERIAL. © 2013 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. If you are a student using this Manual,
you are using it without permission.
1217
PROBLEM 7.165
A counterweight D is attached to a cable that passes over a small pulley at A
and is attached to a support at B. Knowing that L = 45 ft and h = 15 ft,
determine (a) the length of the cable from A to B, (b) the weight per unit
length of the cable. Neglect the weight of the cable from A to D.
SOLUTION
Given:
L = 45 ft
h = 15 ft
TA = 80 lb
xB = 22.5 ft
By symmetry:
TB = TA = Tm = 80 lb
We have
yB = c cosh
and
yB = h + c = 15 + c
Then
or
Solve by trial for c:
(a)
xB
22.5
= c cosh
c
c
c cosh
22.5
= 15 + c
c
cosh
22.5 15
= +1
c
c
c = 18.9525 ft
sB = c sinh
xB
c
= (18.9525 ft) sinh
22.5
18.9525
= 28.170 ft
Length = 2sB = 2(28.170 ft) = 56.3 ft 
(b)
Tm = wyB = w(h + c)
80 lb = w(15 ft + 18.9525 ft)
w = 2.36 lb/ft 
PROPRIETARY MATERIAL. © 2013 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. If you are a student using this Manual,
you are using it without permission.
1218