CHAPTER 37
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 6.79.
Solution
D = 400 N Ø
From Problem 6.79: On JD
FBD of JD:
≠ SFy = 0 : V - 400 N = 0 Æ SFx = 0 : F = 0 +
SM J = 0 : M - ( 400 N )(0.15 m) = 0 V = 400 N ≠ 
F =0 
M = 60.0 N ◊ m

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3
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 6.80.
Solution
D = 200 N Ø
From Problem 6.80: On JD
FBD of JD:
≠ SFy = 0 : V - 200 N = 0 Æ SFx = 0 : F = 0 +
SM J = 0 : M - (200 N )(0.15 m) = 0 V = 200 N ≠ 
F =0 
M = 30.0 N ◊ m

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4
PROBLEM 7.3
Determine the internal forces at Point J when a = 90°.
Solution
Free body: Entire bracket
SM D = 0 : (1.4 kN )(0.6 m) - A(0.175 m) = 0
+
A = + 4.8 kN A = 4.8 kN ¨
+
+
SFx = 0 : Dx - 4.8 = 0
D x = 4.8 kN Æ
SFy = 0 : D y - 1.4 = 0
D y = 1.4 kN ≠
Free body: JCD
+
+
+
SFx = 0 : 4.8 kN - F = 0 F = 4.80 kN ¨ 
SFy = 0 : 1.4 kN - V = 0 V = 1.400 kN Ø 
SM J = 0 : ( 4.8 kN )(0.2 m) + (1.4 kN )(0.3 m) - M = 0
M = + 1.38 kN ◊ m M = 1.380 kN ◊ m

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5
PROBLEM 7.4
Determine the internal forces at Point J when a = 0.
Solution
Free body: Entire bracket
+
+
SFy = 0 : D y = 0
Dy = 0
SM A = 0 : (1.4 kN )(0.375 m) - Dx (0.175 m) = 0
Dx = + 3 kN
D x = 3 kN Æ
Free body: JCD
+
+
+
SFx = 0 : 3 kN - F = 0 F = 3.00 kN ¨ 
SFy = 0 : - V = 0 V =0
SM J = 0 : (3 kN )(0.2 m) - M = 0
M = + 0.6 kN ◊ m M = 0.600 kN ◊ m

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6
PROBLEM 7.5
Determine the internal forces at Point J of the structure shown.
Solution
FBD ABC:
SM D = 0 : (0.375 m)( 400 N ) - (0.24 m)C y = 0
C y = 625 N ≠
SM B = 0 : - (0.45 m)C x + (0.135 m)( 400 N ) = 0
C x = 120 N Æ
FBD CJ:
≠ SFy = 0 : 625 N - F = 0 F = 625 N Ø 
Æ SFx = 0 : 120 N - V = 0 V = 120.0 N ¨ 
SM J = 0 : M - (0.225 m)(120 N ) = 0
M = 27.0 N ◊ m

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7
PROBLEM 7.6
Determine the internal forces at Point K of the structure shown.
Solution
FBD AK:
Æ SFx = 0 : V = 0
V=0
≠ SFy = 0 : F - 400 N = 0
F = 400 N ≠ 
SM K = 0 : (0.135 m)( 400 N ) - M = 0
M = 54.0 N ◊ m

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8
PROBLEM 7.7
A semicircular rod is loaded as shown. Determine the internal forces at Point J.
Solution
FBD Rod:
SM B = 0 : Ax (2r ) = 0
Ax = 0
SFx¢ = 0 : V - (120 N ) cos 60∞ = 0
V = 60.0 N

F = 103.9 N

M = 18.71

FBD AJ:
SFy¢ = 0 : F + (120 N ) sin 60∞ = 0
F = -103.923 N
SM J = 0 : M - [(0.180 m) sin 60∞](120 N ) = 0
M = 18.7061
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9
PROBLEM 7.8
A semicircular rod is loaded as shown. Determine the internal forces at Point K.
Solution
FBD Rod:
≠ SFy = 0 : By - 120 N = 0 B y = 120 N ≠
SM A = 0 : 2rBx = 0 B x = 0
SFx¢ = 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:
SFy¢ = 0 : F + (120 N ) sin 30∞ = 0
F = - 60 N
SM K = 0 : M - [(0.180 m) sin 30∞](120 N ) = 0
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10
PROBLEM 7.9
An archer aiming at a target is pulling with a 200-N 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
500
Ê3 ˆ
Æ SFx = 0 : 2 Á T1 ˜ - 200 N = 0 T1 = T2 =
N = 166.667 N Ë5 ¯
3
Curve CJB is parabolic: x = ay 2
FBD BJ:
At B : x = 200 mm
y = 800 mm
200 mm
1
=
a=
2
3200
mm
(800 mm)
x=
y2
3200
Slope of parabola = tanq =
At J:
So
dx
2y
y
=
=
dy 3200 1600
È 400 ˘
= 14.036∞ q J = tan -1 Í
Î1600 ˙˚
4
a = tan -1 - 14.036∞ = 39.094∞
3
Ê 500 ˆ
N cos(39.094∞) = 0
SFx¢ = 0 : V - Á
Ë 3 ˜¯
V = 129.352 N
V = 129.4 N

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11
Problem 7.9 (Continued)
Ê 500 ˆ
N sin (39.094∞) = 0
SFy¢ = 0 : F + Á
Ë 3 ˜¯
F = -105.099 N F = 105.1 N

È 3 Ê 500 ˆ ˘
È 4 Ê 500 ˆ ˘
S MJ = 0 : M + (0.4 m) Í Á
N ˜ ˙ + (200 - 50) ¥ 10 -3 m Í Á
N˜ ˙ = 0
¯
Ë
Î5 3
˚
Î5 Ë 3 ¯ ˚
M = 60.0 N ◊ m

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12
PROBLEM 7.10
For the bow of Problem 7.9, determine the magnitude and location of the
maximum (a) axial force, (b) shearing force, (c) bending moment.
PROBLEM 7.9 An archer aiming at a target is pulling with a 200 N 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
Free body: Point A
Ê3 ˆ
SFx = 0 : 2 Á T ˜ - 200 N = 0
Ë5 ¯
+
T=
Free body: Portion of bow BC
+
400
N=0
3
SFy = 0 : FC -
+
SFx = 0 : VC - 100 N = 0 500
N = 166
3
T = 166.7 N v
FC = 133.3 N ≠ v
VC = 100.0 N Æ v
Ê 400 ˆ
≠ SM C = 0 : (100 N )(0.8 m) + Á
N (0.2 m) - M C = 0
Ë 3 ˜¯
ˆ
Ê 320
N ◊ m˜
=Á
¯
Ë 3
MC = 106.7 N ◊ m
v
Equation of parabola
x = ky 2
At B:
Therefore, equation is
200 = k (800)2
x=
y2
3200
k=
1
3200
(1)
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13
Problem 7.10 (continued)
The slope at J is obtained by differentiating (1):
dx =
(a)
2 y dy
dx
y
, tanq =
=
3200
dy 1600
(2)
Maximum axial force
+
Ê 400 ˆ
N cos q - (100 N )sin q = 0 SFV = 0 : - F + Á
Ë 3 ˜¯
Free body: Portion bow CK
F=
400
cos q - 100 sin q 3
F is largest at C(q = 0)
Fm = 133.3 N at C 
(b)
Maximum shearing force
+
Ê 400 ˆ
N sin q + (100 N ) cos q = 0 SFV = 0 : -V + Á
Ë 3 ˜¯
Ê 400 ˆ
sin q + 100 cos q
V =Á
Ë 3 ˜¯
V is largest at B (and D)
Ê 1ˆ
q = q max = tan -1 Á ˜ = 26.56∞
Ë 2¯
400
Vm =
sin 26.56∞ + 100 cos 26.56∞
3
= 149.065 N
Where
(c)
Vm = 149.1 N at B and D 
Maximum bending moment
+
Ê 400 ˆ
Ê 320 ˆ
N◊m+Á
N x + (100 N ) y = 0 SM K = 0 : M - Á
Ë 3 ˜¯
Ë 3 ˜¯
M=
320 400 x
- 100 y
3
3
M is largest at C, where x = y = 0. M m = 106.7 N ◊ m at C 
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14
PROBLEM 7.11
Two members, each consisting of a straight and a
quarter-circular portion of rod, are connected as shown
and support a 375-N load at A. Determine the internal
forces at Point J.
Solution
Free body: Entire frame
SM C = 0 : (375 N )(300 mm) - F (225 mm) = 0
+
F = 500 N Ø v
+
SFx = 0 : C x = 0
+
SFy = 0 : C y - 375 N - 500 N = 0
C y = +875 N C = 875 N ≠ v
Free body: Member BEDF
SM B = 0 : D(300 mm) - (500 N )(375 mm) = 0 +
D = 625 N ≠ v
+
SFx = 0 : Bx = 0
+
SFy = 0 : By + 625 N - 500 N = 0
B = 125.0 N Ø v
By = -125 N Free body: BJ
+
SFx = 0 : F - (125 N )sin 30∞ = 0
F = 62.5 N
+
SFy = 0 : V - (125 N ) cos 30∞ = 0
V = 108.3 N
+
30.0° 
60.0° 
ˆ
Ê 75
m =0
SM J = 0 : - M + (125 N ) Á
Ë 1000 ˜¯
M = 9.375 N ◊ m
M = 9.38 N ◊ m

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15
PROBLEM 7.12
Two members, each consisting of a straight and a
quarter circular portion of rod, are connected as shown
and support a 375-N load at A. Determine the internal
forces at Point K.
Solution
Free body: Entire frame
SM C = 0 : (375 N )(300 mm) - F (225 mm) = 0
+
F = 500 N ¯ v
+
SFx = 0 : C x = 0
+
SFy = 0 : C y - 375 N - 500 N = 0
C y = +875 N C = 875 N ≠ v
Free body: Member BEDF
SM B = 0 : D(300 mm) - (500 N )(375 mm) = 0 +
D = 625 N ≠ v
+
SFx = 0 : Bx = 0
+
SFy = 0 : By + 625 N - 500 N = 0
By = -125 N B = 125.0 N Ø v
Free body: DK
We found in Problem 7.11 that
D = 625 N ≠ on BEDF.
D = 625 N Ø on DK. v
Thus
+
SFx = 0 : F - (625 N ) cos 30∞ = 0
F = 541.266 N
F = 541 N
60.0° 
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16
Problem 7.12 (Continued)
+
SFy = 0 : V - (625 N )sin 30∞ = 0
V = 312.5 N +
V = 313 N
30.0° 
SM K = 0 : M - (625 N )d = 0
Ê 20.0962 m ˆ
M = (625 N )d = (625 N ) Á
Ë 1000 ˜¯
= 12.560 N ◊ m
M = 12.56 N ◊ m

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17
PROBLEM 7.13
A semicircular rod is loaded as shown. Determine the internal forces
at Point J knowing that q = 30∞.
Solution
FBD AB:
Ê3 ˆ
Ê4 ˆ
SM A = 0 : r Á C ˜ + r Á C ˜ - 2r (280 N ) = 0
Ë5 ¯
Ë5 ¯
C = 400 N
4
Æ SFx = 0 : - Ax + ( 400 N ) = 0
5
A x = 320 N ¨
3
≠ SFy = 0 : Ay + ( 400 N ) - 280 N = 0
5
A y = 40.0 N ≠
FBD AJ:
SFx¢ = 0 : F - (320 N )sin 30∞ - ( 40.0 N ) cos 30∞ = 0
F = 194.641 N
F = 194.6 N
60.0∞ 
SFy¢ = 0 : V - (320 N ) cos 30∞ + ( 40 N )sin 30∞ = 0
V = 257.13 N
V = 257 N
30.0∞ 
SM 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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18
PROBLEM 7.14
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
Ê4
ˆ
Ê3
ˆ
+ SM A = 0 : Á FCD ˜ ( 0.16 m) + Á FCD ˜ ( 0.16 m)
Ë5
¯
Ë5
¯
- (280 N )(0.32 m) = 0
FCD = 400 N
v
4
SFx = 0 : Ax + ( 400 N ) = 0
5
+
Ax = - 320 N A x = 320 N ¨ v
3
+ SFy = 0 : Ay + ( 400 N ) - 280 N = 0
5
Ay = + 40.0 N A y = 40.0 N ≠ v
Free body: AJ(For q = 90∞)
+
SM J = 0 : (320 N )(0.16 m)sin q - ( 40.0 N )(0.16 m)(1 - cos q ) - M = 0
M = 51.2 sin q + 6.4 cos q - 6.4 (1)
For maximum value between A and C:
dM
= 0 : 51.2 cos q - 6.4 sin q = 0
dq
51.2
tan q =
=8
6.4
q = 82.87∞ v
Carrying into (1):
M = 51.2 sin 82.87∞ + 6.4 cos 82.87∞ - 6.4 = + 45.20 N ◊ m v
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19
Problem 7.14 (continued)
Free body: BJ(For q > 90∞)
+ SM J = 0 : M - ( 280 N )( 0.16 m)(1 - cos f ) = 0
M = ( 44.8 N ◊ m)(1 - cos f )
Largest value occurs for f = 90∞, that is, at C, and is
M C = 44.8 N ◊ m v
We conclude that
M max = 45.2 N ◊ m
for
q = 82.9∞ 
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20
PROBLEM 7.15
Knowing that the radius of each pulley is 150 mm, that
a = 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
SFx = 0 : 500 N - F = 0
+
F = 500 N +
SFy = 0 : V - 500 N = 0
V = 500 N +
V = 500 N ≠ 
SM J = 0 : (500 N )(0.6 m) = 0
M = 300 N ◊ m (b)
F = 500 N ¨ 
M = 300 N ◊ m

Free body: ABK
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21
Problem 7.15 (continued)
SFx = 0 : 500 N - 500 N + (500 N )sin 20∞ - V = 0
+
V = 171.01 N +
SFy = 0 : - 500 N - (500 N ) cos 20∞ + F = 0
F = 969.8 N +
V = 171.0 N ¨ 
F = 970 N ≠ 
SM 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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22
PROBLEM 7.16
Knowing that the radius of each pulley is 150 mm, that a = 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:
SFx = 0 : 500 N - F = 0
+
F = 500 N
+
SFy = 0 : V - 500 N = 0
V = 500 N +
V = 500 N ≠ 
SM J = 0 : (500 N )(0.6 m) = 0
M = 300 N ◊ m (b)
F = 500 N ¨ 
M = 300 N ◊ m

FBD: Portion ABK
SFx = 0 : 500 N - 500 N + (500 N ) sin 30∞ - V SFy = 0 : - 500 N - (500 N ) cos 30∞ + F = 0 +
SM K = 0 : (500 N )(1.2 m) - (500 N ) sin 30∞(0.9 m) - M = 0 V = 250 N ¨ 
F = 933 N ≠ 
M = 375 N ◊ m

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23
PROBLEM 7.17
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
S MA = 0 : - Bx (1.8 m) - (360 N )(2.6 m) = 0
+
Bx = - 520 N B x = 520 N ¨ v
SFx = 0 : Ax - 520 N = 0
+
Ax = + 520 N +
A x = 520 N Æ v
SFy = 0 : Ay + By - 360 N = 0
Ay + By = 360 N (1)
Free body: Member AE
+
S ME = 0 : - Ay (2.4 m) - (360 N )(1.6 m) = 0
Ay = - 240 N By = 360 N + 240 N
By = + 600 N From (1):
A y = 240 N Ø v
B y = 600 N ≠ v
Free body: BJ
We recall that the forces applied to a pulley may be applied directly to its axle.
+ SFy
= 0:
3
4
(600 N ) + (520 N )
5
5
3
- 360 N - (360 N ) - F = 0
5
F = + 200 N F = 200 N

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24
Problem 7.17 (continued)
+
SFx = 0 :
4
3
4
(600 N ) - (520 N ) - (360 N ) + V = 0
5
5
5
V = +120.0 NV = 120.0 N
+

SM 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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25
PROBLEM 7.18
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
S MA = 0 : - Bx (1.8 m) - (360 N )(2.6 m) = 0
+
Bx = - 520 N B x = 520 N ¨ v
SFx = 0 : Ax - 520 N = 0
+
Ax = + 520 N A x = 520 N Æ v
SFy = 0 : Ay + By - 360 N = 0
+
Ay + By = 360 N (1)
Free body: Member AE
S ME = 0 : - Ay (2.4 m) - (360 N )(1.6 m) = 0 +
Ay = - 240 N A y = 240 N Ø v
By = 360 N + 240 N
From (1):
By = + 600 N B y = 600 N ≠ v
Free body: AK
SFx = 0 : 520 N - F = 0
+
F = + 520 N +
SFy = 0 : 360 N - 240 N - V = 0
V = +120.0 N +
F = 520 N ¨ 
V = 120.0 N Ø 
S MK = 0 : (240 N )(1.6 m) - (360 N )(0.8 m) - M = 0
M = + 96.0 N ◊ m M = 96.0 N ◊ m

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26
PROBLEM 7.19
A 100 mm-diameter pipe is supported every 3 m by a small frame
consisting of two members as shown. Knowing that the combined
weight of the pipe and its contents is 150 N/m and neglecting the
effect of friction, determine the magnitude and location of the
maximum bending moment in member AC.
Solution
Free body: 3-m section of pipe
+
4
SFx = 0 : D - ( 450 N ) = 0 5
3
+ SFy = 0 : E - ( 450 N ) = 0 5
D = 360 N
v
E = 270 N
v
Free body: Frame
SM B = 0 : - Ay (375 mm) + (360 N )(50 mm) + (270 N )(175 mm) = 0
+
Ay = +174 N +
A y = 174.0 N ≠ v
4
3
SFy = 0 : By + 174 N - (360 N ) - (270 N ) = 0 5
5
By = + 276 N B y = 276 N ≠ v
3
4
SFx = 0 : Ax + Bx - (360 N ) + (270 N ) = 0 5
5
+
Ax + Bx = 0 (1)
Free body: Member AC
+
SM C = 0 : (360 N )(50 mm) - (174 N )(240 mm) - Ax (180 mm) = 0
Ax = -132 N From (1):
A x = 132.0 N ¨ v
Bx = - Ax = +132 N B x = 132.0 N Æ v
Free body: Portion AJ
For x £ 250 mm (AJ £ AD ):
+
3
4
SM J = 0 : (132 N ) x - (174 N ) x + M = 0
5
5
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27
Problem 7.19 (continued)
M = 60 x
M max = 15 N ◊ m for x = 0.25 m
M max = 15 N ◊ m at D v
For x > 250 mm (AJ > AD ):
+
3
4
SMJ = 0 : (132 N ) x - (174 N ) x + (360 N )( x - 0.25) + M = 0
5
5
(x in m, M in N · m)
M = 90 - 300 x
M max = 15 N ◊ m for x = 0.25 m
M max = 15.00 N ◊ m at D 
Thus:
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28
PROBLEM 7.20
For the frame of Problem 7.19, determine the magnitude and ­location
of the maximum bending moment in member BC.
PROBLEM 7.19 A 100 mm-diameter pipe is supported every 3 m
by a small frame consisting of two members as shown. Knowing
that the combined weight of the pipe and its contents is 150 N / m
and neglecting the effect of friction, determine the magnitude and
location of the maximum bending moment in member AC.
Solution
Free body: 3-m section of pipe
+
4
SFx = 0 : D - ( 450 N ) = 0 5
D = 360 N
v
3
= 0 : E - ( 450 N ) = 0 5
E = 270 N
v
+ SFy
Free body: Frame
SM B = 0 : - Ay (375 mm) + (360 N )(50 mm) + (270 N )(175 mm) = 0
+
Ay = +174 N A y = 174.0 N ≠ v
4
3
SFy = 0 : By + 174 N - (360 N ) - (270 N ) = 0 5
5
+
By = + 276 N B y = 276 N ≠ v
3
4
SFx = 0 : Ax + Bx - (360 N ) + (270 N ) = 0 5
5
+
Ax + Bx = 0 (1)
Free body: Member AC
+
SM C = 0 : (360 N )(50 mm) - (174 N )(240 mm) - Ax (180 mm) = 0
Ax = -132 N From (1):
A x = 132.0 N ¨ v
Bx = - Ax = +132 N Free body: Portion BK
B x = 132.0 N Æ v
For x £ 175 mm (BK £ BE ):
3
4
+ SM K = 0 : ( 276 N ) x - (132) x - M = 0
5
5
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29
Problem 7.20 (continued)
M = 60 x
M max = 10.5 N ◊ m for
x = 0.175 m
M max = 10.50 N ◊ m at E v
For x > 175 mm (BK > BE ):
+
3
4
SM K = 0 : (276 N ) x - (132 N ) x - (270 N )( x - 0.175 m) - M = 0
5
5
M = 47.25 - 210 x
M max = 10.5 N ◊ m for
x = 0.175 m
M max = 10.50 N ◊ m at E 
Thus
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30
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:
Æ SFx = 0 :
Ax = 0
SM D = 0 : aP - 2aAy = 0
Ay =
P
2
Æ SFx = 0 : V = 0 
FBD AJ:
≠ SFy = 0 :
P
-F =0
2
FBD Rod:
P
Ø
2
SM J = 0 : M = 0 
(b)
F=
Ê3 ˆ
Ê4 ˆ
SM A = 0 : 2a Á D ˜ + 2a Á D ˜ - aP = 0
Ë5 ¯
Ë5 ¯
Æ SFx = 0 : Ax -
4 5
P=0
5 14
≠ SFy = 0 : Ay - P +
3 5
P=0
5 14
D=
5P
14
Ax =
2P
7
Ay =
11P
14
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31
Problem 7.21 (Continued)
FBD AJ:
Æ SFx = 0:
2
P -V = 0 7
V=
2P
¨
7
≠ SFy = 0:
11P
- F = 0
14
F=
11P
Ø
14
SM J = 0 : a
(c)
2P
-M =0
7
M=
2
aP
7

FBD Rod:
SM A = 0:
a Ê 4D ˆ
˜ - aP = 0
Á
2Ë 5 ¯
D=
5P
2
Ax -
4 5P
=0
5 2
Ax = 2 P
≠ SFy = 0 : Ay - P -
3 5P
=0
5 2
Ay =
Æ SFx = 0:
5P
2
FBD AJ:
Æ SFx = 0:
2P - V = 0 ≠ SFy = 0:
5P
-F =0
2
SM J = 0 : a(2 P ) - M = 0 V = 2P ¨ 
F=
5P
Ø
2
M = 2aP

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32
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:
SMD = 0 : aP - 2aA = 0
A=
Æ SFx = 0 : V -
FBD AJ:
P
=0
2
V=
≠ SFy = 0: P
Æ
2
F=0 
SM J = 0 : M - a
(b)
P
¨
2
P
=0
2
M=
aP
2

FBD Rod:
aÊ4 ˆ
SM D = 0 : aP - Á A˜ = 0
2Ë5 ¯
A=
5P
2
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33
Problem 7.22 (Continued)
FBD AJ:
Æ SFx = 0 :
3 5P
-V = 0 5 2
≠ SFy = 0 :
4 5P
-F =0
5 2
V=
F = 2P Ø 
M=
(c)
FBD Rod:
3P
¨
2
3
aP
2

Ê4 ˆ
Ê3 ˆ
SM D = 0 : aP - 2a Á A˜ - 2a Á A˜ = 0 Ë5 ¯
Ë5 ¯
A=
Ê 3 5P ˆ
= 0
Æ SFx = 0 : V - Á
Ë 5 14 ˜¯
≠ SFy = 0 :
4 5P
-F =0
5 14
Ê 3 5P ˆ
=0
SM J = 0 : M - a Á
Ë 5 14 ˜¯
5P
14
V=
3P
Æ
14
F=
2P
Ø
7
M=
3
aP
14

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34
PROBLEM 7.23
A semicircular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when q = 60∞.
Solution
FBD Rod:
SM A = 0 :
2r
W - 2rB = 0 p
B=
SFy¢ = 0 : F +
W
Æ
p
W
W
sin 60∞ - cos 60∞ = 0 3
p
F = - 0.12952 W FBD BJ:
W ˆ 3r Ê W ˆ
Ê
SM 0 = 0 : r Á F - ˜ +
Á ˜ +M =0
Ë
p ¯ 2p Ë 3 ¯
1 1ˆ
Ê
M = Wr Á 0.12952 + - ˜ = 0.28868 Wr
Ë
p 2p ¯
On BJ
M J = 0.289 Wr

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35
PROBLEM 7.24
A semicircular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when q = 150∞.
Solution
FBD Rod:
≠ SFy = 0 : Ay - W = 0
SM B = 0 :
A y = W ≠
2r
W - 2rAx = 0 p
W
A x = ¨
p
FBD AJ:
SFx¢ = 0 :
W
5W
cos 30∞ +
sin 30∞ - F = 0 F = 0.69233W
p
6
Wˆ
Ê
ÊW ˆ
SM 0 = 0 : 0.25587r Á ˜ + r Á F - ˜ - M = 0 Ë
Ë 6¯
p¯
1˘
È 0.25587
M = Wr Í
+ 0.69233 - ˙
6
p
Î
˚
M = Wr (0.4166)
On AJ
M = 0.417 Wr

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36
PROBLEM 7.25
A quarter-circular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when q = 30∞.
Solution
FBD Rod:
Æ SFx = 0 : A x = 0 SM B = 0 :
FBD AJ:
2r
W - rAy = 0
p
Ay =
a = 15∞, weight of segment = W
r=
2W
≠
p
30∞ W
=
90∞ 3
r
r
sin a = p sin 15∞ = 0.9886r a
12
SFy¢ = 0 :
2W
W
cos 30∞ - cos 30∞ - F = 0 p
3
F=
W 3 Ê 2 1ˆ
Á - ˜
2 Ë p 3¯
W
2W ˆ
Ê
+ r cos15∞ = 0 SM 0 = M + r Á F ˜
Ë
p ¯
3
M = 0.0557Wr

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37
PROBLEM 7.26
A quarter-circular rod of weight W and uniform cross section is supported as shown.
Determine the bending moment at Point J when q = 30∞.
Solution
FBD Rod:
SM A = 0 : rB -
2r
W = 0
p
B=
a = 15∞ =
FBD BJ:
r=
r
p
12
Weight of segment = W
p
12
sin 15∞ = 0.98862r 30∞ W
= 90∞ 3
SFy¢ = 0 : F -
SM 0 = 0 : rF - ( r cos15∞)
2W
¨
p
W
2W
cos 30∞ sin 30∞ = 0 3
p
Ê 3 1ˆ
F=Á
+ ˜W
Ë 6 p¯
W
-M =0
3
Ê 3 1ˆ Ê
cos15∞ ˆ
M = rW Á
+ ˜ - Á 0.98862
˜ Wr 3 ¯
Ë 6 p¯ Ë
M = 0.289 Wr

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38
PROBLEM 7.27
For the rod of Problem 7.25, determine the magnitude and location of the maximum
bending moment.
PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at Point J when q = 30∞.
Solution
FBD Rod:
Æ SFx = 0 : Ax = 0
2W
p
r
r = sin a a
2a 4a
W
Weight of segment = W p =
p
2
SM B = 0 :
2r
W - rAy = 0
p
q
a= ,
2
SFx¢ = 0 : - F F=
FBD AJ:
Ay =
4a
2W
W cos 2a +
cos 2a = 0 p
p
2W
2W
(1 - 2a ) cos 2a =
(1 - q ) cos q p
p
2W ˆ
4a
Ê
SM 0 = 0 : M + Á F ˜¯ r + ( r cos a ) W = 0
Ë
p
p
2W
4aW r
(1 + q cos q - cos q )r sin a cos a
p a
p
1
1
sin a cos a = sin 2a = sin q
2
2
M=
But,
so
M=
2r
W (1 - cos q + q cos q - sin q )
p
dM 2rW
=
(sin q - q sin q + cos q - cos q ) = 0
dq
p
(1 - q )sin q = 0
dM
= 0 for q = 0, 1, np ( n = 1, 2,L)
dq
Only 0 and 1 in valid range
At
q = 0 M = 0, at q = 1 rad
for
at q = 57.3∞ M = M max = 0.1009Wr 
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39
PROBLEM 7.28
For the rod of Problem 7.26, determine the magnitude and location of the maximum
bending moment.
PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at Point J when q = 30∞.
Solution
FBD Bar:
SM A = 0 : rB -
2r
W =0
p
B=
2W
¨
p
q
so
2
r
r = sin a a
2a
Weight of segment = W p a=
0a 
p
4
2
=
SFx¢ = 0 : F -
4a
W
p
4a
2W
W cos 2a sin 2a = 0 p
p
2W
(sin 2a + 2a cos 2a )
p
2W
=
(sin q + q cos q )
p
F=
FBD BJ:
SM 0 = 0 : rF - ( r cos a )
M=
But,
so
4a
W -M =0
p
2
ˆ 4a
Êr
Wr (sin q + q cos q ) - Á sin a cos a ˜
W
¯ p
Ëa
p
1
1
sin a cos a = sin 2a = sin q
2
2
M=
2Wr
(sin q + q cos q - sin q )
p
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40
Problem 7.28 (Continued)
or
M=
2
Wrq cos q
p
dM 2
= Wr (cos q - q sin q ) = 0 at q tan q = 1 dq p
Solving numerically
q = 0.8603 rad
M = 0.357Wr
and

at q = 49.3∞ 
(Since M = 0 at both limits, this is the maximum)
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41
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
(a)
From A to B:
+
+
SFy = 0:
V = -P SM1 = 0 : M = - Px From B to C:
+
+
(b)
SFy = 0 : - P - P - V = 0
V = -2 P SM 2 = 0 : Px + P ( x - a) + M = 0
|V | max = 2 P;
M = -2 Px + Pa | M | max = 3Pa 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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42
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
(a)
Reactions:
2P
≠
3
P
C = ≠
3
A=
From A to B:
+
+
From B to C:
+
+
(b)
2P
3
2P
SM1 = 0 : M = +
x
3
SFy = 0 : V = +
P
3
P
SM 2 = 0 : M = + ( L - x ) 3
SFy = 0 : V = -
|V | max =
2P
;
3
| M | max =
2 PL
9

PROPRIETARY MATERIAL. © 2010 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
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43
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
FBD beam:
(a)
1
L
Ay = D = ( w )
2
2
By symmetry:
Ay = D =
wL
≠
4
Along AB:
≠ SFy = 0 :
wL
-V = 0
4
SM J = 0 : M - x
V=
wL
4
wL
=0
4
wL
M=
x (straight)
4
Along BC:
≠ SFy = 0 :
wL
- wx1 - V = 0
4
wL
V=
- wx1
4
V = 0 at
Straight with
SM k = 0 : M +
Parabola with
x1 =
L
4
x1
ˆ wL
ÊL
=0
wx1 - Á + x1 ˜
¯ 4
Ë4
2
M=
ˆ
w Ê L2 L
+ x1 - x12 ˜ Á
2Ë 8 2
¯
M=
3
L
wL2 at x1 = 32
4
Section CD by symmetry
(b)
|V |max =
From diagrams:
wL
on AB and CD 
4
| M |max =
3wL2
at center 
32
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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44
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)
Along AB:
≠ SFy = 0 : - wx - V = 0
Straight with
V =SM J = 0 : M +
Parabola with
wL
2
V = - wx at
x=
L
2
x
1
wx = 0 M = - wx 2 2
2
M =-
wL2
8
at
x=
L
2
Along BC:
L
1
-V = 0
V = - wL 2
2
Lˆ L
Ê
SM k = 0 : M + Á x1 + ˜ w = 0 Ë
4¯ 2
≠ SFy = 0 : - w
M =Straight with
(b)
wL Ê L
ˆ
Á + x1 ˜¯ 2 Ë4
3
M = - wL2
8
From diagrams:
at
x1 =
L
2
|V |max =
wL
on BC 
2
|M |max =
3wL2
at C 
8
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45
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: Portion AJ
+
+
(a)
SFy = 0 : - P - V = 0 SM J = 0 : M + Px - PL = 0 V = -P v
M = P( L - x) v
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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46
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
(a)
FBD Beam:
SM C = 0 : LAy - M 0 = 0
Ay =
M0
Ø
L
≠ SFy = 0 : - Ay + C = 0
C=
Along AB:
≠ SFy = 0 : -
M0
≠
L
M0
-V = 0
L
V =-
SM J = 0 : x
M0
L
M0
+M =0
L
M0
x
L
M
M = - 0 at B
2
M =Straight with
Along BC:
≠ SFy = 0 : -
M0
-V = 0
L
SM K = 0 : M + x
M=
Straight with
(b)
V =-
M0
- M0 = 0
L
M0
at B
2
M0
L
Ê
M = M 0 Á1 Ë
xˆ
˜
L¯
M = 0 at C
|V |max = P everywhere 
From diagrams:
| M |max =
M0
at B 
2
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47
PROBLEM 7.35
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)
Just to the right of A:
+
SFy = 0 V1 = +15 kN
M1 = 0
Just to the left of C:
V2 = +15 kN
M 2 = +15 kN ◊ m
V3 = +15 kN
M 3 = +5 kN ◊ m
V4 = -15 kN
M 4 = +12.5 kN ◊ m
V5 = -35 kN
M 5 = +5 kN ◊ m
Just to the right of C:
Just to the right of D:
Just to the right of E:
At B:
(b)
M B = -12.5 kN ◊ m
| M |max = 12.50 kN ◊ m 
|V |max = 35.0 kN PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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48
PROBLEM 7.36
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
+
SM 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 ≠ v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay + 37.5 kN - 40 kN - 32 kN - 16 kN = 0
A = 50.5 kN ≠ v
Ay = +50.5 kN (a)
Shear and bending moment.
V1 = 50.5 kN
Just to the right of A:
M1 = 0 v
Just to the right of C:
+
+
SFy = 0 : 50.5 kN - 40 kN - V2 = 0 SM 2 = 0 : M 2 - (50.5 kN )(0.6 m) = 0 V2 = +10.5 kN v
M 2 = +30.3 kN ◊ m v
Just to the right of D:
+
+
SFy = 0 : 50.5 - 40 - 32 - V3 = 0 SM 3 = 0 : M 3 - (50.5)(1.5) + ( 40)(0.9) = 0 V3 = -21.5 kN v
M 3 = +39.8 kN ◊ m v
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49
Problem 7.36 (Continued)
Just to the right of E:
+
+
SFy = 0 : V4 + 37.5 = 0 V4 = -37.5 kN v
SM 4 = 0 : - M 4 + (37.5)(0.2) = 0 VB = M B = 0 At B:
(b)
M 4 = +7.50 kN ◊ m v
v
|V |max = 50.5 kN 
|M |max = 39.8 kN ◊ m 
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50
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
+
SM A = 0 : E (3 m) - (30 kN )(1 m) - (60 kN )(2 m) - (22.5 kN )( 4 m) = 0
E = +80 kN E = 80.0 kN ≠ v
+
SFx = 0 : Ax = 0 +
SFy = 0 : Ay + 80 kN - 30 kN - 60 kN - 22.5 kN = 0 A = 32.5 kN ≠ v
Ay = +32.5 kN (a)
Shear and bending moment
Just to the right of A:
V1 = +32.5 kN
M1 = 0 v
Just to the right of C:
+
+
SFy = 0 : 32.5 kN - 30 kN - V2 = 0 SM 2 = 0 : M 2 - (32.5 kN )(1 m) = 0 V2 = +2.5 kN v
M 2 = +32.5 kN ◊ m v
Just to the right of D:
+
+
SFy = 0 : 32.5 - 30 - 60 - V3 = 0 SM 3 = 0 : M 3 - (32.50)(2) + (30)(1) = 0 V3 = -57.5 kN v
M 3 = +35 kN ◊ m v
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51
Problem 7.37 (Continued)
Just to the right of E:
+
+
SFy = 0 : V4 - 22.5 kN = 0 V4 = +22.5 kN v
SM 4 = 0 : - M 4 - (22.5)1 = 0 M 4 = -22.5 kN ◊ m v
VB = M B = 0 v
At B:
(b)
|V |max = 57.5 kN 
| M |max = 35.0 kN ◊ m 
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52
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
+
SM C = 0 : (600 N )(200 mm) - (1500 N )(500 mm) + E (900 mm) - (600 N )(1200 mm) = 0
E = +1500 N E = 1500 N ≠ v
SFx = 0 : C x = 0
+
SFy = 0 : C y + 1500 N - 600 N - 1500 N - 600 N = 0
C y = +1200 N (a)
C = 1200 N ≠ v
Shear and bending moment
Just to the right of A:
+
SFy = 0 : - 600 N - V1 = 0 V1 = -600 N, M1 = 0 v
+
SFy = 0 : 1200 N - 600 N - V2 = 0 V2 = +600 N v
SM C = 0 : M 2 + (600 N )(0.2 m) = 0 M 2 = -120 N ◊ m v
Just to the right of C:
+
Just to the right of D:
+
+
SFy = 0 : 1200 - 600 - 1500 - V3 = 0 V3 = -900 N v
SM 3 = 0 : M 3 + (600)(0.7) - (1200)(0.5) = 0, M 3 = +180 N ◊ m v
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53
Problem 7.38 (Continued)
Just to the right of E:
+ SFy = 0 : V4 - 600 N = 0 +
SM 4 = 0 : - M 4 - (600 N )(0.3 m) = 0 M 4 = -180 N ◊ m v
VB = M B = 0 v
At B:
(b)
V4 = +600 N v
|V |max = 900 N 
| M |max = 180.0 N ◊ m 
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54
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
SM A = 0 : B(5 m) - (60 kN )(2 m) - (50 kN )( 4 m) = 0
+
B = +64.0 kN B = 64.0 kN ≠ v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay + 64.0 kN - 6.0 kN - 50 kN = 0
Ay = +46.0 kN (a)
A = 46.0 kN ≠ v
Shear and bending-moment diagrams
From A to C:
+
+
SFy = 0 : 46 - V = 0 SM y = 0 : M - 46 x = 0 V = +46 kN v
M = ( 46 x ) kN ◊ m v
From C to D:
+
+
SFy = 0 : 46 - 60 - V = 0 V = -14 kN v
SM 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 v
For
x = 3 m : M D = +78.0 kN ◊ m v
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55
Problem 7.39 (Continued)
From D to B:
+
SFy = 0 : V + 64 - 25m = 0 V = (25m - 64) kN +
Ê mˆ
SM j = 0 : 64 m - (25 m ) Á ˜ - M = 0 Ë 2¯
M = (64 m - 12.5m 2 ) kN ◊ m
For
m = 0 : VB = -64 kN MB = 0 v
(b)
|V |max = 64.0 kN 
| M |max = 92.0 kN ◊ m 
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56
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
SM B = 0 : (60 kN )(6 m) - C (5 m) + (80 kN )(2 m) = 0 +
C = +104 kN
From A to C:
C = 104 kN ≠
+
SFy = 0 : 104 - 60 - 80 + B = 0
B = 36 kN ≠
+
SFy = 0 : - 30 x - V = 0
V = -30 x
+
Ê xˆ
SM1 = 0 : (30 x ) Á ˜ + M = 0
Ë 2¯
M = -15 x 2
From C to D:
+
+
SFy = 0 : 104 - 60 - V = 0
SM 2 = 0 : (60)( x - 1) - (104)( x - 2) + M = 0
V = +44 kN
M = 44 x - 148
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57
Problem 7.40 (Continued)
From D to B:
+
+
SFy = 0 : V = -36 kN
SM 3 = 0 : (36)(7 - x ) - M = 0 M = -36 x + 252
(b)
|V |max = 60.0 kN | M |max = 72.0 kN ◊ m 
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58
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:
1
Ay = B = 40 kN + (60 kN/m)(1.5 m) A y = B = 85 kN ≠
2
Along AC:
≠ SFy = 0 : 85 kN - V = 0 V = 85 kN
SM J = 0 : M - x(85 kN ) M = (85 kN ) x
M = 51 kN ◊ m at C ( x = 0.6 m)
Along CD:
≠ SFy = 0 : 85 kN - 40 kN - (60 kN/m) x1 - V = 0 V = 45 kN - (60 kN/m) x1
V = 0 at x1 = 0.75 m (at center)
SM K = 0 : M +
x1
(60 kN/m) x1 + ( 40 kN ) x1 - (0.6 m + x1 )(85 kN ) = 0
2
M = 51 kN ◊ m + ( 45 kN ) x1 - (30 kN/m) x12
M = 67.875 kN ◊ m at x1 = 0.75 m
Complete diagram by symmetry
(b)
|V |max = 85 kN on AC and DB 
From diagrams:
|M |max = 67.875 kN ◊ m at center 
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59
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
SM A = 0 : B(5 m) - (120 kN )(1.5 m) - (60 kN )(3 m) = 0
+
B = +72.0 kN B = 72.0 kN ≠ 
SFx = 0 : Ax = 0
+
SFy = 0 : Ay - 120 - 60 + 72 = 0
Ay = +108 kN (a)
A = 108.0 kN ≠ 
Shear and bending-moment diagrams
From A to C:
+
SFy = 0 : 108 - 40 x - V = 0
V = (108 - 40 x ) kN
+
Ê xˆ
SM J = 0 : M + ( 40 x ) Á ˜ - 108 x = 0
Ë 2¯
M = 108 x - 20 x 2
For x = 0:
VA = +108 kN For x = 3m:
VC = -12 kN MA = 0 
M C = +144 kN ◊ m 
V = 0 at x = 2.7 m (D)
At x = 2.7 m, M =145.8 kN ◊ m
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60
PROBLEM 7.42 (Continued)
From C to B:
+
+
SFy = 0 : V + 72 = 0 V = -72 kN 
SM J = 0 : 72a - M = 0
M = (72a) kN ◊ m
For a = 2 m, M C = +144 kN ◊ m 
For a = 0: MB = 0 
(b)
|V |max = 108.0 kN 
|M |max = 145.8 kN ◊ m 
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61
PROBLEM 7.43
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
(a)
FBD Beam:
≠ SFy = 0 : w (1.5 m) - 2(3.0 kN ) = 0
w = 4.0 kN/m
Along AC:
≠ SFy = 0 : ( 4.0 kN/m) x - V = 0
V = ( 4.0 kN/m) x
x
SM J = 0 : M - ( 4.0 kN/m) x = 0
2
M = (2.0 kN/m) x 2
Along CD:
≠ SFy = 0 : ( 4.0 kN/m) x - 3.0 kN - V = 0
V = ( 4.0 kN/m) x - 3.0 kN
x
SM K = 0 : M + ( x - 0.3 m)(3.0 kN ) - ( 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 at C and D 
(b)
|M |max = 0.225 kN ◊ m at center 
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62
PROBLEM 7.44
Solve Problem 7.43 knowing that a = 0.5 m.
PROBLEM 7.43 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
+
(a)
SFy = 0 : w g (1.5 m) - 3 kN - 3 kN = 0 w g = 4 kN/m v
Shear and bending moment
From A to C:
+
+
SFy = 0 : 4 x - V = 0 V = ( 4 x ) kN
x
SM J = 0 : M - ( 4 x ) = 0, M = (2 x 2 ) kN ◊ m
2
For x = 0: VA = M A = 0 v
For x = 0.5 m :
From C to D:
VC = 2 kN, +
+
M C = 0.500 kN ◊ m v
SFy = 0 : 4 x - 3 kN - V = 0
V = ( 4 x - 3) kN
x
SM J = 0 : M + (3 kN )( x - 0.5) - ( 4 x ) = 0
2
M = (2 x 2 - 3x + 1.5) kN ◊ m
For x = 0.5 m : VC = -1.00 kN, M C = 0.500 kN ◊ m v
For x = 0.75 m : VC = 0, M C = 0.375 kN ◊ m v
For x = 1.0 m : VC = 1.00 kN, M C = 0.500 kN ◊ m v
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63
PROBLEM 7.44 (Continued)
From D to B:
+
+
SFy = 0 : V + 4 m = 0 V = -( 4 m ) kN
m
SM J = 0 : ( 4 m ) - M = 0, M = 2m 2
2
For m = 0: For m = 0.5 m :
VB = M B = 0 v
VD = -2 kN, M D = 0.500 kN ◊ m v
(b)
|V |max = 2.00 kN 
|M |max = 0.500 kN ◊ m 
PROPRIETARY MATERIAL. © 2010 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
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64
PROBLEM 7.45
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
+
(a)
SFy = 0 : w g ( 4 m) - ( 45 kN/m)(2 m) = 0 w g = 22.5 kN/m 
Shear and bending-moment diagrams
From A to C:
+
SFy = 0 : 22.5 x - 45 x - V = 0
V = ( -22.5 x ) kN
+
x
x
SM J = 0 : M + ( 45 x ) - (22.5 x ) = 0
2
2
M = ( -11.25 x 2 ) kN ◊ m
For x = 0: VA = M A = 0 
For x = 1m, VC = -22.5 kN M C = -11.25 kN ◊ m 
From C to D:
+
+
SFy = 0 : 22.5 x = 45 - V = 0, V = (22.5 x - 45) kN SM J = 0 : M + 45( x - 0.5) - (22.5 x )
x
=0 2
M = 11.25 x 2 - 45 x + 22.5
For x = 1m,
VC = -22.5 kN For x = 2 m
VC = 0, M C = -11.25 kN ◊ m 
M C = -22.5 kN ◊ m 
PROPRIETARY MATERIAL. © 2010 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
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65
PROBLEM 7.45 (Continued)
For x = 3m VD = 22.5 kN, M D = -11.25 kN ◊ m 
VB = M B = 0 
At B:
(b)
|V |max = 22.5 kN 
Bending moment diagram consists of three distinct arcs of parabola.
|M |max = 22.5 kN ◊ m 
M £ 0 everywhere 
Since entire diagram is below the x axis:
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66
PROBLEM 7.46
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
+
SFy = 0 : w g ( 4 m) - ( 45 kN/m)(2 m) = 0
w g = 22.5 kN/m v
(a)
Shear and bending-moment diagrams from A to C:
SFy = 0 : 22.5 x - V = 0
x
+ SM J = 0 : M - ( 22.5 x )
2
For x = 0: +
V = (22.5 x )kN
M = (11.25 x 2 ) kN ◊ m
VA = M A = 0 v
For x = 1 m : VC = 22.5 kN, M C = 11.25 kN ◊ m v
From C to D:
+
+
SFy = 0 : 22.5 x - 45( x - 1) - V = 0
V = ( 45 - 22.5 x ) kN
x -1
x
SM J = 0 : M + 45( x - 1)
- (22.5 x ) = 0
2
2
M = [11.25 x 2 - 22.5( x - 1)2 ]kN ◊ m
For x = 1 m : VC = 22.5 kN, For x = 2 m: VL = 0, For x = 3 m : VD = -22.5 kN, M C = 11.25 kN ◊ m v
M L = 22.5 kN ◊ m v
M D = 11.25 kN ◊ m v
VB = M B = 0 v
At B:
|V |max = 22.5 kN 
(b)
| M | max = 22.5 kN ◊ m 
Bending-moment diagram consists of three distinct arcs of
­parabola, all located above the x axis.
Thus: M ≥ 0 everywhere 
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67
PROBLEM 7.47
Assuming the upward reaction of the ground on beam AB to be ­uniformly
distributed and knowing that P = wa, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear
and bending moment.
Solution
Free body: Entire beam
SFy = 0 : w g ( 4 a) - 2wa - wa = 0
+
wg =
(a)
3
w v
4
Shear and bending-moment diagrams
From A to C:
SFy = 0 :
+
3
wx - wx - V = 0
4
1
V = - wx
4
x Ê3 ˆ x
SM J = 0 : M + ( wx ) - Á wx˜ = 0
2 Ë4 ¯ 2
1
M = - wx 2
8
For x = 0: 1
For x = a :
VC = - wa 4
From C to D:
+
+
+
SFy = 0 :
VA = M A = 0 v
1
M C = - wa2 v
8
3
wx - wa - V = 0
4
ˆ
Ê3
V = Á x - a˜ w
¯
Ë4
aˆ 3 Ê x ˆ
Ê
SM J = 0 : M + wa Á x - ˜ - wx Á ˜ = 0
Ë
2¯ 4 Ë 2¯
3
aˆ
Ê
M = wx 2 - wa Á x - ˜ Ë
8
2¯
(1)
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68
PROBLEM 7.47 (Continued)
1
1
VC = - wa M C = - wa2 v
4
8
1
VD = + wa MD = 0 v
For x = 2a :
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.
For x = a :
|V |max =
(b)
To find | M |max , we differentiate Eq. (1) and set
dM
dx
1
wa 
2
= 0:
4
dM 3
= wx - wa = 0, x = a
3
dx 4
2
3 Ê4 ˆ
wa
Ê 4 1ˆ
M = w Á a˜ - wa2 Á - ˜ = Ë 3 2¯
8 Ë3 ¯
6
2
1
|M |max = wa2 
6
Bending-moment diagram consists of four distinct arcs of parabola.
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69
PROBLEM 7.48
Solve Problem 7.47 knowing that P = 3wa.
PROBLEM 7.47 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
+
SFy = 0 : w g ( 4 a) - 2wa - 3wa = 0
wg =
(a)
5
w v
4
Shear and bending-moment diagrams
From A to C:
+
SFy = 0 :
5
wx - wx - V = 0
4
1
V = + wx
4
x Ê5 ˆ x
SM J = 0 : M + ( wx ) - Á wx˜ = 0
2 Ë4 ¯ 2
1
M = + wx 2
8
For x = 0: +
1
VC = + wa 4
For x = a :
VA = M A = 0 v
1
M C = + wa2 v
8
From C to D:
+
+
SFy = 0 :
5
wx - wa - V = 0
4
ˆ
Ê5
V = Á x - a˜ w
¯
Ë4
aˆ 5 Ê x ˆ
Ê
SM J = 0 : M + wa Á x - ˜ - wx Á ˜ = 0
Ë
2¯ 4 Ë 2¯
5
aˆ
Ê
M = wx 2 - wa Á x - ˜ Ë
8
2¯
(1)
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70
PROBLEM 7.48 (Continued)
For x = a : For x = 2a : 1
1
VC = + wa, M C = + wa2 v
4
8
3
VD = + wa, M D = + wa2 v
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.
3
wa 
2
= 0:
|V |max =
(b)
To find | M |max , we differentiate Eq. (1) and set
dM
dx
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 = wa2 
Bending-moment diagram consists of four distinct arcs of parabola.
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71
PROBLEM 7.49
Draw the shear and bending-moment diagrams for the beam AB, and ­determine
the shear and bending moment (a) just to the left of C, (b) just to the right of C.
Solution
Free body: Entire beam
+
SM A = 0 : B(0.6 m) - (600 N )(0.2 m) = 0
B = +200 N B = 200 N ≠ v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay - 600 N + 200 N = 0
Ay = +400 N A = 400 N ≠ v
We replace the 600-N load by an equivalent force-couple system at C
Just to the right of A:
V1 = +400 N, M1 = 0 v
(a) Just to the left of C:
V2 = +400 N 
M 2 = ( 400 N )(0.4 m) (b) Just to the right of C:
M 3 = (200 N )(0.2 m) Just to the left of B: M 2 = +160.0 N ◊ m 
V3 = -200 N 
M 3 = + 40.0 N ◊ m 
V4 = -200 N, M 4 = 0 v
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72
PROBLEM 7.50
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 q = 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)
By symmetry:
T1 = T2 = T
≠ SFy = 0 : 2T sin 60∞ - 3 kN = 0
3
kN
3
3
T1x =
2 3
3
T1 y = kN
2
T=
FBD Beam:
M = (0.5 m)
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
ˆ
Ê dV
= -0.6 kN/m˜ with 1.5 kN
ÁË
¯
dx
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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73
PROBLEM 7.50 (Continued)
Moment diagram: M is piecewise parabolic
ˆ
Ê dM
decreasing with V ˜
ÁË
¯
dx
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+
1
M G = -0.242 kN ◊ m + (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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74
PROBLEM 7.51
Solve Problem 7.50 when q = 60∞.
PROBLEM 7.50 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 q = 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
E x = Dx = D y tanq Thus:
+
(1)
SFy = 0 : D y + E y - 3 kN = 0 D y = E y = 1.5 kN ≠ v
D x = (1.5 kN ) tan q Æ
From (1):
E = (1.5 kN ) tan q ¨ v
We replace the forces at D and E by equivalent force-couple systems at F and H, where
M 0 = (1.5 kN tan q )(0.5 m) = (750 N ◊ m) tan q (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:
SFy = 0 : - V - 600 x = 0 V = ( -600 x )N
x
SM J = 0 : M + (600 x ) = 0, M = ( -300 x 2 )N ◊ m
2
+
+
For x = 0: For x = 1.5 m: VA = M A = 0 v
VF = -900 N, M F = -675 N ◊ m v
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75
PROBLEM 7.51 (Continued)
From F to H:
+
SFy = 0 : 1500 - 600 x - V = 0
V = (1500 - 600 x ) N
+
x
SM J = 0 : M + (600 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 v
For x = 2.5 m: VG = 0, M G = ( M 0 - 375) N ◊ m v
From G to B, The V and M diagrams will be obtained by symmetry,
(b) |V |max = 900 N 
Making q = 60∞ in Eq. (2):
M 0 = 750 tan 60∞ = 1299 N ◊ m
M = 1299 - 675 = 624 N ◊ m v
Thus, just to the right of F:
M G = 1299 - 375 = 924 N ◊ m v
and
( b)
|V |max = 900 N 
| M |max = 924 N ◊ m 
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76
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
FBD whole:
SM D = 0 : (0.45 m)(7.5 kN ) - (0.45 m)(30 kN )
- (1.35 m)(7.5 kN ) + (0.6 m)G = 0
G = 33.75 kN Æ
(Dimensions in m, loads in kN, moments in kN.m)
Æ SFx = 0 : -C x + G = 0 C x = 33.75 kN ¨
≠ SFy = 0 : C y - 7.5 kN - 30 kN - 7.5 kN = 0 C y = 45 kN ≠
Beam AB, with forces C and G replaced by equivalent force/couples at D and F. Couple at D due to
Cx = 33.75 × 0.3 = 10.125 kN . m
Along AD:
Ø SFy = 0 : - 7.5 kN - V = 0 V = -7.5 kN
SM J = 0 : x(7.5 kN ) + M = 0 M = -(7.5 kN ) x
M = -3.375 kN ◊ m at x = 0.45 m
Along DE:
≠ SFy = 0 : - 7.5 kN + 45 kN - V = 0
V = 37.5 kN
SM K = 0 : M + 10.125 kN ◊ m - x1 ( 45 kN )
+ (0.45 + x1 )(7.5 kN ) = 0
M = (37.5 kN)x1 - 13.5
M = 3.375 kN ◊ m at x1 = 0.45 m
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77
PROBLEM 7.52 (Continued)
Along EF:
≠ SFy = 0 : V - 7.5 kN = 0 V = 7.5 kN
SM N = 0 : - M + 10.125 kN ◊ m - ( x4 + 0.45 m)(7.5 kN )
M = 6.75 kN ◊ m - (7.5 kN ) x4
M = 3.375 kN ◊ m at x4 = 0.45 m
M = 6.75 kN ◊ m at x4 = 0
Along FB:
≠ SFy = 0 : V - 7.5 kN = 0
V = 7.5 kN
SM L = 0 : - M - x3 (7.5 kN ) = 0
M = ( -7.5 kN ) x3
M = -3.375 kN ◊ m at x3 = 0.45 m
|V |max = 37.5 kN on DE 
From diagrams:
| M |max = 13.5 kN ◊ m at D + 
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78
PROBLEM 7.53
Draw the shear and bending-moment diagrams for the
beam AB, and determine the maximum absolute values of
the shear and bending moment.
Solution
SM G = 0 : T (225 mm) - (200 N )(225 mm) - (500 N )(525 mm) = 0
+
+
+
SFx = 0 : -
4100
+ Gx = 0 3
SFy = 0 : G y - 200 N - 500 N = 0
4100
N
3
4100
Gx =
NÆ
3
T=
G y = 700 N ≠
Equivalent loading on straight part of beam AB
From A to E:
SFy = 0 : V = +700 N
+
SM1 = 0 : + 170.833 N ◊ m - (700 N ) x + M = 0
M = -170.833 + 700 x
From E to F:
+
+
SFy = 0 : 700 - 200 - V = 0
V = +500 N
SM 2 = 0 : + 170.833 N ◊ m - (700 N ) x + (200 N )( x - 0.225) + M = 0
M = -125.833 + 500 x
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79
PROBLEM 7.53 (Continued)
SFy = 0 V = +500 N
From F to B:
+
SM 3 = 0 : - (500)(0.525 - x ) - M = 0
M = -262.5 + 500 x
|V |max = 700 N 
| M |max = 170.8 N ◊ m 
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80
PROBLEM 7.54
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
+
SM 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 ≠ v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay + 800 N - 3( 400 N ) = 0
Ay = +400 N A = 400 N ≠ v
We replace the loads by equivalent force-couple systems at C, D, and E.
We consider successively the following F-B diagrams
V1 = +400 N
M1 = 0
V2 = +400 N
M 2 = +60 N ◊ m
V3 = 0
M 3 = +120 N ◊ m
V4 = 0
M 4 = +120 N ◊ m
V5 = -400 N
M 5 = +180 N ◊ m
V6 = -400 N
M 6 = +60 N ◊ m
V7 = -800 N
M 7 = +120 N ◊ m
V8 = -800 N
M8 = 0
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81
PROBLEM 7.54 (Continued)
(b)
|V |max = 800 N 
|M |max = 180.0 N ◊ m 
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82
PROBLEM 7.55
For the structural member of Problem 7.50, determine (a) the
­angle  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 7.50 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  = 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
See solution of Problem 7.50 for reduction of loading or beam AB to the following:
M 0 = (750 N ◊ m) tan q v
where
[Equation (2)]
The largest negative bending moment occurs Just to the left of F:
+
Ê 1.5 m ˆ
SM1 = 0 : M1 + (900 N ) Á
=0 Ë 2 ˜¯
M1 = -675 N ◊ m v
The largest positive bending moment occurs
At G:
+
SM 2 = 0 : M 2 - M 0 + (1500 N )(1.25 m - 1 m) = 0
M 2 = M 0 - 375 N ◊ m v
Equating M 2 and -M1 :
M 0 - 375 = +675
M 0 = 1050 N ◊ m
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83
PROBLEM 7.55 (Continued)
(a)
From Equation (2):
tan q =
1050
= 1.400 750
q = 54.5∞ 
|M |max = 675 N ◊ m 
(b)
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84
PROBLEM 7.56
For the beam of Problem 7.43, 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 7.43 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:
+
a
SM1 = 0 : M1 = ( 4a) 2
M1 = 2a2 v
The largest negative bending moment occurs
At the centerline:
+
SM 2 = 0 : M 2 + 3(0.75 - a) - 3(0.375) = 0 M 2 = -(1.125 - 3a) v
2a2 = 1.125 - 3a
Equating M1 and -M 2 :
a2 + 1.5a - 0.5625 = 0
(a)
Solving the quadratic equation:
(b)
Substituting:
a = 0.31066, | M |max = M1 = 2(0.31066)2
a = 0.311 m 
| M |max = 193.0 N ◊ m
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85
PROBLEM 7.57
For the beam of Problem 7.47, 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 7.47 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
+
SFy = 0 : w g ( 4 a) - 2wa - kwa = 0
w
w g = (2 + k )
4
wg
= a w
k = 4a - 2 Setting
We have
(1)
(2)
Minimum value of B.M. For M to have negative values, we must have w g < 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:
+
aˆ
Ê xˆ
Ê
SM J = 0 : M + wa Á x - ˜ - a wx Á ˜ = 0
Ë 2¯
¯
Ë
2
M=
We differentiate and set
dM
= 0:
dx
1
w (a x 2 - 2ax + a2 ) 2
ax - a = 0
Substituting in (3):
M min =
xmin =
a
a
(3)
(4)
1 2Ê 1 2 ˆ
wa Á - + 1˜
Ëa a ¯
2
M min = - wa2
1- a
2a
(5)
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86
PROBLEM 7.57 (continued)
Maximum value of bending moment occurs at D
+
Ê 3a ˆ
SM D = 0 : M D + wa Á ˜ - (2a wa)a = 0
Ë 2¯
3ˆ
Ê
M max = M D = wa2 Á 2a - ˜ Ë
2¯
Equating -M min and M max :
wa2
(6)
1- a
3ˆ
Ê
= wa2 Á 2a - ˜
Ë
2a
2¯
4a 2 - 2a - 1 = 0
a=
(a)
Substitute in (2):
(b)
Substitute for a in (5):
2 + 20
8
a=
1+ 5
= 0.809
4
k = 4(0.809) - 2 | M |max = - M min = - wa2
Substitute for a in (4):
1 - 0.809
2(0.809)
k = 1.236 
| M |max = 0.1180wa2 
xmin =
a
1.236a v
0.809
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87
PROBLEM 7.58
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:
+
Ê xˆ
SM1 = 0 : M + wx Á ˜ = 0
Ë 2¯
1
M = - wx 2
2
1 2
M A = - wa
2
Between A and B:
+
Ê1 ˆ
Ê1 ˆ
SM 2 = 0 : M - Á wL˜ ( x - a) + ( wx ) Á x ˜ = 0
Ë2 ¯
Ë2 ¯
1
1
1
M = - wx 2 + wLx - wLa 2
2
2
At center C:
x=
L
2
2
1 Ê Lˆ
1
Ê Lˆ 1
M C = - w Á ˜ + wL Á ˜ - wLa
Ë 2¯ 2
2 Ë 2¯
2
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88
PROBLEM 7.58 (continued)
We set
| M A| = | MC | :
1
1
1
1
1
1
- wa2 = wL2 - wLa + wa2 = wL2 - wLa
2
8
2
2
8
2
a2 + La - 0.25L2 = 0
1
1
a = ( L ± L2 + L2 ) = ( 2 - 1) L
2
2
1
M max = w (0.207 L)2 = 0.0214 wL2
2
a = 0.207 L 
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89
PROBLEM 7.59
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 . (See hint for
Problem 7.55.)
Solution
By symmetry:
Ay = B = 300 kN - P
Along AC:
SM J = 0 : M - x(300 kN - P ) = 0
M = (300 - P ) x
M = (300 - P ) kN ◊ m at
x =1m
Along CD:
Along DE:
SM K = 0 : M + ( x - 1)(300 kN ) - x(300 - P ) = 0
M = 300 - Px
M = (300 - 2 P ) kN ◊ m at x = 2 m
SM L = 0 : M - ( x - 2 m) P + ( x - 1 m)(300 kN )
- x(300 - P ) = 0
M = (300 - 2 P ) kN ◊ m (consst)
Complete diagram by symmetry
For minimum |M |max , set M max = - M min
300 kN ◊ m - P = - [300 kN ◊ m - 2 P ]
M min = 300 - 2 P
(a)
P = 200 kN 
(b)
| M |max = 100 kN ◊ m 
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90
PROBLEM 7.60
Knowing that P = Q = 750 N, 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
+
SM A = 0 : Da - (750)(0.75) - (750)(1.5) = 0
D=
1687.5
, a in m. v
a
Free body: CB
+
1687.5
( a - 0.75) = 0 a
Ê 1.125 ˆ
M C = 1125 Á1 ˜
Ë
a ¯
SM C = 0 : - M C - (750)(0.75) +
v
Free body: DB
+
SM D = 0 : - M D - (750)(1.5 - a) = 0
M D = -750(1.5 - a) v
(a) We set
M max = | M min | or
Ê 1.125 ˆ
M C = - M D : 1125 Á1 ˜ = 750(1.5 - a)
Ë
a ¯
1.5 -
1.6875
= 1.5 - a
a
a2 = 1.6875 a = 1.299 m a = 1.299 m 
(b) | M |max = - M D = 750(1.5 - 1.299)
| M |max = 150.75 N ◊ m
| M | max = 150.8 N ◊ m 
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91
PROBLEM 7.61
Solve Problem 7.60 assuming that P = 1500 N and Q = 750 N.
PROBLEM 7.60 Knowing that P = Q = 750 N, 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
+
SM A = 0 : Da - (1500)(0.75) - (750)(1.5) = 0
D=
(a in m),
2250
v
a
Free body: CB
+
SM C = 0 : - M C - (750)(0.75) +
2250
( a - 0.75) = 0
a
Ê 1ˆ
M C = 1687.5 Á1 - ˜ v
Ë a¯
Free body: DB
+
SM D = 0 : - M D - (750)(1.5 - a) = 0
M D = - 750(1.5 - a) v
(a) We set
M max = | M min | or
Ê 1ˆ
M C = - M D : 1687.5 Á1 - ˜ = 750(1.5 - a)
Ë a¯
2.25
2.25 = 1.5 - a
a
a2 + 0.75a - 2.25 = 0
a = 1.17116 m
a = 1.171 m 
| M |max = - M D = 750(1.5 - 1.17116)
(b)
= 246.63 N ◊ m
|M | max = 247 N ◊ m 
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92
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:
SM J = 0 : - M -
x
wx = 0
2
1
M = - wx 2
2
M due to counterweight:
SM 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
1
So M min = WL - wL2 . For minimum | M |max set M max = - M min ,
2
1
W2
= - WL + wL2 or W 2 + 2wLW - w 2 L2 = 0
so
2w
2
W = - wL ± 2w 2 L2 ( need +) W = ( 2 - 1)wL = 0.414 wL 
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93
PROBLEM 7.62* (continued)
(b)
w may be removed
M max =
Without w,
W 2 ( 2 - 1)2 2
wL =
2w
2
M max = 0.858 wL2 
M = Wx
M max = WL at A
With w (see Part a)
M = Wx -
w 2
x
2
W2
W
at x =
w
2w
1
= WL - wL2 at x = L
2
M max =
M min
For minimum M max , set M max ( no w ) = - M min ( with w )
1
1
WL = - WL + wL2 Æ W = wL Æ 2
4
With
1 2
wL 
4
1
W = wL 
4
M max =
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94
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
+
SFy = 0 : C - P - P = 0
C = 2P ≠
+
SM C = 0 : P (2a) + P ( a) - M C = 0
MC = 3Pa
Shear diagram
VA = - P
At A:
|V | max = 2 P 
Bending-moment diagram
MA = 0
At A:
| M | max = 3Pa 
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95
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
+
Ê 2L ˆ
S M C = 0 : P Á ˜ - A( L) = 0
Ë 3¯
2
A= P≠
3
2
+ S FY = 0 :
P - P +C = 0
3
1
C= P ≠
3
Shear diagram
At A:
VA =
2
P
3
|V | max =
2
P 
3
Bending-moment diagram
MA = 0
At A:
|M | max =
2
PL 
9
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96
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
Reactions at A and D
Because of the symmetry of the supports and loading.
1 Ê Lˆ 1
A = D = Á w ˜ = wL
2Ë 2¯ 4
A=D=
Shear diagram
1
wL ≠ v
4
1
VA = + wL
4
At A:
From B to C:
|V | max =
Oblique straight line
1
wL 
4
Bending-moment diagram
MA = 0
At A:
3
wL2 
32
Since V has no discontinuity at B nor C, the slope of the parabola at these
points is the same as the slope of the adjoining straight line segment.
From B to C: ARC of parabola
|M | max =
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97
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
+
+
SFy = 0 : C - w
L
=0
2
1
C = wL ≠
2
Ê 1 ˆ Ê 3L ˆ
SM C = 0 : Á wL˜ Á ˜ = M C = 0
Ë2 ¯Ë 4 ¯
3
MC = wL2
8
Shear diagram
VA = 0
At A:
|V | max =
1
wL 
2
Bending-moment diagram
M =0
At A:
dM
=V = 0
dx
3
|M | max = wL2 
8
From A to B: ARC of parabola
Since V has no discontinuity at B, the slope of the parabola at B is equal to
the slope of the straightline segment.
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98
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
+
SFy = 0 : B - P = 0
B=P≠
+
S 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 v
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99
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
SFy = 0: A = C
+
SM C = 0 : Al - M 0 = 0
A=C =
M0
L
Shear diagram
At A:
VA = -
M0
L
|V | max =
M0

L
Bending-moment diagram
At A:
MA = 0
At B, M increases by M0 on account of applied couple.
|M | max = M 0 /2 v
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100
PROBLEM 7.69
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
+
SM A = 0 : (8)(2) + ( 4)(3.2) - 4C = 0
C = 7.2 kN ≠
SFy = 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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101
PROBLEM 7.70
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
+
SM D = 0 : (500 N )(0.8 m) + (500 N )(0.4 m)
- (2400 N/m)(0.3 m)(0.15 m) - A(1.2 m) = 0
A = 410 N ≠
+
SFy = 0 : 410 - 2(500) - 2400(0.3) + D = 0
D = 1310 N ≠
Shear diagram
At A:
VA = + 410 N |V | max = 720 N 
Bending-moment diagram
At A:
MA = 0 | M | max = 164.0 N ◊ m 
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102
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
SFx = 0 : Ax = 0
SM B = 0 : (60 kN )(3 m) + (50 kN )(1 m) - Ay (5 m) = 0
+
Ay = + 46.0 kN v
+
SFy = 0 : B + 46.0 kN - 60 kN - 50 kN = 0
B = + 64.0 kN v
Shear diagram
VA = Ay = + 46.0 kN
At A:
|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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103
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
+
SM B = 0 : (30 kN/m)(2 m)(6 m) - C (5 m) + (80 kN )(2 m) = 0
C = 104 kN ≠
+
SFy = 0 : 104 - 30(2) - 80 + B = 0
B = 36 kN ≠
Shear diagram
VA = 0
At A:
|V |max = 60.0 kN 
Bending-moment diagram
MA = 0
At A:
| M |max = 72.0 kN ◊ m 
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104
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:
1
A = B = ( 40 + 40 + 1.5 ¥ 60) kN
2
A = B = 85 kN ≠ v
Shear diagram
At A:
VA = +85 kN
|V |max = 85 kN 
Bending-moment diagram
At A:
MA = 0
| M |max = 67.88 kN ◊ m 
Discontinuities in slope at C and D, due to the discontinuities of V.
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105
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
+
SM A = 0 : B(5 m) - (120 kN )(1.5 m) - (60 kN )(3 m) = 0
B = +72.0 kN B = 72.0 kN ≠ v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay - 120 - 60 + 72 = 0
Ay = +108 kN A = 108.0 kN ≠ v
Shear diagram
At A:
VA = Ay = 108 kN
|V |max = 108.0 kN 
Bending-moment diagram
At A:
MA = 0
V = 0
At D:
| M |max = 145.8 kN ◊ m 
MD=Mmax=145.8 kN · m v
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106
PROBLEM 7.75
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
+
SM B = 0 : 8 kN ◊ m + (10 kN )( 4 m) + (20 kN )(2 m) - Ay (6 m) = 0
44
kN 
3
= 14.667 kN
Ay =
SFx = 0 : Ax = 0
Shear diagram
At D:
VD = -
46
kN = - 15.33 kN
3
|V | max = -15.33 kN 
Bending-moment diagram
At D:
92
kN◊m
3
= 30.667 kN◊m
MD =
| M |max = 30.7 kN ◊ m 
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107
PROBLEM 7.76
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
+
SM B = 0 : 10 kN ◊ m + 20 kN ◊ m + (10 kN )(6 m)
+ (15 kN )( 4 m) + (20 kN )(2 m) - Ay (8 m)) = 0
Ay = + 23.75 kN v
SFx = 0 : Ax = 0
Shear diagram
At A:
VA = Ay = +23.75 kN
|V |max = 23.8 kN 
Bending-moment diagram
At D:
MD = 65 kN ◊ m
| M |max = 65.0 kN ◊ m 
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108
PROBLEM 7.77
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
SFx = 0 : Ax = 0
+
SM B = 0 : (19.2 kN )(1.2 m) - Ay (3.2 m) = 0
Ay = + 7.20 kN v
Shear diagram
VA = VC = Ay = + 7.20 kN
To determine Point D where V = 0, we write
VD - VC = wu
0 - 7.20 kN = - (8 kN/m)u u = 0.9 m v
We next compute all areas
Bending-moment diagram
At A:
MA = 0
Largest value occurs at D with
AD = 0.8 + 0.9 = 1.700 m
| M |max = 9.00 kN ◊ m 
1.700 m from A 
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109
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 bending
moment.
Solution
Free body: Beam
SFx = 0 : Ax = 0
+
SM B = 0 : ( 40 kN )(1.5 m) - Ay (3.75 m) = 0
Ay = +16.00 kN v
Shear diagram
VA = VC = Ay = +16.00 kN
To determine Point E where V = 0, we write
VE - VC = - wu
0 - 16 kN = - (20 kN/m)u u = 0.800 m v
We next compute all areas
Bending-moment diagram
At A:
MA = 0
Largest value occurs at E with
AE = 1.25 + 0.8 = 2.05 m | M |max = 26.4 kN ◊ m 
2.05 m from A 
From A to C and D to B: Straight line segments. From C to D: Parabola
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110
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: Entire beam
+
SM C = 0 : - ( 40 kN ◊ m) + A(7.5 m) - (15 kN/m)(6 m)( 4.5 m) = 0
A=
+
SFy = 0 :
178
kN =59.33kN ≠
3
178
- (90) + C = 0
3
92
C=
kN =30.67 kN ≠
3
Shear diagram
VA = +
At A:
178
kN
3
To locate Point D (where V = 0)
VD - VA = - wu
178
0kN = - 15 kN/m u
3
u = 3.9556 m
Bending-moment diagram
MD = 77.35 kN ◊ m
| M |max = 77.4 kN ◊ m 
3.96 m from A 
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111
PROBLEM 7.80
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
+
SM A = 0 : (30 kN )(5 m) - (75 kN )(3.5 m) + B(2 m) = 0
B = + 56.25 kN
B = 56.25 kN ≠
+
SFy = 0 : A + 56.25 kN - 75 kN + 30 kN = 0
A = -11.25 kN
V(kN)
(–18)
(40.5)
A
(–22.5)
–11.25
B
1.2 m
D
–30
1.8 m
BD =
VB 45
=
= 1.8 m
10 25
m(kN.m)
18
A
–22.5
+18
B
D
–22.5
M max = 22.5 kN ◊ m 
2 m from A
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112
PROBLEM 7.81
(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
1
A = B = (5 ¥ 4 + 1.5)
2
A = B = 10.75 kN ≠ v
Load diagram for AB
The 1.5 kN force at D is replaced by an equivalent force-couple system at C.
Shear diagram
VA = A = 21.5 kN
At A:
To determine Point E where V = 0:
VE - VA = - wx
0 - 10.75 kN = - (5 kN/m) x
10.75
x=
5
x = 2.15 m v
We compute all areas
Bending-moment diagram
MA = 0
At A:
Note 1.5 kN◊m drop at C due to couple
| M |max = 11.56 kN ◊ m 
2.15 m from A 
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113
PROBLEM 7.82
Solve Problem 7.81 assuming that the 1.5 kN force applied at D is directed
upward.
PROBLEM 7.81 (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:
1
A = B = (5 ¥ 4 - 1.5)
2
A = B = 9.25 kN ≠ v
Load diagram
The 1.5 kN force at D is replaced by an equivalent force-couple
system at C.
Shear diagram
VA = A = 9.25 kN
At A:
To determine Point E where V = 0:
VE - VA = - wx
0 - 9.25 kN = - (5 kN/m) x
9.25
x=
= 1.85 m
5
x = 1.85 m v
We compute all areas
Bending-moment diagram
MA = 0
At A:
Note 1.5 kN ◊ m increase at C due to couple
| M |max = 8.56 kN ◊ m 
1.85 m from A 
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114
PROBLEM 7.83
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
+
SM A = 0 : B( 4 m) - (100 kN )(2 m) - 20 kN◊m = 0
B = + 55 kN v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay + 55 - 100 = 0
Ay = + 45 kN v
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 v
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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115
PROBLEM 7.84
Solve Problem 7.83 assuming that the 20-kN · m couple applied at B
is counterclockwise.
PROBLEM 7.83 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
+
SM A = 0 : B( 4 m) - (100 kN )(2 m) - 20 kN ◊ m = 0
B = + 45 kN v
SFx = 0 : Ax = 0
+
SFy = 0 : Ay + 45 - 100 = 0
Ay = + 55 kN v
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 v
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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116
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
Distributed load
xˆ
˜
L¯
Ê
w = w0 Á 1 Ë
SM A = 0 :
L
3
1
ˆ
Ê
ÁË Total = w0 L˜¯
2
ˆ
Ê1
ÁË w0 L˜¯ - LB = 0
2
B=
w0 L
≠
6
w L
w L
1
≠ SFy = 0 : Ay - w0 L + 0 = 0 A y = 0 ≠
2
6
3
w0 L
3
Shear:
VA = Ay =
Then
dV
= -w Æ V
dx
x
Ê
= V A - Ú w0 Á 1 0
Ë
xˆ
˜ dx
L¯
1 w0 2
Ê w Lˆ
x
V = Á 0 ˜ - w0 x +
Ë 3 ¯
2 L
È1 x 1 Ê x ˆ 2 ˘
= w0 L Í - + Á ˜ ˙
ÍÎ 3 L 2 Ë L ¯ ˙˚
Note: At
x=L
V =-
w0 L
6
2
Ê xˆ 2
Ê xˆ
V = 0 at Á ˜ - 2 Á ˜ +
Ë L¯ 3
Ë L¯
=0Æ
1
x
=13
L
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117
problem 7.85 (continued)
Moment:
Then
MA = 0
x
x/L Ê x ˆ Ê x ˆ
Ê dM ˆ
˜¯ = V Æ M = Ú0 Vdx = L Ú0 V ÁË ˜¯ d ÁË ˜¯
ÁË
dx
L
L
M = w0 L2 Ú
x/ L È1
0
È1 Ê x ˆ
M = w0 L2 Í Á ˜
ÍÎ 3 Ë L ¯
2˘
Ê xˆ
˙dÁ ˜
˙˚ Ë L ¯
2
3
1Ê xˆ ˘
1Ê xˆ
- Á ˜ + Á ˜ ˙
6 Ë L ¯ ˙˚
2 Ë L¯
x 1Ê xˆ
Í - + Á ˜
ÍÎ 3 L 2 Ë L ¯
Ê x
1ˆ
M max Á at = 1 = 0.06415w0 L2
˜
3¯
Ë L
È1 x 1 Ê x ˆ 2 ˘
V = w0 L Í - + Á ˜ ˙ 
ÍÎ 3 L 2 Ë L ¯ ˙˚
(a)
È1 Ê x ˆ 1 Ê x ˆ 2 1 Ê x ˆ 3 ˘
M = w0 L2 Í Á ˜ - Á ˜ + Á ˜ ˙ 
6 Ë L ¯ ˙˚
ÍÎ 3 Ë L ¯ 2 Ë L ¯
M max = 0.0642 w0 L2 
(c)
at x = 0.423L 
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118
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
(a)
We note that at
Load:
B( x = L) : VB = 0, M B = 0 (1)
xˆ 1 Ê xˆ
Ê 4x ˆ
Ê
w ( x ) = w 0 Á 1 - ˜ - w0 Á ˜ = w0 Á 1 - ˜
Ë 3L ¯
Ë
L¯ 3 Ë L¯
Shear: We use Eq. (7.2) between C ( x = x ) and B( x = L):
L
L
x
x
VB - VC = - Ú w ( x )dx 0 - V ( x ) = - Ú w ( x )dx
LÊ
4x ˆ
V ( x ) = w0 Ú Á1 - ˜ dx
x Ë
3L ¯
L
Ê
È
2x2 ˘
2L
2x2 ˆ
= w0 Í x =
+
w
L
x
˙
0Á
3L ˚ x
3
3L ˜¯
Ë
Î
w
V ( x ) = 0 (2 x 2 - 3Lx + L2 ) 3L
(2) 
Bending moment: We use Eq. (7.4) between C ( x = x ) and B( x = L) :
L
M B - M C = Ú V ( x )dx 0 - M ( x )
x
w L
= 0 Ú (2 x 2 - 3Lx + L2 )dx
3L x
L
M ( x) = -
w0 È 2 3 3 2
˘
x - Lx + L2 x ˙
Í
3L Î 3
2
˚x
w0
[4 x 3 - 9 Lx 2 + 6 L2 x ]Lx
18 L
w
= - 0 [( 4 L3 - 9 L3 + 6 L3 ) - ( 4 x 3 - 9 Lx 2 + 6 L2 x )]
18 L
=-
M ( x) =
w0
( 4 x 3 - 9 Lx 2 + 6 L2 x - L3 ) 18L
(3) 
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119
problem 7.86 (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
x=
w0 L2
18
x=0
At
L
2

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120
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
p x
dv
= - w = w0 cos
2L
dx
px
Ê 2L ˆ
+ C1 V = - Ú wdx = - w0 Á ˜ sin
Ë p ¯
2L
(1)
dM
px
Ê 2L ˆ
+ C1
= V = - w0 Á ˜ sin
¯
Ë
p
dx
2L
2
px
Ê 2L ˆ
+ C1 x + C2 M = Ú Vdx = + w0 Á ˜ cos
Ë p ¯
2L
(2)
Boundary conditions
At x = 0:
V = C1 = 0 C1 = 0
At x = 0:
Ê 2L ˆ
M = + w0 Á ˜ cos(0) + C2 = 0
Ë p ¯
2
Ê 2L ˆ
C 2 = - w0 Á ˜
Ë p ¯
2
px
Ê 2L ˆ
V = - w0 Á ˜ sin

Ë p ¯
2L
Eq. (1)
2
p xˆ
Ê 2L ˆ Ê
M = w0 Á ˜ Á -1 + cos ˜ 
Ë p ¯ Ë
2L ¯
2
4
Ê 2L ˆ
| M max | = w0 Á ˜ | - 1 + 0 |= 2 w0 L2 
Ë p ¯
p
M max at x = L : PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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121
PROBLEM 7.88
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
L
4 w0
0
L2
≠ SFy = 0 : w g L - Ú
(a)
( Lx - x 2 )dx = 0
4 w0 Ê 1 2 1 3 ˆ 2
Á LL - L ˜¯ = w0 L
3
3
L2 Ë 2
2w
wg = 0
3
wg L =
È x Ê xˆ2˘ 2
Í - Á ˜ ˙ - w0
ÍÎ L Ë L ¯ ˙˚ 3
x
dx
so dx =
® net load w = 4 w0
L
L
Define
x=
or
ˆ
Ê 1
w = 4 w0 Á - + x - x 2 ˜
¯
Ë 6
x
ˆ
Ê 1
V = V ( 0 ) - Ú 4 w0 L Á - + x - x 2 ˜ d x
0
¯
Ë 6
Ê1
1
1
ˆ
= 0 + 4 w0 L Á x + x 2 - x 3 ˜
Ë6
2
3 ¯
V=
x
x
2
M = M 0 + Ú Vdx = 0 + w0 L2 Ú (x - 3x 2 + 2x 3 )dx
0
0
3
2
1 ˆ 1
Ê1
= w0 L2 Á x 2 - x 3 + x 4 ˜ = w0 L2 (x 2 - 2x 3 + x 4 ) Ë2
3
2 ¯ 3
(b)
Max M occurs where
V = 0 ®1 - 3x + 2x 2 = 0 ® x =
2
w0 L (x - 3x 2 + 2x 3 ) 
3

1
2
2
1ˆ 1
Ê1 2 1 ˆ w L
Ê
M Á x = ˜ = w0 L2 Á - + ˜ = 0
Ë 4 8 16 ¯
Ë
2¯ 3
48
M max =
w0 L2
at center of beam 
48
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122
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
SFx = 0 : C x = 0
+
SM 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 ≠ v
Free body: Portion EB
+
SM E = 0 : B(0.3 m) - 1.300 kN ◊ m = 0
B = 4.333 kN ≠ v
Free body: Entire beam
+
SM 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 Ø 
+
SM y = 0 : 11.667 kN + 4.333 kN
-6 kN - P - 6 kN = 0
P = 4.00 kN Ø 
Load diagram
(b) Shear diagram
At A: VA = 0
|V |max = 6 kN v
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123
Problem 7.89 (continued)
Bending-moment diagram
MA = 0
At A:
| M |max = 1300 N ◊ m v
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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124
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
SFx = 0 : C x = 0 +
SM 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 ≠ v
Free body: Portion EB
+
SM E = 0 : B(0.3 m) - 1.450 kN ◊ m = 0 B = 4.833 kN ≠ v
Free body: Entire beam
+
SM 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 Ø 
+
SM y = 0 : 11.167 kN + 4.833 kN
- 6 kN - P - 7.50 kN = 0
P = 2.50 kN Ø 
Load diagram
(b)
Shear diagram
VA = 0
At A:
|V |max = 6 kN v
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125
Problem 7.90 (continued)
Bending-moment diagram
MA = 0
At A:
| M |max = 1450 N ◊ m v
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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126
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 +20 kN◊m at D and +19.0 kN ◊ m at E, (a) determine
P and Q, (b) draw the shear and bending-moment diagrams for the beam.
Solution
(a)
Free body: Portion DE
+
SM E = 0 : 19 kN ◊ m - 20 kN ◊ m + (8kN)(1 m) - VD (2 m) = 0 VD = +3.5 kN +
SFy = 0 : 3.5 kN - 8 kN - VE = 0 VE = -4.5 kN Free body: Portion AD
+
SM A = 0 : 20 kN ◊ m - P (1 m) - (8 kN)(1 m) - (3.5 kN)(2 m) = 0 P = 5.00 kN Ø 
SFx = 0 : Ax = 0 +
SFy = 0 : Ay - 8 kN - 5 kN - 3.5 kN = 0 Ay = +16.5 kN A = 16.5 kN ≠ 
Free body: Portion EB
+
SM B = 0 : ( 4.5 kN)(2 m) + (8 kN)(1 m) + Q(1 m) - 19.0 kN ◊ m = 0
Q = 2.00 kN Ø 
+
SFy = 0 : B - 2 kN - 8 kN - 4.5 kN = 0
B = 14.5 kN ≠ 
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127
Problem 7.91* (continued)
(b)
Load diagram
Shear diagram
VA = A = +16.5 kN
At A:
To determine Point G where V = 0, we write
VG - VC = - w m
0 - 7.5 kN = -4 m fi m =
7.5
4
m = 1.875 m 
CG = 1.875 m
GF = 4 - 1.875 = 2.125 m
|V |max = 16.5 kN at A 
We next compute all areas
Bending-moment diagram
At A: M A = 0
Largest value occurs at G with
AG = 1 + 1.875 = 2.875 m | M |max = 21.5 kN ◊ m 
2.88 m from A 
Bending-moment diagram consists of 3 distinct arcs of parabolas.
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128
PROBLEM 7.92*
Solve Problem 7.91 assuming that the bending moment was found to be
+21 kN ◊ m at D and +23 kN ◊ m 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 +20 kN · m at D and
+19.0 kN · m at E, (a) determine P and Q, (b) draw the shear and ­bendingmoment diagrams for the beam.
Solution
(a)
Free body: Portion DE
+
SM E = 0 : 23 kN ◊ m - 21 kN ◊ m + (8 kN)(1 m) - VD (2 m) = 0 VD = +5 kN
+
SFy = 0 : 5 kN - 8 kN - VE = 0 VE = -3 kN
Free body: Portion AD
SM A = 0 : 21 kN ◊ m - P (1 m) - (8 kN)(1 m) - (5 kN)(2 m) = 0 +
P = 3 kN P = 3.00 kN Ø 
SFx = 0 : Ax = 0
+
SFy = 0 : Ay - 8 kN - 3 kN - 5 kN = 0 Ay = 16 kN A = 16 kN ≠ v
Free body: Portion EB
+
SM B = 0 : (3 kN)(2 m) + (8 kN)(1 m) + Q(1 m) - 23 kN ◊ m = 0 Q = 9 kN Q = 9.00 kN Ø 
+
SFy = 0 : B - 8 - 9 - 3 = 0 B = 20 kN ≠ 
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129
Problem 7.92* (continued)
(b)
Load diagram
Shear diagram
At A:
VA = A = 16 kN
To determine Point G where V = 0, we write
VG - VC = - w m 0 - 9 = -4 m
9
m = = 2.25 m
4
CG = 2.25 m
GF = 1.75 m
m = 2.25 m v
|V |max = 20 kN at B 
We next compute all areas
Bending-moment diagram
At A: M A = 0
Largest value occurs at G with
AG = 1 + 2.25 = 3.25 m | M |max = 24.1 kN ◊ m 
3.25 m from A 
Bending-moment diagram consists of 3 distinct arcs of parabolas.
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130
PROBLEM 7.93
Two loads are suspended as shown from the cable ABCD. Knowing
that hB = 1.8 m, determine (a) the distance hC , (b) the components
of the reaction at D, (c) the maximum tension in the cable.
Solution
® SFx = 0 : - Ax + Dx = 0
FBD Cable:
Ax = Dx
SM A = 0 : (10 m)D y - (6 m)(10 kN) - (3 m)(6 kN) = 0
D y = 7.8 kN ≠
≠ SFy = 0 : Ay - 6 kN - 10 kN + 7.8 kN = 0
A y = 8.2 kN≠
FDB AB:
SM B = 0 : (1.8 m) Ax - (3 m)(8.2 kN) = 0
Ax =
From above
FBD CD:
Dx = Ax =
41
kN¨
3
41
kN
3
ˆ
Ê 41
kN ˜ = 0
SM C = 0 : ( 4 m)(7.8 kN) - hC Á
¯
Ë 3
hC = 2.283 m
(a)
hC = 2.28 m 
(b)
D x = 13.67 kN ® 
D y = 7.80 kN ≠ 
Since Ax = Bx and Ay > By , max T is TAB
2
TAB =
ˆ
Ê 41
Ax2 + Ay2 = Á
kN ˜ + (8.2 kN)2 ¯
Ë 3
Tmax = 15.94 kN 
(c)
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131
PROBLEM 7.94
Knowing that the maximum tension in cable ABCD is 15 kN,
determine (a) the distance hB , (b) the distance hC .
Solution
® SFx = 0 : - Ax + Dx = 0
FBD Cable:
Ax = Dx
SM A = 0 : (10 m) D y - (6 m)(10 kN) - (3 m)(6 kN) = 0
D y = 7.8 kN ≠
≠ SFy = 0 : Ay - 6 kN - 10 kN + 7.8 kN = 0
A y = 8.2 kN ≠
Since
Ax = Dx
and
Ay > D y , Tmax = TAB FBD Pt A:
≠ SFy = 0 : 8.2 kN - (15 kN) sin q A = 0 q A = sin -1
8.2 kN
= 33.139∞
15 kN
® SFx = 0 : - Ax + (15 kN) cos q A = 0
Ax = (15 kN) cos(33.139∞) = 12.56 kN FBD CD:
From FBD cable:
hB = (3 m) tan q A
= (3 m) tan(33.139∞)
(a)
hB = 1.959 m 
SM C = 0 : ( 4 m)(7.8 kN) - hC (12.56 kN) = 0 hC = 2.48 m 
(b)
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132
PROBLEM 7.95
If dC = 4 m, determine (a) the reaction at A, (b) the reaction at E.
Solution
4m
3m
4m
Ay
A
Ax
4m
dc = ym
4m
D
(a)
Ey
FBD cable:
SM E = 0 : ( 4 m)(1500 N) + (8 m)(1000 N) + (12 m)(1500 N)
E E
x
- (3 m)Ax - (16 m)Ay = 0
C
B
1500 N
1000 N
1500 N
4m
Ax
4m
Ax + 8 Ay = -6000 N ICD
A
B
1500 N
(1)
FBD ABC:
SM C = 0 : ( 4 m)(1500 N) + (1m)Ax - (8 m)Ay = 0
4m
Ay
3m
3 Ax + 16 Ay = 32000 N (2)
C
1000 N
Solving (1) and (2)
Ax = 4000 N
Ay = 1250 N
So A = 4190 N
17.35° 
cable: Æ SFx = 0 : - Ax + E x = 0
(b)
E x = Ax = 4000 N
≠ SFx = 0 : Ay - (1500 + 1000 + 1500) kN + E y = 0
E y = 4000 N - Ay = 4000 - 1250 = 2750 N
So E = 4850 N
34.5∞ 
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133
PROBLEM 7.96
If dC = 2.25 m, determine (a) the reaction at A, (b) the reaction at E.
Solution
FBD Cable:
4m
3m
Ay
A
Ax
4m
4m
dc
4m
D
C
B
Ey
E
Ex
dc =2.25m
SM E = 0 : ( 4 m)(1500 N) + (8 m)(1000 N)
(a)
+(12 m)(1500 N) - (3m)Ax - (16 m)Ay = 0
3 Ax + 16 Ay = 32000 N 1500 N
1000 N
1500 N
(1)
FBD ABC:
S M C = 0 : ( 4 m)(1500 N) - (0.75 m)Ax - (8 m)Ay = 0
Ay
3m
Ax
2.25 m
B
A
4m
C
4m
1000 N
1500 N
0.75 Ax + 8 Ay = 6000 N ICD
Solving (1) and (2)
Ax
40000
N,
3
(2)
Ay = -500 N
A = 13340 N
So
2.15° 
Note: this implies dB < 3 m (in fact dB = 2.85 m)
(b)
FBD cable: Æ SFx = 0 : -
40000
N + Ex = 0
3
Ex =
40000
N
3
≠ SFy = 0 : 500 N - 1500 N - 1000 N - 1500 N + E y = 0
E y = 4500 N
E = 14070 N
18.65∞ 
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134
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
+
SM C = 0 : 3 Ax - 4 Ay + (5 kN)(2 m) = 0
Ax =
4
10
Ay - 3
3
(1)
Free body: Entire cable
+
SM 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 ˆ
Ê4
4 Á Ay - ˜ - 10 Ay + 100 = 0
Ë3
3¯
Ay = +18.571 kN
4
10
Ax = (18.511) - = +21.429 kN
3
3
Eq. (1)
+
+
SFx = 0 : - Ax + E x = 0 - 21.429 + E x = 0
SFy = 0 : 18.571 kN + E y + 5 kN + 5 kN + 10 kN = 0
A y = 18.571 kN ≠
A x = 21.429 kN ¨
E x = 21.429 kN ®
E y = 1.429 kN ≠
E = 21.5 kN
3.81∞ 
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135
PROBLEM 7.97 (Continued)
Portion AB
+
SM 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
dD = 4.20 m 
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136
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
+
+
SFy = 0 : Ay - 5 kN - 5 kN - 10 kN = 0
A y = 20 kN ≠
SM A = 0 : E ( 4 m) - (5 kN )(2 m) - (5 kN )( 4 m) - (10 kN )(7m) = 0
E = +25 kN E = 25.0 kN ® 
SFx = 0 : - Ax + 25 kN = 0
A x = 25 kN ¨
A = 32.0 kN
38.7∞ 
Free body: Portion ABC
+
SM 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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137
PROBLEM 7.99
If dC = 5 m, determine (a) the distances dB and dD, (b) the
maximum tension in the cable.
Solution
Free body: Entire cable
+
SM A = 0 : E y (10 m) - E x (2.5 m) - (10 kN )(2 m) - (10 kN )(5 m) - (10 kN )(7 m) = 0
2.5E x - 10 E y + 140 = 0 (1)
Free body: Portion CDE
+
SM C = 0 : E y (5 m) - E x (5 m) - (10 kN )(2 m) = 0
5E x - 5E y + 20 = 0 (2)
Solving (1) and (2)
40
kN
3
52
Ey =
kN
3
Ex =
2
Ê 52 ˆ
Ê 40 ˆ
Tm = E x2 + E y2 = Á ˜ + Á ˜
Ë 3¯
Ë 3¯
Tm = 21.8683 kN
2
Tm = 21.9 kN 
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138
PROBLEM 7.99 (Continued)
Free body: Portion DE
+
Ê 40 ˆ
Ê 52 ˆ
SM D = 0 : Á kN ˜ (3 m) - Á kN˜ d D = 0 ¯
Ë 3
¯
Ë 3
Return to free body of entire cable
+
SFy = 0 : Ay - 3(10 kN ) +
SFx = 0 :
dD = 3.90 m 
Ê
Ê 52 ˆ ˆ
Ê 40 ˆ
ÁË with E x = ÁË 3 kN ˜¯ , E y = ÁË 3 kN ˜¯ ˜¯
52
kN = 0
3
40
- Ax = 0
3
Ay =
38
kN
3
Ax =
40
kN
3
+
Free body: Portion AB
+
SM B = 0 : Ax ( d B - 2.5) - Ay (2) = 0
76
Ê 40 ˆ
=0
ÁË ˜¯ ( d B - 2.5) 3
3
d B - 2.5 = 1.9
d B = 4.4 m
d B = 4.40 m 
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139
PROBLEM 7.100
Determine (a) the distance dC for which portion BC of the
cable is horizontal, (b) the corresponding components of
the reaction at E.
Solution
Free body: Portion CDE
+
+
SFy = 0 : E y - 2(10 kN ) = 0 E y = 20 kN
SM C = 0 : (20 kN )(5 m) - E x dC - (10 kN )(2 m) = 0
E x dC = 80 kN ◊ m (1)
Free body: Entire cable
+
SM A = 0 : (20 kN )(10 m) - E x (2.5 m) - (10 kN )(2 m) - (10 kN )(5 m) - (10 kN )(7 m)) = 0
E x = 24 kN
From Eq. (1):
dC =
80 80 10
=
=
E x 24 3
dC = 3.33 m 
E x = 24 kN ®;
Components of reaction at E:
E y = 20 kN ≠ 
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140
PROBLEM 7.101
Cable ABC supports two loads as shown. Knowing that b = 1.2 m, determine
(a) the required magnitude of the horizontal force P, (b) the corresponding
distance a.
Solution
FBD ABC:
≠ SFy = 0 : - 200 N - 400 N + C y = 0
C y = 600 N ≠
FBD BC:
SM B = 0 : (1.2 m)(600 N ) - (3 m)C x = 0
C x = 240 N ®
From ABC:
® SFx = 0 : - P + C x = 0
P = C x = 240 N
(a)
P = 240 N 
SM C = 0 : (1.2 m)( 400 N ) + a(200 N ) - ( 4.5 m)(240 N ) = 0
(b)
a = 3.00 m 
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141
PROBLEM 7.102
Cable ABC supports two loads as shown. Determine the distances a and b when
a horizontal force P of magnitude 300 N is applied at A.
Solution
FBD ABC:
® SFx = 0 : C x - P = 0
C x = 300 N ®
≠ SFy = 0 : C y - 200 N - 400 N = 0
C y = 600 N
FBD BC:
SM B = 0 : b(600 N ) - (3 m)(300 N ) = 0 : b =
3
m
2
b = 1.500 m 
FBD AB:
SM B = 0 : ( a - b)(200 N ) - (1.5 m)300 N = 0
a - b = 2.25 m
a = b + 2.25 m
= 1.5 m + 2.25 m
a = 3.75 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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142
PROBLEM 7.103
Knowing that mB = 70 kg and mC = 25 kg, determine the
­ agnitude of the force P required to maintain equilibrium.
m
Solution
Free body: Portion CD
+
SM C = 0 : D y ( 4 m) - Dx (3 m) = 0
Dy =
3
Dx
4
Free body: Entire cable
+
SM A = 0 :
3
Dx (14 m) - WB ( 4 m) - WC (10 m) - P (5 m) = 0 4
+
SM 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/s2 :
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 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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143
PROBLEM 7.104
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
+
SM C = 0 : D y ( 4 m) - Dx (3 m) = 0
Dy =
3
Dx
4
Free body: Entire cable
+
SM A = 0 :
3
Dx (14 m) - WB ( 4 m) - WC (10 m) - P (5 m) = 0 4
+
SM B = 0 :
3
Dx (10 m) - Dx (5 m) - WC (6 m) = 0
4
(1)
Free body: Portion BCD
Dx = 2.4WC (2)
For mB = 18 kg mC = 10 kg
g = 9.81 m/s2 :
WB = 18 g
Eq. (2):
Dx = 2.4WC = 2.4(10 g ) = 24 g
3
24 g (14) - (18 g )( 4) - (10 g )(10) - 5P = 0
4
80 g - 5P : P = 16 g
Eq. (1):
WC = 10 g
P = 16(9.81) = 156.96 N P = 157.0 N 
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144
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
+
SM D = 0 : E y ( 4 m) - E x (5 m) = 0
Ey =
5
Ex
4
Free body: Portion CDE
+
SM C = 0 :
5
E x (8 m) - E x (7 m) - P (2 m) = 0
4
2
Ex = P 3
(1)
Free body: Portion BCDE
+
SM B = 0 :
5
E x (12 m) - E x (5 m) - (120 kN )( 4 m) = 0
4
10 E x - 480 = 0; E x = 48 kN
48 kN =
Eq. (1):
2
P
3
P = 72.0 kN 
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145
PROBLEM 7.105 (Continued)
Free body: Entire cable
+
SM A = 0 :
5
E x (16 m) - E x (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 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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146
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
+
SM D = 0 : E y ( 4 m) - E x (6 m) = 0
Ey =
3
Ex
2
Free body: Portion CDE
+
SM C = 0 :
3
E x (8 m) - E x (8 m) - P (2 m) = 0
2
1
Ex = P 2
(1)
Free body: Portion BCDE
+
SM B = 0 :
3
E x (12 m) - E x (6 m) + (120 kN )( 4 m) = 0
2
12 E x = 480
Eq (1):
Ex =
E x = 40 kN
1
P;
2
40 kN =
1
P
2
P = 80.0 kN 
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147
PROBLEM 7.106 (Continued)
Free body: Entire cable
+
SM A = 0 :
3
E x (16 m) - E x (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 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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148
PROBLEM 7.107
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
y B = 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 = 2sB = 2(69.0478 m) = 138.0956 m Eq. 7.18:
L = 138.1 m 
Tm = wy B = w ( h + c)
= (0.65 kg/m)(9.81 m/s2 )(30 m + 64.459 m)
Tm = 602.32 N Tm = 602 N 
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149
PROBLEM 7.108
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 = wx B
+
SM B = 0 : T0 y B - ( wx B )
Horiz. comp. = T0 =
(a)
Cable AB
w ( 45 m)2
2h
x B = 30 m,
T0 =
Equate T0 = T0
wx B2
2 yB
x B = 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
x B = 45 m,
T0 =
y B = h = 6.75 m
w x B2 (3.924 N/m)( 45 m)2
=
= 588.6 N
2 yB
2(6.75 m)
W = wx B = (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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150
PROBLEM 7.108 (Continued)
Cable BC
x B = 30 m,
yB = 3 m
wx B2 (3.924 N/m)(30 m)2
=
= 588.6 N (Checks)
2 yB
2(3 m)
T0 =
W = wx B = (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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151
PROBLEM 7.109
Each cable of the Golden Gate Bridge supports a load w = 180 kN/m along the horizontal. Knowing that the
span L is 1245 m and that the sag h is 140 m, determine (a) the maximum tension in each cable, (b) the length
of each cable.
Solution
Eq. (7.8) Page 386:
At B:
(a)
yB =
wx B2
2T0
T0 =
wx B2 (180 kN/m)(622.5 m)2
=
2 yB
2(140 m)
T0 = 249,111 kN
W = wx B = (180 kN/m)(622.5 m) = 112,050 kN
Tm = T02 + W 2 = (249,111)2 + (112, 050 kN )2 = 273,151 kN
Tm = 273,000 kN 
(b)
˘
È 2 Ê y ˆ2 2 Ê y ˆ4
sB = x B Í1 + Á B ˜ - Á B ˜ + L˙
5 Ë yB ¯
˙˚
ÍÎ 3 Ë x B ¯
y B 140 m
=
= 0.22490
x B 6225 m
2
È 2
˘
sB = (622.5 m) Í1 + (0.2249)2 - (0.2249)4 + L˙ = 642.854 m
3
5
Î
˚
Length = 2sB = 2(642.854 m) =1285.708 m
Length = 1286 m 
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152
PROBLEM 7.110
The center span of the George Washington Bridge, as originally constructed, consisted of a uniform ­roadway
suspended from four cables. The uniform load supported by each cable was w = 160 kN/m along the ­horizontal.
Knowing that the span L is 1050 m and that the sag h is 95 m, determine for the original configuration
(a) the maximum tension in each cable, (b) the length of each cable.
Solution
W = wx B = (160 kN/m)(525 m)
W = 84, 000 kN
+
SM B = 0 : T0 (95 m) - (84, 000 kN)(262.5 m) = 0
T0 = 232,105 kN
(a)
Tm = T02 + W 2
= (84, 000)2 + (232,105)2 = 246, 837 kN
Tm = 247, 000 kN 
(b)
˘
È 2 Ê y ˆ2 2 Ê y ˆ4
sB = x B Í1 + Á B ˜ - Á B ˜ + L˙
5 Ë xB ¯
˙˚
ÍÎ 3 Ë x B ¯
95 m
yB
=
= 0.18095
x B 525 m
2
È 2
˘
= (525 m) Í1+ (0.18095)2 - (0.18095) 4 + L˙
5
Î 3
˚
sB = 536.235 m; Length = 2sB = 1072.47 m Length = 1072 m 
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153
PROBLEM 7.111
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:
Block:
+
m = 25 kg
W = 25 (9.81)
= 245.25 N
m = 450 kg
W = 4414.5 N
SM B = 0 : (245.25)(2.5) + ( 4414.5)(3) - C x (2.5) = 0
C x = 5543 N
SFx = 0 : Ax = C x = 5543 N
+
SM A = 0 : C y (5) - (5543)(2.5) - (245.25)(2.5) = 0
C y = 2894 N ≠
+
SFy = 0 : C y - Ay - 245.25N = 0
2894 N - Ay - 245.25 N = 0
Point A:
tan f A =
Point C: tan fC =
Ay
Ax
Cy
Cx
A y = 2649 N Ø
=
2649
= 0.4779; 5543
f A = 25.5∞ 
=
2894
= 0.5221; 5543
fC = 27.6∞ 
W = (12.5 kg) g = 122.6
Free body: Half cable
+
SM 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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154
PROBLEM 7.112
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
sB = (50.5 m) = 25.25 m,
2
1
x B = (50 m) = 25 m
2
yB = h
(a)
È 2 Ê y ˆ2˘
sB = x B Í1 + Á B ˜ ˙
ÍÎ 3 Ë x B ¯ ˙˚
È 2 Ê y ˆ2˘
25.25 m = 25 m Í1 - Á B ˜ ˙
ÍÎ 3 Ë x B ¯ ˙˚
2
2
Ê yB ˆ
Ê 3ˆ
=
0
.
01
ÁË ˜¯ = 0.015
ÁË x ˜¯
2
B
yB
= 0.12247
xB
h
= 0.12247
25 m
h = 3.0619 m (b)
h = 3.06 m v
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
+ SM 0 = 0 : T0 (3.0619 m) - (185.78 N)(12.5 m) = 0
T0 = 758.4 N
Bx = T0 = 758.4 N
+
SFy = 0 : By - 185.78 N = 0 By = 185.78 N
Tm = Bx2 + By2 = (758.4 N )2 + (185.78 N )2 Tm = 781 N v
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155
PROBLEM 7.113
A cable of length L +  is suspended between two points that are at the same elevation and a distance L apart.
(a) Assuming that  is small compared to L and that the cable is parabolic, determine the approximate sag in
terms of L and . (b) If L = 30 m and  = 1.2 m, determine the approximate sag. [Hint: Use only the first two
terms of Eq. (7.10).
Solution
Eq. 7.10
(a)
(First two terms)
sB
xB
sB
yB
È 2 Ê y ˆ2˘
= x B Í1 + Á B ˜ ˙
ÍÎ 3 Ë x B ¯ ˙˚
= L /2
1
= ( L + D)
2
=h
2
L ÈÍ 2 Ê h ˆ ˘˙
1
( L + D) = 1 + Á L ˜
2
2 Í 3Ë 2¯ ˙
Î
˚
D 4 h2
3
=
; h2 = L D; 2 3 L
8
(b)
L = 30 m, D = 1.2 m
h=
h=
3
3 6
(30)(1.2) =
m; 8
2
3
LD 
8
h = 3.67 m 
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156
PROBLEM 7.114
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 = 116 m in winter to hs = 118 m in summer. Knowing that the span is L = 1278 m,
determine the change in length of the cables due to extreme temperature changes.
Solution
Eq. 7.10.
Winter:
Summer:
˘
È 2 Ê y ˆ2 2 Ê y ˆ4
sB = x B Í1 + Á B ˜ - Á B ˜ + L˙
5 Ë xB ¯
˙˚
ÍÎ 3 Ë x B ¯
1
y B = h = 116 m, x B = L = 639 m
2
È 2 Ê 116 ˆ 2 2 Ê 116 ˆ 4
˘
+ L˙ = 652.761 m
- Á
sB = (639) Í1 + Á
˜
˜
5 Ë 639 ¯
ÍÎ 3 Ë 639 ¯
˙˚
1
y B = h = 118 m, x B = L = 639 m
2
È 2 Ê 118 ˆ 2 2 Ê 118 ˆ 4
˘
+ L˙ = 635.230 m
- Á
sB = (639) Í1 + Á
˜
˜
5 Ë 639 ¯
ÍÎ 3 Ë 639 ¯
˙˚
D = 2( D sB ) = 2(653.230 m - 652.761 m) = 2(0.469 m)
Change in length = 0.938 m 
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157
PROBLEM 7.115
Each cable of the side spans of the Golden Gate Bridge supports
a load w = 170 kN/m along the horizontal. Knowing that for the
side spans the maximum vertical distance h from each cable to the
chord AB is 9 m and occurs at midspan, determine (a) the maximum
tension in each cable, (b) the slope at B.
Solution
FBD AB:
SM A = 0 : (330 m)TBy - (150 m)TBx - (165 m)W = 0
0.33TBy - 0.15TBx = 0.165W (1)
W
=0
2
= 0.04125W (2)
FBD CB:
SM C = 0 : (165 m)TBy - (84 m)TBx - (82.5 m)
0.165TBy - 0.084 TBx
Solving (1) and (2)
TBy = 144925 kN
TBx = 257125 kN
Tmax = TB = TB2x + TB2y
tan q B =
TBy
TBx
: Tmax = 295155 N
q B = 29.407∞
So that
(a)
Tmax = 295000 kN 
(b)
q B = 29.4∞ 
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158
PROBLEM 7.116
A steam pipe weighing 75 kg/m that passes between two
buildings 12 m 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 7.5 N/m,
determine (a) the location of the lowest Point C of the cable,
(b) the maximum tension in the cable.
Solution
w = {(75 ¥ 9.81) + 7.5} N/m = 743.25 N/m
Note:
x B - x A = 12 m
or
x A = x B - 12 m
(a)
(b)
Use Eq. 7.8
Point A:
yA =
wx 2A
w ( x B - 12)2
; 2.7 =
2T0
2T0
(1)
Point B:
yB =
wx B2
wx 2
; 1.2 = B 2T0
2T0
(2)
Dividing (1) by (2):
2.7 ( x B - 12)2
=
; x B = 4.8 m 1.2
x B2
Maximum slope and thus Tmax is at A
Point C is 4.8 m to left of B 
(Use –ve root to get + ve xB)
x A = x B - 12 = 4.8 - 12 = -7.2 m
yA =
(743.25 N/m)( -7.2)2
wx 2A
; 2.7 =
; T0 = 7135.2 N
2T0
2T0
W AC = (743.25 N/m)(7.2 m) = 5351. 4 N
Tmax = 8919 N
Tmax = 8920 N 
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159
PROBLEM 7.117
Cable AB supports a load uniformly distributed along the
h­ orizontal as shown. Knowing that at B the cable forms an angle
B = 35° with the horizontal, determine (a) the maximum tension
in the cable, (b) the vertical distance a from A to the lowest point
of the cable.
Solution
Free body: Entire cable
By = Bx tan 35∞
W = ( 45 kg/m)(12 m)(9.81 m/s2 )
W = 5297.4 N
SM A = 0 : W (6 m) + Bx (1.8 m) - By (12 m) = 0
+
(5297.4)(6) + 1.8 Bx - Bx tan 35∞(12) = 0
Bx = 4814 N
By = ( 4814 N ) tan 35∞ = 3370.8 N
Free body: Portion CB
+
SFy = 0 : By - WBC = 0
WBC = By = 3370.8 N
WBC = ( 45 kg/m)(9.81 m/s2 )b
3370.8 N = ( 441.45 N/m)b
b = 7.6357 m
+
Ê1 ˆ
SM B = 0 : T0 dC - WBC Á b˜ = 0
Ë2 ¯
1
( 4814 N )dC - (3370.8 N ) (7.6357 m) = 0
2
dC = 2.6733 m
(a)
dC = 1.8 m + a; 2.6733 m = 1.8 m + a; (b)
Tm = B = Bx2 + By2 = ( 4814 N )2 + (3370.8 N )
Tm = 5877 N a = 0.873 m 
Tm = 5880 N 
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160
PROBLEM 7.118
Cable AB supports a load uniformly distributed along the ­horizontal
as shown. Knowing that the lowest point of the cable is located
at a distance a = 0.6 m below A, determine (a) the maximum
­tension in the cable, (b) the angle B that the cable forms with the
horizontal at B.
Solution
Note: x B - x A = 12 m
or x A = x B - 12 m
Point A:
yA =
wx 2A
w ( x B - 12)2
; 0.6 =
2T0
2T0
(1)
Point B:
yB =
wx B2
wx 2
; 2.4 = B 2T0
2T0
(2)
Dividing (1) by (2):
0.6 ( x B - 12)2
=
; xB = 8 m
2.4
x B2
(a)
2.4 =
Eq. (2):
w (8)2
; T0 = 13.333w
2T0
Free body: Portion CB
SFy = 0 By = wx B
By = 8w
Tm2 = Bx2 + By2 ; Tm2 = (13.333w )2 + (8w )2
Tm = 15.549w = 15.549( 45)(9.81) q B = tan -1 By /Bx = tan -1 8w /13.333w Tm = 6860 N 
q B = 31.0∞ 
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161
PROBLEM 7.119*
A cable AB of span L and a simple beam A´B´ of the same span are
s­ ubjected 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
SM B = 0 : LACy + aT0 - SM B loads = 0 (1)
FBD Cable:
(Where SM B loads includes all applied loads)
xˆ
Ê
SM C = 0 : xACy - Á h - a ˜ T0 - SM C left = 0 Ë
L¯
(2)
FBD AC:
(Where SM C left includes all loads left of C)
x
x
(1) - (2): hT0 - SM B loads + SM C left = 0
L
L
(3)
FBD Beam:
SM B = 0 : LABy - SM B loads = 0 (4)
SM C = 0 : xABy - SM C left - M C = 0 (5)
FBD AC:
Comparing (3) and (6)
x
x
( 4) - (5): - SM B loads + SM C left + M C = 0 L
L
M C = hT0
(6)
Q.E.D.
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162
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 (a) Knowing that the maximum tension in cable
ABCD is 15 kN, determine the distance hB .
Solution
+
SM B = 0 : A(10 m) - (6 kN )(7 m) - (10 kN )( 4 m) = 0
A = 8.2 kN
+
SFy = 0 : 8.2 kN - 6 kN - 10 kN + B = 0
B = 7.8 kN
At A:
Tm2 = T02 + A2
152 = T02 + 8.2
T0 = 12.56 kN
At B:
M B = T0 hB ; 24.6 kN ◊ m = (12.56 kN )hB ; hB = 1.959 m 
At C:
M C = T0 hC ; 31.2 kN ◊ m = (12.56 kN )hC ; hC = 2.48 m 
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163
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
+
SM 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
1.4 m
hB
h
hB
hD
= C = D ;
=
=
20 kN ◊ m 30 kN ◊ m 30 kN ◊ m
M B MC M D
Ê 20 ˆ
hB = 1.4 Á ˜ = 0.9333 m
Ë 30 ¯
P
Ê 30 ˆ
hD = 1.4 Á ˜ = 1.4 m
Ë 30 ¯
d B = 0.8 m + 0.9333 m
P
dD = 2.8 m + 1.4 m
d B = 1.733 m 
dD = 4.20 m 
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164
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 (a) If dC = 5 m, determine the distances
dB and dD.
Solution
Free body: Beam AE
+
SM E = 0 : - A(10) + 10(8) + 10(5) + 10(3) = 0
A = 16 kN ≠
+ SFy = 0 : 16 - 3(10) + B = 0
B = 14 kN ≠
Geometry:
Given:
dC = 5 m
Then,
hC = dC - 1.25 m = 3.75 m
Since M = T0 h, h is proportional to M. Thus,
h
hB
h
= C = D
M B MC M D
or,
or,
Then,
hB 3.75 m hD
=
=
32
50
42
hB = 2.4 m hD = 3.15 m
d B = 2 + hB = 2 + 2.4 = 4.4 m d B = 4.40 m 
dD = 0.75 + hD = 0.75 + 3.15 = 3.90 m dD = 3.90 m 
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165
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 (a) Determine the distance dC for which
portion BC of the cable is horizontal.
Solution
Free body: Beam AE
SM E = 0 : - A(10) + 10(8) + 10(5) + 10(3) = 0
+
A = 16 kN ≠
+ SFy = 0 : 16 - 3(10) + B = 0
B = 14 kN ≠
Geometry:
Given:
Then,
dC = d B
1.25 + hC = 2 + hB
hC = 0.75 + hB (1)
Since M = T0h, h is proportional to M. Thus,
h
hB
= C
M B MC
Substituting (2) into (1):
or,
hB hC
=
32 50
hB = 0.64hC 25
m =2.083m
12
25 10
dC = 1.25 + hC = 1.25 +
= m
12 3
hC = 0.75 + 0.64hC
Then,
(2)
hC =
dC = 3.33 m 
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166
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:
≠ SFy = 0 : Ty ( x + D x ) - Ty ( x ) - w ( x ) D x = 0
So
Ty ( x + D x )
T0
-
Ty ( x )
Ty
But
So
In lim :
D x Æ0
T0
T0
dy
dx
x +Dx
Dx
dy
dx
x
=
w( x)
Dx
T0
=
dy
dx
=
w( x)
T0
d 2 y w( x)
=
T0
dx 2
Q.E.D.
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167
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 (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
px
L
From Problem 7.124
px
d 2 y w ( x ) w0
=
=
cos
2
T
T
L
dx
0
0
So
Ê
ˆ
dy
ÁË using dx = 0˜¯
0
px
dy W0 L
sin
=
dx T0p
L
y=
But
And
w L2 Ê
pˆ
Ê Lˆ
y Á ˜ = h = 0 2 Á1 - cos ˜
Ë
Ë 2¯
2¯
T0p
T0 = Tmin
so T0 =
w0 L2
p 2h
Tmin =
so
Tmax = TA = TB :
TBy =
w0 L2 Ê
p xˆ
1 - cos ˜ [ using y(0) = 0] 
2 Á
Ë
L¯
T0p
TBy
T0
w0 L p
=
dy
dx
=
x = L/2
w0 L2
p 2h

w0 L
T0p
TB = TB2y + T02 =
2
w0 L
Ê Lˆ
1+ Á ˜ 
Ë p h¯
p
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168
PROBLEM 7.126
If the weight per unit length of the cable AB is w0/cos2 , 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*
But
w ( x )dx =
w0
cos2 q
ds
dx = cos q ds, so w ( x ) =
w0
cos3 q
From Problem 7.119
w0
d 2 y w( x)
=
=
2
T0
dx
T0 cos3 q
In general
d 2 y d Ê dy ˆ d
dq
= Á ˜ = (tan q ) = sec2 q
2
¯
Ë
dx
dx
dx
dx
dx
So
or
w0
w0
dq
=
=
3
2
dx T0 cos q sec q T0 cos q
T0
cos q dq = dx = rdq cos q
w0
T0
= constant. So curve is circular arc
w0
*For large sag, it is not appropriate to approximate ds by dx.
Giving r =
Q.E.D.
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169
PROBLEM 7.127
A 30-m cable is strung as shown between two buildings. The
maximum tension is found to be 500 N, and the lowest point
of the cable is observed to be 4 m above the ground. Determine
(a) the horizontal distance between the buildings, (b) the total
mass of the cable.
Solution
sB = 15 m
Tm = 500 N
Eq. 7.17:
Eq. 7.15:
36 + 12c + c 2 - 225 = c 2
12c = 189 c = 15.75 m
x
x
sB = c sinh B ; 15 = (15.75)sinh B
c
c
xB
xB
sinh
= 0.95238
= 0.8473
c
c
x B = 0.8473(15.75) = 13.345 m; L = 2 x B (a)
(b)
y B2 - sB2 = c 2 ; (6 + c)2 - 152 = c 2
Eq. 7.18:
L = 26.7 m 
Tm = wy B ; 500 N = w (6 + 15.75)
w = 22.99 N/m
W = 2sB w = (30 m)(22.99 N/m)
= 689.7 N
689.7 N
W
m=
=
g 9.81 m/s2
Total mass = 70.3 kg 
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170
PROBLEM 7.128
A 60 m steel surveying tape weighs 20 N. If the tape is stretched between two points at the same elevation and
pulled until the tension at each end is 80 N, determine the horizontal distance between the ends of the tape.
Neglect the elongation of the tape due to the tension.
Solution
sB = 30 m
Eq. 7.18:
Eq. 7.17:
Eq. 7.15:
Ê 20 N ˆ 1
w=Á
= N/m Tm = 80 N
Ë 60 m ˜¯ 3
1
Tm = wy B ; 80 N = ( N/m) y B ; y B = 240 m
3
y B2 - sB2 = c 2 ; (240)2 - (30)2 = c 2 ; c = 238.118 m
xB
x
; 30 = 238.118 sinh B
c
c
x
x
Ê ˆ
Ê ˆ
sinh Á B ˜ = 0.12599; Á B ˜ = 0.12566
Ë c ¯
Ë c ¯
x B = 0.12566 ¥ 238.118 m = 29.922 m
sB = c sinh
L = 2 x B = 59.844 m L = 59.8 m 
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171
PROBLEM 7.129
A 200-m-long aerial tramway cable having a mass per unit length of 3.5 kg/m is suspended between two points
at the same elevation. Knowing that the sag is 50 m, find (a) the horizontal distance between the supports,
(b) the maximum tension in the cable.
Solution
Given:
Length = 200 m
Unit mass = 3.5 kg/m
h = 50 m
Then,
w = (3.5 kg/m)(9.81 m/s2 ) = 34.335 N/m
sB = 100 m
y B = h + c = 50 m + c
y B2 - sB2 = c 2 ; (50 + c)2 - (100)2 = c 2
502 + 100c + c 2 - 1002 = c 2
c = 75 m
xB
x
sB = c sinh ; 100 = 75 sinh B
c
75
x B = 82.396 m
span = L = 2 x B = 2(82.396 m) Tm = wy B = (34.335 N/m)(50 m + 75 m) L = 164.8 m 
Tm = 4290 N 
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172
PROBLEM 7.130
An electric transmission cable of length 120 m weighing 4 kg/m is suspended between two points at the
same elevation. Knowing that the sag is 30 m, determine the horizontal distance between the supports and the
maximum tension.
Solution
sB = 60 m
Eq. 7.17:
y B2 - sB2 = c 2 ; (30 + c)2 - 602 = c 2
900 + 60c + c 2 - 3600 = c 2 ; c = 45 m
sB = c sinh
Eq. 7.15:
sinh
xB
Êx ˆ
; 60 = 45 sinh Á B ˜
Ë c ¯
c
xB 4 xB
= ;
= 1.0986
3
c
c
x B = ( 45)(1.0986) = 49.437 m
L = 2 x B = 2( 49.437 m) = 98.874 m Eq. 7.18:
Tm = wy B = ( 4 ¥ 9.81 N/m)(30 + 45) m = 2943 N
L = 98.9 m 
Tm = 2940 N 
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173
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
ˆ
Ê
sT = 20 m Á so sB =
= 10 m˜
¯
Ë
2
w = (0.2 kg/m)(9.81 m/s2 )
= 1.96200 N/m
hB = 8 m
y B2 = (c + hB )2 = c 2 + sB2
So
c=
sB2 - hB2
2hB
(10 m)2 - (8 m)2
2(8 m)
= 2.250 m
c=
Now
xB
s
Æ x B = c sinh -1 B
c
c
Ê 10 m ˆ
= (2.250 m)sinh -1 Á
Ë 2.250 m ˜¯
sB = c sinh
x B = 4.9438 m
P = T0 = wc = (1.96200 N/m)(2.250 m)
(a)
P = 4.41 N Æ 
L = 2 x B = 2( 4.9438 m)
(b)
L = 9.89 m 
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174
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/s2 ) = 1.96200 N/m
P
P = T0 = wc c =
w
20 N
= 10.1937 m
c=
1.9620 N/m
y B2 = ( hB + c)2 = c 2 + sB2
20 m
= 10 m
h2 + 2ch - sB2 = 0 sB =
2
h2 + 2(10.1937 m)h - 100 m2 = 0
h = 4.0861 m
sB = c sinh
(a)
h = 4.09 m 
(b)
L = 17.69 m 
xA
s
Ê 10 m ˆ
Æ x B = c sinh -1 B = (10.1937 m)sinh -1 Á
Ë 10.1937 m ˜¯
c
c
= 8.8468 m
L = 2 x B = 2(8.8468 m)
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175
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:
sT = 20 m Æ sB =
20 m
= 10 m
2
w = (0.2 kg/m)(9.81 m/s2 ) = 1.96200 N/m
L = 15 m
L
xB
= c sinh 2
c
c
7.5 m
10 m = c sinh
c
c = 5.5504 m
sB = c sinh
Solving numerically:
Êx ˆ
Ê 7.5 ˆ
y B = c cosh Á B ˜ = (5.5504) cosh Á
Ë 5.5504 ˜¯
Ë c ¯
y B = 11.4371 m
hB = y B - c = 11.4371 m - 5.5504 m
P = wc = (1.96200 N/m)(5.5504 m)
(a)
hB = 5.89 m 
(b)
P = 10.89 N Æ 
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176
PROBLEM 7.134
Determine the sag of a 9 m chain that is attached to two points at the same elevation that are 6 m apart.
Solution
9m
= 4.5 m L = 6 m
2
L
xB = = 3 m
2
x
sB = c sinh B
c
3m
4.5 m = c sinh
c
c = 1.84941 m
sB =
Solving numerically:
y B = c cosh
xB
c
= (1.84941 m) cosh
3m
1.84941 m
= 4.8652 m
hB = y B - c = 4.8652 - 1.8494 = 3.0158 m hB = 3.02 m 
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177
PROBLEM 7.135
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
x
y B = c cosh B
c
30
y B = (18.494) cosh
18.494
y B = 48.652 m
sB = c sinh
yB = h + c
48.652 = h + 18.494
h = 30.158 m Eq. 7.18:
h = 30.2 m 
Tm = wy B
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 g
9.81 m/s
m = 56.6 kg 
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178
PROBLEM 7.136
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 =15 m and h = 5 m, 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 = 15 m
h=5m
TA = 400 N
x B = 7.5 m
By symmetry:
TB = TA = Tm = 400 N
We have
y B = c cosh
and
yB = h + c = 5 + c
Then
or
Solve by trial for c:
(a)
xB
7.5
= c cosh
c
c
7.5
=5+c
c
7.5 5
cosh
= +1
c
c
c cosh
c = 6.3175 m
x
sB = c sinh B
c
= (6.3175 m)sinh
7.5
6.3175
= 9.3901 m
Length = 2sB = 2(9.3901 m) = 18.78 m 
(b)
Tm = wy B = w ( h + c)
400 N = w (5 m + 6.3175 m)
w = 35.343 N/m
w = 35.3 N/m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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179
PROBLEM 7.137
A uniform cord 1-25 m long passes over a pulley at B and is attached to a pin
support at A. Knowing that L = 0.5 m and neglecting the effect of friction,
­determine the smaller of the two values of h for which the cord is in equilibrium.
Solution
b = 1250 mm - 2sB
Length of overhang:
Weight of overhang equals max. tension
Tm = TB = wb = w (1250 mm - 2sB )
xB
c
xB
y B = c cosh
c
Tm = wy B
sB = c sinh
Eq. 7.15:
Eq. 7.16:
Eq. 7.18:
w (1250 mm - 2sB ) = wy B
x ˆ
x
Ê
w Á1250 mm - 2c sinh B ˜ = wc cosh B
Ë
c ¯
c
x B = 250 : 1250 - 2c sinh
Solve by trial and error:
For c =138.725 mm
c = 138.725 mm
and
250
250
= c cosh
c
c
c = 693.55 mm
250 mm
= 431.953 mm
138.725 mm
y B = h + c; 431.953 mm = h + 138.725 mm
y B = (138.725 mm) cosh
h = 293.228 mm For c = 693.55 mm
h = 293 mm 
250
= 739.098 mm
693.55
y B = h + c; 739.098 = h + 693.55
y B = (693.55 mm) cosh
h = 45.548 mm h = 45.5 mm 
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180
PROBLEM 7.138
A cable weighing 3 kg/m is suspended between two points at the same elevation that are 48 m apart. Determine
the smallest allowable sag of the cable if the maximum tension is not to exceed 1800 N.
Solution
Eq. 7.18:
Eq. 7.16:
Solve for c:
Tm = wy B ; 1800 N = (3 ¥ 9.81 N/m) y B ;
x
y B = c cosh B
c
24 m
61.162 m = c cosh
c
y B = 61.162 m
c = 55.9335 m and c = 9.35868 m
y B = h + c; h = y B - c
For
c = 55.9335 m; h = 61.162 - 55.9335 = 5.23 m 
For
c = 9.35868 m; h = 61.162 - 9.35868 = 51.8 m 
For
smallest h = 5.23 m 
Tm 1800 N: PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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181
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
x
y B = c cosh B
c
L
hB + c = c cosh
2c
5m ˆ
Ê
5 m = c Á cosh
- 1˜
Ë
¯
c
Solving numerically:
c = 3.0938 m
y B = hB + c = 5 m + 3.0938 m
= 8.0938 m
Tmax = TB = wy B
= (0.4 kg/m)(9.81 m/s2 )(8.0938 m)
Tmax = 31.8 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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182
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
x
L
y B = hB + c = c cosh B = c cosh
c
2c
5m ˆ
Ê
3 m = c Á c cosh
– 1˜
Ë
¯
c
Solving numerically:
c = 4.5945 m
y B = hB + c = 3 m + 4.5945 m
= 7.5945 m
Tmax = TB = wy B
= (0.4 kg/m)(9.81 m/s2 )(7.5945 m)
Tmax = 29.8 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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183
PROBLEM 7.141
A uniform cable weighing 4.5 kg/m is held in the position shown by a
horizontal force P applied at B. Knowing that P = 900 N and A = 60°,
determine (a) the location of Point B, (b) the length of the cable.
Solution
Eq. 7.18:
T0 = P = cw
c=
c = 20.3874 m
P
cos 60∞
cw
=
= 2cw
0.5
Tm =
At A:
(a)
900 N
P
=
w ( 4.5 ¥ 9.81) N/m
Tm = w ( h + c)
Eq. 7.18:
2cw = w ( h + c)
2c = h + c h = b = c b = 20.4 m 
xA
c
xA
h + c = c cosh
c
y A = c cosh
Eq. 7.16:
(2 ¥ 20.3874) = (20.3874) cosh
cosh
xA
=2
20.3874
xA
20.3874
xA
= 1.3170
20.3874
x A = 26.8502 m (b)
Eq. 7.15:
a = 26.9 m 
xB
26.8502
= (20.3874)sinh
c
20.3874
s A = 35.314 m
s A = c sinh
length = s A s A = 35.3 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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184
PROBLEM 7.142
A uniform cable weighing 4.5 kg/m is held in the position shown by a
horizontal force P applied at B. Knowing that P = 750 N and A = 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)
750 N
P
=
= 16.9895 m
w 4.5 ¥ 9.81 N/m
Tm = w ( h + c)
Eq. 7.18:
2cw = w ( h + c)
2c = h + c h = c = b Eq. 7.16:
b = 16.99 m 
xA
c
xA
h + c = c cosh
c
y A = c cosh
Êx ˆ
2 ¥ 16.9895 m = 16.9895 cosh Á A ˜
Ë c¯
cosh
xA
=2
c
xA
= 1.3170
c
x A = 1.3170(16.9895 m) = 22.375 m (b)
Eq. 7.15:
s A = c sinh
a = 22.4 m 
xA
22.375
= (16.9895)sinh
c
16.9895
s A = 29.428 m length = s A = 29.4 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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185
PROBLEM 7.143
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
Solving numerically
Then
xD = a = 3.6 m h = 4 m
x
yD = c cosh D
c
a
h + c = c cosh
c
3.6 m ˆ
Ê
4 m = c Á cosh
- 1˜
Ë
¯
c
c = 2.0712 m
y B = h + c = 4 m + 2.0712 m = 6.0712 m
F = Tmax = wy B = (2 kg/m)(9.81 m/s2 )(6.0712 m) F = 119.1 N Æ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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186
PROBLEM 7.144
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
x
yD = c cosh D
c
a
h + c = c cosh
c
6m ˆ
Ê
4 m = c Á cosh
- 1˜
Ë
¯
c
Solving numerically
c = 5.054 m
y B = h + c = 4 m + 5.054 m = 9.054 m
F = TD = wyD = (2 kg/m)(9.81 m/s2 )(9.054 m) F = 177.6 N Æ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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187
PROBLEM 7.145
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
x B - x A = 12 m
Note:
- x A = 12 m - x B
or,
- xA
12 - x B
; c + 0.6 = c cosh
c
c
x
x
y B = c cosh B ; c + 2.4 = c cosh B c
c
y A = c cosh
Point A:
Point B:
(1)
(2)
From (1):
12 x B
Ê c + 0.6 ˆ
= cosh -1 Á
Ë c ˜¯
c
c
(3)
From (2):
xB
Ê c + 2.4 ˆ
= cosh -1 Á
Ë c ˜¯
c
(4)
Add (3) + (4):
12
Ê c + 2.4 ˆ
Ê c + 0.6 ˆ
+ cosh -1 Á
= cosh -1 Á
˜
Ë c ˜¯
Ë c ¯
c
Solve by trial and error:
Eq. (2)
c = 13.6214 m
13.6214 + 2.4 = 13.6214 cosh
cosh
xB
= 1.1762;
c
xB
c
xB
= 0.58523
c
x B = 0.58523(13.6214 m) = 7.9717 m
Point C is 7.97 m to left of B 
y B = c + 2.4 = 13.6214 + 2.4 = 16.0214 m
Eq. 7.18:
Tm = wy B = (0.45 kg/m)(9.81 m/s2 )(16.0214 m)
Tm = 70.726 N Tm = 70.7 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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188
PROBLEM 7.146
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:
x B - x A = 12 m
or
- x A = 12 m - x B
12 - x B
- xA
; c + 2 = c cosh
c
c
x
x
y B = c cosh B ; c + 3.8 = c cosh B c
c
12 x B
Ê c + 2ˆ
= cosh -1 Á
Ë c ˜¯
c
c
xB
Ê c + 3.8 ˆ
= cosh -1 Á
Ë c ˜¯
c
y A = c cosh
Point A:
Point B:
From (1):
From (2):
(2)
(3)
(4)
12
Ê c + 3.8 ˆ
Ê c + 2ˆ
+ cosh -1 Á
= cosh -1 Á
Ë c ˜¯
Ë c ˜¯
c
c = 6.8154 m
Add (3) + (4):
Solve by trial and error:
Eq. (2):
(1)
6.8154 m + 3.8 m = (6.8154 m) cosh
cosh
xB
c
xB
xB
= 1.5576
= 1.0122
c
c
x B = 1.0122(6.8154 m) = 6.899 m
Point C is 6.90 m to left of B 
y B = c + 3.8 = 6.8154 + 3.8 = 10.6154 m
Eq. (7.18):
Tm = wy B = (0.45 kg/m)(9.81 m/s2 )(10.6154 m)
Tm = 46.86 N Tm = 46.9 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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189
PROBLEM 7.147*
The 3 m 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 m = 0, cable ^ rod
x dy
x
y = c cosh ;
= sinh
c dx
c
x
dy
tan q =
= sinh A
c
dx A
xA
-1
= sinh (tan (90∞ - q ))
c
x A = c sinh -1 (tan (90∞ - q )) Point A:
3 m = AC + CB
x
x
3 = c sinh A + c sinh B
c
c
xB 3
xA
sinh
= - sinh
c
c
c
x ˘
3
È
x B = c sinh -1 Í - sinh A ˙ c ˚
Îc
xA
x
y A = c cosh
y B = c cosh B c
c
yB - y A
In  ABD:
tanq =
xB + x A
(1)
Length of cable = 3 m (2)
(3)
(4)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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190
PROBLEM 7.147* (Continued)
Method of solution:
For given value of , 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 
Choose new trial value of  and repeat above procedure until calculated value of  is equal to given value of .
For q = 30∞
Result of trial and error procedure:
c = 0.5409 m
x A = 0.71234 m
x B = 1.1081 m
y A = 1.0818 m
y B = 2.1329 m
a = yB - y A
= 2.1329 - 1.0818 m
= 1.0511 m
a = 1.051 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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191
PROBLEM 7.148*
Solve Problem 7.147 assuming that the angle q formed by the rod and
the horizontal is 45°.
PROBLEM 7.147 The 3 m 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 m = 0, cable ^ rod
Point A:
Length of cable = 3 m
In  ABD:
x dy
x
y = c cosh ;
= sinh
c dx
c
x
dy
tan q =
= sinh A
c
dx A
xA
-1
= sinh (tan (90∞ - q ))
c
x A = c sinh -1 (tan (90∞ - q )) 3 m = AC + CB
x
x
3 = c sinh A + c sinh B
c
c
xB 3
xA
sinh
= - sinh
c
c
c
x ˘
3
È
x B = c sinh -1 Í - sinh A ˙ c ˚
Îc
xA
x
y A = c cosh
y B = c cosh B c
c
yB - y A
tanq =
xB + x A
(1)
(2)
(3)
(4)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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192
PROBLEM 7.148* (Continued)
Method of solution:
For given value of q, 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 q
Choose new trial value of q and repeat above procedure until calculated value of q is equal to given value of q.
For q = 30∞
Result of trial and error procedure:
c = 0.55956 m
x A = 0.49318 m
x B = 1.21918 m
y A = 0.79134 m
y B = 2.50377 m
a = yB - y A
= 2.50377 m - 0.79134 m
= 1.71243 m
a = 1.712 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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193
PROBLEM 7.149
Denoting by q the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan q,
(b) y = c sec q.
Solution
dy
x
= sinh
dx
c
x
s = c sinh = c tan q
c
tan q =
(a)
(b)
Q.E.D.
Also
y 2 = s2 + c 2 (cosh 2 x = sinh 2 x + 1)
so
y 2 = c 2 (tan 2 q + 1)c 2 sec2 q
and
y = c secq
Q.E.D.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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194
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 = 3.75 N/m and Tm = 36 kN.
Solution
Tm = wy B
(a)
xB
c
ˆ
x
1 ˜
cosh B
xB ˜
c
˜
c ¯
= wc cosh
We shall find ratio
( ) for when T
xB
c
m
Ê
Á
= wx B Á
Á
Ë
is minimum
2
˘
È
ˆ
Ê
Í
˙
d Tm
x
x
1
Á 1 ˜
= wx B Í
sinh B - Á
cosh B ˙ = 0
˜
x
x
x
c
c ˙
Í B
Á B˜
d ÊÁ B ˆ˜
Í c
˙
Ë c ¯
Ë c ¯
Î
˚
xB
c
xB
cosh
c
x
tanh B
c
xB
Solve by trial and error for:
c
xB
sB
sB = c sinh
= c sinh(1.200):
c
c
sinh
Eq. 7.17:
1
xB
c
c
=
xB
=
= 1.200 (1)
= 1.509
y B2 - sB2 = c 2
È Ê sB ˆ 2 ˘
= c Í1 + Á ˜ ˙ = c 2 (1 + 1.5092 )
ÍÎ Ë c ¯ ˙˚
y B = 1.810c
y B2
2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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195
PROBLEM 7.150* (Continued)
Eq. 7.18:
Tm = wy B
= 1.810 wc
c=
Eq. (1):
Span:
(b)
Tm
1.810 w
x B = 1.200c = 1.200
Tm
T
= 0.6630 m
w
1.810 w
L = 2 x B = 2(0.6630)
Tm
w
L = 1.326
Tm

w
For w = 3.75 N/m and Tm = 36000 N
36000 N
3.75 N/m
= 12, 729 .6 m
L = 1.326
L = 12.73 km 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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196
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/s2 ) = 29.43 N/m
Tmax = wymax
ymax = c cosh
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 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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197
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 = wy B = 2wsB
yB
L
c cosh
2c
L
tanh
2c
L
2c
hB
c
= 2 sB
= 2c sinh
=
L
2c
1
2
1
= 0.549306
2
y -c
L
= B
= cosh - 1 = 0.154701
c
2c
= tanh -1
hB
=
L 2
hB
c
L
2c
( )
=
0.5(0.154701)
= 0.14081 0.549306
hB
= 0.1408 
L
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198
PROBLEM 7.153*
A cable of weight w per unit length 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 B and Tm.
Solution
L
2c
L
L
Lˆ
Ê
= w Á cosh - sinh ˜
Ë
2c 2c
2c ¯
(a) Tmax = wy B = wc cosh
dTmax
dc
dTmax
=0
dc
L 2c
L
Æ
tanh =
= 1.1997
2c L
2c
For min Tmax ,
yB
L
= cosh = 1.8102
2c
c
h yB
=
- 1 = 0.8102
c
c
0.8102
h È 1 h Ê 2c ˆ ˘
= Í
= 0.3375 ÁË ˜¯ ˙ =
L Î 2 c L ˚ 2(1.1997)
(b) T0 = wc Tmax = wc cosh
But T0 = Tmax cos q B
L
2c
Tmax
L y
= cosh = B
2c
T0
c
Tmax
= sec q B
T0
Êy ˆ
So q B = sec-1 Á B ˜ = sec-1 (1.8102) = 56.46∞ Ë c ¯
Tmax = wy B = w
h
= 0.338 
L
y B Ê 2c ˆ Ê L ˆ
L
Á ˜ Á ˜ = w (1.8102)
2(1.1997)
c Ë L ¯ Ë 2¯
q B = 56.5∞ 
Tmax = 0.755 wL 
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199
PROBLEM 7.154
It has been experimentally determined that the bending moment at Point K of
the frame shown is 300 N · m. Determine (a) the tension in rods AE and FD,
(b) the corresponding internal forces at Point J.
Solution
Free body: ABK
(a)
SM K = 0 : 300 N ◊ m -
+
8
15
T (0.2 m) - T (0.12 m) = 0 17
17
T = 1500 N 
Free body: AJ
8
8
T = (1500 N )
17
17
= 705.88 N
15
15
Ty = T = (1500 N )
17
17
= 1323.53 N
Tx =
Internal forces on ABJ
SFx = 0 : 705.88 N - V = 0
+
(b) V = + 705.88 N +
SFy = 0 : F - 1323.53 N = 0
F = +1323.53 N +
V = 706 N ¨ 
F = 1324 N ≠ 
SM J = 0 : M - (705.88 N )(0.1 m) - (1323.53 N )(0.12 m) = 0 M = + 229.4 N ◊ m
M = 229 N ◊ m

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200
PROBLEM 7.155
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:
 M 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.
 M E = 0 : (1.4 m)(360 N ) + (1.8 m)(560 N )
- (2.4 m) By = 0
B y = 630 N ≠
FBD BJ:
3
 Fx ¢ = 0 : F + 360 N - (630 N - 360 N )
5
4
- (560 N ) = 0
5
F = 250 N

V = 120.0 N

4
3
 Fy ¢ = 0 : V + (630 N - 360 N ) - (560 N ) = 0
5
5
 M 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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201
PROBLEM 7.156
A steel channel of weight per unit length w = 30 kg/m forms one side of a flight of stairs. Determine the
internal forces at the center C of the channel due to its own weight for each of the support conditions shown.
Solution
(a)
Free body: AB
AB = 32 + 42 = 5 m
W = (30 ¥ 9.81)N/m ¥ 5 m = 1471.5 N
+ SM B = 0 : A( 4 m) - (1471.5 N)(2 m) = 0
A = +735.75 N
A = 735.75 N ≠
Free body: AC
(735.75 N forces form a couple)
+
SF = 0 F=0 
SF V=0 
SM C = 0 : M - (735.75 N)(1 m) = 0
M = +735.75 N ◊ m
M = 736 N ◊ m
(b)

Free body: AB
+ SM A = 0 : B(3 m) - (1471.5 N)(2 m) = 0
B = +981 N
B = 981 N ¨
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202
PROBLEM 7.156 (Continued)
Free body: CB
+
SM C = 0 : (981 N)(1.5 m) - (735.75 N)(1 m) - M = 0
M = +735.75 N ◊ m
M = 736 N ◊ m
+

3
3
4
tan q = ; sin q = ; cos q =
4
5
5
3
4
SF = 0 : F - (735.75 N) - (981 N) = 0
5
5
F = +1226.25 N F = 1226 N
+ SF
4
3
= 0 : V - (735.75 N) + (981 N) = 0
5
5
V =0 V=0
M = 736 N ◊ m
On portion AC internal forces are
, F = 1226 N [, V = 0 
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203
PROBLEM 7.157
For the beam shown, determine (a) the magnitude P of the two concentrated
loads for which the maximum absolute value of the bending moment is as
small as possible, (b) the corresponding value of | M |max .
Solution
Free body: Entire beam
By symmetry
1
D = E = (3000 N/m)(0.75 m) + P
2
D = 1125 N + P
Free body: Portion AD
+
SM D = 0 : M D = -0.25P
Free body: Portion ADC
+
SM C = 0 : P (0.625 m) - (1125 + P )(0.375 m)
+ (1125 N)(0.1875 m) + M C = 0
M C = 210.9375 N ◊ m - (0.25 m)P
(a)
We equate:
| M D | = | MC |
| - 0.25P | = | 210.9375 - 0.25P |
(b)
For P = 421.875 N:
0.25P = 210.9375 - 0.25P
P = 421.875 N
M D = -0.25 ¥ (421.875 N) P = 422 N 
| M |max = 105.5 N ◊ m 
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204
PROBLEM 7.158
Knowing that the magnitude of the concentrated loads P is 375 N, (a) draw
the shear and bending-moment diagrams for beam AB, (b) determine the
maximum absolute values of the shear and bending moment.
Solution
By symmetry:
D=E
1
= (3000 N/m)(0.75 m) + 375 N
2
= 1500 N
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205
PROBLEM 7.159
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
+
SM B = 0: (24 kN)(3 m)
+ (24 kN)(2.4 m) + (21.6 kN)(0.9 m)
- Ay (3.6 m) = 0
Ay = +91.4 kN
SFx = 0 : Ax = 0
Shear diagram
At A:
VA = Ay = +41.4 kN |V |max = 41.4 kN 
Bending-moment diagram
At A:
MA = 0 | M |max = 35.3 kN ◊ m 
The slope of the parabola at E is the same as that of the segment DE
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206
PROBLEM 7.160
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 = 30 kN, (b) P = 15 kN.
Solution
Free body: Beam
SFx = 0 : Ax = 0
+
SM A = 0 : C (3 m) - (90 kN)(1.5 m) - P ( 4 m) = 0
C = 45 kN +
4
P
3
(1) v
4 ˆ
Ê
SFy = 0 : Ay + Á 45 + P ˜ - 90 - P = 0
Ë
3 ¯
1
Ay = 45 kN - P 3
(a)
(2) v
P = 30 kN
Load diagram
Substituting for P in Eqs. (2) and (1):
Shear diagram
1
Ay = 45 - (30) = 35 kN
3
4
C = 45 + (30) = 85 kN
3
VA = Ay = +35 kN
To determine Point D where V = 0:
VD - VA = - wx
0 - 35 kN = ( -30 kN/m)x fi x =
7
m
6
x=
7
m v
6
We compute all areas
Bending-moment diagram
At A:
MA = 0
| M |max = 30 kN ◊ m, at C 
Parabola from A to C
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207
PROBLEM 7.160 (Continued)
(b)
P = 15 kN
Load diagram
Substituting for P in Eqs. (2) and (1):
1
A = 45 - (15) = 40 kN
3
4
C = 45 + (15) = 65 kN
3
Shear diagram
VA = Ay = +40 kN
To determine D where V = 0:
VD - VA = - wx
0 - ( 40 kN) = -(30 kN/m) x
4
fi x = m
3
We compute all areas
x=
4
m v
3
Bending-moment diagram
At A:
MA = 0
M max =
80
kN ◊ m = 26.667 kN ◊ m
3
| M |max = 26.7 kN ◊ m 
4
m from A 
3
Parabola from A to C.
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208
PROBLEM 7.161
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)
1
W
A = B = W , where
= Total load
2
L
Reactions at supports:
L
LÊ
p xˆ
W = Ú wdx = w0 Ú Á1 - sin ˜ dx
0
0 Ë
L¯
L
px˘
L
È
= w0 Í x + cos ˙
x
L ˚0
Î
2ˆ
Ê
= w0 L Á 1 - ˜
Ë p¯
1
1
2ˆ
Ê
V A = A = W = w0 L Á 1 - ˜
Ë
2
2
p¯
Thus
Load:
Shear:
MA = 0 (1)
p xˆ
Ê
w ( x ) = w0 Á1 - sin ˜
Ë
L¯
From Eq. (7.2):
x
V ( x ) - VA = - Ú w ( x )dx
0
xÊ
p xˆ
= - w0 Ú Á1 - sin ˜ dx
0Ë
L¯
Integrating and recalling first of Eqs. (1),
x
1
2ˆ
px˘
L
È
Ê
V ( x ) - w0 L Á1 - ˜ = - w0 Í x + cos ˙
¯
Ë
2
p
p
L ˚0
Î
L
L
p xˆ
1
2ˆ
Ê
Ê
w0 L Á1 - ˜ - w0 Á 2 + cos ˜ + w0
¯
Ë
Ë
¯
L
p
p
p
2
L
p xˆ
ÊL
V ( x ) = w0 Á - x - cos ˜
Ë2
L¯
p
V ( x) =
(2) 
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209
PROBLEM 7.161 (Continued)
Bending moment: From Eq. (7.4) and recalling that M A = 0.
x
M ( x ) - M A = Ú V ( x )dx
0
x
2
ÈL
1
px˘
Ê Lˆ
= w0 Í x - x 2 - Á ˜ sin ˙
Ëp¯
2
L ˙˚
ÍÎ 2
0
M ( x) =
(b)
1 Ê
2 L2
p xˆ
w0 Á Lx - x 2 - 2 sin ˜
2 Ë
L¯
p
(3) 
Maximum bending moment
dM
= V = 0.
dx
This occurs at x =
L
2
as we may check from (2):
pˆ
ÊL L L
Ê Lˆ
V Á ˜ = w0 Á - - cos ˜ = 0
Ë2 2 p
Ë 2¯
2¯
From (3):
2
pˆ
L2 2 L2
Ê Lˆ 1 Ê L
- 2 sin ˜
M Á ˜ = w0 Á Ë 2¯ 2 Ë 2
4 p
2¯
1
8ˆ
Ê
= w0 L2 Á1 - 2 ˜
Ë p ¯
8
= 0.0237w0 L2
M max = 0.0237w0 L2 , at
x=
L
2

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210
PROBLEM 7.162
An oil pipeline is supported at 2 m 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 2 kN. Knowing
that dC = 4 m, 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.
SM F = 0: (2 m + 4 m + 6 m + 8 m)(2 kN)
- (10 m)Ay - (1.5 m) Ax = 0
3 Ax + 20 Ay = 80 kN FBD ABC:
(1)
SM C = 0 : (2.5 m)Ax - ( 4 m)Ay + (2 m)(2 kN) = 0
5Ax - 8 Ay = -8
(2)
Solving (1) and (2)
From FBD Cable:
Ax =
120
N = 3.8710 kN ¨
31
Ay =
106
N = 3.4194 kN ≠
31
Æ SFx = 0 : - 3.871 kN + Fx = 0
Fx = 3.871 kN Æ
FBD DEF:
106
N - 4 (2 kN) + Fy = 0
31
142
Fy =
N≠
31
≠ SFy = 0 :
Fy = 4.5806 kN ≠
2
Since Ax = Fx and Fy  Ay ,
Tmax = TEF
2
Ê 142 ˆ
Ê 120 ˆ
+Á
= Á
.
Ë 31 ˜¯
Ë 31 ˜¯
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211
PROBLEM 7.162 (Continued)
(a)
Tmax = 5.9972 kN, Tmax = 5.60 kN 
Ê 120 ˆ
Ê 142 ˆ
kN ˜ - d D Á
kN ˜ - (2 m)(2 kN) = 0
SM D = 0 : ( 4 m) Á
¯
Ë 31
¯
Ë 31
dD = 3.70 m 
(b)
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212
PROBLEM 7.163
Solve Problem 7.162 assuming that dC = 3 m.
PROBLEM 7.162 An oil pipeline is supported at 2 m 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 2 kN.
Knowing that dC = 4 m, determine (a) the maximum tension in
the cable, (b) the distance dD.
Solution
FBD CDEF:
SM C = 0 : (6 m)Fy - (3 m)Fx - (2 m + 4 m)(2 kN) = 0
Fx - 2 Fy = -4 kN (1)
SM A = 0 : (10 m)Fy - (1.5 m)Fx
FBD Cable:
- (2 m)(1 + 2 + 3 + 4)(2 kN) = 0
3Fx - 20 Fy = -80 kN (2)
Solving (1) and (2),
Fx =
40
34
kN Æ , Fy =
kN ≠
7
7
Æ SFx = 0 : - Ax +
40
40
kN = 0, A x =
kN Æ
7
7
Point F:
34
22
- 4(2 kN) = 0, A y =
kN ≠
7
7
22
Ay =
kN
7
≠ SFy = 0 : Ay +
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213
PROBLEM 7.163 (Continued)
Since
Ax = Ay
and
2
Fy  Ay , Tmax = TEF
Ê 34 ˆ
Ê 40 ˆ
Tmax = Á ˜ + Á ˜
Ë 7¯
Ë 7¯
2
= 7.4997 kN
Tmax = 7.50 kN 
(a)
FBD DEF:
Ê 40 ˆ
Ê 34 ˆ
SM D = 0 : ( 4 m) Á kN ˜ - dD Á kN ˜
¯
Ë 7
¯
Ë 7
- (2 m)(2 kN) = 0
dD = 2.70 m 
(b)
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214
PROBLEM 7.164
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/s2 )
= 7.848 N/m
W = (7.848 N/m)(37.5 m)
W = 294.3 N
(a)
+
ˆ
Ê1
SM B = 0 : T0 (2 m) - W Á 37.5 m˜ = 0
¯
Ë2
1
T0 (2 m) - (294.3 N) (37.5 m) = 0
2
T0 = 2759 N
Tm2 = (294.3 N)2 + (2759 N)2 Tm = 2770 N 
(b)
˘
È 2 Ê y ˆ2
sB = x B Í1 + Á B ˜ + L˙
˙˚
ÍÎ 3 Ë x B ¯
˘
È 2 Ê 2 m ˆ2
= 37.5 m Í1+ Á
+ L˙
˜
˙˚
ÍÎ 3 Ë 37.5 m ¯
= 37.57 m
Length = 2sB = 2(37.57 m) Length = 75.14 m 
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215
PROBLEM 7.165
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
+
SFy = 0 : Ay - 9w = 0
A y = 9w ≠
+
SM A = 0 : T0 a - (9w )( 4.5 m) = 0 (1)
Free body: Portion CB
+
SFy = 0 : By - 6w = 0
B y = 6w ≠
Free body: Entire cable
+
SM A = 0 : 15w (7.5 m) - 6w (15 m) - T0 (2.25 m) = 0
T0 = 10 w
(a)
Eq. (1):
T0 a - (9w )( 4.5 m) = 0
10 wa = (9w )( 4.5) = 0 a = 4.05 m 
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216
PROBLEM 7.165 (Continued)
(b)
Length = AC + CB
Portion AC: x A = 9 m,
Portion CB:
y A = a = 4.05 m;
y A 4.05
=
= 0.45
xA
9
˘
È 2 Ê y ˆ2 2 Ê y ˆ4
s AC = x B Í1 + Á A ˜ - Á B ˜ + L˙
5 Ë xA ¯
˙˚
ÍÎ 3 Ë x A ¯
2
ˆ
Ê 2
s AC = 9 m Á1+ 0.452 - 0.454 + L˜ = 10.0067 m
¯
Ë 3
5
yB
x B = 6 m, y B = 4.05 - 2.25 = 1.8 m;
= 0.3
xB
2
Ê 2
ˆ
sCB = 6 m Á1 + 0.32 - 0.34 + L˜ = 6.341 m
¯
Ë 3
5
Total length = 10.067 m + 6.341 m (c)
Total length = 16.41 m 
Components of reaction at A
Ay = 9w = 9(60 kg/m)(9.81 m/s2 )
= 5297.4 N
Ax = T0 = 10 w = 10(60 kg/m)(9.81 m/s2 )
= 5886 N
A x = 5890 N ¨ 
A y = 5300 N ≠ 
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