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 Prob. 6.77.
SOLUTION
ΣFx = 0: − F = 0
FBD JD:
F=0
ΣFy = 0: V − 20 lb − 20 lb = 0
V = 40.0 lb
ΣM J = 0: M − ( 2 in.)( 20 lb ) − ( 6 in.)( 20 lb ) = 0
M = 160.0 lb ⋅ in.
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 Prob. 6.76.
SOLUTION
FBD AJ:
ΣFx = 0: 60 lb − V = 0
V = 60.0 lb
ΣFy = 0: − F = 0
F=0
ΣM J = 0: M − (1 in.)( 60 lb ) = 0
M = 60.0 lb ⋅ in.
PROBLEM 7.3
For the frame and loading of Prob. 6.80, determine the internal forces
at a point J located halfway between points A and B.
SOLUTION
FBD Frame:
ΣFy = 0: Ay − 80 kN = 0
A y = 80 kN
ΣM E = 0: (1.2 m ) Ax − (1.5 m )( 80 kN ) = 0
A x = 100 kN
 0.3 m 
 = 21.801°
 0.75 m 
θ = tan −1 
FBD AJ:
ΣFx′ = 0: F − ( 80 kN ) sin 21.801° − (100 kN ) cos 21.801° = 0
F = 122.6 kN
ΣFy′ = 0: V + ( 80 kN ) cos 21.801° − (100 kN ) sin 21.801° = 0
V = 37.1 kN
ΣM J = 0: M + (.3 m )(100 kN ) − (.75 m )( 80 kN ) = 0
M = 30.0 kN ⋅ m
PROBLEM 7.4
For the frame and loading of Prob. 6.101, determine the internal forces
at a point J located halfway between points A and B.
SOLUTION
FBD Frame:
ΣFy = 0: Ay − 100 N = 0
A y = 100 N
ΣM F = 0:  2 ( 0.32 m ) cos 30°  Ax − ( 0.48 m )(100 N ) = 0
A x = 86.603 N
FBD AJ:
ΣFx′ = 0: F − (100 N ) cos 30° − ( 86.603 N ) sin 30° = 0
F = 129.9 N
ΣFy′ = 0: V + (100 N ) sin 30° − ( 86.603 N ) cos 30° = 0
V = 25.0 N
ΣM J = 0: ( 0.16 m ) cos 30° ( 86.603 N )
− ( 0.16 m ) sin 30° (100 N ) − M = 0
M = 4.00 N ⋅ m
PROBLEM 7.5
Determine the internal forces at point J of the structure shown.
SOLUTION
FBD Frame:
AB is two-force member, so
Ay
Ax
=
0.36 m 0.15 m
Ay =
5
Ax
12
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0
A x = 624 N
Ay =
5
Ax = 260 N or A y = 260 N
12
ΣFx = 0: F − 624 N = 0
F = 624 N
FBD AJ:
ΣFy = 0: 260 N − V = 0
V = 260 N
ΣM J = 0: M − ( 0.2 m )( 260 N ) = 0
M = 52.0 N ⋅ m
PROBLEM 7.6
Determine the internal forces at point K of the structure shown.
SOLUTION
FBD Frame:
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0
A x = 624 N
AB is two-force member, so
Ay
Ax
5
=
→ Ay =
Ax
0.36 m 0.15 m
12
ΣFx = 0: − Ax + C x = 0
A y = 260 N
C x = A x = 624 N
ΣFy = 0: Ay + C y − 390 N = 0
C y = 390 N − 260 N = 130 N or C y = 130 N
FBD CK:
ΣFx′ = 0: F +
12
5
( 624 N ) + (130 N ) = 0
13
13
F = −626 N
ΣFy′ = 0:
F = 626 N
12
5
(130 N ) − ( 624 N ) − V = 0
13
13
V = −120 N
V = 120.0 N
ΣM K = 0: ( 0.1 m )( 624 N ) − ( 0.24 m )(130 N ) − M = 0
M = 31.2 N ⋅ m
PROBLEM 7.7
A semicircular rod is loaded as shown. Determine the internal forces
at point J.
SOLUTION
FBD Rod:
ΣM B = 0: Ax ( 2r ) = 0
Ax = 0
ΣFx′ = 0: V − ( 30 lb ) cos 60° = 0
V = 15.00 lb
FBD AJ:
ΣFy′ = 0: F + ( 30 lb ) sin 60° = 0
F = −25.98 lb
F = 26.0 lb
ΣM J = 0: M − [ (9 in.) sin 60°] ( 30 lb ) = 0
M = −233.8 lb ⋅ in.
M = 234 lb ⋅ in.
PROBLEM 7.8
A semicircular rod is loaded as shown. Determine the internal forces
at point K.
SOLUTION
FBD Rod:
ΣFy = 0: By − 30 lb = 0
ΣM A = 0: 2rBx = 0
B y = 30 lb
Bx = 0
ΣFx′ = 0: V − ( 30 lb ) cos 30° = 0
FBD BK:
V = 25.98 lb
V = 26.0 lb
ΣFy′ = 0: F + ( 30 lb ) sin 30° = 0
F = −15 lb
F = 15.00 lb
ΣM K = 0: M − ( 9 in.) sin 30°  ( 30 lb ) = 0
M = 135.0 lb ⋅ in.
PROBLEM 7.9
An archer aiming at a target is pulling with a 210-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
3 
ΣFx = 0: 2  T1  − 210 N = 0
5 
T1 = T2 = 175 N
Curve CJB is parabolic: y = ax 2
FBD BJ:
At B :
x = 0.64 m,
y = 0.16 m
a=
0.16 m
( 0.64 m )
2
=
1
2.56 m
1
( 0.32 m )2 = 0.04 m
2.56 m
So, at J : yJ =
Slope of parabola = tan θ =
dy
= 2ax
dx
 2
At J : θ J = tan −1 
( 0.32 m ) = 14.036°
 2.56 m

So
α = tan −1
4
− 14.036° = 39.094°
3
ΣFx′ = 0: V − (175 N ) cos ( 39.094° ) = 0
V = 135.8 N
ΣFy′ = 0: F + (175 N ) sin ( 39.094° ) = 0
F = −110.35 N
F = 110.4 N
PROBLEM 7.9 CONTINUED
3

Σ M J = 0 : M + ( 0.32 m )  (175 N ) 
5

4

+ ( 0.16 − 0.04 ) m   (175 N )  = 0
5


M = 50.4 N ⋅ m
PROBLEM 7.10
For the bow of Prob. 7.9, determine the magnitude and location of
the maximum (a) axial force, (b) shearing force, (c) bending moment.
SOLUTION
By symmetry
T1 = T2 = T
3
ΣFx = 0: 2T1   − 210 N = 0
5
FBD Point A:
4
(175 N ) = 0
5
FC = 140 N
3
(175 N ) − VC = 0
5
VC = 105 N
ΣFy = 0: FC −
ΣFx = 0:
FBD BC:
T1 = 175 N
3

4

ΣM C = 0: M C − ( 0.64 m )  (175 N )  − ( 0.16 m )  (175 N )  = 0
5

5

M C = 89.6 N ⋅ m
Also: if y = ax 2 and, at B, y = 0.16 m, x = 0.64 m
0.16 m
Then
a=
And
θ = tan −1
( 0.64 m )
FBD CK:
2
=
1
;
2.56 m
dy
= tan −1 2ax
dx
ΣFx′ = 0: (140 N ) cos θ − (105 N ) sin θ + F = 0
So
F = (105 N ) sin θ − (140 N ) cos θ
dF
= (105 N ) cos θ + (140 N ) sin θ
dθ
ΣFy′ = 0: V − (105 N ) cos θ − (140 N ) sin θ = 0
So
V = (105 N ) cos θ + (140 N ) sin θ
PROBLEM 7.10 CONTINUED
And
dV
= − (105 N ) sin θ + (140 N ) cos θ
dθ
ΣM K = 0: M + x (105 N ) + y (140 N ) − 89.6 N ⋅ m = 0
M = − (105 N ) x −
(140 N ) x 2 + 89.6 N ⋅ m
( 2.56 m )
dM
= − (105 N ) − (109.4 N/m ) x + 89.6 N ⋅ m
dx
Since none of the functions, F, V, or M has a vanishing derivative in
the valid range of 0 ≤ x ≤ 0.64 m ( 0 ≤ θ ≤ 26.6° ) , the maxima are at
the limits ( x = 0, or x = 0.64 m ) .
Therefore,
(a)
Fmax = 140.0 N
at C
(b)
Vmax = 156.5 N
at B
(c)
M max = 89.6 N ⋅ m
at C
PROBLEM 7.11
A semicircular rod is loaded as shown. Determine the internal forces at
point J knowing that θ = 30o.
SOLUTION
FBD AB:
4 
3 
ΣM A = 0: r  C  + r  C  − 2r ( 70 lb ) = 0
5 
5 
C = 100 lb
ΣFx = 0: − Ax +
4
(100 lb ) = 0
5
A x = 80 lb
ΣFy = 0: Ay +
3
(100 lb ) − 70 lb = 0
5
A y = 10 lb
FBD AJ:
ΣFx′ = 0: F − ( 80 lb ) sin 30° − (10 lb ) cos 30° = 0
F = 48.66 lb
F = 48.7 lb
60°
ΣFy′ = 0: V − ( 80 lb ) cos 30° + (10 lb ) sin 30° = 0
V = 64.28 lb
V = 64.3 lb
30°
ΣM 0 = 0: ( 8 in.)( 48.66 lb ) − ( 8 in.)(10 lb ) − M = 0
M = 309.28 lb ⋅ in.
M = 309 lb ⋅ in.
PROBLEM 7.12
A semicircular rod is loaded as shown. Determine the magnitude and
location of the maximum bending moment in the rod.
SOLUTION
4 
3 
ΣM A = 0: r  C  + r  C  − 2r ( 70 lb ) = 0
5 
5 
FBD AB:
C = 100 lb
ΣFx = 0: − Ax +
4
(100 lb ) = 0
5
A x = 80 lb
ΣFy = 0: Ay +
3
(100 lb ) − 70 lb = 0
5
A y = 10 lb
ΣM J = 0: M − ( 8 in.)(1 − cosθ )(10 lb ) − ( 8 in.)( sin θ )( 80 lb ) = 0
FBD AJ:
M = ( 640 lb ⋅ in.) sin θ + ( 80 lb ⋅ in.) ( cos θ − 1)
dM
= ( 640 lb ⋅ in.) cos θ − ( 80 lb ⋅ in.) sin θ = 0
dθ
for
θ = tan −1 8 = 82.87° ,
where
d 2M
= − ( 640 lb ⋅ in.) sin θ − ( 80 lb ⋅ in.) cos θ < 0
dθ 2
M = 565 lb ⋅ in. at θ = 82.9° is a max for AC
So
FBD BK:
ΣM K = 0: M − ( 8 in.)(1 − cos β )( 70 lb ) = 0
M = ( 560 lb ⋅ in.)(1 − cos β )
dM
= ( 560 lb ⋅ in.) sin β = 0
dβ
So, for β =
π
2
for β = 0, where M = 0
, M = 560 lb ⋅ in. is max for BC
∴ M max = 565 lb ⋅ in. at θ = 82.9°
PROBLEM 7.13
Two members, each consisting of straight and 168-mm-radius quartercircle portions, are connected as shown and support a 480-N load at D.
Determine the internal forces at point J.
SOLUTION
FBD Frame:
 24 
ΣM A = 0: ( 0.336 m )  C  − ( 0.252 m )( 480 N ) = 0
 25 
C = 375 N
ΣFy = 0: Ax −
24
C =0
25
Ax =
24
( 375 N ) = 360 N
25
A x = 360 N
ΣFy = 0: Ay − 480 N +
FBD CD:
7
( 375 N ) = 0
24
A y = 375 N
ΣM C = 0: ( 0.324 m )( 480 N ) − ( 0.27 m ) B = 0
B = 576 N
ΣFx = 0: C x −
24
( 375 N ) = 0
25
C x = 360 N
ΣFy = 0: −480 N +
FBD CJ:
7
( 375 N ) + ( 576 N ) − C y = 0
25
C y = 201 N
ΣFx′ = 0: V − ( 360 N ) cos 30° − ( 201 N ) sin 30° = 0
V = 412 N
ΣFy′ = 0: F + ( 360 N ) sin 30° − ( 201 N ) cos 30° = 0
F = −5.93 N
F = 5.93 N
ΣM 0 = 0: ( 0.168 m )( 201 N + 5.93 N ) − M = 0
M = 34.76 N ⋅ m
M = 34.8 N ⋅ m
PROBLEM 7.14
Two members, each consisting of straight and 168-mm-radius quartercircle portions, are connected as shown and support a 480-N load at D.
Determine the internal forces at point K.
SOLUTION
FBD CD:
ΣFx = 0: C x = 0
ΣM B = 0: ( 0.054 m )( 480 N ) − ( 0.27 m ) C y = 0
C y = 96 N
ΣFy = 0: B − C y = 0
FBD CK:
B = 96 N
ΣFy′ = 0: V − ( 96 N ) cos 30° = 0
V = 83.1 N
ΣFx′ = 0: F − ( 96 N ) sin 30° = 0
F = 48.0 N
ΣM K = 0: M − ( 0.186 m )( 96 N ) = 0
M = 17.86 N ⋅ m
PROBLEM 7.15
Knowing that the radius of each pulley is 7.2 in. and neglecting friction,
determine the internal forces at point J of the frame shown.
SOLUTION
FBD Frame:
Note: Tension T in cord is 90 lb at any cut. All radii = 0.6 ft
ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0
B x = 140 lb
ΣM E = 0: ( 5.4 ft )(140 lb ) − ( 7.2 ft ) By
+ ( 4.8 ft ) 90 lb − ( 0.6 ft ) 90 lb = 0
FBD BCE with pulleys and
cord:
B y = 157.5 lb
ΣFx = 0: Ex − 140 lb = 0
E x = 140 lb
ΣFy = 0: 157.5 lb − 90 lb − 90 lb + E y = 0
E y = 22.5 lb
FBD EJ:
ΣFx′ = 0: −V +
3
4
(140 lb ) + ( 22.5 lb − 90 lb ) = 0
5
5
V = 30 lb
ΣFy′ = 0: F + 90 lb −
V = 30.0 lb
4
3
(140 lb ) − ( 90 lb − 22.5 lb ) = 0
5
5
F = 62.5lb
F = 62.5 lb
ΣM J = 0: M + (1.8 ft )(140 lb ) + ( 0.6 ft )( 90 lb )
+ ( 2.4 ft )( 22.5 lb ) − ( 3.0 ft )( 90 lb ) = 0
M = − 90 lb ⋅ ft
M = 90.0 lb ⋅ ft
PROBLEM 7.16
Knowing that the radius of each pulley is 7.2 in. and neglecting
friction, determine the internal forces at point K of the frame shown.
SOLUTION
FBD Whole:
Note: T = 90 lb
ΣM B = 0: ( 5.4 ft ) Ax − ( 6 ft )( 90 lb ) − ( 7.8 ft )( 90 lb ) = 0
A x = 2.30 lb
FBD AE:
Note: Cord tensions moved to point D as per Problem 6.91
ΣFx = 0: 230 lb − 90 lb − Ex = 0
E x = 140 lb
ΣM A = 0: (1.8 ft )( 90 lb ) − ( 7.2 ft ) E y = 0
E y = 22.5 lb
FBD KE:
ΣFx = 0: F − 140 lb = 0
F = 140.0 lb
ΣFy = 0: V − 22.5 lb = 0
V = 22.5 lb
ΣM K = 0: M − ( 2.4 ft )( 22.5 lb ) = 0
M = 54.0 lb ⋅ ft
PROBLEM 7.17
Knowing that the radius of each pulley is 7.2 in. and neglecting friction,
determine the internal forces at point J of the frame shown.
SOLUTION
FBD Whole:
ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) = 0
B x = 130 lb
FBD BE with pulleys and cord:
ΣM E = 0: ( 5.4 ft )(130 lb ) − ( 7.2 ft ) By
+ ( 4.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0
B y = 150 lb
ΣFx = 0: Ex − 130 lb = 0
E x = 130 lb
ΣFy = 0: E y + 150 lb − 90 lb − 90 lb = 0
E y = 30 lb
FBD JE and pulley:
ΣFx′ = 0: − F − 90 lb +
4
3
(130 lb ) + ( 90 lb − 30 lb ) = 0
5
5
F = 50.0 lb
ΣFy′ = 0: V +
3
4
(130 lb ) + ( 30 lb − 90 lb ) = 0
5
5
V = −30 lb
V = 30.0 lb
ΣM J = 0: − M + (1.8 ft )(130 lb ) + ( 2.4 ft )( 30 lb ) + ( 0.6 ft )( 90 lb )
− ( 3.0 ft )( 90 lb ) = 0
M = 90.0 lb ⋅ ft
PROBLEM 7.18
Knowing that the radius of each pulley is 7.2 in. and neglecting
friction, determine the internal forces at point K of the frame shown.
SOLUTION
FBD Whole:
ΣM B = 0: ( 5.4 ft ) Ax − ( 7.8 ft )( 90 lb ) = 0
A x = 130 lb
FBD AE:
ΣM E = 0: − ( 7.2 ft ) Ay − ( 4.8 ft )( 90 lb ) = 0
Ay = −60 lb
A y = 60 lb
ΣFx = 0:
FBD AK:
F=0
ΣFy = 0: −60 lb + 90 lb − V = 0
V = 30.0 lb
ΣM K = 0: ( 4.8 ft )( 60 lb ) − ( 2.4 ft )( 90 lb ) − M = 0
M = 72.0 lb ⋅ ft
PROBLEM 7.19
A 140-mm-diameter pipe is supported every 3 m by a small frame
consisting of two members as shown. Knowing that the combined
mass per unit length of the pipe and its contents is 28 kg/m and
neglecting the effect of friction, determine the internal forces at
point J.
SOLUTION
FBD Whole:
(
)
W = ( 3 m )( 28 kg/m ) 9.81 m/s 2 = 824.04 N
ΣM A = ( 0.6 m ) Cx − ( 0.315 m )( 824.04 N ) = 0
C x = 432.62 N
FBD pipe:
By symmetry: N1 = N 2
ΣFy = 0: 2
N1 =
21
N1 − W = 0
29
29
(824.04 N )
42
= 568.98 N
Also note:
 20 
a = r tan θ = 70 mm  
 21 
a = 66.67 mm
FBD BC:
ΣM B = 0: ( 0.3 m )( 432.62 N ) − ( 0.315 m ) C y
+ ( 0.06667 m )( 568.98 N ) = 0
C y = 532.42 N
PROBLEM 7.19 CONTINUED
FBD CJ:
ΣFx′ = 0: F −
21
20
( 432.62 N ) − ( 532.42 N ) = 0
29
29
F = 680 N
ΣFy′ = 0:
21
20
( 532.42 N ) − ( 432.62 N ) − V = 0
29
29
V = 87.2 N
ΣM J = 0: ( 0.15 m )( 432.62 N ) − ( 0.1575 m )( 532.42 N ) + M = 0
M = 18.96 N ⋅ m
PROBLEM 7.20
A 140-mm-diameter pipe is supported every 3 m by a small frame
consisting of two members as shown. Knowing that the combined mass
per unit length of the pipe and its contents is 28 kg/m and neglecting the
effect of friction, determine the internal forces at point K.
SOLUTION
FBD Whole:
(
)
W = ( 3 m )( 28 kg/m ) 9.81 m/s 2 = 824.04 N
ΣM C = 0: (.6 m ) Ax − (.315 m )( 824.04 N ) = 0
A x = 432.62 N
FBD pipe
By symmetry: N1 = N 2
ΣFy = 0: 2
N2 =
21
N1 − W = 0
29
29
824.04 N
42
= 568.98 N
Also note:
FBD AD:
a = r tan θ = ( 70 mm )
20
21
a = 66.67 mm
ΣM B = 0: ( 0.3 m )( 432.62 N ) − ( 0.315 m ) Ay
− ( 0.06667 m )( 568.98 N ) = 0
A y = 291.6 N
PROBLEM 7.20 CONTINUED
FBD AK:
ΣFx′ = 0:
21
20
( 432.62 N ) + ( 291.6 N ) − F = 0
29
29
F = 514 N
ΣFy′ = 0:
21
20
( 291.6 N ) − ( 432.62 N ) + V = 0
29
29
V = 87.2 N
ΣM K = 0: ( 0.15 m )( 432.62 N ) − ( 0.1575 m )( 291.6 N ) − M = 0
M = 18.97 N ⋅ m
PROBLEM 7.21
A force P is applied to a bent rod which is supported by a roller and
a pin and bracket. For each of the three cases shown, determine the
internal forces at point J.
SOLUTION
(a) FBD Rod:
ΣM D = 0: aP − 2aA = 0
A=
P
2
ΣFx = 0: V −
P
=0
2
V =
FBD AJ:
ΣFy = 0:
F=0
ΣM J = 0: M − a
P
=0
2
M =
(b) FBD Rod:
ΣM D = 0: aP −
A=
5P
2
P
2
a4 
 A = 0
25 
aP
2
PROBLEM 7.21 CONTINUED
FBD AJ:
ΣFx = 0:
3 5P
−V = 0
5 2
V =
ΣFy = 0:
3P
2
4 5P
−F =0
5 2
F = 2P
M =
(c) FBD Rod:
3
aP
2
3 
4 
ΣM D = 0: aP − 2a  A  − 2a  A  = 0
5 
5 
A=
5P
14
 3 5P 
ΣFx = 0: V − 
=0
 5 14 
V =
ΣFy = 0:
3P
14
4 5P
−F =0
5 14
F=
2P
7
 3 5P 
ΣM J = 0: M − a 
=0
 5 14 
M =
3
aP
14
PROBLEM 7.22
A force P is applied to a bent rod which is supported by a roller and
a pin and bracket. For each of the three cases shown, determine the
internal forces at point J.
SOLUTION
(a) FBD Rod:
ΣFx = 0: Ax = 0
ΣM D = 0: aP − 2aAy = 0
Ay =
P
2
ΣFx = 0: V = 0
FBD AJ:
ΣFy = 0:
P
−F =0
2
F=
P
2
ΣM J = 0: M = 0
(b) FBD Rod:
ΣM A = 0
4 
3 
2a  D  + 2a  D  − aP = 0
5 
5 
D=
5P
14
ΣFx = 0: Ax −
4 5
P=0
5 14
Ax =
2P
7
ΣFy = 0: Ay − P +
3 5
P=0
5 14
Ay =
11P
14
PROBLEM 7.22 CONTINUED
FBD AJ:
ΣFx = 0:
2
P −V = 0
7
V =
2P
7
11P
−F =0
14
ΣFy = 0:
F=
ΣM J = 0: a
2P
−M =0
7
M =
(c) FBD Rod:
FBD AJ:
ΣM A = 0:
11P
14
a  4D 

 − aP = 0
2 5 
D=
2
aP
7
5P
2
ΣFx = 0: Ax −
4 5P
=0
5 2
Ax = 2P
ΣFy = 0: Ay − P −
3 5P
=0
5 2
Ay =
5P
2
ΣFx = 0: 2 P − V = 0
V = 2P
ΣFy = 0:
5P
−F =0
2
F=
5P
2
ΣM J = 0: a ( 2 P ) − M = 0
M = 2aP
PROBLEM 7.23
A rod of weight W and uniform cross section is bent into the circular
arc of radius r shown. Determine the bending moment at point J
when θ = 30°.
SOLUTION
FBD CJ:
Note α =
180° − 60°
π
= 60° =
2
3
r =
r
α
sin α =
3r
3
3 3
r
=
π 2
2π
Weight of section = W
ΣFy′ = 0: F −
120
4
= W
270 9
4
W cos 30° = 0
9
ΣM 0 = 0: rF − ( r sin 60° )
F =
2 3
W
9
4W
−M =0
9
2 3 3 3 3 4
2 3 1 
−
−  Wr
M = r
W = 
π 
2π 2 9 
 9
 9
M = 0.0666Wr
PROBLEM 7.24
A rod of weight W and uniform cross section is bent into the circular
arc of radius r shown. Determine the bending moment at point J
when θ = 120°.
SOLUTION
(a) FBD Rod:
ΣFx = 0: Ax = 0
ΣM B = 0: rAy +
2r W 2r 2W
−
=0
π 3
π 3
Ay =
2W
3π
FBD AJ:
α =
Note:
60°
π
= 30° =
2
6
Weight of segment = W
F =
r
α
sin α =
ΣM J = 0: ( r cos α − r sin 30° )
M =
60
2W
=
270
9
r
3r
sin 30° =
π /6
π
2W
2W
+ ( r − r sin 30° )
−M =0
9
3π
 3 1
2W  3r 3 r
3r 
1 
− +
− +

 = Wr 

9 π 2
2 2π 
9 3π 
 3π
M = 0.1788Wr
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
θ = 30o.
SOLUTION
FBD Rod:
ΣFx = 0: A x = 0
ΣM B = 0:
2r
π
W − rAy = 0
α = 15°, weight of segment = W
FBD AJ:
r =
r
α
sin α =
ΣFy′ = 0:
2W
π
F=
Ay =
2W
π
30° W
=
90°
3
r
sin15° = 0.9886r
π /12
cos 30° −
W
cos 30° − F = 0
3
W 3  2 1
 − 
2 π 3
2W 
W

ΣM 0 = M + r  F −
=0
 + r cos15°
π
3


M = 0.0557 Wr
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
θ = 30o.
SOLUTION
FBD Rod:
ΣM A = 0: rB −
B=
r =
r
π /12
π
π
π
12
sin15° = 0.98862r
Weight of segment = W
ΣFy′ = 0: F −
W =0
2W
α = 15° =
FBD BJ:
2r
30° W
=
90°
3
W
2W
cos 30° −
sin 30° = 0
π
3
 3 1
+ W
F=
 6
π 

ΣM 0 = 0: rF − ( r cos15° )
W
−M =0
3
 3 1 
cos15° 
M = rW 
+  −  0.98862
 Wr
 6

π 
3 

M = 0.289Wr
PROBLEM 7.27
For the rod of Prob.7.26, determine the magnitude and location of the
maximum bending moment.
SOLUTION
FBD Bar:
ΣM A = 0: rB −
α =
2r
π
θ
W =0
r =
r
α
=
F =
=
2W
π
2W
π
4α
π
M =
π
W
2W
π
sin 2α = 0
( sin 2α + 2α cos 2α )
( sin θ + θ cosθ )
4α
π
W −M =0
r
 4α
Wr ( sin θ + θ cos θ ) −  sin α cos α 
W
π
α
 π
2
But,
sin α cos α =
so
M =
or
4
2α
π /2
4α
W cos 2α −
ΣM 0 = 0: rF − ( r cos α )
FBD BJ:
π
sin α ,
Weight of segment = W
ΣFx′ = 0: F −
2W
π
0≤α ≤
so
2
B=
2Wr
π
1
1
sin 2α = sin θ
2
2
( sin θ + θ cosθ − sin θ )
M =
2
π
Wrθ cosθ
dM
2
= Wr ( cos θ − θ sin θ ) = 0 at θ tan θ = 1
dθ
π
PROBLEM 7.27 CONTINUED
Solving numerically
θ = 0.8603 rad
and
M = 0.357Wr
at θ = 49.3°
(Since M = 0 at both limits, this is the maximum)
PROBLEM 7.28
For the rod of Prob.7.25, determine the magnitude and location of the
maximum bending moment.
SOLUTION
FBD Rod:
ΣFx = 0: Ax = 0
2r
ΣM B = 0:
W − rAy = 0
π
α =
θ
2
r =
,
r
α
F =
2W
π
4α
π
π
2α
4α
W
=
π /2
π
2W
W cos 2α +
(1 − 2α ) cos 2α =
2W
sin α
Weight of segment = W
ΣFx′ = 0: − F −
Ay =
π
2W
π
cos 2α = 0
(1 − θ ) cosθ
2W 
4α

W =0
ΣM 0 = 0: M +  F −
 r + ( r cos α )
π 
π

FBD AJ:
M =
2W
π
(1 + θ cosθ − cosθ ) r −
sin α cos α =
But,
M =
so
2r
π
4αW r
π
α
sin α cos α
1
1
sin 2α = sin θ
2
2
W (1 − cosθ + θ cosθ − sin θ )
dM
2rW
=
( sin θ − θ sin θ + cosθ − cosθ ) = 0
dθ
π
(1 − θ ) sin θ = 0
for
dM
=0
dθ
for
θ = 0, 1, nπ ( n = 1, 2,")
Only 0 and 1 in valid range
At
θ = 0 M = 0,
at θ = 57.3°
at θ = 1 rad
M = M max = 0.1009 Wr
PROBLEM 7.29
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
FBD beam:
(a) By symmetry: Ay = D =
1
L
( w)
2
2
Ay = D =
wL
4
Along AB:
ΣFy = 0:
ΣM J = 0: M − x
wL
−V = 0
4
wL
=0
4
M =
V =
wL
x (straight)
4
Along BC:
ΣFy = 0:
wL
− wx1 − V = 0
4
V =
ΣM k = 0: M +
M =
wL
− wx1
4
V =0
straight with
wL
4
at
x1 =
L
4
x1
L
 wL
wx1 −  + x1 
=0
2
4
 4

w  L2 L
+ x1 − x12 

2 8
2

PROBLEM 7.29 CONTINUED
Parabola with
M =
3
L
wL2 at x1 =
32
4
Section CD by symmetry
(b) From diagrams:
V max =
wL
on AB and CD
4
M max =
3wL2
at center
32
PROBLEM 7.30
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) Along AB:
ΣFy = 0: −wx − V = 0
straight with
V =−
ΣM J = 0: M +
wL
2
x=
at
x
wx = 0
2
M =−
parabola with
V = − wx
L
2
1
M = − wx 2
2
wL2
L
at x =
8
2
Along BC:
ΣFy = 0: − w
L
−V = 0
2
1
V = − wL
2
L L

ΣM k = 0: M +  x1 +  w = 0
4 2

M =−
straight with
(b) From diagrams:
wL  L

 + x1 
2 4

3
L
M = − wL2 at x1 =
8
2
V max =
wL
on BC
2
M max =
3wL2
at C
8
PROBLEM 7.31
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) Along AB:
ΣFy = 0: P − V = 0
V = P
ΣM J = 0: M − Px = 0
M = Px
straight with M =
PL
at B
2
Along BC:
ΣFy = 0: P − P − V = 0
V =0
L

ΣM K = 0: M + Px1 − P  + x1  = 0
2

M =
(b) From diagrams:
PL
2
(constant)
V max = P along AB
M max =
PL
along BC
2
PROBLEM 7.32
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
ΣM C = 0: LAy − M 0 = 0
(a) FBD Beam:
Ay =
M0
L
ΣFy = 0: − Ay + C = 0
M0
L
C=
Along AB:
ΣFy = 0: −
M0
−V = 0
L
ΣM J = 0: x
V =−
M0
+M =0
L
M0
L
M =−
M0
x
L
straight with M = −
M0
at B
2
M0
−V = 0
L
V =−
M0
− M0 = 0
L
x

M = M 0 1 − 
L


Along BC:
ΣFy = 0: −
ΣM K = 0: M + x
straight with M =
(b) From diagrams:
M0
at B
2
M0
L
M = 0 at C
V max = P everywhere
M max =
M0
at B
2
PROBLEM 7.33
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣM B = 0:
(.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay = 0
A y = 12.625 kips
ΣFy = 0: 12.625 kips − 10 kips − 8 kips − 4 kips + B = 0
B = 9.375 kips
Along AC:
ΣFy = 0: 12.625 kips − V = 0
V = 12.625 kips
ΣM J = 0: M − x (12.625 kips ) = 0
M = (12.625 kips ) x
M = 22.725 kip ⋅ ft at C
Along CD:
ΣFy = 0: 12.625 kips − 10 kips − V = 0
V = 2.625 kips
ΣM K = 0: M + ( x − 1.8 ft )(10 kips ) − x (12.625 kips ) = 0
M = 18 kip ⋅ ft + ( 2.625 kips ) x
M = 29.8125 kip ⋅ ft at D ( x = 4.5 ft )
PROBLEM 7.33 CONTINUED
Along DE:
Along DE:
ΣFy = 0: (12.625 − 10 − 8 ) kips − V = 0
V = −5.375 kips
ΣM L = 0: M + x1 ( 8 kips ) + ( 2.7 ft + x1 )(10 kips )
− ( 4.5 ft + x1 )(12.625 kips ) = 0
M = 29.8125 kip ⋅ ft − ( 5.375 kips ) x1
M = 5.625 kip ⋅ ft at E
Along EB:
( x1 = 4.5 ft )
Along EB:
ΣFy = 0: V + 9.375 kips = 0
V = 9.375 kips
ΣM N = 0: x2 ( 9.375 kip ) − M = 0
M = ( 9.375 kips ) x2
M = 5.625 kip ⋅ ft at E
(b) From diagrams:
V max = 12.63 kips on AC
M max = 29.8 kip ⋅ ft at D
PROBLEM 7.34
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣM C = 0:
(1.2 m )( 4 kN ) − (1 m )(16 kN ) + ( 2 m )(8 kN ) + ( 3.2 m ) B = 0
B = −1.5 kN
ΣFy = 0: −4 kN + C y − 16 kN + 8 kN − 1.5 kN = 0
C y = 13.5 kN
Along AC:
ΣFy = 0: −4 kN − V = 0
V = −4 kN
ΣM J = 0: M + x ( 4 kN ) = 0
M = −4 kN x
M = −4.8 kN ⋅ m at C
Along CD:
ΣFy = 0: −4 kN + 13.5 kN − V = 0
V = 9.5 kN
ΣM K = 0: M + (1.2 m + x1 )( 4 kN ) − x1 (13.5 kN ) = 0
M = −4.8 kN ⋅ m + ( 9.5 kN ) x1
M = 4.7 kN ⋅ m at D ( x1 = 1 m )
PROBLEM 7.34 CONTINUED
Along DE:
ΣFy = 0: V + 8 kN − 1.5 kN = 0
V = −6.5 kN
ΣM L = 0: M − x3 ( 8 kN ) + ( x3 + 1.2 m ) (1.5 kN ) = 0
M = −1.8 kN ⋅ m + ( 6.5 kN ) x3
M = 4.7 kN ⋅ m at D ( x3 = 1 m )
Along EB:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM N = 0: x2 (1.5 kN ) + M = 0
M = − (1.5 kN ) x2
(b) From diagrams:
M = −1.8 kN ⋅ m at E
V max = 9.50 kN ⋅ on CD
M max = 4.80 kN ⋅ m at C
PROBLEM 7.35
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) Along AC:
ΣFy = 0: −3 kip − V = 0
V = −3 kips
ΣM J = 0: M + x ( 3 kips ) = 0
M = ( 3 kips ) x
M = −9 kip ⋅ ft at C
Along CD:
ΣFy = 0: −3 kips − 5 kips − V = 0
V = −8 kips
ΣM K = 0: M + ( x − 3 ft )( 5 kips ) + x ( 3 kips ) = 0
M = +15 kip ⋅ ft − ( 8 kips ) x
M = −16.2 kip ⋅ ft at D ( x = 3.9 ft )
Along DE:
ΣFy = 0: −3 kips − 5 kips + 6 kips − V = 0
V = −2 kips
ΣM L = 0: M − x1 ( 6 kips ) + (.9 ft + x1 )( 5 kips )
+ ( 3.9 ft + x1 )( 3 kips ) = 0
M = −16.2 kip ⋅ ft − ( 2 kips ) x1
M = −18.6 kip ⋅ ft at E
( x1 = 1.2 ft )
PROBLEM 7.35 CONTINUED
Along EB:
ΣFy = 0: −3 kips − 5 kips + 6 kips − 4 kips − V = 0
V = −6 kips
ΣM N = 0: M + ( 4 kips ) x2 + ( 2.1 ft + x2 )( 5 kips )
+ ( 5.1 ft + x2 )( 3 kips ) − (1.2 ft + x2 )( 6 kips ) = 0
M = −18.6 kip ⋅ ft − ( 6 kips ) x2
M = −33 kip ⋅ ft at B ( x2 = 2.4 ft )
(b) From diagrams:
V max = 8.00 kips on CD
M max = 33.0 kip ⋅ ft at B
PROBLEM 7.36
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
ΣM E = 0:
(a) FBD Beam:
(1.1 m )( 540 N ) − ( 0.9 m ) C y + ( 0.4 m )(1350 N ) − ( 0.3 m )( 540 N ) = 0
C y = 1080 N
ΣFy = 0: −540 N + 1080 N − 1350 N
−540 N + E y = 0
E y = 1350 N
Along AC:
ΣFy = 0: −540 N − V = 0
V = −540 N
ΣM J = 0: x ( 540 N ) + M = 0
M = − ( 540 N ) x
Along CD:
ΣFy = 0: −540 N + 1080 N − V = 0
V = 540 N
ΣM K = 0: M + ( 0.2 m + x1 )( 540 N ) − x1 (1080 N ) = 0
M = −108 N ⋅ m + ( 540 N ) x1
M = 162 N ⋅ m at D ( x1 = 0.5 m )
PROBLEM 7.36 CONTINUED
Along DE:
ΣFy = 0: V + 1350 N − 540 N = 0
V = −810 N
ΣM N = 0: M + ( x3 + 0.3 m )( 540 N ) − x3 (1350 N ) = 0
M = −162 N ⋅ m + ( 810 N ) x3
M = 162 N ⋅ m at D ( x3 = 0.4 )
Along EB:
ΣFy = 0: V − 540 N = 0
V = 540 N
ΣM L = 0: M + x2 ( 540 N ) = 0
M = −162 N ⋅ m at E
(b) From diagrams
M = −540 N x2
( x2 = 0.3 m )
V max = 810 N on DE W
M max = 162.0 N ⋅ m at D and E W
PROBLEM 7.37
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣFy = 0: Ay + ( 6 ft )( 2 kips/ft ) − 12 kips − 2 kips = 0
A y = 2 kips
ΣM A = 0: M A + ( 3 ft )( 6 ft )( 2 kips/ft )
− (10.5 ft )(12 kips ) − (12 ft )( 2 kips ) = 0
M A = 114 kip ⋅ ft
Along AC:
ΣFy = 0: 2 kips + x ( 2 kips/ft ) − V = 0
V = 2 kips + ( 2 kips/ft ) x
V = 14 kips at C ( x = 6 ft )
ΣM J = 0: 114 kip ⋅ ft − x ( 2 kips )
−
x
x ( 2 kips/ft ) + M = 0
2
M = (1 kip/ft ) x 2 + ( 2 kips ) x − 114 kip ⋅ ft
M = −66 kip ⋅ ft at C ( x = 6 ft )
Along CD:
ΣFy = 0: V − 12 kips − 2 kips = 0
V = 14 kips
ΣM k = 0: −M − x1 (12 kips ) − (1.5 ft + x1 )( 2 kips ) = 0
PROBLEM 7.37 CONTINUED
M = −3 kip ⋅ ft − (14 kips ) x1
Along DB:
M = −66 kip ⋅ ft at C
ΣFy = 0:
( x1 = 4.5 ft )
V − 2 kips = 0
V = + 2 kips
ΣM L = 0: −M − 2 kip x3 = 0
M = − ( 2 kips ) x3
M = −3 kip ⋅ ft at D ( x = 1.5 ft )
(b) From diagrams:
V max = 14.00 kips on CD W
M max = 114.0 kip ⋅ ft at A W
PROBLEM 7.38
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣM A = (15 ft ) B − (12 ft )( 2 kips/ft )( 6 ft ) − ( 6 ft )(12 kips ) = 0
B = 14.4 kips
ΣFy = 0: Ay − 12 kips − ( 2 kips/ft )( 6 ft ) + 14.4 kips
A y = 9.6 kips
Along AC:
ΣFy = 0: 9.6 kips − V = 0
V = 9.6 kips
ΣM J = 0: M − x ( 9.6 kips ) = 0
M = ( 9.6 kips ) x
M = 57.6 kip ⋅ ft at C ( x = 6 ft )
Along CD:
ΣFy = 0: 9.6 kips − 12 kips − V = 0
V = −2.4 kips
ΣM K = 0: M + x1 (12 kips ) − ( 6 ft + x1 )( 9.6 kips ) = 0
M = 57.6 kip ⋅ ft − ( 2.4 kips ) x1
M = 50.4 kip ⋅ ft at D
PROBLEM 7.38 CONTINUED
Along DB:
ΣFy = 0: V − x3 ( 2 kips/ft ) + 14.4 kips = 0
V = −14.4 kips + ( 2 kips/ft ) x3
V = −2.4 kips at D
ΣM L = 0: M +
x3
( 2 kips/ft )( x3 ) − x3 (14.4 kips ) = 0
2
M = (14.4 kips ) x3 − (1 kip/ft ) x32
M = 50.4 kip ⋅ ft at D ( x3 = 6 ft )
(b) From diagrams:
V max = 14.40 kips at B W
M max = 57.6 kip ⋅ ft at C W
PROBLEM 7.39
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a) By symmetry:
Ay = B = 8 kN +
1
( 4 kN/m )( 5 m )
2
A y = B = 18 kN
Along AC:
ΣFy = 0: 18 kN − V = 0
ΣM J = 0: M − x (18 kN )
V = 18 kN
M = (18 kN ) x
M = 36 kN ⋅ m at C ( x = 2 m )
Along CD:
ΣFy = 0: 18 kN − 8 kN − ( 4 kN/m ) x1 − V = 0
V = 10 kN − ( 4 kN/m ) x1
V = 0 at x1 = 2.5 m ( at center )
ΣM K = 0: M +
x1
( 4 kN/m ) x1 + (8 kN ) x1 − ( 2 m + x1 )(18 kN ) = 0
2
M = 36 kN ⋅ m + (10 kN/m ) x1 − ( 2 kN/m ) x12
M = 48.5 kN ⋅ m at x1 = 2.5 m
Complete diagram by symmetry
(b) From diagrams:
V max = 18.00 kN on AC and DB W
M max = 48.5 kN ⋅ m at center W
PROBLEM 7.40
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
ΣM D = 0: ( 2 m )( 2 kN/m )( 2 m ) − ( 2.5 m )( 4 kN/m )( 3 m )
− ( 4 m ) F − ( 5 m )( 22 kN ) = 0
F = 22 kN
ΣFy = 0: − ( 2 m )( 2 kN/m ) + D y
− ( 3 m )( 4 kN/m ) − 22 kN + 22 kN = 0
D y = 16 kN
Along AC:
ΣFy = 0: − x ( 2 kN/m ) − V = 0
V = − ( 2 kN/m ) x
V = −4 kN at C
ΣM J = 0: M +
x
( 2 kN/m )( x ) ≠ 0
2
M = − (1 kN/m ) x 2
M = −4 kN ⋅ m at C
Along CD:
ΣFy = 0: − ( 2 m )( 2 kN/m ) − V = 0
V = −4 kN
ΣM K = 0: (1 m + x1 )( 2 kN/m )( 2 m ) = 0
M = −4 kN ⋅ m − ( 4 kN/m ) x1
M = −8 kN ⋅ m at D
PROBLEM 7.40 CONTINUED
Along DE:
ΣFy = 0: − ( 2 kN/m )( 2 m ) + 16 kN − V = 0
V = 12 kN
ΣM L = 0: M − x2 (16 kN ) + ( x2 + 2 m )( 2 kN/m )( 2 m ) = 0
M = −8 kN ⋅ m + (12 kN ) x2
M = 4 kN ⋅ m at E
Along EF:
ΣFy = 0: V − x4 ( 4 kN/m ) − 22 kN + 22 kN = 0
V = ( 4 kN/m ) x4
ΣM 0 = 0: M +
V = 12 kN at E
x4
( 4 kN/m ) x4 − (1 m )( 22 kN ) = 0
2
M = 22 kN ⋅ m − ( 2 kN/m ) x42
M = 4 kN ⋅ m at E
Along FB:
ΣFy = 0: V + 22 kN = 0
V = 22 kN
ΣM N = 0: M − x3 ( 22 kN ) = 0
M = ( 22 kN ) x3
M = 22 kN ⋅ m at F
(b) From diagrams:
V max = 22.0 kN on FB W
M max = 22.0 kN ⋅ m at F W
PROBLEM 7.41
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
ΣFy = 0: (12 m ) w − ( 6 m )( 3 kN/m ) = 0
(a)
w = 1.5 kN/m
Along AC:
ΣFy = 0: x (1.5 kN/m ) − V = 0
V = (1.5 kN/m ) x
V = 4.5 kN at C
ΣM J = 0: M −
x
(1.5 kN/m )( x ) = 0
2
M = ( 0.75 kN/m ) x 2
M = 6.75 N ⋅ m at C
Along CD:
ΣFy = 0: x (1.5 kN/m ) − ( x − 3 m )( 3 kN/m ) − V = 0
V = 9 kN − (1.5 kN/m ) x
V = 0 at x = 6 m
x
 x − 3m
ΣM K = 0: M + 
 ( 3 kN/m )( x − 3 m ) − (1.5 kN/m ) x = 0
2
2


M = −13.5 kN ⋅ m + ( 9 kN ) x − ( 0.75 kN/m ) x 2
M = 13.5 kN ⋅ m at center ( x = 6 m )
Finish by symmetry
(b) From diagrams:
V max = 4.50 kN at C and D
M max = 13.50 kN ⋅ m at center
PROBLEM 7.42
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
(a) FBD Beam:
ΣFy = 0: ( 4 m )( w ) − ( 2 m )(12 kN/m ) = 0
w = 6 kN/m
Along AC:
ΣFy = 0: − x ( 6 kN/m ) − V = 0
V = − ( 6 kN/m ) x
V = −6 kN at C ( x = 1 m )
ΣM J = 0: M +
x
( 6 kN/m )( x ) = 0
2
M = − ( 3 kN/m ) x 2
M = −3 kN ⋅ m at C
Along CD:
ΣFy = 0: − (1 m )( 6 kN/m ) + x1 ( 6 kN/m ) − v = 0
V = ( 6 kN/m )(1 m − x1 )
V = 0 at x1 = 1 m
ΣM K = 0: M + ( 0.5 m + x1 )( 6 kN/m )(1 m ) −
x1
( 6 kN/m ) x1 = 0
2
M = −3 kN ⋅ m − ( 6 kN ) x1 + ( 3 kN/m ) x12
M = −6 kN ⋅ m at center ( x1 = 1 m )
Finish by symmetry
(b) From diagrams:
V max = 6.00 kN at C and D
M max = 6.00 kN at center
PROBLEM 7.43
Assuming the upward reaction of the ground on beam AB to be
uniformly distributed and knowing that a = 0.9 ft, (a) draw the
shear and bending-moment diagrams, (b) determine the maximum
absolute values of the shear and bending moment.
SOLUTION
(a) FBD Beam:
ΣFy = 0: ( 4.5 ft ) w − 900 lb − 900 lb = 0
w = 400 lb/ft
Along AC:
ΣFy = 0: x ( 400 lb ) − V = 0
V = 360 lb at C
ΣM J = 0: M −
V = ( 400 lb ) x
( x = 0.9 ft )
x
( 400 lb/ft ) x = 0
2
M = ( 200 lb/ft ) x 2
M = 162 lb ⋅ ft at C
Along CD:
ΣFy = 0: ( 0.9 ft + x1 )( 400 lb/ft ) − 900 lb − V = 0
V = −540 lb + ( 400 lb/ft ) x1
ΣM K = 0: M + x1 ( 900 lb ) −
V = 0 at x1 = 1.35 ft
0.9 ft + x1
( 400 lb/ft )( 0.9 ft + x1 ) = 0
2
M = 162 lb ⋅ ft − ( 540 lb ) x1 + ( 200 lb/ft ) x12
M = −202.5 lb ⋅ ft at center ( x1 = 1.35 ft )
Finish by symmetry
(b) From diagrams:
V max = 540 lb at C + and D −
M max = 203 lb ⋅ ft at center
PROBLEM 7.44
Solve Prob. 7.43 assuming that a = 1.5 ft.
SOLUTION
(a) FBD Beam:
ΣFy = 0: ( 4.5 ft ) w − 900 lb − 900 lb = 0
w = 400 lb/ft
Along AC:
x ( 400 lb/ft ) − V = 0
ΣFy = 0:
V = ( 400 lb/ft ) x
V = 600 lb at C
ΣM J = 0: M −
( x = 1.5 ft )
x
( 400 lb/ft ) x = 0
2
M = ( 200 lb/ft ) x 2
M = 450 lb ⋅ ft at C
Along CD:
ΣFy = 0: x ( 400 lb/ft ) − 900 lb − V = 0
V = −900 lb + ( 400 lb/ft ) x
V = −300 at x = 1.5 ft
V = 0 at x = 2.25 ft
ΣM K = 0: M + ( x − 1.5 ft )( 900 lb ) −
x
( 400 lb/ft ) x = 0
2
M = 1350 lb ⋅ ft − ( 900 lb ) x + ( 200 lb/ft ) x 2
M = 450 lb ⋅ ft at x = 1.5 ft
M = 337.5 lb ⋅ ft at x = 2.25 ft ( center )
Finish by symmetry
(b) From diagrams:
V max = 600 lb at C − and D +
M max = 450 lb ⋅ ft at C and D
PROBLEM 7.45
Two short angle sections CE and DF are bolted to the uniform
beam AB of weight 3.33 kN, and the assembly is temporarily
supported by the vertical cables EG and FH as shown. A second
beam resting on beam AB at I exerts a downward force of 3 kN
on AB. Knowing that a = 0.3 m and neglecting the weight of the
ngle 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 angle CE:
T =
(a) By symmetry:
3.33 kN + 3 kN
= 3.165 kN
2
ΣFy = 0: T − PC = 0
PC = T = 3.165 kN
ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0
By symmetry:
M C = 0.3165 kN ⋅ m
PD = 3.165 kN; M D = 0.3165 kN ⋅ m
Along AC:
ΣFy = 0: − x (1.11 kN/m ) − V = 0
V = − (1.11 kN/m ) x
V = −1.332 kN at C
ΣM J = 0: M +
M = ( 0.555 kN/m ) x 2
( x = 1.2 m )
x
(1.11 kN/m ) x = 0
2
M = − 0.7992 kN ⋅ m at C
Along CI:
ΣFy = 0: − (1.11 kN/m ) x + 3.165 kN − V = 0
V = 3.165 kN − (1.11 kN/m ) x
V = 1.5 kN at I
( x = 1.5 m )
ΣM k = 0:
M + (1.11 kN/m ) x − ( x − 1.2 m )( 3.165 kN ) − ( 0.3165 kN ⋅ m ) = 0
PROBLEM 7.45 CONTINUED
M = 3.4815 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2
M = − 0.4827 kN ⋅ m at C
M = 0.01725 kN ⋅ m at I
Note: At I, the downward 3 kN force will reduce the shear V by 3
kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B,
the diagram can be completed by symmetry.
(b) From diagrams:
V max = 1.833 kN at C and D
M max = 799 N ⋅ m at C and D
PROBLEM 7.46
Solve Prob. 7.45 when a = 0.6 m.
SOLUTION
FBD angle CE:
T =
(a) By symmetry:
3.33 kN + 3 kN
= 3.165 kN
2
ΣFy = 0: T − PC = 0
PC = T = 3.165 kN
ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0
By symmetry:
PD = 3.165 kN
M C = 0.3165 kN ⋅ m
M D = 0.3165 kN ⋅ m
Along AC:
ΣFy = 0: − (1.11 kN/m ) x − V = 0
V = − (1.11 kN/m ) x
V = − 0.999 kN at C
ΣM J = 0: M +
( x = 0.9 m )
x
(1.11 kN/m ) x = 0
2
M = − ( 0.555 kN/m ) x 2
M = − 0.44955 kN ⋅ m at C
Along CI:
ΣFy = 0: − x (1.11 kN/m ) + 3.165 kN − V = 0
V = 3.165 kN − (1.11 kN/m ) x
V = 1.5 kN at I
V = 2.166 kN at C
( x = 1.5 m )
ΣM K = 0:
M − 0.3165 kN ⋅ m + ( x − 0.9 m )( 3.165 kN ) +
x
(1.11 kN/m ) x = 0
2
PROBLEM 7.46 CONTINUED
M = −2.532 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2
M = − 0.13305 kN ⋅ m at C
M = 0.96675 kN ⋅ m at I
Note: At I, the downward 3 kN force will reduce the shear V by 3
kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B,
the diagram can be completed by symmetry.
(b) From diagrams:
V max = 2.17 kN at C and D
M max = 967 N ⋅ m at I
PROBLEM 7.47
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
FBD CD:
ΣFy = 0: −1.2 kN + C y = 0
ΣM C = 0: ( 0.4 m )(1.2 kN ) − M C = 0
C y = 1.2 kN
M C = 0.48 kN ⋅ m
FBD Beam:
ΣM A = 0: (1.2 m ) B + 0.48 kN ⋅ m − ( 0.8 m )(1.2 kN ) = 0
B = 0.4 kN
ΣFy = 0: Ay − 1.2 kN + 0.4 kN = 0
A y = 0.8 kN
Along AC:
ΣFy = 0: 0.8 kN − V = 0
ΣM J = 0: M − x ( 0.8 kN ) = 0
V = 0.8 kN
M = ( 0.8 kN ) x
M = 0.64 kN ⋅ m at x = 0.8 m
Along CB:
ΣFy = 0: V + 0.4 kN = 0
ΣM K = 0: x1 ( 0.4 kN ) − M = 0
V = −0.4 kN
M = ( 0.4 kN ) x1
M = 0.16 kN ⋅ m at x1 = 0.4 m
(a)
Just left of C:
V = 800 N
M = 640 N ⋅ m
(b)
Just right of C:
V = −400 N
M = 160.0 N ⋅ m
PROBLEM 7.48
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 angle:
ΣFy = 0: Fy − 600 N = 0
Fy = 600 N
ΣM Base = 0: M − ( 0.3 m )( 600 N ) = 0
M = 180 N ⋅ m
All three angles are the same.
FBD Beam:
ΣM A = 0: (1.8 m ) B − 3 (180 N ⋅ m )
− ( 0.3 m + 0.9 m + 1.5 m )( 600 N ) = 0
B = 1200 N
ΣFy = 0: Ay − 3 ( 600 N ) + 1200 N = 0
A y = 600 N
Along AC:
ΣFy = 0: 600 N − V = 0
V = 600 N
ΣM J = 0: M − x ( 600 N ) = 0
M = ( 600 N ) x = 180 N ⋅ m at x = .3 m
Along CD:
ΣFy = 0: 600 N − 600 N − V = 0
V =0
ΣM K = 0: M + ( x − 0.3 m )( 600 N ) − 180 N ⋅ m − x ( 600 N ) = 0
M = 360 N ⋅ m
PROBLEM 7.48 CONTINUED
Along DE:
ΣFy = 0: V − 600 N + 1200 N = 0
V = −600 N
ΣM N = 0: −M − 180 N ⋅ m − x2 ( 600 N ) + ( x2 + 0.3 m )(1200 N ) = 0
M = 180 N ⋅ m + ( 600 N ) x2 = 540 N ⋅ m at D, x2 = 0.6 m
M = 180 N ⋅ m at E (x2 = 0)
Along EB:
ΣFy = 0: V + 1200 N = 0
ΣM L = 0: x1 (1200 N ) − M = 0
V = −1200 N
M = (1200 N ) x1
M = 360 N ⋅ m at x1 = 0.3 m
From diagrams:
V max = 1200 N on EB
M max = 540 N ⋅ m at D +
PROBLEM 7.49
Draw the shear and bending-moment diagrams for the beam AB,
and determine the maximum absolute values of the shear and
bending moment.
SOLUTION
FBD Whole:
ΣM D = 0: (1.2 m )(1.5 kN ) − (1.2 m )( 6 kN )
− ( 3.6 m )(1.5 kN ) + (1.6 m ) G = 0
G = 6.75 kN
ΣFx = 0: − Dx + G = 0
Beam AB, with forces D and G
replaced by equivalent
force/couples at C and F
D x = 6.75 kN
ΣFy = 0: Dy − 1.5 kN − 6 kN − 1.5 kN = 0
D y = 9 kN
Along AD:
ΣFy = 0: −1.5 kN − V = 0
ΣM J = 0: x (1.5 kN ) + M = 0
V = −1.5 kN
M = − (1.5 kN ) x
M = −1.8 kN at x = 1.2 m
Along DE:
ΣFy = 0: −1.5 kN + 9 kN − V = 0
V = 7.5 kN
ΣM K = 0: M + 5.4 kN ⋅ m − x1 ( 9 kN ) + (1.2 m + x1 )(1.5 kN ) = 0
M = 7.2 kN ⋅ m + ( 7.5 kN ) x1
M = 1.8 kN ⋅ m at x1 = 1.2 m
PROBLEM 7.49 CONTINUED
Along EF:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM N = 0: − M + 5.4 kN ⋅ m − ( x4 + 1.2 m )(1.5 kN )
M = 3.6 kN ⋅ m − (1.5 kN ) x4
M = 1.8 kN ⋅ m at x4 = 1.2 m;
M = 3.6 kN ⋅ m at x4 = 0
Along FB:
ΣFy = 0: V − 1.5 kN = 0
ΣM L = 0: − M − x3 (1.5 kN ) = 0
V = 1.5 kN
M = ( −1.5 kN ) x3
M = −1.8 kN ⋅ m at x3 = 1.2 m
From diagrams:
V max = 7.50 kN on DE
M max = 7.20 kN ⋅ m at D +
PROBLEM 7.50
Neglecting the size of the pulley at G, (a) draw the shear and
bending-moment diagrams for the beam AB, (b) determine the
maximum absolute values of the shear and bending moment.
SOLUTION
FBD Whole:
(a)
ΣM A = 0: ( 0.5 m )
12
5
D + (1.2 m ) D − ( 2.5 m )( 480 N ) = 0
13
13
D = 1300 N
ΣFy = 0: Ay +
5
(1300 N ) − 480 N = 0
13
Ay = −20 N
A y = 20 N
Beam AB with pulley forces and Along AE:
force at D replaced by equivalent
force-couples at B, F, E
ΣFy = 0: −20 N − V = 0
V = −20 N
ΣM J = 0: M + x ( 20 N )
M = − ( 20 N ) x
M = −24 N ⋅ m at x = 1.2 m
Along EF:
ΣFy = 0: V − 288 N − 480 N = 0
V = 768 N
ΣM L = 0: −M − x2 ( 288 N ) − ( 28.8 N ⋅ m ) − ( x2 + 0.6 m )( 480 N ) = 0
M = −316.8 N ⋅ m − ( 768 N ) x2
M = −316.8 N ⋅ m at x2 = 0 ; M = −624 N ⋅ m at x2 = 0.4 m
Along FB:
ΣFy = 0: V − 480 N = 0
ΣM K = 0: − M − x1 ( 480 N ) = 0
V = 480 N
M = − ( 480 N ) x1
M = −288 N ⋅ m at x1 = 0.6 m
PROBLEM 7.50 CONTINUED
(b) From diagrams:
V max = 768 N along EF
M max = 624 N ⋅ m at E +
PROBLEM 7.51
For the beam of Prob. 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 .
(Hint: Draw the bending-moment diagram and then equate the absolute
values of the largest positive and negative bending moments obtained.)
SOLUTION
FBD Beam:
ΣFy = 0: Lw − 2 P = 0
w=2
ΣM J = 0: M −
M =
Along AC:
P
L
x  2P 
x = 0

2 L 
P 2
x
L
P 2
a at x = a
L
ΣM K = 0: M + ( x − a ) P −
M = P (a − x) +
M =
x  2P 
x = 0

2 L 
P 2 Pa 2
x =
L
L
at
x=a
L
L

M = P  a −  at x =
2
4


Along CD:
This is M min by symmetry–see moment diagram completed by
symmetry.
For minimum M max , set M max = −M min :
P
a2
L

= −P  a − 
L
4

or
a 2 + La −
Solving:
a=
Positive answer
L2
=0
4
−1 ± 2
L
2
(a)
a = 0.20711L = 0.932 ft
(b)
M max = 0.04289PL = 173.7 lb ⋅ ft
PROBLEM 7.52
For the assembly of Prob. 7.45, 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
Prob. 7.51.)
SOLUTION
FBD Angle:
By symmetry of whole arrangement:
T =
3.33 kN + 3 kN
= 3.165 kN
2
ΣFy = 0: T − F = 0
F = 3.165 kN
ΣM 0 = 0: M − ( 0.1 m )( 3.165 kN ) = 0
ΣM J = 0: M +
M = 0.3165 kN ⋅ m
x
(1.11 kN/m ) x = 0
2
M = − ( 0.555 kN/m ) x 2 = − ( 0.555 kN/m )(1.5 m − a )
Along AC:
2
at C ( this is M min )
x
(1.11 kN/m ) x
2
−  x − (1.5 m − a )  ( 3.165 kN ) = 0
ΣM K = 0: M − 0.3165 kN ⋅ m +
Along CI:
M = −4.431 kN ⋅ m + ( 3.165 kN )( x + a ) − ( 0.555 kN/m ) x 2
M max ( at x = 1.5 m ) = − 0.93225 kN ⋅ m + ( 3.165 kN ) a
For minimum M max , set M max = −M min :
− 0.93225 kN ⋅ m + ( 3.165 kN ) a = ( 0.555 kN/m )(1.5 m − a )
Yielding:
a 2 − ( 8.7027 m ) a + 3.92973 m 2 = 0
Solving:
a = 4.3514 ± 13.864 = 0.4778 m, 8.075 m
Second solution out of range, so
(a)
2
a = 0.478 m
M max = 0.5801 kN ⋅ m
(b)
M max = 580 N ⋅ m
PROBLEM 7.53
For the beam shown, determine (a) the magnitude P of the two
upward forces for which the maximum value of the bending
moment is as small as possible, (b) the corresponding value of
M max . (See hint for Prob. 7.51.)
SOLUTION
By symmetry:
Ay = B = 60 kN − P
Along AC:
ΣM J = 0: M − x ( 60 kN − P ) = 0
M = ( 60 kN − P ) x
M = 120 kN ⋅ m − ( 2 m ) P at x = 2 m
Along CD:
ΣM K = 0: M + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0
M = 120 kN ⋅ m − Px
M = 120 kN ⋅ m − ( 4 m ) P at x = 4 m
Along DE:
ΣM L = 0: M − ( x − 4 m ) P + ( x − 2 m )( 60 kN )
− x ( 60 kN − P ) = 0
M = 120 kN ⋅ m − ( 4 m ) P
(const)
Complete diagram by symmetry
For minimum M max , set M max = −M min
120 kN ⋅ m − ( 2 m ) P = − 120 kN ⋅ m − ( 4 m ) P 
M min = 120 kN ⋅ m − ( 4 m) P
(a)
P = 40.0 kN
(b)
M max = 40.0 kN ⋅ m
PROBLEM 7.54
For the beam and loading shown, determine (a) the distance a for
which the maximum absolute value of the bending moment in the
beam is as small as possible, (b) the corresponding value of
M max . (See hint for Prob. 7.51.)
SOLUTION
ΣM A = 0: M A − (1.5 ft )(1 kip ) − ( 3.5 ft )( 4 kips )
FBD Beam:
+ ( 3.5 ft + a )( 2 kips ) = 0
M A = 8.5 kip ⋅ ft − ( 2 kips ) a 
ΣFy = 0: Ay − 1 kip − 4 kips + 2 kips = 0
A y = 3 kips
Along AC:
ΣM J = 0: M − x ( 3 kips ) + 8.5 kip ⋅ ft − ( 2 kips ) a = 0
M = ( 3 kips ) x + ( 2 kips ) a − 8.5 kip ⋅ ft
M = ( 2 kips ) a − 4 kip ⋅ ft at C ( x = 1.5 ft )
M = ( 2 kips ) a − 8.5 kip ⋅ ft at A ( M min )
Along DB:
ΣM K = 0: − M + x1 ( 2 kips ) = 0
M = ( 2 kips ) x1
M = ( 2 kips ) a at D
ΣM L = 0: ( x2 + a )( 2 kips ) − x2 ( 4 kips ) − M = 0
M = ( 2 kips ) a − ( 2 kips ) x2
Along CD:
M = ( 2 kips ) a − 4 kip ⋅ ft at C
( see above )
For minimum M max , set M max ( at D ) = −M min ( at A)
( 2 kips ) a = − ( 2 kips ) a − 8.5 kip ⋅ ft 
4a = 8.5 ft
a = 2.125 ft
a = 2.13 ft
(a)
So
M max = ( 2 kips ) a = 4.25 kip ⋅ ft
(b)
M max = 4.25 kip ⋅ ft
PROBLEM 7.55
Knowing that P = Q = 375 lb, determine (a) the distance a for
which the maximum absolute value of the bending moment in
beam AB is as small as possible, (b) the corresponding value of
M max . (See hint for Prob. 7.51.)
SOLUTION
ΣM A = 0: ( a ft ) D − ( 4 ft )( 375 lb ) − ( 8 ft )( 375 lb ) = 0
D=
FBD Beam:
4500
lb
a
ΣFy = 0: Ay − 2 ( 375 lb ) +
4500
lb = 0
a
4500 

A y =  750 −
 lb
a 

It is apparent that M = 0 at A and B, and that all segments of the M
diagram are straight, so the max and min values of M must occur
at C and D
4500 

ΣM C = 0: M − ( 4 ft )  750 −
 lb = 0
a 

Segment AC:
18000 

M =  3000 −
 lb ⋅ ft
a 

ΣM D = 0: − ( 8 − a ) ft  ( 375 lb ) − M = 0
Segment DB:
M = −375 ( 8 − a ) lb ⋅ ft
For minimum M max , set M max = −M min
So
3000 −
18000
= 375 ( 8 − a )
a
a 2 = 48
a = 6.9282 ft
a = 6.93 ft
(a)
M max = 375 ( 8 − a ) = 401.92 lb ⋅ ft
(b)
M max = 402 lb ⋅ ft
PROBLEM 7.56
Solve Prob. 7.55 assuming that P = 750 lb and Q = 375 lb.
SOLUTION
ΣM D = 0: − ( a ft ) Ay + ( a − 4 ) ft  ( 750 lb )
FBD Beam:
− ( 8 − a ) ft  ( 375 lb ) = 0
6000 

A y = 1125 −
 lb
a 

It is apparent that M = 0 at A and B, and that all segments of the
M-diagram are straight, so M max and M min occur at C and D.
6000 

ΣM C = 0: M − ( 4 ft ) 1125 −
 lb = 0
a 

Segment AC:
24000 

M =  4500 −
 lb ⋅ ft
a 

ΣM D = 0: − M − ( 8 − a ) ft  ( 375 lb ) = 0
Segment DB:
M = −375 ( 8 − a ) lb ⋅ ft
For minimum M max , set M max = −M min
4500 −
24000
= 375 ( 8 − a )
a
a 2 + 4a − 64 = 0
a = 6.2462 ft
Then
a = −2 ± 68 ( need + )
a = 6.25 ft
(a)
M max = 375 ( 8 − a ) = 657.7 lb ⋅ ft
(b)
M max = 658 lb ⋅ ft
PROBLEM 7.57
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
x
wx = 0
2
ΣM J = 0: −M −
M due to distributed load:
1
M = − wx 2
2
ΣM J = 0: −M + xw = 0
M due to counter weight:
(a) Both applied:
M = wx
M = Wx −
w 2
x
2
And here M =
dM
W
= W − wx = 0 at x =
dx
w
W2
> 0 so M max ; M min must be at x = L
2w
So M min = WL −
1 2
wL . For minimum M max set M max = −M min , so
2
W2
1
= −WL + wL2 or W 2 + 2wLW − w2 L2 = 0
2w
2
W = −wL ± 2w2 L2 (need +)
W2
M max =
=
2w
(b) w may be removed:
Without w,
With w (see part a)
W =
( 2 − 1) wL = 0.414wL
( 2 − 1) wL
2
2
M max = 0.858wL2
2
M = Wx, M max = WL at A
M = Wx −
M min = WL −
w 2
x ,
2
M max =
1 2
wL at x = L
2
W2
W
at x =
2w
w
PROBLEM 7.57 CONTINUED
For minimum M max , set M max ( no w ) = −M min ( with w )
WL = −WL +
With
1 2
1
wL → W = wL →
2
4
M max =
1 2
wL
4
W =
1
wL
4
PROBLEM 7.58
Using the method of Sec. 7.6, solve Prob. 7.29.
SOLUTION
(a) and (b)
By symmetry: Ay = D =
Shear Diag:
1  L  wL
w  =
2 2 
4
V jumps to Ay =
or A y = D =
wL
4
wL
at A,
4
and stays constant (no load) to B. From B to C, V is linear
wL
L
wL
 dV

−w =−
= −w  , and it becomes
at C.

4
2
4
 dx

(Note: V = 0 at center of beam. From C to D, V is again constant.)
Moment Diag: M starts at zero at A
wL 
 dM
=
and increases linearly 

4  to B.
 dV
MB = 0 +
L  wL  wL2
.

=
4 4 
16
From B to C M is parabolic
wL
 dM

= V , which decreases to zero at center and −
at C  ,

4
 dx

M is maximum at center.
M max =
Then, M is linear with
wL2 1  L  wL 
+  

16
2  4  4 
dM
wL
=−
to D
dy
4
MD = 0
V max =
M max =
wL
4
3wL2
32
Notes: Symmetry could have been invoked to draw second half.
Smooth transitions in M at B and C, as no discontinuities in V.
PROBLEM 7.59
Using the method of Sec. 7.6, solve Prob. 7.30.
SOLUTION
(a) and (b)
Shear Diag:
V = 0 at A and is linear
wL
 dV

L
 dV

= −w  to − w   = −
at B. V is constant 
= 0  from

2
 dx

 dx

2
B to C.
V max =
wL
2
Moment Diag: M = 0 at A and is
 dM

decreasing with V  to B.
parabolic 
 dx

MB =
1  L  wL 
wL2
−
=
−
 

2  2  2 
8
wL 
 dM
=−
From B to C, M is linear 

dx
2 

MC = −
wL2  L  wL 
3wL2
−  
=
−

8
8
 2  2 
M max =
Notes: Smooth transition in M at B, as no discontinuity in V.
It was not necessary to predetermine reactions at C.
In fact they are given by −VC and − M C .
3wL2
8
PROBLEM 7.60
Using the method of Sec. 7.6, solve Prob. 7.31.
SOLUTION
(a) and (b)
Shear Diag:
 dV

= 0  to B. V jumps down P
V jumps to P at A, then is constant 
 dx

to zero at B, and is constant (zero) to C.
V max = P
Moment Diag:
 dM

M is linear 
= V = P  to B.
dy


PL
L
M B = 0 +  ( P) =
.
2
2
PL
 dM

= 0  at
to C
M is constant 
2
 dx

M max =
PL
2
Note: It was not necessary to predetermine reactions at C. In fact they
are given by −VC and − M C .
PROBLEM 7.61
Using the method of Sec. 7.6, solve Prob. 7.32.
SOLUTION
(a) and (b)
ΣM C = 0: LAy − M 0 = 0
Ay =
M0
L
Shear Diag:
V jumps to −
M0
 dV

at A and is constant 
= 0  all the way to C
L
dx


V max =
M0
W
L
Moment Diag:
M 
 dM
M is zero at A and linear 
= V = − 0  throughout.
L 
 dx
LM 
M
M B− = −  0  = − 0 ,
2 L 
2
M
but M jumps by + M 0 to + 0 at B.
2
M
LM 
MC = 0 −  0  = 0
2
2 L 
M max =
M0
W
2
PROBLEM 7.62
Using the method of Sec. 7.6, solve Prob. 7.33.
SOLUTION
(a) and (b)
ΣM B = 0: ( 0.6 ft )( 4 kips ) + ( 5.1 ft )( 8 kips )
+ ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay = 0
A y = 12.625 kips
Shear Diag:
 dV

V is piecewise constant, 
= 0  with discontinuities at each
 dx

concentrated force. (equal to force)
V max = 12.63 kips W
Moment Diag:
 dM

= V  throughout.
M is zero at A, and piecewise linear 
 dx

M C = (1.8 ft )(12.625 kips ) = 22.725 kip ⋅ ft
M D = 22.725 kip ⋅ ft + ( 2.7 ft )( 2.625 kips )
= 29.8125 kip ⋅ ft
M E = 29.8125 kip ⋅ ft − ( 4.5 ft )( 5.375 kips )
= 5.625 kip ⋅ ft
M B = 5.625 kip ⋅ ft − ( 0.6 ft )( 9.375 kips ) = 0
M max = 29.8 kip ⋅ ft W
PROBLEM 7.63
Using the method of Sec. 7.6, solve Prob. 7.36.
SOLUTION
(a) and (b)
FBD Beam:
ΣM E = 0: (1.1 m )( 0.54 kN ) − ( 0.9 m ) C y
+ ( 0.4 m )(1.35 kN ) − ( 0.3 m )( 0.54 kN ) = 0
C y = 1.08 kN
ΣFy = 0: − 0.54 kN + 1.08 kN − 1.35 kN + E − 0.54 kN = 0
E = 1.35 kN
Shear Diag:
 dV

V is piecewise constant, 
= 0 everywhere  with discontinuities at
dx


each concentrated force. (equal to the force)
V max = 810 N W
Moment Diag:
M is piecewise linear starting with M A = 0
M C = 0 − 0.2 m ( 0.54 kN ) = 0.108 kN ⋅ m
M D = 0.108 kN ⋅ m + ( 0.5 m )( 0.54 kN ) = 0.162 kN ⋅ m
M E = 0.162 kN ⋅ m − ( 0.4 m )( 0.81 kN ) = − 0.162 kN ⋅ m
M B = 0.162 kN ⋅ m + ( 0.3 m )( 0.54 kN ) = 0
M max = 0.162 kN ⋅ m = 162.0 N ⋅ m W
PROBLEM 7.64
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) and (b)
Shear Diag:
 dV

V = 0 at A and linear 
= −2 kN/m  to C
 dx

VC = −1.2 m ( 2 kN/m ) = −2.4 kN.
At C, V jumps 6 kN to 3.6 kN, and is constant to D. From there, V is
 dV

= +3 kN/m  to B
linear 
 dx

VB = 3.6 kN + (1 m )( 3 kN/m ) = 6.6 kN
V max = 6.60 kN W
M A = 0.
Moment Diag:
 dM

decreasing with V  .
From A to C, M is parabolic, 
 dx

MC = −
1
(1.2 m )( 2.4 kN ) = −1.44 kN ⋅ m
2
 dM
From C to D, M is linear 
 dx

= 3.6 kN 

M D = −1.44 kN ⋅ m + ( 0.6 m )( 3.6 kN )
= 0.72 kN ⋅ m.
 dM

From D to B, M is parabolic 
increasing with V 
 dx

M B = 0.72 kN ⋅ m +
1
(1 m )( 3.6 + 6.6 ) kN
2
= 5.82 kN ⋅ m
M max = 5.82 kN ⋅ m W
Notes: Smooth transition in M at D. It was unnecessary to predetermine
reactions at B, but they are given by −VB and −M B
PROBLEM 7.65
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
ΣM B = 0: ( 3 ft )(1 kip/ft )( 6 ft ) + ( 8 ft )( 6 kips )
(a) and (b)
+ (10 ft )( 6 kips ) − (12 ft ) Ay = 0
A y = 10.5 kips
Shear Diag:
V is piecewise constant from A to E, with discontinuities at A, C, and E
equal to the forces. VE = −1.5 kips. From E to B, V is linear
 dV

= −1 kip/ft  ,

dx


so
VB = −1.5 kips − ( 6 ft )(1 kip/ft ) = −7.5 kips
V max = 10.50 kips W
Moment Diag: M A = 0, then M is piecewise linear to E
M C = 0 + 2 ft (10.5 kips ) = 21 kip ⋅ ft
M D = 21 kip ⋅ ft + ( 2 ft )( 4.5 kips ) = 30 kip ⋅ ft
M E = 30 kip ⋅ ft − ( 2 ft )(1.5 kips ) = 27 kip ⋅ ft
 dM

decreasing with V  , and
From E to B, M is parabolic 
 dx

M B = 27 kip ⋅ ft −
1
( 6 ft )(1.5 kips + 7.5 kips ) = 0
2
M max = 30.0 kip ⋅ ft W
PROBLEM 7.66
Using the method of Sec. 7.6, solve Prob. 7.37.
SOLUTION
(a) and (b)
FBD Beam:
ΣFy = 0: Ay + ( 6 ft )( 2 kips/ft ) − 12 kips − 2 kips = 0
A y = 2 kips
ΣM A = 0: M A + ( 3 ft )( 2 kips/ft )( 6 ft )
− (10.5 ft )(12 kips ) − (12 ft )( 2 kips ) = 0
M A = 114 kip ⋅ ft
Shear Diag:
 dV

VA = Ay = 2 kips. Then V is linear 
= 2 kips/ft  to C, where
dx


VC = 2 kips + ( 6 ft )( 2 kips/ft ) = 14 kips.
V is constant at 14 kips to D, then jumps down 12 kips to 2 kips and is
constant to B
V max = 14.00 kips W
Moment Diag:
M A = −114 kip ⋅ ft.
 dM

increasing with V  and
From A to C, M is parabolic 
dx


M C = −114 kip ⋅ ft +
1
( 2 kips + 14 kips )( 6 ft )
2
M C = −66 kip ⋅ ft.
Then M is piecewise linear.
M D = −66 kip ⋅ ft + (14 kips )( 4.5 ft ) = −3 kip ⋅ ft
M B = −3 kip ⋅ ft + ( 2 kips )(1.5 ft ) = 0
M max = 114.0 kip ⋅ ft W
PROBLEM 7.67
Using the method of Sec. 7.6, solve Prob. 7.38.
SOLUTION
(a) and (b)
FBD Beam:
 kips 
ΣM B = 0: ( 3 ft )  2
 ( 6 ft ) + ( 9 ft )(12 kips ) − (15 ft ) Ay = 0
 ft 
A y = 9.6 kips
Shear Diag:
V jumps to Ay = 9.6 kips at A, is constant to C, jumps down 12 kips
to −2.4 kips at C, is constant to D, and then is linear
 dV

= −2 kips/ft  to B

 dx

VB = −2.4 kips − ( 2 kips/ft )( 6 ft )
= −14.4 kips
V max = 14.40 kips W
Moment Diag:
 dM

= 9.6 kips/ft 

dx


M is linear from A to C
M C = 9.6 kips ( 6 ft ) = 57.6 kip ⋅ ft,
M is linear from C to D
 dM

 dx

= −2.4 kips/ft 

M D = 57.6 kip ⋅ ft − 2.4 kips ( 3 ft )
M D = 50.4 kip ⋅ ft.
 dM

M is parabolic 
decreasing with V  to B.
 dx

M B = 50.4 kip ⋅ ft −
1
( 2.4 kips + 14.4 kips )( 6 ft ) = 0
2
=0
M max = 57.6 kip ⋅ ft W
PROBLEM 7.68
Using the method of Sec. 7.6, solve Prob. 7.39.
SOLUTION
(a) and (b)
FBD Beam:
By symmetry:
1
( 5 m )( 4 kN/m ) + 8 kN
2
or A y = B = 18 kN
Ay = B =
Shear Diag:
V jumps to 18 kN at A, and is constant to C, then drops 8 kN to 10 kN.
 dV

= −4 kN/m  , reaching −10 kN at
After C, V is linear 
 dx

D VD = 10 kN − ( 4 kN/m )( 5 m )  passing through zero at the beam
center. At D, V drops 8 kN to −18 kN and is then constant to B
V max = 18.00 kN W
Moment Diag:
 dM

M A = 0. Then M is linear 
= 18 kN/m  to C
 dx

M C = (18 kN )( 2 m ) = 36 kN ⋅ m, M is parabolic to D
 dM

decreases with V to zero at center 

dx


M center = 36 kN ⋅ m +
1
(10 kN )( 2.5 m ) = 48.5 kN ⋅ m = M max
2
M max = 48.5 kN ⋅ m W
Complete by invoking symmetry.
PROBLEM 7.69
Using the method of Sec. 7.6, solve Prob. 7.40.
SOLUTION
(a) and (b)
FBD Beam:
ΣM F = 0: (1 m )( 22 kN ) + (1.5 m )( 4 kN/m )( 3 m )
− ( 4 m ) Dy + ( 6 m )( 2 kN/m )( 2 m ) = 0
D y = 16 kN
ΣFy = 0: 16 kN + 22 kN − Fy − ( 2 kN/m )( 2 m )
− ( 4 kN/m )( 3 m ) = 0
Fy = 22 kN
Shear Diag:
 dV

VA = 0, then V is linear 
= −2 kN/m  to C;
 dx

VC = −2 kN/m ( 4 m ) = −4 kN
V is constant to D, jumps 16 kN to 12 kN and is constant to E.
 dV

= −4 kN/m  to F.
Then V is linear 
 dx

VF = 12 kN − ( 4 kN/m )( 3 m ) = 0.
V jumps to −22 kN at F, is constant to B, and returns to zero.
V max = 22.0 kN W
Moment Diag:
 dM

M A = 0, M is parabolic 
decreases with V  to C.
 dx

1
M C = − ( 4 kN )( 2 m ) = −4 kN ⋅ m.
2
PROBLEM 7.69 CONTINUED
 dM

Then M is linear 
= −4 kN  to D.
dx


M D = −4 kN ⋅ m − ( 4 kN )(1 m ) = −8 kN ⋅ m
 dM

= 12 kN  , and
From D to E, M is linear 
 dx

M E = −8 kN ⋅ m + (12 kN )(1m )
M E = 4 kN ⋅ m
 dM

M is parabolic 
decreasing with V  to F.
dx


M F = 4 kN ⋅ m +
1
(12 kN )( 3 m ) = 22 kN ⋅ m.
2
 dM

Finally, M is linear 
= −22 kN  , back to zero at B.
dx


M max = 22.0 kN ⋅ m W
PROBLEM 7.70
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of
the shear and bending moment.
SOLUTION
(a) and (b)
FBD Beam:
ΣM B = 0: (1.5 m )(16 kN )
+ ( 3 m )( 8 kN ) + 6 kN ⋅ m − ( 4.5 m ) Ay = 0
A y = 12 kN
Shear Diag:
V is piecewise constant with discontinuities equal to the concentrated
forces at A, C, D, B
V max = 12.00 kN W
Moment Diag:
 dM

=V
After a jump of −6 kN ⋅ m at A, M is piecewise linear 
 dx

So
M C = −6 kN ⋅ m + (12 kN )(1.5 m ) = 12 kN ⋅ m
M D = 12 kN ⋅ m + ( 4 kN )(1.5 m ) = 18 kN ⋅ m
M B = 18 kN ⋅ m − (12 kN )(1.5 m ) = 0
M max = 18.00 kN ⋅ m W
PROBLEM 7.71
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the
shear and bending moment.
SOLUTION
(a)
FBD Beam:
ΣM A = 0: ( 8 m ) F + (11 m )( 2 kN ) + 10 kN ⋅ m − ( 6 m )( 8 kN )
− 12 kN ⋅ m − ( 2 m )( 6 kN ) = 0 F = 5 kN
ΣFy = 0: Ay − 6 kN − 8 kN + 5 kN + 2 kN = 0
A y = 7 kN
Shear Diag:
V is piecewise constant with discontinuities equal to the concentrated
forces at A, C, E, F, G
Moment Diag:
M is piecewise linear with a discontinuity equal to the couple at D.
M C = ( 7 kN )( 2 m ) = 14 kN ⋅ m
M D− = 14 kN ⋅ m + (1 kN )( 2 m ) = 16 kN ⋅ m
M D+ = 16 kN ⋅ m + 12 kN ⋅ m = 28 kN ⋅ m
M E = 28 kN ⋅ m + (1 kN )( 2 m ) = 30 kN ⋅ m
M F = 30 kN ⋅ m − ( 7 kN )( 2 m ) = 16 kN ⋅ m
M G = 16 kN ⋅ m − ( 2 kN )( 3 m ) = 10 kN ⋅ m
(b)
V max = 7.00 kN
M max = 30.0 kN
PROBLEM 7.72
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
(a)
FBD Beam:
ΣM B = 0: ( 3 ft )(1.2 kips/ft )( 6 ft ) − ( 8 ft ) Ay = 0
A y = 2.7 kips
Shear Diag:
V = Ay = 2.7 kips at A, is constant to C, then linear
 dV

= −1.2 kips/ft  to B.

dx


VB = 2.7 kips − (1.2 kips/ft )( 6 ft )
VB = −4.5 kips
V = 0 = 2.7 kips − (1.2 kips/ft ) x1 at
x1 = 2.25 ft
Moment Diag:
 dM

M A = 0, M is linear 
= 2.7 kips  to C.
 dx

M C = ( 2.7 kips )( 2 ft ) = 5.4 kip ⋅ ft
 dM

decreasing with V 
Then M is parabolic 
 dx

(b)
M max occurs where
dM
=V =0
dx
M max = 5.4 kip ⋅ ft +
1
( 2.7 kips ) x1;
2
x1 = 2.25 m
M max = 8.4375 kip ⋅ ft
M max = 8.44 kip ⋅ ft, 2.25 m right of C
Check:
M B = 8.4375 kip ⋅ ft −
1
( 4.5 kips )( 3.75 ft ) = 0
2
PROBLEM 7.73
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
FBD Beam:
(a)
ΣM B = 0: ( 6 ft )( 2 kips/ft )( 8 ft ) − (15 ft ) Ay = 0
A y = 6.4 kips
Shear Diag:
V = Ay = 6.4 kips at A, and is constant to C, then linear
 dV

= −2 kips/ft  to D,

 dx

VD = 6.4 kips − ( 2 kips/ft )( 8 ft ) = −9.6 kips
V = 0 = 6.4 kips − ( 2 kips/ft ) x1 at x1 = 3.2 ft
Moment Diag:
 dM

M A = 0, then M is linear 
= 6.4 kips  to M C = ( 6.4 kips )( 5 ft ) .
dx


dM


M C = 32 kip ⋅ ft. M is then parabolic 
decreasing with V  .
dx


(b)
M max occurs where
M max = 32 kip ⋅ ft +
dM
= V = 0.
dx
1
( 6.4 kips ) x1;
2
x1 = 3.2 ft
M max = 42.24 kip ⋅ ft
M max = 42.2 kip ⋅ ft, 3.2 ft right of C
M D = 42.24 kip ⋅ ft −
M is linear from D to zero at B.
1
( 9.6 kips )( 4.8 ft ) = 19.2 kip ⋅ ft
2
PROBLEM 7.74
For the beam shown, draw the shear and bending-moment diagrams and
determine the maximum absolute value of the bending moment knowing
that (a) P = 14 kN, (b) P = 20 kN.
SOLUTION
(a)
FBD Beam:
ΣFy = 0: Ay − (16 kN/m )( 2 m ) − 6 kN + P = 0
Ay = 38 kN − P
(a)
A y = 24 kN
(b)
A y = 18 kN
ΣM A = 0: ( 5 m ) P − ( 3.5 m )( 6 kN ) − 1 m (16 kN/m )( 2 m ) − M A = 0
M A = ( 5 m ) P − 53 kN ⋅ m
(b)
(a)
M A = 17 kN ⋅ m
(b)
M A = 47 kN ⋅ m
Shear Diags:
 dV

VA = Ay . Then V is linear 
= −16 kN/m  to C.
 dx

VC = VA − (16 kN/m )( 2 m ) = VA − 32 kN
(a)
VC = −8 kN
(b)
VC = −14 kN
V = 0 = VA − (16 kN/m ) x1
(a)
x1 = 1.5 m
(b)
x1 = 1.125 m
V is constant from C to D, decreases by 6 kN at D and is constant to
B (at − P)
PROBLEM 7.74 CONTINUED
Moment Diags:
 dM

M A = M A reaction. Then M is parabolic 
decreasing with V  .
 dx

The maximum occurs where V = 0. M max = M A +
(a)
M max = 17 kN ⋅ m +
1
VA x1.
2
1
( 24 kN )(1.5 m ) = 35.0 kN ⋅ m
2
1.5 ft from A
(b)
M max = 47 kN ⋅ m +
1
(18 kN )(1.125 m ) = 57.125 kN ⋅ m
2
M max = 57.1 kN ⋅ m 1.125 ft from A
M C = M max +
(a)
M C = 35 kN ⋅ m −
(b)
M C = 57.125 kN ⋅ m −
1
VC ( 2 m − x1 )
2
1
(8 kN )( 0.5 m ) = 33 kN ⋅ m
2
1
(14 kN )( 0.875 m ) = 51 kN ⋅ m
2
M is piecewise linear along C, D, B, with M B = 0 and
M D = (1.5 m ) P
(a)
M D = 21 kN ⋅ m
(b)
M D = 30 kN ⋅ m
PROBLEM 7.75
For the beam shown, draw the shear and bending-moment diagrams,
and determine the magnitude and location of the maximum absolute
value of the bending moment knowing that (a) M = 0 ,
(b) M = 12 kN ⋅ m.
SOLUTION
FBD Beam:
ΣM A = 0: ( 4 m ) B − (1 m )( 20 kN/m )( 2 m ) − M = 0
B = 10 kN +
(a)
M
4m
(a)
B = 10 kN
(b)
B = 13 kN
ΣFy = 0: Ay − ( 20 kN/m )( 2 m ) + B = 0
Ay = 40 kN − B
(a)
A y = 30 kN
(b)
A y = 27 kN
Shear Diags:
(b)
 dV

= −20 kN/m  to C.
VA = Ay , then V is linear 
 dx

VC = Ay − ( 20 kN/m )( 2 m ) = Ay − 40 kN
(a)
VC = −10 kN
(b)
VC = −13 kN
V = 0 = Ay − ( 20 kN/m ) x1 at x1 =
(a)
x1 = 1.5 m
(b)
x1 = 1.35 m
V is constant from C to B.
Ay m
20 kN
PROBLEM 7.75 CONTINUED
Moment Diags:
 dM

M A = applied M . Then M is parabolic 
decreases with V 
 dx

M is max where V = 0. M max = M +
(a)
M max =
1
Ay x1.
2
1
( 30 kN )(1.5 m ) = 22.5 kN ⋅ m
2
1.500 m from A
(b)
M max = 12 kN ⋅ m +
1
( 27 kN )(1.35 m ) = 30.225 kN ⋅ m
2
M max = 30.2 kN 1.350 m from A
M C = M max −
1
VC ( 2 m − x1 )
2
(a)
M C = 20 kN ⋅ m
(b)
M C = 26 kN ⋅ m
 dM

Finally, M is linear 
= VC  to zero at B.
 dx

PROBLEM 7.76
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
FBD Beam:
(a)
ΣM B = 0: ( 3 m )( 40 kN/m )( 6 m ) − ( 30 kN ⋅ m ) − ( 6 m ) Ay = 0
A y = 115 kN
Shear Diag:
 dM

= − 40 kN/m  to B.
VA = Ay = 115 kN, then V is linear 
 dx

VB = 115 kN − ( 40 kN/m )( 6 m ) = −125 kN.
V = 0 = 115 kN − ( 40 kN/m ) x1 at x1 = 2.875 m
Moment Diag:
 dM

M A = 0. Then M is parabolic 
decreasing with V  . Max M occurs
 dx

where V = 0,
M max =
1
(115 kN/m )( 2.875 m ) = 165.3125 kN ⋅ m
2
1
(125 kN )( 6 m − x1 )
2
1
= 165.3125 kN ⋅ m − (125 kN )( 6 − 2.875 ) m
2
= −30 kN ⋅ m as expected.
M B = M max −
(b)
M max = 165.3 kN ⋅ m ( 2.88 m from A )
PROBLEM 7.77
Solve Prob. 7.76 assuming that the 30 kN ⋅ m couple applied at B is
counterclockwise
SOLUTION
(a)
FBD Beam:
ΣM B = 0: 30 kN ⋅ m + ( 3 m )( 40 kN/m )( 6 m ) − ( 6 m ) Ay = 0
A y = 125 kN
Shear Diag:
 dV

VA = Ay = 125 kN, V is linear 
= −40 kN/m  to B.
 dx

VB = 125 kN − ( 40 kN/m )( 6 m ) = −115 kN
V = 0 = 115 kN − ( 40 kN/m ) x1 at x1 = 3.125 m
Moment Diag:
 dM

M A = 0. Then M is parabolic 
decreases with V  . Max M occurs
 dx

where V = 0,
M max =
1
(125 kN )( 3.125 m ) = 195.3125 kN ⋅ m
2
(b)
M max = 195.3 kN ⋅ m ( 3.125 m from A )
M B = 195.3125 kN ⋅ m −
1
(115 kN )( 6 − 3.125) m
2
M B = 30 kN ⋅ m as expected.
PROBLEM 7.78
For beam AB, (a) draw the shear and bending-moment diagrams, (b)
determine the magnitude and location of the maximum absolute value
of the bending moment.
SOLUTION
(a)
Replacing the load at E with
equivalent force-couple at C:
ΣM A = 0: ( 6 m ) D − ( 8 m )( 2 kN ) − ( 3 m )( 4 kN )
− (1.5 m )( 8 kN/m )( 3 m ) − 4 kN ⋅ m = 0
D = 10 kN
ΣFy = 0: Ay + 10 kN − 2 kN − 4 kN − ( 8 kN/m )( 3 m ) = 0
A y = 20 kN
Shear Diag:
 dV

VA = Ay = 20 kN, then V is linear 
= −8 kN/m  to C.
 dx

VC = 20 kN − ( 8 kN/m )( 3 m ) = −4 kN
V = 0 = 20 kN − ( 8 kN/m ) x1 at x1 = 2.5 m
At C, V decreases by 4 kN to −8 kN.
At D, V increases by 10 kN to 2 kN.
Moment Diag:
 dM

M A = 0, then M is parabolic 
decreasing with V  . Max M occurs
 dx

where V = 0.
M max =
1
( 20 kN )( 2.5 m ) = 25 kN ⋅ m
2
(b)
M max = 25.0 kN ⋅ m, 2.50 m from A
PROBLEM 7.78 CONTINUED
M C = 25 kN ⋅ m −
1
( 4 kN )( 0.5 m ) = 24 kN ⋅ m.
2
At C, M decreases by 4 kN ⋅ m to 20 kN ⋅ m. From C to B, M is piecewise
dM
dM
linear with
= +2 kN to B.
= −8 kN to D, then
dx
dx
M D = 20 kN ⋅ m − ( 8 kN )( 3 m ) = −4 kN ⋅ m.
MB = 0
PROBLEM 7.79
Solve Prob. 7.78 assuming that the 4-kN force applied at E is directed
upward.
SOLUTION
(a)
Replacing the load at E with
equivalent force-couple at C.
ΣM A = 0: ( 6 m ) D − ( 8 m )( 2 kN ) + ( 3 m )( 4 kN )
− 4 kN ⋅ m − (1.5 m )( 8 kN/m )( 3 m ) = 0
D=
ΣFy = 0: Ay +
22
kN
3
22
kN − ( 8 kN/m )( 3 m ) + 4 kN − 2 kN = 0
3
Ay =
44
kN
3
Shear Diag:
VA = Ay =
44
 dV

kN, then V is linear 
= −8 kN/m  to C.
3
 dx

VC =
44
28
kN − ( 8 kN/m )( 3 m ) = −
kN
3
3
V =0=
44
11
kN − ( 8 kN/m ) x1 at x1 =
m.
3
6
At C, V increases 4 kN to −
At D, V increases
16
kN.
3
22
kN to 2 kN.
3
PROBLEM 7.79 CONTINUED
Moment Diag:
 dM

M A = 0. Then M is parabolic 
decreasing with V  . Max M occurs
 dx

where V = 0.
M max =
1  44
 11  121
kN 
m =
kN ⋅ m

2 3
9
 6

(b)
MC =
M max = 13.44 kN ⋅ m at 1.833 m from A
121
1  28
 7 
kN ⋅ m − 
kN  m  = 8 kN ⋅ m.
9
2 3
 6 
At C, M increases by 4 kN ⋅ m to 12 kN ⋅ m. Then M is linear
16
 dM

=−
kN  to D.

3
 dx

 16

kN  ( 3 m ) = −4 kN ⋅ m. M is again linear
M D = 12 kN ⋅ m − 
3


 dM

= 2 kN  to zero at B.

dx


PROBLEM 7.80
For the beam and loading shown, (a) derive the equations of the shear and
bending-moment curves, (b) draw the shear and bending-moment diagrams,
(c) determine the magnitude and location of the maximum bending moment.
SOLUTION
x

Distributed load w = w0 1 − 
L

ΣM A = 0:
1


 total = w0 L 
2


L 1

 w0 L  − LB = 0
3 2

B=
w0 L
6
1
wL
w0 L + 0 = 0
2
6
Ay =
w0 L
3
ΣFy = 0: Ay −
Shear:
VA = Ay =
w0 L
,
3
x 
dV
x
= − w → V = VA − ∫ w0 1 −  dx
0 
L
dx
Then
 1 x 1  x 2 
1 w0 2
w L
V =  0  − w0 x +
x = w0 L  − +   
2 L
 3 
 3 L 2  L  
Note: At x = L, V = −
w0 L
;
6
2
x
x
x 2
V = 0 at   − 2   + = 0 →
= 1−
L
L
L 3
1
3
Moment:
M A = 0,
Then
x
x/ L  x   x 
 dM 

 = V → M = ∫0 Vdx = L ∫0 V   d  
 dx 
L L
x / L  1 x 1  x 2   x 
 − +
d
M = w0 L ∫
0  3 L 2  L    L 
2


 1  x  1  x  2 1  x 3 
M = w0 L2    −   +   
6  L  
 3  L  2  L 
PROBLEM 7.80 CONTINUED
 x
M max  at
=1−
 L

1
2
 = 0.06415w0 L
3
 1 x 1  x 2 
V = w0 L  − +   
 3 L 2  L  
(a)
 1  x  1  x  2 1  x 3 
M = w0 L    −   +   
6  L  
 3  L  2  L 
2
(c)
M max = 0.0642 w0 L2
at x = 0.423L
PROBLEM 7.81
For the beam and loading shown, (a) derive the equations of the shear
and bending-moment curves, (b) draw the shear and bending-moment
diagrams, (c) determine the magnitude and location of the maximum
bending moment.
SOLUTION
 x

w = w0  4   − 1
 L

Distributed load
dV
= −w, and V ( 0 ) = 0, so
dx
Shear:
x
V = ∫ − wdx = −
0
x/L
∫0
x
Lwd  
L
2
 x
x/L

 x   x 
  x 

V =∫
wo L 1 − 4    d   = w L   − 2 
0  L   L  
0
 L   L 



Notes: At x = L, V = −w0 L
x
x
  = 2 
L
L
And V = 0 at
2
or
x
1
=
L
2
M ( 0 ) = 0,
dM
=V
dx
1
x
Also V is max where w = 0  = 
4
L
Vmax =
Moment:
1
w0 L
8
x
x/ L  x   x 
M = ∫ vdx = L ∫
V  d  
0
0
L L
2
x / L  x 
  − 2  x   d  x 
M = w0 L2 ∫
0
 L 
L  L


(a)
2
 x 
x 
V = w0 L   − 2   
 L  
 L 
 1  x  2 2  x 3 
M = w0 L2    −   
3  L  
 2  L 
PROBLEM 7.81 CONTINUED
M max =
1
L
w0 L2 at x =
24
2
1
M min = − w0 L2 at x = L
6
M max =
w0 L2
L
at x =
24
2
(c)
M max = −M min =
w0 L2
at B
6
PROBLEM 7.82
For the beam shown, (a) draw the shear and bending-moment
diagrams, (b) determine the magnitude and location of the
maximum bending moment. (Hint: Derive the equations of the
shear and bending-moment curves for portion CD of the beam.)
SOLUTION
(a)
FBD Beam:
ΣM B = 0:
( 3a )  w0 ( 3a ) − 5aAy = 0
1
2

ΣFy = 0: 0.9w0a −
A y = 0.9w0a
1
w0 ( 3a ) + B = 0
2
B = 0.6w0a
Shear Diag:
V = Ay = 0.9w0a from A to C and V = B = − 0.6w0a from B to D.
x1
. If x1 is measured right to left,
3a
dV
x w
dM
= + w and
= − V . So, from D, V = − 0.6w0a + ∫0 1 0 x1dx1,
3a
dx1
dx1
Then from D to C, w = w0
2

1  x1  
V = w0a  − 0.6 +   
6  a  

2
x 
Note: V = 0 at  1  = 3.6, x1 =
a
3.6 a
Moment Diag:
 dM

M = 0 at A, increasing linearly 
= 0.9w0a  to M C = 0.9w0a 2.
 dx1

 dM

= 0.6w0a  to
Similarly M = 0 at B increasing linearly 
 dx

M D = 0.6w0a 2. Between C and D
2
x1 
1  x1  

M = 0.6w0a + w0a ∫ 0.6 −   dx1,
0 
6 a  


2
3

x  1 x  
M = w0a 2 0.6 + 0.6  1  −  1  
 a  18  a  

PROBLEM 7.82 CONTINUED
(b)
At
x1
=
a
3.6, M = M max = 1.359w0a 2
x1 = 1.897a left of D
PROBLEM 7.83
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
(
wg L =
Define ξ =
or
)
L 4w0
Lx − x 2 dx = 0
ΣFy = 0: wg L − ∫
0 L2
(a)
4w0  1 2 1 3  2
 LL − L  = w0 L
3  3
L2  2
wg =
2w0
3
 x  x 2  2
x
dx
so dξ =
→ net load w = 4w0  −    − w0
L
L
 L  L   3
 1

w = 4w0  − + ξ − ξ 2 
6


ξ
 1

V = V ( 0 ) − ∫ 4w0 L  − + ξ − ξ 2  d ξ =
0
 6

1

1


1
0 + 4w0 L  ξ + ξ 2 − ξ 3 
6
2
3
2
w0 L ξ − 3ξ 2 + 2ξ 3
3
x
ξ
2
M = M 0 + ∫ Vdx = 0 + w0 L2 ∫
0
0
3
(ξ − 3ξ + 2ξ ) dξ
=
(b)
(
V =
2
)
3
(
2
1  1
1
w0 L2  ξ 2 − ξ 3 + ξ 4  = w0 L2 ξ 2 − 2ξ 3 + ξ 4
3
2  3
2
Max M occurs where V = 0 → 1 − 3ξ + 2ξ 2 = 0 → ξ =
)
1
2
1 1
1  w L2

1 2
M  ξ =  = w0 L2  − +  = 0
2 3
48

 4 8 16 
M max =
w0 L2
at center of beam
48
PROBLEM 7.84
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 +325 lb ft
at D and +800 lb ft at E, (a) determine P and Q, (b) draw the shear and
bending-moment diagrams for the beam.
SOLUTION
FBD ACD:
(a)
ΣM D − = 0: 0.325 kip ⋅ ft − (1 ft ) C y + (1.5 ft )( 2 kips/ft )(1 ft ) = 0
C y = 3.325 kips
FBD EB:
ΣM E = 0: (1 ft ) B − 0.8 kip ⋅ ft = 0
B = 0.8 kip
FBD Beam:
ΣM D = 0: (1.5 ft )( 2 kips/ft )(1 ft ) − (1 ft )( 3.325 kips )
− (1 ft ) Q + 2 ft ( 0.8 kips ) = 0
Q = 1.275 kips
ΣFy = 0: 3.325 kips + 0.8 kips − 1.275 kips
− ( 2 kips/ft )(1ft ) − P = 0
(a)
P = 0.85 kip
P = 850 lb
Q = 1.275 kips
(b) Shear Diag:
 dV

= −2 kips/ft  from 0 at A to
V is linear 
 dx

− ( 2 kips/ft )(1 ft ) = −2 kips at C. Then V is piecewise constant with
discontinuities equal to forces at C, D, E, B
Moment Diag:
 dM

decreasing with V  from 0 at A to
M is parabolic 
 dx

1
− ( 2 kips )(1 ft ) = −1 kip ⋅ ft at C. Then M is piecewise linear with
2
PROBLEM 7.84 CONTINUED
M D = −1 kip ⋅ ft + (1.325 kips )(1 ft ) = 0.325 kip ⋅ ft
M E = 0.325 kip ⋅ ft + ( 0.475 kips )(1 ft ) = 0.800 kip ⋅ ft
M B = 0.8 kip ⋅ ft − ( 0.8 kip )(1 ft ) = 0
PROBLEM 7.85
Solve Prob. 7.84 assuming that the bending moment was found to be
+260 lb ft at D and +860 lb ft at E.
SOLUTION
FBD ACD:
(a)
ΣM D = 0: 0.26 kip ⋅ ft − (1 ft ) C y + (1.5 ft )( 2 kips/ft )(1 ft ) = 0
C y = 3.26 kips
FBD DB:
ΣM E = 0: (1 ft ) B − 0.86 kip ⋅ ft
B = 0.86 kip
FBD Beam:
ΣM D = 0: (1.5 ft )( 2 kips/ft )(1 ft )
− (1 ft )( 3.26 kips ) + (1 ft ) Q + ( 2 ft )( 0.86 kips ) = 0
Q = 1.460 kips
Q = 1.460 kips
ΣFy = 0: 3.26 kips + 0.86 kips − 1.460 kips
− P − ( 2 kips/ft )(1 ft ) = 0
P = 0.66 kips
P = 660 lb
(b) Shear Diag:
 dV

V is linear 
= −2 kips/ft  from 0 at A to
 dx

− ( 2 kips/ft )(1 ft ) = −2 kips at C. Then V is piecewise
constant with discontinuities equal to forces at C, D, E, B.
Moment Diag:
 dM

decreasing with V  from 0 at A to
M is parabolic 
 dx

1
− ( 2 kips/ft )(1 ft ) = −1 kip ⋅ ft at C. Then M is piecewise linear with
2
PROBLEM 7.85 CONTINUED
M 0 = 0.26 kip ⋅ ft
M E = 0.86 kip ⋅ ft, M B = 0
PROBLEM 7.86
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 +7 kN · m
at D and +5 kN · m at E, (a) determine P and Q, (b) draw the shear
and bending-moment diagrams for the beam.
SOLUTION
FBD AD:
ΣM D = 0: 7 kN ⋅ m + (1 m )( 0.6 kN/m )( 2 m )
(a)
− ( 2 m ) Ay = 0
2 Ay − P = 8.2 kN
(1)
ΣM E = 0: ( 2 m ) B − (1 m ) Q − (1 m )( 0.6 kN/m )( 2 m )
− 5 kN ⋅ m = 0
FBD EB:
2B − Q = 6.2 kN
(2)
( 6 m ) B − (1 m ) P − ( 5 m ) Q
− ( 3 m )( 0.6 kN/m )( 6 m ) = 0
ΣM A = 0:
6 B − P − 5Q = 10.8 kN
(3)
ΣM B = 0: (1 m ) Q + ( 5 m ) P + ( 3 m )( 0.6 kN/m )( 6 m )
− (6 m) A = 0
6 A − Q − 5P = 10.8 kN
P = 6.60 kN , Q = 600 N
Solving (1)–(4):
A y = 7.4 kN ,
(4)
B = 3.4 kN
(b) Shear Diag:
dV
= − 0.6 kN/m throughout, and
dx
discontinuities equal to forces at A, C, F, B.
V is piecewise linear with
Note V = 0 = 0.2 kN − ( 0.6 kN/m ) x at x =
1
m
3
PROBLEM 7.86 CONTINUED
Moment Diag:
 dM

decreasing with V  with “breaks” in
M is piecewise parabolic 
 dx

slope at C and F.
MC =
M max = 7.1 kN ⋅ m +
1
( 7.4 + 6.8) kN (1 m ) = 7.1 kN ⋅ m
2
1
1
( 0.2 kN )  m  = 7.133 kN ⋅ m
2
3 
1
 2 
M F = 7.133 kN ⋅ m − ( 2.2 kN )  3 m  = 3.1 kN ⋅ m
2
 3 
PROBLEM 7.87
Solve Prob. 7.86 assuming that the bending moment was found to be
+3.6 kN · m at D and +4.14 kN · m at E.
SOLUTION
FBD AD:
ΣM D = 0: 3.6 kN ⋅ m + (1 m ) P + (1 m )( 0.6 kN/m )( 2 m )
(a)
− ( 2 m ) Ay = 0
2 Ay − P = 4.8 kN
FBD EB:
ΣM E = 0:
(1)
( 2 m ) B − (1 m ) Q − (1 m )( 0.6 kN/m )( 2 m )
− 4.14 kN ⋅ m = 0
2B − Q = 5.34 kN
(2)
ΣM A = 0: ( 6 m ) B − ( 5 m ) Q − (1 m ) P − ( 3 m )( 0.6 kN/m )( 6 m ) = 0
6 B − P − 5Q = 10.8 kN
(3)
By symmetry:
6 A − Q − 5P = 10.8 kN
Solving (1)–(4)
(4)
P = 660 N , Q = 2.28 kN
Ay = 2.73 kN , B = 3.81 kN
(b) Shear Diag:
 dV

= − 0.6 kN/m  throughout, and
V is piecewise linear with 
 dx

discontinuities equal to forces at A, C, F, B.
Note that V = 0 = 1.47 kN − ( 0.6 kN/m ) x at x = 2.45 m
Moment Diag:
 dM

decreasing with V  , with “breaks” in
M is piecewise parabolic 
 dx

slope at C and F.
PROBLEM 7.87 CONTINUED
MC =
1
( 2.73 + 2.13) kN (1 m ) = 2.43 kN ⋅ m
2
M max = 2.43 kN ⋅ m +
1
(1.47 kN )( 2.45 m ) = 4.231 kN ⋅ m
2
M F = 4.231 kN ⋅ m −
1
( 0.93 kN )(1.55 m ) = 3.51 kN ⋅ m
2
PROBLEM 7.88
Two loads are suspended as shown from cable ABCD. Knowing
that dC = 1.5 ft, determine (a) the distance dB, (b) the components
of the reaction at A, (c) the maximum tension in the cable.
SOLUTION
ΣM A = 0: (10 ft ) Dy − 8 ft ( 450 lb ) − 4 ft ( 600 lb ) = 0
FBD cable:
Dy = 600 1b
ΣFy = 0: Ay + 600 lb − 600 lb − 450 lb = 0
Ay = 450 lb
ΣFx = 0: Ax − Dx = 0
FBD pt D:
(1)
600 lb
D
T
= x = CD : Dx = 800 lb
3
4
5
= Ax
So
TCD = 1000 lb
And
800 lb 450 lb
=
dB
4 ft
FBD pt A:
d B = 2.25 ft
(a)
(b)
A x = 800 lb
A y = 450 lb
TAB =
So
(800 lb )2 + ( 450 lb )2 = 918 lb
(c)
Tmax = TCD = 1000 lb
Note: TCD is Tmax as cable slope is largest in section CD.
PROBLEM 7.89
Two loads are suspended as shown from cable ABCD. Knowing that the
maximum tension in the cable is 720 lb, determine (a) the distance dB,
(b) the distance dC.
SOLUTION
ΣM A = 0: (10 ft ) Dy − ( 8 ft )( 450 lb ) − ( 4 ft )( 600 lb ) = 0
FBD cable:
D y = 600 1b
ΣFy = 0: Ay + 600 lb − 600 lb − 450 lb = 0
A y = 450 lb
FBD pt D:
ΣFx = 0: Ax − Bx = 0
Since Ax = Bx ; And Dy > Ay , Tension TCD > TAB
So
TCD = Tmax = 720 lb
Dx =
( 720 lb )2 − ( 600 lb )2 = 398 lb = Ax
dC
2ft
=
600 lb 398lb
FBD pt. A:
dB
4ft
=
450 lb 398lb
dC = 3.015 ft
(a)
d B = 4.52 ft
(b)
dC = 3.02 ft
PROBLEM 7.90
Knowing that dC = 4 m, determine (a) the reaction at A, (b) the reaction
at E.
SOLUTION
(a) FBD cable:
ΣM E = 0:
( 4 m )(1.2 kN ) + (8 m )( 0.8 kN ) + (12 m )(1.2 kN )
− ( 3 m ) Ax − (16 m ) Ay = 0
3 Ax + 16 Ay = 25.6 kN
(1)
FBD ABC:
ΣM C = 0: ( 4 m )(1.2 kN ) + (1 m ) Ax − ( 8 m ) Ay = 0
Ax − 8 Ay = −4.8 kN
(2)
Ax = 3.2 kN
Solving (1) and (2)
Ay = 1 kN
So A = 3.35 kN
(b)
cable:
17.35°
ΣFx = 0: − Ax + Ex = 0
Ex = Ax = 3.2 kN
ΣFy = 0: Ay − (1.2 + 0.8 + 1.2 ) kN + E y = 0
E y = 3.2 kN − Ay = ( 3.2 − 1) kN = 2.2 kN
So E = 3.88 kN
34.5°
PROBLEM 7.91
Knowing that dC = 2.25 m, determine (a) the reaction at A, (b) the
reaction at E.
SOLUTION
FBD Cable:
(a)
ΣM E = 0: ( 4 m )(1.2 kN ) + ( 8 m )( 0.8 kN )
+ (12 m )(1.2 kN ) − (3 m) Ax − (16 m ) Ay = 0
3 Ax + 16 Ay = 25.6 kN
(1)
ΣM C = 0: ( 4 m )(1.2 kN ) − ( 0.75 m ) Ax − ( 8 m ) Ay = 0
FBD ABC:
0.75 Ax + 8 Ay = 4.8 kN
Solving (1) and (2)
Ax =
(2)
32
kN,
3
Ay = − 0.4 kN
So
A = 10.67 kN
2.15°
Note: this implies d B < 3 m (in fact d B = 2.85 m)
(b) FBD cable:
ΣFx = 0: −
32
kN + Ex = 0
3
Ex =
32
kN
3
ΣFy = 0: − 0.4 kN − 1.2 kN − 0.8 kN − 1.2 kN + E y = 0
E y = 3.6 kN
E = 11.26 kN
18.65°
PROBLEM 7.92
Cable ABCDE supports three loads as shown. Knowing that
dC = 3.6 ft, determine (a) the reaction at E, (b) the distances
dB and dD.
SOLUTION
FBD Cable:
ΣM A = 0: ( 2.4 ft ) Ex + ( 8 ft ) E y − ( 2 ft )( 360 )
(a)
− ( 4 ft )( 720 lb ) − ( 6 ft )( 240 lb ) = 0
0.3Ex + E y = 630 lb
FBD CDE:
(1)
ΣM C = 0: − (1.2 ft ) Ex + ( 4 ft ) E y − ( 2 ft )( 240 lb ) = 0
− 0.3Ex + E y = +120 lb
Solving (1) and (2)
Ex = 850 lb
(a)
(b) cable:
ΣFx = 0: − Ax + Ex = 0
(2)
E y = 375 lb
E = 929 lb
23.8°
Ax = Ex = 850 lb
ΣFy = 0: Ay − 360 lb − 720 lb − 240 lb + 375 lb = 0
Point A:
Ay = 945 lb
dB
945 lb
=
2 ft 850 lb
d B = 2.22 ft
PROBLEM 7.92 CONTINUED
ΣM D = 0: ( 2 ft )( 375 lb ) − ( d D − 2.4 ft )( 850 lb ) = 0
Segment DE:
d D = 3.28 ft
PROBLEM 7.93
Cable ABCDE supports three loads as shown. Determine (a) the
distance dC for which portion CD of the cable is horizontal, (b) the
corresponding reactions at the supports.
SOLUTION
Segment DE:
ΣFy = 0: E y − 240 lb = 0
E y = 240 lb
ΣM A = ( 2.4 ft ) Ex + ( 8 ft )( 240 lb ) − ( 6 ft )( 240 lb )
FBD Cable:
− ( 4 ft )( 720 lb ) − ( 2 ft )( 360 lb ) = 0
E x = 1300 lb
So
From Segment DE:
ΣM D = 0: ( 2 ft ) E y − ( dC − 2.4 ft ) Ex = 0
dC = 2.4 ft +
Ey
Ex
( 2 ft ) = ( 2.4 ft ) +
240 lb
( 2 ft ) = 2.7692 ft
1300 lb
(a)
dC = 2.77 ft
From FBD Cable:
ΣFx = 0: − Ax + Ex = 0
A x = 1300 lb
ΣFy = 0: Ay − 360 lb − 720 lb − 240 lb + E y = 0
A y = 1080 lb
(b)
A = 1.690 kips
39.7°
E = 1.322 kips
10.46°
PROBLEM 7.94
An oil pipeline is supported at 6-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 4 kN. Knowing
that dC = 12 m, determine (a) the maximum tension in the cable,
(b) the distance d D .
SOLUTION
FBD Cable:
Note: A y and Fy shown are forces on cable, assuming the 4 kN loads
at A and F act on supports.
ΣM F = 0: ( 6 m ) 1( 4 kN ) + 2 ( 4 kN ) + 3 ( 4 kN ) + 4 ( 4 kN ) 
− ( 30 m ) Ay − ( 5 m ) Ax = 0
Ax + 6 Ay = 48 kN
FBD ABC:
(1)
ΣM C = 0: ( 6 m )( 4 kN ) + ( 7 m ) Ax − (12 m ) Ay = 0
7 Ax − 12 Ay = −24 kN
Solving (1) and (2)
A x = 8 kN
(2)
20
kN
3
Ay =
From FBD Cable:
ΣFx = 0: − Ax + Fx = 0
Fx = Ax = 8 kN
ΣFy = 0: Ay − 4 ( 4 kN ) + Fy = 0
20 
28

Fy = 16 kN − Ay = 16 −
kN > Ay
 kN =
3 
3

So
TEF > TAB
Tmax = TEF =
Fx2 + Fy2
FBD DEF:
(a)
Tmax =
(18 kN )
2
2
 28

+
kN  = 12.29 kN
 3

 28

ΣM D = 0: (12 m ) 
kN  − d D ( 8 kN ) − ( 6 m )( 4 kN ) = 0
 3

(b)
d D = 11.00 m
PROBLEM 7.95
Solve Prob. 7.94 assuming that dC = 9 m.
SOLUTION
FBD Cable:
Note: 4 kN loads at A and F act directly on supports, not on cable.
ΣM A = 0: ( 30 m ) Fy − ( 5 m ) Fx
− ( 6 m ) 1( 4 kN ) + 2 ( 4 kN ) + 3 ( 4 kN ) + 4 ( 4 kN )  = 0
Fx − 6 Fy = −48 kN
(1)
ΣM C = 0: (18 ) Fy − ( 9 m ) Fx − (12 m )( 4 kN ) − ( 6 m )( 4 kN ) = 0
FBD CDEF:
Fx − 2Fy = −8 kN
(2)
Fx = 12 kN
Solving (1) and (2)
TEF =
Fy = 10 kN
(10 kN )2 + (12 kN )2 = 15.62 kN
Since slope EF > slope AB this is Tmax
Tmax = 15.62 kN
(a)
Also could note from FBD cable
ΣFy = 0: Ay + Fy − 4 ( 4 kN ) = 0
Ay = 16 kN − 12 kN = 4 kN
Thus
FBD DEF:
(b)
Ay < Fy
and
TAB < TEF
ΣM D = 0: (12 m )(10 kN ) − d D (12 kN ) − ( 6 m )( 4 kN ) = 0
d D = 8.00 m
PROBLEM 7.96
Cable ABC supports two boxes as shown. Knowing that b = 3.6 m,
determine (a) the required magnitude of the horizontal force P,
(b) the corresponding distance a.
SOLUTION
(
)
W = ( 8 kg ) 9.81 m/s 2 = 78.48 N
FBD BC:
ΣM A = 0: ( 3.6 m ) P − ( 2.4 m )
3W
− aW = 0
2
a 

P = W 1 +

3.6 m 

ΣFx = 0: −T1x + P = 0
ΣFy = 0: T1y − W −
T1 y
But
T1x
=
3
W =0
2
2.8 m
a
P=
So
T1x = P
5W
2
T1 y =
5W
2.8 m
=
a
2P
so
5Wa
5.6 m
(2)
a = 1.6258 m,
Solving (1) and (2):
(1)
P = 1.4516W
(a)
P = 1.4516 ( 78.48 ) = 113.9 N
(b)
a = 1.626 m
PROBLEM 7.97
Cable ABC supports two boxes as shown. Determine the distances
a and b when a horizontal force P of magnitude 100 N is applied at C.
SOLUTION
FBD pt C:
Segment BC:
2.4 m − a
0.8 m
=
100 N
117.72 N
a = 1.7204 m
a = 1.720 m
ΣM A = 0: b (100 N ) − ( 2.4 m )(117.72 N )
2

− (1.7204 m )  117.72 N  = 0
3

b = 4.1754 m
b = 4.18 m
PROBLEM 7.98
Knowing that WB = 150 lb and WC = 50 lb, determine the magnitude
of the force P required to maintain equilibrium.
SOLUTION
FBD CD:
ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0
3Dx = 4 Dy
(1)
ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 50 lb ) = 0
FBD BCD:
2D y − Dx = 60 lb
Solving (1) and (2)
FBD Cable:
D x = 120 lb
(2)
D y = 90 lb
ΣM A = 0: ( 42 ft )( 90 lb ) − ( 30 ft )( 50 lb )
− (12 ft )(150 lb ) − (15 ft ) P = 0
P = 32.0 lb
PROBLEM 7.99
Knowing that WB = 40 lb and WC = 22 lb, determine the magnitude
of the force P required to maintain equilibrium.
SOLUTION
FBD CD:
ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0
4D y = 3Dx
FBD BCD:
(1)
ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 22 lb ) = 0
10 Dy − 5Dx = 132 lb
Solving (1) and (2)
D x = 52.8 lb
(2)
D y = 39.6 lb
FBD Whole:
ΣM A = 0: ( 42 ft )( 39.6 lb ) − ( 30 ft )( 22 lb )
− (12 ft )( 40 lb ) − (15 ft ) P = 0
P = 34.9 lb
PROBLEM 7.100
If a = 4.5 m, determine the magnitudes of P and Q required to
maintain the cable in the shape shown.
SOLUTION
FBD pt C:
By symmetry:
TBC = TCD = T
 1 
ΣFy = 0: 2 
T  − 180 kN = 0
 5 
Tx = 180 kN
T = 90 5 kN
Ty = 90 kN
Segment DE:
ΣM E = 0: ( 7.5 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0
P = 108.0 kN
Segment AB:
ΣM A = 0: ( 4.5 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0
Q = 45.0 kN
PROBLEM 7.101
If a = 6 m, determine the magnitudes of P and Q required to maintain the
cable in the shape shown.
SOLUTION
FBD pt C:
TBC = TCD = T
By symmetry:
 1 
ΣFy = 0: 2 
T  − 180 kN = 0
 5 
Tx = 180 kN
T = 90 5 kN
Ty = 90 kN
FBD DE:
ΣM E = 0: ( 9 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0
P = 120.0 kN
FBD AB:
ΣM A = 0: ( 6 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0
Q = 90.0 kN
PROBLEM 7.102
A transmission cable having a mass per unit length of 1 kg/m is strung
between two insulators at the same elevation that are 60 m apart. Knowing
that the sag of the cable is 1.2 m, determine (a) the maximum tension in
the cable, (b) the length of the cable.
SOLUTION
(a) Since h = 1.2 m L = 30 m we can approximate the load as evenly distributed in the horizontal
direction.
(
)
w = 1 kg/m 9.81 m/s 2 = 9.81 N/m.
w = ( 60 m )( 9.81 N/m )
w = 588.6 N
Also we can assume that the weight of half the cable acts at the
FBD half-cable:
1
chord point.
4
ΣM B = 0: (15 m )( 294.3 N ) − (1.2 m ) Tmin = 0
Tmin = 3678.75 N = Tmax x
ΣFy = 0: Tmax y − 294.3 N = 0
Tmax y = 294.3 N
Tmax = 3690.5 N
Tmax = 3.69 kN
(b)
2
4


2 y 
2 y 
sB = xB 1 +  B  −  B  + "
3  xB 
5  xB 




2
4


2  1.2 
2  1.2 
= ( 30 m ) 1 + 
 − 
 + " = 30.048 m
3  30 
5  30 


so
s = 2sB = 60.096 m
s = 60.1 m
Note: The more accurate methods of section 7.10, which assume the load is evenly distributed along the length
instead of horizontally, yield Tmax = 3690.5 N and s = 60.06 m. Answers agree to 3 digits at least.
PROBLEM 7.103
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
Half-cable FBDs:
T1x = T2 x to create zero horizontal force on tower → thus T01 = T02
ΣM B = 0: (15 m )  w ( 30 m )  − h1T0 = 0
FBD I:
( 450 m ) w
h =
2
1
FBD II:
T0
ΣM B = 0: ( 2 m ) T0 − (10 m )  w ( 20 m )  = 0
T0 = (100 m ) w
( 450 m ) w = 4.50 m
h =
2
(a)
1
ΣFx = 0: T1x − T0 = 0
T1x = (100 m ) w
FBD I:
ΣFy = 0: T1y − ( 30 m ) w = 0
T1y = ( 30 m ) w
T1 =
(100 m )2 + ( 30 m )2 w
(
= (104.4 m )( 0.4 kg/m ) 9.81 m/s 2
= 409.7 N
)
(100 m ) w
PROBLEM 7.103 CONTINUED
ΣFy = 0: T2y − ( 20 m ) w = 0
FBD II:
T2y = ( 20 m ) w
T2 x = T1x = (100 m ) w
T2 =
(100 m )2 + ( 20 m )2 w = 400.17 N
(b)
T1 = 410 N
T2 = 400 N
*
Since h L it is reasonable to approximate the cable weight as being distributed uniformly along the
horizontal. The methods of section 7.10 are more accurate for cables sagging under their own weight.
PROBLEM 7.104
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 = 9.75
kips/ft along the horizontal. Knowing that the span L is 3500 ft and
that the sag h is 316 ft, determine for the original configuration (a)
the maximum tension in each cable, (b) the length of each cable.
SOLUTION
FBD half-span:
W = ( 9.75 kips/ft )(1750 ft ) = 17, 062.5 kips
ΣM B = 0: ( 875 ft )(17, 065 kips ) − ( 316 ft ) T0 = 0
T0 = 47, 246 kips
Tmax = T02 + W 2 =
( 47, 246 kips )2 + (17, 063 kips )2
(a)
Tmax = 50, 200 kips
2
4


2 y
2 y
s = x 1 +   −   + "
3 x 
5 x


2
4


2  316 ft 
2  316 ft 
−
sB = (1750 ft ) 1 + 


 + "
3  1750 ft 
5  1750 ft 


= 1787.3 ft
(b)
*
To get 3-digit accuracy, only two terms are needed.
l = 2sB = 3575 ft
PROBLEM 7.105
Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft
along the horizontal. Knowing that the span L is 4150 ft and that the
sag h is 464 ft, determine (a) the maximum tension in each cable,
(b) the length of each cable.
SOLUTION
FBD half-span:
(a)
 2075 ft 
ΣM B = 0: 
 ( 23032.5 kips ) − ( 464 ft ) T0 = 0
 2 
T0 = 47, 246 kips
Tmax = T02 + W 2 =
(b)
( 47, 246 kips )2 + ( 23,033 kips )2 = 56, 400 kips
2
4


2 y
2 y
s = x 1 +   −   + "
3 x 
5 x


2
4


2  464 ft 
2  464 ft 
sB = ( 2075 ft ) 1 + 
 − 
 + "
3  2075 ft 
5  2075 ft 


sB = 2142 ft
l = 2s B
l = 4284 ft
PROBLEM 7.106
To mark the positions of the rails on the posts of a fence, a homeowner
ties a cord to the post at A, passes the cord over a short piece of pipe
attached to the post at B, and ties the free end of the cord to a bucket
filled with bricks having a total mass of 20 kg. Knowing that the mass
per unit length of the rope is 0.02 kg/m and assuming that A and B are
at the same elevation, determine (a) the sag h, (b) the slope of the cable
at B. Neglect the effect of friction.
SOLUTION
FBD pulley:
ΣM P = 0: (Tmax − WB ) r = 0
Tmax = WB = 196.2 N
FBD half-span:*
2
T0 = Tmax
−W2 =
(196.2 N )2 − ( 4.91 N )2 = 196.139 N
 25 m 
ΣM B = 0: 
 ( 4.905 N ) − h (196.139 N ) = 0
 2 
h = 0.3126 m = 313 mm
(a)
(b)
*See note Prob. 7.103
θ B = sin −1
W
 4.905 N 
= sin −1 
 = 1.433°
Tmax
 196.2 N 
PROBLEM 7.107
A small ship is tied to a pier with a 5-m length of rope as shown.
Knowing that the current exerts on the hull of the ship a 300-N force
directed from the bow to the stern and that the mass per unit length of the
rope is 2.2 kg/m, determine (a) the maximum tension in the rope,
(b) the sag h. [Hint: Use only the first two terms of Eq. (7.10).]
SOLUTION
(a) FBD ship:
ΣFx = 0: T0 − 300 N = 0
T0 = 300 N
FBD half-span:*
Tmax = T02 + W 2 =
(b)
ΣM A = 0: hTx −
L
W =0
4
2


2 4
s = x 1 +   + "
3 x


h=
but
( 300 N )2 = ( 54 N )2 = 305 N
LW
4Tx
yA = h =
LW
4Tx
so
yA
W
=
2Tx
xA
2

2  53.955 N 
L
( 2.5 m ) = 1 + 
 − " → L = 4.9732 m
2 
3  600 N 

So h =
*See note Prob. 7.103
LW
= 0.2236 m
4Tx
h = 224 mm
PROBLEM 7.108
The center span of the Verrazano-Narrows Bridge consists of two
uniform roadways suspended from four cables. The design of the bridge
allowed for the effect of extreme temperature changes which cause the
sag of the center span to vary from hw = 386 ft in winter to hs = 394 ft in
summer. Knowing that the span is L = 4260 ft, determine the change in
length of the cables due to extreme temperature changes.
SOLUTION
2
4


2 y
2 y
s = x  1 +   −   + "
3 x
5 x


Knowing
2
2


2 h 
2 h 
l = 2sTOT = L 1 + 
 − 
 + "
3  L/2 
5  L/2 




Winter:
2
4


2  386 ft 
2  386 ft 
lw = ( 4260 ft ) 1 + 
"
−
+
 = 4351.43 ft



3  2130 ft 
5  2130 ft 


Summer:
2
4


2  394 ft 
2  394 ft 
ls = ( 4260 ft )  1 + 
 − 
 + " = 4355.18 ft
3  2130 ft 
5  2130 ft 


∆l = ls − lw = 3.75 ft
PROBLEM 7.109
A cable of length L + ∆ is suspended between two points which 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
(a)
2


2 y
s = x 1 +   − "
3 x


2


2 h 
L + ∆ = 2sTOT = L 1 + 
 − "
3  L/2 




2
2
∆
2  2h 
8 h 
=   =   →h=
L
3 L 
3 L 
(b)
For
L = 30 m,
∆ = 1.2 m
3
L∆
8
h = 3.67 m
PROBLEM 7.110
Each cable of the side spans of the Golden Gate Bridge supports a load
w = 10.2 kips/ft along the horizontal. Knowing that for the side spans the
maximum vertical distance h from each cable to the chord AB is 30 ft and
occurs at midspan, determine (a) the maximum tension in each cable,
(b) the slope at B.
SOLUTION
FBD AB:
ΣM A = 0: (1100 ft ) TBy − ( 496 ft ) TBx − ( 550 ft )W = 0
11TBy − 4.96TBx = 5.5W
(1)
FBD CB:
ΣM C = 0: ( 550 ft ) TBy − ( 278 ft ) TBx − ( 275 ft )
W
=0
2
11TBy − 5.56TBx = 2.75W
Solving (1) and (2)
TBy = 28,798 kips
Solving (1) and (2)
TBx = 51, 425 kips
Tmax = TB = TB2x + TB2y
So that
(2)
tan θ B =
TBy
TBx
(a)
Tmax = 58,900 kips
(b)
θ B = 29.2°
PROBLEM 7.111
A steam pipe weighting 50 lb/ft that passes between two buildings 60 ft
apart is supported by a system of cables as shown. Assuming that the
weight of the cable is equivalent to a uniformly distributed loading of 7.5
lb/ft, determine (a) the location of the lowest point C of the cable, (b) the
maximum tension in the cable.
SOLUTION
FBD AC:
FBD CB:
ΣM A = 0: (13.5 ft ) T0 −
a
( 57.5 lb/ft ) a = 0
2
(
)
T0 = 2.12963 lb/ft 2 a 2
(1)
60 ft − a
( 57.5 lb/ft )( 60 ft − a ) − ( 6 ft ) T0 = 0
2
ΣM B = 0:
(
)
6T0 = 28.75 lb/ft 2 3600 ft 2 − (120 ft ) a + a 2 
Using (1) in (2)
(2)
0.55a 2 − (120 ft ) a + 3600 ft 2 = 0
Solving: a = (108 ± 72 ) ft
a = 36 ft (180 ft out of range)
So
(a)
C is 36 ft from A
C is 6 ft below and 24 ft left of B
T0 = 2.1296 lb/ft 2 ( 36 ft ) = 2760 lb
2
W1 = ( 57.5 lb/ft )( 36 ft ) = 2070 lb
(b)
Tmax = TA = T02 + W12 =
( 2760 lb )2 + ( 2070 lb )2 = 3450 lb
PROBLEM 7.112
Chain AB supports a horizontal, uniform steel beam having a mass
per unit length of 85 kg/m. If the maximum tension in the cable is
not to exceed 8 kN, determine (a) the horizontal distance a from A
to the lowest point C of the chain, (b) the approximate length of the
chain.
SOLUTION
ΣM A = 0: y AT0 −
yA =
a
wa = 0
2
ΣM B = 0: − yBT0 +
wa 2
2T0
yB =
d = ( y B − yB ) =
w 2
b − a2
2T0
)
2
2
2
2
− ( wb )  = w2 ( b 2 − a 2 ) = L2 w2 ( 4b 2 − 4 Lb + L2 )
( 2d )2 Tmax

2
or
Using
wb 2
2T0
2
T0 = TB2 − ( wb ) = Tmax
− ( wb )
But
∴
(
b
wb = 0
2

T2 
4 L2 + d 2 b 2 − 4L3b +  L4 − 4d 2 max
=0
w2 

(
)
L = 6 m,
d = 0.9 m,
Tmax = 8 kN,
yields
b = ( 2.934 ± 1.353) m
(
)
w = ( 85 kg/m ) 9.81 m/s2 = 833.85 N/m
b = 4.287 m
( since b > 3 m )
(a)
a = 6 m − b = 1.713 m
PROBLEM 7.112 CONTINUED
2
T0 = Tmax
− ( wb ) = 7156.9 N
2
yA
wa
=
= 0.09979
xA
2T0
yB
wb
=
= 0.24974
xB
2T0
2
2




2  yA 
2  yB 



l = s A + sB = a 1 + 
 +" + b 1+ 
 + "
3  xA 
3  xB 








2
2
2
2


= (1.713 m ) 1 + ( 0.09979 )  + ( 4.287 m ) 1 + ( 0.24974 )  = 6.19 m
3
3




(b)
l = 6.19 m
PROBLEM 7.113
Chain AB of length 6.4 m supports a horizontal, uniform steel beam
having a mass per unit length of 85 kg/m. Determine (a) the horizontal
distance a from A to the lowest point C of the chain, (b) the maximum
tension in the chain.
SOLUTION
ΣM P = 0:
x
wx − yT0 = 0
2
wx 2
2T0
so
y
wx
=
x
2T0
and d = yB − y A =
w 2
b − a2
2T0
Geometry:
y =
FBD Segment:
(
)
2
2


2 y  
2 y  
l = s A + sB = a 1 +  A   + b 1 +  B  
3  a  
3  b  


Also
l−L=
2
2
2  y A 
w2 3
 yB  
+
a + b3



  =
2
3  a 
 b   6T0
(
(
)
2
3
3
1
4d 2
2d a +b
3
3
a +b =
=
6 b2 − a 2 2
3 b2 − a 2 2
(
)
(
)
(
)
)
Using l = 6.4 m, L = 6 m, d = 0.9 m, b = 6 m − a, and solving for a,
knowing that a < 3 ft
a = 2.2196 m
(a)
T0 =
Then
w 2
b − a2
2d
(
)
(
)
a = 2.22 m
w = ( 85 kg/m ) 9.81 m/s 2 = 833.85 N/m
And with
b = 6 m − a = 3.7804 m
And
Tmax = TB = T02 + ( wb )
=
T0 = 4338 N
2
( 4338 N )2 + (833.85 N/m )2 ( 3.7804 m )2
Tmax = 5362 N
(b)
Tmax = 5.36 kN
PROBLEM 7.114
A cable AB of span L and a simple beam A′B′ of the same span
are subjected to identical vertical loadings as shown. Show that the
magnitude of the bending moment at a point C ′ in the beam is
equal to the product , T0h where T0 is the magnitude of the horizontal
component of the tension force in the cable and h is the vertical
distance between point C and the chord joining the points of support
A and B.
SOLUTION
FBD Cable:
ΣM B = 0: LACy + aT0 − ΣM B loads = 0
(1)
(Where ΣM B loads includes all applied loads)
x

ΣM C = 0: xACy −  h − a  T0 − ΣM C left = 0
L


FBD AC:
(Where ΣM C left includes all loads left of C)
x
(1) − ( 2 ):
L
FBD Beam:
FBD AC:
(2)
hT0 −
x
ΣM B loads + ΣM C left = 0
L
(3)
ΣM B = 0: LABy − ΣM B loads = 0
(4)
ΣM C = 0: xABy − ΣM C left − M C = 0
(5)
x
( 4 ) − ( 5):
L
Comparing (3) and (6)
−
x
ΣM B loads + ΣM C left + M C = 0
L
M C = hT0
Q.E.D.
(6)
PROBLEM 7.115
Making use of the property established in Prob. 7.114, solve the
problem indicated by first solving the corresponding beam problem.
Prob. 7.89a.
SOLUTION
FBD Beam:
ΣM D = 0: ( 2 ft )( 450 lb ) + ( 6 ft )( 600 lb ) − (10 ft ) A By = 0
A By = 450 lb
ΣM B = 0: M B − ( 4 ft )( 450 lb ) = 0
Section AB:
M B = 1800 lb ⋅ ft
ΣM A = 0: (10 ft ) DCy − ( 8 ft )( 450 lb ) − ( 4 ft )( 600 lb ) = 0
Cable:
DCy = 600 lb
( Note: Dy > Ay so Tmax = TCD )
2
2
T0 = Tmax
− DCy
T0 =
( 720 lb )2 − ( 600 lb )2
T0 = 398 lb
dB =
M B 1800 lb ⋅ ft
=
= 4.523 ft
T0
398 lb
d B = 4.52 ft
PROBLEM 7.116
Making use of the property established in Prob. 7.114, solve the problem
indicated by first solving the corresponding beam problem. Prob. 7.92b.
SOLUTION
FBD Beam:
ΣM E = 0: ( 2 ft )( 240 lb ) + ( 4 ft )( 720 lb )
+ ( 6 ft )( 360 lb ) − ( 8 ft ) ABy = 0
A By = 690 lb
ΣM B = 0: M B − ( 2 ft )( 690 lb ) = 0
M B = 1380 lb ⋅ ft
ΣM B = 0: M C + ( 2 ft )( 360 lb ) − ( 4 ft )( 690 lb ) = 0
M C = 2040 lb ⋅ ft
ΣM D = 0: M D + ( 2 ft )( 720 lb ) + ( 4 ft )( 360 lb )
− ( 6 ft )( 690 lb ) = 0
M D = 1260 lb ⋅ ft
hC = dC − 1.2 ft = 3.6 ft − 1.2 ft = 2.4 ft
T0 =
MC
2040 lb ⋅ ft
=
= 850 lb
hC
2.4 ft
hB =
M B 1380 lb ⋅ ft
=
= 1.6235 ft
T0
850 lb
Cable:
d B = hB + 0.6 ft
h0 =
d B = 2.22 ft
M D 1260 lb ⋅ ft
=
= 1.482 ft
T0
850 lb
d B = h0 + 1.8 ft
d D = 3.28 ft
PROBLEM 7.117
Making use of the property established in Prob. 7.114, solve the
problem indicated by first solving the corresponding beam problem.
Prob. 7.94b.
SOLUTION
FBD Beam:
A By = F = 8 kN
By symmetry:
M B = M E;
MC = M D
AC:
ΣM C = 0: M C + ( 6 m )( 4 kN ) − (12 m )( 8 kN ) = 0
M C = 72 kN ⋅ m
so
M D = 72 kN ⋅ m
Cable:
Since
M D = MC
hD = hC = 12 m − 3 m = 9 m
d D = hD + 2 m = 11 m
d D = 11.00 m
PROBLEM 7.118
Making use of the property established in Prob. 7.114, solve the
problem indicated by first solving the corresponding beam problem.
Prob. 7.95b.
SOLUTION
FBD Beam:
By symmetry: M B = M E
and
MC = M D
Cable:
Since
M D = M C , hD = hC
hD = hC = dC − 3 m = 9 m − 3 m = 6 m
Then
d D = hD + 2 m = 6 m + 2 m = 8 m
d D = 8.00 m
PROBLEM 7.119
Show that the curve assumed by a cable that carries a distributed load
w( x) is defined by the differential equation d 2 y / dx 2 = w( x ) / T0 ,
where T0 is the tension at the lowest point.
SOLUTION
ΣFy = 0: Ty ( x + ∆x ) − Ty ( x ) − w ( x ) ∆ x = 0
FBD Elemental segment:
Ty ( x + ∆x )
So
T0
−
Ty ( x )
T0
Ty
But
T0
=
=
w( x)
∆x
T0
dy
dx
dy
dy
−
w( x)
dx x + ∆x dx x
=
∆x
T0
So
In
lim :
∆x → 0
w( x)
d2y
=
2
T0
dx
Q.E.D.
PROBLEM 7.120
Using the property indicated in Prob. 7.119, determine the curve
assumed by a cable of span L and sag h carrying a distributed load
w = w0 cos(π x / L ) , where x is measured from mid-span. Also
determine the maximum and minimum values of the tension in
the cable.
SOLUTION
w ( x ) = w0 cos
πx
L
From Problem 7.119
w( x)
πx
d2y
w
=
= 0 cos
2
T0
T0
L
dx
So


dy
= 0
 using
dx 0


dy W0 L
πx
=
sin
dx
T0π
L
y =
But
w L2 
π
L
y   = h = 0 2 1 − cos 
2
T0π 
2
And
T0 = Tmin
Tmax = TA = TB :
TBy =
w0 L2 
πx
1 − cos
  using y ( 0 ) = 0 
2 
L  
T0π 
T0 =
so
w0 L2
π 2h
Tmin =
so
TBy
T0
=
w0 L2
π 2h
dy
wL
= 0
dx x = L
T0π
2
w0 L
π
2
TB = TBy
+ T02 =
w0 L
π
 L 
1+ 

πh 
2
PROBLEM 7.121
If the weight per unit length of the cable AB is w0 / cos 2 θ , prove
that the curve formed by the cable is a circular arc. (Hint: Use the
property indicated in Prob. 7.119.)
SOLUTION
Elemental Segment:
Load on segment*
But
dx = cosθ ds,
w0
ds
cos 2 θ
w0
w( x) =
cos3 θ
w ( x ) dx =
so
From Problem 7.119
d2y
w( x)
w0
=
=
T0
dx 2
T0 cos3 θ
In general
d2y
d  dy 
d
dθ
=
( tan θ ) = sec2 θ
 =
2
dx  dx  dx
dx
dx
So
dθ
w0
w0
=
=
3
2
dx
T0 cosθ
T0 cos θ sec θ
or
T0
cosθ dθ = dx = rdθ cosθ
w0
Giving r =
T0
= constant. So curve is circular arc
w0
*For large sag, it is not appropriate to approximate ds by dx.
Q.E.D.
PROBLEM 7.122
Two hikers are standing 30-ft apart and are holding the ends of a
35-ft length of rope as shown. Knowing that the weight per unit length
of the rope is 0.05 lb/ft, determine (a) the sag h, (b) the magnitude
of the force exerted on the hand of a hiker.
SOLUTION
Half-span:
w = 0.05 lb/ft,
L = 30 ft,
sB = c sinh
sB =
35
ft
2
yB
xB
 15 ft 
17.5 ft = c sinh 

 c 
c = 15.36 ft
Solving numerically,
Then
yB = c cosh
xB
15 ft
= (15.36 ft ) cosh
= 23.28 ft
c
15.36 ft
(a)
hB = yB − c = 23.28 ft − 15.36 ft = 7.92 ft
(b)
TB = wyB = ( 0.05 lb/ft )( 23.28 ft ) = 1.164 lb
PROBLEM 7.123
A 60-ft chain weighing 120 lb is suspended between two points at
the same elevation. Knowing that the sag is 24 ft, determine (a) the
distance between the supports, (b) the maximum tension in the chain.
SOLUTION
sB = 30 ft,
w=
hB = 24 ft,
120 lb
= 2 lb/ft
60 ft
xB =
yB2 = c 2 + sB2 = ( hB + c )
L
2
2
= hB2 + 2chB + c 2
( 30 ft ) − ( 24 ft )
sB2 − hB2
=
2hB
2 ( 24 ft )
2
c=
2
c = 6.75 ft
Then
sB = c sinh
xB
s
→ xB = c sinh −1 B
c
c
 30 ft 
xB = ( 6.75 ft ) sinh −1 
 = 14.83 ft
 6.75 ft 
(a)
L = 2 xB = 29.7 ft
Tmax = TB = wyB = w ( c + hB ) = ( 2 lb/ft )( 6.75 ft + 24 ft ) = 61.5 lb
(b)
Tmax = 61.5 lb
PROBLEM 7.124
A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched
between two points at the same elevation and pulled until the tension
at each end is 16 lb, determine the horizontal distance between the
ends of the tape. Neglect the elongation of the tape due to the tension.
SOLUTION
sB = 100 ft,
w=
4 lb
= 0.02 lb/ft
200 ft
Tmax = 16 lb
Tmax = TB = wyB
yB =
TB
16 lb
=
= 800 ft
w
0.02 lb/ft
c 2 = yB2 − sB2
c=
But
(800 ft )2 − (100 ft )2 = 793.73 ft
yB = xB cosh
xB
y
→ xB = c cosh −1 B
c
c
 800 ft 
= ( 793.73 ft ) cosh −1 
 = 99.74 ft
 793.73 ft 
L = 2 xB = 2 ( 99.74 ft ) = 199.5 ft
PROBLEM 7.125
An electric transmission cable of length 130 m and mass per unit
length of 3.4 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 = 65 m,
(
hB = 30 m
)
w = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m
yB2 = c 2 + s 2B
( c + hB )2 = c 2 + sB2
( 65 m ) − ( 30 m )
sB2 − hB2
=
2hB
2 ( 30 m )
2
c=
2
= 55.417 m
Now
sB = c sinh
xB
s
 65 m 
→ xB = c sinh −1 B = ( 55.417 m ) sinh −1 

c
c
 55.417 m 
= 55.335 m
L = 2 xB = 2 ( 55.335 m ) = 110.7 m
Tmax = wyB = w ( c + hB ) = ( 33.35 N/m )( 55.417 m + 30 m ) = 2846 N
Tmax = 2.85 kN
PROBLEM 7.126
A 30-m length of wire having a mass per unit length of 0.3 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 = 12 m,
(b) the corresponding span L.
SOLUTION
FBD Cable:
s = 30 m
30 m


= 15 m 
 so sB =
2


(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
hB = 12 m
yB2 = ( c + hB ) = c 2 + s 2B
2
c=
So
sB2 − hB2
2hB
2
2
15 m ) − (12 m )
(
c=
= 3.375 m
2 (12 m )
Now
sB = c sinh
xB
s
 15 m 
→ xB = c sinh −1 B = ( 3.375 m ) sinh −1 

c
c
 3.375 m 
xB = 7.4156 m
P = T0 = wc = ( 2.943 N/m )( 3.375 m )
L = 2 xB = 2 ( 7.4156 m )
(a)
(b)
P = 9.93 N
L = 14.83 m
PROBLEM 7.127
A 30-m length of wire having a mass per unit length of 0.3 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 = 30 N,
determine (a) the sag h, (b) the corresponding span L.
SOLUTION
FBD Cable:
sT = 30 m,
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
P = T0 = wc
c=
c=
P
w
30 N
= 10.1937 m
2.943 N/m
yB2 = ( hB + c ) = c 2 + sB2
2
h 2 + 2ch − sB2 = 0
sB =
30 m
= 15 m
2
h 2 + 2 (10.1937 m ) h − 225 m 2 = 0
h = 7.9422 m
sB = c sinh
(a)
h = 7.94 m
xA
s
 15 m 
→ xB = c sinh −1 B = (10.1937 m ) sinh −1 

c
c
 10.1937 m 
= 12.017 m
L = 2 xB = 2 (12.017 m )
(b)
L = 24.0 m
PROBLEM 7.128
A 30-m length of wire having a mass per unit length of 0.3 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 = 22.5 m,
(b) the corresponding force P.
SOLUTION
FBD Cable:
sT = 30 m → sB =
30 m
= 15 m
2
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m
L = 22.5 m
sB = c sinh
xB
L/2
= c sinh
c
c
15 m = c sinh
11.25 m
c
Solving numerically: c = 8.328 m
yB2 = c 2 + sB2 = ( 8.328 m ) + (15 m ) = 294.36 m 2
2
2
yB = 17.157 m
hB = yB − c = 17.157 m − 8.328 m
(a)
P = wc = ( 2.943 N/m )( 8.328 m )
(b)
hB = 8.83 m
P = 24.5 N
PROBLEM 7.129
A 30-ft wire is suspended from two points at the same elevation
that are 20 ft apart. Knowing that the maximum tension is 80 lb,
determine (a) the sag of the wire, (b) the total weight of the wire.
SOLUTION
L = 20 ft
sB =
xB =
30 ft
= 15 ft
2
sB = c sinh
xB
10 ft
= c sinh
c
c
Solving numerically:
yB = c cosh
20 ft
= 10 ft
2
c = 6.1647 ft
xB
 10 ft 
= ( 6.1647 ft ) cosh 

c
 1.1647 ft 
yB = 16.217 ft
hB = yB − c = 16.217 ft − 6.165 ft
Tmax = wyB
So
W =
and
(a)
hB = 10.05 ft
(b)
Wm = 148.0 lb
W = w ( 2 sB )
Tmax
80 lb
( 2sB ) =
( 30 ft )
yB
16.217 ft
PROBLEM 7.130
Determine the sag of a 45-ft chain which is attached to two points at
the same elevation that are 20 ft apart.
SOLUTION
sB =
45 ft
= 22.5 ft
2
xB =
L = 20 ft
L
= 10 ft
2
sB = c sinh
xB
c
22.5 ft = c sinh
Solving numerically:
yB = c cosh
10 ft
c
c = 4.2023 ft
xB
c
= ( 4.2023 ft ) cosh
10 ft
= 22.889 ft
4.2023 ft
hB = yB − c = 22.889 ft − 4.202 ft
hB = 18.69 ft
PROBLEM 7.131
A 10-m rope is attached to two supports A and B as shown.
Determine (a) the span of the rope for which the span is equal to
the sag, (b) the corresponding angle θB.
SOLUTION
y = c cosh
We know
x
c
At B,
yB = c + h = c cosh
or
1 = cosh
So
c=
h
h
−
2c c
h
= 4.933
c
Solving numerically
sB = c sinh
h
2c
xB
s
h
→ T = c sinh
c
2
2c
sT
10 m
=
= 0.8550 m
 h 
 4.933 
2sinh   2 sinh 

 2c 
 2 
h = 4.933c = 4.933 ( 0.8550 ) m = 4.218 m
h = 4.22 m
(a)
From
x
y = c cosh ,
c
At B,
tan θ =
L = h = 4.22 m W
dy
x
= sinh
dx
c
dy
L
4.933
= sinh
= sinh
= 5.848
dx B
2c
2
θ = tan −1 5.848
(b)
θ = 80.3° W
PROBLEM 7.132
A cable having a mass per unit length of 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
(
)
w = ( 3 kg/m ) 9.81 m/s 2 = 29.43 N/m
L = 48 m,
Tmax ≤ 1800 N
Tmax = wyB → yB =
yB ≤
yB = c cosh
xB
c
Tmax
w
1800 N
= 61.162 m
29.43 N/m
61.162 m = c cosh
Solving numerically
48m / 2
*
c
c = 55.935 m
h = yB − c = 61.162 m − 55.935 m
h = 5.23 m W
*Note: There is another value of c which will satisfy this equation. It is much smaller, thus
corresponding to a much larger h.
PROBLEM 7.133
An 8-m length of chain having a mass per unit length of 3.72 kg/m
is attached to a beam at A and passes over a small pulley at B as
shown. Neglecting the effect of friction, determine the values of
distance a for which the chain is in equilibrium.
SOLUTION
TB = wa
Neglect pulley size and friction
But
TB = wyB
so
yB = c cosh
c cosh
But sB = c sinh
So
yB = a
xB
c
1m
=a
c
8m − a
1m
= c sinh
c
2
xB
c
4 m = c sinh
(
1m c
1m
+ cosh
c
c
2
16 m = c 3e1/c − e−1/c
Solving numerically
)
c = 0.3773 m, 5.906 m
1m

( 0.3773 m ) cosh 0.3773 m = 2.68 m W
1m 
=
a = c cosh
c
 ( 5.906 m ) cosh 1 m = 5.99 m W

5.906 m
PROBLEM 7.134
A motor M is used to slowly reel in the cable shown. Knowing that
the weight per unit length of the cable is 0.5 lb/ft, determine the
maximum tension in the cable when h = 15 ft.
SOLUTION
w = 0.5 lb/ft
L = 30 ft
yB = c cosh
hB = 15 ft
xB
c
hB + c = c cosh
L
2c
15 ft


− 1
15 ft = c  cosh
c


Solving numerically c = 9.281 ft
yB = ( 9.281 ft ) cosh
15 ft
= 24.281 ft
9.281 ft
Tmax = TB = wyB = ( 0.5 lb/ft )( 24.281 ft )
Tmax = 12.14 lb W
PROBLEM 7.135
A motor M is used to slowly reel in the cable shown. Knowing that
the weight per unit length of the cable is 0.5 lb/ft, determine the
maximum tension in the cable when h = 9 ft.
SOLUTION
w = 0.5 lb/ft,
L = 30 ft,
yB = hB + c = c cosh
hB = 9 ft
xB
L
= c cosh
2c
c
15 ft


9 ft = c  cosh
− 1
c


Solving numerically c = 13.783 ft
yB = hB + c = 9 ft + 13.783 ft = 21.783 ft
Tmax = TB = wyB = ( 0.5 lb/ft )( 21.78 ft )
Tmax = 11.39 lb W
PROBLEM 7.136
To the left of point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the weight per unit length
of the cable is 1.5 lb/ft, determine the force F when a = 10.8 ft.
SOLUTION
yD = c cosh
xD
c
h + c = c cosh
a
c
10.8 m


− 1
12 m = c  cosh
c


Solving numerically
Then
c = 6.2136 m
yB = ( 6.2136 m ) cosh
10.8 m
= 18.2136 m
6.2136 m
F = Tmax = wyB = (1.5 lb/ft )(18.2136 m )
F = 27.3 lb
W
PROBLEM 7.137
To the left of point B the long cable ABDE rests on the rough
horizontal surface shown. Knowing that the weight per unit length
of the cable is 1.5 lb/ft, determine the force F when a = 18 ft.
SOLUTION
yD = c cosh
xD
c
c + h = c cosh
a
c
a


h = c  cosh − 1
c


18 ft


12 ft = c  cosh
− 1
c


Solving numerically
c = 15.162 ft
yB = h + c = 12 ft + 15.162 ft = 27.162 ft
F = TD = wyD = (1.5 lb/ft )( 27.162 ft ) = 40.74 lb
F = 40.7 lb
W
PROBLEM 7.138
A uniform cable has a mass per unit length of 4 kg/m and is held in
the position shown by a horizontal force P applied at B. Knowing that
P = 800 N and θA = 60°, determine (a) the location of point B, (b) the
length of the cable.
SOLUTION
(
)
w = 4 kg/m 9.81 m/s 2 = 39.24 N/m
P = T0 = wc
c=
P
800 N
=
w 39.24 N/m
c = 20.387 m
y = c cosh
x
c
dy
x
= sinh
dx
c
tan θ = −
dy
−a
a
= − sinh
= sinh
dx −a
c
c
a = c sinh −1 ( tan θ ) = ( 20.387 m ) sinh −1 ( tan 60° )
a = 26.849 m
y A = c cosh
a
26.849 m
= ( 20.387 m ) cosh
= 40.774 m
c
20.387 m
b = y A − c = 40.774 m − 20.387 m = 20.387 m
So
s = c sinh
(a)
a
26.849 m
= ( 20.387 m ) sinh
= 35.31 m
c
20.387 m
B is 26.8 m right and 20.4 m down from A W
(b)
s = 35.3 m W
PROBLEM 7.139
A uniform cable having a mass per unit length of 4 kg/m is held in
the position shown by a horizontal force P applied at B. Knowing
that P = 600 N and θA = 60°, determine (a) the location of point B,
(b) the length of the cable.
SOLUTION
(
)
w = ( 4 kg/m ) 9.81 m/s 2 = 39.24 N/m
P = T0 = wc
c=
P
600 N
=
w 39.24 N/m
c = 15.2905 m
y = c cosh
At A:
So
tan θ = −
x
c
dy
x
= sinh
dx
c
dy
−a
a
= − sinh
= sinh
dx −a
c
c
a = c sinh −1 ( tan θ ) = (15.2905 m ) sinh −1 ( tan 60° ) = 20.137 m
a
c
yB = h + c = c cosh
a


h = c  cosh − 1
c




20.137 m
= (15.2905 m )  cosh
− 1
15.2905 m


= 15.291 m
So
s = c sinh
(a)
B is 20.1 m right and 15.29 m down from A
a
20.137 m
= (15.291 m ) sinh
= 26.49 m
c
15.291 m
(b)
s = 26.5 m
PROBLEM 7.140
The cable ACB weighs 0.3 lb/ft. Knowing that the lowest point of the
cable is located at a distance a = 1.8 ft below the support A, determine
(a) the location of the lowest point C, (b) the maximum tension in the
cable.
SOLUTION
y A = c cosh
−a
= c + 1.8 ft
c
1.8 ft 

a = c cosh −1 1 +

c 

yB = c cosh
b
= c + 7.2 ft
c
7.2 ft 

b = c cosh −1 1 +

c 

But

1.8 ft 
7.2 ft  

−1 
a + b = 36 ft = c cosh −1 1 +
 + cosh  1 +

c 
c  



Solving numerically
Then
c = 40.864 ft

7.2 ft 
b = ( 40.864 ft ) cosh −1  1 +
 = 23.92 ft
40.864 ft 

(a)
Tmax = wyB = ( 0.3 lb/ft )( 40.864 ft + 7.2 ft )
C is 23.9 ft left of and 7.20 ft below B W
(b)
Tmax = 14.42 lb W
PROBLEM 7.141
The cable ACB weighs 0.3 lb/ft. Knowing that the lowest point of
the cable is located at a distance a = 6 ft below the support A,
determine (a) the location of the lowest point C, (b) the maximum
tension in the cable.
SOLUTION
y A = c cosh
−a
= c + 6 ft
c
6 ft 

a = c cosh −1 1 +

c 

yB = c cosh
b
= c + 11.4 ft
c
11.4 ft 

b = c cosh −1 1 +

c 

So
Solving numerically

6 ft 
11.4 ft  

−1 
a + b = c cosh −1  1 +
 + cosh  1 +
 = 36 ft
c 
c  



c = 20.446 ft
11.4 ft 

b = ( 20.446 ft ) cosh −1 1 +
 = 20.696 ft
20.446 ft 

(a) C is 20.7 ft left of and 11.4 ft below B
 20.696 ft 
Tmax = wyB = ( 0.3 lb/ft )( 20.446 ft ) cosh 
 = 9.554 lb
 20.446 ft 
(b)
Tmax = 9.55 lb
PROBLEM 7.142
Denoting by θ the angle formed by a uniform cable and the horizontal,
show that at any point (a) s = c tan θ , (b) y = c sec θ .
SOLUTION
tan θ =
(a)
s = c sinh
(b)
Also
dy
x
= sinh
dx
c
x
= c tan θ Q.E.D.
c
(
)
y 2 = s 2 + c 2 cosh 2 x = sinh 2 x + 1
(
)
So
y 2 = c 2 tan 2 θ + 1 = c 2 sec2 θ
And
y = c sec θ Q.E.D.
PROBLEM 7.143
(a) Determine the maximum allowable horizontal span for a uniform
cable of mass per unit length m′ 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 m′ = 0.34 kg/m and
Tm = 32 kN.
SOLUTION
TB = Tmax = wyB
= wc cosh
Let
ξ =
L
2c
so
xB
L  2c 
L
= w   cosh
c
2 L 
2c
Tmax =
wL
cosh ξ
2ξ

dTmax
wL 
1
=
 sinh ξ − cosh ξ 
dξ
ξ
2ξ 

For
min Tmax ,
tanh ξ −
1
ξ
=0
Solving numerically ξ = 1.1997
(Tmax )min =
wL
cosh (1.1997 ) = 0.75444wL
2 (1.9997 )
(a)
If
Lmax =
(
Tmax
T
= 1.3255 max
0.75444w
w
)
Tmax = 32 kN and w = ( 0.34 kg/m ) 9.81 m/s 2 = 3.3354 N/m
Lmax = 1.3255
32.000 N
= 12 717 m
3.3354 N/m
(b)
Lmax = 12.72 km
PROBLEM 7.144
A cable has a weight per unit length of 2 lb/ft and is supported as shown.
Knowing that the span L is 18 ft, determine the two values of the sag h
for which the maximum tension is 80 lb.
SOLUTION
ymax = c cosh
Tmax = wymax
ymax =
ymax =
Tmax
w
80 lb
= 40 ft
2 lb/ft
c cosh
Solving numerically
L
=h+c
2c
9 ft
= 40 ft
c
c1 = 2.6388 ft
c2 = 38.958 ft
h = ymax − c
h1 = 40 ft − 2.6388 ft
h1 = 37.4 ft
h2 = 40 ft − 38.958 ft
h2 = 1.042 ft
PROBLEM 7.145
Determine the sag-to-span ratio for which the maximum tension in
the cable is equal to the total weight of the entire cable AB.
SOLUTION
Tmax = wyB = 2wsB
y B = 2sB
c cosh
L
L
= 2c sinh
2c
2c
tanh
L
1
=
2c
2
L
1
= tanh −1 = 0.549306
2c
2
hB
y −c
L
= B
= cosh
−1
c
c
2c
= 0.154701
hB
hB / c
=
L
2( L / 2c)
=
0.5 ( 0.154701)
= 0.14081
0.549306
hB
= 0.1408
L
PROBLEM 7.146
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 sagto-span ratio for which the maximum tension is as small as possible,
(b) the corresponding values of θ B and Tm .
SOLUTION
Tmax = wyB = wc cosh
(a)
L
2c
dTmax
L
L
L

sinh 
= w  cosh
−
dc
2c 2c
2c 

min Tmax ,
For
tanh
dTmax
=0
dc
L
L
2c
= 1.1997
=
→
2c
2c
L
yB
L
= cosh
= 1.8102
c
2c
h
y
= B − 1 = 0.8102
c
c
h  1 h  2c  
0.8102
=
= 0.3375
  =
L  2 c  L   2 (1.1997 )
T0 = wc
(b)
Tmax = wc cosh
But
T0 = Tmax cosθ B
So
θ B = sec−1 
L
2c
Tmax
L
y
= cosh
= B
2c
T0
c
Tmax
= secθ B
T0
 yB 
−1
 = sec (1.8102 )
c


= 56.46°
Tmax = wyB = w
h
= 0.338
L
θ B = 56.5°
yB  2c  L 
L
   = w (1.8102 )
c  L  2 
2 (1.1997 )
Tmax = 0.755wL
PROBLEM 7.147
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)
ΣM A = 0: ( 6 ft ) E − ( 8 ft )( 4.5 kips )
− ( 4 ft )(12 kips ) − ( 2 ft )( 6 kips ) = 0
E = 16 kips
ΣM E = 0: − ( 6 ft ) Ay + ( 4 ft )( 6 kips )
+ ( 2 ft )(12 kips ) − ( 2 ft )( 4.5 kips ) = 0
A y = 6.5 kips
Shear Diag: V is piece wise constant with discontinuities equal to the
forces at A, C, D, E, B
Moment Diag: M is piecewise linear with slope changes at C, D, E
MA = 0
M C = ( 6.5 kips )( 2 ft ) = 13 kip ⋅ ft
M C = 13 kip ⋅ ft + ( 0.5 kips )( 2 ft ) = 14 kip ⋅ ft
M D = 14 kip ⋅ ft − (11.5 kips )( 2 ft ) = −9 kip ⋅ ft
M B = − 9 kip ⋅ ft + ( 4.5 kips )( 2 ft ) = 0
(b)
V max = 11.50 kips on DE
M max = 14.00 kip ⋅ ft at D
PROBLEM 7.148
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)
ΣM B = 0: ( 4 ft )(12 kips ) + ( 7 ft )( 2.5 kips/ft )( 6 ft )
− (10 ft ) Ay = 0
A y = 15.3 kips
Shear Diag: VA = Ay = 15.3 kips, then V is linear
 dV

= −2.5 kips/ft  to C.

dx


VC = 15.3 kips − ( 2.5 kips/ft )( 6 ft ) = 0.3 kips
At C , V decreases by 12 kips to − 11.7 kips and is constant to B.
Moment Diag: M A = 0 and M is parabolic
 dM

decreasing with V  to C

dx


MC =
1
(15.3 kips + 0.3 kip )( 6 ft ) = 46.8 kip ⋅ ft
2
M B = 46.8 kip ⋅ ft − (11.7 kips )( 4 ft ) = 0
(b)
V max = 15.3 kips
M max = 46.8 kip ⋅ ft
PROBLEM 7.149
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
FBD Cable:
ΣFx = 0: − Ax + Dx = 0
Ax = Dx
ΣM A = 0: (10 m ) D y − ( 6 m )(10 kN ) − ( 3 m )( 6 kN ) = 0
Dy = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
FBD AB:
ΣM B = 0: (1.8 m ) Ax − ( 3 m )( 8.2 kN ) = 0
Ax =
From above
FBD CD:
41
kN
3
Dx = Ax =
41
kN
3
 41

kN  = 0
ΣM C = 0: ( 4 m )( 7.8 kN ) − hC 
 3

hC = 2.283 m
hC = 2.28 m
(a)
Dx = 13.67 kN
(b)
Dy = 7.80 kN
Since Ax = Bx and Ay > By , max T is TAB
2
TAB =
Ax2 + Ay2 =
2
 41

kN  + ( 8.2 kN )

 3

(c)
Tmax = 15.94 kN
PROBLEM 7.150
Knowing that the maximum tension in cable ABCD is 15 kN,
determine (a) the distance hB, (b) the distance hC.
SOLUTION
FBD Cable:
ΣFx = 0: − Ax + Dx = 0
Ax = Dx
ΣM A = 0: (10 m ) D y − ( 6 m )(10 kN ) − ( 3 m )( 6 kN ) = 0
D y = 7.8 kN
ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0
A y = 8.2 kN
Since
Ax = Dx
and
Ay > Dy ,
Tmax = TAB
ΣFy = 0: 8.2 kN − (15 kN ) sin θ A = 0
FBD pt A:
θ A = sin −1
8.2 kN
= 33.139°
15 kN
ΣFx = 0: − Ax + (15 kN ) cos θ A = 0
Ax = (15 kN ) cos ( 33.139° ) = 12.56 kN
FBD CD:
From FBD cable:
hB = ( 3 m ) tan θ A = ( 3 m ) tan ( 33.139° )
(a)
hB = 1.959 m
ΣM C = 0: ( 4 m )( 7.8 kN ) − hC (12.56 kN ) = 0
(b)
hC = 2.48 m
PROBLEM 7.151
A semicircular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at point J
when θ = 60°.
SOLUTION
FBD Rod:
ΣM A = 0:
2r
π
B=
ΣFy′ = 0: F +
FBD BJ:
W − 2rB = 0
W
π
W
W
sin 60° − cos 60° = 0
π
3
F = − 0.12952W
W  3r  W 

ΣM 0 = 0: r  F −  +
+M =0
π  2π  3 

1
1 

M = Wr  0.12952 + −
= 0.28868Wr
π 2π 

On BJ
M J = 0.289Wr
PROBLEM 7.152
A semicircular rod of weight W and uniform cross section is
supported as shown. Determine the bending moment at point J
when θ = 150°.
SOLUTION
FBD rod:
ΣFy = 0: Ay − W = 0
Ay = W
2r
ΣM B = 0:
W − 2rAx = 0
W π
Ax =
π
FBD AJ:
ΣFx′ = 0:
W
π
cos 30° +
5W
sin 30° − F = 0 F = 0.69233W
6
W
W 

ΣM 0 = 0: 0.25587r   + r  F −  − M = 0
π 
 6

1
 0.25587
M = Wr 
+ 0.69233 − 
π
 6
M = Wr ( 0.4166 )
On AJ
M = 0.417Wr
PROBLEM 7.153
Determine the internal forces at point J of the structure shown.
SOLUTION
FBD ABC:
ΣM D = 0: ( 0.375 m )( 400 N ) − ( 0.24 m ) C y = 0
C y = 625 N
ΣM B = 0: − ( 0.45 m ) C x + ( 0.135 m )( 400 N ) = 0
C x = 120 N
FBD CJ:
ΣFy = 0: 625 N − F = 0
F = 625 N
ΣFx = 0: 120 N − V = 0
V = 120.0 N
ΣM J = 0: M − ( 0.225 m )(120 N ) = 0
M = 27.0 N ⋅ m
PROBLEM 7.154
Determine the internal forces at point K of the structure shown.
SOLUTION
FBD AK:
ΣFx = 0: V = 0
V =0
ΣFy = 0: F − 400 N = 0
F = 400 N
ΣM K = 0: ( 0.135 m )( 400 N ) − M = 0
M = 54.0 N ⋅ m
PROBLEM 7.155
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 the
θ = 30° and neglecting the weight of the channel sections, (a) draw the
shear and bending-moment diagrams for beam AB, (b) determine the
maximum absolute values of the shear and bending moment in the beam.
SOLUTION
FBD Beam + channels:
(a)
T1 = T2 = T
By symmetry:
ΣFy = 0: 2T sin 60° − 3 kN = 0
T =
FBD Beam:
With cable force replaced by
equivalent force-couple system at
F and G
3
kN
3
M = ( 0.5 m )
T1x =
3
2 3
3
T1y =
2 3
3
kN
2
kN = 0.433 kN ⋅ m
Shear Diagram: V is piecewise linear
 dV

= − 0.6 kN/m  with 1.5 kN

dx


discontinuities at F and H.
V F − = − ( 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
Moment Diagram: M is piecewise parabolic
 dM

decreasing with V  with discontinuities of .433 kN at F and H.

dx


1
M F − = − ( 0.9 kN )(1.5 m ) = − 0.675 kN ⋅ m
2
M increases by 0 .433 kN ⋅ m to − 0.242 kN ⋅ m at F +
M G = − 0.242 kN ⋅ m +
1
( 0.6 kN )(1 m ) = 0.058 kN ⋅ m
2
Finish by invoking symmetry
(b)
V max = 900 N
at F − and G +
M max = 675 N ⋅ m
at F and G
PROBLEM 7.156
(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.
PROBLEM 7.157
Cable ABC supports two loads as shown. Knowing that b = 4 ft, determine
(a) the required magnitude of the horizontal force P, (b) the corresponding
distance a.
SOLUTION
FBD ABC:
ΣFy = 0: −40 lb − 80 lb + C y = 0
C y = 120 lb
FBD BC:
ΣM B = 0: ( 4 ft )(120 lb ) − (10 ft ) Cx = 0
C x = 48 lb
From ABC:
ΣFx = 0: − P + Cx = 0
P = C x = 48 lb
(a)
P = 48.0 lb
ΣM C = 0: ( 4 ft )( 80 lb ) + a ( 40 lb ) − (15 ft )( 48 lb ) = 0
(b)
a = 10.00 ft
PROBLEM 7.158
Cable ABC supports two loads as shown. Determine the distances a and b
when a horizontal force P of magnitude 60 lb is applied at A.
SOLUTION
FBD ABC:
ΣFx = 0: C x − P = 0
Cx = 60 lb
ΣFy = 0: C y − 40 lb − 80 lb = 0
C y = 120 lb
FBD BC:
ΣM B = 0: b (120 lb ) − (10 ft )( 60 lb ) = 0
b = 5.00 ft
FBD AB:
ΣM B = 0: ( a − b )( 40 lb ) − ( 5 ft ) 60 lb = 0
a − b = 7.5 ft
a = b + 7.5 ft
= 5 ft + 7.5 ft
a = 12.50 ft