7–7.
Determine the internal normal force, shear force, and
moment at points C and D in the simply-supported beam.
Point D is located just to the left of the 2500-lb force.
2500 lb
500 lb/ft
A
3 ft
SOLUTION
With reference to Fig. a, we have
a + ©MA = 0;
By(12) - 500(6)(3) - 2500(9) = 0
By = 2625 lb
a + ©MB = 0;
+ ©F = 0;
:
x
2500(3) + 500(6)(9) - A y(12) = 0
A y = 2875 lb
Ax = 0
Using these results and referring to Fig. b, we have
+ ©F = 0;
:
x
NC = 0
+ c ©Fy = 0;
2875 - 500(3) - VC = 0
VC = 1375 lb
Ans.
a + ©MC = 0;
MC + 500(3)(1.5) - 2875(3) = 0
MC = 6375 lb # ft
Ans.
Ans.
Also, by referring to Fig. c, we have
+ ©F = 0;
:
x
+ c ©Fy = 0;
a + ©MD = 0;
ND = 0
Ans.
VD + 2625 - 2500 = 0
VD = -125 lb
Ans.
2625(3) - MD = 0
MD = 7875 lb # ft
Ans.
The negative sign indicates that VD acts in the opposite sense to that shown on the
free-body diagram.
B
D
C
3 ft
3 ft
3 ft
*7–24.
Determine the internal normal force, shear force, and
bending moment at point C.
40 kN
8 kN/m
60°
B
A
3m
C
3m
3m
0.3 m
SOLUTION
Free body Diagram: The support reactions at A need not be computed.
Internal Forces: Applying equations of equilibrium to segment BC, we have
+ ©F = 0;
:
x
+ c ©Fy = 0;
-40 cos 60° - NC = 0
NC = -20.0 kN
VC - 24.0 - 12.0 - 40 sin 60° = 0
VC = 70.6 kN
a + ©MC = 0;
Ans.
Ans.
-24.011.52 - 12.0142 - 40 sin 60°16.32 - MC = 0
MC = - 302 kN # m
Ans.
*7–28.
Determine the normal force, shear force, and moment at
sections passing through points E and F. Member BC is
pinned at B and there is a smooth slot in it at C. The pin at C
is fixed to member CD.
500 lb
80 lb/ ft
350 lb ft
E
60
B
F
D
A
C
2 ft
SOLUTION
a + ©MB = 0;
-120(2) - 500 sin 60°(3) + Cy (5) = 0
Cy = 307.8 lb
+ ©F = 0;
:
x
Bx - 500 cos 60° = 0
Bx = 250 lb
+ c ©Fy = 0;
By - 120 - 500 sin 60° + 307.8 = 0
By = 245.2 lb
+ ©F = 0;
:
x
-NE - 250 = 0
NE = -250 lb
Ans.
+ c ©Fy = 0;
VE = 245 lb
Ans.
a + ©ME = 0;
- ME - 245.2(2) = 0
ME = - 490 lb # ft
Ans.
+ ©F = 0;
:
x
NF = 0
Ans.
+ c ©Fy = 0;
- 307.8 - VF = 0
VF = -308 lb
a + ©MF = 0;
Ans.
307.8(4) + MF = 0
MF = - 1231 lb # ft = - 1.23 kip # ft
Ans.
1 ft
2 ft
3 ft
2 ft
4 ft
2 ft
7–57.
Draw the shear and bending-moment diagrams for each of the
two segments of the compound beam.
150 lb/ft
A
C
SOLUTION
10 ft
Support Reactions: From FBD (a),
a + ©MA = 0; By 1122 - 2100172 = 0
By = 1225 lb
A y + 1225 - 2100 = 0
A y = 875 lb
+ c ©Fy = 0;
From FBD (b),
a + ©MD = 0; 1225162 - Cy182 = 0
Cy = 918.75 lb
Dy = 306.25 lb
+ c ©Fy = 0; Dy + 918.75 - 1225 = 0
Shear and Moment Functions: Member AB.
For 0 ◊ x<12 ft [FBD (c)],
+ c ©Fy = 0;
875 - 150x - V = 0
V = 5875 - 150x6 lb
a + ©M = 0;
Ans.
x
M + 150xa b - 875x = 0
2
M = 5875x - 75.0x26 lb # ft
Ans.
For 12 ft<x ◊ 14 ft [FBD (d)],
+ c ©Fy = 0;
V - 150114 - x2 = 0
V = 52100 - 150x6 lb
a + ©M = 0;
-150114 - x2 a
Ans.
14 - x
b - M = 0
2
M = 5 -75.0x2 + 2100x - 147006 lb # ft
Ans.
For member CBD, 0 ◊ x<2 ft [FBD (e)],
+ c ©Fy = 0;
a + ©M = 0;
918.75 - V = 0
918.75x - M = 0
V = 919 lb
M = 5919x6 lb # ft
Ans.
Ans.
For 2 ft<x ◊ 8 ft [FBD (f)],
+ c ©Fy = 0;
+ ©M = 0;
V + 306.25 = 0
V = 306 lb
Ans.
306.2518 - x2 - M = 0
M = 52450 - 306x6 lb # ft
Ans.
D
B
2 ft 2 ft
4 ft
7–61.
Draw the shear and moment diagrams for the beam.
C
3m
B
SOLUTION
A
Support Reactions: From FBD (a),
a + ©MB = 0;
Ans.
9.00122 - A y162 = 0
6m
A y = 3.00 kN
Shear and Moment Functions: For 0 … x … 6 m [FBD (b)],
+ c ©Fy = 0;
3.00 -
x2
- V = 0
4
V = b 3.00 -
x2
r kN
4
Ans.
The maximum moment occurs when V = 0, then
0 = 3.00 -
a + ©M = 0;
M + ¢
x2
4
x = 3.464 m
x2 x
≤ a b - 3.00x = 0
4
3
M = b 3.00x -
x3
r kN # m
12
Thus,
Mmax = 3.0013.4642 -
3.464 3
= 6.93 kN # m
12
3 kN/m
Ans.
7–87.
w0
Draw the shear and moment diagrams for the beam.
L
––
2
SOLUTION
Support Reactions:
a + ©MA = 0;
MA -
w0L L
w0L 2L
a b a
b = 0
2
4
4
3
MA =
+ c ©Fy = 0;
Ay -
7w0L2
24
w0L
w0L
= 0
2
4
Ay =
3w0L
4
L
––
2
*7–92.
Draw the shear and moment diagrams for the beam.
5 kip/ ft
15 kip ft
15 kip ft
B
A
6 ft
SOLUTION
Support Reactions: From FBD (a),
a + ©MA = 0;
By (10) + 15.0(2) + 15 - 50.0(5) - 15.0(12) - 15 = 0
By = 40.0 kip
+ c ©Fy = 0;
Ay + 40.0 - 15.0 - 50.0 - 15.0 = 0
Ay = 40.0 kip
Shear and Moment Diagrams: The value of the moment at supports A and B can be
evaluated using the method of sections [FBD (c)].
a + ©M = 0;
M + 15.0(2) + 15 = 0
M = -45.0 kip # ft
10 ft
6 ft
7–105.
Determine the maximum uniform loading w, measured in
lb>ft, that the cable can support if it is capable of sustaining
a maximum tension of 3000 lb before it will break.
50 ft
6 ft
w
SOLUTION
y =
1
a wdx bdx
FH L L
At x = 0,
dy
= 0
dx
At x = 0, y = 0
C1 = C2 = 0
y =
w 2
x
2FH
At x = 25 ft, y = 6 ft
FH = 52.08 w
dy
w 2
2
= tan umax =
x
dx max
FH x = 25 ft
umax = tan - 1 (0.48) = 25.64°
Tmax =
FH
= 3000
cos umax
FH = 2705 lb
w = 51.9 lb/ft
Ans.
7–106.
The cable is subjected to a uniform loading of w = 250 lb>ft.
Determine the maximum and minimum tension in the cable.
50 ft
6 ft
w
SOLUTION
From Example 7–12:
FH =
250 (50)2
w0 L2
=
= 13 021 lb
8h
8 (6)
umax = tan - 1 a
Tmax =
250 (50)
w0 L
b = 25.64°
b = tan - 1 a
2FH
2(13 021)
FH
13 021
= 14.4 kip
=
cos umax
cos 25.64°
Ans.
The minimum tension occurs at u = 0°.
Tmin = FH = 13.0 kip
Ans.
7–111.
y
If the slope of the cable at support A is zero, determine the
deflection curve y = f(x) of the cable and the maximum
tension developed in the cable.
12 m
B
4.5 m
SOLUTION
A
x
Using Eq. 7–12,
y =
1
a w(x)dxb dx
FH L L
y =
1
p
a 4 cos
* dxb dx
FH L L
24
y =
24
p
1
c4(103) d sin x + C1
FH L p
24
y = -
4 kN/m
p
24 96(103)
c
cos x d + C1x + C2
p
pFH
24
dy
= 0 at x = 0 results in C1 = 0.
dx
Applying the boundary condition y = 0 at x = 0, we have
Applying the boundary condition
0 = -
24 96(103)
c
cos 0° d + C2
p pFH
2304(103)
C2 =
p2FH
Thus,
y =
2304(103)
p
c1 - cos
xd
24
p2FH
Applying the boundary condition y = 4.5 m at x = 12 m, we have
4.5 =
2304(103)
p
c1 - cos
(12) d
2
24
p FH
FH = 51.876(103) N
Substituting this result into Eqs. (1) and (2), we obtain
96(103)
dy
p
sin x
=
dx
24
p(51.876)(103)
p
= 0.5890 sin x
24
and
y =
2304(103)
p 2(51.876)(103)
= 4.5 a1 - cos
c1 - cos
p
xd
24
p
xb m
24
Ans.
The maximum tension occurs at point B where the cable makes the greatest angle
with the horizontal. Here,
umax = tan-1 a
dy
p
b = tan-1 c0.5890 sina (12)b d = 30.50°
`
dx x = 12 m
24
Thus,
Tmax
51.876(103)
FH
=
= 60.207(103) N = 60.2 kN
= cos u
cos 30.50°
max
Ans.
p
w ⫽ 4 cos –– x
24
*7–120.
The power transmission cable weighs 10 lb>ft. If the
resultant horizontal force on tower BD is required to be
zero, determine the sag h of cable BC.
300 ft
A
SOLUTION
FH
w0
Bcosh ¢ x ≤ - 1 R
w0
FH
y =
FH
10
Bcosh ¢ x ≤ - 1 R ft
10
FH
Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft,
10 =
(FH)AB
10(150)
Bcosh ¢
≤ - 1R
10
(FH)AB
Solving by trial and error yields
(FH)AB = 11266.63 lb
Since the resultant horizontal force at B is required to be zero,
(FH)BC = (FH)AB = 11266.62 lb. Applying the boundary condition of cable BC
y = h at x = -100 ft to Eq. (1), we obtain
h =
10(-100)
11266.62
c cosh B
R - 1s
10
11266.62
= 4.44 ft
10 ft
D
The origin of the x, y coordinate system is set at the lowest point of the cables. Here,
w0 = 10 lb>ft. Using Eq. 4 of Example 7–13,
y =
200 ft
B
Ans.
h
C
7–121.
The power transmission cable weighs 10 lb>ft. If h = 10 ft,
determine the resultant horizontal and vertical forces the
cables exert on tower BD.
300 ft
A
SOLUTION
y =
FH
w0
Bcosh ¢
w0
x≤ - 1R
FH
FH
10
Bcosh ¢ x ≤ - 1 R ft
10
FH
Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft,
10 =
10(150)
(FH)AB
Bcosh ¢
≤ - 1R
10
(FH)AB
Solving by trial and error yields
(FH)AB = 11266.63 lb
Applying the boundary condition of cable BC, y = 10 ft at x = -100 ft to Eq. (2),
we have
10 =
10(100)
(FH)BC
Bcosh ¢
≤ - 1R
10
(FH)BC
Solving by trial and error yields
(FH)BC = 5016.58 lb
Thus, the resultant horizontal force at B is
(FH)R = (FH)AB - (FH)BC = 11266.63 - 5016.58 = 6250 lb = 6.25 kip
Using Eq. (1), tan (uB)AB = sin h B
sin h B
10 ft
D
The origin of the x, y coordinate system is set at the lowest point of the cables. Here,
w0 = 10 lb>ft. Using Eq. 4 of Example 7–13,
y =
200 ft
B
10(150)
R = 0.13353 and
11266.63
Ans.
tan (uB)BC =
10(-100)
R = 0.20066. Thus, the vertical force of cables AB and BC acting
5016.58
on point B are
(Fv)AB = (FH)AB tan (uB)AB = 11266.63(0.13353) = 1504.44 lb
(Fv)BC = (FH)BC tan (uB)BC = 5016.58(0.20066) = 1006.64 lb
The resultant vertical force at B is therefore
(Fv)R = (Fv)AB + (Fv)BC = 1504.44 + 1006.64
= 2511.07 lb = 2.51 kip
Ans.
h
C
7–134.
Determine the normal force, shear force, and moment at
points B and C of the beam.
7.5 kN
5m
3m
1m
Free body Diagram: The support reactions need not be computed for this case .
Internal Forces: Applying the equations of equilibrium to segment DC [FBD (a)],
we have
+ ©F = 0;
:
x
NC = 0
+ c ©Fy = 0;
VC - 3.00 - 6 = 0
Ans.
VC = 9.00 kN
Ans.
- MC - 3.00(1.5) - 6(3) - 40 = 0
MC = -62.5 kN # m
Ans.
Applying the equations of equilibrium to segment DB [FBD (b)], we have
+ ©F = 0;
:
x
NB = 0
+ c ©Fy = 0;
VB - 10.0 - 7.5 - 4.00 - 6 = 0
Ans.
VB = 27.5 kN
a + ©MB = 0;
C
40 kN m
5m
SOLUTION
1 kN/m
B
A
a + ©MC = 0;
6 kN
2 kN/m
Ans.
- MB - 10.0(2.5) - 7.5(5)
- 4.00(7) - 6(9) - 40 = 0
MB = -184.5 kN # m
Ans.