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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–1. Determine the tension in each segment of the cable
and the cable’s total length.
A
4 ft
D
7 ft
B
C
Applying method of joints, we have
Equations of Equilibrium:
50 lb
Joint B:
4 ft
+
: a Fx = 0;
FBC cos u - FBA a
+ c a Fy = 0;
FBA a
+
: a Fx = 0;
FCD cos f - FBC cos u = 0
[3]
+ c a Fy = 0;
FBC sin u + FCD sin f - 100 = 0
[4]
7
265
4
265
b = 0
b - FBC sin u - 50 = 0
[1]
[2]
Geometry:
y
2y2 + 25
sin f =
2y + 6y + 18
3 + y
2
5
cos u =
2y2 + 25
cos f =
2y + 6y + 18
3
2
Substitute the above results into Eqs. [1], [2], [3] and [4] and solve. We have
FBC = 46.7 lb
FBA = 83.0 lb
FCD = 88.1 lb
Ans.
y = 2.679 ft
The total length of the cable is
l = 272 + 42 + 252 + 2.6792 + 232 + (2.679 + 3)2
= 20.2 ft
Ans.
125
3 ft
100 lb
Joint C:
sin u =
5 ft
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5–2. Cable ABCD supports the loading shown. Determine
the maximum tension in the cable and the sag of point B.
D
A
yB
2m
C
Referring to the FBD in Fig. a,
a + a MA = 0;
TCD a
4
217
B
b(4) + TCD a
1
217
1m
b(2) - 6(4) - 4(1) = 0
TCD = 6.414 kN = 6.41 kN(Max)
4 kN
Ans.
Joint C: Referring to the FBD in Fig. b,
+
: a Fx = 0;
6.414 a
217
+ c a Fy = 0;
6.414a
217
1
b - TBC cos u = 0
4
b - 6 - TBC sin u = 0
Solving,
TBC = 1.571 kN = 1.57 kN
(6TCD)
u = 8.130°
Joint B:
Referring to the FBD in Fig. c,
+
: a Fx = 0;
1.571 cos 8.130° - TAB cos f = 0
+ c a Fy = 0;
TAB sin f + 1.571 sin 8.130° - 4 = 0
Solving,
TAB = 4.086 kN = 4.09 kN
(6TCD)
f = 67.62°
Then, from the geometry,
yB
= tan f;
1
yB = 1 tan 67.62°
= 2.429 m = 2.43 m
Ans.
126
3m
0.5 m
6 kN
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5–3. Determine the tension in each cable segment and the
distance yD.
A
yD
D
7m
B
Joint B:
2m
Referring to the FBD in Fig. a,
+
: a Fx = 0;
TBC a
+ c a Fy = 0;
TAB a
229
b - TAB a
265
7
b - TBC a
2
5
265
4
229
2 kN
b = 0
C
4m
b - 2 = 0
TBC = 1.596 kN = 1.60 kN
Ans.
Joint C: Referring to the FBD in Fig. b,
+
: a Fx = 0;
TCD cos u - 1.596 a
5
229
b = 0
+ c a Fy = 0;
TCD sin u + 1.596 a
2
b - 4 = 0
229
3m
4 kN
Solving,
TAB = 2.986 kN = 2.99 kN
5m
Solving,
TCD = 3.716 kN = 3.72 kN
Ans.
u = 66.50°
From the geometry,
yD + 3 tan u = 9
yD = 9 - 3 tan 66.50° = 2.10 m
Ans.
127
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*5–4. The cable supports the loading shown. Determine the
distance xB the force at point B acts from A. Set P = 40 lb.
xB
A
5 ft
At B
+
: a Fx = 0;
+ c a Fy = 0;
B
xB
2x2B + 25
40 -
5
2x2B + 25
xB – 3
2(xB - 3)2 + 64
TAB -
8
2(xB - 3)2 + 64
TAB -
13xB - 15
2(xB - 3)2 + 64
TBC = 0
8 ft
TBC = 0
C
(1) 2 ft
TBC = 200
D
+ c a Fy = 0;
5
3
4
At C
+
: a Fx = 0;
P
30 lb
3 ft
xB - 3
4
3
(30) +
TBC TCD = 0
5
2(xB - 3)2 + 64
213
8
2(xB - 3) + 64
2
30 - 2xB
2(xB - 3)2 + 64
TBC +
2
213
TCD -
3
(30) = 0
5
TBC = 102
(2)
Solving Eqs. (1) & (2)
13xB - 15
200
=
30 - 2xB
102
xB = 4.36 ft
Ans.
5–5. The cable supports the loading shown. Determine the
magnitude of the horizontal force P so that xB = 6 ft.
xB
A
5 ft
At B
+
: a Fx = 0;
+ c a Fy = 0;
B
P 5
261
5P -
6
261
TAB -
TAB 63
273
8
273
3
273
TBC = 0
8 ft
TBC = 0
C
(1) 2 ft
TBC = 0
+ c a Fy = 0;
3
3 ft
4
3
3
(30) +
TBC TCD = 0
5
273
213
8
273
18
273
TBC -
5
4
At C
+
: a Fx = 0;
D
2
213
TCD -
3
(30) = 0
5
TBC = 102
(2)
Solving Eqs. (1) & (2)
5P
63
=
18
102
P = 71.4 lb
Ans.
128
30 lb
P
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5–6. Determine the forces P1 and P2 needed to hold the
cable in the position shown, i.e., so segment CD remains
horizontal. Also find the maximum loading in the cable.
1.5 m
A
E
B
1m
C
D
5 kN
P1
2m
Method of Joints:
Joint B:
+
: a Fx = 0;
FBC a
+ c a Fy = 0;
FAB a
4
217
b - FAB a
2
b = 0
2.5
[1]
1.5
1
b - 5 = 0
b - FBC a
2.5
217
[2]
Solving Eqs. [1] and [2] yields
FBC = 10.31 kN
FAB = 12.5 kN
Joint C:
+
: a Fx = 0;
FCD - 10.31 a
+ c a Fy = 0;
10.31 a
217
+
: a Fx = 0;
FDE a
222.25
4
b - 10 = 0
[1]
+ c a Fy = 0;
FDE a
25
b - P2 = 0
[2]
1
4
217
b = 0
FCD = 10.00 kN
b - P1 = 0 P1 = 2.50 kN
Ans.
Joint D:
222.25
Solving Eqs. [1] and [2] yields
P2 = 6.25 kN
Ans.
FDE = 11.79 kN
Thus, the maximum tension in the cable is
Fmax = FAB = 12.5 kN
Ans.
129
4m
P2
5m
4m
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5–7. The cable is subjected to the uniform loading. If the
slope of the cable at point O is zero, determine the equation
of the curve and the force in the cable at O and B.
y
A
B
8 ft
O
x
500 lb/ ft
15 ft
From Eq. 5–9.
y =
15 ft
h 2
8
x =
x2
L2
(15)2
y = 0.0356x2
Ans.
From Eq. 5–8
To = FH =
500(15)2
woL2
=
= 7031.25 lb = 7.03 k
2h
2(8)
Ans.
From Eq. 5–10.
TB = Tmax = 2(FH)2 + (woL)2 = 2(7031.25)2 + [(500)(15)]2
= 10 280.5 lb = 10.3 k
Ans.
Also, from Eq. 5–11
15 2
L 2
TB = Tmax = woL 1 + a b = 500(15) 1 + a
b = 10 280.5 lb = 10.3 k
2h
A
2(8)
A
Ans.
*5–8. The cable supports the uniform load of w0 = 600 lb>ft.
Determine the tension in the cable at each support A and B.
B
A
15 ft
y =
10 ft
wo 2
x
2 FH
15 =
600 2
x
2 FH
w0
25 ft
600
10 =
(25 - x)2
2 FH
600 3
600
x =
(25 - x)2
2(15)
2(10)
x2 = 1.5(625 - 50x + x2)
0.5x2 - 75x + 937.50 = 0
Choose root 6 25 ft
x = 13.76 ft
FH =
wo 2
600
x =
(13.76)2 = 3788 lb
2y
2(15)
130
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5–8. Continued
At B:
y =
wo 2
600
x =
x2
2 FH
2(3788)
dy
= tan uB = 0.15838 x 2
= 2.180
dx
x = 13.76
uB = 65.36°
FH
3788
=
= 9085 lb = 9.09 kip
cos uB
cos 65.36°
TB =
Ans.
At A:
y =
wo 2
600
x =
x2
2 FH
2(3788)
dy
= tan uA = 0.15838 x 2
= 1.780
dx
x = (25 - 13.76)
uA = 60.67°
TA =
FH
3788
=
= 7734 lb = 7.73 kip
cos uA
cos 60.67°
Ans.
5–9. Determine the maximum and minimum tension in the
cable.
y
10 m
10 m
B
A
2m
x
16 kN/m
The minimum tension in the cable occurs when u = 0°. Thus, Tmin = FH.
With wo = 16 kN>m, L = 10 m and h = 2 m,
Tmin = FH =
And
(16 kN>m)(10 m)2
woL2
=
= 400 kN
2h
2(2 m)
Ans.
Tmax = 2FH2 + (woL)2
= 24002 + [16(10)]2
= 430.81 kN
= 431 kN
Ans.
131
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–10. 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
y =
1
a wdxbdx
FH L L
At x = 0,
dy
= 0
dx
At x = 0,
y=0
C1 = C2 = 0
y =
w 2
x
2 FH
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.
5–11. 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
FH =
250(50)2
woL2
=
= 13 021 lb
8h
8(6)
umax = tan - 1 a
Tmax =
250(50)
woL
b = tan - 1 a
b = 25.64°
2 FH
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.
132
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–12. The cable shown is subjected to the uniform load w0 .
Determine the ratio between the rise h and the span L that
will result in using the minimum amount of material for the
cable.
L
h
w0
From Eq. 5–9,
h
4h
x2 = 2 x2
L
L 2
a b
2
y =
dy
8h
= 2x
dx
L
From Eq. 5–8,
FH =
L 2
b
woL2
2
=
2h
8h
wo a
Since
FH = Ta
dx
b,
ds
then
woL2 ds
a b
8h dx
T =
Let sallow be the allowable normal stress for the cable. Then
T
= sallow
A
T
= A
sallow
dV = A ds
dV =
T
sallow
ds
The volume of material is
V =
sallow L0
2
1
2
2
T ds =
2
sallow
1
2
woL2 (ds)2
c
d
dx
L0 8h
dy 2
dx + dy
dx2 + dy2
ds
ddx
=
c1
+
a
=
= c
b ddx
dx
dx
dx
dx2
2
2
1
2
=
dy 2
woL2
c 1 + a b ddx
dx
L0 4hsallow
=
2
woL2
h2x2
c 1 + 64 a 4 b ddx
4hsallow L0
L
=
woL2 L
woL2 L
8h2
16 h
c +
d =
c +
a bd
4hsallow 2
3L
8 sallow h
3 L
1
Require,
woL2
16
L
dV
=
c- 2 +
d = 0
dh
8sallow h
3L
h = 0.433 L
Ans.
133
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–13. The trusses are pin connected and suspended from
the parabolic cable. Determine the maximum force in the
cable when the structure is subjected to the loading shown.
D
E
14 ft
6 ft
K
J
I
16 ft
A
C
F
G H B
5k
4 @ 12 ft ! 48 ft
4k
4 @ 12 ft ! 48 ft
Entire structure:
a + a MC = 0;
4(36) + 5(72) + FH(36) - FH(36) - (Ay + Dy(96) = 0
(A y + Dy) = 5.25
(1)
Section ABD:
a + a MB = 0;
FH(14) - (Ay + Dy)(48) + 5(24) = 0
Using Eq. (1):
FH = 9.42857 k
From Eq. 5–8:
wo =
2FHh
L
2
=
2(9.42857)(14)
482
= 0.11458 k>ft
From Eq. 5–11:
48 2
L 2
Tmax = woL 1 + a b = 0.11458(48) 1 + c
d = 10.9 k
2h
A
2(14)
A
5–14. Determine the maximum and minimum tension in
the parabolic cable and the force in each of the hangers. The
girder is subjected to the uniform load and is pin connected
at B.
E
9 ft
1 ft
D
Member BC:
+
: a Fx = 0;
Ans.
2 k/ ft
10 ft
Bx = 0
C
A
Member AB:
+
: a Fx = 0;
10 ft
Ax = 0
FBD 1:
a + a MA = 0;
FH(1) - By(10) - 20(5) = 0
134
B
30 ft
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5–14. Continued
FBD 2:
a + a MC = 0;
-FH(9) - By(30) + 60(15) = 0
Solving,
By = 0 ,
FH = Fmin = 100 k
Ans.
Max cable force occurs at E, where slope is the maximum.
From Eq. 5–8.
Wo =
2FHh
L2
=
2(100)(9)
302
= 2 k>ft
From Eq. 5–11,
30 2
L 2
Fmax = woL 1 + a b = 2(30) 1 + a
b
A
2h
2(9)
A
Fmax = 117 k
Ans.
Each hanger carries 5 ft of wo.
T = (2 k>ft)(5 ft) = 10 k
Ans.
5–15. Draw the shear and moment diagrams for the pinconnected girders AB and BC. The cable has a parabolic
shape.
E
9 ft
1 ft
D
a + a MA = 0;
T(5) + T(10) + T(15) + T(20) + T(25)
2 k/ ft
10 ft
+ T(30) + T(35) + Cy(40) – 80(20) = 0
C
A
Set T = 10 k (See solution to Prob. 5–14)
+ c a Fy = 0;
10 ft
Cy = 5 k
7(10) + 5 – 80 + A y = 0
Ay = 5 k
Mmax = 6.25 k # ft
Ans.
135
B
30 ft
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*5–16. The cable will break when the maximum tension
reaches Tmax = 5000 kN . Determine the maximum
uniform distributed load w required to develop this
maximum tension.
100 m
12 m
w
With Tmax = 80(103) kN, L = 50 m and h = 12 m,
L 2
Tmax = woL 1 + a b
A
2h
8000 = wo(50)c
A
1 + a
50 2
b d
24
wo = 69.24 kN>m = 69.2 kN>m
Ans.
5–17. The cable is subjected to a uniform loading of
w = 60 kN> m. Determine the maximum and minimum
tension in cable.
100 m
12 m
w
The minimum tension in cable occurs when u = 0°. Thus, Tmin = FH.
Tmin = FH =
(60 kN>m)(50 m)2
woL2
=
= 6250 kN
2h
2(12 m)
= 6.25 MN
And,
Ans.
Tmax = 2F2H + (woL)2
= 262502 + [60(50)]2
= 6932.71 kN
= 6.93 MN
Ans.
136
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–18. The cable AB is subjected to a uniform loading of
200 N>m. If the weight of the cable is neglected and the
slope angles at points A and B are 30° and 60°, respectively,
determine the curve that defines the cable shape and the
maximum tension developed in the cable.
y
B
60"
A
Here the boundary conditions are different from those in the text.
30"
x
Integrate Eq. 5–2,
200 N/ m
T sin u = 200x + C1
15 m
Divide by by Eq. 5–4, and use Eq. 5–3
dy
1
=
(200x + C1)
dx
FH
y =
1
(100x2 + C1x + C2)
FH
At x = 0, y = 0; C2 = 0
At x = 0,
y =
dy
= tan 30°;
dx
C1 = FH tan 30°
1
(100x2 + FH tan 30°x)
FH
dy
1
=
(200x + FH tan 30°)
dx
FH
At x = 15 m,
dy
= tan 60°;
dx
FH = 2598 N
y = (38.5x2 + 577x)(10 - 3) m
Ans.
umax = 60°
Tmax =
FH
2598
=
= 5196 N
cos umax
cos 60°
Tmax = 5.20 kN
Ans.
137
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–19. The beams AB and BC are supported by the cable
that has a parabolic shape. Determine the tension in the cable
at points D, F, and E, and the force in each of the equally
spaced hangers.
E
D
3m
3m
F
9m
A
C
B
+
: a Fx = 0;
Bx = 0
a + a MA = 0;
3 kN
(Member BC)
FF(12) - FF(9) - By(8) - 3(4) = 0
3FF - By(8) = 12
+
: a Fx = 0;
2m 2m 2m 2m 2m 2m 2m 2m
(1)
A x = 0 (Member AB)
a + a MC = 0;
-FF(12) + FF(9) - By(8) + 5(6) = 0
-3FF - By(8) = -30
(2)
Soving Eqs. (1) and (2),
By = 1.125 kN ,
FF = 7.0 kN
Ans.
From Eq. 5–8.
wo =
2FHh
L
2
=
2(7)(3)
82
= 0.65625 kN>m
From Eq. 5–11,
8 2
L 2
Tmax = woL 1 + a b = 0.65625(8) 1 + a
b
2h
A
2(3)
A
Tmax = TE = TD = 8.75 kN
Ans.
Load on each hanger,
T = 0.65625(2) = 1.3125 kN = 1.31 kN
Ans.
138
5 kN
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–20. Draw the shear and moment diagrams for beams
AB and BC. The cable has a parabolic shape.
E
D
3m
3m
F
9m
A
C
B
a + a MA = 0;
5 kN
3 kN
Member ABC:
T(2) + T(4) + T(6) + T(8) + T(10)
2m 2m 2m 2m 2m 2m 2m 2m
+ T(12) + T(14) + Cy(16) - 3(4) - 5(10) = 0
Set T = 1.3125 kN (See solution to Prob 5–19).
+ c a Fy = 0;
Cy = -0.71875 kN
7(1.3125) - 8 - 0.71875 + A y = 0
Ay = -0.46875 kN
Mmax = 3056 kN # m
Ans.
5–21. The tied three-hinged arch is subjected to the
loading shown. Determine the components of reaction at
A and C and the tension in the cable.
15 kN
10 kN
B
2m
Entire arch:
+
: a Fx = 0;
Ax = 0
a + a MA = 0;
Cy(5.5) - 15(0.5) - 10(4.5) = 0
A
Ans.
C
2m
Cy = 9.545 kN = 9.55 kN
+ c a Fy = 0;
Ans.
9.545 - 15 - 10 + A y = 0
A y = 15.45 kN = 15.5 kN
Ans.
Section AB:
a + a MB = 0;
-15.45(2.5) + T(2) + 15(2) = 0
T = 4.32 kN
Ans.
139
0.5 m
2m
1m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–22. Determine the resultant forces at the pins A, B, and
C of the three-hinged arched roof truss.
4 kN
3 kN
2 kN
4 kN
5 kN
B
5m
A
C
3m
3m
1m1m
Member AB:
a + a MA = 0;
Bx(5) + By(8) - 2(3) - 3(4) - 4(5) = 0
Member BC:
a + a MC = 0;
-Bx(5) + By(7) + 5(2) + 4(5) = 0
Soving,
By = 0.533 k,
Member AB:
+
: a Fx = 0;
+ c a Fy = 0;
Bx = 6.7467 k
A x = 6.7467 k
A y - 9 + 0.533 = 0
A y = 8.467 k
Member BC:
+
: a Fx = 0;
+ c a Fy = 0;
Cx = 6.7467 k
Cy - 9 + 0.533 = 0
Cy = 9.533 k
FB = 2(0.533)2 + (6.7467)2 = 6.77 k
FA = 2(6.7467)2 + (8.467)2 = 10.8 k
FC = 2(6.7467) + (9.533) = 11.7 k
2
2
Ans.
Ans.
Ans.
140
2m
3m
2m
© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–23. The three-hinged spandrel arch is subjected to the
loading shown. Determine the internal moment in the arch
at point D.
8 kN 8 kN
6 kN 6 kN
3 kN
3 kN
2m 2m 2m
4 kN
4 kN
2m 2m 2m
B
D
5m
3m
A
C
Member AB:
a + a MA = 0;
3m
Bx(5) + By(8) - 8(2) - 8(4) - 4(6) = 0
Bx + 1.6By = 14.4
(1)
Member CB:
a + a MC = 0;
B(y)(8) - Bx(5) + 6(2) + 6(4) + 3(6) = 0
-Bx + 1.6By = -10.8
(2)
Soving Eqs. (1) and (2) yields:
By = 1.125 kN
Bx = 12.6 kN
Segment BD:
a + a MD = 0;
-MD + 12.6(2) + 1.125(5) - 8(1) - 4(3) = 0
MD = 10.825 kN # m = 10.8 kN # m
Ans.
141
5m
8m
© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*5–24. The tied three-hinged arch is subjected to the
loading shown. Determine the components of reaction at A
and C, and the tension in the rod
5k
3k
4k
B
15 ft
C
A
6 ft
Entire arch:
a + a MA = 0;
+ c a Fy = 0;
+
: a Fx = 0;
8 ft
6 ft
10 ft
10 ft
-4(6) - 3(12) - 5(30) + Cy(40) = 0
Cy = 5.25 k
Ans.
A y + 5.25 - 4 - 3 - 5 = 0
A y = 6.75 k
Ans.
Ax = 0
Ans.
Section BC:
a + a MB = 0;
-5(10) - T(15) + 5.25(20) = 0
T = 3.67 k
Ans.
5–25. The bridge is constructed as a three-hinged trussed
arch. Determine the horizontal and vertical components of
reaction at the hinges (pins) at A, B, and C. The dashed
member DE is intended to carry no force.
60 k
40 k 40 k
20 k 20 k
D 10 ft E
B
100 ft
h1
A
h2
h3
C
30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft
Member AB:
a + a MA = 0;
Bx(90) + By(120) - 20(90) - 20(90) - 60(30) = 0
9Bx + 12By = 480
(1)
Member BC:
a + a MC = 0;
-Bx(90) + By(120) + 40(30) + 40(60) = 0
-9Bx + 12By = -360
(2)
Soving Eqs. (1) and (2) yields:
Bx = 46.67 k = 46.7 k
By = 5.00 k
142
Ans.
© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–25. Continued
Member AB:
+
: a Fx = 0;
+ c a Fy = 0;
A x - 46.67 = 0
A x = 46.7 k
Ans.
Ay - 60 - 20 - 20 + 5.00 = 0
A y = 95.0 k
Member BC:
+
: a Fx = 0;
+ c a Fy = 0;
Ans.
-Cx + 46.67 = 0
Cx = 46.7 k
Ans.
Cy - 5.00 - 40 - 40 = 0
Cy = 85 k
Ans.
5–26. Determine the design heights h1, h2, and h3 of the
bottom cord of the truss so the three-hinged trussed arch
responds as a funicular arch.
60 k
40 k 40 k
20 k 20 k
D 10 ft E
B
100 ft
h1
A
h2
h3
C
30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft 30 ft
y = -Cx 2
-100 = -C(120)2
C = 0.0069444
Thus,
y = -0.0069444x2
y1 = -0.0069444(90 ft)2 = -56.25 ft
y2 = -0.0069444(60 ft)2 = -25.00 ft
y3 = -0.0069444(30 ft)2 = -6.25 ft
h1 = 100 ft - 56.25 ft = 43.75 ft
Ans.
h2 = 100 ft - 25.00 ft = 75.00 ft
Ans.
h3 = 100 ft - 6.25 ft = 93.75 ft
Ans.
143
© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–27. Determine the horizontal and vertical components
of reaction at A, B, and C of the three-hinged arch. Assume
A, B, and C are pin connected.
4k
B
5 ft
Member AB:
a + a MA = 0;
2 ft
3k
A
8 ft
Bx(5) + By(11) - 4(4) = 0
C
Member BC:
a + a MC = 0;
4 ft
7 ft
10 ft
5 ft
-Bx(10) + By(15) + 3(8) = 0
Soving,
By = 0.216 k, Bx = 2.72 k
Member AB:
+
: a Fx = 0;
+ c a Fy = 0;
Ans.
A x - 2.7243 = 0
A x = 2.72 k
Ans.
A y - 4 + 0.216216 = 0
A y = 3.78 k
Member BC:
+
: a Fx = 0;
+ c a Fy = 0;
Ans.
Cx + 2.7243 - 3 = 0
Cx = 0.276 k
Ans.
Cy - 0.216216 = 0
Cy = 0.216 k
Ans.
*5–28. The three-hinged spandrel arch is subjected to the
uniform load of 20 kN>m. Determine the internal moment
in the arch at point D.
20 kN/m
B
Member AB:
a + a MA = 0;
5m
D
Bx(5) + By(8) - 160(4) = 0
3m
A
C
Member BC:
a + a MC = 0;
3m
-Bx(5) + By(8) + 160(4) = 0
Solving,
Bx = 128 kN,
By = 0
Segment DB:
a + a MD = 0;
128(2) - 100(2.5) - MD = 0
MD = 6.00 kN # m
Ans.
144
5m
8m
© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5–29. The arch structure is subjected to the loading
shown. Determine the horizontal and vertical components
of reaction at A and D, and the tension in the rod AD.
2 k/ft
B
3 ft
3k
E
C
3 ft
A
D
8 ft
+
: a Fx = 0;
-A x + 3 k = 0;
a + a MA = 0;
-3 k (3 ft) - 10 k (12 ft) + Dy(16 ft) = 0
Ax = 3 k
Ans.
Dy = 8.06 k
+ c a Fy = 0;
Ans.
A y - 10 k + 8.06 k = 0
Ay = 1.94 k
a + a MB = 0;
Ans.
8.06 k (8 ft) - 10 k (4 ft) - TAD(6 ft) = 0
TAD = 4.08 k
Ans.
145
4 ft
4 ft
6 ft