# Solutions (8th ed structural analysis) chapter 5

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5–1. Determine the tension in each segment of the cable
and the cable’s total length.

A
4 ft
D

7 ft

B

Equations of Equilibrium:
Joint B:

50 lb
4 ft

+

: a Fx = 0;

FBC cos u - FBA a

+ c a Fy = 0;
Joint C:

C

Applying method of joints, we have

FBA a

7

265

4
265

b = 0

b - FBC sin u - 50 = 0

[1]
[2]

+
: a Fx = 0;

FCD cos f - FBC cos u = 0

[3]

+ c a Fy = 0;

FBC sin u + FCD sin f - 100 = 0

[4]

Geometry:
sin u =
sin f =

y

cos u =

2y2 + 25

3 + y

2

2y + 6y + 18

cos f =

5
2y2 + 25
3

2

2y + 6y + 18

Substitute the above results into Eqs. [1], [2], [3] and [4] and solve. We have
FBC = 46.7 lb

FBA = 83.0 lb

Ans.

FCD = 88.1 lb

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

125

Ans.

5 ft

3 ft
100 lb

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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;

Joint C:

TCD a

4
217

B

b(4) + TCD a

1
217

b (2) - 6(4) - 4(1) = 0

TCD = 6.414 kN = 6.41 kN(Max)

1m
4 kN

Ans.

Referring to the FBD in Fig. b,

+
: a Fx = 0;

+ c a Fy = 0;
Solving,

6.414 a
6.414 a

1
217
4

217

b - TBC cos u = 0
b - 6 - TBC sin u = 0

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°
Ans.

= 2.429 m = 2.43 m

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:

+
: a Fx = 0;

+ c a Fy = 0;
Solving,

TBC a
TAB a

5
229
7

265

b - TAB a
b - TBC a

TAB = 2.986 kN = 2.99 kN
Joint C:

2m

Referring to the FBD in Fig. a,
4
265
2

229

2 kN

b = 0

C
4m

b - 2 = 0

TBC = 1.596 kN = 1.60 kN

+ c a Fy = 0;
Solving,

TCD cos u - 1.596 a
TCD sin u + 1.596 a

5
229
2

229

3m
4 kN

Ans.

Referring to the FBD in Fig. b,

+
: a Fx = 0;

5m

b = 0
b - 4 = 0
Ans.

TCD = 3.716 kN = 3.72 kN
u = 66.50°
From the geometry,
yD + 3 tan u = 9

Ans.

yD = 9 - 3 tan 66.50° = 2.10 m

127

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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

40 -

2x2B + 25

5

2x2B + 25

TAB -

2(xB - 3)2 + 64

+
: a Fx = 0;

+ c a Fy = 0;

2(xB - 3)2 + 64

P

TBC = 0
8 ft

TBC = 0
C

(1) 2 ft

TBC = 200

D

5

3

4

30 lb

3 ft

xB - 3
4
3
(30) +
TBC TCD = 0
5
2(xB - 3)2 + 64
213
8

2

2(xB - 3) + 64
30 - 2xB

Solving Eqs. (1) & (2)

2(xB - 3)2 + 64
8

TAB -

13xB - 15

At C

xB – 3

2(xB - 3)2 + 64

TBC +

2

213

TCD -

3
(30) = 0
5

TBC = 102

(2)

13xB - 15
200
=
30 - 2xB
102
Ans.

xB = 4.36 ft

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 At C

+
: a Fx = 0;

+ c a Fy = 0;

261

TAB -

TAB 63
273

8
273

3
273

TBC = 0
8 ft

TBC = 0
C

(1) 2 ft

TBC = 0

8

273
273

TBC -

D

3

5
4

3 ft

4
3
3
(30) +
TBC TCD = 0
5
273
213

18

Solving Eqs. (1) & (2)

6

2

213

TCD -

3
(30) = 0
5

TBC = 102

(2)

63
5P
=
18
102
Ans.

P = 71.4 lb
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

1.5 m

A

E
B

1m

C

D

5 kN

P1
2m

Method of Joints:
Joint B:

+
: a Fx = 0;

+ c a Fy = 0;

FBC a
FAB a

4
217

b - FAB a

2
b = 0
2.5

[1]

1.5
1
b - FBC a
b - 5 = 0
2.5
217

[2]

Solving Eqs. [1] and [2] yields
FBC = 10.31 kN

FAB = 12.5 kN

Joint C:

+
: a Fx = 0;

+ c a Fy = 0;
Joint D:

+
: a Fx = 0;

+ c a Fy = 0;

FCD - 10.31 a
10.31 a
FDE a
FDE a

1

217

4
217

b = 0

FCD = 10.00 kN
Ans.

b - P1 = 0 P1 = 2.50 kN

4
222.25
25

222.25

b - 10 = 0

[1]

b - P2 = 0

[2]

Solving Eqs. [1] and [2] yields
Ans.

P2 = 6.25 kN
FDE = 11.79 kN
Thus, the maximum tension in the cable is

Ans.

Fmax = FAB = 12.5 kN

129

4m

P2
5m

4m

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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
L 2
15 2
TB = Tmax = woL 1 + a b = 500(15) 1 + a
b = 10 280.5 lb = 10.3 k
A
2h
A
2(8)

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

y =

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 =

15 ft

10 ft

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 =

(16 kN>m)(10 m)2
woL2
=
= 400 kN
2h
2(2 m)

Ans.

And
Tmax = 2FH2 + (woL)2

= 24002 + [16(10)]2
= 430.81 kN

Ans.

= 431 kN

131

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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 wdxb dx
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
Ans.

w = 51.9 lb>ft

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°
Ans.

Tmin = FH = 13.0 kip

132

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*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

woL2 ds
a b
8h dx

T =

dx
b,
ds

then

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 =

sallow

1
2

woL2 (ds)2
c
d
dx
L0 8h

dx + dy
dx2 + dy2
dy 2
ds
=
= c
ddx
=
c1
+
a
b ddx
dx
dx
dx
dx2
2

=

2

2

1
2

dy 2
woL2
c 1 + a b ddx
dx
L0 4hsallow
1

=
=

2
woL2
h2x2
c 1 + 64 a 4 b ddx
4hsallow L0
L

woL2 L
woL2 L
8h2
16 h
c +
c +
d =
a bd
4hsallow 2
3L
8 sallow h
3 L

Require,

woL2
dV
L
16
=
c- 2 +
d = 0
dh
8sallow h
3L

Ans.

h = 0.433 L

133

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5–13. The trusses are pin connected and suspended from
the parabolic cable. Determine the maximum force in the

D

E

14 ft
6 ft

K

J

I

16 ft
A
G H B
5k
4 @ 12 ft ϭ 48 ft

C

F

4k
4 @ 12 ft ϭ 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:
L 2
48 2
Tmax = woL 1 + a b = 0.11458(48) 1 + c
d = 10.9 k
A
2h
A
2(14)

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
1 ft

Member BC:

+
: a Fx = 0;

2 k/ft

Bx = 0
A
10 ft

Ax = 0

FBD 1:
a + a MA = 0;

9 ft
D

10 ft

Member AB:

+
: a Fx = 0;

Ans.

FH(1) - By(10) - 20(5) = 0

134

C

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 ,

Ans.

FH = Fmin = 100 k

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,
L 2
30 2
Fmax = woL 1 + a b = 2(30) 1 + a
b
A
2h
A
2(9)

Ans.

Fmax = 117 k

Each hanger carries 5 ft of wo.
Ans.

T = (2 k>ft)(5 ft) = 10 k

5–15. Draw the shear and moment diagrams for the pinconnected girders AB and BC. The cable has a parabolic
shape.
a + a MA = 0;

E
1 ft

T(5) + T(10) + T(15) + T(20) + T(25)

9 ft
D

2 k/ft

10 ft

+ T(30) + T(35) + Cy(40) – 80(20) = 0
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

C

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

Ans.

wo = 69.24 kN>m = 69.2 kN>m

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

Ans.

And,
Tmax = 2F2H + (woL)2

= 262502 + [60(50)]2
= 6932.71 kN

Ans.

= 6.93 MN

136

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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
60Њ

A

Here the boundary conditions are different from those in the text.

30Њ
x

Integrate Eq. 5–2,
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°
Ans.

Tmax = 5.20 kN

137

200 N/ m

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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

+
: a Fx = 0;

Bx = 0

a + a MA = 0;

3 kN

(Member BC)

FF(12) - FF(9) - By(8) - 3(4) = 0

+
: a Fx = 0;

A x = 0 (Member AB)

a + a MC = 0;

-FF(12) + FF(9) - By(8) + 5(6) = 0
(2)

-3FF - By(8) = -30
Soving Eqs. (1) and (2),
By = 1.125 kN ,

Ans.

FF = 7.0 kN

From Eq. 5–8.
2FHh
L

2

=

2(7)(3)
82

= 0.65625 kN>m

From Eq. 5–11,
L 2
8 2
Tmax = woL 1 + a b = 0.65625(8) 1 + a
b
A
2h
A
2(3)

Ans.

Tmax = TE = TD = 8.75 kN

Ans.

T = 0.65625(2) = 1.3125 kN = 1.31 kN

138

5 kN

2m 2m 2m 2m 2m 2m 2m 2m

(1)

3FF - By(8) = 12

wo =

C

B

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*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

Member ABC:

a + a MA = 0;

5 kN

3 kN

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
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

Ans.

Cy = 9.545 kN = 9.55 kN
+ c a Fy = 0;

9.545 - 15 - 10 + A y = 0
Ans.

A y = 15.45 kN = 15.5 kN
Section AB:

a + a MB = 0;

-15.45(2.5) + T(2) + 15(2) = 0
Ans.

T = 4.32 kN

139

0.5 m

2m

1m

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

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;

3m
1m1m

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
2

2

FC = 2(6.7467) + (9.533) = 11.7 k

140

Ans.
Ans.
Ans.

2m

3m

2m

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
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
A

5m
3m

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

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
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
Ans.

Cy = 5.25 k
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
Ans.

T = 3.67 k

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
A

Bx(90) + By(120) - 20(90) - 20(90) - 60(30) = 0
9Bx + 12By = 480

(1)

Member BC:

a + a MC = 0;

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;

h1

-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.

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
Ans.

A x = 46.7 k
Ay - 60 - 20 - 20 + 5.00 = 0

Ans.

A y = 95.0 k
Member BC:
+
: a Fx = 0;
+ c a Fy = 0;

-Cx + 46.67 = 0
Ans.

Cx = 46.7 k
Cy - 5.00 - 40 - 40 = 0

Ans.

Cy = 85 k

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
A

h1

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

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;

8 ft

Bx(5) + By(11) - 4(4) = 0
C

Member BC:

a + a MC = 0;

2 ft

3k

A

4 ft

7 ft

10 ft

5 ft

-Bx(10) + By(15) + 3(8) = 0

Soving,
Ans.

By = 0.216 k, Bx = 2.72 k
Member AB:
+
: a Fx = 0;
+ c a Fy = 0;

A x - 2.7243 = 0
Ans.

A x = 2.72 k
A y - 4 + 0.216216 = 0

Ans.

A y = 3.78 k
Member BC:
+
: a Fx = 0;
+ c a Fy = 0;

Cx + 2.7243 - 3 = 0
Ans.

Cx = 0.276 k
Cy - 0.216216 = 0

Ans.

Cy = 0.216 k

*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;

D
A

Bx(5) + By(8) - 160(4) = 0

5m
3m

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

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 + a MA = 0;
+ c a Fy = 0;
a + a MB = 0;

-A x + 3 k = 0;

Ans.

Ax = 3 k

-3 k (3 ft) - 10 k (12 ft) + Dy(16 ft) = 0
Ans.

Dy = 8.06 k
A y - 10 k + 8.06 k = 0

Ans.

Ay = 1.94 k
8.06 k (8 ft) - 10 k (4 ft) - TAD(6 ft) = 0

Ans.

145

4 ft

4 ft

6 ft

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