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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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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;

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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*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

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

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

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;

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

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

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

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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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 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

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°

Ans.

Tmin = FH = 13.0 kip

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

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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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

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

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

*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

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

Ans.

And,

Tmax = 2F2H + (woL)2

= 262502 + [60(50)]2

= 6932.71 kN

Ans.

= 6.93 MN

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

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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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

+

: 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

Load on each hanger,

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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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

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

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

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

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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

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

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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.

5–23. 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

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

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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.

*5–24. 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

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.

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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.

5–25. 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

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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

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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.

5–29. 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 + 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.

TAD = 4.08 k

145

4 ft

4 ft

6 ft

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

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

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;

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

*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

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

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

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;

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

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

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

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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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 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

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°

Ans.

Tmin = FH = 13.0 kip

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

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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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

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

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

*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

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

Ans.

And,

Tmax = 2F2H + (woL)2

= 262502 + [60(50)]2

= 6932.71 kN

Ans.

= 6.93 MN

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

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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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

+

: 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

Load on each hanger,

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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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

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

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

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

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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

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

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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.

5–23. 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

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

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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.

*5–24. 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

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.

© 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

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

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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.

5–27. 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

© 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 + 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.

TAD = 4.08 k

145

4 ft

4 ft

6 ft

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