Chapter 5

© 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.

5–1. 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


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


5–3. Determine the tension in each cable segment and the distance yD.

A

yD D

7m

B

Joint B:

+ : a Fx = 0;


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


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


*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


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


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


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


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


*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


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 ! 48 ft

C

F

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: 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


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


*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


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, 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


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


*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


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


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


*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.


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


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


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

Chapter 5

Recommend Documents