HW4 Solutions

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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

B

+ : a Fx = 0; + c a Fy = 0;

xB

40 -

2x2B + 25 5

2x2B + 25

xB – 3

TAB -

8

TAB -

13xB - 15 2(xB - 3)2 + 64

2(xB - 3)2 + 64

2(xB - 3)2 + 64

TBC = 0 8 ft

TBC = 0 C

(1) 2 ft

TBC = 200

D

5

3

4

At C

30 lb

3 ft

+ : a Fx = 0; + c a Fy = 0;

P

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) 63 5P = 18 102 P = 71.4 lb

Ans. 128

30 lb

P

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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 =

2(9.42857)(14)

2FHh 2

=

L

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 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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5–18. 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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*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;

+ : a Fx = 0;

10 ft

10 ft

-4(6) - 3(12) - 5(30) + Cy(40) = 0 Cy = 5.25 k

+ c a Fy = 0;

8 ft

6 ft

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.

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5–25.

Continued

Member AB: + : a Fx = 0;

A x - 46.67 = 0 A x = 46.7 k

+ c a Fy = 0;

Ans.

Ay - 60 - 20 - 20 + 5.00 = 0 A y = 95.0 k

Member BC: + : a Fx = 0;

Ans.

-Cx + 46.67 = 0 Cx = 46.7 k

+ c a Fy = 0;

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

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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 2 ft

3k 5 ft

Member AB: a + a MA = 0;

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;

Ans.

A x - 2.7243 = 0 A x = 2.72 k

+ c a Fy = 0;

Ans.

A y - 4 + 0.216216 = 0 A y = 3.78 k

Member BC: + : a Fx = 0;

Ans.

Cx + 2.7243 - 3 = 0 Cx = 0.276 k

+ c a Fy = 0;

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 3m

A

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

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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6–22. Where should the beam ABC be loaded with a 300 lb兾ft uniform distributed live load so it causes (a) the largest moment at point A and (b) the largest shear at D? Calculate the values of the moment and shear. Assume the support at A is fixed, B is pinned and C is a roller.

A

B D 8 ft

(a) (MA) max =

1 (36)( -16)(0.3) = -86.4 k # ft 2

(b) (VD) max = c(1)(8) +

C

8 ft

20 ft

Ans.

1 (1)(20) d(0.3) = 5.40 k 2

Ans.

6–23. The beam is used to support a dead load of 800 N兾m, a live load of 4 kN兾m, and a concentrated live load of 20 kN. Determine (a) the maximum positive (upward) reaction at B, (b) the maximum positive moment at C, and (c) the maximum negative shear at C. Assume B and D are pins.

A

C 4m

Referring to the influence line for the vertical reaction at B, the maximum positive reaction is 1 1 (By) max (+) = 1.5(20) + c (16 - 0)(1.5) d(4) + c (16 - 0)(1.5) d(0.8) 2 2 Ans.

= 87.6 kN

Referring to the influence line for the moment at C shown in Fig. b, the maximum positive moment is 1 1 (Mc) max (+) = 2(20) + c (8 - 0)(2) d (4) + c (8 - 0)(2) d(0.8) 2 2 1 + c (16 - 8)( -2) d (0.8) 2 = 72.0 kN # m

Ans.

Referring to the influence line for the shear at C shown in, the maximum negative shear is 1 (VC) max (-) = -0.5(20) + c (4 - 0)( -0.5) d(4) 2 1 1 + c (16 - 8)( -0.5) d(4) + c (4 - 0)( -0.5) d(0.8) 2 2 1 1 + c (8 - 4)(0.5) d(0.8) + c (16 - 8)( -0.5) d(0.8) 2 2 = -23.6 kN

Ans.

166

4m

D

E

B 4m

4m

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6–37. A uniform live load of 1.75 kN兾m and a single concentrated live force of 8 kN are placed on the floor beams. If the beams also support a uniform dead load of 250 N兾m, determine (a) the maximum negative shear in panel BC of the girder and (b) the maximum positive moment at B.

1.5 m

3m

1.5 m D

A

By referring to the influence line for the shear in panel BC shown in Fig. a, the maximum negative shear is (VBC) max (-) = -0.6667(8) 1 + c (4.5 - 0)(-0.6667) d(1.75 + 0.25) 2 1 + c (6 - 4.5)(0.6667) d(0.25) 2 = -8.21 kN

Ans.

By referring to the influence line for the moment at B shown in Fig. b, the maximum positive moment is 1 (MB) max (+) = 1(8) + c (4.5 - 0)(1) d (1.75 + 0.25) 2 1 + c (6 - 4.5)(-1) d(0.25) 2 = 12.3 kN # m

Ans.

177

B

C

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6–57. Draw the influence line for the force in member CD, and then determine the maximum force (tension or compression) that can be developed in this member due to a uniform live load of 800 lb兾ft which acts along the bottom cord of the truss.

H

G

F

10 ft E

A B 10 ft

C 10 ft

D 10 ft

10 ft

Referring to the influence line for the force of member CD, the maximum tensile force is 1 (FCD) max (+) = c (40 - 0)(0.75) d(0.8) = 12.0 k (T) 2

Ans.

6–58. Draw the influence line for the force in member CF, and then determine the maximum force (tension or compression) that can be developed in this member due to a uniform live load of 800 lb兾ft which is transmitted to the truss along the bottom cord.

H

G

F

10 ft E

A B 10 ft

Referring to the influence line for the force in member CF, the maximum tensile and compressive force are 1 (FCF) max (+) = c (26.67 - 0)(0.7071) d(0.8) = 7.54 k (T) 2

Ans.

1 (FCF) max (-) = c (40 - 26.67)(-0.3536) d(0.8) 2 = -1.89 k = 1.89 k (C)

184

C 10 ft

D 10 ft

10 ft

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6–67. Draw the influence line for the force in member BC of the bridge truss. Determine the maximum force (tension or compression) that can be developed in the member due to a 5-k truck having the wheel loads shown. Assume the truck can travel in either direction along the center of the deck, so that half the load shown is transferred to each of the two side trusses. Also assume the members are pin connected at the gusset plates.

8 ft 3k 2k

J

I

H

15 ft B

20 ft

3(1) + 2(0.867) = 2.37 k (T) 2

C

20 ft

D

20 ft

20 ft

Ans.

*6–68. Draw the influence line for the force in member IC of the bridge truss. Determine the maximum force (tension or compression) that can be developed in the member due to a 5-k truck having the wheel loads shown. Assume the truck can travel in either direction along the center of the deck, so that half the load shown is transferred to each of the two side trusses. Also assume the members are pin connected at the gusset plates.

8 ft 3k 2k

J

I

H

G

F 15 ft E

B

A

20 ft

(FIC) max =

F

E A

(FBC) max =

G

3(0.833) + 2(0.667) = 1.92 k (T) 2

Ans.

190

C

20 ft

D

20 ft

20 ft

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6–78. Determine the absolute maximum moment in the girder due to the loading shown.

10 k

8k

3 k4 k

3 ft 2 ft 2 ft 25 ft

Referring to Fig. a, the location of FR for the moving load is + T FR = a Fy;

FR = 10 + 8 + 3 + 4 = 25 k

a + FRx = a MC;

-25x = -8(3) - 3(5) - 4(7) x = 2.68 ft.

Assuming that the absolute maximum moment occurs under 10 k load, Fig. b, a + a MB = 0;

4(6.84) + 3(8.84) + 8(10.84) + 10(13.84) - Ay(25) = 0 Ay = 11.16 k

Referring to Fig. c, a + a MS = 0;

MS - 11.16(11.16) = 0 MS = 124.55 k # ft

Assuming that the absolute maximum moment occurs under the 8-k force, Fig. d, a + a MB = 0;

4(8.34) + 3(10.34) + 8(12.34) + 10(15.34) - Ay(25) = 0 Ay = 12.66 k

Referring to Fig. e, a + a MS = 0;

MS + 10(3) - 12.66(12.66) = 0 MS = 130.28 k # ft = 130 k # ft (Abs. Max.)

196

Ans.