hw4 solutions

128

At B

(1)

At C

(2)

Solving Eqs. (1) & (2)

Ans.xB = 4.36 ft

13xB - 15

30 - 2xB=

200102

30 - 2xB

2(xB - 3)2+ 64

TBC = 102

8

2(xB - 3)2+ 64

TBC +

2

213 TCD -

35

(30) = 0+ caFy = 0;

45

(30) +

xB - 3

2(xB - 3)2+ 64

TBC -

3

213 TCD = 0:+ aFx = 0;

13xB - 15

2(xB - 3)2+ 64

TBC = 200

5

2x2B + 25

TAB -

8

2(xB - 3)2+ 64

TBC = 0+ caFy = 0;

40 -

xB

2x2B + 25

TAB -

xB – 3

2(xB - 3)2+ 64

TBC = 0:+ aFx = 0;

*5–4. The cable supports the loading shown. Determine thedistance the force at point B acts from A. Set .P = 40 lbxB

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5 ft

2 ft

3 ft30 lb

D

C

B

A

xB

5

43

8 ft

P

5 ft

2 ft

3 ft30 lb

D

C

B

A

xB

5

43

8 ft

P

At B

(1)

At C

(2)

Solving Eqs. (1) & (2)

Ans.P = 71.4 lb

6318

=

5P

102

18

273 TBC = 102

8

273 TBC -

2

213 TCD -

35

(30) = 0+ caFy = 0;

45

(30) +

3

273 TBC -

3

213 TCD = 0:+ aFx = 0;

5P -

63

273 TBC = 0

5

261 TAB -

8

273 TBC = 0+ caFy = 0;

P -

6

261 TAB -

3

273 TBC = 0:+ aFx = 0;

5–5. The cable supports the loading shown. Determine themagnitude of the horizontal force P so that .xB = 6 ft

134


Entire structure:

a

(1)

Section ABD:

a

Using Eq. (1):

From Eq. 5–8:

From Eq. 5–11:

Ans.Tmax = woLA1 + a L

2hb2

= 0.11458(48)A1 + c 482(14)

d2 = 10.9 k

wo =

2FHh

L2 =

2(9.42857)(14)

482 = 0.11458 k>ft

FH = 9.42857 k

FH(14) - (Ay + Dy)(48) + 5(24) = 0+aMB = 0;

(Ay + Dy) = 5.25

4(36) + 5(72) + FH(36) - FH(36) - (Ay + Dy(96) = 0+aMC = 0;

5–13. The trusses are pin connected and suspended fromthe parabolic cable. Determine the maximum force in thecable when the structure is subjected to the loading shown.

4 k5 k

AF G H B

C

IJK

16 ft

4 @ 12 ft � 48 ft 4 @ 12 ft � 48 ft

D E

6 ft

14 ft

Member BC:

Member AB:

FBD 1:

a FH(1) - By(10) - 20(5) = 0+aMA = 0;

Ax = 0:+ aFx = 0;

Bx = 0:+ aFx = 0;

5–14. Determine the maximum and minimum tension inthe parabolic cable and the force in each of the hangers.Thegirder is subjected to the uniform load and is pin connectedat B.

A

D

BC

E

30 ft

9 ft1 ft

10 ft

10 ft

2 k/ ft

137


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

Integrate Eq. 5–2,

Divide by by Eq. 5–4, and use Eq. 5–3

At

At

At

Ans.

Ans.Tmax = 5.20 kN

Tmax =

FH

cos umax=

2598 cos 60°

= 5196 N

umax = 60°

y = (38.5x2+ 577x)(10- 3) m

x = 15 m, dy

dx= tan 60°; FH = 2598 N

dy

dx=

1FH

(200x + FH tan 30°)

y =

1FH

(100x2+ FH tan 30°x)

x = 0, dy

dx= tan 30°; C1 = FH tan 30°

x = 0, y = 0; C2 = 0

y =

1FH

(100x2+ C1x + C2)

dy

dx=

1FH

(200x + C1)

T sin u = 200x + C1

5–18. The cable AB is subjected to a uniform loading of. If the weight of the cable is neglected and the

slope angles at points A and B are and , respectively,determine the curve that defines the cable shape and themaximum tension developed in the cable.

60°30°200 N>m

15 m200 N/m

y

xA

B

60�

30�

142


Entire arch:

a

Ans.

Ans.

Ans.

Section BC:

a

Ans.T = 3.67 k

-5(10) - T(15) + 5.25(20) = 0+aMB = 0;

Ax = 0 :+ aFx = 0;

Ay = 6.75 k

Ay + 5.25 - 4 - 3 - 5 = 0+ caFy = 0;

Cy = 5.25 k

-4(6) - 3(12) - 5(30) + Cy(40) = 0+aMA = 0;

*5–24. The tied three-hinged arch is subjected to theloading shown. Determine the components of reaction at Aand C, and the tension in the rod

AC

B4 k

3 k5 k

6 ft 6 ft8 ft 10 ft 10 ft

15 ft

Member AB:

a

(1)

Member BC:

a

(2)

Soving Eqs. (1) and (2) yields:

Ans.By = 5.00 kBx = 46.67 k = 46.7 k

-9Bx + 12By = -360

-Bx(90) + By(120) + 40(30) + 40(60) = 0+aMC = 0;

9Bx + 12By = 480

Bx(90) + By(120) - 20(90) - 20(90) - 60(30) = 0+aMA = 0;

5–25. The bridge is constructed as a three-hinged trussedarch. Determine the horizontal and vertical components ofreaction at the hinges (pins) at A, B, and C. The dashedmember DE is intended to carry no force.

30 ft 30 ft30 ft 30 ft 30 ft

10 ftD E20 k20 k60 k 40 k40 k

B

A C

100 ft

30 ft

h1h2

h3

30 ft 30 ft

143


Member AB:

Ans.

Ans.

Member BC:

Ans.

Ans.Cy = 85 k

Cy - 5.00 - 40 - 40 = 0+ caFy = 0;

Cx = 46.7 k

-Cx + 46.67 = 0:+ aFx = 0;

Ay = 95.0 k

Ay - 60 - 20 - 20 + 5.00 = 0+ caFy = 0;

Ax = 46.7 k

Ax - 46.67 = 0:+ aFx = 0;

Thus,

Ans.

Ans.

Ans.h3 = 100 ft - 6.25 ft = 93.75 ft

h2 = 100 ft - 25.00 ft = 75.00 ft

h1 = 100 ft - 56.25 ft = 43.75 ft

y3 = -0.0069444(30 ft)2= -6.25 ft

y2 = -0.0069444(60 ft)2= -25.00 ft

y1 = -0.0069444(90 ft)2= -56.25 ft

y = -0.0069444x2

C = 0.0069444

-100 = -C(120)2

y = -Cx 2

5–26. Determine the design heights h1, h2, and h3 of thebottom cord of the truss so the three-hinged trussed archresponds as a funicular arch.

30 ft 30 ft30 ft 30 ft 30 ft

10 ftD E20 k20 k60 k 40 k40 k

B

A C

100 ft

30 ft

h1h2

h3

30 ft 30 ft

5–25. Continued

144


Member AB:

a

Member BC:

a

Soving,

Ans.

Member AB:

Ans.

Ans.

Member BC:

Ans.

Ans.Cy = 0.216 k

Cy - 0.216216 = 0+ caFy = 0;

Cx = 0.276 k

Cx + 2.7243 - 3 = 0:+ aFx = 0;

Ay = 3.78 k

Ay - 4 + 0.216216 = 0+ caFy = 0;

Ax = 2.72 k

Ax - 2.7243 = 0:+ aFx = 0;

Bx = 2.72 kBy = 0.216 k,

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

Bx(5) + By(11) - 4(4) = 0+aMA = 0;

5–27. Determine the horizontal and vertical componentsof reaction at A, B, and C of the three-hinged arch. AssumeA, B, and C are pin connected.

5 ft

4 ft 7 ft 10 ft 5 ft

8 ft

A

C

B

4 k

3 k 2 ft

Member AB:

a

Member BC:

a

Solving,

Segment DB:

a

Ans.MD = 6.00 kN # m

128(2) - 100(2.5) - MD = 0 +aMD = 0;

By = 0Bx= 128 kN,

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

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

*5–28. The three-hinged spandrel arch is subjected to theuniform load of . Determine the internal momentin the arch at point D.

20 kN>m

A C

D

B

5 m

3 m

3 m 8 m

20 kN/m

5 m

166


(a) Ans.

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

12

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

(MA) max =

12

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

Referring to the influence line for the vertical reaction at B, the maximum positivereaction is

Ans.

Referring to the influence line for the moment at C shown in Fig. b, the maximumpositive moment is

Ans.

Referring to the influence line for the shear at C shown in, the maximum negativeshear is

Ans.= -23.6 kN

+ c12

(8 - 4)(0.5) d(0.8) + c12

(16 - 8)(-0.5) d(0.8)

+ c12

(16 - 8)(-0.5) d(4) + c12

(4 - 0)(-0.5) d(0.8)

(VC) max (-) = -0.5(20) + c12

(4 - 0)(-0.5) d(4)

= 72.0 kN # m

+ c12

(16 - 8)(-2) d(0.8)

(Mc) max (+) = 2(20) + c12

(8 - 0)(2) d(4) + c12

(8 - 0)(2) d(0.8)

= 87.6 kN

(By) max (+) = 1.5(20) + c12

(16 - 0)(1.5) d(4) + c12

(16 - 0)(1.5) d(0.8)

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

A B C

8 ft 8 ft 20 ft

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

EBC DA

177


By referring to the influence line for the shear in panel BC shown in Fig. a, themaximum negative shear is

Ans.

By referring to the influence line for the moment at B shown in Fig. b, the maximumpositive moment is

Ans.= 12.3 kN # m

+ c12

(6 - 4.5)(-1) d(0.25)

(MB) max (+) = 1(8) + c12

(4.5 - 0)(1) d(1.75 + 0.25)

= -8.21 kN

+ c12

(6 - 4.5)(0.6667) d(0.25)

+ c12

(4.5 - 0)(-0.6667) d(1.75 + 0.25)

(VBC) max (-) = -0.6667(8)

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

3 m 1.5 m1.5 m

AB

D

184


Referring to the influence line for the force of member CD, the maximum tensileforce is

Ans.= 12.0 k (T)(FCD) max (+) = c12

(40 - 0)(0.75) d(0.8)

6–57. Draw the influence line for the force in memberCD, and then determine the maximum force (tension orcompression) that can be developed in this member due toa uniform live load of 800 lb�ft which acts along the bottomcord of the truss.

Referring to the influence line for the force in member CF, the maximum tensile and compressive force are

Ans.

= -1.89 k = 1.89 k (C)

(FCF) max (-) = c12

(40 - 26.67)(-0.3536) d(0.8)

= 7.54 k (T)(FCF) max (+) = c12

(26.67 - 0)(0.7071) d(0.8)

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

A E

B

H

CC

G

D

F

10 ft 10 ft 10 ft 10 ft

10 ft

A E

B

H

CC

G

D

F

10 ft 10 ft 10 ft 10 ft

10 ft

190


Ans.(FBC) max =

3(1) + 2(0.867)

2= 2.37 k (T)

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

Ans.(FIC) max =

3(0.833) + 2(0.667)

2= 1.92 k (T)

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

J I H G

DCBE

F

15 ft

2 k3 k8 ft

A

20 ft 20 ft 20 ft 20 ft

J I H G

DCBE

F

15 ft

2 k3 k8 ft

A

20 ft 20 ft 20 ft 20 ft

196


Referring to Fig. a, the location of FR for the moving load is

a

.

Assuming that the absolute maximum moment occurs under 10 k load, Fig. b,

a

Referring to Fig. c,

a

Assuming that the absolute maximum moment occurs under the 8-k force, Fig. d,

a

Referring to Fig. e,

a

Ans.MS = 130.28 k # ft = 130 k # ft (Abs. Max.)

MS + 10(3) - 12.66(12.66) = 0+ aMS = 0;

Ay = 12.66 k

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

MS = 124.55 k # ft

MS - 11.16(11.16) = 0+ aMS = 0;

Ay = 11.16 k

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

x = 2.68 ft

-25x = -8(3) - 3(5) - 4(7)+ FRx = aMC;

FR = 10 + 8 + 3 + 4 = 25 k+ T FR = aFy;

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

25 ft

10 k8 k

3 k

2 ft2 ft3 ft

4 k

hw4 solutions

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