Solutions_Chapter_4(WǪ).pdf

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4–1. The A992 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 60 mm2, determine the displacement of B and A, Neglect the size of the couplings at B, C, and D.

D 0.75 m C 60⬚

60⬚

1.50 m 3.30 kN

3.30 kN B 5

3

3

3

4

3

10.4(10 )(1.50) 16.116(10 )(0.75) PL + dB = © = -6 9 AE 60(10 )(200)(10 ) 60(10 - 6)(200)(109)

2 kN

= 0.00231 m = 2.31 mm

Ans.

5 4

A

0.50 m 2 kN

8 kN

3

dA = dB +

8(10 )(0.5) 60(10 - 6)(200)(109)

= 0.00264 m = 2.64 mm

Ans.

Ans:

dB = 2.31 mm, dA = 2.64 mm 183


Aluminum Eal = 70 GPa

*4–4. Determine the displacement of B with respect to C of the composite shaft in Prob. 4–3.

AAB = 58 mm2

Copper Ecu = 126 GPa ABC = 77 mm2

A

B 450 mm

dB>C =

PL = AE

( - 23)(300) (77)(126)

= - 0.7111 mm

Ans.

The negative sign indicates end B moves towards end C.

186

16 kN

ACD = 39 mm2 8 kN

16 kN

9 kN

Steel Est = 200 GPa

C

300 mm

8 kN 400 mm

7 kN D


4–7. If P1 = 225 kN and P2 = 675 kN, determine the vertical displacement of end A of the high strength precast concrete column.

P1

P1 A

3m

a P2

a P2

150 mm 150 mm Section a-a

B

Internal Loading: The normal forces developed in segments AB and BC are shown 3 m on the free-body diagrams of these segments in Figs. a and b, respectively. Displacement: The cross-sectional area of segments AB and BC are AAB = 150(150) = 22500 mm2 and ABC = 250(250) = 62500 mm2

dA>C = ©

b

b C

250 mm 250 mm Section b-b

PAB LAB PBC LBC PL = + AE AAB Econ ABC Econ

(- 450)(3)(1000) +

= (22500)(29.4)

(- 1800)(3)(1000) (62500)(29.4)

= - 4.979 mm

Ans.

The negative sign indicates that end A is moving towards C.

Ans: dA = - 4.979 mm 189


4–9. The assembly consists of two 10-mm diameter red brass C83400 copper rods AB and CD, a 15-mm diameter 304 stainless steel rod EF, and a rigid bar G. If P = 5 kN, determine the horizontal displacement of end F of rod EF.

300 mm A

450 mm B

P E

4P F

C

DG

P

Internal Loading: The normal forces developed in rods EF, AB, and CD are shown on the free-body diagrams in Figs. a and b. p Displacement: The cross-sectional areas of rods EF and AB are AEF = (0.0152) = 4 p 56.25(10 - 6)p m2 and AAB = (0.012) = 25(10 - 6)p m2. 4 dF = ©

PEF LEF PAB LAB PL = + AE AEF Est AAB Ebr 20(103)(450)

=

-6

5(103)(300) 9

56.25(10 )p(193)(10 )

+

25(10 - 6)p(101)(109) Ans.

= 0.453 mm The positive sign indicates that end F moves away from the fixed end.

Ans: dF = 0.453 mm 191


4–13. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm2 and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied.

C 300 N/m 1.5 m D A

B 2m

a+ ©MA = 0;

2m

1200(2) - TCB(0.6)(2) = 0 TCB = 2000 N

dB>C =

(2000)(2.5) PL = 0.0051835 = AE 14(10 - 6)(68.9)(109)

(2.5051835)2 = (1.5)2 + (2)2 - 2(1.5)(2) cos u u = 90.248° u = 90.248° - 90° = 0.2478° = 0.004324 rad

dD = u r = 0.004324(4000) = 17.3 mm

Ans.

Ans:

dD = 17.3 mm 195


20 kN

4–15. The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 0 at y = 0 to w = 3 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.

A y

B

Equation of Equilibrium: For entire post [FBD (a)] + c ©Fy = 0;

F + 3.00 - 20 = 0

w

2m

F

F = 17.0 kN

Ans.

Internal Force: FBD (b) + c ©Fy = 0;

-F(y) +

1 3y a b y - 20 = 0 2 2

3 F(y) = e y2 - 20 f kN 4 Displacement: L

dA>B =

2m F(y) dy 1 3 = a y2 - 20b dy A(y)E AE 4 L0 L0

=

2m y3 1 a - 20y b 2 AE 4 0

= -

38.0 kN # m AE 38.0(103)

= -

p 2 4 (0.06 )

13.1 (109)

= -1.026 A 10 - 3 B m Ans.

= -1.03 mm Negative sign indicates that end A moves toward end B.

Ans: F = 17.0 kN, dA>B = - 1.03 mm 197


*4–20. The A992 steel drill shaft of an oil well extends 3600 m into the ground. Assuming that the pipe used to drill the well is suspended freely from the derrick at A, determine the maximum average normal stress in each pipe segment and the elongation of its end D with respect to the fixed end at A. The shaft consists of three different sizes of pipe, AB, BC, and CD, each having the length, weight per unit length, and cross-sectional area indicated.

A 2

AAB = 1600 mm wAB = 50 N/m

B 2

ABC = 1125 mm wBC = 40 N/m ACD = 800 mm2 wCD = 30 N/m

sA =

50(1500) + 78000 P = = 95.625 MPa A 1600

Ans.

sB =

40(1500) + 18000 P = = 69.33 MPa A 1125

Ans.

sC =

30(600) P = = 22.5 MPa A 1800

Ans.

dD = ©

1500 m

600 1500 1500 (40x + 18000)dx (50x + 78000)dx P(x) dx 30x dx + + = 3 3 (1800)(203)(10 ) L0 (1125)(203)(10 ) L0 1600(203)(103) L A(x) E L0

= 1.2 m

Ans.

202

1500 m C D

600 m


r1 12 mm

4–26. Determine the elongation of the tapered A992 steel shaft when it is subjected to an axial force of 80 kN. Hint: Use the result of Prob. 4–23.

80 kN

r1 12 mm 80 kN

500 mm 100 mm

d = (2)

r2 50 mm

100 mm

PL1 PL2 + p E r2r1 AE

(2)(80)(100) =

80(500) +

p(203)(50)(12)

p(50 )2 (203)

= 0.0669 mm

Ans.

Ans: d = 0.0669 mm 208


P

*4–28. Bone material has a stress–strain diagram that can be defined by the relation s = E[P>(1 + kEP)], where k and E are constants. Determine the compression within the length L of the bone, where it is assumed the cross-sectional area A of the bone is constant.

L

P dx s = ; P = A dx

s = Ea

P b; 1 + kEP

P = A

Ea

dx b dx

1 + kE a

dx b dx P

PkE dx dx P + a b = Ea b A A dx dx PkE dx P = aE ba b A A dx L

d

L0

dx =

L0

P dx AE a1 -

Pk b A

PL PL AE = d = E(A - Pk) Pk a1 b A

Ans.

210


4–30. The weight of the kentledge exerts an axial force of P = 1500 kN on the 300-mm diameter high-strength concrete bore pile. If the distribution of the resisting skin friction developed from the interaction between the soil and the surface of the pile is approximated as shown, determine the resisting bearing force F for equilibrium. Take p0 = 180 kN>m. Also, find the corresponding elastic shortening of the pile. Neglect the weight of the pile.

P p0

12 m

F

Internal Loading: By considering the equilibrium of the pile with reference to its entire free-body diagram shown in Fig. a. We have + c ©Fy = 0;

F +

1 (180)(12) - 1500 = 0 2

F = 420 kN

Ans.

Also, p(y) =

180 y = 15y kN>m 12

The normal force developed in the pile as a function of y can be determined by considering the equilibrium of the sectional of the pile with reference to its free-body diagram shown in Fig. b. + c ©Fy = 0;

1 (15y)y + 420 - P(y) = 0 2

P(y) = (7.5y2 + 420) kN

p Displacement: The cross-sectional area of the pile is A = (0.32) = 0.0225p m2. 4 We have 12 m P(y)dy (7.5y2 + 420)(103)dy = 0.0225p(29.0)(109) L0 A(y)E L0 L

d =

12 m

=

L0

c 3.6587(10 - 6)y2 + 0.2049(10 - 3) d dy

= c1.2196(10 - 6)y3 + 0.2049(10 - 3)y d

12 m 0

= 4.566(10 - 3) m = 4.57 mm

Ans.

Ans: F = 420 kN, d = 4.57 mm 212


4–33. The steel pipe is filled with concrete and subjected to a compressive force of 80 kN. Determine the average normal stress in the concrete and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm. Est = 200 GPa, Ec = 24 GPa.

80 kN

500 mm

Pst + Pcon - 80 = 0

+ c ©Fy = 0;

(1)

dst = dcon Pst L p 2 4 (0.08

2

9

=

- 0.07 ) (200) (10 )

Pcon L p 2 4 (0.07 ) (24)

(109)

Pst = 2.5510 Pcon

(2)

Solving Eqs. (1) and (2) yields Pst = 57.47 kN sst =

scon =

Pst = Ast

Pcon = 22.53 kN 57.47 (103)

p 4

(0.082 - 0.072)

= 48.8 MPa

Ans.

22.53 (103) Pcon = 5.85 MPa = p 2 Acon 4 (0.07 )

Ans.

Ans: sst = 48.8 MPa, scon = 5.85 MPa 215


4–35. If column AB is made from high strength pre-cast concrete and reinforced with four 20 mm diameter A-36 steel rods, determine the maximum allowable floor loadings P. The allowable normal stress for the high strength concrete and the steel are (sallow)con = 18 MPa and (sallow)st = 170 MPa, respectively.

P

P

A

a

225 mm

a 225 mm

3m

Section a-a

B

Equation of Equilibrium: Referring to the free-body diagram of the cut part of the concrete column shown in Fig. a, + c ©Fy = 0;

Pcon + 4Pst - 2P = 0

(1)

Compatibility Equation: Since the steel bars and the concrete are firmly bonded, their deformation must be the same. Thus, dcon = dst Pcon(3)(1000) 2 p c 225(225) - a b (20) d 29.4 4

=

Pst(3)(1000) p 2 (20) 203 4

Pcon = 22.77Pst

(2)

Solving Eqs. (1) and (2), Pst = 0.0747P

Pcon = 1.70099 P 9

Allowable Normal Stress: (scon)allow =

Pcon ; Acon

18 =

1.7009P p 2 225(225) - 4 a b (20) 4

P = 522.45 kN = 522 kN (controls) (sst)allow =

Pst ; Ast

170 =

Ans.

0.0747P 2 p (20) 4 P = 714.591 kN

Ans: P = 522 kN 217


4–39. The load of 12600 N is to be supported by the two essentially vertical A-36 steel wires. If originally wire AB is 1500 mm long and wire AC is 1000 mm long, determine the cross-sectional area of AB if the load is to be shared equally between both wires. Wire AC has a cross-sectional area of 13 mm2.

B

C 1500 mm 1000 mm A

TAC = TAB =

12600 = 6300 kN 2 dAC = dAB

6300(1000)

6300(1500) =

13 (203)

AAB ( 203 )

AAB = 19.5 mm2

Ans.

Ans: AAB = 19.5 mm2 221


4–42. The 2014-T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. When no load is applied to the assembly, the gap between end C and the rigid support is 0.5 mm. Determine the support reactions when the axial force of 400 kN is applied.

D A 400 mm

400 kN B

A992 steel

800 mm

50 mm

a

a 25 mm

2014–T6 aluminum alloy

Section a–a C

Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, FD + (FC)al + (FC)st - 400(103) = 0

+ c ©Fy = 0;

(1)

Compatibility Equation: Using the method of superposition, Fig. b, dC = dP - dFC

(+ T)

0.5 = +

400(103)(400) 2

9

p(0.025 )(73.1)(10 )

- ≥

(FC)al (800) 2

9

+

[(FC)al + (FC)st](400)

p(0.025 )(73.1)(10 )

p(0.0252)(73.1)(109)

220.585(103) = 3(FC)al + (FC)st

¥ (2)

Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their deformation must be the same. Thus, (dBC)st = (dBC)al (FC)st (800) 2

2

9

=

p(0.05 - 0.025 )(200)(10 )

(FC)al (800) p(0.0252)(73.1)(109)

(FC)st = 8.2079(FC)al

(3)

Solving Eqs. (2) and (3), (FC)al = 19.681 kN

(FC)st = 161.54 kN

Substituting these results into Eq. (1), FD = 218.777 kN = 219 kN

Ans.

Also, FC = (FC)al + (FC)st = 19.681 + 161.54 = 181 kN

Ans.

Ans: FD = 219 kN, FC = 181 kN 224


4–45. The bolt has a diameter of 20 mm and passes through a tube that has an inner diameter of 50 mm and an outer diameter of 60 mm. If the bolt and tube are made of A-36 steel, determine the normal stress in the tube and bolt when a force of 40 kN is applied to the bolt. Assume the end caps are rigid.

160 mm

40 kN

40 kN 150 mm

Referring to the FBD of left portion of the cut assembly, Fig. a + ©F = 0; : x

40(103) - Fb - Ft = 0

(1)

Here, it is required that the bolt and the tube have the same deformation. Thus dt = db Ft(150)

p 2 4 (0.06

- 0.052) C 200(109) D

=

Fb(160) p 2 4 (0.02 )

C 200(109) D

Ft = 2.9333 Fb

(2)

Solving Eqs (1) and (2) yields Fb = 10.17 (103) N

Ft = 29.83 (103) N

Thus, sb =

10.17(103) Fb = 32.4 MPa = p 2 Ab 4 (0.02 )

st =

Ft = At

29.83 (103) p 2 4 (0.06

- 0.052)

Ans.

= 34.5 MPa

Ans.

Ans: sb = 32.4 MPa, st = 34.5 MPa 227


4–47. The support consists of a solid red brass C83400 copper post surrounded by a 304 stainless steel tube. Before the load is applied the gap between these two parts is 1 mm. Given the dimensions shown, determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials.

P A

1 mm

0.25 m

60 mm 80 mm

Require,

10 mm

dst = dbr + 0.001 Fst(0.25) 2

2

9

=

p[(0.05) - (0.04) ]193(10 )

Fbr(0.25) p(0.03)2(101)(109)

+ 0.001

0.45813 Fst = 0.87544 Fbr + 106

(1)

Fst + Fbr - P = 0

+ c ©Fy = 0;

(2)

Assume brass yields, then (Fbr)max = sg Abr = 70(106)(p)(0.03)2 = 197 920.3 N (Pg)br = sg>E =

70.0(106) 101(109)

= 0.6931(10 - 3) mm>mm

dbr = (eg)brL = 0.6931(10 - 3)(0.25) = 0.1733 mm < 1 mm Thus only the brass is loaded. P = Fbr = 198 kN

Ans.

Ans: P = 198 kN 229


4–51. The rigid bar supports the uniform distributed load kN/m.Determine Determinethe theforce force in in each each cable cable if each cable of 90 6 kip>ft. 3 mm 200 GPa. has a cross-sectional cross-sectional area areaof of36 andEE= = 31110 0.05 in22, ,and 2 ksi.

C

ft 26 m

6 kip/ft 90 kN/m A D

B ft 13 m

a + ©MA = 0; u = tan - 1

TCB a

62 = 45° 62

2 25

(1) –-270(1.5) b (3) 54(4.5) + TCD a

2 25

(3)==00 b9

ft 13 m

ft 13 m

(1)

270 kN

1.5 m

1.5 m

L2B¿C¿ = (3) (8.4853)22– -2(1)(2.8284) 2(3)(8.4853) cos (1)2 ++ (2.8284) cos u9 u¿ Also, 2.8284 m

(3)2 ++ (2.8284) cos u9 u¿ L2D¿C¿ = (9) (8.4853)22– -2(3)(2.8284) 2(9)(8.4853) cos

(2)

Thus, eliminating cos u¿ . (0.019642) - L2B¿C¿(0.176778)

2m 5m

5m 1m

1m

2 1.5910 == -–L L2D9C9 1.001735 ++ 1.590978 (0.058926) + +1.001735 D¿C¿(0.0065473)

0.0065473 L2D¿C¿+ 0.589243 + 0.589256 L2B¿C¿(0.019642) (0.176778) == 0.058926 L2D9C9 L2B¿C¿ = 0.333 L2D¿C¿ + 30 3.333 But, 5 + d+BC9d,BC¿ , LB¿C = 245

, 5 + d+DC9 LD¿C = 245 dDC¿

Neglect squares or d¿ B since small strain occurs. 2 2 L2D¿C = ( 245 245 dBC =) 5 =+ 45 2 5+ d2BC 5 + d+BCd)BC

2 2 245 dDC L2D¿C = ( 245 =) 5=+ 45 2 5+ d2DC 5 + d+DCd)DC

= 0.333(5 0.333(45+ 2+ 2245 ) + 30 45 5++ 22245 5 + ddBC 5 dDC)d+DC3.333 BC = 2 245 = =0.333(2245 2 5 d+BCdBC 0.333(2 5 ddDC DC))

dDC = 3dBC Thus,

T CD 245 TCBTCB 5 5 245 5 3 = 3 AE AE AE AE TCD = 3 TCB From Eq. (1). kip = 27.2 kip TCD = 27.1682 135.84 kN

Ans.

TCB = 9.06 45.28kip kN

Ans.

233

Ans: T CD 135.84 kN, T CB 45.28 kN


4–59. Two identical rods AB and CD each have a length L and diameter d, and are used to support the rigid beam, which is pinned at F. If a vertical force P is applied at the end of the beam, determine the angle of rotation of the beam. The rods are made of material that has a modulus of elasticity of E.

D

P C

F A a

Equation of Equilibrium: Referring to the free-body diagram of the rigid beam shown in Fig. a, a + ©MF = 0;

a

2a

B

FAB(a) + FCD(a) - P(3a) = 0 FAB + FCD = 3P

(1)

Compatibility Equation: Referring to the geometry of the deformation diagram of the rods shown in Fig. b, dAB = dCD FABL FCDL = AE AE FAB = FCD

(2)

Solving Eqs. (1) and (2), FAB = FCD =

3 P 2

Displacement: Using these results,

dAB

FABLAB = = AE

3 a P bL 2 p a d2 b E 4

=

6PL pd2E

Referring to Fig. b, the angle of tilt u of the beam is u =

dAB a

=

6PL>pd2E 6PL = a pd2Ea

Ans.

Ans: u =

242

6PL pd2Ea


4–63. If the supporting rods of equal diameter are made from A992 steel, determine the required diameter to the nearest 3.175 mm of each rod when P = 450 kN. The allowable normal stress of the steel is sallow = 170 MPa.

P

A

600 mm

600 mm 30

Equation of Equilibrium: Referring to the free-body diagram of joint A shown in Fig. a, + : ©Fx = 0;

FAB sin 30° - FAC sin 30° = 0

+ c ©Fy = 0;

2F cos 30° + FAD - 450 = 0

B

30

D

C

FAB = FAC = F (1)

Compatibility Equation: Due to symmetry, joint A will displace vertically. Referring to the geometry shown in Fig. b, we have

dF = d FAD cos 30° F600 AEst

= e

FAD 600 cos 30° AEst

f cos 30°

F = 0.75FAD

(2)

Solving Eqs. (1) and (2), FAD = 195.75 kN

F = 146.79 kN

Normal Stress: Since all of the rods have the same diameter and rod AD is subjected to the greatest load, it is the critical member. sallow =

FAD AAD

170 =

195.75 p 2 d 4

Use d = 38.29 mm

Ans.

Ans: Use d = 38.29 mm 246


4–67. The assembly consists of a 6061-T6-aluminum member and a C83400-red-brass member that rest on the rigid plates. Determine the distance d where the vertical load P should be placed on the plates so that the plates remain horizontal when the materials deform. Each member has a width of 200 mm and they are not bonded together.

P d

750 mm Aluminum

150 mm

+ c ©Fy = 0; a + ©MO = 0;

Red brass

75 mm

- P + Fal + Fbr = 0 75 Fal + 187.5 Fbr - Pd = 0

d = dbr = dal Fbr L Fal L = Abr Ebr Aal Eal Fbr = Fal a

(75)(200)(102) Abr Ebr b = 0.730 Fal b = Fal a Aal Eal 150(200)(70)

Thus, P = 1.730 Fal 75 Fal + 187.5(0.730 Fal ) = (1.730 Fal)d d = 122.47 mm

Ans.

Ans: d = 122.47 mm 250


4–70. The rod is made of A992 steel and has a diameter of 6 mm. If the rod is 1.2 m long when the springs are compressed 12 mm. and the temperature of the rod is T = 5C, determine the force in the rod when its temperature is T = 70C.

k 1250 N/m

k 1250 N/m

1.2 m

Compatibility: +) (:

x = dT - dF x = 12(10 - 6)(70 - 5)(0.6)(1000) 1.25(x + 12)(0.6)(1000) -

p 2 4 (6 )(203)

x = 0.966 mm F = (1.25)(0.966 + 12) = 16.20 kN

Ans.

Ans: F = 16.20 kN 253


4–75. 40-ft-long A-36 4–75. The 12-m-long A-36steel steelrails rails on on a train track are laid with a small gap between them to allow for thermal expansion. Determine the required gap d so that the rails just touch one another when the temperature is increased from Using thisgap, gap,what whatwould wouldbe be the T11 ==–30°C -20°FtotoTT 90°F. T 30°C. Using this 2 2= = axial force in the rails if the temperature were to rise to T The cross-sectional area of of each railrail is 3200 mm The cross-sectional area each is 5.10 in2. T33 ==40°C? 110°F?

d

d

40m ft 12

Thermal Expansion: Note that since adjacent rails expand, each rail will be d required to expand on each end, or d for the entine rail. 2

12 m

8.64 mm

3 -6 12(10–6)[30 – (–30)](12)(10 ) )[90 - (- 20)](40)(12) d = a¢TL = 6.60(10

8.640 mm = 0.34848 in. = 0.348 in.

Ans.

Compatibility: + B A:

8.64 mm

= dTT –- ddFF 80.34848 .64 mm = 3 F(40)(12) F(12)(10 ) 3 -6 0.34848 8.64 = 12(10–6)[40 – (–30)(12)(10 )– )[110 - (- 20)](40)(12) = 6.60(10 3 3 3200(200)(10 ) 5.10(29.0)(10 )

N = 76.80 kN FF = = 76800 19.5 kip

Ans.

Ans: d = 8.640 mm, F = 76.80 kN 258


4–78. When the temperature is at 30°C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, the temperatures at ends A and B rise to 130°C and 80°C, respectively. If the temperature drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume each tank provides a rigid support at A and B.

150 mm 10 mm Section a - a 6m x

A

a

a

B

Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a, the temperature gradient can be expressed as a function of x as T(x) = 80 +

50 50 (6 - x) = a130 x b °C 6 6

Thus, the change in temperature as a function of x is ¢T = T(x) - 30° = a 130 -

50 50 xb - 30 = a 100 xb°C 6 6

Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of 6m

dT = a

L

¢Tdx = 12(10 - 6)

L0

a 100 -

50 xbdx = 0.0054 m = 5.40 mm 6

Using the method of superposition, Fig. b, +) (:

0 = dT - dF 0 = 5.40 -

F(6000) p(0.162 - 0.152)(200)(109)

F = 1 753 008 N Normal Stress: s =

1 753 008 F = = 180 MPa A p(0.162 - 0.152)

Ans.

Ans: s = 180 MPa 261


4–83. The The wires wires AB AB and and AC AC are are made made of of steel, steel, and wire 4–83. 150-lb force is applied, AD is made of copper. Before the 750-N in.mm longlong and and AD AD is 40isin. long. If AB and AC are each 60 1500 1000 mm the temperature is increased by 80°F, determine the long. If the temperature is increased by 25°C, determine forceforce in each wire the in each wireneeded neededtotosupport support the the load. load. Take 3 –6)>°F, Estst == 29(10 ksi, EE 17(10GPa, ) ksi, ast ast= =4.5(10 8(10-6 E 200 3)GPa, )/°C, cucu == 120 -6 –6 Eachwire wirehas hasaacross-sectional cross-sectional area area of = 5.4(10 9.60(10)/°C. )>°F. acu Each cu = 0.0123 in22.. 7.68 mm

C

D

B

40 in.mm 1000 in. 150060mm

45� 45

45� 45

60 in.mm 1500

A 150 750 lb N

Equations of Equilibrium: + ©F = 0; : x

FAC cos 45° - FAB cos 45° = 0 FAC = FAB = F

+ c ©Fy = 0;

750 = 0 2F sin 45° + FAD - 150

[1]

Compatibility: 4.5(10 0.16875 in. mm (dAC)T = 8.0 A 10–6- 6)(25)(1500) B (80)(60) == 0.03840 (dAC)Tr =

(dAC)T 0.16875 0.03840 ==0.23865 = 0.05431mm in. cos 45° cos 45°

6 0.03072mm in. (dAD)T = 9.60 5.4(10 A 10–6-)(25)(1000) B (80)(40) == 0.13500

d0 = (dAC)Tr - (dAD)T = 0.23865 0.05431 –-0.13500 0.03072= =0.23865 0.02359 in. mm (dAD)F = (dAC)Fr + d0 F(60) F(1000) FAD F(1500) AD (40) = + 0.02359 + 0.10365 3 =6 3 6 7.68(120)(10 7.68(200)(10 ) ) cos 45° 0.0123(17.0)(10 ) 0.0123(29.0)(10 ) cos 45° 0.1913F - 0.2379F = 23.5858 1.0851F – 1.3811F = 103.65 ADAD

[2]

Solving Eq. [1] and [2] yields: 243.6lbN FAC = FAB = F = 10.0

Ans.

lbN FAD = 136 405.5

Ans.

Ans: FAC 243.6 N, FAD 405.5 N 266


4–85. The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance heating. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional area of 375 mm2. Est = 200 GPa, Eal = 70 GPa , ast = 12(10 - 6)>°C and aal = 23(10 - 6)>°C.

0.7 mm B

F C

– 240 mm 300 mm

D A

+ E

dst + (dT)st = (dT)al - dal - 0.0007 Fst(0.3) (125)(10 - 6)(200)(109)

+ 12(10 - 6)(50 - 30)(0.3)

= 23(10 - 6)(180 - 30)(0.24) -

Fal(0.24) 375(10 - 6)(70)(109)

- 0.0007

12.0Fst + 9.14286Fal = 56000 + c ©Fy = 0;

(1)

Fal - 2Fst = 0

(2)

Solving Eqs. (1) and (2) yields: FAB = FEF = Fst = 1.85 kN

Ans.

FCD = Fal = 3.70 kN

Ans: FAB = FEF = 1.85 kN 268


4–89. The steel bar has the dimensions shown. Determine the maximum axial force P that can be applied so as not to exceed an allowable tensile stress of sallow = 150 MPa.

20 mm 60 mm

30 mm mm 30 P

P r ⫽ 15 mm 24 mm

Assume failure occurs at the fillet: w 60 = = 2 h 30 From the text,

and

r 15 = = 0.5 h 30

K = 1.4 smax = sallow = Ksavg 150 (106) = 1.4 c

P d 0.03 (0.02)

P = 64.3 kN Assume failure occurs at the hole: 2r 24 = = 0.4 w 60 From the text,

K = 2.2 smax = sallow = Ksavg 150 (106) = 2.2 c

P d (0.06 - 0.024) (0.02)

P = 49.1 kN (controls!)

Ans.

Ans: P = 49.1 kN 272


4–91. Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and has an allowable stress of sallow = 147 21 ksi. MPa.

40.125 mm in. 25 mm 1.25 in.

37.5 1.875mm in.

P

Assume failure of the fillet. r 0.25 5 = 0.2 = = 0.2 h 1.25 25

P

37.5 w 1.875 ==1.51.5 = 25 h 1.25

15 mm 0.75 in.

r mmin. � 50.25

From Fig. 4-24, K = 1.73 sallow = smax = Ksavg P P = 1.73 b 1 21 47 = 17.3 aa b 1.25 (0.125) 25(4) PP = 8497.1kip N = 1.897 Assume failure of the hole. r 0.375 7.5 ==0.20 = 0.20 w 1.875 37.5 From Fig. 4-25, K = 2.45 sallow = smax = Ksavg P P b = 2.45 1 21 47 = 2.45 aa b (1.875 - 0.75)(0.125) (37.5 – 15)(4) PP = N (controls) = 5400 1.21 kip (controls)

Ans.

= 5.4 kN

Ans.

Ans: P = 5.4 kN 274

Solutions_Chapter_4(WǪ).pdf

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