# d B = PL Ans. = m = 1.59 mm d A = PL AE = 12(10 3 )(3) Ans. = m = 6.14 mm Ans:

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1 4 5. The assembly consists of a steel rod and an aluminum rod, each having a diameter of 12 mm. If the rod is subjected to the axial loadings at and at the couling, determine the dislacement of the couling and the end. The unstretched length of each segment is shown in the figure. Neglect the size of the connections at and, and assume that they are rigid. E st = 200 GPa, E al = 70 GPa. 3 m 6 kn 2 m 18 kn d = PL E = 12(10 3 )(3) 4 (0.012)2 (200)(10 9 ) = m = 1.59 mm d = PL E = 12(10 3 )(3) 4 (0.012)2 (200)(10 9 ) + 18(10 3 )(2) 4 (0.012)2 (70)(10 9 ) = m = 6.14 mm d = 1.59 mm, d = 6.14 mm 187

2 4 10. The assembly consists of two 10-mm diameter red brass coer rods and D, a 15-mm diameter 304 stainless steel rod EF, and a rigid bar G. If the horizontal dislacement of end F of rod EF is 0.45 mm, determine the magnitude of P. 300 mm E P 450 mm F 4P D G P Internal Loading: The normal forces develoed in rods EF,, and D are shown on the free-body diagrams in Figs. a and b. Dislacement: The cross-sectional areas of rods EF and are EF = 56.25(10-6 ) m 2 and = 4 (0.012 ) = 25(10-6 ) m 2. 4 ( ) = d F = PL E = P EF L EF + P L EF E st E br 0.45 = 4P(450) 56.25(10-6 )(193)(10 9 ) + P(300) 25(10-6 )(101)(10 9 ) P = 4967 N = 4.97 kn P = 49.7 kn 192

3 4 13. The rigid bar is suorted by the in-connected rod that has a cross-sectional area of 14 mm 2 and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is alied. 1.5 m 300 N/m 2 m 2 m D a+ M = 0; 1200(2) - T (0.6)(2) = 0 d > = PL E = (2000)(2.5) 14(10-6 )(68.9)(10 9 ) = ( ) 2 = (1.5) 2 + (2) 2-2(1.5)(2) cos u u = T = 2000 N u = = = rad d D = u r = (4000) = 17.3 mm d D = 17.3 mm 195

4 4 17. The hanger consists of three 2014-T6 aluminum alloy rods, rigid beams and D, and a sring. If the vertical dislacement of end F is 5 mm, determine the magnitude of the load P. Rods and D each have a diameter of 10 mm, and rod EF has a diameter of 15 mm. The sring has a stiffness of k = 100 MN>m and is unstretched when P = mm E Internal Loading: The normal forces develoed in rods EF,, and D and the sring are shown in their resective free-body diagrams shown in Figs. a, b, and c. D 450 mm Dislacements: The cross-sectional areas of the rods are EF = and 4 ( ) = 56.25(10-6 ) m 2 = D =. 4 (0.012 ) = 25(10-6 ) m 2 F P d F>E = F EF L EF EF E al = P(450) 56.25(10-6 )(73.1)(10 9 ) = (10-6 )P T d > = F L E al = (P>2)(450) 25(10-6 )(73.1)(10 9 ) = (10-6 )P T The ositive signs indicate that ends F and move away from E and, resectively. lying the sring formula with k = c100(10 3 ) kn m da1000 N 1 kn ba 1 m 1000 mm b = 100(103 ) N>mm. d E> = F s k = -P 100(10 3 ) = -10(10-6 )P = 10(10-6 )P T The negative sign indicates that E moves towards. Thus, the vertical dislacement of F is (+T) d F> = d > + d F>E + d E> 5 = (10-6 )P (10-6 )P + 10(10-6 )P P = N = 59.5 kn P = 59.5 kn 199

5 4 19. ollar can slide freely along the smooth vertical guide. If the vertical dislacement of the collar is in. and the suorting 0.75 in. diameter rod is made of 304 stainless steel, determine the magnitude of P. P 2 ft Internal Loading: The normal force develoed in rod can be determined by considering the equilibrium of collar with reference to its free-body diagram, Fig. a. 1.5 ft + c F y = 0; -F a 4 5 b - P = 0 F = P Dislacements: The cross-sectional area of rod is =, and the initial length of rod is 4 (0.752 ) = in 2 L = = 2.5 ft. The axial deformation of rod is d = F L E st = -1.25P(2.5)(12) (28.0)(10 3 ) = P The negative sign indicates that end moves towards. From the geometry shown in Fig. b, we obtain u = tan - 1 a 1.5. Thus, 2 b = d = (d ) V cos u P = cos P = 9.24 ki P = 9.24 ki 201

6 4 33. The steel ie is filled with concrete and subjected to a comressive force of 80 kn. Determine the average normal stress in the concrete and the steel due to this loading. The ie has an outer diameter of 80 mm and an inner diameter of 70 mm. E st = 200 GPa, E c = 24 GPa. 80 kn 500 mm + c F y = 0; P st + P con - 80 = 0 (1) d st = d con P st L 4 ( ) (200) (10 9 ) = P con L 4 (0.072 ) (24) (10 9 ) P st = P con (2) Solving Eqs. (1) and (2) yields P st = kn P con = kn s st = P st st = (10 3 ) 4 ( ) = 48.8 MPa s con = P con = (103 ) = 5.85 MPa con 4 (0.072 ) s st = 48.8 MPa, s con = 5.85 MPa 215

7 4 39. The load of 2800 lb is to be suorted by the two essentially vertical -36 steel wires. If originally wire is 60 in. long and wire is 40 in. long, determine the crosssectional area of if the load is to be shared equally between both wires. Wire has a cross-sectional area of 0.02 in in. 40 in. T = T = = 1400 lb d = d 1400(40) (0.02)(29)(10 6 ) = 1400(60) (29)(10 6 ) = 0.03 in 2 = 0.03 in 2 221

8 4 45. The bolt has a diameter of 20 mm and asses 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 -36 steel, determine the normal stress in the tube and bolt when a force of 40 kn is alied to the bolt.ssume the end cas are rigid. 40 kn 160 mm 150 mm 40 kn Referring to the FD of left ortion of the cut assembly, Fig. a : + F x = 0; 40(10 3 ) - F b - F t = 0 (1) Here, it is required that the bolt and the tube have the same deformation. Thus d t = d b F t (150) 4 ( )200(10 9 )D = F t = F b F b (160) 4 (0.022 )200(10 9 )D (2) Solving Eqs (1) and (2) yields Thus, F b = (10 3 ) N F t = (10 3 ) N s b = F b = 10.17(103 ) = 32.4 MPa b 4 (0.022 ) s t = F t t = (10 3 ) 4 ( ) = 34.5 MPa s b = 32.4 MPa, s t = 34.5 MPa 227

9 4 47. The suort consists of a solid red brass coer ost surrounded by a 304 stainless steel tube. efore the load is alied the ga between these two arts is 1 mm. Given the dimensions shown, determine the greatest axial load that can be alied to the rigid ca without causing yielding of any one of the materials. P 1 mm 0.25 m Require, 60 mm 80 mm 10 mm d st = d br F st (0.25) [(0.05) 2 - (0.04) 2 ]193(10 9 ) = F br (0.25) (0.03) 2 (101)(10 9 ) F st = F br c F y = 0; F st + F br - P = 0 (1) (2) ssume brass yields, then (F br ) max = s g br = 70(10 6 )()(0.03) 2 = N (P g ) br = s g >E = 70.0(106 ) 101(10 9 ) = (10-3 ) mm>mm d br = (e g ) br L = (10-3 )(0.25) = mm < 1 mm Thus only the brass is loaded. P = F br = 198 kn P = 198 kn 229

10 4 51. The rigid bar suorts the uniform distributed load of 6 ki>ft. Determine the force in each cable if each cable has a cross-sectional area of 0.05 in 2, and E = ksi. 6 ft 6 ki/ft 3 ft 3 ft 3 ft D a + M = 0; T a 2 (1) 25 b(3) - 54(4.5) + T Da 2 25 b9 = 0 u = tan = 45 L 2 = (3) 2 + (8.4853) 2-2(3)(8.4853) cos u lso, L 2 D = (9) 2 + (8.4853) 2-2(9)(8.4853) cos u (2) Thus, eliminating cos u. -L 2 ( ) = -L 2 D ( ) L 2 ( ) = L 2 D L 2 = L 2 D + 30 ut, L = d, L D = d D Neglect squares or d since small strain occurs. L 2 D = (245 + d ) 2 = d L 2 D = (245 + d D ) 2 = d D d = 0.333( d D ) d = 0.333(2245 d D ) Thus, d D = 3d T D 245 E = 3 T 245 E T D = 3 T From Eq. (1). T D = ki = 27.2 ki T = 9.06 ki T D = 27.2 ki, T = 9.06 ki 233

11 4 57. The rigid bar is originally horizontal and is suorted by two -36 steel cables each having a crosssectional area of 0.04 in 2. Determine the rotation of the bar when the 800-lb load is alied. 12 ft 800 lb D Referring to the FD of the rigid bar Fig. a, a + M = 0; F a 12 (1) 13 b(5) + F D a 3 b(16) - 800(10) = ft 5 ft 6 ft The unstretched lengths of wires and D are L = = 13 ft and L D = = 20 ft. The stretch of wires and D are d = F L E = F (13) d E D = F D L D = F D(20) E E Referring to the geometry shown in Fig. b, the vertical dislacement of a oint on d the rigid bar is d. For oints and D, cos u and cos u D = 3 = 12 v =. Thus, cos u 13 5 the vertical dislacements of oints and D are d v = d D v = d = F (13)>E cos u 12>13 The similar triangles shown in Fig. c gives = 169 F 12E d D = F D (20)>E = 100 F D cos u D 3>5 3 E d v 5 = d D v a 169 F 12 E b = 1 16 a 100 F D 3 E b F = F D (2) Solving Eqs (1) and (2), yields Thus, F D = lb F = lb d D v = 100(614.73) 3(0.04)29.0 (10 6 )D = ft Then u = a ft ba ft b = u =

12 4 69. The assembly has the diameters and material makeu indicated. If it fits securely between its fixed suorts when the temerature is T 1 = 70 F, determine the average normal stress in each material when the temerature reaches T 2 = 110 F T6 luminum 304 Stainless ronze steel 12 in. 8 in. 4 in. D 4 ft 6 ft 3 ft F x = 0; F = F = F F(4)(12) d >D = 0; - (6) 2 (10.6)(10 6 ) (10-6 )(110-70)(4)(12) F(6)(12) - (4) 2 (15)(10 6 ) (10-6 )(110-70)(6)(12) F(3)(12) - (2) 2 (28)(10 6 ) (10-6 )(110-70)(3)(12) = 0 F = ki s al = (6) 2 = 2.46 ksi s br = (4) 2 = 5.52 ksi s st = (2) 2 = 22.1 ksi s al = 2.46 ksi, s br = 5.52 ksi, s st = 22.1 ksi 252

13 4 78. When the temerature is at 30, the -36 steel ie fits snugly between the two fuel tanks. When fuel flows through the ie, the temeratures at ends and rise to 130 and 80, resectively. If the temerature dro along the ie is linear, determine the average normal stress develoed in the ie. ssume each tank rovides a rigid suort at and. 150 mm 10 mm Section a - a 6 m x a a Temerature Gradient: Since the temerature varies linearly along the ie, Fig. a, the temerature gradient can be exressed as a function of x as T(x) = (6 - x) = a xb Thus, the change in temerature as a function of x is T = T(x) - 30 = a xb - 30 = a xb omatibility Equation: If the ie is unconstrained, it will have a free exansion of 6m d T = a Tdx = 12(10-6 ) a xbdx = m = 5.40 mm L L 6 0 Using the method of suerosition, Fig. b, ( : + ) 0 = d T - d F F(6000) 0 = ( )(200)(10 9 ) Normal Stress: F = N s = F = ( = 180 MPa ) s = 180 MPa 261

14 4 81. The 50-mm-diameter cylinder is made from m 1004-T61 magnesium and is laced in the clam when the temerature is T 1 = 20. If the 304-stainless-steel carriage bolts of the clam each have a diameter of 10 mm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the force in the cylinder when the temerature rises to T 2 = mm 150 mm +c F y = 0; F st = F mg = F d mg = d st a mg L mg T - F mgl mg E mg mg = a st L st T + F stl st E st st 26(10-6 )(0.1)(110) - F(0.1) 44.7(10 9 ) 4 (0.05)2 F = 904 N = 17(10-6 )(0.150)(110) + F(0.150) 193(10 9 )(2) 4 (0.01)2 F = 904 N 264

15 4 83. The wires and are made of steel, and wire D is made of coer. efore the 150-lb force is alied, and are each 60 in. long and D is 40 in. long. If the temerature is increased by 80 F, determine the force in each wire needed to suort the load. Take E st = 29(10 3 ) ksi, E cu = 17(10 3 ) ksi, a st = 8(10-6 )> F, a cu = 9.60(10-6 )> F. Each wire has a cross-sectional area of in in. D 40 in in. 150 lb Equations of Equilibrium: : + F x = 0; + c F y = 0; F cos 45 - F cos 45 = 0 F = F = F 2F sin 45 + F D = 0 (1) omatibility: (d ) T = 8.0(10-6 )(80)(60) = in. (d ) T2 = (d ) T cos 45 = = in. cos 45 (d D ) T = 9.60(10-6 )(80)(40) = in. d 0 = (d ) T2 - (d D ) T = = in. (d D ) F = (d ) Fr + d 0 F D (40) (17.0)(10 6 ) = F(60) (29.0)(10 6 ) cos F D F = (2) Solving Eq. (1) and (2) yields: F = F = F = 10.0 lb F D = 136 lb F = F = 10.0 lb, F D = 136 lb 266

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