Ans. 4 (32 ) = 2.25p in 2, Ans.

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04 Solutions 46060 5/25/10 3:19 M age 122 4 1. The shi is ushed through the water using an -36 steel roeller shaft that is 8 m long, measured from the roeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the roeller exerts a force on the shaft of 5 kn. The bearings at and are journal bearings. Internal Force: s shown on FD. 5 kn D Dislacement: d E -5.00 (10 3 )(8) 4 (0.42-0.3 2 ) 200(10 9 ) 8 m -3.638(10-6 ) m -3.6410-3 mm Negative sign indicates that end moves towards end D. 4 2. The coer shaft is subjected to the axial loads shown. Determine the dislacement of end with resect to end D. The diameters of each segment are d 3 in., d and d Take E cu 18110 3 2 in., D 1 in. 2 ksi. The normal forces develoed in segment, and D are shown in the FDS of each segment in Fig. a, b and c resectively. 6 ki 50 in. 75 in. 60 in. 2 ki 2 ki 3 ki D 1 ki The cross-sectional area of segment, and D are and D. 4 (12 ) 0.25 in 2 4 (22 ) in 2 4 (32 ) 2.25 in 2, Thus, d >D i i + + D D i E i E u E u D E u 6.00 (50) (2.25)18(10 3 )D + 2.00 (75) 18(10 3 )D + -1.00 (60) (0.25) 18(10 3 )D 0.766(10-3 ) in. The ositive sign indicates that end moves away from D. 122

04 Solutions 46060 5/25/10 3:19 M age 123 4 3. The -36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm 2, determine the dislacement of its end D. Neglect the size of the coulings at,, and D. The normal forces develoed in segments, and D are shown in the FDS of each segment in Fig. a, b and c, resectively. 1 m 1.5 m 1.25 m 9 kn 4 kn D 2 kn The cross-sectional areas of all the segments are 2 50 mm 2 1 m a. 1000 mm b 50.0(10-6 ) m 2 d D i i i E i 1 E S a + + D D b 1 50.0(10-6 ) 200(10 9 )D c -3.00(10 3 )(1) + 6.00(10 3 )(1.5) + 2.00(10 3 )(1.25) d 0.850(10-3 ) m 0.850 mm The ositive sign indicates that end D moves away from the fixed suort. 123

04 Solutions 46060 5/25/10 3:19 M age 124 *4 4. The -36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 50 mm 2, determine the dislacement of. Neglect the size of the coulings at,, and D. 1 m 1.5 m 1.25 m 9 kn 4 kn D 2 kn The normal forces develoed in segments and are shown the FDS of each segment in Fig. a and b, resectively. The cross-sectional area of these two segments are 50 mm 2 2 1 m a. Thus, 10.00 mm b 50.0 (10-6 ) m 2 d i i i E i 1 E S + 1 50.0(10-6 ) 200(10 9 )D c -3.00(10 3 )(1) + 6.00(10 3 )(1.5) d 0.600 (10-3 ) m 0.600 mm The ositive sign indicates that couling moves away from the fixed suort. 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 Ga, E al 70 Ga. 3 m 6 kn 2 m 18 kn d E 12(10 3 )(3) 4 (0.012)2 (200)(10 9 0.00159 m 1.59 mm ) d E 12(10 3 )(3) 4 (0.012)2 (200)(10 9 ) + 18(10 3 )(2) 4 (0.012)2 (70)(10 9 ) 0.00614 m 6.14 mm 4 6. The bar has a cross-sectional area of 3 in 2, and x w 500x 1/3 lb/in. E 35110 3 2 ksi. Determine the dislacement of its end when it is subjected to the distributed loading. 4 ft x x (x) w dx 500 x 1 3 dx 1500 4 0 (x) dx d 0 E 1 (3)(35)(10 6 ) 0 0 4(12) x 4 3 1500 4 x 4 1500 3 dx a (3)(35)(10 8 )(4) ba3 7 b(48)1 3 d 0.0128 in. 124

04 Solutions 46060 5/25/10 3:19 M age 125 4 7. The load of 800 lb is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the vertical dislacement of the load if the members were horizontal before the load was alied. Each wire has a cross-sectional area of 0.05 in 2. Referring to the FD of member,fig.a a+ M 0; F (5) - 800(1) 0 F 160 lb a+ M 0; 800(4) - F H (5) 0 F H 640 lb Using the results of F and F H, and referring to the FD of member D,Fig.b a+ M D 0; F F (7) - 160(7) - 640(2) 0 F F 342.86 lb a+ M 0; 640(5) - F DE (7) 0 F DE 457.14 lb Since E and F are fixed, d D F DE DE E st d F F F E st From the geometry shown in Fig. c, 457.14(4)(2) 0.05 28.0 (10 6 )D 342.86 (4)(12) 0.05 28.0 (10 6 )D 0.01567 in T 0.01176 in T E H D 2 ft 5 ft 4.5 ft 800 lb 1 ft F 4 ft d H 0.01176 + 5 (0.01567-0.01176) 0.01455 in T 7 Subsequently, Thus, d >H F H H E st d > F E st 160(4.5)(12) 0.0528.0(10 6 )D +T d d H + d >H 0.01455 + 0.02469 0.03924 in T +T d d + d > 0.01176 + 0.006171 0.01793 in T From the geometry shown in Fig. d, 640(4.5)(12) 0.05 28.0(10 6 )D 0.02469 in T 0.006171 in T d 0.01793 + 4 (0.03924-0.01793) 0.0350 in T 5 125

04 Solutions 46060 5/25/10 3:19 M age 126 *4 8. The load of 800 lb is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the angle of tilt of each member after the load is alied.the members were originally horizontal, and each wire has a cross-sectional area of 0.05 in 2. Referring to the FD of member,fig.a, a+ M 0; F (5) - 800(1) 0 F 160 lb a+ M 0; 800(4) - F H (5) 0 F H 640 lb Using the results of F and F H and referring to the FD of member D,Fig.b, a+ M D 0; F F (7) - 160(7) - 640(2) 0 F F 342.86 lb a+ M 0; 640(5) - F DE (7) 0 F DE 457.14 lb Since E and F are fixed, d D F DE DE E st d F F F E st From the geometry shown in Fig. c 457.14 (4)(12) 0.05 28.0(10 6 )D 342.86 (4)(12) 0.05 28.0(10 6 )D 0.01567 in T 0.01176 in T E H D 2 ft 5 ft 4.5 ft 800 lb 1 ft F 4 ft d H 0.01176 + 5 (0.01567-0.01176) 0.01455 in T 7 u 0.01567-0.01176 7(12) 46.6(10-6 ) rad Subsequently, d >H F H H E st d > F E st 640 (4.5)(12) 0.05 28.0(10 6 )D 160 (4.5)(12) 0.05 28.0(10 6 )D 0.02469 in T 0.006171 in T Thus, +T d d H + d >H 0.01455 + 0.02469 0.03924 in T +T d d + d > 0.01176 + 0.006171 0.01793 in T From the geometry shown in Fig. d f 0.03924-0.01793 5(12) 0.355(10-3 ) rad 126

04 Solutions 46060 5/25/10 3:19 M age 127 4 8. ontinued 127

04 Solutions 46060 5/25/10 3:19 M age 128 4 9. The assembly consists of three titanium (Ti-61-4V) rods and a rigid bar. The cross-sectional area of each rod is given in the figure. If a force of 6 ki is alied to the ring F, determine the horizontal dislacement of oint F. Internal Force in the Rods: a + M 0; F D (3) - 6(1) 0 F D 2.00 ki : + F x 0; 6-2.00 - F 0 F 4.00 ki D 4 ft D 1 in 2 1.5 in 2 6 ft E 2 ft 1 ft F 6 ki 1 ft EF 2 in 2 Dislacement: d F D D D E 2.00(4)(12) (1)(17.4)(10 3 0.0055172 in. ) d F E 4.00(6)(12) (1.5)(17.4)(10 3 0.0110344 in. ) d F>E F EF EF EF E 6.00(1)(12) (2)(17.4)(10 3 0.0020690 in. ) de œ 2 0.0055172 ; de œ 0.0036782 in. 3 d E d + d œ E 0.0055172 + 0.0036782 0.0091954 in. d F d E + d F>E 0.0091954 + 0.0020690 0.0113 in. 4 10. The assembly consists of three titanium (Ti-61-4V) rods and a rigid bar. The cross-sectional area of each rod is given in the figure. If a force of 6 ki is alied to the ring F, determine the angle of tilt of bar. Internal Force in the Rods: a + M 0; F D (3) - 6(1) 0 F D 2.00 ki : + F x 0; 6-2.00 - F 0 F 4.00 ki D 4 ft D 1 in 2 1.5 in 2 6 ft E 2 ft 1 ft F 6 ki 1 ft EF 2 in 2 Dislacement: d F D D D E 2.00(4)(12) (1)(17.4)(10 3 0.0055172 in. ) d F E 4.00(6)(12) (1.5)(17.4)(10 3 0.0110344 in. ) u tan - 1 d - d 3(12) - 1 0.0110344-0.0055172 tan 3(12) 0.00878 128

04 Solutions 46060 5/25/10 3:19 M age 129 4 11. The load is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the vertical dislacement of the 500-lb load if the members were originally horizontal when the load was alied. Each wire has a cross-sectional area of 0.025 in 2. E F G 3 ft Internal Forces in the wires: FD (b) a + M 0; F (4) - 500(3) 0 F 375.0 lb D 1.8 ft H 1 ft 2 ft I 3 ft 1 ft 500 lb 5 ft + c F y 0; F H + 375.0-500 0 F H 125.0 lb FD (a) a + M D 0; F F (3) - 125.0(1) 0 F F 41.67 lb + c F y 0; F DE + 41.67-125.0 0 F DE 83.33 lb Dislacement: d D F DE DE DE E 83.33(3)(12) 0.0042857 in. 0.025(28.0)(10 6 ) d F F F F E 41.67(3)(12) 0.0021429 in. 0.025(28.0)(10 6 ) dh œ 2 0.0021429 ; dh œ 0.0014286 in. 3 d H 0.0014286 + 0.0021429 0.0035714 in. d >H F H H H E 125.0(1.8)(12) 0.0038571 in. 0.025(28.0)(10 6 ) d d H + d >H 0.0035714 + 0.0038571 0.0074286 in. d F G G G E 375.0(5)(12) 0.0321428 in. 0.025(28.0)(10 6 ) dl œ 3 0.0247143 ; dl œ 0.0185357 in. 4 d l 0.0074286 + 0.0185357 0.0260 in. 129

04 Solutions 46060 5/25/10 3:19 M age 130 *4 12. The load is suorted by the four 304 stainless steel wires that are connected to the rigid members and D. Determine the angle of tilt of each member after the 500-lb load is alied. The members were originally horizontal, and each wire has a cross-sectional area of 0.025 in 2. E F G 3 ft Internal Forces in the wires: FD (b) a + M 0; F G (4) - 500(3) 0 F G 375.0 lb D 1.8 ft H 1 ft 2 ft I 3 ft 1 ft 500 lb 5 ft + c F y 0; F H + 375.0-500 0 F H 125.0 lb FD (a) a + M D 0; F F (3) - 125.0(1) 0 F F 41.67 lb + c F y 0; F DE + 41.67-125.0 0 F DE 83.33 lb Dislacement: d D F DE DE DE E 83.33(3)(12) 0.0042857 in. 0.025(28.0)(10 6 ) d F F F F E 41.67(3)(12) 0.0021429 in. 0.025(28.0)(10 6 ) dh œ 2 0.0021429 ; dh œ 0.0014286 in. 3 d H d œ H + d 0.0014286 + 0.0021429 0.0035714 in. tan a 0.0021429 36 ; a 0.00341 d >H F H H H E 125.0(1.8)(12) 0.0038571 in. 0.025(28.0)(10 6 ) d d H + d >H 0.0035714 + 0.0038571 0.0074286 in. d F G G G E 375.0(5)(12) 0.0321428 in. 0.025(28.0)(10 6 ) tan b 0.0247143 48 ; b 0.0295 130

04 Solutions 46060 5/25/10 3:19 M age 131 4 13. The bar has a length and cross-sectional area. Determine its elongation due to the force and its own weight.the material has a secific weight g (weight>volume) and a modulus of elasticity E. (x) dx d (x) E 1 E 0 (gx + ) dx 1 E a g2 2 + b g2 2E + E 4 14. The ost 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 rovides a frictional resistance that is uniformly distributed along its sides of w 4 kn>m, determine the force F at its bottom needed for equilibrium.lso, what is the dislacement of the to of the ost with resect to its bottom? Neglect the weight of the ost. 2 m y 20 kn w Equation of Equilibrium: For entire ost [FD (a)] + c F y 0; F + 8.00-20 0 F 12.0 kn F Internal Force: FD (b) + c F y 0; -F(y) + 4y - 20 0 F(y) {4y - 20} kn Dislacement: F(y)dy d > 0 (y)e 1 E 0 2 m (4y - 20)dy 1 E 2y2-20y 2 m 0-32.0 kn # m E 32.0(10 3 ) - 4 (0.062 ) 13.1 (10 9 ) - 0.8639 10-3 m - 0.864 mm Negative sign indicates that end moves toward end. 131

04 Solutions 46060 5/25/10 3:20 M age 132 4 15. The ost 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 rovides 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. lso, what is the dislacement of the to of the ost with resect to its bottom? Neglect the weight of the ost. 2 m y 20 kn w F Equation of Equilibrium: For entire ost [FD (a)] + c F y 0; F + 3.00-20 0 F 17.0 kn Internal Force: FD (b) + c F y 0; -F(y) + 1 2 a 3y by - 20 0 2 F(y) e 3 4 y2-20 f kn Dislacement: F(y) dy d > 0 (y)e 1 E 0 2m 1 E a y3 2 m 4-20yb 2-38.0 kn # m E 38.0(10 3 ) - 4 (0.062 ) 13.1 (10 9 ) -1.026 10-3 m a 3 4 y2-20bdy 0-1.03 mm Negative sign indicates that end moves toward end. 132

04 Solutions 46060 5/25/10 3:20 M age 133 *4 16. The linkage is made of two in-connected -36 steel members, each having a cross-sectional area of 1.5 in 2. If a vertical force of 50 ki is alied to oint, determine its vertical dislacement at. 2 ft 1.5 ft 1.5 ft nalysing the equilibrium of Joint by referring to its FD,Fig.a, : + F x 0 ; F a 3 5 b - F a 3 5 b 0 F F F + c F y 0-2Fa 4 b - 50 0 F -31.25 ki 5 The initial length of members and is 21.5 2 + 2 2 12 in (2.50 ft)a b 30 in. The axial deformation of members 1 ft and is d F E (-31.25)(30) (1.5)29.0(10 3 )D -0.02155 in. The negative sign indicates that end moves toward and. From the geometry shown in Fig. b, u tan - 1 a 1.5. Thus, 2 b 36.87 d g d cos u 0.02155 0.0269 in. T cos 36.87 133

04 Solutions 46060 5/25/10 3:20 M age 134 4 17. The linkage is made of two in-connected -36 steel members, each having a cross-sectional area of 1.5 in 2. Determine the magnitude of the force needed to dislace oint 0.025 in. downward. nalysing the equilibrium of joint by referring to its FD,Fig.a : + F x 0; F a 3 5 b - F a 3 5 b 0 F F F + c F y 0; -2Fa 4 5 b - 0 F -0.625 2 ft 1.5 ft 1.5 ft The initial length of members and are 21.5 2 + 2 2 12 in (2.50 ft)a b 30 in. The axial deformation of members 1 ft and is d F E -0.625(30) (1.5)29.0(10 3 )D -0.4310(10-3 ) The negative sign indicates that end moves toward and. From the geometry shown in Fig. b, we obtain u tan - 1 a 1.5. Thus 2 b 36.87 (d ) g d cos u 0.025 0.4310(10-3 ) cos 36.87 46.4 kis 134

04 Solutions 46060 5/25/10 3:20 M age 135 4 18. The assembly consists of two -36 steel rods and a rigid bar D. Each rod has a diameter of 0.75 in. If a force of 10 ki is alied to the bar as shown, determine the vertical dislacement of the load. 2 ft 3 ft Here, F EF 10 ki. Referring to the FD shown in Fig. a, a+ M 0; F D (2) - 10(1.25) 0 F D 6.25 ki 1.25 ft E D 0.75 ft F 1 ft a+ M D 0; 10(0.75) - F (2) 0 F 3.75 ki 10 ki The cross-sectional area of the rods is. Since oints 4 (0.752 ) 0.140625 in 2 and are fixed, d F E st d D F D D E st 3.75 (2)(12) 0.140625 29.0(10 3 )D 6.25(3)(12) 0.140625 29.0(10 3 )D 0.007025 in. T 0.01756 in T From the geometry shown in Fig. b Here, d E 0.007025 + 1.25 2 (0.01756-0.00725) 0.01361 in. T d F>E F EF EF E st 10 (1) (12) 0.140625 29.0(10 3 )D 0.009366 in T Thus, +T d F d E + d F>E 0.01361 + 0.009366 0.0230 in T 135

04 Solutions 46060 5/25/10 3:20 M age 136 4 19. The assembly consists of two -36 steel rods and a rigid bar D. Each rod has a diameter of 0.75 in. If a force of 10 ki is alied to the bar, determine the angle of tilt of the bar. Here, F EF 10 ki. Referring to the FD shown in Fig. a, a+ M 0; F D (2) - 10(1.25) 0 F D 6.25 ki 2 ft 3 ft a+ M D 0; 10(0.75) - F (2) 0 F 3.75 ki E D The cross-sectional area of the rods is. Since oints 4 (0.752 ) 0.140625 in 2 and are fixed then, 1.25 ft 0.75 ft F 1 ft d F E st d D F D D E st From the geometry shown in Fig. b, 3.75 (2)(12) 0.140625 29.0(10 3 )D 6.25 (3)(12) 0.140625 29.0(10 3 )D 0.007025 in T 0.01756 in T 10 ki u 0.01756-0.007025 2(12) 0.439(10-3 ) rad 136

04 Solutions 46060 5/25/10 3:20 M age 137 *4 20. The rigid bar is suorted by the in-connected rod that has a cross-sectional area of 500 mm 2 and is made of -36 steel. Determine the vertical dislacement of the bar at when the load is alied. Force In The Rod. Referring to the FD of member,fig.a 3 m 45 kn/m a+ M 0; F a 3 5 b (4) - 1 2 (45)(4) c 1 3 (4) d 0 F 50.0 kn 4 m Dislacement. The initial length of rod is 23 2 + 4 2 5 m. The axial deformation of this rod is d F E st 50.0(10 3 )(5) 0.5(10-3 ) 200(10 9 )D 2.50 (10-3 ) m From the geometry shown in Fig. b, u tan - 1 a 3. Thus, 4 b 36.87 (d ) g d sin u 2.50(10-3 ) sin 36.87 4.167 (10-3 ) m 4.17 mm 137

04 Solutions 46060 5/25/10 3:20 M age 138 4 21. sring-suorted ie hanger consists of two srings which are originally unstretched and have a stiffness of k 60 kn>m, three 304 stainless steel rods, and D, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the ie and the fluid it carries have a total weight of 4 kn, determine the dislacement of the ie when it is attached to the suort. F k G 0.75 m E D k H 0.75 m Internal Force in the Rods: FD (a) 0.25 m 0.25 m a + M 0; F D (0.5) - 4(0.25) 0 F D 2.00 kn + c F y 0; F + 2.00-4 0 F 2.00 kn FD (b) + c F y 0; F EF - 2.00-2.00 0 F EF 4.00 kn Dislacement: d D d E F EF EF EF E 4.00(10 3 )(750) 4 (0.012)2 (193)(10 9 0.1374 mm ) d > d >D D D D E 2(10 3 )(750) 4 (0.005)2 (193)(10 9 0.3958 mm ) d d D + d >D 0.1374 + 0.3958 0.5332 mm Dislacement of the sring d s F s k 2.00 60 0.0333333 m 33.3333 mm d lat d + d s 0.5332 + 33.3333 33.9 mm 138

04 Solutions 46060 5/25/10 3:20 M age 139 4 22. sring-suorted ie hanger consists of two srings, which are originally unstretched and have a stiffness of k 60 kn>m, three 304 stainless steel rods, and D, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the ie is dislaced 82 mm when it is filled with fluid, determine the weight of the fluid. F k G 0.75 m E D k H 0.75 m Internal Force in the Rods: FD (a) a + M 0; F D (0.5) - W(0.25) 0 F D W 2 0.25 m 0.25 m + c F y 0; F + W 2 - W 0 F W 2 FD (b) + c F y 0; F EF - W 2 - W 2 0 F EF W Dislacement: d D d E F EF EF EF E W(750) 4 (0.012)2 (193)(10 9 ) d > d >D F D D D E Dislacement of the sring d s F s k d lat d + d s 34.35988(10-6 ) W W 2 (750) 4 (0.005)2 (193)(10 9 ) 98.95644(10-6 ) W d d D + d >D 34.35988(10-6 ) W + 98.95644(10-6 ) W 0.133316(10-3 ) W W 2 60(10 3 (1000) 0.008333 W ) 82 0.133316(10-3 ) W + 0.008333W W 9685 N 9.69 kn 139

04 Solutions 46060 5/25/10 3:20 M age 140 4 23. The rod has a slight taer and length. It is susended from the ceiling and suorts a load at its end. Show that the dislacement of its end due to this load is d >1Er 2 r 1 2. Neglect the weight of the material. The modulus of elasticity is E. r 2 r(x) r 1 + r 2 - r 1 x r 1 + (r 2 - r 1 )x (x) 2 (r 1 + (r 2 - r 1 )x) 2 dx d (x)e 2 E 0 dx [r 1 + (r 2 - r 1 )x] 2 r 1-2 E c 1 (r 2 - r 2 )(r 1 + (r 2 - r 1 )x) d ƒ 0-2 E(r 2 - r 1 ) c 1 r 1 + (r 2 - r 1 ) - 1 r 1 d 2 - E(r 2 - r 1 ) c 1 r 2-1 r 1 d - 2 E(r 2 - r 1 ) c r 1 - r 2 r 2 r 1 d 2 E(r 2 - r 1 ) c r 2 - r 1 r 2 r 1 d E r 2 r 1 QED *4 24. Determine the relative dislacement of one end of the taered late with resect to the other end when it is subjected to an axial load. d 2 t w d 1 + d 2 - d 1 h d (x) dx (x)e h E t 0 h E t d 1 h 0 h dx d 1 h + (d 2 - d 1 )x h x d 1 h + (d 2 - d 1 )x h E 0 dx d2 - d1 1 + d 1 h x h E t d 1 h a d 1 h bcln a1 + d 2 - d h 1 d 2 - d 1 d 1 h xbdƒ 0 h E t(d 2 - d 1 ) cln a1 + d 2 - d 1 d 1 bd h dx [d 1h + ( d 2 - d 1 )x ]t h h E t(d 2 - d 1 ) cln a d 1 + d 2 - d 1 bd d 1 d 1 h h E t(d 2 - d 1 ) cln d 2 d 1 d 140

04 Solutions 46060 5/25/10 3:20 M age 141 4 25. Determine the elongation of the -36 steel member when it is subjected to an axial force of 30 kn. The member is 10 mm thick. Use the result of rob. 4 24. 20 mm 30 kn 30 kn 75 mm 0.5 m Using the result of rob. 4-24 by substituting d 1 0.02 m, d 2 0.075 m t 0.01 m and 0.5 m. d 2c E st t(d 2 - d 1 ) ln d 2 d d 1 30(10 3 ) (0.5) 2c 200(10 9 )(0.01)(0.075-0.02) ln a 0.075 0.02 bd 0.360(10-3 ) m 0.360 mm 4 26. The casting is made of a material that has a secific weight g and modulus of elasticity E. If it is formed into a yramid having the dimensions shown, determine how far its end is dislaced due to gravity when it is susended in the vertical osition. b 0 b 0 Internal Forces: + c F z 0; (z) - 1 3 gz 0 (z) 1 3 gz Dislacement: (z) dz d (z) E 0 3 gz E dz 0 1 g z dz 3E 0 g2 6E 141

04 Solutions 46060 5/25/10 3:20 M age 142 4 27. The circular bar has a variable radius of r r 0 e ax and is made of a material having a modulus of elasticity of E. Determine the dislacement of end when it is subjected to the axial force. Dislacements: The cross-sectional area of the bar as a function of x is (x) r 2 r 2 0 e 2ax. We have (x)dx d 0 (x)e r 2 0 E 0 r 2 0 E c - 1 2ae 2ax d 2 0 dx e 2ax x r 0 r r 0 e ax - 2ar 2 0 E a1 - e - 2a b *4 28. The edestal is made in a shae that has a radius defined by the function r 2>12 + y 1>2 2 ft, where y is in feet. If the modulus of elasticity for the material is E 14110 3 2 si, determine the dislacement of its to when it suorts the 500-lb load. y 500 lb 0.5 ft 4 ft r 2 (2 y 1/2 ) d (y) dy (y) E 500 14(10 3 )(144) 0 ( 4 4 0.01974(10-3 ) (4 + 4y 1 2 + y) dy 0 dy 2 2 + y ) 2-1 2 1 ft y r 0.01974(10-3 )c4y + 4a 2 4 1 3 y3 2 b + 2 y2 d 0 0.01974(10-3 )(45.33) 0.8947(10-3 ) ft 0.0107 in. 142

04 Solutions 46060 5/25/10 3:20 M age 143 4 29. The suort is made by cutting off the two oosite sides of a shere that has a radius r 0. If the original height of the suort is r 0 >2, determine how far it shortens when it suorts a load. The modulus of elasticity is E. r 0 r 0 2 Geometry: Dislacement: r 2 (r 0 cos u) 2 r 0 2 cos 2 u (y) dy d (y) E 0 2 E 0 y r 0 sin u; dy r 0 cos u du u r 0 cos u du r 2 0 cos 2 u R 2 r 0 E 0 u du cos u R 2 r 0 E [ln (sec u + tan u)] 2 u 0 2 [ln (sec u + tan u)] r 0 E When y r 0 4 ; u 14.48 d 2 [ln (sec 14.48 + tan 14.48 )] r 0 E 0.511 r 0 E lso, Geometry: (y) x 2 (r 0 2 - y 2 ) Dislacement: (y) dy d (y) E 2 E 0 0 0 dy r 0 2 - y 2 2 E 1 2r 0 ln r 0 + y [ln 1.667 - ln 1] r 0 E 0 r 0 - y R 2 0 0.511 r 0 E 143

04 Solutions 46060 5/25/10 3:20 M age 144 4 30. The weight of the kentledge exerts an axial force of 1500 kn on the 300-mm diameter high strength concrete bore ile. If the distribution of the resisting skin friction develoed from the interaction between the soil and the surface of the ile is aroximated as shown, and the resisting bearing force F is required to be zero, determine the maximum intensity 0 kn>m for equilibrium. lso, find the corresonding elastic shortening of the ile. Neglect the weight of the ile. 0 12 m Internal oading: y considering the equilibrium of the ile with reference to its entire free-body diagram shown in Fig. a. We have 1 (12) - 1500 0 2 0 + c Fy 0; 0 250 kn>m Thus, (y) 250 y 20.83y kn>m 12 The normal force develoed in the ile as a function of y can be determined by considering the equilibrium of a section of the ile shown in Fig. b. 1 (20.83y)y - (y) 0 2 + c Fy 0; (y) 10.42y2 kn Dislacement: The cross-sectional area of the ile is (0.32) 0.0225 m2. 4 We have d 12 m (y)dy 10.42(103)y2dy 0.0225(29.0)(109) 0 0 (y)e 12 m 0 5.0816(10-6)y2dy 1.6939(10-6)y3 冷 0 12 m 2.9270(10-3)m 2.93 mm 144 F

04 Solutions 46060 5/25/10 3:20 M age 145 4 31. The column is constructed from high-strength concrete and six -36 steel reinforcing rods. If it is subjected to an axial force of 30 ki, determine the average normal stress in the concrete and in each rod. Each rod has a diameter of 0.75 in. 4 in. 30 ki Equations of Equilibrium: + c F y 0; 6 st + con - 30 0 [1] 3 ft omatibility: d st d con st (3)(12) 4 (0.752 )(29.0)(10 3 ) con(3)(12) [ 4 (82 ) - 6( 4 )(0.75)2 ](4.20)(10 3 ) st 0.064065 con [2] Solving Eqs. [1] and [2] yields: st 1.388 ki con 21.670 ki verage Normal Stress: s con con con s st st 1.388 3.14 ksi st 4 (0.752 ) 21.670 4 (82 ) - 6 4 (0.75 2 ) 0.455 ksi *4 32. The column is constructed from high-strength concrete and six -36 steel reinforcing rods. If it is subjected to an axial force of 30 ki, determine the required diameter of each rod so that one-fourth of the load is carried by the concrete and three-fourths by the steel. 4 in. 30 ki Equilibrium: The force of 30 ki is required to distribute in such a manner that 3/4 of the force is carried by steel and 1/4 of the force is carried by concrete. Hence st 3 4 (30) 22.5 ki con 1 (30) 7.50 ki 4 omatibility: 3 ft d st d con st st E st con con E con st 22.5 con E con 7.50 E st 6 a 4 bd2 3 4 (82 ) - 6 4 d 2 D (4.20)(10 3 ) 29.0(10 3 ) d 1.80 in. 145

04 Solutions 46060 5/25/10 3:20 M age 146 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 Ga, E c 24 Ga. 80 kn + c F y 0; st + con - 80 0 st 4 (0.082-0.07 2 ) (200) (10 9 ) con 4 (0.072 ) (24) (10 9 ) Solving Eqs. (1) and (2) yields st 57.47 kn con 22.53 kn s st st st d st d con st 2.5510 con 57.47 (10 3 ) 4 (0.082-0.07 2 ) 48.8 Ma s con con 22.53 (103 ) 5.85 Ma con 4 (0.072 ) (1) (2) 500 mm 4 34. The 304 stainless steel ost has a diameter of d 2 in. and is surrounded by a red brass 83400 tube. oth rest on the rigid surface. If a force of 5 ki is alied to the rigid ca, determine the average normal stress develoed in the ost and the tube. Equations of Equilibrium: 8 in. 5 ki 3 in. + c F y 0; st + br - 5 0[1] omatibility: d 0.5 in. d st d br st (8) 4 (22 )(28.0)(10 3 ) br(8) 4 (62-5 2 )(14.6)(10 3 ) st 0.69738 br [2] Solving Eqs. [1] and [2] yields: verage Normal Stress: br 2.9457 ki st 2.0543 ki s br br br 2.9457 4 (62-5 2 ) 0.341 ksi s st st 2.0543 0.654 ksi st 4 (22 ) 146

04 Solutions 46060 5/25/10 3:20 M age 147 4 35. The 304 stainless steel ost is surrounded by a red brass 83400 tube. oth rest on the rigid surface. If a force of 5 ki is alied to the rigid ca, determine the required diameter d of the steel ost so that the load is shared equally between the ost and tube. Equilibrium: The force of 5 ki is shared equally by the brass and steel. Hence 8 in. 5 ki 3 in. st br 2.50 ki omatibility: d 0.5 in. d st d br st E st br E br st bre br E st a 4 bd2 4 (62-5 2 )(14.6)(10 3 ) 28.0(10 3 ) d 2.39 in. *4 36. The comosite bar consists of a 20-mm-diameter -36 steel segment and 50-mm-diameter red brass 83400 end segments D and. Determine the average normal stress in each segment due to the alied load. 250 mm 500 mm 250 mm 50 mm 20 mm 75 kn 100 kn ; + F x 0; F - F D + 75 + 75-100 - 100 0 D 75 kn 100 kn F - F D - 50 0 (1) ; + 0 D - d D 150(0.5) 0 4 (0.02)2 (200)(10 9 ) - 50(0.25) 4 (0.052 )(101)(10 9 ) F D (0.5) - 4 (0.052 )(101)(10 9 ) - F D (0.5) 4 (0.022 )(200)(10 9 ) F D 107.89 kn From Eq. (1), F 157.89 kn s D D 107.89(103 ) 55.0 Ma D 4 (0.052 ) s 42.11(103 ) 134 Ma 4 (0.022 ) s 157.89(103 ) 80.4 Ma 4 (0.052 ) 147

04 Solutions 46060 5/25/10 3:20 M age 148 4 37. The comosite bar consists of a 20-mm-diameter -36 steel segment and 50-mm-diameter red brass 83400 end segments D and. Determine the dislacement of with resect to due to the alied load. 250 mm 500 mm 250 mm 50 mm 20 mm 75 kn 100 kn D 75 kn 100 kn ; + 0 D - d D 0-150(10 3 )(500) 4 (0.022 )(200)(10 9 ) - 50(10 3 )(250) 4 (0.052 )(101)(10 9 ) F D (500) 4 (0.052 )(101)(10 9 ) - F D (500) 4 (0.02)2 (200)(10 9 ) F D 107.89 kn Dislacement: d > 42.11(103 )(500) E st 4 (0.022 )200(10 9 ) 0.335 mm 4 38. The -36 steel column, having a cross-sectional area of 18 in 2, is encased in high-strength concrete as shown. If an axial force of 60 ki is alied to the column, determine the average comressive stress in the concrete and in the steel. How far does the column shorten? It has an original length of 8 ft. 16 in. 60 ki 9 in. + c F y 0; st + con - 60 0 d st d con ; st(8)(12) 18(29)(10 3 ) con (8)(12) [(9)(16) - 18](4.20)(10 3 ) st 0.98639 con Solving Eqs. (1) and (2) yields st 29.795 ki; con 30.205 ki s st st 29.795 1.66 ksi st 18 s con con con 30.205 0.240 ksi 9(16) - 18 Either the concrete or steel can be used for the deflection calculation. d st st E 29.795(8)(12) 18(29)(10 3 0.0055 in. ) (1) (2) 8 ft 148

04 Solutions 46060 5/25/10 3:20 M age 149 4 39. The -36 steel column is encased in high-strength concrete as shown. If an axial force of 60 ki is alied to the column, determine the required area of the steel so that the force is shared equally between the steel and concrete. How far does the column shorten? It has an original length of 8 ft. 16 in. 60 ki 9 in. 8 ft The force of 60 ki is shared equally by the concrete and steel. Hence st con 30 ki d con d st ; con E con st E st st cone con E st [9(16) - st] 4.20(10 3 ) 29(10 3 ) d st st E st 18.2 in 2 30(8)(12) 18.2(29)(10 3 ) 0.00545 in. *4 40. The rigid member is held in the osition shown by three -36 steel tie rods. Each rod has an unstretched length of 0.75 m and a cross-sectional area of 125 mm 2. Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size of the turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, to shorten 1.5 mm when the turnbuckle is rotated one revolution. 0.75 m 0.5 m E 0.5 m D 0.75 m a+ M E 0; -T (0.5) + T D (0.5) 0 F T T D T (1) +T F y 0; T EF - 2T 0 T EF 2T (2) Rod EF shortens 1.5mm causing (and D) to elongate. Thus: 0.0015 d > + d E>F 0.0015 2.25T 37500 T(0.75) (125)(10-6 )(200)(10 9 ) + 2T(0.75) (125)(10-6 )(200)(10 9 ) T 16666.67 N T T D 16.7 kn T EF 33.3 kn 149

04 Solutions 46060 5/25/10 3:20 M age 150 4 41. The concrete ost is reinforced using six steel reinforcing rods, each having a diameter of 20 mm. Determine the stress in the concrete and the steel if the ost is subjected to an axial load of 900 kn. E st 200 Ga, E c 25 Ga. 900 kn 250 mm 375 mm Referring to the FD of the uer ortion of the cut concrete ost shown in Fig. a + c F y 0; con + 6 st - 900 0 (1) Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus con con E con 0 con d st st st E st con 0.25(0.375) - 6( 4 )(0.022 )D25(10 9 )D con 36.552 st Solving Eqs (1) and (2) yields st ( 4 )(0.022 )200(10 9 )D (2) st 21.15 kn con 773.10 kn Thus, s con con con 773.10(10 3 ) 0.15(0.375) - 6( 4 )(0.022 ) 8.42 Ma s st st 21.15(103 ) st 4 (0.022 ) 67.3 Ma 150

04 Solutions 46060 5/25/10 3:20 M age 151 4 42. The ost is constructed from concrete and six -36 steel reinforcing rods. If it is subjected to an axial force of 900 kn, determine the required diameter of each rod so that one-fifth of the load is carried by the steel and four-fifths by the concrete. E st 200 Ga, E c 25 Ga. The normal force in each steel rod is st 1 5 (900) 30 kn 6 900 kn 250 mm 375 mm The normal force in concrete is con 4 (900) 720 kn 5 Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus d con d st con con E con st st E st 720(10 3 ) 0.25(0.375) - 6( 4 d2 )D25(10 9 )D 49.5 d 2 0.09375 30(10 3 ) 4 d2 200(10 9 )D d 0.02455 m 24.6 mm 4 43. The assembly consists of two red brass 83400 coer alloy rods and D of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid ca G. If the suorts at, and F are rigid, determine the average normal stress develoed in rods, D and EF. 300 mm 450 mm 40 kn 30 mm E F 30 mm D G 40 kn 40 mm Equation of Equilibrium: Due to symmetry, F F D F. Referring to the freebody diagram of the assembly shown in Fig. a, : + F x 0; 2F + F EF - 240(10 3 )D 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + 0 -d + d EF 40(10 3 )(300) 0-4 (0.032 )(101)(10 9 ) + c F EF (450) 4 (0.042 )(193)(10 9 ) + F EF>2(300) 4 (0.032 )(101)(10 9 ) d F EF 42 483.23 N Substituting this result into Eq. (1), F 18 758.38 N 151

04 Solutions 46060 5/25/10 3:20 M age 152 4 43. ontinued Normal Stress: We have, s s D F 18 758.38 26.5 Ma D 4 (0.032 ) s EF F EF 42 483.23 33.8 Ma EF 4 (0.042 ) 152

04 Solutions 46060 5/25/10 3:20 M age 153 *4 44. The two ies are made of the same material and are connected as shown. If the cross-sectional area of is and that of D is 2, determine the reactions at and D when a force is alied at the junction. 2 2 D Equations of Equilibrium: ; + F x 0; F + F D - 0 [1] omatibility: : + 0 d - d 0 2 2E - F 2 E 0 4E - 3F 4E + F 2 2E S F 3 From Eq. [1] F D 2 3 153

04 Solutions 46060 5/25/10 3:20 M age 154 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 (0.062-0.05 2 )200(10 9 )D F t 2.9333 F b F b (160) 4 (0.022 )200(10 9 )D (2) Solving Eqs (1) and (2) yields Thus, F b 10.17 (10 3 ) N F t 29.83 (10 3 ) N s b F b 10.17(103 ) 32.4 Ma b 4 (0.022 ) s t F t t 29.83 (10 3 ) 4 (0.062-0.05 2 ) 34.5 Ma 154

04 Solutions 46060 5/25/10 3:20 M age 155 4 46. If the ga between and the rigid wall at D is initially 0.15 mm, determine the suort reactions at and D when the force 200 kn is alied. The assembly is made of 36 steel. 600 mm 600 mm 0.15 mm D 25 mm 50 mm Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, : + F x 0; 200(10 3 ) - F D - F 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + d d - d FD 200(10 3 )(600) 0.15 4 (0.052 )(200)(10 9 ) - F D (600) 4 (0.052 )(200)(10 9 ) + F D (600) 4 (0.0252 )(200)(10 9 ) S F D 20 365.05 N 20.4 kn Substituting this result into Eq. (1), F 179 634.95 N 180 kn 155

04 Solutions 46060 5/25/10 3:20 M age 156 4 47. Two -36 steel wires are used to suort the 650-lb engine. Originally, is 32 in. long and is 32.008 in. long. Determine the force suorted by each wire when the engine is susended from them. Each wire has a crosssectional area of 0.01 in 2. + c F y 0; T + T - 650 0 (1) d d + 0.008 T (32) (0.01)(29)(10 6 ) T (32.008) (0.01)(29)(10 6 ) + 0.008 32T - 32.008T 2320 T 361 lb T 289 lb *4 48. Rod has a diameter d and fits snugly between the rigid suorts at and when it is unloaded. The 0 x modulus of elasticity is E. Determine the suort reactions at and if the rod is subjected to the linearly distributed axial load. x Equation of Equilibrium: Referring to the free-body diagram of rod shown in Fig. a, 0 : + F x 0; 1 2 0 - F - F 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + 0 d - d F (x)dx 0 E 0 0 (x)dx - F 0 - F () E 156

04 Solutions 46060 5/25/10 3:20 M age 157 4 48. ontinued Here, (x) 1 2 a 0 xbx 0 2 x2 0 0 2 0. Thus, 0 0 2 x3 3 ` x 2 dx - F 0 - F F 0 6 Substituting this result into Eq. (1), F 0 3 157

04 Solutions 46060 5/25/10 3:20 M age 158 4 49. The taered member is fixed connected at its ends and and is subjected to a load 7 ki at x 30 in. Determine the reactions at the suorts. The material is 2 in. thick and is made from 2014-T6 aluminum. 6 in. 3 in. x 60 in. y 120 - x 1.5 60 y 3-0.025 x : + F x 0; F + F - 7 0 (1) d > 0-30 0 -F 30 0 F dx 2(3-0.025 x)(2)(e) + 60 dx (3-0.025 x) + F dx (3-0.025x) 0 40 F ln(3-0.025 x) 30 0-40 F ln(3-0.025x) 60 30 0 -F (0.2876) + 0.40547 F 0 30 30 60 F dx 2(3-0.025 x)(2)(e) 0 F 1.40942 F Thus, from Eq. (1). F 4.09 ki F 2.91 ki 158

04 Solutions 46060 5/25/10 3:20 M age 159 4 50. The taered member is fixed connected at its ends and and is subjected to a load. Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed s allow 4 ksi. The member is 2 in. thick. 6 in. 3 in. x 60 in. y 120 - x 1.5 60 y 3-0.025 x : + F x 0; F + F - 0 d > 0 - x 0 F dx 2(3-0.025 x)(2)(e) + x 60 F dx 2(3-0.025 x)(2)(e) 0 x 60 dx -F (3-0.025 x) + F dx (3-0.025 x) 0 0 x F (40) ln (3-0.025 x) x 0 - F (40) ln (3-0.025x) 60 x 0 F ln (1-0.025 x ) -F 3 ln (2-0.025x 1.5 ) For greatest magnitude of require, 4 F 2(3-0.025 x)(2) ; F 48-0.4 x 4 F 2(3) ; F 24 ki Thus, (48-0.4 x) ln a1-0.025 x 3 b -24 ln a2-0.025 x b 1.5 Solving by trial and error, x 28.9 in. Therefore, F 36.4 ki 60.4 ki 159

04 Solutions 46060 5/25/10 3:20 M age 160 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 31110 3 2 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 - 1 6 6 45 2 (3) 2 + (8.4853) 2-2(3)(8.4853) cos u lso, 2 D (9) 2 + (8.4853) 2-2(9)(8.4853) cos u (2) Thus, eliminating cos u. - 2 (0.019642) + 1.5910-2 D (0.0065473) + 1.001735 2 (0.019642) 0.0065473 2 D + 0.589256 2 0.333 2 D + 30 ut, 245 + d, D 245 + d D Neglect squares or d since small strain occurs. 2 D (245 + d ) 2 45 + 2245 d 2 D (245 + d D ) 2 45 + 2245 d D 45 + 2245 d 0.333(45 + 2245 d D ) + 30 2 245 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 27.1682 ki 27.2 ki T 9.06 ki 160

04 Solutions 46060 5/25/10 3:20 M age 161 *4 52. The rigid bar is originally horizontal and is suorted by two cables each having a cross-sectional area of 0.05 in 2, and E 31110 3 2 ksi. Determine the slight rotation of the bar when the uniform load is alied. See solution of rob. 4-51. T D 27.1682 ki 6 ft d D Using Eq. (2) of rob. 4-51, T D 245 0.05(31)(10 3 ) 27.1682245 0.05(31)(10 3 0.1175806 ft ) ( 245 + 0.1175806) 2 (9) 2 + (8.4852) 2-2(9)(8.4852) cos u 3 ft 6 ki/ft 3 ft 3 ft D u 45.838 Thus, u 45.838-45 0.838 4 53. The ress consists of two rigid heads that are held 1 together by the two -36 steel 2 -in.-diameter rods. 6061- T6-solid-aluminum cylinder is laced in the ress and the screw is adjusted so that it just resses u against the cylinder. If it is then tightened one-half turn, determine the average normal stress in the rods and in the cylinder. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead reresents the distance the screw advances along its axis for one comlete turn of the screw. 12 in. 2 in. 10 in. : + F x 0; 2F st - F al 0 d st 0.005 - d al F st (12) ( 4 )(0.5)2 (29)(10 3 ) 0.005 - F al (10) (1) 2 (10)(10 3 ) Solving, F st 1.822 ki F al 3.644 ki s rod F st st 1.822 ( 9.28 ksi 4 )(0.5)2 s cyl F al 3.644 1.16 ksi 2 al (1) 161

04 Solutions 46060 5/25/10 3:20 M age 162 4 54. The ress consists of two rigid heads that are held 12 in. 1 together by the two -36 steel 2 -in.-diameter rods. 6061- T6-solid-aluminum cylinder is laced in the ress and the screw is adjusted so that it just resses u against the cylinder. Determine the angle through which the screw can be turned before the rods or the secimen begin to yield. The single-threaded screw on the bolt has a lead of 0.01 in. Note: The lead reresents the distance the screw advances along its axis for one comlete turn of the screw. 2 in. 10 in. : + F x 0; 2F st - F al 0 d st d - d al F st (12) ( 4 )(0.5)2 (29)(10 3 ) d - F al (10) (1) 2 (10)(10 3 ) (1) ssume steel yields first, s Y 36 Then F al 14.137 ki; F st ( 4 )(0.5)2 ; F st 7.068 ki s al 14.137 (1) 2 4.50 ksi 4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1), Thus, d 0.01940 in. u 360 0.01940 0.01 u 698 162

04 Solutions 46060 5/25/10 3:20 M age 163 4 55. The three susender bars are made of -36 steel and have equal cross-sectional areas of 450 mm 2. Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown. 2 m 50 kn 80 kn D E F 1 m 1 m 1 m 1 m Referring to the FD of the rigid beam, Fig. a, + c F y 0; F D + F E + F F - 50(10 3 ) - 80(10 3 ) 0 (1) a + M D 0; F E (2) + F F (4) - 50(10 3 )(1) - 80(10 3 )(3) 0 (2) Referring to the geometry shown in Fig. b, d E d D + a d F - d D b(2) 4 d E 1 2 d D + d F F E E 1 2 a F D E + F F E b F D + F F 2 F E (3) Solving Eqs. (1), (2) and (3) yields F E 43.33(10 3 ) N F D 35.83(10 3 ) N F F 50.83(10 3 ) N Thus, s E F E 43.33(103 ) 0.45(10-3 96.3 Ma ) s D F D 35.83(103 ) 0.45(10-3 79.6 Ma ) s F 113 Ma 163

04 Solutions 46060 5/25/10 3:20 M age 164 *4 56. The rigid bar suorts the 800-lb load. Determine the normal stress in each -36 steel cable if each cable has a cross-sectional area of 0.04 in 2. 12 ft 800 lb D 5 ft 5 ft 6 ft 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) 0 5 The unstretched length of wires and D are 212 2 + 5 2 13 ft and D 212 2 + 16 2 20 ft. The stretches of wires and D are d F E F (13) d E D F D D F D(20) E E Referring to the geometry shown in Fig. b, the vertical dislacement of the oints on d the rigid bar is dg. For oints and D, cos u and cos u D 3 12. Thus, cos u 13 5 the vertical dislacement of oints and D are d g d D g The similar triangles shown in Fig. c give d F (13)>E 169 F cos u 12>13 12E d D F D (20)>E 100 F D cos u D 3>5 3 E d g 5 d D g 16 1 5 a 169 F 12 E b 1 16 a 100 F D 3E b F 125 169 F D (2) Solving Eqs. (1) and (2), yields Thus, F D 614.73 lb F 454.69 lb s D F D D 614.73 0.04 15.37(103 ) si 15.4 ksi s F 454.69 0.04 11.37(103 ) si 11.4 ksi 164

04 Solutions 46060 5/25/10 3:20 M age 165 4 56. ontinued 165

04 Solutions 46060 5/25/10 3:20 M age 166 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, 5 ft 5 ft 6 ft a + M 0; F a 12 (1) 13 b(5) + F D a 3 b(16) - 800(10) 0 5 The unstretched length of wires and D are 212 2 + 5 2 13 ft and D 212 2 + 16 2 20 ft. The stretch of wires and D are d F E F (13) d D F D D F D(20) E E E Referring to the geometry shown in Fig. b, the vertical dislacement of the oints on d the rigid bar is d. For oints and D, cos u and cos u D 3 12 g. Thus, cos u 13 5 the vertical dislacement of oints and D are d g d D g The similar triangles shown in Fig. c gives d F (13)>E 169 F cos u 12>13 12E d D F D (20)>E 100 F D cos u D 3>5 3 E d g 5 d D g 16 1 5 a 169 F 12 E b 1 16 a 100 F D 3 E b F 125 169 F D (2) Solving Eqs (1) and (2), yields Thus, F D 614.73 lb F 454.69 lb d D g 100(614.73) 3(0.04)29.0 (10 6 )D 0.01766 ft Then u a 0.01766 ft ba 180 16 ft b 0.0633 166

04 Solutions 46060 5/25/10 3:20 M age 167 4 58. The horizontal beam is assumed to be rigid and suorts the distributed load shown. Determine the vertical reactions at the suorts. Each suort consists of a wooden ost having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take E w 12 Ga. 18 kn/m 1.40 m 2 m 1 m a + M 0; F (1) - F (2) 0 (1) + c F y 0; F + F + F - 27 0 (2) d - d 2 d - d 3 ; 3d - d 2d 3F E - F E 2F E ; 3F - F 2F (3) Solving Eqs. (1) (3) yields : F 5.79 kn F 9.64 kn F 11.6 kn 4 59. The horizontal beam is assumed to be rigid and suorts the distributed load shown. Determine the angle of tilt of the beam after the load is alied. Each suort consists of a wooden ost having a diameter of 120 mm and an unloaded (original) length of 1.40 m.take E w 12 Ga. a + M 0; F (1) - F (2) 0 (1) c + F y 0; F + F + F - 27 0 (2) 18 kn/m 1.40 m d - d 2 d - d 3 ; 3d - d 2d 2 m 1 m 3F E - F E 2F E ; 3F - F 2F (3) Solving Eqs. (1) (3) yields : F 5.7857 kn; F 9.6428 kn; F 11.5714 kn d F E 5.7857(103 )(1.40) 4 (0.122 )12(10 9 0.0597(10-3 ) m ) d F E 11.5714(103 )(1.40) 4 (0.122 )12(10 9 0.1194(10-3 ) m ) tan u 0.1194-0.0597 3 (10-3 ) u 0.0199(10-3 ) rad 1.14(10-3 ) 167

04 Solutions 46060 5/25/10 3:20 M age 168 *4 60. The assembly consists of two osts D and F made of -36 steel and having a cross-sectional area of 1000 mm 2, and a 2014-T6 aluminum ost E having a crosssectional area of 1500 mm 2. If a central load of 400 kn is alied to the rigid ca, determine the normal stress in each ost. There is a small ga of 0.1 mm between the ost E and the rigid member. 400 kn 0.5 m 0.5 m 0.4 m D E F Equation of Equilibrium. Due to symmetry, F D F F F. Referring to the FD of the rigid ca, Fig. a, + c F y 0; F E + 2F - 400(10 3 ) 0 (1) omatibility Equation. Referring to the initial and final osition of rods D (F) and E,Fig.b, d 0.1 + d E F(400) 1(10-3 )200(10 9 )D 0.1 + F E (399.9) 1.5(10-3 )73.1(10 9 )D F 1.8235 F E + 50(10 3 ) (2) Solving Eqs (1) and (2) yield Normal Stress. F E 64.56(10 3 ) N F 167.72(10 3 ) N s D s F F st 167.72(103 ) 1(10-3 ) 168 Ma s E F E 64.56(103 ) al 1.5(10-3 43.0 Ma ) 168

04 Solutions 46060 5/25/10 3:20 M age 169 4 61. The distributed loading is suorted by the three susender bars. and EF are made of aluminum and D is made of steel. If each bar has a cross-sectional area of 450 mm 2, determine the maximum intensity w of the distributed loading so that an allowable stress of 1s allow 2 st 180 Ma in the steel and 1s allow 2 al 94 Ma in the aluminum is not exceeded. E st 200 Ga, E al 70 Ga. ssume E is rigid. 1.5 m 1.5 m D F al st al 2 m E w a+ M 0; F EF (1.5) - F (1.5) 0 F EF F F + c F y 0; 2F + F D - 3w 0 (1) omatibility condition : d d F (70)(10 9 ) F D (200)(10 9 ) ; F 0.35 F D (2) ssume failure of and EF: F (s allow ) al 94(10 6 )(450)(10-6 ) 42300 N From Eq. (2) F D 120857.14 N From Eq. (1) w 68.5 kn>m ssume failure of D: F D (s allow ) st 180(10 6 )(450)(10-6 ) 81000 N From Eq. (2) F 28350 N From Eq. (1) w 45.9 kn>m (controls) 169

04 Solutions 46060 5/25/10 3:20 M age 170 4 62. The rigid link is suorted by a in at, a steel wire having an unstretched length of 200 mm and crosssectional area of 22.5 mm 2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm 2. If the link is subjected to the vertical load shown, determine the average normal stress in the wire and the block. E st 200 Ga, E al 70 Ga. Equations of Equilibrium: a+ M 0; 450(250) - F (150) - F D (150) 0 750 - F - F D 0 omatibility: [1] 200 mm 100 mm 150 mm 150 mm 450 N D 50 mm Solving Eqs. [1] and [2] yields: verage Normal Stress: d d D F (200) 22.5(10-6 )200(10 9 ) F D (50) 40(10-6 )70(10 9 ) F 0.40179 F D F D 535.03 N F 214.97 N s D F D 535.03 13.4 Ma D 40(10-6 ) [2] s F 214.97 22.5(10-6 ) 9.55 Ma 170

04 Solutions 46060 5/25/10 3:20 M age 171 4 63. The rigid link is suorted by a in at, a steel wire having an unstretched length of 200 mm and crosssectional area of 22.5 mm 2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm 2. If the link is subjected to the vertical load shown, determine the rotation of the link about the in. Reort the answer in radians. E st 200 Ga, E al 70 Ga. Equations of Equilibrium: a+ M 0; 450(250) - F (150) - F D (150) 0 750 - F - F D 0 omatibility: [1] 200 mm 100 mm 150 mm 150 mm 450 N D 50 mm Solving Eqs. [1] and [2] yields : Dislacement: d d D F (200) 22.5(10-6 )200(10 9 ) F D (50) 40(10-6 )70(10 9 ) F 0.40179 F D F D 535.03 N F 214.97 N [2] d D F D D D E al 535.03(50) 40(10-6 )(70)(10 9 ) 0.009554 mm tan u d D 150 0.009554 150 u 63.7(10-6 ) rad 0.00365 171

04 Solutions 46060 5/25/10 3:20 M age 172 *4 64. The center ost of the assembly has an original length of 124.7 mm, whereas osts and have a length of 125 mm. If the cas on the to and bottom can be considered rigid, determine the average normal stress in each ost. The osts are made of aluminum and have a cross-sectional area of 400 mm 2. E al 70 Ga. 100 mm 100 mm 800 kn/m 125 mm a+ M 0; -F (100) + F (100) 0 F F F + c F y 0; 2F + F - 160 0 d d + 0.0003 F (0.125) 400 (10-6 )(70)(10 6 ) F (0.1247) 400 (10-6 )(70)(10 6 ) + 0.0003 0.125 F - 0.1247F 8.4 Solving Eqs. (2) and (3) F 75.762 kn F 8.547 kn (1) (2) (3) 800 kn/m s s 75.726 (103 ) 400(10-6 ) 189 Ma s 8.547 (103 ) 400 (10-6 ) 21.4 Ma 4 65. The assembly consists of an -36 steel bolt and a 83400 red brass tube. If the nut is drawn u snug against the tube so that 75 mm, then turned an additional amount so that it advances 0.02 mm on the bolt, determine the force in the bolt and the tube.the bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm 2. Equilibrium: Since no external load is alied, the force acting on the tube and the bolt is the same. omatibility: 0.02 d t + d b 0.02 (75) 100(10-6 )(101)(10 9 ) + (75) 4 (0.0072 )(200)(10 9 ) 1164.83 N 1.16 kn 172

04 Solutions 46060 5/25/10 3:20 M age 173 4 66. The assembly consists of an -36 steel bolt and a 83400 red brass tube. The nut is drawn u snug against the tube so that 75 mm. Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm 2. llowable Normal Stress: (s g ) st 250 10 6 st 4 (0.007)2 st 9.621 kn (s g ) br 70.010 6 br 100(10-6 ) br 7.00 kn Since st 7 br, by comarison he brass will yield first. omatibility: a d t + d b 7.00(10 3 )(75) 100(10-6 )(101)(10 9 ) + 7.00(10 3 )(75) 4 (0.007)2 (200)(10 9 ) 0.120 mm 173

04 Solutions 46060 5/25/10 3:20 M age 174 4 67. The three susender bars are made of the same material and have equal cross-sectional areas. Determine the average normal stress in each bar if the rigid beam E is subjected to the force. D F a + M 0; F D (d) + F EF (2d) - a d 2 b 0 F D + 2F EF 2 (1) E + c F y 0; F + F D + F EF - 0 (2) d 2 d 2 d d - d E d d - d E 2d 2d d + d E 2F D E F E + F EF E 2F D - F - F EF 0 (3) Solving Eqs. (1), (2) and (3) yields F 7 12 F D 3 F EF 12 s s D s EF 7 12 3 12 *4 68. steel surveyor s tae is to be used to measure the length of a line. The tae has a rectangular cross section of 0.05 in. by 0.2 in. and a length of 100 ft when T 1 60 F and the tension or ull on the tae is 20 lb. Determine the true length of the line if the tae shows the reading to be 463.25 ft when used with a ull of 35 lb at T 2 90 F. The ground on which it is laced is flat. a st 9.60110-6 2> F, E st 29110 3 2 ksi. 0.2 in. 0.05 in. d T a T 9.6(10-6 )(90-60)(463.25) 0.133416 ft d (35-20)(463.25) 0.023961 ft E (0.2)(0.05)(29)(10 6 ) 463.25 + 0.133416 + 0.023961 463.41 ft 174

04 Solutions 46060 5/25/10 3:20 M age 175 4 69. Three bars each made of different materials are connected together and laced between two walls when the temerature is T 1 12. Determine the force exerted on the (rigid) suorts when the temerature becomes T 2 18. The material roerties and cross-sectional area of each bar are given in the figure. Steel E st 200 Ga a st 12(10 6 )/ st 200 mm 2 rass E br 100 Ga a br 21(10 6 )/ br 450 mm 2 oer E cu 120 Ga a cu 17(10 6 )/ cu 515 mm 2 300 mm 200 mm 100 mm ( ; + ) 0 T - d 0 12(10-6 )(6)(0.3) + 21 (10-6 )(6)(0.2) + 17 (10-6 )(6)(0.1) F(0.3) - 200(10-6 )(200)(10 9 ) - F(0.2) 450(10-6 )(100)(10 9 ) - F(0.1) 515(10-6 )(120)(10 9 ) F 4203 N 4.20 kn 4 70. The rod is made of -36 steel and has a diameter of 0.25 in. If the rod is 4 ft long when the srings are comressed 0.5 in. and the temerature of the rod is T 40 F, determine the force in the rod when its temerature is T 160 F. k 1000 lb/in. 4 ft k 1000 lb/in. omatibility: : + x d T - d F x 6.60(10-6 )(160-40)(2)(12) - 1.00(0.5)(2)(12) 4 (0.252 )(29.0)(10 3 ) x 0.01869 in. F 1.00(0.01869 + 0.5) 0.519 ki 4 71. 6-ft-long steam ie is made of -36 steel with s Y 40 ksi. It is connected directly to two turbines and as shown. The ie has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T 1 70 F. If the turbines oints of attachment are assumed rigid, determine the force the ie exerts on the turbines when the steam and thus the ie reach a temerature of T 2 275 F. 6 ft omatibility: : + 0 d T - d F 0 6.60(10-6 F(6)(12) )(275-70)(6)(12) - 4 (42-3.5 2 )(29.0)(10 3 ) F 116 ki 175

04 Solutions 46060 5/25/10 3:20 M age 176 *4 72. 6-ft-long steam ie is made of -36 steel with s Y 40 ksi. It is connected directly to two turbines and as shown. The ie has an outer diameter of 4 in. and a wall thickness of 0.25 in. The connection was made at T 1 70 F. If the turbines oints of attachment are assumed to have a stiffness of k 80110 3 2 ki>in., determine the force the ie exerts on the turbines when the steam and thus the ie reach a temerature of T 2 275 F. 6 ft omatibility: x d T - d F x 6.60(10-6 80(10 3 )(x)(3)(12) )(275-70)(3)(12) - 4 (42-3.5 2 )(29.0)(10 3 ) x 0.001403 in. F k x 80(10 3 )(0.001403) 112 ki 4 73. The ie is made of -36 steel and is connected to the collars at and. When the temerature is 60 F, there is no axial load in the ie. If hot gas traveling through the ie causes its temerature to rise by T 140 + 15x2 F, where x is in feet, determine the average normal stress in the ie. The inner diameter is 2 in., the wall thickness is 0.15 in. 8 ft omatibility: 0 d T - d F Where d T a T dx 0 8ft 0 6.6010-6 (40 + 15 x) dx - 0 F(8) (29.0)(10 3 ) 0 6.6010-6 40(8) + 15(8)2 R - 2 F(8) (29.0)(10 3 ) F 19.14 verage Normal Stress: s 19.14 19.1 ksi 176

04 Solutions 46060 5/25/10 3:20 M age 177 4 74. The bronze 86100 ie has an inner radius of 0.5 in. and a wall thickness of 0.2 in. If the gas flowing through it changes the temerature of the ie uniformly from T 200 F at to T 60 F at, determine the axial force it exerts on the walls.the ie was fitted between the walls when T 60 F. 8 ft Temerature Gradient: T(x) 60 + a 8 - x b140 200-17.5x 8 omatibility: 0 d T - d F Where d T 1 a Tdx 2ft 0 9.6010-6 [(200-17.5x) - 60] dx - 0 F(8) 4 (1.42-1 2 )15.0(10 3 ) 2ft 0 9.6010-6 (140-17.5x) dx - 0 F(8) 4 (1.42-1 2 ) 15.0(10 3 ) F 7.60 ki 4 75. The 40-ft-long -36 steel rails on a train track are laid with a small ga between them to allow for thermal exansion. Determine the required ga d so that the rails just touch one another when the temerature is increased from T 1-20 F to T 2 90 F. Using this ga, what would be the axial force in the rails if the temerature were to rise to T 3 110 F? The cross-sectional area of each rail is 5.10 in 2. d 40 ft d Thermal Exansion: Note that since adjacent rails exand, each rail will be d required to exand on each end, or d for the entine rail. 2 d a T 6.60(10-6 )[90 - (-20)](40)(12) 0.34848 in. 0.348 in. omatibility: : + 0.34848 d T - d F 0.34848 6.60(10-6 )[110 - (-20)](40)(12) - F(40)(12) 5.10(29.0)(10 3 ) F 19.5 ki 177

04 Solutions 46060 5/25/10 3:20 M age 178 *4 76. The device is used to measure a change in temerature. ars and D are made of -36 steel and 2014-T6 aluminum alloy resectively. When the temerature is at 75 F, E is in the horizontal osition. Determine the vertical dislacement of the ointer at E when the temerature rises to 150 F. 1.5 in. 0.25 in. 3 in. E Thermal Exansion: d T D a al T D 12.8(10-6 )(150-75)(1.5) 1.44(10-3 ) in. d T a st T 6.60(10-6 )(150-75)(1.5) 0.7425(10-3 ) in. From the geometry of the deflected bar E shown Fig. b, d E d T + d T D - d T S(3.25) 0.25 0.7425(10-3 ) + 1.44(10-3 ) - 0.7425(10-3 ) R(3.25) 0.25 D 0.00981 in. 4 77. The bar has a cross-sectional area, length, modulus of elasticity E, and coefficient of thermal exansion a. The temerature of the bar changes uniformly along its length from T at to T at so that at any oint x along the bar T T + x1t - T 2>. Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar and the bar has a temerature of T. T x T : + 0 T - d F (1) However, d T a T dx a(t + T - T T a 0 ac T - T 2 From Eq.(1). T - T 0 a 2 (T - T ) - F E x dx ac T - T 2 d a 2 (T - T ) x - T )dx x2 d 0 F a E 2 (T - T ) 178

04 Solutions 46060 5/25/10 3:20 M age 179 4 78. The -36 steel rod has a diameter of 50 mm and is lightly attached to the rigid suorts at and when T 1 80. If the temerature becomes T 2 20 and an axial force of 200 kn is alied to its center, determine the reactions at and. 0.5 m 0.5 m Referring to the FD of the rod, Fig. a : + F x 0; F - F + 200(10 3 ) 0 (1) When the rod is unconstrained at, it has a free contraction of d T a st T 12(10-6 )(80-20)(1000) 0.72 mm. lso, under force and F with unconstrained at, the deformation of the rod are d F d E 200(10 3 )(500) 4 (0.052 )200(10 9 )D F E F (1000) 4 (0.052 )200(10 9 )D Using the method of suer osition, Fig. b, 0.2546 mm 2.5465(10-6 ) F : + 0 -d T + d + d F 0-0.72 + 0.2546 + 2.5465(10-6 ) F F 182.74(10 3 ) N 183 kn Substitute the result of into Eq (1), F F 382.74(10 3 ) N 383 kn 179

04 Solutions 46060 5/25/10 3:20 M age 180 4 79. The -36 steel rod has a diameter of 50 mm and is lightly attached to the rigid suorts at and when T 1 50. Determine the force that must be alied to the collar at its midoint so that, when T 2 30, the reaction at is zero. 0.5 m 0.5 m When the rod is unconstrained at, it has a free contraction of d T a st T 12(10-6 )(50-30)(1000) 0.24 mm. lso, under force with unconstrained at, the deformation of the rod is d E (500) 4 (0.052 )200(10 9 )D 1.2732(10-6 ) Since F is required to be zero, the method of suerosition, Fig. b, gives : + 0 -d T + d 0-0.24 + 1.2732(10-6 ) 188.50(10 3 ) N 188 kn *4 80. The rigid block has a weight of 80 ki and is to be suorted by osts and, which are made of -36 steel, and the ost, which is made of 83400 red brass. If all the osts have the same original length before they are loaded, determine the average normal stress develoed in each ost when ost is heated so that its temerature is increased by 20 F. Each ost has a cross-sectional area of 8 in 2. Equations of Equilibrium: 3 ft 3 ft a+ M 0; F (3) - F (3) 0 F F F + c F y 0; 2F + F - 80 0 [1] omatibility: ( +T) (d ) F - (d ) T d F F 8(14.6)(10 3 ) - 9.8010-5 (20) F 8(29.0)(10 3 ) 8.5616 F - 4.3103 F 196 [2] Solving Eqs. [1] and [2] yields: average Normal Sress: F 22.81 ki F 34.38 ki s s F 22.81 8 2.85 ksi s F 34.38 4.30 ksi 8 180

04 Solutions 46060 5/25/10 3:20 M age 181 4 81. The three bars are made of -36 steel and form a in-connected truss. If the truss is constructed when T 1 50 F, determine the force in each bar when T Each bar has a cross-sectional area of 2 in 2 2 110 F.. 5 ft 5 ft 4 ft D 3 ft 3 ft (d T ) - (d F ) (d T ) D + (d F ) D (1) However, d d œ cos u; œ d d cos u 5 4 d Substitute into Eq. (1) 5 4 (d T) - 5 4 (d F) (d T ) D + (d F ) D 5 4 c6.60(10-6 )(110-50 )(5)(12) - F (5)(12) 2(29)(10 3 ) d 6.60(10-6 )(110-50 )(4)(12) + F D(4)(12) 2(29)(10 3 ) 620.136 75F + 48F D (2) : + F x 0; 3 5 F - 3 5 F 0; F F + c F y 0; F D - 2a 4 5 F b 0 (3) Solving Eqs. (2) and (3) yields : F D 6.54 ki F F 4.09 ki 181

04 Solutions 46060 5/25/10 3:20 M age 182 4 82. The three bars are made of -36 steel and form a inconnected truss. If the truss is constructed when T 1 50 F, determine the vertical dislacement of joint when T Each bar has a cross-sectional area of 2 in 2 2 150 F.. 5 ft 5 ft (d T ) - (d F ) (d T ) D + (d F ) D (1) 4 ft However, d d cos u; œ D œ d d cos u 5 4 d 3 ft 3 ft Substitute into Eq. (1) 5 4 (d T) - 5 4 (d T) (d T ) D + (d F ) D 5 4 c6.60(10-6 )(150-50 )(5)(12) - F (5)(12) 2(29)(10 3 ) d 6.60(10-6 )(150-50 )(4)(12) + F D(4)(12) 2(29)(10 3 ) 239.25-6.25F 153.12 + 4 F D 4 F D + 6.25F 86.13 (2) : + F x 0; 3 5 F - 3 5 F 0; F F + c F y 0; F D - 2a 4 5 F b 0; F D 1.6F (3) Solving Eqs. (2) and (3) yields: F 6.8086 ki: F D 10.8939 ki (d ) r (d T ) D + (d T ) D 6.60(10-6 )(150-50 )(4)(12) + 10.8939(4)(12) 2(29)(10 3 ) 0.0407 in. c 182

04 Solutions 46060 5/25/10 3:20 M age 183 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 E a st 8(10-6 cu 17(10 3 st 29(10 3 ) ksi, ) ksi, )> F, a cu 9.60(10-6 )> F. Each wire has a cross-sectional area of 0.0123 in 2. 60 in. D 40 in. 45 45 60 in. 150 lb Equations of Equilibrium: : + F x 0; F cos 45 - F cos 45 0 F F F + c F y 0; 2F sin 45 + F D - 150 0 [1] omatibility: (d ) T 8.010-6 (80)(60) 0.03840 in. (d ) Tr (d ) T cos 45 0.03840 0.05431 in. cos 45 (d D ) T 9.6010-6 (80)(40) 0.03072 in. d 0 (d ) Tr - (d D ) T 0.05431-0.03072 0.02359 in. (d D ) F (d ) Fr + d 0 F D (40) 0.0123(17.0)(10 6 ) F(60) 0.0123(29.0)(10 6 ) cos 45 + 0.02359 0.1913F D - 0.2379F 23.5858 [2] Solving Eq. [1] and [2] yields: F F F 10.0 lb F D 136 lb 183

04 Solutions 46060 5/25/10 3:20 M age 184 *4 84. The M1004-T61 magnesium alloy tube is caed with a rigid late E.The ga between E and end of the 6061-T6 aluminum alloy solid circular rod D is 0.2 mm when the temerature is at 30. Determine the normal stress develoed in the tube and the rod if the temerature rises to 80. Neglect the thickness of the rigid ca. a a 25 mm 20 mm E Section a-a 25 mm D 0.2 mm 300 mm 450 mm omatibility Equation: If tube and rod D are unconstrained, they will have a free exansion of d T a mg T 26(10-6 )(80-30)(300) 0.39 mm and d T ) D a al T D 24(10-6 )(80-30)(450) 0.54 mm. Referring to the deformation diagram of the tube and the rod shown in Fig. a, d d T - d F D + d T D - d F D D 0.2 0.39 - F 32 017.60 N Normal Stress: F(300) 0.025 2-0.02 2 (44.7)(10 9 ) S + 0.54 - F(450) 4 0.025 2 (68.9)(10 9 ) S s s D F F D 32 017.60 0.025 2-0.02 2 32 017.60 4 0.025 2 65.2 Ma F 107 442.47 N 45.3 Ma 184

04 Solutions 46060 5/25/10 3:20 M age 185 4 85. The M1004-T61 magnesium alloy tube is caed with a rigid late. The ga between E and end of the 6061-T6 aluminum alloy solid circular rod D is 0.2 mm when the temerature is at 30. Determine the highest temerature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid ca. a a 25 mm 20 mm E Section a-a 25 mm D 0.2 mm 300 mm 450 mm Then s D F 107 442.47 218.88Ma 6 (s Y ) al D 4 0.025 2 (O.K.!) omatibility Equation: If tube and rod D are unconstrained, they will have a free exansion of d 26(10-6 7.8(10-6 T a mg T )(T - 30)(300) ) (T - 30) and d 24(10-6 T D a al T D )(T - 30)(450) 0.0108(T - 30). Referring to the deformation diagram of the tube and the rod shown in Fig. a, d d T - d F D + d T D - d F D D 0.2 7.8(10-3 )(T - 30) - 107 442.47(300) 0.025 2-0.02 2 (44.7)(10 9 ) S 107 442.47(450) + 0.0108(T - 30) - 4 0.025 2 (68.9)(10 9 ) S T 172 185

04 Solutions 46060 5/25/10 3:20 M age 186 4 86. The steel bolt has a diameter of 7 mm and fits through an aluminum sleeve as shown. The sleeve has an inner diameter of 8 mm and an outer diameter of 10 mm. The nut at is adjusted so that it just resses u against the sleeve. If the assembly is originally at a temerature of T 1 20 and then is heated to a temerature of T 2 100, determine the average normal stress in the bolt and the sleeve. E st 200 Ga, 14(10-6 )>, a al 23(10-6 )>. E al 70 Ga, a st omatibility: (d s ) T - (d b ) T (d s ) F + (d b ) F 23(10-6 )(100-20) - 14(10-6 )(100-20) F 4 (0.012-0.008 2 )70(10 9 ) + F 4 (0.0072 )200(10 9 ) F 1133.54 N verage Normal Stress: s s F 1133.54 s 4 (0.012-0.008 2 40.1 Ma ) s b F 1133.54 29.5 Ma b 4 (0.0072 ) 4 87. Determine the maximum normal stress develoed in the bar when it is subjected to a tension of 8 kn. 5 mm 40 mm 20 mm For the fillet: w h 40 20 2 r h 10 20 0.5 r 10 mm 20 mm From Fig. 10-24. K 1.4 For the hole: From Fig. 4-25. K 2.375 s max Ks avg s max Ks avg 8 (10 3 ) 1.4 a 0.02 (0.005) b 112 Ma r w 10 40 0.25 8 (10 3 ) 2.375 a (0.04-0.02)(0.005) b 190 Ma 186

04 Solutions 46060 5/25/10 3:20 M age 187 *4 88. If the allowable normal stress for the bar is s allow 120 Ma, determine the maximum axial force that can be alied to the bar. ssume failure of the fillet. w h 40 20 2; r h 10 20 0.5 5 mm 40 mm 20 mm r 10 mm 20 mm From Fig. 4-24. K 1.4 s allow s max Ks avg 120 (10 6 ) 1.4 a 0.02 (0.005) b 8.57 kn ssume failure of the hole. r w 10 20 0.25 From Fig. 4-25. K 2.375 s allow s max Ks avg 120 (10 4 ) 2.375 a (0.04-0.02) (0.005) b 5.05 kn (controls) 4 89. The member is to be made from a steel late that is 0.25 in. thick. If a 1-in. hole is drilled through its center, determine the aroximate width w of the late so that it can suort an axial force of 3350 lb. The allowable stress is s allow 22 ksi. 3350 lb 3350 lb w 0.25 in. s allow s max Ks avg 3.35 22 Kc (w - 1)(0.25) d 1 in. w 3.35K + 5.5 5.5 y trial and error, from Fig. 4-25, choose r 0.2; K 2.45 w w 3.35(2.45) + 5.5 5.5 2.49 in. r Since w 0.5 2.49 0.2 OK 187

04 Solutions 46060 5/25/10 3:20 M age 188 4 90. The -36 steel late has a thickness of 12 mm. If there are shoulder fillets at and, and s allow 150 Ma, determine the maximum axial load that it can suort. alculate its elongation, neglecting the effect of the fillets. Maximum Normal Stress at fillet: From the text, K 1.4 r h 30 60 0.5 and w h 120 60 2 60 mm r 30 mm 60 mm 120 mm r 30 mm D 200 mm 800 mm 200 mm Dislacement: d E s max s allow Ks avg 77142.86(400) (0.06)(0.012)(200)(10 9 ) + 77142.86(800) (0.12)(0.012)(200)(10 9 ) 0.429 mm 150(10 6 ) 1.4 0.06(0.012) R 77142.86 N 77.1 kn 4 91. Determine the maximum axial force that can be alied to the bar. The bar is made from steel and has an allowable stress of s allow 21 ksi. 1.875 in. 0.125 in. 1.25 in. ssume failure of the fillet. From Fig. 4-24, K 1.73 r h 0.25 1.25 0.2 w h 1.875 1.25 1.5 0.75 in. r 0.25 in. ssume failure of the hole. From Fig. 4-25, K 2.45 s allow s max Ks avg 21 1.73 a 1.25 (0.125) b 1.897 ki r w 0.375 1.875 0.20 s allow s max Ks avg 21 2.45 a (1.875-0.75)(0.125) b 1.21 ki (controls) 188

04 Solutions 46060 5/25/10 3:20 M age 189 *4 92. Determine the maximum normal stress develoed in the bar when it is subjected to a tension of 2 ki. 1.875 in. 0.125 in. 1.25 in. t fillet: From Fig. 4-24, K 1.73 r h 0.25 1.25 0.2 w h 1.875 1.25 1.5 0.75 in. r 0.25 in. s max Ka b 1.73c 2 d 22.1 ksi 1.25(0.125) t hole: r w 0.375 1.875 0.20 From Fig. 4-25, K 2.45 2 s max 2.45c d 34.8 ksi (ontrols) (1.875-0.75)(0.125) 4 93. Determine the maximum normal stress develoed in the bar when it is subjected to a tension of 8 kn. 60 mm 5 mm 30 mm Maximum Normal Stress at fillet: r h 15 30 0.5 and w h 60 30 2 r 15 mm 12 mm From the text, K 1.4 s max Ks avg K h t Maximum Normal Stress at the hole: From the text, K 2.65 8(10 3 ) 1.4 R 74.7 Ma (0.03)(0.005) r w 6 60 0.1 s max K s avg K (w - 2r) t 8(10 3 ) 2.65 (0.06-0.012)(0.005) R 88.3 Ma (ontrols) 189

04 Solutions 46060 5/25/10 3:20 M age 190 4 94. The resulting stress distribution along section for the bar is shown. From this distribution, determine the aroximate resultant axial force alied to the bar. lso, what is the stress-concentration factor for this geometry? 0.5 in. 4 in. 1 in. sd Volume under curve 12 ksi 3 ksi Number of squares 10 10(3)(1)(0.5) 15 ki s avg 15 ki (4 in.)(0.5 in.) 7.5 ksi K s max s avg 12 ksi 7.5 ksi 1.60 4 95. The resulting stress distribution along section for the bar is shown. From this distribution, determine the aroximate resultant axial force alied to the bar. lso, what is the stress-concentration factor for this geometry? 0.6 in. 0.5 in. Number of squares 28 28(6)(0.2)(0.5) 16.8 ki s avg 16.8 28 ksi 2(0.6)(0.5) 0.2 in. 36 ksi 6 ksi 0.8 in. 0.6 in. K s max s avg 36 28 1.29 *4 96. The resulting stress distribution along section for the bar is shown. From this distribution, determine the aroximate resultant axial force alied to the bar. lso, what is the stress-concentration factor for this geometry? 20 mm 80 mm 10 mm Number of squares 19 30 Ma 5 Ma 19(5)(10 6 )(0.02)(0.01) 19 kn s avg 19(103 ) 23.75 Ma 0.08(0.01) K s max s avg 30 Ma 23.75 Ma 1.26 190

04 Solutions 46060 5/25/10 3:20 M age 191 4 97. The 300-ki weight is slowly set on the to of a ost made of 2014-T6 aluminum with an -36 steel core. If both materials can be considered elastic erfectly lastic, determine the stress in each material. 1 in. luminum 2 in. Equations of Equilibrium: Steel + c F y 0; st + al - 300 0 [1] Elastic nalysis: ssume both materials still behave elastically under the load. d st d al st 4 (2)2 (29)(10 3 ) al 4 (42-2 2 )(10.6)(10 3 ) st 0.9119 al Solving Eqs. [1] and [2] yields: al 156.91 ki st 143.09 ki verage Normal Stress: s al al al 156.91 4 (42-2 2 ) 16.65 ksi 6 (s g ) al 60.0 ksi (OK!) s st st 143.09 st 4 (22 ) 45.55 ksi 7 (s g ) st 36.0 ksi Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the steel is s st (s y ) st 36.0 ksi st (s g ) st st 36.0a 4 b 22 113.10 ki From Eq. [1] al 186.90 ki s al al al 186.90 4 (42-2 2 ) 19.83 ksi 6 (s g) al 60.0 ksi Then s al 19.8 ksi 191

04 Solutions 46060 5/25/10 3:20 M age 192 4 98. The bar has a cross-sectional area of 0.5 in 2 and is 8 ki 5 ki made of a material that has a stress strain diagram that can be aroximated by the two line segments shown. 5 ft 2 ft Determine the elongation of the bar due to the alied s(ksi) loading. 40 20 verage Normal Stress and Strain: For segment verage Normal Stress and Strain: For segment Elongation: s 5 10.0 ksi 0.5 10.0 20 e 0.001 ; e 0.001 (10.0) 0.00050 in.>in. 20 s 13 26.0 ksi 0.5 26.0-20 e - 0.001 40-20 0.021-0.001 e 0.0070 in.>in. d e 0.00050(2)(12) 0.0120 in. d e 0.0070(5)(12) 0.420 in. 0.001 0.021 (in./in.) d Tot d + d 0.0120 + 0.420 0.432 in. 192

04 Solutions 46060 5/25/10 3:20 M age 193 4 99. The rigid bar is suorted by a in at and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is s Y 530 Ma, and E st 200 Ga, determine the intensity of the distributed load w that can be laced on the beam and will just cause wire E to yield. What is the dislacement of oint G for this case? For the calculation, assume that the steel is elastic erfectly lastic. E D 800 mm G Equations of Equilibrium: 400 mm 250 mm 150 mm w a+ M 0; F E (0.4) + F D (0.65) - 0.8w (0.4) 0 0.4 F E + 0.65F D 0.32w [1] lastic nalysis: Wire D will yield first followed by wire E. When both wires yield F E F D (s g ) 53010 6 a 4 b 0.0042 6.660 kn Substituting the results into Eq. [1] yields: w 21.9 kn>m Dislacement: When wire E achieves yield stress, the corresonding yield strain is From the geometry e g s g E 530(106 ) 200(10 9 0.002650 mm>mm ) d E e g E 0.002650(800) 2.120 mm d G 0.8 d E 0.4 d G 2d E 2(2.120) 4.24 mm 193

04 Solutions 46060 5/25/10 3:20 M age 194 *4 100. The rigid bar is suorted by a in at and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is s Y 530 Ma, and E st 200 Ga, determine (a) the intensity of the distributed load w that can be laced on the beam that will cause only one of the wires to start to yield and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the calculation, assume that the steel is elastic erfectly lastic. Equations of Equilibrium: a+ M 0; F E (0.4) + F D (0.65) - 0.8w (0.4) 0 E D 800 mm G 400 mm 250 mm w 150 mm 0.4 F E + 0.65 F D 0.32w [1] (a) y observation, wire D will yield first. Then F D s g 53010 6 a. 4 b 0.0042 6.660 kn From the geometry d E 0.4 d D 0.65 ; d D 1.625d E F D E 1.625 F E E F D 1.625 F E [2] Using F D 6.660 kn and solving Eqs. [1] and [2] yields: F E 4.099 kn w 18.7 kn>m (b) When both wires yield F E F D (s g ) 53010 6 a 4 b 0.0042 6.660 kn Substituting the results into Eq. [1] yields: w 21.9 kn>m 194

04 Solutions 46060 5/25/10 3:20 M age 195 4 101. The rigid lever arm is suorted by two -36 steel wires having the same diameter of 4 mm. If a force of 3 kn is alied to the handle, determine the force develoed in both wires and their corresonding elongations. onsider -36 steel as an elastic-erfectly lastic material. 150 mm 150 mm 450 mm 30 300 mm E Equation of Equilibrium. Refering to the free-body diagram of the lever shown in Fig. a, a+ M E 0; F (300) + F D (150) - 310 3 (450) 0 D 2F + F D 910 3 (1) Elastic nalysis. ssuming that both wires and D behave as linearly elastic, the comatibility equation can be written by referring to the geometry of Fig. b. d a 300 150 bd D d 2d D (2) F E 2a F D E b F 2F D (3) Solving Eqs. (1) and (3), F D 1800 N F 3600 N Normal Stress. s D F D D s F 1800 4 0.004 2 3600 4 0.004 2 143.24 Ma 6 (s Y ) st 286.48 Ma 7 (s Y ) st (O.K.) (N.G.) Since wire yields, the elastic analysis is not valid. The solution must be reworked using F (s Y ) st 25010 6 c 4 0.0042 d 3141.59 N 3.14 kn Substituting this result into Eq. (1), F D 2716.81 N 2.72 kn s D F D 2716.81 D 4 0.004 2 216.20 Ma 6 (s Y ) st (O.K.) 195

04 Solutions 46060 5/25/10 3:20 M age 196 4 101. ontinued Since wire D is linearly elastic, its elongation can be determined by d D F D D D E st 2716.81(300) 4 0.004 2 (200)10 9 0.3243 mm 0.324 mm From Eq. (2), d 2d D 2(0.3243) 0.649 mm 196

04 Solutions 46060 5/25/10 3:20 M age 197 4 102. The rigid lever arm is suorted by two -36 steel wires having the same diameter of 4 mm. Determine the smallest force that will cause (a) only one of the wires to yield; (b) both wires to yield. onsider -36 steel as an elastic-erfectly lastic material. 150 mm 150 mm 450 mm 30 300 mm E D Equation of Equilibrium. Refering to the free-body diagram of the lever arm shown in Fig. a, a+ M E 0; F (300) + F D (150) - (450) 0 2F + F D 3 (1) Elastic nalysis. The comatibility equation can be written by referring to the geometry of Fig. b. d a 300 150 bd D d 2d D F E 2a F D E b F D 1 2 F (2) ssuming that wire is about to yield first, F (s Y ) st 25010 6 c 4 0.0042 d 3141.59 N From Eq. (2), F D 1 (3141.59) 1570.80 N 2 Substituting the result of F and F D into Eq. (1), 2618.00 N 2.62 kn lastic nalysis. Since both wires and D are required to yield, F F D (s Y ) st 25010 6 c 4 0.0042 d 3141.59 N Substituting this result into Eq. (1), 197

04 Solutions 46060 5/25/10 3:20 M age 198 4 103. The three bars are inned together and subjected to the load. If each bar has a cross-sectional area, length, and is made from an elastic erfectly lastic material, for which the yield stress is s Y, determine the largest load (ultimate load) that can be suorted by the bars, i.e., the load that causes all the bars to yield. lso, what is the horizontal dislacement of oint when the load reaches its ultimate value? The modulus of elasticity is E. D u u 3141.59 N 3.14 kn When all bars yield, the force in each bar is, F Y s Y : + F x 0; - 2s Y cos u - s Y 0 s Y (2 cos u + 1) ar will yield first followed by bars and D. d d D F Y() E s Y E s Y E d d cos u s Y E cos u *4 104. The rigid beam is suorted by three 25-mm diameter -36 steel rods. If the beam suorts the force of 230 kn, determine the force develoed in each rod. onsider the steel to be an elastic erfectly-lastic material. Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, 600 mm D E F + c F y 0; a + M 0; F D + F E + F F - 23010 3 0 F E (400) + F F (1200) - 23010 3 (800) 0 (1) 400 mm 400 mm 400 mm F E + 3F F 46010 3 (2) Elastic nalysis. Referring to the deflection diagram of the beam shown in Fig. b, the comatibility equation can be written as d E d D + a d F - d D b(400) 1200 d E 2 3 d D + 1 3 d F F E E 2 3 a F D E b + 1 3 a F F E b F E 2 3 F D + 1 3 F F (3) Solving Eqs. (1), (2), and (3) F F 131 428.57 N F E 65 714.29 N F D 32 857.14 N 198

04 Solutions 46060 5/25/10 3:20 M age 199 4 104. ontinued Normal Stress. s F F F 131428.57 267.74 Ma 7 (s Y ) st F 4 0.025 2 s E F E 65714.29 E 4 0.025 2 133.87 Ma 6 (s Y ) st s D F D 32857.14 D 4 0.025 2 66.94 Ma 6 (s Y ) st (N.G.) (O.K.) (O.K.) Since rod F yields, the elastic analysis is not valid. The solution must be reworked using F F (s Y ) st F 25010 6 c 4 0.0252 d 122 718.46 N 123 kn Substituting this result into Eq. (2), F E 91844.61 N 91.8 kn Substituting the result for F F and F E into Eq. (1), F D 15436.93 N 15.4 kn s E F E 91844.61 E 4 0.025 2 187.10 Ma 6 (s Y ) st s D F D 15436.93 D 4 0.025 2 31.45 Ma 6 (s Y ) st (O.K.) (O.K.) 199

04 Solutions 46060 5/25/10 3:20 M age 200 4 105. The rigid beam is suorted by three 25-mm diameter -36 steel rods. If the force of 230 kn is alied on the beam and removed, determine the residual stresses in each rod. onsider the steel to be an elastic erfectly-lastic material. 600 mm D E F 400 mm 400 mm 400 mm Equation of Equilibrium. Referring to the free-body diagram of the beam shown in Fig. a, + c F y 0; a + M 0; F D + F E + F F - 23010 3 0 F E (400) + F F (1200) - 23010 3 (800) 0 F E + 3F F 46010 3 (1) (2) Elastic nalysis. Referring to the deflection diagram of the beam shown in Fig. b, the comatibility equation can be written as d E d D + a d F - d D b(400) 1200 d E 2 3 d D + 1 3 d F (3) F E E 2 3 a F D E b + 1 3 a F F E b F E 2 3 F D + 1 3 F F (4) Solving Eqs. (1), (2), and (4) F F 131428.57 N F E 65714.29 N F D 32857.14 N Normal Stress. s F F F 131428.57 267.74 Ma (T) 7 (s Y ) st F 4 0.025 2 s E F E 65714.29 E 4 0.025 2 133.87 Ma (T) 6 (s Y ) st s D F D 32857.14 D 4 0.025 2 66.94 Ma (T) 6 (s Y ) st (N.G.) (O.K.) (O.K.) Since rod F yields, the elastic analysis is not valid. The solution must be reworked using s F (s Y ) st 250 Ma (T) F F s F F 25010 6 c 4 0.0252 d 122718.46 N 200

04 Solutions 46060 5/25/10 3:20 M age 201 4 105. ontinued Substituting this result into Eq. (2), F E 91844.61 N Substituting the result for F F and F E into Eq. (1), F D 15436.93N s E F E 91844.61 E 4 0.025 2 187.10 Ma (T) 6 (s Y ) st s D F D 15436.93 D 4 0.025 2 31.45 Ma (T) 6 (s Y ) st (O.K.) (O.K.) Residual Stresses. The rocess of removing can be reresented by alying the force, which has a magnitude equal to that of but is oosite in sense, Fig. c. Since the rocess occurs in a linear manner, the corresonding normal stress must have the same magnitude but oosite sense to that obtained from the elastic analysis. Thus, œ s F œ 267.74 Ma () s E œ 133.87 Ma () s D 66.94 Ma () onsidering the tensile stress as ositive and the comressive stress as negative, œ (s F ) r s F + s F œ (s E ) r s E + s E œ (s D ) r s D + s D 250 + (-267.74) -17.7 Ma 17.7 Ma () 187.10 + (-133.87) 53.2 Ma (T) 31.45 + (-66.94) -35.5 Ma 35.5 Ma () 201

04 Solutions 46060 5/25/10 3:20 M age 202 4 106. The distributed loading is alied to the rigid beam, which is suorted by the three bars. Each bar has a cross-sectional area of 1.25 in 2 and is made from a material having a stress strain diagram that can be aroximated by the two line segments shown. If a load of w 25 ki>ft is alied to the beam, determine the stress in each bar and the vertical dislacement of the beam. 60 36 s(ksi) 4 ft 4 ft 5 ft 0.0012 0.2 (in./in.) w a + M 0; F (4) - F (4) 0; F F F + c F y 0; 2F + F - 200 0 (1) Since the loading and geometry are symmetrical, the bar will remain horizontal. Therefore, the dislacement of the bars is the same and hence, the force in each bar is the same. From Eq. (1). F F 66.67 ki Thus, s s s 66.67 1.25 53.33 ksi From the stress-strain diagram: 53.33-36 e - 0.0012 60-36 : e 0.14477 in.>in. 0.2-0.0012 d e 0.14477(5)(12) 8.69 in. 202

04 Solutions 46060 5/25/10 3:20 M age 203 4 107. The distributed loading is alied to the rigid beam, which is suorted by the three bars. Each bar has a cross-sectional area of 0.75 in 2 and is made from a material having a stress strain diagram that can be aroximated by the two line segments shown. Determine the intensity of the distributed loading w needed to cause the beam to be dislaced downward 1.5 in. 60 36 s(ksi) 4 ft 4 ft 5 ft 0.0012 0.2 (in./in.) w a+ M 0; F (4) - F (4) 0; F F F + c F y 0; 2F + F - 8 w 0 (1) Since the system and the loading are symmetrical, the bar will remain horizontal. Hence the dislacement of the bars is the same and the force suorted by each bar is the same. From Eq. (1), F F 2.6667 w (2) From the stress-strain diagram: e 1.5 0.025 in.>in. 5 (12) s - 36 0.025-0.0012 60-36 ; s 38.87 ksi 0.2-0.0012 Hence F s 38.87 (0.75) 29.15 ki From Eq. (2), w 10.9 ki>ft 203

04 Solutions 46060 5/25/10 3:20 M age 204 *4 108. The rigid beam is suorted by the three osts,, and of equal length. osts and have a diameter of 75 mm and are made of aluminum, for which E al 70 Ga and 1s Y 2 al 20 Ma. ost has a diameter of 20 mm and is made of brass, for which E br 100 Ga and 1s Y 2 br 590 Ma. Determine the smallest magnitude of so that (a) only rods and yield and (b) all the osts yield. al 2 m br 2 m 2 m 2 m al M 0; F F F al + c F y 0; F at + 2F at - 2 0 (1) (a) ost and will yield, F al (s t ) al omatibility condition: 20(10 4 )( a )(0.075)2 88.36 kn (E al ) r (s r) al E al 20(104 ) 70(10 4 ) 0.0002857 d br d al 0.0002857() F br () 4 (0.02)2 (100)(10 4 ) 0.0002857 F br 8.976 kn From Eq. (1), s br 8.976(103 ) 4 (0.023 ) 28.6 Ma 6 s r OK. 8.976 + 2(88.36) - 2 0 92.8 kn (b) ll the osts yield: F br (s r ) br (590)(10 4 )( 4 )(0.022 ) 185.35 kn F al 88.36 kn From Eq. (1); 185.35 + 2(88.36) - 2 0 181 kn 204

04 Solutions 46060 5/25/10 3:20 M age 205 4 109. The rigid beam is suorted by the three osts,, and. osts and have a diameter of 60 mm and are made of aluminum, for which E al 70 Ga and 1s Y 2 al 20 Ma. ost is made of brass, for which E br 100 Ga and 1s Y 2 br 590 Ma. If 130 kn, determine the largest diameter of ost so that all the osts yield at the same time. al br al 2 m 2 m 2 m 2 m + c F y 0; 2(F g ) al + F br - 260 0 (1) (F al ) g (s g ) al 20(10 6 )( 4 )(0.06)2 56.55 kn From Eq. (1), 2(56.55) + F br - 260 0 F br 146.9 kn (s g ) br 590(10 6 ) 146.9(103 ) 4 (d ) 3 d 0.01779 m 17.8 mm 205

04 Solutions 46060 5/25/10 3:20 M age 206 4 110. The wire has a diameter of 0.125 in. and the material has the stress strain characteristics shown in the figure. Determine the vertical dislacement of the handle at D if the ull at the gri is slowly increased and reaches a magnitude of (a) 450 lb, (b) 600 lb. 40 in. D Equations of Equilibrium: s (ksi) 50 in. 30 in. a + M 0; F (50) - (80) 0 [1] (a) From Eq. [1] when 450 lb, F 720 lb verage Normal Stress and Strain: From the Stress Strain diagram Dislacement: s F 720 4 (0.1252 ) 58.67 ksi 58.67 e 70 0.007 ; e 0.005867 in.>in. d e 0.005867(40) 0.2347 in. 80 70 0.007 0.12 (in./in.) d D 80 d 50 ; d D 8 (0.2347) 0.375 in. 5 (b) From Eq. [1] when 600 lb, F 960 lb verage Normal Stress and Strain: From Stress Strain diagram Dislacement: s F 78.23-70 e - 0.007 80-70 0.12-0.007 e 0.09997 in.>in. d e 0.09997(40) 3.9990 in. 960 78.23 ksi 4 (0.125)2 d D 80 d 50 ; d D 8 (3.9990) 6.40 in. 5 206

04 Solutions 46060 5/25/10 3:20 M age 207 4 111. The bar having a diameter of 2 in. is fixed connected at its ends and suorts the axial load. If the material is elastic erfectly lastic as shown by the stress strain diagram, determine the smallest load needed to cause segment to yield. If this load is released, determine the ermanent dislacement of oint. 2 ft 3 ft s (ksi) 20 0.001 (in./in.) When is increased, region will become lastic first, then will become lastic. Thus, F F s 20()(1) 2 62.832 ki : + F x 0; F + F - 0 (1) 2(62.832) 125.66 ki 126 ki The deflection of oint is, d e (0.001)(3)(12) 0.036 in. ; onsider the reverse of on the bar. F (2) E F (3) E F 1.5 F So that from Eq. (1) F 0.4 F 0.6 d F E 0.4()(3)(12) E 0.4(125.66)(3)(12) (1) 2 (20>0.001) 0.02880 in. : d 0.036-0.0288 0.00720 in. ; 207

04 Solutions 46060 5/25/10 3:20 M age 208 *4 112. Determine the elongation of the bar in rob. 4 111 when both the load and the suorts are removed. 2 ft 3 ft s (ksi) 20 When is increased, region will become lastic first, then will become lastic. Thus, F F s 20()(1) 2 62.832 ki 0.001 (in./in.) : + F x 0; F + F - 0 (1) 2(62.832) 125.66 ki 126 ki The deflection of oint is, d e (0.001)(3)(12) 0.036 in. ; onsider the reverse of on the bar. F (2) E F (3) E F 1.5 F So that from Eq. (1) F 0.4 F 0.6 The resultant reactions are F F -62.832 + 0.6(125.66) 62.832-0.4(125.66) 12.568 ki When the suorts are removed the elongation will be, d E 12.568(5)(12) (1) 2 0.0120 in. (20>0.001) 208

04 Solutions 46060 5/25/10 3:20 M age 209 4 113. material has a stress strain diagram that can be described by the curve s c 1>2. Determine the deflection d of the end of a rod made from this material if it has a length, cross-sectional area, and a secific weight g. s s c e 1 2 ; s 2 c 2 e s 2 (x) c 2 e(x) (1) d However s(x) (x) dd ; e(x) dx From Eq. (1), 2 (x) 2 c 2 dd dx ; dd dx 2 (x) 2 c 2 d g2 1 2 c 2 2 (x) dx c 2 0 d g3 3 3c 2 x 2 dx g2 c 2 x3 3 0 1 2 c 2 (gx) 2 dx 0 4 114. The 2014-T6 aluminum rod has a diameter of 0.5 in. and is lightly attached to the rigid suorts at and when T 1 70 F. If the temerature becomes T 2-10 F, and an axial force of 16 lb is alied to the rigid collar as shown, determine the reactions at and. /2 /2 5 in. 8 in. : + 0 - T + d 0 0.016(5) 4 (0.52 )(10.6)(10 3 ) - 12.8(10-6 F (13) )[70 - (-10 )](13) + 4 (0.52 )(10.6)(10 3 ) F 2.1251 ki 2.13 ki : + F x 0; 2(0.008) + 2.1251 - F 0 F 2.14 ki 4 115. The 2014-T6 aluminum rod has a diameter of 0.5 in. and is lightly attached to the rigid suorts at and when T 1 70 F. Determine the force that must be alied to the collar so that, when T 0 F, the reaction at is zero. /2 /2 5 in. 8 in. : + 0 - T + d (5) 0 4 (0.52 )(10.6)(10 3 ) - 12.8(10-6 )[(70)(13)] + 0 4.85 ki 209

04 Solutions 46060 5/25/10 3:20 M age 210 *4 116. The rods each have the same 25-mm diameter and 600-mm length. If they are made of -36 steel, determine the forces develoed in each rod when the temerature increases to 50. 600 mm 60 60 Equation of Equilibrium: Referring to the free-body diagram of joint shown in Fig. a, 600 mm D + c F x 0; : + F x 0; F D sin 60 - F sin 60 0 F - 2F cos 60 0 F F D F F F (1) omatibility Equation: If and are unconstrained, they will have a free exansion of d T d T a st T 12(10-6 )(50)(600) 0.36 mm. Referring to the initial and final osition of joint, d F - d T ad T b - d F Due to symmetry, joint will dislace horizontally, and d. Thus, cos 60 2d ad T b 2(d T ) and d F 2d F. Thus, this equation becomes d d F - d T 2d T - 2d F (600) 4 0.025 2 (200)(10 9 ) F + 2F 176 714.59 F(600) - 0.36 2(0.36) - 2 4 0.025 2 (200)(10 9 ) S (2) Solving Eqs. (1) and (2), F F F D 58 904.86 N 58.9 kn 210

04 Solutions 46060 5/25/10 3:20 M age 211 4 117. Two -36 steel ies, each having a crosssectional area of 0.32 in 2, are screwed together using a union at as shown. Originally the assembly is adjusted so that no load is on the ie. If the union is then tightened so that its screw, having a lead of 0.15 in., undergoes two full turns, determine the average normal stress develoed in the ie. ssume that the union at and coulings at and are rigid. Neglect the size of the union. Note: The lead would cause the ie, when unloaded, to shorten 0.15 in. when the union is rotated one revolution. 3 ft 2 ft The loads acting on both segments and are the same since no external load acts on the system. 0.3 d > + d > 0.3 (3)(12) 0.32(29)(10 3 ) + (2)(12) 0.32(29)(10 3 ) 46.4 ki s s 46.4 145 ksi 0.32 4 118. The brass lug is force-fitted into the rigid casting. The uniform normal bearing ressure on the lug is estimated to be 15 Ma. If the coefficient of static friction between the lug and casting is m s 0.3, determine the axial force needed to ull the lug out. lso, calculate the dislacement of end relative to end just before the lug starts to sli out. E br 98 Ga. 15 Ma 100 mm 150 mm 20 mm Equations of Equilibrium: : + F x 0; - 4.50(10 6 )(2)()(0.02)(0.1) 0 56.549 kn 56.5 kn Dislacement: d > a E 56.549(103 )(0.15) (0.02 2 )(98)(10 9 ) + 0 0.1 m 0.56549(10 6 ) x dx (0.02 2 )(98)(10 9 ) 0.00009184 m 0.0918 mm 211

04 Solutions 46060 5/25/10 3:20 M age 212 4 119. The assembly consists of two bars and D of the same material having a modulus of elasticity E 1 and coefficient of thermal exansion a 1, and a bar EF having a modulus of elasticity E 2 and coefficient of thermal exansion a 2. ll the bars have the same length and cross-sectional area. If the rigid beam is originally horizontal at temerature T 1, determine the angle it makes with the horizontal when the temerature is increased to T 2. D F E Equations of Equilibrium: a+ M 0; F F EF F d d + c F y 0; F D - 2F 0 [1] omatibility: d (d ) T - (d ) F d D (d D ) T + (d D ) F d EF (d EF ) T - (d EF ) F From the geometry d D - d d d EF - d 2d 2d D d EF + d 2(d D ) T + (d D ) F D (d EF ) T - (d EF ) F + (d ) T - (d ) F 2a 1 (T 2 - T 1 ) + F D () R E 1 a 2 (T 2 - T 1 ) - F() E 2 + a 1 (T 2 - T 1 ) - F() E 1 [2] Substitute Eq. [1] into [2]. 2a 1 (T 2 - T 1 ) + 4F E 1 a 2 (T 2 - T 1 ) - F E 2 + a 1 (T 2 - T 1 ) - F E 1 5F E 1 + F E 2 a 2 (T 2 - T 1 ) - a 1 (T 2 - T 1 ) F 5E 2 + E 1 E 1 E 2 b (T 2 - T 1 )(a 2 - a 1 ) ; F E 1E 2 (T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 (d EF ) T a 2 (T 2 - T 1 ) (d EF ) F E 1E 2 (T 2 - T 1 )(a 2 - a 1 )() E 2 (5E 2 + E 1 ) E 1 (T 2 - T 1 )(a 2 - a 1 )() 5E 2 + E 1 d EF (d EF ) T - (d EF ) F a 2 (T 2 - T 1 )(5E 2 - E 1 ) - E 1 (T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 (d ) T a 1 (T 2 - T 1 ) (d ) F E 1E 2 (T 2 - T 1 )(a 2 - a 1 )() E 1 (5E 2 + E 1 ) E 2 (T 2 - T 1 )(a 2 - a 1 )() 5E 2 + E 1 d (d ) T - (d ) F a 1 (5E 2 + E 1 )(T 2 - T 1 ) - E 2 (T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 212

04 Solutions 46060 5/25/10 3:20 M age 213 4 119. ontinued d EF - d (T 2 - T 1 ) 5E 2 + E 1 [a 2 (5E 2 + E 1 ) - E 1 (a 2 - a 1 ) - a 1 (5E 2 + E 1 ) + E 2 (a 2 - a 1 )] (T 2 - T 1 ) 5E 2 + E 1 (5E 2 + E 1 )(a 2 - a 1 ) + (a 2 - a 1 )(E 2 - E 1 )D (T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 (5E 2 + E 1 + E 2 - E 1 ) (T 2 - T 1 )(a 2 - a 1 )(6E 2 ) 5E 2 + E 1 u d EF - d 2d 3E 2(T 2 - T 1 )(a 2 - a 1 ) d(5e 2 + E 1 ) *4 120. The rigid link is suorted by a in at and two -36 steel wires, each having an unstretched length of 12 in. and cross-sectional area of 0.0125 in 2. Determine the force develoed in the wires when the link suorts the vertical load of 350 lb. 5 in. 12 in. Equations of Equilibrium: a + M 0; -F (9) - F (4) + 350(6) 0 [1] 4 in. omatibility: 6 in. d 4 F () 4E d 9 F () 9E 350 lb 9F - 4F 0 [2] Solving Eqs. [1] and [2] yields: F 86.6 lb F 195 lb 213

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