CONTENTS. To the Instructor. 1 Stress 1. 2 Strain Mechanical Properties of Materials Axial Load Torsion 214.

Transcription

1 FM_TO /22/10 11:26 M Page iii ONTENTS To the Instructor iv 1 Stress 1 2 Strain 73 3 Mechanical Proerties of Materials 92 4 xial Load Torsion ending Transverse Shear ombined Loadings Stress Transformation Strain Transformation Design of eams and Shafts Deflection of eams and Shafts uckling of olumns Energy Methods 1159

2 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 PL E (10 3 )(8) 4 ( ) 200(10 9 ) 8 m (10-6 ) m mm Negative sign indicates that end moves towards end D 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 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 P il i P L + P L + P D L D i E i E u E u D E u 6.00 (50) (2.25)18(10 3 )D (75) 18(10 3 )D (60) (0.25) 18(10 3 )D 0.766(10-3 ) in. The ositive sign indicates that end moves away from D. 122

3 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 mm b 50.0(10-6 ) m 2 d D P il i i E i 1 E S ap L + P L + P D L D b (10-6 ) 200(10 9 )D c -3.00(10 3 )(1) (10 3 )(1.5) (10 3 )(1.25) d 0.850(10-3 ) m mm The ositive sign indicates that end D moves away from the fixed suort. 123

4 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 m a. Thus, mm b 50.0 (10-6 ) m 2 d P il i i E i 1 E S P L + P L (10-6 ) 200(10 9 )D c -3.00(10 3 )(1) (10 3 )(1.5) d (10-3 ) m mm The ositive sign indicates that couling moves away from the fixed suort 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)( 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 4 6. The bar has a cross-sectional area of 3 in 2, and x w 500x 1/3 lb/in. E ksi. Determine the dislacement of its end when it is subjected to the distributed loading. 4 ft x x P(x) w dx 500 x 1 3 dx 1500 L L 4 0 L P(x) dx d L 0 E 1 (3)(35)(10 6 ) L0 0 4(12) x x dx a (3)(35)(10 8 )(4) ba3 7 b(48)1 3 d in. 124

5 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 lb a+ M 0; 640(5) - F DE (7) 0 F DE lb Since E and F are fixed, d D F DE L DE E st d F F L F E st From the geometry shown in Fig. c, (4)(2) (10 6 )D (4)(12) (10 6 )D in T in T E H D 2 ft 5 ft 4.5 ft 800 lb 1 ft F 4 ft d H ( ) in T 7 Subsequently, Thus, d >H F H L H E st d > F L E st 160(4.5)(12) (10 6 )D +T d d H + d >H in T +T d d + d > in T From the geometry shown in Fig. d, 640(4.5)(12) (10 6 )D in T in T d P ( ) in T 5 125

6 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 lb a+ M 0; 640(5) - F DE (7) 0 F DE lb Since E and F are fixed, d D F DE L DE E st d F F L F E st From the geometry shown in Fig. c (4)(12) (10 6 )D (4)(12) (10 6 )D in T in T E H D 2 ft 5 ft 4.5 ft 800 lb 1 ft F 4 ft d H ( ) in T 7 u (12) 46.6(10-6 ) rad Subsequently, d >H F H L H E st d > F L E st 640 (4.5)(12) (10 6 )D 160 (4.5)(12) (10 6 )D in T in T Thus, +T d d H + d >H in T +T d d + d > in T From the geometry shown in Fig. d f (12) 0.355(10-3 ) rad 126

7 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued 127

8 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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; F 0 F 4.00 ki D 4 ft D 1 in in 2 6 ft E 2 ft 1 ft F 6 ki 1 ft EF 2 in 2 Dislacement: d F D L D D E 2.00(4)(12) (1)(17.4)( in. ) d F L E 4.00(6)(12) (1.5)(17.4)( in. ) d F>E F EF L EF EF E 6.00(1)(12) (2)(17.4)( in. ) de œ ; de œ in. 3 d E d + d œ E in. d F d E + d F>E in 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; F 0 F 4.00 ki D 4 ft D 1 in in 2 6 ft E 2 ft 1 ft F 6 ki 1 ft EF 2 in 2 Dislacement: d F D L D D E 2.00(4)(12) (1)(17.4)( in. ) d F L E 4.00(6)(12) (1.5)(17.4)( in. ) u tan - 1 d - d 3(12) tan 3(12)

9 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 in 2. E F G 3 ft Internal Forces in the wires: FD (b) a + M 0; F (4) - 500(3) 0 F 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 F H lb FD (a) a + M D 0; F F (3) (1) 0 F F lb + c F y 0; F DE F DE lb Dislacement: d D F DEL DE DE E 83.33(3)(12) in (28.0)(10 6 ) d F FL F F E 41.67(3)(12) in (28.0)(10 6 ) dh œ ; dh œ in. 3 d H in. d >H F HL H H E 125.0(1.8)(12) in (28.0)(10 6 ) d d H + d >H in. d F GL G G E 375.0(5)(12) in (28.0)(10 6 ) dl œ ; dl œ in. 4 d l in. 129

10 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 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 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 F H lb FD (a) a + M D 0; F F (3) (1) 0 F F lb + c F y 0; F DE F DE lb Dislacement: d D F DEL DE DE E 83.33(3)(12) in (28.0)(10 6 ) d F FL F F E 41.67(3)(12) in (28.0)(10 6 ) dh œ ; dh œ in. 3 d H d œ H + d in. tan a ; a d >H F HL H H E 125.0(1.8)(12) in (28.0)(10 6 ) d d H + d >H in. d F GL G G E 375.0(5)(12) in (28.0)(10 6 ) tan b ; b

11 04 Solutions /25/10 3:19 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The bar has a length L and cross-sectional area. Determine its elongation due to the force P and its own weight.the material has a secific weight g (weight>volume) and a modulus of elasticity E. P(x) dx d L (x) E 1 E L0 L (gx + P) dx L 1 E a gl2 2 + PLb gl2 2E + PL E P 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 F 12.0 kn F Internal Force: FD (b) + c F y 0; -F(y) + 4y F(y) {4y - 20} kn Dislacement: L F(y)dy d > L 0 (y)e 1 E L0 2 m (4y - 20)dy 1 E 2y2-20y 2 m kn # m E 32.0(10 3 ) - 4 (0.062 ) 13.1 (10 9 ) m mm Negative sign indicates that end moves toward end. 131

12 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 F 17.0 kn Internal Force: FD (b) + c F y 0; -F(y) a 3y by F(y) e 3 4 y2-20 f kn Dislacement: L F(y) dy d > L 0 (y)e 1 E L0 2m 1 E a y3 2 m 4-20yb kn # m E 38.0(10 3 ) - 4 (0.062 ) 13.1 (10 9 ) m a 3 4 y2-20bdy mm Negative sign indicates that end moves toward end. 132

13 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 P 50 ki is alied to oint, determine its vertical dislacement at. P 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 F ki 5 The initial length of members and is L in (2.50 ft)a b 30 in. The axial deformation of members 1 ft and is d FL E (-31.25)(30) (1.5)29.0(10 3 )D 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 d g d cos u in. T cos

14 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P needed to dislace oint 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 - P 0 F P 2 ft 1.5 ft P 1.5 ft The initial length of members and are L in (2.50 ft)a b 30 in. The axial deformation of members 1 ft and is d FL E P(30) (1.5)29.0(10 3 )D (10-3 ) P 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 (d ) g d cos u (10-3 ) P cos P 46.4 kis 134

15 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 ) in 2 and are fixed, d F L E st d D F D L D E st 3.75 (2)(12) (10 3 )D 6.25(3)(12) (10 3 )D in. T in T From the geometry shown in Fig. b Here, d E ( ) in. T d F>E F EF L EF E st 10 (1) (12) (10 3 )D in T Thus, +T d F d E + d F>E in T 135

16 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 ) in 2 and are fixed then, 1.25 ft 0.75 ft F 1 ft d F L E st d D F D L D E st From the geometry shown in Fig. b, 3.75 (2)(12) (10 3 )D 6.25 (3)(12) (10 3 )D in T in T 10 ki u (12) 0.439(10-3 ) rad 136

17 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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) (45)(4) c 1 3 (4) d 0 F 50.0 kn 4 m Dislacement. The initial length of rod is L m. The axial deformation of this rod is d F L 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 (d ) g d sin u 2.50(10-3 ) sin (10-3 ) m 4.17 mm 137

18 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 F 2.00 kn FD (b) + c F y 0; F EF F EF 4.00 kn Dislacement: d D d E F EFL EF EF E 4.00(10 3 )(750) 4 (0.012)2 (193)( mm ) d > d >D P DL D D E 2(10 3 )(750) 4 (0.005)2 (193)( mm ) d d D + d >D mm Dislacement of the sring d s F s k m mm d lat d + d s mm 138

19 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 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 EFL EF EF E W(750) 4 (0.012)2 (193)(10 9 ) d > d >D F DL D D E Dislacement of the sring d s F s k d lat d + d s (10-6 ) W W 2 (750) 4 (0.005)2 (193)(10 9 ) (10-6 ) W d d D + d >D (10-6 ) W (10-6 ) W (10-3 ) W W 2 60(10 3 (1000) W ) (10-3 ) W W W 9685 N 9.69 kn 139

20 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The rod has a slight taer and length L. It is susended from the ceiling and suorts a load P at its end. Show that the dislacement of its end due to this load is d PL>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 L x r 1L + (r 2 - r 1 )x L L (x) L 2 (r 1L + (r 2 - r 1 )x) 2 Pdx d L (x)e PL2 E L0 L dx [r 1 L + (r 2 - r 1 )x] 2 r 1 P - PL2 E c 1 (r 2 - r 2 )(r 1 L + (r 2 - r 1 )x) d L ƒ 0 - PL 2 E(r 2 - r 1 ) c 1 r 1 L + (r 2 - r 1 )L - 1 r 1 L d PL 2 - E(r 2 - r 1 ) c 1 r 2 L - 1 r 1 L d - PL 2 E(r 2 - r 1 ) c r 1 - r 2 r 2 r 1 L d PL 2 E(r 2 - r 1 ) c r 2 - r 1 r 2 r 1 L d PL 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 P. P d 2 t w d 1 + d 2 - d 1 h d L P(x) dx (x)e Ph E t L0 Ph E t d 1 h L0 h dx d 1 h + (d 2 - d 1 )x h x d 1 h + (d 2 - d 1 )x h P E L0 dx d2 - d1 1 + d 1 h x Ph 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 Ph 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 Ph E t(d 2 - d 1 ) cln a d 1 + d 2 - d 1 bd d 1 d 1 P h Ph E t(d 2 - d 1 ) cln d 2 d 1 d 140

21 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 Prob mm 30 kn 30 kn 75 mm 0.5 m Using the result of rob by substituting d m, d m t 0.01 m and L 0.5 m. PL 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)( ) ln a bd 0.360(10-3 ) m mm 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 L Internal Forces: + c F z 0; P(z) gz 0 P(z) 1 3 gz Dislacement: L P(z) dz d L (z) E 0 L 3 gz L E dz 0 1 L g z dz 3E L0 gl2 6E 141

22 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P. L Dislacements: The cross-sectional area of the bar as a function of x is (x) r 2 r 2 0 e 2ax. We have L P(x)dx d L 0 (x)e P r 2 0 E L0 P r 2 0 E c - 1 L 2ae 2ax d 2 0 L dx e 2ax x r 0 r r 0 e ax P P - 2ar 2 0 E a1 - e - 2aL 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 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 L P(y) dy (y) E (10 3 )(144) L0 ( (10-3 ) (4 + 4y y) dy L 0 dy y ) ft y r (10-3 )c4y + 4a y3 2 b + 2 y2 d (10-3 )(45.33) (10-3 ) ft in. 142

23 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P. The modulus of elasticity is E. P r 0 r 0 2 Geometry: Dislacement: r 2 (r 0 cos u) 2 r 0 2 cos 2 u L P(y) dy d L (y) E 0 y r 0 sin u; dy r 0 cos u du u 2 P r 0 cos u du E L0 r 2 0 cos 2 u R 2 P r 0 E L0 u du cos u R 2P r 0 E [ln (sec u + tan u)] 2 u 0 2P [ln (sec u + tan u)] r 0 E When y r 0 4 ; u d 2P [ln (sec tan )] r 0 E 0.511P r 0 E lso, Geometry: (y) x 2 (r y 2 ) Dislacement: L P(y) dy d L (y) E 2P E L0 0 0 dy r y 2 2P E 1 2r 0 ln r 0 + y P [ln ln 1] r 0 E 0 r 0 - y R P r 0 E 143

24 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The weight of the kentledge exerts an axial force of P 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. P 0 12 m Internal Loading: y considering the equilibrium of the ile with reference to its entire free-body diagram shown in Fig. a. We have 1 (12) c Fy 0; 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 - P(y) c Fy 0; P(y) 10.42y2 kn Dislacement: The cross-sectional area of the ile is (0.32) m2. 4 We have L d 12 m P(y)dy 10.42(103)y2dy (29.0)(109) L0 L0 (y)e 12 m L (10-6)y2dy (10-6)y3 冷 0 12 m (10-3)m 2.93 mm 144 F

25 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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; 6P st + P con [1] 3 ft omatibility: d st d con P st (3)(12) 4 (0.752 )(29.0)(10 3 ) P con(3)(12) [ 4 (82 ) - 6( 4 )(0.75)2 ](4.20)(10 3 ) P st P con [2] Solving Eqs. [1] and [2] yields: P st ki P con ki verage Normal Stress: s con P con con s st P st ksi st 4 (0.752 ) (82 ) ( ) 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 P st 3 4 (30) 22.5 ki P con 1 (30) 7.50 ki 4 omatibility: 3 ft d st d con P st L st E st P con L con E con st 22.5 con E con 7.50 E st 6 a 4 bd2 3 4 (82 ) d 2 D (4.20)(10 3 ) 29.0(10 3 ) d 1.80 in. 145

26 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 + c F y 0; P st + P con P st L 4 ( ) (200) (10 9 ) P con L 4 (0.072 ) (24) (10 9 ) Solving Eqs. (1) and (2) yields P st kn P con kn s st P st st d st d con P st P con (10 3 ) 4 ( ) 48.8 MPa s con P con (103 ) 5.85 MPa con 4 (0.072 ) (1) (2) 500 mm The 304 stainless steel ost has a diameter of d 2 in. and is surrounded by a red brass 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; P st + P br - 5 0[1] omatibility: d 0.5 in. d st d br P st (8) 4 (22 )(28.0)(10 3 ) P br(8) 4 ( )(14.6)(10 3 ) P st P br [2] Solving Eqs. [1] and [2] yields: verage Normal Stress: P br ki P st ki s br P br br ( ) ksi s st P st ksi st 4 (22 ) 146

27 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The 304 stainless steel ost is surrounded by a red brass 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. P st P br P 2.50 ki omatibility: d 0.5 in. d st d br PL st E st PL br E br st bre br E st a 4 bd2 4 ( )(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 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 D 75 kn 100 kn F - F D (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 kn From Eq. (1), F kn s D P D (103 ) 55.0 MPa D 4 (0.052 ) s P 42.11(103 ) 134 MPa 4 (0.022 ) s P (103 ) 80.4 MPa 4 (0.052 ) 147

28 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The comosite bar consists of a 20-mm-diameter -36 steel segment and 50-mm-diameter red brass 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 kn Dislacement: d > P L 42.11(103 )(500) E st 4 (0.022 )200(10 9 ) mm 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; P st + P con d st d con ; P st(8)(12) 18(29)(10 3 ) P con (8)(12) [(9)(16) - 18](4.20)(10 3 ) P st P con Solving Eqs. (1) and (2) yields P st ki; P con ki s st P st ksi st 18 s con P con con ksi 9(16) - 18 Either the concrete or steel can be used for the deflection calculation. d P stl st E (8)(12) 18(29)( in. ) (1) (2) 8 ft 148

29 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P st P con P 30 ki d con d st ; PL con E con PL st E st st cone con E st [9(16) - st] 4.20(10 3 ) 29(10 3 ) d P stl st E st 18.2 in 2 30(8)(12) 18.2(29)(10 3 ) 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 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: d > + d E>F T T(0.75) (125)(10-6 )(200)(10 9 ) + 2T(0.75) (125)(10-6 )(200)(10 9 ) T N T T D 16.7 kn T EF 33.3 kn 149

30 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 GPa, E c 25 GPa. 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; P con + 6P st (1) Since the steel rods and the concrete are firmly bonded, their deformation must be the same. Thus P con L con E con 0 con d st P st L st E st P con L 0.25(0.375) - 6( 4 )(0.022 )D25(10 9 )D P con P st Solving Eqs (1) and (2) yields P st L ( 4 )(0.022 )200(10 9 )D (2) P st kn P con kn Thus, s con P con con (10 3 ) 0.15(0.375) - 6( 4 )(0.022 ) 8.42 MPa s st P st 21.15(103 ) st 4 (0.022 ) 67.3 MPa 150

31 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 GPa, E c 25 GPa. The normal force in each steel rod is P st 1 5 (900) 30 kn kn 250 mm 375 mm The normal force in concrete is P 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 P con L con E con P st L st E st 720(10 3 ) L 0.25(0.375) - 6( 4 d2 )D25(10 9 )D 49.5 d (10 3 )L 4 d2 200(10 9 )D d m 24.6 mm The assembly consists of two red brass 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 P + 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 N Substituting this result into Eq. (1), F N 151

32 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Normal Stress: We have, s s D F MPa D 4 (0.032 ) s EF F EF MPa EF 4 (0.042 ) 152

33 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 P is alied at the junction. P L 2 L 2 D Equations of Equilibrium: ; + F x 0; F + F D - P 0 [1] omatibility: : + 0 d P - d 0 PL 2 2E - F L 2 E 0 PL 4E - 3F L 4E + F L 2 2E S F P 3 From Eq. [1] F D 2 3 P 153

34 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 154

35 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P 200 kn is alied. The assembly is made of 36 steel. 600 mm 600 mm 0.15 mm P 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 P - d FD 200(10 3 )(600) (0.052 )(200)(10 9 ) - F D (600) 4 (0.052 )(200)(10 9 ) + F D (600) 4 ( )(200)(10 9 ) S F D N 20.4 kn Substituting this result into Eq. (1), F N 180 kn 155

36 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Two -36 steel wires are used to suort the 650-lb engine. Originally, is 32 in. long and is 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 (1) d d T (32) (0.01)(29)(10 6 ) T (32.008) (0.01)(29)(10 6 ) T T 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 L 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, L 0 : + F x 0; 1 2 0L - F - F 0 (1) omatibility Equation: Using the method of suerosition, Fig. b, : + 0 d P - d F L P(x)dx 0 L E 0 L 0 P(x)dx - F L L 0 - F (L) E 156

37 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Here, P(x) 1 2 a 0 L xbx 0 2L x L L0. Thus, L 0 0 2L x3 3 ` L x 2 dx - F L 0 - F L F 0L 6 Substituting this result into Eq. (1), F 0L 3 157

38 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The taered member is fixed connected at its ends and and is subjected to a load P 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. P 3 in. x 60 in. y x y x : + F x 0; F + F (1) d > 0 - L F L 30 0 F dx 2( x)(2)(e) + L 60 dx ( x) + F dx L ( x) 0 40 F ln( x) F ln( x) F (0.2876) F F dx 2( x)(2)(e) 0 F F Thus, from Eq. (1). F 4.09 ki F 2.91 ki 158

39 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The taered member is fixed connected at its ends and and is subjected to a load P. 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. P 3 in. x 60 in. y x y x : + F x 0; F + F - P 0 d > 0 - L x 0 F dx 2( x)(2)(e) + Lx 60 F dx 2( x)(2)(e) 0 x 60 dx -F L ( x) + F dx L ( x) 0 0 x F (40) ln ( x) x 0 - F (40) ln ( x) 60 x 0 F ln ( x ) -F 3 ln ( x 1.5 ) For greatest magnitude of P require, 4 F 2( x)(2) ; F x 4 F 2(3) ; F 24 ki Thus, ( x) ln a x 3 b -24 ln a x b 1.5 Solving by trial and error, x 28.9 in. Therefore, F 36.4 ki P 60.4 ki 159

40 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 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 L 2 D + 30 ut, L d, L D d D Neglect squares or d since small strain occurs. L 2 D (245 + d ) d L 2 D (245 + d D ) 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 160

41 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 ksi. Determine the slight rotation of the bar when the uniform load is alied. See solution of Prob T D ki 6 ft d D Using Eq. (2) of Prob. 4-51, T D (31)(10 3 ) (31)( ft ) ( ) 2 (9) 2 + (8.4852) 2-2(9)(8.4852) cos u 3 ft 6 ki/ft 3 ft 3 ft D u Thus, u The ress consists of two rigid heads that are held 1 together by the two -36 steel 2 -in.-diameter rods 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 d al F st (12) ( 4 )(0.5)2 (29)(10 3 ) F al (10) (1) 2 (10)(10 3 ) Solving, F st ki F al ki s rod F st st ( 9.28 ksi 4 )(0.5)2 s cyl F al ksi 2 al (1) 161

42 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The ress consists of two rigid heads that are held 12 in. 1 together by the two -36 steel 2 -in.-diameter rods 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 ki; F st ( 4 )(0.5)2 ; F st ki s al (1) ksi 4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1), Thus, d in. u u

43 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 L E 1 2 a F DL E + F F L 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( MPa ) s D F D 35.83(103 ) 0.45( MPa ) s F 113 MPa 163

44 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 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 L ft and L D ft. The stretches 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 the oints on d the rigid bar is dg. For oints and D, cos u and cos u D 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 a 169 F 12 E b 1 16 a 100 F D 3E b F F D (2) Solving Eqs. (1) and (2), yields Thus, F D lb F lb s D F D (10 3 ) si 15.4 ksi D 0.04 s F (10 3 ) si 11.4 ksi

45 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued 165

46 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 L ft and L D ft. The stretch of wires and D are d F L E F (13) d D F D L 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 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 g 100(614.73) 3(0.04)29.0 (10 6 )D ft Then u a ft ba ft b

47 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 GPa. 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 (2) d - d 2 d - d 3 ; 3d - d 2d 3F L E - F L E 2F L E ; 3F - F 2F (3) Solving Eqs. (1) (3) yields : F 5.79 kn F 9.64 kn F 11.6 kn 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 GPa. a + M 0; F (1) - F (2) 0 (1) c + F y 0; F + F + F (2) 18 kn/m 1.40 m d - d 2 d - d 3 ; 3d - d 2d 2 m 1 m 3F L E - F L E 2F L E ; 3F - F 2F (3) Solving Eqs. (1) (3) yields : F kn; F kn; F kn d F L E (103 )(1.40) 4 (0.122 )12( (10-3 ) m ) d F L E (103 )(1.40) 4 (0.122 )12( (10-3 ) m ) tan u (10-3 ) u (10-3 ) rad 1.14(10-3 ) 167

48 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 d E F(400) 1(10-3 )200(10 9 )D F E (399.9) 1.5(10-3 )73.1(10 9 )D F F E + 50(10 3 ) (2) Solving Eqs (1) and (2) yield Normal Stress. F E 64.56(10 3 ) N F (10 3 ) N s D s F F st (103 ) 1(10-3 ) 168 MPa s E F E 64.56(103 ) al 1.5( MPa ) 168

49 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 MPa in the steel and 1s allow 2 al 94 MPa in the aluminum is not exceeded. E st 200 GPa, E al 70 GPa. 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 FL (70)(10 9 ) F D L (200)(10 9 ) ; F 0.35 F D (2) ssume failure of and EF: F (s allow ) al 94(10 6 )(450)(10-6 ) N From Eq. (2) F D N From Eq. (1) w 68.5 kn>m ssume failure of D: F D (s allow ) st 180(10 6 )(450)(10-6 ) N From Eq. (2) F N From Eq. (1) w 45.9 kn>m (controls) 169

50 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 GPa, E al 70 GPa. Equations of Equilibrium: a+ M 0; 450(250) - F (150) - F D (150) 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 F D F D N F N s D F D MPa D 40(10-6 ) [2] s F (10-6 ) 9.55 MPa 170

51 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 GPa, E al 70 GPa. Equations of Equilibrium: a+ M 0; 450(250) - F (150) - F D (150) 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 F D F D N F N [2] d D F DL D D E al (50) 40(10-6 )(70)(10 9 ) mm tan u d D u 63.7(10-6 ) rad

52 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *4 64. The center ost of the assembly has an original length of 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 GPa. 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 d d F (0.125) 400 (10-6 )(70)(10 6 ) F (0.1247) 400 (10-6 )(70)(10 6 ) F F 8.4 Solving Eqs. (2) and (3) F kn F kn (1) (2) (3) 800 kn/m s s (103 ) 400(10-6 ) 189 MPa s (103 ) 400 (10-6 ) 21.4 MPa The assembly consists of an -36 steel bolt and a red brass tube. If the nut is drawn u snug against the tube so that L 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. L 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 P(75) 100(10-6 )(101)(10 9 ) + P(75) 4 ( )(200)(10 9 ) P N 1.16 kn 172

53 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The assembly consists of an -36 steel bolt and a red brass tube. The nut is drawn u snug against the tube so that L 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. L llowable Normal Stress: (s g ) st P st 4 (0.007)2 P st kn (s g ) br P br 100(10-6 ) P br 7.00 kn Since P st 7 P 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 ) (10 3 )(75) 4 (0.007)2 (200)(10 9 ) mm 173

54 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P. D F a + M 0; F D (d) + F EF (2d) - Pa d 2 b 0 P L F D + 2F EF P 2 (1) E + c F y 0; F + F D + F EF - P 0 (2) d 2 d 2 d d - d E d d - d E 2d 2d d + d E 2F D L E F L E + F EFL E 2F D - F - F EF 0 (3) Solving Eqs. (1), (2) and (3) yields F 7P 12 F D P 3 F EF P 12 s s D s EF 7P 12 P 3 P 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 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 > F, E st ksi. P 0.2 in in. P d T a TL 9.6(10-6 )(90-60)(463.25) ft d PL E (35-20)(463.25) ft (0.2)(0.05)(29)(10 6 ) L ft 174

55 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Three bars each made of different materials are connected together and laced between two walls when the temerature is T Determine the force exerted on the (rigid) suorts when the temerature becomes T The material roerties and cross-sectional area of each bar are given in the figure. Steel E st 200 GPa a st 12(10 6 )/ st 200 mm 2 rass E br 100 GPa a br 21(10 6 )/ br 450 mm 2 oer E cu 120 GPa a cu 17(10 6 )/ cu 515 mm 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 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) (0.5)(2)(12) 4 (0.252 )(29.0)(10 3 ) x in. F 1.00( ) ki 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 F. 6 ft omatibility: : + 0 d T - d F (10-6 F(6)(12) )(275-70)(6)(12) - 4 ( )(29.0)(10 3 ) F 116 ki 175

56 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * 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 ki>in., determine the force the ie exerts on the turbines when the steam and thus the ie reach a temerature of T F. 6 ft omatibility: x d T - d F x 6.60( (10 3 )(x)(3)(12) )(275-70)(3)(12) - 4 ( )(29.0)(10 3 ) x in. F k x 80(10 3 )( ) 112 ki 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 x2 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: L 0 d T - d F Where d T a T dx L 0 8ft ( x) dx - L 0 F(8) (29.0)(10 3 ) (8) + 15(8)2 R - 2 F(8) (29.0)(10 3 ) F verage Normal Stress: s ksi 176

57 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The bronze 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 b x 8 omatibility: 0 d T - d F Where d T 1 a Tdx 2ft [( x) - 60] dx - L 0 F(8) 4 ( )15.0(10 3 ) 2ft ( x) dx - L 0 F(8) 4 ( ) 15.0(10 3 ) F 7.60 ki 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 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 TL 6.60(10-6 )[90 - (-20)](40)(12) in in. omatibility: : d T - d F (10-6 )[110 - (-20)](40)(12) - F(40)(12) 5.10(29.0)(10 3 ) F 19.5 ki 177

58 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 in. 3 in. E Thermal Exansion: d T D a al TL D 12.8(10-6 )(150-75)(1.5) 1.44(10-3 ) in. d T a st TL 6.60(10-6 )(150-75)(1.5) (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) (10-3 ) (10-3 ) (10-3 ) R(3.25) 0.25 D in The bar has a cross-sectional area, length L, 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>L. 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 L T a L L 0 ac T - T 2 From Eq.(1). T - T L 0 al 2 (T - T ) - FL E x dx ac T - T 2L L d al 2 (T - T ) x - T )dx L x2 d 0 F a E 2 (T - T ) 178

59 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel rod has a diameter of 50 mm and is lightly attached to the rigid suorts at and when T If the temerature becomes T 2 20 and an axial force of P 200 kn is alied to its center, determine the reactions at and. P 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 TL 12(10-6 )(80-20)(1000) 0.72 mm. lso, under force P and F with unconstrained at, the deformation of the rod are d F d P PL E 200(10 3 )(500) 4 (0.052 )200(10 9 )D F L E F (1000) 4 (0.052 )200(10 9 )D Using the method of suer osition, Fig. b, mm (10-6 ) F : + 0 -d T + d P + d F (10-6 ) F F (10 3 ) N 183 kn F Substitute the result of into Eq (1), F (10 3 ) N 383 kn 179

60 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel rod has a diameter of 50 mm and is lightly attached to the rigid suorts at and when T Determine the force P that must be alied to the collar at its midoint so that, when T 2 30, the reaction at is zero. P 0.5 m 0.5 m When the rod is unconstrained at, it has a free contraction of d T a st TL 12(10-6 )(50-30)(1000) 0.24 mm. lso, under force P with unconstrained at, the deformation of the rod is d P PL E P(500) 4 (0.052 )200(10 9 )D (10-6 ) P Since F is required to be zero, the method of suerosition, Fig. b, gives : + 0 -d T + d P (10-6 )P P (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 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 [1] omatibility: ( +T) (d ) F - (d ) T d F F L 8(14.6)(10 3 ) (20)L FL 8(29.0)(10 3 ) F F 196 [2] Solving Eqs. [1] and [2] yields: average Normal Sress: F ki F ki s s F ksi s F ksi 8 180

61 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 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) (d F) (d T ) D + (d F ) D 5 4 c6.60(10-6 )( )(5)(12) - F (5)(12) 2(29)(10 3 ) d 6.60(10-6 )( )(4)(12) + F D(4)(12) 2(29)(10 3 ) F + 48F D (2) : + F x 0; 3 5 F 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

62 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 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) (d T) (d T ) D + (d F ) D 5 4 c6.60(10-6 )( )(5)(12) - F (5)(12) 2(29)(10 3 ) d 6.60(10-6 )( )(4)(12) + F D(4)(12) 2(29)(10 3 ) F F D 4 F D F (2) : + F x 0; 3 5 F 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 ki: F D ki (d ) r (d T ) D + (d T ) D 6.60(10-6 )( )(4)(12) (4)(12) 2(29)(10 3 ) in. c 182

63 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 in in. D 40 in 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 [1] omatibility: (d ) T (80)(60) in. (d ) Tr (d ) T cos in. cos 45 (d D ) T (80)(40) in. d 0 (d ) Tr - (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 183

64 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *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 TL 26(10-6 )(80-30)(300) 0.39 mm and d T ) D a al TL 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 F N Normal Stress: F(300) (44.7)(10 9 ) S F(450) (68.9)(10 9 ) S s s D F F D MPa F N 45.3 MPa 184

65 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 MPa 6 (s Y ) al D (O.K.!) omatibility Equation: If tube and rod D are unconstrained, they will have a free exansion of d 26( (10-6 T a mg TL )(T - 30)(300) ) (T - 30) and d 24(10-6 T D a al TL D )(T - 30)(450) (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 (10-3 )(T - 30) (300) (44.7)(10 9 ) S (450) (T - 30) (68.9)(10 9 ) S T

66 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 GPa, 14(10-6 )>, a al 23(10-6 )>. E al 70 GPa, a st omatibility: (d s ) T - (d b ) T (d s ) F + (d b ) F 23(10-6 )(100-20)L - 14(10-6 )(100-20)L FL 4 ( )70(10 9 ) + FL 4 ( )200(10 9 ) F N verage Normal Stress: s s F s 4 ( MPa ) s b F MPa b 4 ( ) Determine the maximum normal stress develoed in the bar when it is subjected to a tension of P 8 kn. 5 mm 40 mm 20 mm For the fillet: P P w h r h r 10 mm 20 mm From Fig K 1.4 For the hole: From Fig K s max Ks avg s max Ks avg 8 (10 3 ) 1.4 a 0.02 (0.005) b 112 MPa r w (10 3 ) a ( )(0.005) b 190 MPa 186

67 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *4 88. If the allowable normal stress for the bar is s allow 120 MPa, determine the maximum axial force P that can be alied to the bar. ssume failure of the fillet. w h ; r h P 5 mm 40 mm 20 mm r 10 mm 20 mm P From Fig K 1.4 s allow s max Ks avg 120 (10 6 P ) 1.4 a 0.02 (0.005) b P 8.57 kn ssume failure of the hole. r w From Fig K s allow s max Ks avg 120 (10 4 P ) a ( ) (0.005) b P 5.05 kn (controls) 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 lb 3350 lb w 0.25 in. s allow s max Ks avg Kc (w - 1)(0.25) d 1 in. w 3.35K y trial and error, from Fig. 4-25, choose r 0.2; K 2.45 w w 3.35(2.45) in. r Since w OK 187

68 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel late has a thickness of 12 mm. If there are shoulder fillets at and, and s allow 150 MPa, determine the maximum axial load P 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 and w h mm P r 30 mm 60 mm 120 mm P r 30 mm D 200 mm 800 mm 200 mm Dislacement: d PL E s max s allow Ks avg (400) (0.06)(0.012)(200)(10 9 ) (800) (0.12)(0.012)(200)(10 9 ) mm 150(10 6 P ) (0.012) R P N 77.1 kn Determine the maximum axial force P that can be alied to the bar. The bar is made from steel and has an allowable stress of s allow 21 ksi in in in. ssume failure of the fillet. P P From Fig. 4-24, K 1.73 r h w h in. r 0.25 in. ssume failure of the hole. From Fig. 4-25, K 2.45 s allow s max Ks avg P a 1.25 (0.125) b P ki r w s allow s max Ks avg P a ( )(0.125) b P 1.21 ki (controls) 188

69 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *4 92. Determine the maximum normal stress develoed in the bar when it is subjected to a tension of P 2 ki in in in. t fillet: From Fig. 4-24, K 1.73 r h w h P 0.75 in. r 0.25 in. P s max Ka P b 1.73c 2 d 22.1 ksi 1.25(0.125) t hole: r w From Fig. 4-25, K s max 2.45c d 34.8 ksi (ontrols) ( )(0.125) Determine the maximum normal stress develoed in the bar when it is subjected to a tension of P 8 kn. 60 mm 5 mm 30 mm Maximum Normal Stress at fillet: P P r h and w h r 15 mm 12 mm From the text, K 1.4 P s max Ks avg K h t Maximum Normal Stress at the hole: From the text, K (10 3 ) 1.4 R 74.7 MPa (0.03)(0.005) r w P s max K s avg K (w - 2r) t 8(10 3 ) 2.65 ( )(0.005) R 88.3 MPa (ontrols) 189

70 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The resulting stress distribution along section for the bar is shown. From this distribution, determine the aroximate resultant axial force P alied to the bar. lso, what is the stress-concentration factor for this geometry? P 0.5 in. 4 in. 1 in. P L sd Volume under curve 12 ksi 3 ksi Number of squares 10 P 10(3)(1)(0.5) 15 ki s avg P 15 ki (4 in.)(0.5 in.) 7.5 ksi K s max s avg 12 ksi 7.5 ksi The resulting stress distribution along section for the bar is shown. From this distribution, determine the aroximate resultant axial force P alied to the bar. lso, what is the stress-concentration factor for this geometry? 0.6 in. 0.5 in. Number of squares 28 P 28(6)(0.2)(0.5) 16.8 ki s avg P ksi 2(0.6)(0.5) 0.2 in. 36 ksi 6 ksi 0.8 in. 0.6 in. P K s max s avg *4 96. The resulting stress distribution along section for the bar is shown. From this distribution, determine the aroximate resultant axial force P alied to the bar. lso, what is the stress-concentration factor for this geometry? 20 mm 80 mm 10 mm P Number of squares MPa 5 MPa P 19(5)(10 6 )(0.02)(0.01) 19 kn s avg P 19(103 ) MPa 0.08(0.01) K s max s avg 30 MPa MPa

71 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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; P st + P al [1] Elastic nalysis: ssume both materials still behave elastically under the load. d st d al P st L 4 (2)2 (29)(10 3 ) P al L 4 ( )(10.6)(10 3 ) P st P al Solving Eqs. [1] and [2] yields: P al ki P st ki verage Normal Stress: s al P al al ( ) ksi 6 (s g ) al 60.0 ksi (OK!) s st P st st 4 (22 ) 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 P st (s g ) st st 36.0a 4 b ki From Eq. [1] P al ki s al P al al ( ) ksi 6 (s g) al 60.0 ksi Then s al 19.8 ksi 191

72 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 verage Normal Stress and Strain: For segment verage Normal Stress and Strain: For segment Elongation: s P ksi e ; e (10.0) in.>in. 20 s P ksi e e in.>in. d e L (2)(12) in. d e L (5)(12) in P (in./in.) d Tot d + d in. 192

73 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 MPa, and E st 200 GPa, 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) F E F D 0.32w [1] Plastic nalysis: Wire D will yield first followed by wire E. When both wires yield F E F D (s g ) a 4 b 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( mm>mm ) d E e g L E (800) mm d G 0.8 d E 0.4 d G 2d E 2(2.120) 4.24 mm 193

74 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * 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 MPa, and E st 200 GPa, 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 F D 0.32w [1] (a) y observation, wire D will yield first. Then F D s g a. 4 b kn From the geometry d E 0.4 d D 0.65 ; d D 1.625d E F D L E F EL E F D F E [2] Using F D kn and solving Eqs. [1] and [2] yields: F E kn w 18.7 kn>m (b) When both wires yield F E F D (s g ) a 4 b kn Substituting the results into Eq. [1] yields: w 21.9 kn>m 194

75 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The rigid lever arm is suorted by two -36 steel wires having the same diameter of 4 mm. If a force of P 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 P 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) (450) 0 D 2F + F D (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 bd D d 2d D (2) F L E 2a F D L 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 MPa 6 (s Y ) st MPa 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 c d N 3.14 kn Substituting this result into Eq. (1), F D N 2.72 kn s D F D D MPa 6 (s Y ) st (O.K.) 195

76 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Since wire D is linearly elastic, its elongation can be determined by d D F D L D D E st (300) (200) mm mm From Eq. (2), d 2d D 2(0.3243) mm 196

77 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The rigid lever arm is suorted by two -36 steel wires having the same diameter of 4 mm. Determine the smallest force P 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 P 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) - P(450) 0 2F + F D 3P (1) Elastic nalysis. The comatibility equation can be written by referring to the geometry of Fig. b. d a bd D d 2d D F L E 2a F D L E b F D 1 2 F (2) ssuming that wire is about to yield first, F (s Y ) st c d N From Eq. (2), F D 1 ( ) N 2 Substituting the result of F and F D into Eq. (1), P N 2.62 kn Plastic nalysis. Since both wires and D are required to yield, F F D (s Y ) st c d N Substituting this result into Eq. (1), 197

78 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The three bars are inned together and subjected to the load P. If each bar has a cross-sectional area, length L, 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 P 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 L L u u L P P N 3.14 kn When all bars yield, the force in each bar is, F Y s Y : + F x 0; P - 2s Y cos u - s Y 0 P s Y (2 cos u + 1) ar will yield first followed by bars and D. d d D F Y(L) E s YL E s YL E d d cos u s YL E cos u * The rigid beam is suorted by three 25-mm diameter -36 steel rods. If the beam suorts the force of P 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 P F + c F y 0; a + M 0; F D + F E + F F F E (400) + F F (1200) (800) 0 (1) 400 mm 400 mm 400 mm F E + 3F F (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 d F F E L E 2 3 a F DL E b a F F L E b F E 2 3 F D F F (3) Solving Eqs. (1), (2), and (3) F F N F E N F D N 198

79 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Normal Stress. s F F F MPa 7 (s Y ) st F s E F E E MPa 6 (s Y ) st s D F D D MPa 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 c d N 123 kn Substituting this result into Eq. (2), F E N 91.8 kn Substituting the result for F F and F E into Eq. (1), F D N 15.4 kn s E F E E MPa 6 (s Y ) st s D F D D MPa 6 (s Y ) st (O.K.) (O.K.) 199

80 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The rigid beam is suorted by three 25-mm diameter -36 steel rods. If the force of P 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 P 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 F E (400) + F F (1200) (800) 0 F E + 3F F (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 d F (3) F E L E 2 3 a F D L E b a F F L E b F E 2 3 F D F F (4) Solving Eqs. (1), (2), and (4) F F N F E N F D N Normal Stress. s F F F MPa (T) 7 (s Y ) st F s E F E E MPa (T) 6 (s Y ) st s D F D D MPa (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 MPa (T) F F s F F c d N 200

81 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Substituting this result into Eq. (2), F E N Substituting the result for F F and F E into Eq. (1), F D N s E F E E MPa (T) 6 (s Y ) st s D F D D MPa (T) 6 (s Y ) st (O.K.) (O.K.) Residual Stresses. The rocess of removing P can be reresented by alying the force P, which has a magnitude equal to that of P 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 MPa () s E œ MPa () s D MPa () 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 ( ) MPa 17.7 MPa () ( ) 53.2 MPa (T) (-66.94) MPa 35.5 MPa () 201

82 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 s(ksi) 4 ft 4 ft 5 ft (in./in.) w a + M 0; F (4) - F (4) 0; F F F + c F y 0; 2F + F (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 ki Thus, s s s ksi From the stress-strain diagram: e : e in.>in d el (5)(12) 8.69 in. 202

83 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 s(ksi) 4 ft 4 ft 5 ft (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 w (2) From the stress-strain diagram: e in.>in. 5 (12) s ; s ksi Hence F s (0.75) ki From Eq. (2), w 10.9 ki>ft 203

84 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * The rigid beam is suorted by the three osts,, and of equal length. Posts and have a diameter of 75 mm and are made of aluminum, for which E al 70 GPa and 1s Y 2 al 20 MPa. Post has a diameter of 20 mm and is made of brass, for which E br 100 GPa and 1s Y 2 br 590 MPa. Determine the smallest magnitude of P so that (a) only rods and yield and (b) all the osts yield. al 2 m P P br 2 m 2 m 2 m al M 0; F F F al + c F y 0; F at + 2F at - 2P 0 (1) (a) Post and will yield, F al (s t ) al omatibility condition: 20(10 4 )( a )(0.075) kn (E al ) r (s r) al E al 20(104 ) 70(10 4 ) d br d al (L) F br (L) 4 (0.02)2 (100)(10 4 ) L F br kn From Eq. (1), s br 8.976(103 ) 4 (0.023 ) 28.6 MPa 6 s r OK (88.36) - 2P 0 P 92.8 kn (b) ll the osts yield: F br (s r ) br (590)(10 4 )( 4 )(0.022 ) kn F al kn From Eq. (1); (88.36) - 2P 0 P 181 kn 204

85 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The rigid beam is suorted by the three osts,, and. Posts and have a diameter of 60 mm and are made of aluminum, for which E al 70 GPa and 1s Y 2 al 20 MPa. Post is made of brass, for which E br 100 GPa and 1s Y 2 br 590 MPa. If P 130 kn, determine the largest diameter of ost so that all the osts yield at the same time. al P br P al 2 m 2 m 2 m 2 m + c F y 0; 2(F g ) al + F br (1) (F al ) g (s g ) al 20(10 6 )( 4 )(0.06) kn From Eq. (1), 2(56.55) + F br F br kn (s g ) br 590(10 6 ) 146.9(103 ) 4 (d ) 3 d m 17.8 mm 205

86 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The wire has a diameter of 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) P 450 lb, (b) P 600 lb. 40 in. D Equations of Equilibrium: s (ksi) 50 in. 30 in. P a + M 0; F (50) - P(80) 0 [1] (a) From Eq. [1] when P 450 lb, F 720 lb verage Normal Stress and Strain: From the Stress Strain diagram Dislacement: s F ( ) ksi e ; e in.>in. d e L (40) in P (in./in.) d D 80 d 50 ; d D 8 (0.2347) in. 5 (b) From Eq. [1] when P 600 lb, F 960 lb verage Normal Stress and Strain: From Stress Strain diagram Dislacement: s F e e in.>in. d e L (40) in ksi 4 (0.125)2 d D 80 d 50 ; d D 8 (3.9990) 6.40 in

87 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The bar having a diameter of 2 in. is fixed connected at its ends and suorts the axial load P. If the material is elastic erfectly lastic as shown by the stress strain diagram, determine the smallest load P needed to cause segment to yield. If this load is released, determine the ermanent dislacement of oint. 2 ft P 3 ft s (ksi) P (in./in.) When P is increased, region will become lastic first, then will become lastic. Thus, F F s 20()(1) ki : + F x 0; F + F - P 0 (1) P 2(62.832) ki P 126 ki The deflection of oint is, d el (0.001)(3)(12) in. ; onsider the reverse of P on the bar. F (2) E F (3) E F 1.5 F So that from Eq. (1) F 0.4P F 0.6P d F L E 0.4(P)(3)(12) E 0.4(125.66)(3)(12) (1) 2 (20>0.001) in. : d in. ; 207

88 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * Determine the elongation of the bar in Prob when both the load P and the suorts are removed. P 2 ft 3 ft s (ksi) 20 When P is increased, region will become lastic first, then will become lastic. Thus, F F s 20()(1) ki P (in./in.) : + F x 0; F + F - P 0 (1) P 2(62.832) ki P 126 ki The deflection of oint is, d el (0.001)(3)(12) in. ; onsider the reverse of P on the bar. F (2) E F (3) E F 1.5 F So that from Eq. (1) F 0.4P F 0.6P The resultant reactions are F F (125.66) (125.66) ki When the suorts are removed the elongation will be, d PL E (5)(12) (1) in. (20>0.001) 208

89 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher material has a stress strain diagram that can be described by the curve s cp 1>2. Determine the deflection d of the end of a rod made from this material if it has a length L, cross-sectional area, and a secific weight g. L s s c e 1 2 ; s 2 c 2 e s 2 (x) c 2 e(x) (1) d P However s(x) P(x) dd ; e(x) dx From Eq. (1), P 2 (x) 2 c 2 dd dx ; dd dx P2 (x) 2 c 2 d g2 1 2 c 2 L P2 (x) dx c 2 L0 d g3 L 3 3c 2 L x 2 dx g2 c 2 x3 3 L 0 L 1 2 c 2 (gx) 2 dx L 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 P 16 lb is alied to the rigid collar as shown, determine the reactions at and. P/2 P/2 5 in. 8 in. : T + d (5) 4 (0.52 )(10.6)(10 3 ) (10-6 F (13) )[70 - (-10 )](13) + 4 (0.52 )(10.6)(10 3 ) F ki 2.13 ki : + F x 0; 2(0.008) F 0 F 2.14 ki 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 P that must be alied to the collar so that, when T 0 F, the reaction at is zero. P/2 P/2 5 in. 8 in. : T + d P(5) 0 4 (0.52 )(10.6)(10 3 ) (10-6 )[(70)(13)] + 0 P 4.85 ki 209

90 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * 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 mm 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 TL 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 d 2(d T ) and d F 2d F. Thus, this equation becomes d F - d T 2d T - 2d F (600) (200)(10 9 ) F + 2F F(600) (0.36) (200)(10 9 ) S (2) Solving Eqs. (1) and (2), F F F D N 58.9 kn 210

91 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 P(3)(12) 0.32(29)(10 3 ) + P(2)(12) 0.32(29)(10 3 ) P 46.4 ki s s P ksi The brass lug is force-fitted into the rigid casting. The uniform normal bearing ressure on the lug is estimated to be 15 MPa. If the coefficient of static friction between the lug and casting is m s 0.3, determine the axial force P 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 GPa. 15 MPa 100 mm 150 mm 20 mm P Equations of Equilibrium: : + F x 0; P (10 6 )(2)()(0.02)(0.1) 0 P kn 56.5 kn Dislacement: d > a PL E (103 )(0.15) ( )(98)(10 9 ) + L0 0.1 m (10 6 ) x dx ( )(98)(10 9 ) m mm 211

92 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher 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 L 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 L 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 ) L + F D (L) R E 1 a 2 (T 2 - T 1 ) L - F(L) E 2 + a 1 (T 2 - T 1 ) L - F(L) E 1 [2] Substitute Eq. [1] into [2]. 2a 1 (T 2 - T 1 ) L + 4FL E 1 a 2 (T 2 - T 1 )L - FL E 2 + a 1 (T 2 - T 1 )L - FL 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 ) L (d EF ) F E 1E 2 (T 2 - T 1 )(a 2 - a 1 )(L) E 2 (5E 2 + E 1 ) E 1 (T 2 - T 1 )(a 2 - a 1 )(L) 5E 2 + E 1 d EF (d EF ) T - (d EF ) F a 2 L(T 2 - T 1 )(5E 2 - E 1 ) - E 1 L(T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 (d ) T a 1 (T 2 - T 1 ) L (d ) F E 1E 2 (T 2 - T 1 )(a 2 - a 1 )(L) E 1 (5E 2 + E 1 ) E 2 (T 2 - T 1 )(a 2 - a 1 )(L) 5E 2 + E 1 d (d ) T - (d ) F a 1 L(5E 2 + E 1 )(T 2 - T 1 ) - E 2 L(T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 212

93 04 Solutions /25/10 3:20 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued d EF - d L(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 )] L(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 L(T 2 - T 1 )(a 2 - a 1 ) 5E 2 + E 1 (5E 2 + E 1 + E 2 - E 1 ) L(T 2 - T 1 )(a 2 - a 1 )(6E 2 ) 5E 2 + E 1 u d EF - d 2d 3E 2L(T 2 - T 1 )(a 2 - a 1 ) d(5e 2 + E 1 ) * 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 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 (L) 4E d 9 F (L) 9E 350 lb 9F - 4F 0 [2] Solving Eqs. [1] and [2] yields: F 86.6 lb F 195 lb 213

94 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher shaft is made of a steel alloy having an allowable shear stress of t allow 12 ksi. If the diameter of the shaft is 1.5 in., determine the maximum torque T that can be transmitted. What would be the maximum torque T if a 1-in.-diameter hole is bored through the shaft? Sketch the shear-stress distribution along a radial line in each case. T T llowable Shear Stress: lying the torsion formula t max t allow Tc J 12 T (0.75) 2 (0.754 ) T 7.95 ki # in. llowable Shear Stress: lying the torsion formula t max t allow T c J T (0.75) 12 2 ( ) T ki # in ki # in. t r 0.5 in T r J 6.381(0.5) 2 ( ksi ) 214

95 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid shaft of radius r is subjected to a torque T. Determine the radius r of the inner core of the shaft that resists one-half of the alied torque 1T>22. Solve the roblem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution. r r T a) t max Tc J t (T 2 )r Tr 2T r4 r 3 2 T 2 (r )4 (r ) 3 Since t r r t max ; T (r ) 3 r r a 2T r 3 b r r r b) r 2 r dt 2 tr 2 dr L0 L 0 r 2 r r dt 2 L0 L 0 r t max r 2 dr r 2 r r dt 2 L0 L 0 T 2 4T r 4 L0 r r 3 dr r a 2T r 3 br2 dr r r r 215

96 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid shaft is fixed to the suort at and subjected to the torsional loadings shown. Determine the shear stress at oints and and sketch the shear stress on volume elements located at these oints. 10 kn m 75 mm 4 kn m 50 mm 75 mm The internal torques develoed at ross-sections ass through oint and are shown in Fig. a and b, resectively. The olar moment of inertia of the shaft is J. For 2 ( ) 49.70(10-6 ) m 4 oint, r Thus, t T c J 4(103 )(0.075) 6.036(10 6 ) Pa 6.04 MPa 49.70(10-6 ) From oint, r 0.05 m. t T r J 6(103 )(0.05) (10-6 ) 6.036(106 ) Pa 6.04 MPa. 216

97 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 4. The tube is subjected to a torque of 750 N # m. 75 mm Determine the amount of this torque that is resisted by the gray shaded section. Solve the roblem two ways: (a) by using the torsion formula, (b) by finding the resultant of the 100 mm shear-stress distribution. 750 N m a) lying Torsion Formula: 25 mm t max Tc J 750(0.1) 2 ( MPa ) t max T (0.1) 2 ( ) T 515 N # m b) Integration Method: t a r c b t max and d 2r dr dt rt d rt(2r dr) 2tr 2 dr 0.1m T 2tr 2 dr 2 t max a r L L c br2 dr 2t max c 0.075m 0.1m L 0.075m 2(0.4793)(106 ) 0.1 r 3 dr 0.1 m 4 d 2 c r m 515 N # m 5 5. The coer ie has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at and three torques are alied to it as shown, determine the absolute maximum shear stress develoed in the ie. t max T max c J 90(0.02) 2 ( ) 20 N m 30 N m 26.7 MPa. 80 N m 217

98 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress develoed in regions and DE of the shaft. The bearings at and F allow free rotation of the shaft. (t ) max T c J 35(12)(0.375) 2 (0.375) si 5.07 ksi F E D 25 lb ft 40 lb ft 20 lb ft 35 lb ft (t DE ) max T DE c J 25(12)(0.375) 2 (0.375) si 3.62 ksi 5 7. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress develoed in regions D and EF of the shaft. The bearings at and F allow free rotation of the shaft. (t EF ) max T EF c J 0 F E D 25 lb ft 40 lb ft 20 lb ft 35 lb ft (t D ) max T D c J 15(12)(0.375) 2 (0.375) si 2.17 ksi 218

99 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 8. The solid 30-mm-diameter shaft is used to transmit the torques alied to the gears. Determine the absolute maximum shear stress on the shaft. 300 N m 500 N m 200 N m Internal Torque: s shown on torque diagram. Maximum Shear Stress: From the torque diagram T max 400 N # m. Then, alying torsion Formula. t abs T max c max J 300 mm 400 mm D 400 N m 400(0.015) 2 ( ) 75.5 MPa 500 mm 5 9. The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown. If a torque of T 800 N # m is alied to the rigid disk fixed to its end, determine the maximum shear stress in the shaft. 2 m T 800 N m r i 20 mm r o 25 mm J 2 ((0.038)4 - (0.032) 4 ) + 2 ((0.030)4 - (0.026) 4 ) + 2 ((0.025)4 - (0.020) 4 ) J 2.545(10-6 ) m 4 r i 26 mm r o 30 mm t max Tc J 800(0.038) 11.9 MPa 2.545(10-6 ) r i 32 mm r o 38 mm 219

100 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The couling is used to connect the two shafts together. ssuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. T R n is the number of bolts and F is the shear force in each bolt. T - nfr 0; F T nr T r t avg F T nr ( 4 )d2 4T nrd 2 Maximum shear stress for the shaft: t max T c J T r 2T r4 r 3 2 t avg t max ; 4T nrd 2 2T r 3 n 2 r3 Rd The assembly consists of two sections of galvanized steel ie connected together using a reducing couling at. The smaller ie has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger ie has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the ie is tightly secured to the wall at, determine the maximum shear stress develoed in each section of the ie when the coule shown is alied to the handles of the wrench. 15 lb 6 in. t Tc J 210(0.375) 2 ( ksi ) t Tc J 210(0.5) 2 ( ksi ) 8 in. 15 lb 220

101 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 12. The motor delivers a torque of 50 N # m to the shaft. This torque is transmitted to shaft D using the gears at E and F. Determine the equilibrium torque T on shaft D and the maximum shear stress in each shaft. The bearings,, and D allow free rotation of the shafts. Equilibrium: a+ M E 0; 50 - F(0.05) 0 F 1000 N a+ M F 0; T -1000(0.125) 0 50 mm 30 mm E 35 mm 125 mm T D F T 125 N # m Internal Torque: s shown on FD. Maximum Shear Stress: lying torsion Formula. (t ) max T c J (t D ) max T Dc J 50.0(0.015) 2 ( ) 125(0.0175) 2 ( ) 9.43 MPa 14.8 MPa If the alied torque on shaft D is T 75 N # m, determine the absolute maximum shear stress in each shaft. The bearings,, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating. Equilibrium: a+ M F 0; 75 - F(0.125) 0; F 600 N a+ M E 0; 600(0.05) - T 0 T 30.0 N # m 50 mm 30 mm E 35 mm 125 mm T D F Internal Torque: s shown on FD. Maximum Shear Stress: lying the torsion formula (t E ) max T E c J (t D ) max T Dc J 30.0(0.015) 2 ( ) 75.0(0.0175) 2 ( ) 5.66 MPa 8.91 MPa 221

102 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid 50-mm-diameter shaft is used to transmit the torques alied to the gears. Determine the absolute maximum shear stress in the shaft. 250 N m 75 N m 325 N m 150 N m 500 mm The internal torque develoed in segments, and D of the shaft are shown in Figs. a, b and c. 400 mm 500 mm D The maximum torque occurs in segment. Thus, the absolute maximum shear stress occurs in this segment. The olar moment of inertia of the shaft is J. Thus, 2 ( ) (10-6 )m 4 t max abs T c J 250(0.025) (10-6 ) 10.19(106 )Pa 10.2 MPa 222

103 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid shaft is made of material that has an allowable shear stress of t allow 10 MPa. Determine the required diameter of the shaft to the nearest mm. 15 N m 25 N m 30 N m 60 N m 70 N m D E The internal torques develoed in each segment of the shaft are shown in the torque diagram, Fig. a. Segment DE is critical since it is subjected to the greatest internal torque. The olar moment of inertia of the shaft is J. Thus, 2 a d 4 2 b 32 d4 t allow T 70a d DE c J ; 2 b 10(106 ) 32 d4 d m mm 33 mm 223

104 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 16. The solid shaft has a diameter of 40 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is maximum. The internal torque develoed in each segment of the shaft are shown in the torque diagram, Fig. a. Since segment DE subjected to the greatest torque, the absolute maximum shear stress occurs here. The olar moment of inertia of the shaft is J 80(10-9 ) m 4 2 (0.024 ). Thus, 15 N m 25 N m 30 N m D 60 N m E 70 N m t max T DE c J 70(0.02) 80(10-9 ) 5.57(106 ) Pa 5.57 MPa The shear stress distribution along the radial line is shown in Fig. b The rod has a diameter of 1 in. and a weight of 10 lb/ft. Determine the maximum torsional stress in the rod at a section located at due to the rod s weight. Here, we are only interested in the internal torque. Thus, other comonents of the internal loading are not indicated in the FD of the cut segment of the rod, Fig. a. 4.5 ft 1.5 ft 1.5 ft M x 0; T - 10(4)(2) 0 T 80 lb # ft a 12in 1ft b 960 lb # in. 4 ft The olar moment of inertia of the cross section at is J. 2 (0.54 ) in 4 Thus t max T c J 960 (0.5) si 4.89 ksi 224

105 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The rod has a diameter of 1 in. and a weight of 15 lb/ft. Determine the maximum torsional stress in the rod at a section located at due to the rod s weight. 4.5 ft 1.5 ft 1.5 ft 4 ft Here, we are only interested in the internal torque. Thus, other comonents of the internal loading are not indicated in the FD of the cut segment of the rod, Fig. a. M x 0; T - 15(4)(2) 0 T 120 lb # ft a 12 in 1ft b 1440 lb # in. The olar moment of inertia of the cross-section at is J in 4. Thus, 2 (0.54 ) t max T c J 1440(0.5) si 7.33 ksi 225

106 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Two wrenches are used to tighten the ie. If P 300 N is alied to each wrench, determine the maximum torsional shear stress develoed within regions and. The ie has an outer diameter of 25 mm and inner diameter of 20 mm. Sketch the shear stress distribution for both cases. 250 mm P Internal Loadings: The internal torque develoed in segments and of the ie can be determined by writing the moment equation of equilibrium about the x axis by referring to their resective free - body diagrams shown in Figs. a and b. 250 mm M x 0; T - 300(0.25) 0 T 75 N # m P nd M x 0; T - 300(0.25) - 300(0.25) 0 T 150 N # m llowable Shear Stress: The olar moment of inertia of the ie is J (10-9 )m 4 t max T c J 75(0.0125) ( MPa ) t r 0.01 m T r J 75(0.01) (10-9 ) 33.1 MPa t max T c J 150(0.0125) ( MPa ) t r 0.01 m T r J 150(0.01) (10-9 ) 66.2 MPa The shear stress distribution along the radial line of segments and of the ie is shown in Figs. c and d, resectively. 226

107 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 20. Two wrenches are used to tighten the ie. If the ie is made from a material having an allowable shear stress of t allow 85 MPa, determine the allowable maximum force P that can be alied to each wrench. The ie has an outer diameter of 25 mm and inner diameter of 20 mm. 250 mm P Internal Loading: y observation, segment of the ie is critical since it is subjected to a greater internal torque than segment. Writing the moment equation of equilibrium about the x axis by referring to the free-body diagram shown in Fig. a, we have M x 0; T - P(0.25) - P(0.25) 0 T 0.5P P 250 mm llowable Shear Stress: J (10-9 )m 4 The olar moment of inertia of the ie is t allow T c J ; 85(10 6 ) 0.5P(0.0125) (10-9 ) P N 308 N 227

108 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the outer surface of the shaft and secify their locations, measured from the fixed end N m 2 kn m/m 1.5 m 0.8 m The internal torque for segment is onstant T 1200 N # m, Fig. a. However, the internal for segment varies with x,fig.b. T x T (2000x ) N # m The minimum shear stress occurs when the internal torque is zero in segment. y setting T 0, x x 0.6 m nd d 1.5 m m 0.9 m t min 0 The maximum shear stress occurs when the internal torque is the greatest. This occurs at fixed suort where d 0 t this location, (T ) max 2000(1.5) N # m The olar moment of inertia of the rod is J. Thus, 2 (0.034 ) 0.405(10-6 ) t max (T ) max c J 1800(0.03) 0.405(10-6 ) 42.44(106 )Pa 42.4 MPa 228

109 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft to the nearest mm if the allowable shear stress for the material is t allow 50 MPa. 2 kn m/m 1200 N m 0.8 m 1.5 m The internal torque for segment is constant T 1200 N # m, Fig. a. However, the internal torque for segment varies with x,fig.b. T x T (2000x ) N # m For segment, the maximum internal torque occurs at fixed suort where x 1.5 m. Thus, T max 2000(1.5) N # m Since T max 7 T, the critical cross-section is at. The olar moment of inertia of the rod is J. Thus, 2 a d 4 2 b d4 32 t allow Tc J ; 50(106 ) 1800(d>2) d 4 >32 d m mm 57 mm 229

110 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 24. The coer ie has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at and a uniformly distributed torque is alied to it as shown, determine the shear stress develoed at oints and. These oints lie on the ie s outer surface. Sketch the shear stress on volume elements located at and. 125 lb ft/ft 9 in. 4 in. 12 in. Internal Torque: s shown on FD. Maximum Shear Stress: lying the torsion formula t T c J 125.0(12)(1.25) 2 ( ksi ) t T c J (12)(1.25) 2 ( ksi ) The coer ie has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at and it is subjected to the uniformly distributed torque along its entire length, determine the absolute maximum shear stress in the ie. Discuss the validity of this result. Internal Torque: The maximum torque occurs at the suort. T max (125 lb # ft>ft)a 25 in. 12 in.>ft b lb # ft 125 lb ft/ft 12 in. 9 in. 4 in. Maximum Shear Stress: lying the torsion formula t abs T max c max J (12)(1.25) 2 ( ksi ) ccording to Saint-Venant s rincile, alication of the torsion formula should be as oints sufficiently removed from the suorts or oints of concentrated loading. 230

111 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher cylindrical sring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is alied to the shaft, determine the maximum shear stress in the rubber. t F T r 2 r h T 2 r 2 h T r o ri h Shear stress is maximum when r is the smallest, i.e. r r i. Hence, t max T 2 r i 2 h The -36 steel shaft is suorted on smooth bearings that allow it to rotate freely. If the gears are subjected to the torques shown, determine the maximum shear stress develoed in the segments and. The shaft has a diameter of 40 mm. 300 N m 100 N m The internal torque develoed in segments and are shown in their resective FDs, Figs. a and b. 200 N m The olar moment of inertia of the shaft is J. Thus, 2 (0.024 ) 80(10-9 ) m 4 t max T c J 300(0.02) 80(10-9 ) 23.87(106 )Pa 23.9 MPa t max T c J 200(0.02) 80(10-9 ) 15.92(106 ) Pa 15.9 MPa 231

112 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 28. The -36 steel shaft is suorted on smooth bearings that allow it to rotate freely. If the gears are subjected to the torques shown, determine the required diameter of the shaft to the nearest mm if t allover 60 MPa. 300 N m 100 N m The internal torque develoed in segments and are shown in their resective FDs, Fig. a and b Here, segment is critical since its internal torque is the greatest. The olar moment of inertia of the shaft is J. Thus, 2 a d 4 2 b d4 32 t allow T J ; 60(106 ) 300(d>2) d 4 > N m d m 30 mm When drilling a well at constant angular velocity, the bottom end of the drill ie encounters a torsional resistance T. lso, soil along the sides of the ie creates a distributed frictional torque along its length, varying uniformly from zero at the surface to t at. Determine the minimum torque T that must be sulied by the drive unit to overcome the resisting torques, and comute the maximum shear stress in the ie. The ie has an outer radius r o and an inner radius r i. T L T t L - T 0 T 2T + t L 2 t T Maximum shear stress: The maximum torque is within the region above the distributed torque. t max Tc J t max [(2T + t L) 2 ] (r 0 ) 2 (r r i 4 ) (2T + t L)r 0 (r r i 4 ) 232

113 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft is subjected to a distributed torque along its length of t 110x 2 2 N # m>m, where x is in meters. If the maximum stress in the shaft is to remain constant at 80 MPa, determine the required variation of the radius c of the shaft for 0 x 3 m. x T t dx 10 x 2 dx 10 L L t Tc J ; 80(106 ) ( 3 )x3 c 2 c x3 x c 3 m t (10x 2 ) N m/m c (10-9 ) x 3 c (2.98 x) mm The solid steel shaft has a diameter of 25 mm and is suorted by smooth bearings at D and E. It is couled to a motor at, which delivers 3 kw of ower to the shaft while it is turning at 50 rev>s. If gears and remove 1 kw and 2 kw, resectively, determine the maximum shear stress develoed in the shaft within regions and. The shaft is free to turn in its suort bearings D and E. 1 kw D 2 kw 25 mm E 3 kw T P v 3(103 ) 50(2) N # m T 1 3 T N # m (t ) max T J (0.0125) 2 ( ) 1.04 MPa (t ) max T J (0.0125) 2 ( ) 3.11 MPa 233

114 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 32. The um oerates using the motor that has a ower of 85 W. If the imeller at is turning at 150 rev>min, determine the maximum shear stress develoed in the 20-mm-diameter transmission shaft at. 150 rev/min Internal Torque: v 150 rev min P 85 W 85 N # m>s T P v N # m Maximum Shear Stress: lying torsion formula t max T c J 2 rad rev 1 min 60 s 5.00 rad>s (0.01) 2 (0.014 ) 3.44 MPa The gear motor can develo 2 h when it turns at 450 rev>min. If the shaft has a diameter of 1 in., determine the maximum shear stress develoed in the shaft. The angular velocity of the shaft is and the ower is Then v 450 rev 2 rad min 1 rev 1 min 15 rad>s 60 s P 2 h 550 ft # lb>s 1100 ft # lb>s 1 h T P v lb # ft a 12 in 1ft b lb # in The olar moment of inertia of the shaft is J. Thus, 2 (0.54 ) in 4 t max T c J (0.5) si 1.43 ksi 234

115 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The gear motor can develo 3 h when it turns at 150 rev>min. If the allowable shear stress for the shaft is t allow 12 ksi, determine the smallest diameter of the shaft 1 to the nearest 8 in. that can be used. The angular velocity of the shaft is and the ower is Then v a150 rev min ba2 rad 1 rev min ba1 b 5 rad>s 60 s P (3 h)a 550 ft # lb>s b 1650 ft # lb>s 1 h T P v ( lb # ft)a 12 in 1 ft b lb # in The olar moment of inertia of the shaft is J 2 a d 4 2 b d4 32. Thus, t allow T c J ; (d>2) 12(103 ) d 4 >32 d in. 7 8 in The 25-mm-diameter shaft on the motor is made of a material having an allowable shear stress of t allow 75 MPa. If the motor is oerating at its maximum ower of 5 kw, determine the minimum allowable rotation of the shaft. llowable Shear Stress: J (10-9 ) m 4 The olar moment of inertia of the shaft is t allow Tc J ; 75(106 ) Internal Loading: T N # m T(0.0125) (10-9 ) T P v ; (103 ) v v 21.7 rad>s 235

116 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 36. The drive shaft of the motor is made of a material having an allowable shear stress of t allow 75 MPa. If the outer diameter of the tubular shaft is 20 mm and the wall thickness is 2.5 mm, determine the maximum allowable ower that can be sulied to the motor when the shaft is oerating at an angular velocity of 1500 rev>min. Internal Loading: The angular velocity of the shaft is We have v a1500 rev min T P v P 50 ba2 rad 1 rev min ba1 b 50 rad>s 60 s llowable Shear Stress: The olar moment of inertia of the shaft is J (10-9 ) m 4 t allow Tc J ; 75(106 ) a P 50 b(0.01) (10-9 ) P W 12.7 kw shi has a roeller drive shaft that is turning at 1500 rev>min while develoing 1800 h. If it is 8 ft long and has a diameter of 4 in., determine the maximum shear stress in the shaft caused by torsion. Internal Torque: v 1500 rev 2 rad a min 1 rev b 1 min 50.0 rad>s 60 s P 1800 ha 550 ft # lb>s b ft # lb>s 1 h T P v lb # ft Maximum Shear Stress: lying torsion formula t max Tc J (12)(2) 2 (24 ) 6018 si 6.02 ksi 236

117 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The motor develos a ower of 300 W and turns its connected ulley at 90 rev>min. Determine the required diameters of the steel shafts on the ulleys at and if the allowable shear stress is t allow 85 MPa. 60 mm 90 rev/min 150 mm Internal Torque: For shafts and v 90 rev min P 300 W 300 N # m>s 2 rad a rev b 1 min 3.00 rad>s 60 s T P v N # m v v a r b 3.00a 0.06 b 1.20 rad>s r 0.15 P 300 W 300 N # m>s llowable Shear Stress: For shaft t max t allow T c J d 2 For shaft d m 12.4 mm t max t allow T c J T P v N # m 2 d d 2 2 d 2 4 d m 16.8 mm 237

118 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid steel shaft DF has a diameter of 25 mm and is suorted by smooth bearings at D and E. It is 3 kw couled to a motor at F, which delivers 12 kw of ower to the shaft while it is turning at 50 rev>s. If gears,, and remove 3 kw, 4 kw, and 5 kw resectively, determine the D maximum shear stress develoed in the shaft within regions F and. The shaft is free to turn in its suort bearings D and E. 4 kw 5 kw 12 kw 25 mm E F v 50 rev s c 2 rad rev d 100 rad>s T F P v 12(103 ) N # m T P v 3(103 ) N # m T P v 4(103 ) N # m (t max ) F T F c J (t max ) T c J 38.20(0.0125) 2 ( ) (0.0125) 2 ( ) 12.5 MPa 7.26 MPa *5 40. Determine the absolute maximum shear stress develoed in the shaft in Prob kw 4 kw 5 kw 12 kw 25 mm v 50 rev s c 2 rad d 100 rad>s rev D E F T F P v 12(103 ) N # m T P v 3(103 ) N # m T P v 4(103 ) N # m From the torque diagram, T max 38.2 N # m tabs Tc max J 38.2(0.0125) 2 ( ) 12.5 MPa 238

119 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel tubular shaft is 2 m long and has an outer diameter of 50 mm. When it is rotating at 40 rad> s, it transmits 25 kw of ower from the motor M to the um P. Determine the smallest thickness of the tube if the allowable shear stress is t allow 80 MPa. P M The internal torque in the shaft is T P v 25(103 ) N # m The olar moment of inertia of the shaft is J. Thus, 2 ( i ) So that t allow Tc J ; 80(106 ) t (0.025) 2 ( i 4 ) i m m mm 2.5 mm The -36 solid tubular steel shaft is 2 m long and has an outer diameter of 60 mm. It is required to transmit 60 kw of ower from the motor M to the um P. Determine the smallest angular velocity the shaft can have if the allowable shear stress is t allow 80 MPa. P M The olar moment of inertia of the shaft is J. Thus, 2 (0.034 ) 0.405(10-6 ) m 4 t allow Tc J ; 80(106 ) T(0.03) 0.405(10-6 ) T N # m P Tv ; 60(10 3 ) v v rad>s 17.7 rad>s 239

120 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher steel tube having an outer diameter of 2.5 in. is used to transmit 35 h when turning at 2700 rev>min. Determine the inner diameter d of the tube to the nearest 1 8 in. if the allowable shear stress is t allow 10 ksi. v 2700(2) 60 P Tv rad>s d 2.5 in. 35(550) T(282.74) T lb # ft t max t allow Tc J 10(10 3 ) (12)(1.25) 2 ( c i 4 ) c i in. d 2.48 in. Use d in. *5 44. The drive shaft of an automobile is made of a steel having an allowable shear stress of t allow 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 h to the shaft when it is turning at 1140 rev>min, determine the minimum required thickness of the shaft s wall. v 1140(2) rad>s P Tv 200(550) T(119.38) T lb # ft t allow Tc J 8(10 3 ) (12)(1.25) 2 ( r i 4 ) t r o - r i , r i in. t in. 240

121 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The drive shaft of an automobile is to be designed as a thin-walled tube. The engine delivers 150 h when the shaft is turning at 1500 rev>min. Determine the minimum thickness of the shaft s wall if the shaft s outer diameter is 2.5 in. The material has an allowable shear stress of t allow 7 ksi. v 1500(2) rad>s t allow Tc J P Tv 150(550) T(157.08) T lb # ft 7(10 3 ) (12)(1.25) 2 ( r i 4 ), r i in. t r o - r i t in The motor delivers 15 h to the ulley at while turning at a constant rate of 1800 rm. Determine to the 1 nearest 8 in. the smallest diameter of shaft if the allowable shear stress for steel is t allow 12 ksi. The belt does not sli on the ulley. 3 in. The angular velocity of shaft can be determined using the ulley ratio that is The ower is Thus, v a r b v r a ba1800 rev min P (15 h)a 550 ft # n>s b 8250 ft # lb>s 1 h ba2 rad 1 rev T P v (87.54 lb # ft)a 12 in. 1 ft b lb # in min ba1 b 30 rad>s 60 s 1.5 in. The olar moment of inertia of the shaft is J 2 a d 4 2 b d4 32. Thus, t allow Tc J ; 12(103 ) (d>2) d 4 >32 d in 7 8 in. 241

122 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The roellers of a shi are connected to a -36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm. If the ower outut is 4.5 MW when the shaft rotates at 20 rad>s, determine the maximum torsional stress in the shaft and its angle of twist. T P v 4.5(106 ) (10 3 ) N # m t max Tc J 225(10 3 )(0.170) 44.3 MPa 2 [(0.170)4 - (0.130) 4 ] f TL JG (60) 2 [(0.170)4 - (0.130) 4 )75(10 9 ) rad 11.9 *5 48. shaft is subjected to a torque T. omare the effectiveness of using the tube shown in the figure with that of a solid section of radius c. To do this, comute the ercent increase in torsional stress and angle of twist er unit length for the tube versus the solid section. Shear stress: c 2 c T T For the tube, (t t ) max T c J t c For the solid shaft, (t s ) max T c J s % increase in shear stress (t Tc s) max - (t t ) max J t - Tc J s (100) (100) (t t ) Tc max J s J s - J t 2 (100) c4 - [ 2 [c4 - ( 2 )4 ]] J t 2 [c4 - ( (100) 2 )4 ] 6.67 % ngle of twist: For the tube, f t TL J t (G) For the shaft, f s TL J s (G) % increase in f f t - f s f s (100%) J s - J t 2 (100%) c4 - [ 2 [c4 - ( 2 )4 ]] J t 2 [c4 - ( (100%) 2 )4 ] 6.67 % TL J t (G) - TL J s (G) TL J s (G) (100%) 242

123 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel axle is made from tubes and D and a solid section. It is suorted on smooth bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N # m torques, determine the angle of twist of gear relative to gear D. The tubes have an outer diameter of 30 mm and an inner diameter of 20 mm. The solid section has a diameter of 40 mm. f ND TL JG 400 mm 250 mm 400 mm D 85 N m 2(85)(0.4) 2 ( )(75)(10 9 ) + (85)(0.25) 2 (0.024 )(75)(10 9 ) rad N m The hydrofoil boat has an -36 steel roeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum ower of 2500 h and causes the shaft to rotate at 1700 rm. If the outer 3 diameter of the shaft is 8 in. and the wall thickness is 8 in., determine the maximum shear stress develoed in the shaft. lso, what is the wind u, or angle of twist in the shaft at full ower? 100 ft Internal Torque: v 1700 rev min 2 rad a rev b 1 min rad>s 60 s P 2500 h a 550 ft # lb>s b ft # lb>s 1 h T P v lb # ft Maximum Shear Stress: lying torsion Formula. t max T c J (12)(4) 2 ( ksi ) ngle of Twist: f TL JG (12)(100)(12) 2 ( )11.0(10 6 ) rad

124 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The engine of the helicoter is delivering 600 h to the rotor shaft when the blade is rotating at 1200 rev>min. 1 Determine to the nearest 8 in. the diameter of the shaft if the allowable shear stress is t allow 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made from L2 steel. v 1200(2)() 60 P Tv rad>s 600(550) T(125.66) T lb # ft Shear - stress failure ngle of twist limitation t allow Tc J 8(10 3 ) (12)c 2 c4 f TL JG c in (12)(2)(12) 2 c4 (11.0)(10 6 ) c in. Shear - stress failure controls the design. d 2c 2 (1.3586) 2.72 in. Use d 2.75 in. 244

125 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 52. The engine of the helicoter is delivering 600 h to the rotor shaft when the blade is rotating at 1200 rev>min. 1 Determine to the nearest 8 in. the diameter of the shaft if the allowable shear stress is t allow 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made from L2 steel. v 1200(2)() 60 P Tv rad>s 600(550) T(125.66) T lb # ft Shear - stress failure ngle of twist limitation t allow 10.5(10) (12)c 2 c4 c in. f TL JG (12)(2)(12) 2 c4 (11.0)(10 6 ) c in. Shear stress failure controls the design d 2c 2 (1.2408) 2.48 in. Use d 2.50 in. 245

126 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The 20-mm-diameter -36 steel shaft is subjected to the torques shown. Determine the angle of twist of the end. Internal Torque: s shown on FD. D ngle of Twist: f a TL JG 1 2 (0.014 )(75.0)(10 9 [-80.0(0.8) + (-60.0)(0.6) + (-90.0)(0.2)] ) 80 N m 30 N m 600 mm 20 N m 800 mm 200 mm rad The assembly is made of -36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at. Determine the angle of twist at D. The tube has an outer diameter of 40 mm and wall thickness of 5 mm. The internal torques develoed in segments and D of the assembly are shown in Fig. a and b The olar moment of inertia of solid rod and tube are J 2 ( ) (10-9 ) m 4 and J D. Thus, 2 (0.014 ) 5(10-9 ) m m 0.1 m 0.3 m 150 N m D 60 N m f D T i L i T L + T D L D J i G i J G st J D G st 90(0.4) (10-9 ) [75(10 9 )] (0.4) 5(10-9 ) [75(10 9 )] rad

127 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The assembly is made of -36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at. Determine the angle of twist at. The tube has an outer diameter of 40 mm and wall thickness of 5 mm. The internal torques develoed in segments and of the assembly are shown in Figs. a and b. The olar moment of inertia of the tube is J (10-9 ) m 4 2 ( ). Thus, 0.4 m 0.1 m 0.3 m 150 N m D 60 N m f T i L i T L + T L J i G i JG st J G st (10-9 ) [75( (0.4) + 150(0.1)D )] rad *5 56. The slined ends and gears attached to the -36 steel shaft are subjected to the torques shown. Determine the angle of twist of end with resect to end. The shaft has a diameter of 40 mm. 300 N m 500 N m 200 N m f > TL JG -300(0.3) JG + 200(0.4) JG 190 JG (0.024 )(75)(10 9 ) rad (0.5) JG 300 mm 400 mm D 400 N m 500 mm 247

128 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The motor delivers 40 h to the 304 stainless steel shaft while it rotates at 20 Hz. The shaft is suorted on smooth bearings at and, which allow free rotation of the shaft.the gears and D fixed to the shaft remove 25 h and 15 h, resectively. Determine the diameter of the 1 shaft to the nearest 8 in. if the allowable shear stress is t allow 8 ksi and the allowable angle of twist of with resect to D is in. 8 in. 6 in. D External lied Torque: lying T P, we have 2f T M 40(550) 2(20) lb # ft T 25(550) 2(20) lb # ft T D 15(550) 2(20) lb # ft Internal Torque: s shown on FD. llowable Shear Stress: ssume failure due to shear stress. y observation, section is the critical region. t max t allow Tc J 8(10 3 ) (12)d 2 d in. ngle of Twist: ssume failure due to angle of twist limitation. f >D T DL D JG 2 d () (12)(8) 2 d 2 4 (11.0)(10 6 ) d in. (controls!) Use d in. 248

129 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The motor delivers 40 h to the 304 stainless steel solid shaft while it rotates at 20 Hz.The shaft has a diameter of 1.5 in. and is suorted on smooth bearings at and, which allow free rotation of the shaft. The gears and D fixed to the shaft remove 25 h and 15 h, resectively. Determine the absolute maximum stress in the shaft and the angle of twist of gear with resect to gear D. 10 in. 8 in. 6 in. D External lied Torque: lying T P, we have 2f T M 40(550) 2(20) lb # ft T 25(550) 2(20) lb # ft T D 15(550) 2(20) lb # ft Internal Torque: s shown on FD. llowable Shear Stress: The maximum torque occurs within region of the shaft where T max T lb # ft. ngle of Twist: t abs T max c max J (12)(0.75) 2 (0.754 ) 3.17 ksi f >D T D L D JG 65.65(12)(8) 2 (0.754 )(11.0)(10 6 ) rad

130 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft is made of -36 steel. It has a diameter of 1 in. and is suorted by bearings at and D, which allow free rotation. Determine the angle of twist of with resect to D. The internal torques develoed in segments and D are shown in Figs. a and b. The olar moment of inertia of the shaft is J. Thus, 2 (0.54 ) in 4 2 ft 2.5 ft 60 lb ft 3 ft 60 lb ft D F /D T il i a T L + T D L D J i G i J G st J G st -60(12)(2.5)(12) ( )[11.0(10 6 )] rad 1.15 *5 60. The shaft is made of -36 steel. It has a diameter of 1 in. and is suorted by bearings at and D, which allow free rotation. Determine the angle of twist of gear with resect to. The internal torque develoed in segment is shown in Fig. a The olar moment of inertia of the shaft is J. Thus, 2 (0.54 ) in 4 2 ft 2.5 ft 60 lb ft 3 ft 60 lb ft D f > T L J G st -60(12)(2.5)(12) ( )[11.0(10 6 )] rad

131 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The two shafts are made of -36 steel. Each has a diameter of 1 in., and they are suorted by bearings at,, and, which allow free rotation. If the suort at D is fixed, determine the angle of twist of end when the torques are alied to the assembly as shown. 10 in. D 80 lb ft 30 in. 40 lb ft 8 in. 10 in. 12 in. 4 in. 6 in. Internal Torque: s shown on FD. ngle of Twist: f E a TL JG 1 2 (0.54 )(11.0)(10 5 ) [-60.0(12)(30) (12)(10)] rad rad f F 6 4 f E 6 ( ) rad 4 Since there is no torque alied between F and then f f F rad

132 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The two shafts are made of -36 steel. Each has a diameter of 1 in., and they are suorted by bearings at,, and, which allow free rotation. If the suort at D is fixed, determine the angle of twist of end when the torques are alied to the assembly as shown. 10 in. D 80 lb ft 30 in. 40 lb ft 8 in. 10 in. 12 in. 4 in. 6 in. Internal Torque: s shown on FD. ngle of Twist: f E a TL JG 1 2 (0.54 )(11.0)(10 6 ) rad rad [-60.0(12)(30) (12)(10)] f F 6 4 f E 6 ( ) rad 4 f >F T GF L GF JG -40(12)(10) 2 (0.54 )(11.0)(10 6 ) rad rad f f F + f >F rad

133 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The device serves as a comact torsional sring. It is made of -36 steel and consists of a solid inner shaft which is surrounded by and attached to a tube using a rigid ring at. The ring at can also be assumed rigid and is fixed from rotating. If a torque of T 2 ki # in. is alied to the shaft, determine the angle of twist at the end and the maximum shear stress in the tube and shaft. 12 in. 12 in in. T 1 in. Internal Torque: s shown on FD. Maximum Shear Stress: ngle of Twist: (t ) max T c J (t ) max T c J 2.00(0.5) 2 (0.54 ) 10.2 ksi 2.00(1) 2 ( ksi ) 0.5 in. f T L JG (2.00)(12) 2 ( )11.0( rad ) f > T L JG 2.00(24) 2 (0.54 )11.0( rad ) f f + f > rad

134 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 64. The device serves as a comact torsion sring. It is made of -36 steel and consists of a solid inner shaft which is surrounded by and attached to a tube using a rigid ring at. The ring at can also be assumed rigid and is fixed from rotating. If the allowable shear stress for the material is t allow 12 ksi and the angle of twist at is limited to f allow 3, determine the maximum torque T that can be alied at the end. T 12 in. 12 in. 1 in in. Internal Torque: s shown on FD. llowable Shear Stress: ssume failure due to shear stress. t max t allow T c J 12.0 T (0.5) 2 (0.54 ) T ki # in t max t allow T c J T (1) ( ) T ki # in ngle of Twist: ssume failure due to angle of twist limitation. 0.5 in. f T L JG T(12) 2 ( ) 11.0(10 3 ) T f > T L JG T(24) 2 (0.54 )11.0(10 3 ) T (f ) allow f + f > 3() T T T 2.25 ki # in (controls!) 254

135 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel assembly consists of a tube having an outer radius of 1 in. and a wall thickness of in. Using a rigid late at, it is connected to the solid 1-in-diameter shaft. Determine the rotation of the tube s end if a torque of 200 lb # in. is alied to the tube at this end. The end of the shaft is fixed suorted. 200 lb in. 4 in. 6 in. f T L JG 200(10) rad 2 (0.5)4 (11.0)(10 6 ) f > T L JG -200(4) rad 2 ( )(11.0)(10 6 ) f f + f > rad

136 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The 60-mm diameter shaft is suorted by two journal bearings, while the 80-mm diameter shaft EH is fixed at E and suorted by a journal bearing at H. If T1 2 kn # m and T 2 4 kn # m, determine the angle of twist of gears and. The shafts are made of -36 steel. T 2 E D 600 mm 100 mm H 600 mm 75 mm 900 mm Equilibrium: Referring to the free - body diagram of shaft shown in Fig. a M x 0; F(0.075) - 4(10 3 ) - 2(10 3 ) 0 F 80(10 3 ) N Internal Loading: Referring to the free - body diagram of gear D in Fig. b, M x 0; 80(10 3 )(0.1) - T DH 0 T DH 8(10 3 )N # m lso, from the free - body diagram of gear,fig.c, M x 0; T - 4(10 3 ) 0 T N # m nd from the free - body diagram of gear,fig.d, M x 0; -T T -2(10 3 ) N # m ngle of Twist: The olar moment of inertia of segments, and DH of the shaft are J J and J DH (10-6 ) m 4. We have (10-6 ) m 4 T 1 f D T DH L DH J DH G st 8(10 3 )(0.6) 1.28(10-6 )(75)(10 9 ) rad Then, using the gear ratio, f f D a r D b a 100 b rad r 75 lso, f > T L J G st f > T L J G st -2(10 3 )(0.9) 0.405(10-6 )(75)(10 9 ) 4(10 3 )(0.6) 0.405(10-6 )(75)(10 9 ) rad rad rad Thus, f f + f > f rad 2.66 f f + f > f rad

137 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued 257

138 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The 60-mm diameter shaft is suorted by two journal bearings, while the 80-mm diameter shaft EH is fixed at E and suorted by a journal bearing at H. If the angle of twist at gears and is required to be 0.04 rad, determine the magnitudes of the torques T 1 and T 2.The shafts are made of -36 steel. T mm E 75 mm D 600 mm 100 mm H 900 mm Equilibrium: Referring to the free - body diagram of shaft shown in Fig. a M x 0; F(0.075) - T 1 - T 2 0 F T 1 + T 2 Internal Loading: Referring to the free - body diagram of gear D in Fig. b, M x 0; T 1 + T 2 (0.1) - T DE 0 T DE 1.333T 1 + T 2 lso, from the free - body diagram of gear,fig.c, T 1 M x 0; T - T 2 0 T T 2 and from the free - body diagram of gear,fig.d M x 0; T - T 1 0 T T 1 ngle of Twist: The olar moments of inertia of segments, and DH of the shaft are J J and J DH (10-6 )m 4. We have (10-6 )m 4 f D T DE L DH J DE G st 1.333T 1 + T 2 (0.6) 1.28(10-6 ) (75)(10 9 ) (10-6 )T 1 + T 2 Then, using the gear ratio, lso, f f D r D r (10-6 )T 1 + T 2 a b (10-6 )T 1 + T 2 f > T L J G st f > T L J G st Here, it is required that f f 0.04 rad. Thus, f f + f > T 1 (0.9) 0.405(10-6 )(75)(10 9 ) (10-6 )T 1 T 2 (0.6) 0.405(10-6 )(75)(10 9 ) (10-6 )T (10-6 )T 1 + T (10-6 )T 2 T T (1) f f + f > (10-6 )T 1 + T (10-6 )T T 1 + T (2) 258

139 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Solving Eqs. (1) and (2), T N # m 2.19 kn # m T N # m 3.28 kn # m *5 68. The 30-mm-diameter shafts are made of L2 tool steel and are suorted on journal bearings that allow the shaft to rotate freely. If the motor at develos a torque of T 45 N # m on the shaft, while the turbine at E is fixed from turning, determine the amount of rotation of gears and. Internal Torque: s shown on FD. ngle of Twist: 1.5 m 0.5 m 45 N m D 75 mm 0.75 m 50 mm E f T E L E JG 67.5(0.75) 2 ( )75.0(10 3 ) rad f f

140 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shafts are made of -36 steel and each has a diameter of 80 mm. Determine the angle of twist at end E. 0.6 m 150 mm 10 kn m Equilibrium: Referring to the free - body diagram of shaft DE shown in Fig. a, M x 0; 10(10 3 ) - 2(10 3 ) - F(0.2) 0 F 40(10 3 ) N Internal Loading: Referring to the free - body diagram of gear,fig.b, 200 mm 0.6 m 150 mm 0.6 m D E 2 kn m M x 0; -T - 40(10 3 )(0.15) 0 T -6(10 3 ) N # m Referring to the free - body diagram of gear D,Fig.c, M x 0; 10(10 3 ) - 2(10 3 ) - T D 0 T D 8(10 3 ) N # m Referring to the free - body diagram of shaft DE,Fig.d, M x 0; -T DE - 2(10 3 ) 0 T DE -2(10 3 ) N # m ngle of Twist: The olar moment of inertia of the shafts are J (10-6 ) m 4 We have f T L JG st -6(10 3 )(0.6) 1.28(10-6 )(75)(10 9 ) rad rad Using the gear ratio, lso, Thus, f f r a 150 b rad r 200 f E> T il i T D L D + T DE L DE J i G i JG st JG st (10-6 )(75)(10 9 ) b 8(103 ) + c -2(10 3 ) d r rad f E f + f E> f E rad

141 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued 261

142 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shafts are made of -36 steel and each has a diameter of 80 mm. Determine the angle of twist of gear D. 0.6 m Equilibrium: Referring to the free-body diagram of shaft DE shown in Fig. a, M x 0; 10(10 3 ) - 2(10 3 ) - F(0.2) 0 F 40(10 3 ) N Internal Loading: Referring to the free - body diagram of gear,fig.b, M x 0; -T - 40(10 3 )(0.15) 0 T -6(10 3 ) N # m Referring to the free - body diagram of gear D,Fig.c, M x 0; 10(10 3 ) - 2(10 3 ) - T D 0 T D 8(10 3 ) N # m 200 mm 0.6 m 150 mm 10 kn m D 150 mm 0.6 m E 2 kn m ngle of Twist: The olar moment of inertia of the shafts are J. We have (10-6 ) m 4 f T L JG st -6(10 3 )(0.6) 1.28(10-6 )(75)(10 9 ) rad rad Using the gear ratio, f f r a 150 b rad r 200 lso, f D> T D L D JG st 8(10 3 )(0.6) 1.28(10-6 )(75)(10 9 ) rad Thus, f D f + f D> f D rad

143 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 72. The 80-mm diameter shaft is made of 6061-T6 aluminum alloy and subjected to the torsional loading shown. Determine the angle of twist at end. 0.6 m 0.6 m 10 kn m/m 2 kn m Equilibrium: Referring to the free - body diagram of segment shown in Fig. a, M x 0; -T - 2(10 3 ) 0 T -2(10 3 )N # m nd the free - body diagram of segment,fig.b, M x 0; -T - 10(10 3 )x - 2(10 3 ) 0 T -10(10 3 )x + 2(10 3 )D N # m ngle of Twist: The olar moment of inertia of the shaft is J. We have (10-6 ) m 4 f T il i T L L T dx + J i G i JG al L 0 JG al -2(10 3 )(0.6) 0.6 m - 10(10 3 )x + 2(10 3 )Ddx 1.28(10-6 )(26)(10 9 ) + L0 1.28(10-6 )(26)(10 9 ) (10-6 )(26)(10 9 ) b (103 )x 2 + 2(10 3 )xd 2 0.6m 0 r rad

144 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The taered shaft has a length L and a radius r at end and 2r at end. If it is fixed at end and is subjected to a torque T, determine the angle of twist of end.the shear modulus is G. Geometry: r(x) r + r L x r L + r x L r T 2r L J(x) 2 a r L + r x 4 b L r4 (L + x)4 4 2L ngle of Twist: f L L 0 T dx J(x)G 2TL4 r 4 G L0 L dx (L + x) 4 2TL4 L r 4 G c - 1 3(L + x) 3 d 2 0 7TL 12 r 4 G The rod of radius c is embedded into a medium where the distributed torque reaction varies linearly from zero at to t 0 at. If coule forces P are alied to the lever arm, determine the value of t 0 for equilibrium. lso, find the angle of twist of end. The rod is made from material having a shear modulus of G. Equilibrium: Referring to the free-body diagram of the entire rod shown in Fig. a, L 2 t 0 L 2 d 2 d 2 M x 0; Pd (t 0)a L 2 b 0 P P t o 4Pd L Internal Loading: The distributed torque exressed as a function of x, measured from the left end, is t t o 4Pd>L 8Pd x x. Thus, the resultant L>2 L>2 L 2 x torque within region x of the shaft is T R 1 2 tx 1 2 8Pd 4Pd 2 xrx L L 2 x2 Referring to the free - body diagram shown in Fig. b, M x 0; T - 4Pd L 2 x2 0 T 4Pd L 2 x2 Referring to the free - body diagram shown in Fig. c, M x 0; Pd - T 0 T Pd 264

145 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued ngle of Twist: f T il i J i G i L L 0 T dx JG + T L JG L L>2 0 4Pd L 2 x2 dx 2 c4 G L>2 8Pd c 4 L 2 G x Pd(L>2) 2 c4 G + PLd c 4 G 4PLd 3c 4 G 265

146 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher When drilling a well, the dee end of the drill ie is assumed to encounter a torsional resistance T. Furthermore, soil friction along the sides of the ie creates a linear distribution of torque er unit length, varying from zero at the surface to t 0 at. Determine the necessary torque T that must be sulied by the drive unit to turn the ie. lso, what is the relative angle of twist of one end of the ie with resect to the other end at the instant the ie is about to turn? The ie has an outer radius r o and an inner radius r i. The shear modulus is G. T L t t 0 L + T - T 0 T T t 0L + 2T 2 T(x) + t 0 2L x2 - t 0L + 2T 2 0 T(x) t 0 L + 2T 2 - t 0 2L x2 f L T(x) dx J G 1 J G L0 L ( t 0L + 2T 2 - t 0 2L x2 ) dx 1 JG c t 0 L + 2T x - t L 0 2 6L x3 d ƒ t 0 L 2 + 3T L 3JG 0 However, J 2 (r o 4 - r 4 i ) f 2L(t 0 L + 3T ) 3(r o 4 - r i 4 )G 266

147 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 76. cylindrical sring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is alied to the rigid shaft, determine the angle of twist of the shaft. The shear modulus of the rubber is G. Hint: s shown in the figure, the deformation of the element at radius r can be determined from rdu drg. Use this exression along with t T>12r 2 h2 from Prob. 5 26, to obtain the result. T r o r r i h dr gdr rdu r du g dr du g r du gdr r (1) From Prob. 5-26, t T 2 r 2 h and g t G g T 2 r 2 hg From (1), du T u 2 hg L T dr 2 hg r 3 r i r o T 2 hg c r + 1 o 2r 2 d i dr r 3 T 2 hg c r 2 d ro r i T 4 hg c 1 2 r - 1 i r 2 d o 267

148 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel shaft has a diameter of 50 mm and is fixed at its ends and. If it is subjected to the torque, determine the maximum shear stress in regions and of the shaft. 0.4 m 300 N m 0.8 m Equilibrium: T + T [1] omatibility: f > f > T (0.4) JG T (0.8) JG T 2.00T [2] Solving Eqs. [1] and [2] yields: Maximum Shear stress: T 200 N # m T 100 N # m (t ) max T c J 200(0.025) 8.15 MPa 2 ( ) (t ) max T c J 100(0.025) 4.07 MPa 2 ( ) 268

149 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel shaft has a diameter of 60 mm and is fixed at its ends and.if it is subjected to the torques shown, determine the absolute maximum shear stress in the shaft. 200 N m 500 N m D 1 m 1.5 m 1 m Referring to the FD of the shaft shown in Fig. a, M x 0; T + T (1) Using the method of suerosition, Fig. b f (f ) T - (f ) T 0 T (3.5) JG - c 500 (1.5) JG (1) JG d Substitute this result into Eq (1), T N # m T N # m Referring to the torque diagram shown in Fig. c, segment is subjected to maximum internal torque. Thus, the absolute maximum shear stress occurs here. tbs T c J (0.03) 2 (0.03) MPa 269

150 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The steel shaft is made from two segments: has a diameter of 0.5 in, and has a diameter of 1 in. If it is fixed at its ends and and subjected to a torque of determine the maximum shear stress in the shaft. G st ksi. 5 in. 0.5 in. D 500 lb ft 8 in. 1 in. 12 in. Equilibrium: T + T (1) omatibility condition: f D> f D> T (5) 2 (0.254 )G + T (8) 2 (0.54 )G T (12) 2 (0.54 )G 1408 T 192 T (2) Solving Eqs. (1) and (2) yields T 60 lb # ft T 440 lb # ft t T J 60(12)(0.25) 2 (0.254 ) t D T J 440(12)(0.5) 2 (0.54 ) 29.3 ksi (max) 26.9 ksi 270

151 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 80. The shaft is made of -36 steel, has a diameter of 80 mm, and is fixed at while is loose and can rotate rad before becoming fixed. When the torques are alied to and D, determine the maximum shear stress in regions and D of the shaft. 4 kn m 2 kn m D 600 mm 600 mm Referring to the FD of the shaft shown in Fig. a, 600 mm M x 0; T + T (1) Using the method of suerosition, Fig. b, f (f ) T - (u ) T 4(10 3 )(0.6) (0.044 )75(10 9 )D + 2(10 3 )(0.6) 2 (0.044 )75(10 9 )D R - T (1.8) 2 (0.044 )75(10 9 )D T N # m kn # m Substitute this result into Eq (1), T kn # m Referring to the torque diagram shown in Fig. c, segment D is subjected to a maximum internal torque. Thus, the absolute maximum shear stress occurs here. t $$$ T D c J (103 )(0.04) 2 (0.04) (10 6 ) Pa 28.2 MPa 271

152 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft is made of -36 steel and has a diameter of 80 mm. It is fixed at and the suort at has a torsional stiffness of k 0.5 MN # m>rad. If it is subjected to the gear torques shown, determine the absolute maximum shear stress in the shaft. 4 kn m 2 kn m D 600 mm 600 mm 600 mm Referring to the FD of the shaft shown in Fig. a, M x 0; T + T (1) Using the method of suerosition, Fig. b, f (f ) T - (f ) T T 0.5(10 6 ) D 4(10 3 )(0.6) 2 (0.044 )75(10 9 )D + 2(10 3 )(0.6) T (1.8) 2 (0.044 )75(10 9 )D T - 2 (0.044 )75(10 9 )D T N # m kn # m Substituting this result into Eq (1), T kn # m Referring to the torque diagram shown in Fig. c, segment D subjected to maximum internal torque. Thus, the maximum shear stress occurs here. t $$$ T D J 2.502(103 )(0.04) 2 (0.04)4 $$$ 24.9 MPa 272

153 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft is made from a solid steel section and a tubular ortion made of steel and having a brass core. If it is fixed to a rigid suort at, and a torque of T 50 lb # ft is alied to it at, determine the angle of twist that occurs at and comute the maximum shear stress and maximum shear strain in the brass and steel. Take G st ksi, G br ksi. 2 ft 3 ft 0.5 in. Equilibrium: 1 in. T 50 lb ft T br + T st (1) oth the steel tube and brass core undergo the same angle of twist f > f > TL JG T st (2)(12) 2 (0.54 )(5.6)(10 4 ) T st (2)(12) 2 ( )(11.5)(10 6 ) T br T st (2) Solving Eqs. (1) and (2) yields: T st lb # ft; T br lb # ft f TL JG 1.572(12)(2)(12) 2 (0.54 )(5.6)(10 6 ) + 50(12)(3)(12) 2 (14 )(11.5)(10 6 ) (t st ) max T c J (t st ) max T st c J (12)(1) 2 ( si 395 si (Max) ) (g st ) max (t st) max G (t br ) max T br c J (g br ) max (t br) max G rad (12)(1) 2 (14 ) (10 6 ) 343.(10-6 ) rad 1.572(12)(0.5) 2 (0.54 ) 382 si (10 6 ) 17.2(10-6 ) rad si 96.1 si (Max) 273

154 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The motor develos a torque at gear of 450 lb # ft, which is alied along the axis of the 2-in.-diameter steel shaft D. This torque is to be transmitted to the inion gears at E and F. If these gears are temorarily fixed, determine the maximum shear stress in segments and D of the shaft. lso, what is the angle of twist of each of these segments? The bearings at and D only exert force reactions on the shaft and do not resist torque. G st ksi. E 450 lb ft F 4 ft 3 ft D Equilibrium: T + T D (1) omatibility condition: f > f >D T (4) JG T D(3) JG T 0.75 T D (2) Solving Eqs. (1) and (2), yields T D lb # ft T lb # ft (t ) max (12)(1) 2 (14 ) (t D ) max (12)(1) 2 (14 ) f (12)(4)(12) 2 (14 )(12)(10 6 ) 1.47 ksi 1.96 ksi rad

155 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 84. ortion of the -36 steel shaft is subjected to a linearly distributed torsional loading. If the shaft has the dimensions shown, determine the reactions at the fixed suorts and. Segment has a diameter of 1.5 in. and segment has a diameter of 0.75 in. 300 lb in./in. 60 in. 48 in. Equilibrium: T + T (1) T R t x (300 - t)x 150x + t x 2 ut t 60 - x 300 ; t 5(60 - x) 60 T R 150 x + 1 [5(60 - x)]x 2 (300x - 2.5x 2 ) lb # in. omatibility condition: f > f > T(x) dx f > L JG 1 JG L JG [T x - 150x x 3 ] 60 60T JG [T - (300x - 2.5x 2 )] dx 0 60T (0.754 )G T (48) 2 ( )G 60T - 768T (2) Solving Eqs. (1) and (2) yields: T lb # in lb # ft T lb # in. 732 lb # ft 275

156 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Determine the rotation of joint and the absolute maximum shear stress in the shaft in Prob lb in./in. Equilibrium: T + T (1) 60 in. T R tx (300 - t)x 150x + tx 2 ut t 60 - x 300 ; t 5(60 - x) in. T R 150x + 1 [5(60 - x)]x 2 (300x - 2.5x 2 ) lb # in. omatibility condition: f > f > T(x) dx f > L JG 1 JG L JG [T x - 150x x 3 ] 60 60T JG [T - (300x - 2.5x 2 )] dx 0 60T (0.754 )G T (48) 2 ( )G 60T - 768T (2) Solving Eqs. (1) and (2) yields: T lb # in lb # ft T lb # in. 732 lb # ft For segment : f f > T L JG 217.4(48) rad 2 (0.375)4 (11.0)(10 6 ) f 1.75 t max T J 217.4(0.375) 2.62 ksi 2 (0.375)4 For segment, t max T J (0.75) 13.3 ksi 2 (0.75)4 t abs 13.3 ksi max 276

157 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The two shafts are made of -36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed suorts at and. They are also suorted by journal bearings at and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is alied to the gear at E as shown, determine the reactions at and. F 50 mm 100 mm 1.5 m D E 500 N m 0.75 m Equilibrium: T + F(0.1) T - F(0.05) 0 [1] [2] From Eqs. [1] and [2] T + 2T [3] omatibility: 0.1f E 0.05f F f E 0.5f F T (1.5) JG 0.5c T (0.75) JG d T 0.250T [4] Solving Eqs. [3] and [4] yields: T 222 N # m T 55.6 N # m 277

158 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Determine the rotation of the gear at E in Prob mm 100 mm F D 500 N m 0.75 m E 1.5 m Equilibrium: T + F(0.1) T - F(0.05) 0 [1] [2] From Eqs. [1] and [2] T + 2T [3] omatibility: 0.1f E 0.05f F f E 0.5f F T (1.5) JG 0.5c T (0.75) JG d T 0.250T [4] Solving Eqs. [3] and [4] yields: ngle of Twist: T N # m T N # m f E T L JG 55.56(1.5) 2 ( )(75.0)(10 9 ) rad

159 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 88. The shafts are made of -36 steel and have the same diameter of 4 in. If a torque of 15 ki # ft is alied to gear, determine the absolute maximum shear stress develoed in the shaft. 2.5 ft 2.5 ft Equilibrium: Referring to the free - body diagrams of shafts and DE shown in Figs. a and b, resectively, we have M x 0; T + F(0.5) and M x 0; F(1) - T E 0 (1) (2) 15 ki ft D 12 in. 6 in. E Internal Loadings: The internal torques develoed in segments and of shaft and shaft DE are shown in Figs. c, d, and e, resectively. 3 ft omatibility Equation: f r f D r D T L JG st + T L JG st r T DE L DE r JG D st -T (2.5) + F(0.5)(2.5)D(0.5) -T E (3)(1) T - 0.5F 2.4T E (3) Solving Eqs. (1), (2), and (3), we have F ki T E ki # ft T ki # ft Maximum Shear Stress: y insection, segment of shaft is subjected to the greater torque. t max abs T c 12.79(12)(2) 12.2 ksi J st 2 a24 b

160 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shafts are made of -36 steel and have the same diameter of 4 in. If a torque of 15 ki # ft is alied to gear, determine the angle of twist of gear. 2.5 ft Equilibrium: Referring to the free - body diagrams of shafts and DE shown in Figs. a and b, resectively, 2.5 ft M x 0; T + F(0.5) and (1) 15 ki ft D 6 in. M x 0; F(1) - T E 0 (2) 12 in. E Internal Loadings: The internal torques develoed in segments and of shaft and shaft DE are shown in Figs. c, d, and e, resectively. 3 ft omatibility Equation: It is required that f r f D r D T L JG st + T L JG st r T DE L DE r JG D st -T (2.5) + F(0.5)(2.5)D(0.5) -T E (3)(1) T - 0.5F 2.4T E (3) Solving Eqs. (1), (2), and (3), F ki T E ki # ft T ki # ft ngle of Twist: Here, T -T ki # ft f T L JG st (12)(2.5)(12) 2 a24 b(11.0)(10 3 ) rad

161 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The two 3-ft-long shafts are made of 2014-T6 aluminum. Each has a diameter of 1.5 in. and they are connected using the gears fixed to their ends. Their other ends are attached to fixed suorts at and. They are also suorted by bearings at and D, which allow free 600 lb ft rotation of the shafts along their axes. If a torque of 600 lb # ft is alied to the to gear as shown, determine the maximum shear stress in each shaft. 4 in. E D 3 ft 2 in. F T + Fa 4 12 b T - Fa 2 12 b 0 (1) (2) From Eqs. (1) and (2) T + 2T (3) 4(f E ) 2(f F ); f E 0.5f F T L JG 0.5a T L JG b; T 0.5T (4) Solving Eqs. (3) and (4) yields: T 240 lb # ft; T 120 lb # ft (t D ) max T c J 240(12)(0.75) 2 (0.754 ) 4.35 ksi (t ) max T c J 120(12)(0.75) 2 (0.754 ) 2.17 ksi 281

162 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel shaft is made from two segments: has a diameter of 0.5 in. and has a diameter of 1 in. If the shaft is fixed at its ends and and subjected to a uniform distributed torque of 60 lb # in.>in. along segment, 5 in. determine the absolute maximum shear stress in the shaft. 0.5 in. 20 in. 60 lb in./in. 1 in. Equilibrium: T + T - 60(20) 0 (1) omatibility condition: f > f > f > L T(x) dx JG L0 20 (T - 60x) dx 2 (0.54 )(11.0)(10 6 ) 18.52(10-6 )T (10-6 )T T (5) 2 (0.254 )(11.0)(10 6 ) 18.52(10-6 )T (10-6 )T T T (2) Solving Eqs. (1) and (2) yields: T lb # in. ; T 1080 lb # in. (t max ) T c J (t max ) T c J t abs 5.50 ksi max 1080(0.5) 2 (0.54 ) 120.0(0.25) 2 (0.254 ) 5.50 ksi 4.89 ksi 282

163 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 92. If the shaft is subjected to a uniform distributed torque of t 20 kn # m>m, determine the maximum shear stress develoed in the shaft. The shaft is made of 2014-T6 aluminum alloy and is fixed at and. 400 mm 80 mm 20 kn m/m 600 mm a Equilibrium: Referring to the free - body diagram of the shaft shown in Fig. a, we have 60 mm Section a a a M x 0; T + T - 20(10 3 )(0.4) 0 (1) omatibility Equation: The resultant torque of the distributed torque within the region x of the shaft is T R 20(10 3 )x N # m. Thus, the internal torque develoed in the shaft as a function of x when end is free is T(x) 20(10 3 )x N # m, Fig. b. Using the method of suerosition, Fig. c, f f t - f T 0.4 m T(x)dx 0 L 0 JG 0.4 m 20(10 3 )xdx 0 L JG (10 3 ) x2 0.4 m T T 1600 N # m Substituting this result into Eq. (1), T 6400 N # m 0 - T L JG - T (1) JG Maximum Shear Stress: y insection, the maximum internal torque occurs at suort. Thus, t max abs T c J 6400(0.04) 2 a b 93.1 MPa 283

164 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The taered shaft is confined by the fixed suorts at and. If a torque T is alied at its mid-oint, determine the reactions at the suorts. Equilibrium: Section Proerties: T + T - T 0 [1] L/2 2c L/2 T c r(x) c + c L x c (L + x) L J(x) 2 c c L (L + x) d 4 c4 (L + x)4 4 2L ngle of Twist: Tdx f T L J(x)G L 2 L 2TL4 c 4 G L 2 Tdx c 4 2L 4 (L + x)4 G L dx (L + x) 4-2TL4 L 3c 4 G c 1 (L + x) 3 d TL 324 c 4 G Tdx f L J(x)G L0 L 2T L 4 c 4 G L0-2T L 4 L 3c 4 G c 1 (L + x) 3 d 2 7T L 12c 4 G T dx c 4 2L 4 (L + x) 4 G L dx (L + x) 4 0 omatibility: 0 f T - f 0 37TL 324c 4 G - 7T L 12c 4 G T T Substituting the result into Eq. [1] yields: T T 284

165 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the reactions at the fixed suorts and. t 0 t t 0 ( 1 ( x L 2 ) ) x L y suerosition: x T(x) t 0 a1 + x2 L L 2 b dx t 0 ax + 0 x2 3L 2 b (1) 2t 0 0 f - f 0 L L 0 t 0 ax + x3 3L 2 b dx JG - T (L) JG 7t 2 0L 12 - T (L) T 7t 0 L 12 From Eq. (1), T t 0 al + L3 3L 2 b 4t 0 L 3 T + 7t 0 L 12-4t 0 L 3 0 T 3t 0 L 4 285

166 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher omare the values of the maximum elastic shear stress and the angle of twist develoed in 304 stainless steel shafts having circular and square cross sections. Each shaft has the same cross-sectional area of 9 in length of 36 in., and is subjected to a torque of 4000 lb # 2, in. r a a Maximum Shear Stress: For circular shaft c 2 9; c a 9 b 1 2 (t c ) max Tc J Tc 2T 2 c4 c 3 2(4000) 9 x si For rectangular shaft a 2 9 ; a 3 in. (t r ) max 4.81T a (4000) si ngle of Twist: For circular shaft f c TL JG 4000(36) (10 6 ) rad For rectangular shaft f r 7.10 TL a 4 G 7.10(4000)(36) 3 4 (11.0)(10 6 ) rad The rectangular shaft has a greater maximum shear stress and angle of twist. 286

167 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. *5 96. If a 25 mm and b 15 mm, determine the maximum shear stress in the circular and ellitical shafts when the alied torque is T 80 N # m. y what ercentage is the shaft of circular cross section more efficient at withstanding the torque than the shaft of ellitical cross section? a b For the circular shaft: a (t max ) c T c J 80(0.025) 2 ( ) For the ellitical shaft: 3.26 MPa (t max ) c 2T a b 2 2(80) (0.025)( MPa ) % more efficient (t max) c - (t max ) c (t max ) c (100%) (100%) 178 % It is intended to manufacture a circular bar to resist torque; however, the bar is made ellitical in the rocess of manufacturing, with one dimension smaller than the other by a factor k as shown. Determine the factor by which the maximum shear stress is increased. kd d For the circular shaft: (t max ) c Tc J Td 2 2 d 2 16T 4 d 3 d For the ellitical shaft: (t max ) c 2T a b 2 2T d 2 kd 2 16T 2 k 2 d 3 Factor of increase in shear stress (t max) c (t max ) c 16T k2 d3 16T d 3 1 k 2 287

168 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft is made of red brass and has an ellitical cross section. If it is subjected to the torsional loading shown, determine the maximum shear stress within regions and, and the angle of twist f of end relative to end. 50 N m 20 N m 30 N m Maximum Shear Stress: (t ) max 2T a b 2 2(30.0) (0.05)( ) 2 m 1.5 m 50 mm 20 mm MPa (t ) max 2T a b 2 2(50.0) (0.05)( ) 1.59 MPa ngle of Twist: f > a (a 2 + b 2 )T L a 3 b 3 G ( ) ( )( )(37.0)(10 9 [(-30.0)(1.5) + (-50.0)(2)] ) rad Solve Prob for the maximum shear stress within regions and, and the angle of twist f of end relative to. 50 N m 20 N m 30 N m Maximum Shear Stress: (t ) max 2T a b 2 2(30.0) (0.05)( ) 2 m 1.5 m 50 mm 20 mm MPa (t ) max 2T a b 2 2(50.0) (0.05)( ) 1.59 MPa ngle of Twist: f > (a2 + b 2 ) T L a 3 b 3 G ( )(-30.0)(1.5) ( )( )(37.0)(10 9 ) rad

169 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * Segments and of the shaft have circular and square cross sections, resectively. If end is subjected to a torque of T 2 kn # m, determine the absolute maximum shear stress develoed in the shaft and the angle of twist of end.the shaft is made from -36 steel and is fixed at. 600 mm 600 mm 90 mm 90 mm 30 mm Internal Loadings: The internal torques develoed in segments and are shown in Figs. a, and b, resectively. T Maximum Shear Stress: For segment, t max T c J - 2(103 )(0.03) 2 a0.034 b 47.2 MPa (max) For segment, t max 4.81T a (10 3 )D (0.09) MPa ngle of Twist: f T L JG T L a 4 G 2(10 3 )(0.6) 2 a0.034 b(75)(10 9 ) (2)(103 )(0.6) (0.09) 4 (75)(10 9 ) rad

170 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Segments and of the shaft have circular and square cross sections, resectively. The shaft is made from -36 steel with an allowable shear stress of t allow 75 MPa, and an angle of twist at end which is not allowed to exceed 0.02 rad. Determine the maximum allowable torque T that can be alied at end.the shaft is fixed at. 600 mm 600 mm Internal Loadings: The internal torques develoed in segments and are shown in Figs. a, and b, resectively. llowable Shear Stress: For segment, t allow T c J ; 75(106 ) T(0.03) 2 a0.034 b 90 mm 90 mm T 30 mm T N # m For segment, t allow 4.81T a 3 ; 75(10 6 ) 4.81T (0.09) 3 T N # m ngle of Twist: f T L JG T L a 4 G 0.02 T(0.6) 2 a0.034 b(75)(10 9 ) T(0.6) (0.09) 4 (75)(10 9 ) T N # m 2.80 kn # m (controls) The aluminum strut is fixed between the two walls at and. If it has a 2 in. by 2 in. square cross section, and it is subjected to the torque of 80 lb # ft at, determine the reactions at the fixed suorts. lso, what is the angle of twist at? G al ksi. 2 ft y suerosition: 0 f - f 80 lb ft 3 ft (80)(2) a 4 G (T )(5) a 4 G T 32 lb # ft T T 48 lb # ft f 7.10(32)(12)(3)(12) (2 4 )(3.8)(10 6 ) rad

171 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The square shaft is used at the end of a drive cable in order to register the rotation of the cable on a gauge. If it has the dimensions shown and is subjected to a torque of 8 N # m, determine the shear stress in the shaft at oint. Sketch the shear stress on a volume element located at this oint. 5 mm Maximum shear stress: 8 N m 5 mm (t max ) 4.81T a (8) 308 MPa 3 (0.005) * The 6061-T6 aluminum bar has a square cross section of 25 mm by 25 mm. If it is 2 m long, determine the maximum shear stress in the bar and the rotation of one end relative to the other end. 20 N m 1.5 m Maximum Shear Stress: 0.5 m t max 4.81T max a (80.0) ( MPa ) 60 N m 25 mm 80 N m 25 mm ngle of Twist: f > a 7.10TL a 4 G 7.10(-20.0)(1.5) ( )(26.0)(10 9 ) (-80.0)(0.5) ( )(26.0)(10 9 ) rad

172 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The steel shaft is 12 in. long and is screwed into the wall using a wrench. Determine the largest coule forces F that can be alied to the shaft without causing the steel to yield. t Y 8 ksi. F(16) - T 0 (1) 1 in. 12 in. t max t Y 4.81T a 3 F 8 in. 1 in. 8(10 3 ) 4.81T (1) 3 8 in. T lb # in. F From Eq. (1), F 104 lb The steel shaft is 12 in. long and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft and the amount of dislacement that each coule force undergoes if the coule forces have a magnitude of F 30 lb, G st ksi. 1 in. 12 in. T - 30(16) 0 T 480 lb # in. F 8 in. 1 in. t max 4.18T a (480) (1) 3 8 in. f 7.10TL a 4 G 7.10(480)(12) rad (1) 4 (10.8)(10 6 ) d F 8( ) in ksi F 292

173 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed 12 ksi when a torque of T 20 ki # in. is alied to the tube. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown. T m 2(4) 8 in 2 4 in. t avg T 2 t m 2 in t (8) t in. * Determine the torque T that can be alied to the rectangular tube if the average shear stress is not to exceed 12 ksi. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown and the tube has a thickness of in. T m 2(4) 8 in 2 4 in. t avg T 2 t m ; 12 T 2(0.125)(8) 2 in. T 24 ki # in. 2 ki # ft 293

174 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher For a given maximum shear stress, determine the factor by which the torque carrying caacity is increased if the half-circular section is reversed from the dashed-line osition to the section shown. The tube is 0.1 in. thick. m (1.10)(1.75) - (0.552 ) in in in. 0.6 in. 0.5 in. m (1.10)(1.75) + (0.552 ) in 2 t max T 2t m T 2 t m t max Factor 2t m t max 2t m t max m m For a given average shear stress, determine the factor by which the torque-carrying caacity is increased if the half-circular sections are reversed from the dashed-line ositions to the section shown. The tube is 0.1 in. thick in in. 0.6 in. Section Proerties: 0.5 in. m œ (1.1)(1.8) - (0.552 ) R(2) in 2 2 m (1.1)(1.8) + (0.552 ) R(2) in 2 2 verage Shear Stress: t avg T 2 t m ; T 2 t m t avg Hence, T 2 t m œ The factor of increase T T t avg m œ m

175 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher torque T is alied to two tubes having the cross sections shown. omare the shear flow develoed in each tube. t t ircular tube: t a Square tube: q ct T 2 m T 2 (a>2) 2 2T a 2 a a q st T 2 m T 2a 2 q st T>(2a2 ) q ct 2T>( a 2 ) 4 Thus; q st 4 q ct * Due to a fabrication error the inner circle of the tube is eccentric with resect to the outer circle. y what ercentage is the torsional strength reduced when the eccentricity e is one-fourth of the difference in the radii? verage Shear Stress: For the aligned tube a a b 2 e 2 e 2 b t avg For the eccentric tube T 2 t m T t avg (2)(a - b)()a a + b 2 b 2 T 2(a - b)() a + b 2 2 t avg T 2 t m t a - e 2 - a e 2 + bb a - e - b a (a - b) - b 3 (a - b) 4 T t avg (2)c a + b (a - b) d()a b 2 Factor T T t avg (2) 3 4 (ab)d()a + b 2 2 t avg (2)(a - b)() a + b Percent reduction in strength a1-3 b * 100 % 25 % 4 295

176 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The mean dimensions of the cross section of an airlane fuselage are shown. If the fuselage is made of 2014-T6 aluminum alloy having allowable shear stress of t allow 18 ksi, and it is subjected to a torque of 6000 ki # ft, determine the required minimum thickness t of the cross section to the nearest 1>16 in. lso, find the corresonding angle of twist er foot length of the fuselage. 4.5 ft 3 ft t 3 ft Section Proerties: Referring to the geometry shown in Fig. a, m (6) ft in2 1 ft in 2 F 12 in. ds 2(3) + 2(4.5) fta b in. 1 ft llowable verage Shear Stress: t avg allow T ; (12) 2t m 2t( ) t in in. ngle of Twist: Using the result of t 5, 16 in TL ds f 4 2 m G F t 6000(12)(1)(12) 4( )(3.9)(10 3 ) > (10-3 ) rad

177 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The mean dimensions of the cross section of an airlane fuselage are shown. If the fuselage is made from 2014-T6 aluminum alloy having an allowable shear stress of t allow 18 ksi and the angle of twist er foot length of fuselage is not allowed to exceed rad>ft, determine the maximum allowable torque that can be sustained by the fuselage. The thickness of the wall is t 0.25 in. 4.5 ft 3 ft t 3 ft Section Proerties: Referring to the geometry shown in Fig. a, m (6) ft in2 1 ft in 2 F 12 in. ds 2(3) + 2(4.5) fta b in. 1 ft llowable verage Shear Stress: t avg allow T 2t m ; 18 T 2(0.25)( ) T ki # ina 1ft 12 in. b 5970 ki # ft ngle of Twist: f TL 4 m 2 G F ds t T(1)(12) 4( )(3.9)(10 3 ) a b T ki # ina 1ft 12 in. b 5134 ki # ft (controls) 297

178 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The tube is subjected to a torque of 750 N # m. Determine the average shear stress in the tube at oints and. Referring to the geometry shown in Fig. a, Thus, (t avg ) T 2t m m 0.06 (0.1) m (0.004)(0.006) 15.63(106 )Pa 15.6 MPa 4 mm 6 mm 100 mm 750 N m 4 mm 60 mm 6 mm (t avg ) T 2t m 750 2(0.006)(0.006) 10.42(106 )Pa 10.4 MPa * The tube is made of lastic, is 5 mm thick, and has the mean dimensions shown. Determine the average shear stress at oints and if it is subjected to the torque of T 5 N # m. Show the shear stress on volume elements located at these oints. m (0.11)(0.08) + 1 (0.08)(0.03) 0.01 m mm t t t avg T 2t m 5 50 kpa 2(0.005)(0.01) 60 mm 30 mm T 40 mm 40 mm 298

179 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The mean dimensions of the cross section of the leading edge and torsion box of an airlane wing can be aroximated as shown. If the wing is made of 2014-T6 aluminum alloy having an allowable shear stress of t allow 125 MPa and the wall thickness is 10 mm, determine the maximum allowable torque and the corresonding angle of twist er meter length of the wing. 10 mm 0.5 m 10 mm 2 m 10 mm 0.25 m 0.25 m Section Proerties: Referring to the geometry shown in Fig. a, m 2 a0.52 b (2) m2 2 F ds (0.5) m llowable verage Shear Stress: t avg allow T 2t m ; 125(10 6 ) T 2(0.01)(1.8927) T (10 6 )N # m 4.73 MN # m ngle of Twist: f TL 4 m 2 G F ds t (10 6 )(1) 4( )(27)(10 9 ) (10-3 ) rad >m 299

180 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The mean dimensions of the cross section of the leading edge and torsion box of an airlane wing can be aroximated as shown. If the wing is subjected to a torque of 4.5 MN # m and the wall thickness is 10 mm, determine the average shear stress develoed in the wing and the angle of twist er meter length of the wing. The wing is made of 2014-T6 aluminum alloy. 10 mm 0.5 m 10 mm 2 m 10 mm 0.25 m 0.25 m Section Proerties: Referring to the geometry shown in Fig. a, m 2 a0.52 b (2) m2 2 F ds (0.5) m verage Shear Stress: t avg ngle of Twist: T 2t m 4.5(10 6 ) 2(0.01)(1.8927) 119 MPa f TL 4 m 2 G F ds t 4.5(10 6 )(1) 4( )(27)(10 9 ) (10-3 ) rad >m 300

181 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The symmetric tube is made from a high-strength steel, having the mean dimensions shown and a thickness of 5 mm. If it is subjected to a torque of T 40 N # m, determine the average shear stress develoed at oints and. Indicate the shear stress on volume elements located at these oints. 30 mm 20 mm 60 mm m 4(0.04)(0.06) + (0.04) m 2 40 N m t avg T 2 t m (t avg ) (t avg ) kpa 2(0.005)(0.0112) * The steel used for the shaft has an allowable shear stress of t allow 8 MPa. If the members are connected with a fillet weld of radius r 4 mm, determine the maximum torque T that can be alied. 50 mm 20 mm 20 mm T 2 T T 2 llowable Shear Stress: From the text, K 1.25 D d and r d t max t allow K Tc J t 8(10) (0.01) 2 (0.014 ) R T 20.1 N # m 301

182 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The built-u shaft is to be designed to rotate at 720 rm while transmitting 30 kw of ower. Is this ossible? The allowable shear stress is t allow 12 MPa. 75 mm v 720 rev 2 rad a min 1 rev b 1 min 24 rad>s 60 s 60 mm T P v 30(103 ) N # m Tc t max K J ; 12(106 ) Kc (0.03) 2 (0.034 ) D d d ; K 1.28 r From Fig. 5-32, d r ; r 7.98 mm 60 heck: D - d mm mm No, it is not ossible The built-u shaft is designed to rotate at 540 rm. If the radius of the fillet weld connecting the shafts is r 7.20 mm, and the allowable shear stress for the material is t allow 55 MPa, determine the maximum ower the shaft can transmit. D d ; r d From Fig. 5-32, K 1.30 Tc t max K J ; 55(106 ) 1.30 c[ T(0.03) 2 (0.034 ) d; T N # m v 540 rev 2 rad a min 1 rev b 1 min 18 rad>s 60 s 60 mm 75 mm P Tv (18) W 101 kw 302

183 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The steel shaft is made from two segments: and, which are connected using a fillet weld having a radius of 2.8 mm. Determine the maximum shear stress develoed in the shaft. (t max ) D T Dc J 4.07 MPa 100(0.025) 2 ( ) D 50 mm 20 mm 100 N m 40 N m 60 N m For the fillet: D d ; r d From Fig. 5-32, K (t max ) f K T c J c 60(0.01) 2 (0.014 ) d 50.6 MPa (max) * The steel used for the shaft has an allowable shear stress of t allow 8 MPa. If the members are connected together with a fillet weld of radius r 2.25 mm, determine the maximum torque T that can be alied. 30 mm 30 mm 15 mm T 2 T T 2 llowable Shear Stress: From the text, K 1.30 D d and r d Tc t max t allow K J r 8( (0.0075) ) ( ) S T 8.16 N # m 303

184 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The assembly is subjected to a torque of 710 lb # in. If the allowable shear stress for the material is t allow 12 ksi, determine the radius of the smallest size fillet that can be used to transmit the torque in. 710 lb in. D d From Fig. 5-32, t max t allow K Tc J 12(10 3 ) K(710)(0.375) 2 ( ) K lb ft 1.5 in. r 0.1; r 0.1(0.75) in. d heck: D - d in. OK solid shaft is subjected to the torque T, which causes the material to yield. If the material is elastic lastic, show that the torque can be exressed in terms of the angle of twist f of the shaft as T 4 3 T Y11 - f 3 Y>4f 3 2, where T Y and f Y are the torque and angle of twist when the material begins to yield. f gl r g Y r Y L r Y g YL f (1) When r Y c, f f Y From Eq. (1), c g YL f Y (2) Dividing Eq. (1) by Eq. (2) yields: r Y c f Y f (3) Use Eq from the text. T t Y 6 (4 c3 - r 3 Y ) 2 t Yc 3 a1 - r Y c 3)b Use Eq. 5-24, T Y from the text and Eq. (3) 2 t Yc 3 T 4 3 T Ya1 - f Y 3 4 f 3 b QED 304

185 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher solid shaft having a diameter of 2 in. is made of elastic-lastic material having a yield stress of t and shear modulus of G Y 16 ksi 2 ksi. Determine the torque required to develo an elastic core in the shaft having a diameter of 1 in. lso, what is the lastic torque? Use Eq from the text: T t Y 6 (4 c3 - r 3 Y ) (16) [4(1 3 ) ] ki # in ki # ft Use Eq from the text: T P 2 3 t Yc (16)(13 ) ki # in ki # ft * Determine the torque needed to twist a short 3-mm-diameter steel wire through several revolutions if it is made from steel assumed to be elastic lastic and having a yield stress of t Y 80 MPa. ssume that the material becomes fully lastic. When the material becomes fully lastic then, from Eq in the text, T P 2 t Y 3 c 3 2 (80)(106 ) ( ) N # m 3 305

186 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The solid shaft is made of an elastic-erfectly lastic material as shown. Determine the torque T needed to form an elastic core in the shaft having a radius of r Y 20 mm. If the shaft is 3 m long, through what angle does one end of the shaft twist with resect to the other end? When the torque is removed, determine the residual stress distribution in the shaft and the ermanent angle of twist. T 80 mm T t (MPa) 160 Elastic-Plastic Torque: lying Eq from the text g (rad) T t Y 4c 3 - r 6 Y3 (160)(106 ) D N # m 20.8 kn # m ngle of Twist: f g Y L a b(3) rad 34.4 r Y 0.02 When the reverse T N # m is alied, G 160(106 ) GPa f TL JG (3) 2 (0.044 )(40)(10 9 ) rad The ermanent angle of twist is, f r f - f rad 12.2 Residual Shear Stress: (t ) r c Tc J (0.04) MPa 2 (0.044 ) (t ) r 0.02 m Tc J (0.02) MPa 2 (0.044 ) (t r ) r c MPa (t r ) r 0.02m MPa 306

187 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shaft is subjected to a maximum shear strain of rad. Determine the torque alied to the shaft if the material has strain hardening as shown by the shear stress strain diagram. From the shear - strain diagram, 2 in. r Y ; r Y 0.25 in. From the shear stress strain diagram, t r 24r t (ksi) 12 6 T t 2-6 r ; t r g (rad) c T 2 t r 2 dr L r 3 dr + 2 (3.4286r )r 2 dr L L [6r r 4 ] + 2 c r3 3 d ki # in ki # ft n 80-mm diameter solid circular shaft is made of an elastic-erfectly lastic material having a yield shear stress of t Y 125 MPa. Determine (a) the maximum elastic torque T Y ; and (b) the lastic torque T. Maximum Elastic Torque. T Y 1 2 c3 t Y 1 2 a0.043 b 125 a10 6 b N # m 12.6 kn # m Plastic Torque. T P 2 3 c3 t Y 2 3 a0.043 b 125 a10 6 b N # m 16.8 kn # m 307

188 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * The hollow shaft has the cross section shown and is made of an elastic-erfectly lastic material having a yield shear stress of t Y. Determine the ratio of the lastic torque T to the maximum elastic torque. c T Y Maximum Elastic Torque. In this case, the torsion formula is still alicable. t Y T Y c J c 2 T Y J c t Y Plastic Torque. Using the general equation, with t t Y, The ratio is 2 c4 - a c 4 2 b Rt Y c3 t Y c c T P 2t Y r 2 dr L c>2 2t Y r3 3 ` 7 12 c3 t Y c c>2 T P T Y 7 12 c3 t Y c3 t Y The shaft consists of two sections that are rigidly connected. If the material is elastic lastic as shown, determine the largest torque T that can be alied to the shaft. lso, draw the shear-stress distribution over a radial line for each section. Neglect the effect of stress concentration in. 1 in. T 0.75 in. diameter segment will be fully lastic. From Eq of the text: T T T 2 t Y (c 3 ) 3 t (ksi) 2 (12)(103 ) ( ) lb # in. 110 lb # ft g (rad) For 1 in. diameter segment: t max Tc J (0.5) 2 (0.5) ksi 6 t Y 308

189 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The hollow shaft is made of an elastic-erfectly lastic material having a shear modulus of G and a yield shear stress of t Y. Determine the alied torque T when the material of the inner surface is about to yield (lastic torque). lso, find the corresonding angle of twist and the maximum shear strain. The shaft has a length of L. ci c0 Plastic Torque. Using the general equation with t t Y, c o T P 2t Y r 2 dr L c i 2t Y r3 3 ` co c i 2 3 t Yc o 3 - c i 3 ngle of Twist. When the material is about to yield at the inner surface, g g Y at r r Y c i. lso, Hooke s Law is still valid at the inner surface. g Y t Y G f g Y L t Y>G L t YL r Y c i G c i Shear Strain. Since the shear strain varies linearly along the radial line, Fig. a, g max c o g Y c i g max c o g c Y c o a t Y i c i G b c ot Y c i G 309

190 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The hollow shaft has inner and outer diameters of 60 mm and 80 mm, resectively. If it is made of an elasticerfectly lastic material, which has the t-g diagram shown, determine the reactions at the fixed suorts and. t (MPa) 150 mm 450 mm 15 kn m Equation of Equilibrium. Refering to the free - body diagram of the shaft shown in Fig. a, M x 0; T + T Elastic nalysis. It is required that f > f >. Thus, the comatibility equation is (1) g (rad) f > f > T L JG T L JG T (0.45) T (0.15) T 3T (2) Solving Eqs. (1) and (2), T 3750 N # m T N # m The maximum elastic torque and lastic torque in the shaft can be determined from T Y J c t Y D T(120) N # m 0.04 c o T P 2t Y r 2 dr L c i 2(120)10 6 r3 3 ` 0.04 m 0.03 m N # m Since T 7 T Y, the results obtained using the elastic analysis are not valid. Plastic nalysis. ssuming that segment is fully lastic, T T P N # m 9.3kN # m Substituting this result into Eq. (1), T 5700 N # m 5.70 kn # m 310

191 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Since T 6 T Y, segment of the shaft is still linearly elastic. Here, G GPa f > f > T L JG (0.45) (75)10 9 f > g i c i L ; g i 0.03 (0.15) g i rad Since g i 7 g Y, segment of the shaft is indeed fully lastic rad 311

192 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * The tubular shaft is made of a strain-hardening material having a t-g diagram as shown. Determine the torque T that must be alied to the shaft so that the maximum shear strain is 0.01 rad. 0.5 in. T 0.75 in. t (ksi) g (rad) From the shear strain diagram, g ; g rad 0.75 From the shear stress strain diagram, t ; t ksi t r ; t r + 5 c o T 2 tr 2 dr L c i (13.333r + 5) r 2 dr L (13.333r 3 + 5r 2 ) dr L c r r3 3 d ki # in. 702 lb # ft 312

193 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The shear stress strain diagram for a solid 50-mm-diameter shaft can be aroximated as shown in the figure. Determine the torque T required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 1.5 m long, what is the corresonding angle of twist? T 1.5 m T t (MPa) 125 Strain Diagram: Stress Diagram: r g ; r g m The Ultimate Torque: t 1 50(106 ) r 8(109 ) r t 2-50(10 6 ) r (106 ) - 50(10 6 ) t r c T 2 t r 2 dr L m r 3 dr L m r Dr 2 dr L m r 4 D m r (106 )r m R m g (rad) N # m 3.27 kn # m ngle of Twist: f g max c L a 0.01 b(1.5) 0.60 rad

194 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher tube is made of elastic-erfectly lastic material, which has the t-g diagram shown. If the radius of the elastic core is r Y 2.25 in., determine the alied torque T. lso, find the residual shear-stress distribution in the shaft and the ermanent angle of twist of one end relative to the other when the torque is removed. 3 ft 3 in. T T 6 in. Elastic - Plastic Torque. The shear stress distribution due to T is shown in Fig. a.the linear ortion of this distribution can be exressed as t 10. Thus, 2.25 r 4.444r t r 1.5 in (1.5) ksi. t (ksi) 10 T 2 L tr 2 dr 2.25 in. 3 in rr 2 dr + 2(10) r 2 dr L L 1.5 in r in r3 3 in in in in g (rad) ki # in 39.2 ki # ft ngle of Twist. f g Y L (3)(12) rad r Y 2.25 The rocess of removing torque T is equivalent to the alication of T, which is equal magnitude but oosite in sense to that of T. This rocess occurs in a linear manner and G ksi f T L JG (3)(2) (2.5)10 3 t œ r c o t œ r r Y t œ r c i T c o J (3) T r Y J T c i J (1.5) ksi Thus, the ermanent angle of twist is f P f - f (2.25) ksi ksi rad rad

195 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued nd the residual stresses are (t r ) r co t r c + tr œ c ksi (t r ) r ry t r ry + tr œ r Y ksi (t r ) r ci t r ci + tr œ c i ksi The residual stress distribution is shown in Fig. a. 315

196 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The tube is made of elastic-erfectly lastic material, which has the t-g diagram shown. Determine the torque T that just causes the inner surface of the shaft to yield. lso, find the residual shear-stress distribution in the shaft when the torque is removed. 3 ft 3 in. T T 6 in. Plastic Torque. When the inner surface of the shaft is about to yield, the shaft is about to become fully lastic. t (ksi) 10 T 2 L tr 2 dr 3 in. 2t Y r 2 dr L 1.5 in. 2(10)a r3 3 in. 3 b in g (rad) ki # in ki # ft ngle of Twist. f g Y L (3)(12) rad r Y 1.5 The rocess of removing torque T is equivalent to the alication of T, which is equal magnitude but oosite in sense to that of T. This rocess occurs in a linear manner and G ksi f T L JG (3)(12) (2.5) rad t œ r c o t œ r c i T c o J (3) T c i J (1.5) ksi ksi 316

197 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued nd the residual stresses are (t r ) r co t r c + t œ r c ksi (t r ) r ci t r ci + t œ r c i ksi The shear stress distribution due to T and T and the residual stress distribution are shown in Fig. a. 317

198 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * The 2-m-long tube is made of an elastic-erfectly lastic material as shown. Determine the alied torque T that subjects the material at the tube s outer edge to a shear strain of g max rad. What would be the ermanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube. 35 mm 30 mm T 210 t (MPa) Plastic Torque: The tube is fully lastic if g i Ú g r rad. Therefore the tube is fully lastic. g ; g rad c o T P 2 t g r 2 dr L c i 2 t g 3 c o 3 - c i3 2(210)(106 ) g (rad) N # m 6.98 kn # m ngle of Twist: f P g max L a b(2) rad c o When a reverse torque of T P N # m is alied, G t Y 210(106 ) 70 GPa g Y fp œ T PL JG (2) 2 ( )(70)(10 9 ) rad Permanent angle of twist, f r f P - f P œ rad 9.11 Residual Shear Stress: t œ P o t œ P i T P c J T P r J (0.035) 2 ( MPa ) (0.03) 2 ( MPa ) (t P ) o -t g + t œ P o MPa (t P ) i -t g + t œ P i MPa 318

199 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher steel alloy core is bonded firmly to the coer alloy tube to form the shaft shown. If the materials have the t-g diagrams shown, determine the torque resisted by the core and the tube. 100 mm 450 mm 60 mm t (MPa) 15 kn m Equation of Equilibrium. Refering to the free - body diagram of the cut art of the assembly shown in Fig. a, 180 M x 0; T c + T t (1) Elastic nalysis. The shear modulus of steel and coer are G st and G q GPa. omatibility requires that GPa t (MPa) Steel lloy g (rad) f f t T c L J c G st T tl J t G q g (rad) T c (75)10 9 T t (18)10 9 oer lloy T c T t (2) Solving Eqs. (1) and (2), T t N # m T c N # m The maximum elastic torque and lastic torque of the core and the tube are (T Y ) c 1 2 c3 (t Y ) st (180) N # m (T P ) c 2 3 c3 (t Y ) st (180) N # m and (T Y ) t J c t Y D T c(36)10 6 d N # m 0.05 c o (T P ) t 2(t Y ) q r 2 dr 2(36)10 6 r 3 L 3 2 c i 0.05 m 0.03 m N # m Since T t 7 (T Y ) t, the results obtained using the elastic analysis are not valid. 319

200 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher ontinued Plastic nalysis. ssuming that the tube is fully lastic, T t (T P ) t N # m 7.39 kn # m Substituting this result into Eq. (1), T c N # m 7.61 kn # m Since T c 6 (T Y ) c, the core is still linearly elastic. Thus, f t f tc T cl J c G st (0.45) 2 (0.034 )(75)(10 9 ) f t g i c i L; g i rad g i 0.03 (0.45) rad Since g i 7 (g Y ) q rad, the tube is indeed fully lastic. 320

201 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher torque is alied to the shaft of radius r. If the material has a shear stress strain relation of t kg 1>6, where k is a constant, determine the maximum shear stress in the shaft. g r c g max r r g max r T t kg 1 6 ka g max r r 6 b r T 2 tr 2 dr L 0 2 L r 0 ka g max r g max a 19T 6 12 kr 3 b b 1 6 r 13 6 dr 2ka g max r b a 19 b r kg1 6 max r 3 19 t max kg 1 6 max 19T 12 r onsider a thin-walled tube of mean radius r and thickness t. Show that the maximum shear stress in the tube due to an alied torque T aroaches the average shear stress comuted from Eq as r>t : q. r o r + t 2 2r + t ; r 2 i r - t 2 2r - t 2 J 2 ca2r + t 4 b 2 - a 2r - t 4 b d 2 t r 32 [(2r + t)4 - (2r - t) 4 ] 32 [64 r3 t + 16 r t 3 ] t max Tc J ; c r o 2r + t 2 s r t : q t max T( 2 r + t 2 r + t 2 ) 32 [64 r3 t + 16 r t 3 ] T( 2 ) 2 r t[r t2 ] T( 2r 2r 2 + t 2 r 2) 2 r tc r2 r + 1 t2 2 4 r d 2, then t r : 0 T( 1 r + 0) 2 r t(1 + 0) T 2 r 2 t T 2 t m QED 321

202 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm. When it is rotating at 60 rad>s, it transmits 30 kw of ower from the engine E to the generator G. Determine the smallest thickness of the shaft if the allowable shear stress is t allow 150 MPa and the shaft is restricted not to twist more than 0.08 rad. E G Internal Torque: P 30(10 3 ) Wa 1 N # m>s W b 30(103 ) N # m>s T P v 30(103 ) N # m llowable Shear Stress: ssume failure due to shear stress. t max t allow Tc J r i m mm ngle of Twist: ssume failure due to angle of twist limitation. f TL JG ( (0.03) ) 2 ( r 4 i ) 500(3) r i4 (75.0)(10 9 ) r i m mm hoose the smallest value of r i mm t r o - r i mm 322

203 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The -36 steel circular tube is subjected to a torque of 10 kn # m. Determine the shear stress at the mean radius r 60 mm and comute the angle of twist of the tube if it is 4 m long and fixed at its far end. Solve the roblem using Eqs. 5 7 and 5 15 and by using Eqs and r 60 mm 4 m Shear Stress: lying Eq. 5-7, t 5 mm 10 kn m r o m r i m t r 0.06 m Tr J 10(10 3 )(0.06) 2 ( MPa ) lying Eq. 5-18, t avg ngle of Twist: T 2 t m lying Eq. 5-15, f TL JG 10(10 3 ) 29(0.005)()( ) MPa 10(10 3 )(4) 2 ( )(75.0)(10 9 ) rad lying Eq. 5-20, f TL 4 2 mg L ds t TL 4 2 mg t L ds Where ds 2r L 2TLr 4 2 mg t 2(10)(10 3 )(4)(0.06) 4[()( )] 2 (75.0)(10 9 )(0.005) rad

204 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher Rod is made of -36 steel with an allowable shear stress of 1t allow 2 st 75 MPa, and tube is made of M1004-T61 magnesium alloy with an allowable shear stress of 1t allow 2 mg 45 MPa. The angle of twist of end is not allowed to exceed 0.05 rad. Determine the maximum allowable torque T that can be alied to the assembly. 0.3 m 0.4 m a 60 mm T Internal Loading: The internal torque develoed in rod and tube are shown in Figs. a and b, resectively. 50 mm a llowable Shear Stress: The olar moment of inertia of rod and tube are J and J 2 a b 2 a b (10-9 ) m (10-6 ) m 4. We have 30 mm Section a a t allow st T c J ; 75(10 6 ) T(0.015) (10-9 ) T N # m and t allow mg T c J ; 45(10 6 ) T(0.03) (10-6 ) T N # m ngle of Twist: f > T L J G st -T(0.7) (10-9 )(75)(10 9 ) (10-3 )T (10-3 )T and f > T L J G mg T(0.4) (10-6 )(18)(10 9 ) (10-3 )T It is required that f > 0.05 rad. Thus, f > f > + f > (10-3 )T (10-3 )T T 331 N # m controls 324

205 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher shaft has the cross section shown and is made of 2014-T6 aluminum alloy having an allowable shear stress of t allow 125 MPa. If the angle of twist er meter length is not allowed to exceed 0.03 rad, determine the required minimum wall thickness t to the nearest millimeter when the shaft is subjected to a torque of T 15 kn # m. 75 mm t Section Proerties: Referring to the geometry shown in Fig. a, m 1 2 (0.15) tan m 2 llowable Shear Stress: t avg allow ngle of Twist: ds 2(0.15) + (0.075) m T 2t m ; 125(10 6 ) 15(10 3 ) 2t( ) t m 3.23 mm f TL 4 m 2 G ds t (10 3 )(1) 4( )(27)(10 9 ) a b t t m 7.18 mm (controls) Use t 8 mm 325

206 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher. * The motor develos a torque at gear of 500 lb # ft, which is alied along the axis of the 2-in.- diameter -36 steel shaft D. This torque is to be transmitted to the inion gears at E and F. If these gears are temorarily fixed, determine the maximum shear stress in segments and D of the shaft. lso, what is the angle of twist of each of these segments? The bearings at and D only exert force reactions on the shaft. E 2 ft 1.5 ft 500 lb ft F D Equilibrium: T + T D [1] omatibility: f > f >D T (2) JG T D(1.5) JG T 0.75T D [2] Solving Eqs. [1] and [2] yields: Maximum Shear Stress: T D lb # ft T lb # ft (t ) max T c J (12)(1) 1.64 ksi 2 (14 ) (t D ) max T Dc J (12)(1) 2.18 ksi 2 (14 ) ngle of Twist: f f D T D L D JG (12)(1.5)(12) 2 (14 )(11.0)(10 6 ) rad

207 05 Solutions /25/10 3:53 PM Page Pearson Education, Inc., Uer Saddle River, NJ. ll rights reserved. This material is rotected under all coyright laws as they currently exist. No ortion of this material may be reroduced, in any form or by any means, without ermission in writing from the ublisher The couling consists of two disks fixed to searate shafts, each 25 mm in diameter. The shafts are suorted on journal bearings that allow free rotation. In order to limit the torque T that can be transmitted, a shear in P is used to connect the disks together. If this in can sustain an average shear force of 550 N before it fails, determine the maximum constant torque T that can be transmitted from one shaft to the other. lso, what is the maximum shear stress in each shaft when the shear in is about to fail? 25 mm 130 mm P 25 mm T T Equilibrium: M x 0; T - 550(0.13) 0 T 71.5 N # m Maximum Shear Stress: t max Tc J 71.5(0.0125) 2 ( ) 23.3 MPa The rotating flywheel and shaft is brought to a sudden sto at D when the bearing freezes. This causes the flywheel to oscillate clockwise counterclockwise, so that a oint on the outer edge of the flywheel is dislaced through a 10-mm arc in either direction. Determine the maximum shear stress develoed in the tubular 304 stainless steel shaft due to this oscillation. The shaft has an inner diameter of 25 mm and an outer diameter of 35 mm. The journal bearings at and allow the shaft to rotate freely. 2 m D ngle of Twist: 80 mm f TL Where f rad JG 80 T(2) ( )(75.0)(10 9 ) T N # m Maximum Shear Stress: t max Tc J (0.0175) 2 ( ) 82.0 MPa 327