CCHHAAPPTTEERR 33PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 261 T18 mm PROBLEM 3.1 Determine the torque T that causes a maximum shearing stress of 70 MPa in the steel cylindrical shaft shown. SOLUTION 4max3max36;22(0.018 m) (70 10 Pa)2641.26 N mTcJcJTc 641 N mT PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 262 T18 mm PROBLEM 3.2 For the cylindrical shaft shown, determine the maximum shearing stress caused by a torque of magnitude T  800 N  m. SOLUTION 4maxmax336;222(800 N m)(0.018 m)87.328 10 PaTcJcJTc max87.3 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 263 2.4 m30 mm45 mmT PROBLEM 3.3 (a) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical steel shaft shown. (b) Determine the maximum shearing stress caused by the same torque T in a solid cylindrical shaft of the same cross-sectional area. SOLUTION (a) Given shaft: 442144 64 6466332(45 30 ) 5.1689 10 mm 5.1689 10 m2(5.1689 10 )(45 10 )5.1689 10 N m45 10JccJTc JTJcT  5.17 kN mT  (b) Solid shaft of same area: 22 2 2 3 221243363(45 30 ) 3.5343 10 mmor 33.541 mm2,2(2)(5.1689 10 )87.2 10 Pa(0.033541)AccAcA cTc TJcJc   87.2 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 264 3 in.4 ftT PROBLEM 3.4 (a) Determine the maximum shearing stress caused by a 40-kip  in. torque T in the 3-in.-diameter solid aluminum shaft shown. (b) Solve part a, assuming that the solid shaft has been replaced by a hollow shaft of the same outer diameter and of 1-in. inner diameter. SOLUTION (a) 4(40 kip in.)(1.5 in.)(1.5 in.)27.5451 ksiTcJ 7.55 ksi (b) 44(40 kip in.)(1.5 in.)[(1.5 in.) (0.5 in.) ]27.6394 ksiTcJ 7.64 ksi PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 265 T = 40 kip · in.T'3 in.T'T = 40 kip · in.T4 in.(b)(a) PROBLEM 3.5 (a) For the 3-in.-diameter solid cylinder and loading shown, determine the maximum shearing stress. (b) Determine the inner diameter of the 4-in.-diameter hollow cylinder shown, for which the maximum stress is the same as in part a. SOLUTION (a) Solid shaft: 411(3.0 in.) 1.5 in.22 2 cdJc max3322(40 kip in.)(1.5 in.)7.5451 ksiTcJTc max7.55 ksi (b) Hollow shaft: 11(4.0 in.) 2.0 in.22 ocd 442max44max4422(40 kip in.)(2.0 in.)(2.0 in.)(7.5451 ksi)9.2500 in 1.74395 in.and 2 3.4879 in.oioooioiiiccJTccTccccdc 3.49 in.id PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 266 60 mm30 mmD200 mmT ⫽ 3 kN · m PROBLEM 3.6 A torque 3 kN mT  is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shearing stress at point D which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c) the percent of the torque carried by the portion of the cylinder within the 15-mm radius. SOLUTION (a) 3434 6433366130 mm 30 10 m2(30 10 ) 1.27235 10 m223 kN 3 10 N(3 10 )(30 10 )70.736 10 Pa1.27235 10mcdJcTTcJ       70.7 MPam  (b) 315 mm 15 10 mD 363(15 10 )(70.736 10 )(30 10 )DDc 35.4 MPaD  (c) 32DD DDDD DDDDTJTJ 33 63(15 10 ) (35.368 10 ) 187.5 N m2187.5100% (100%) 6.25%310DDTTT    6.25% PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 267 4 in.8 in.dst 5in.143 in.DCABT PROBLEM 3.7 The solid spindle AB is made of a steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) the corresponding required value of the diameter sd of spindle AB. SOLUTION (a) Analysis of sleeve CD: 21244 4 4 42133211(3) 1.5 in.221.5 0.25 1.25 in.= (1.5 1.25 ) 4.1172 in22(4.1172)(7 10 )= 19.21 10 lb in.1.5ocdcc tJccJTc     19.21 kip in.T (b) Analysis of solid spindle AB: 33333=19.21 101.601in212 10(2)(1.601)1.006 in. 2sTcJJTcccdc   2.01 in.d PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material