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CCHHAAPPTTEERR 33 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. 261 T 18 mm PROBLEM 3.1 Determine the torque T that causes a maximum shearing stress of 70 MPa in the steel cylindrical shaft shown. SOLUTION 4max 3 max 3 6 ; 2 2 (0.018 m) (70 10 Pa) 2 641.26 N m \uf070\uf074 \uf070 \uf074 \uf070 \uf03d \uf03d \uf03d \uf03d \uf0b4 \uf03d \uf0d7 Tc J c J T c 641 N m\uf03d \uf0d7T \uf074\uf020 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 T 18 mm PROBLEM 3.2 For the cylindrical shaft shown, determine the maximum shearing stress caused by a torque of magnitude T \uf03d 800 N \uf0d7 m. SOLUTION 4max max 3 3 6 ; 2 2 2(800 N m) (0.018 m) 87.328 10 Pa \uf070\uf074 \uf074 \uf070 \uf070 \uf03d \uf03d \uf03d \uf0d7\uf03d \uf03d \uf0b4 Tc J c J T c max 87.3 MPa\uf074 \uf03d \uf074\uf020 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 m 30 mm 45 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: \uf028 \uf0294 42 1 4 4 6 4 6 4 6 6 3 3 2 (45 30 ) 5.1689 10 mm 5.1689 10 m 2 (5.1689 10 )(45 10 ) 5.1689 10 N m 45 10 J c c J Tc JT J c T \uf070 \uf070 \uf074\uf074 \uf02d \uf02d \uf02d \uf03d \uf02d \uf03d \uf02d \uf03d \uf0b4 \uf03d \uf0b4 \uf03d \uf03d \uf0b4 \uf0b4\uf03d \uf03d \uf0b4 \uf0d7\uf0b4 5.17 kN mT \uf03d \uf0d7 \uf074\uf020 (b) Solid shaft of same area: \uf028 \uf0292 2 2 2 3 22 1 2 4 3 3 6 3 (45 30 ) 3.5343 10 mm or 33.541 mm 2, 2 (2)(5.1689 10 ) 87.2 10 Pa (0.033541) A c c Ac A c Tc TJ c J c \uf070 \uf070 \uf070 \uf070 \uf070 \uf074 \uf070 \uf074 \uf070 \uf03d \uf02d \uf03d \uf02d \uf03d \uf0b4 \uf03d \uf03d \uf03d \uf03d \uf03d \uf03d \uf0b4\uf03d \uf03d \uf0b4 87.2 MPa\uf074 \uf03d \uf074\uf020 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 ft T PROBLEM 3.4 (a) Determine the maximum shearing stress caused by a 40-kip \uf0d7 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.) 2 7.5451 ksi \uf074 \uf070 \uf03d \uf0d7\uf03d \uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8 \uf03d Tc J 7.55 ksi\uf074 \uf03d \uf074\uf020 (b) 4 4 (40 kip in.)(1.5 in.) [(1.5 in.) (0.5 in.) ] 2 7.6394 ksi \uf074 \uf070 \uf03d \uf0d7\uf03d \uf02d \uf03d Tc J 7.64 ksi\uf074 \uf03d \uf074\uf020 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. T 4 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: 4 1 1 (3.0 in.) 1.5 in. 2 2 2 \uf070 \uf03d \uf03d \uf03d \uf03d c d J c max 3 3 2 2(40 kip in.) (1.5 in.) 7.5451 ksi \uf074 \uf070 \uf070 \uf03d \uf03d \uf0d7\uf03d \uf03d Tc J T c max 7.55 ksi\uf074 \uf03d \uf074\uf020 (b) Hollow shaft: 1 1 (4.0 in.) 2.0 in. 2 2 \uf03d \uf03d \uf03doc d \uf028 \uf0294 42 max 4 4 max 4 4 2 2(40 kip in.)(2.0 in.)(2.0 in.) (7.5451 ksi) 9.2500 in 1.74395 in. and 2 3.4879 in. \uf070 \uf074 \uf070\uf074 \uf070 \uf02d\uf03d \uf03d \uf03d \uf02d \uf0d7\uf03d \uf02d \uf03d \uf05c \uf03d \uf03d \uf03d o i o o o i o i i i c cJ T c c Tcc c c d c 3.49 in.id \uf03d \uf074\uf020 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 mm 30 mm D 200 mmT \ufffd 3 kN · m PROBLEM 3.6 A torque 3 kN mT \uf03d \uf0d7 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) 3 4 3 4 6 4 3 3 3 6 6 1 30 mm 30 10 m 2 (30 10 ) 1.27235 10 m 2 2 3 kN 3 10 N (3 10 )(30 10 ) 70.736 10 Pa 1.27235 10m c d J c T Tc J \uf070 \uf070 \uf074 \uf02d \uf02d \uf02d \uf02d \uf02d \uf03d \uf03d \uf03d \uf0b4 \uf03d \uf03d \uf0b4 \uf03d \uf0b4 \uf03d \uf03d \uf0b4 \uf0b4 \uf0b4\uf03d \uf03d \uf03d \uf0b4\uf0b4 70.7 MPam\uf074 \uf03d \uf074 (b) 315 mm 15 10 mD\uf072 \uf02d\uf03d \uf03d \uf0b4 3 6 3 (15 10 )(70.736 10 ) (30 10 ) D D c \uf072\uf074 \uf074 \uf02d \uf02d \uf02d \uf0b4 \uf0b4\uf03d \uf03d \uf0b4 35.4 MPaD\uf074 \uf03d \uf074 (c) 3 2 D D D D D D D D D D T JT J \uf072 \uf074 \uf070\uf074 \uf072 \uf074\uf072\uf03d \uf03d \uf03d 3 3 6 3 (15 10 ) (35.368 10 ) 187.5 N m 2 187.5100% (100%) 6.25% 3 10 D D T T T \uf070 \uf02d\uf03d \uf0b4 \uf0b4 \uf03d \uf0d7 \uf0b4 \uf03d \uf03d\uf0b4 6.25% \uf074\uf020 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. ds t 5 in.14 3 in. D C A B T 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: \uf028 \uf029 2 1 2 4 4 4 4 4 2 1 3 3 2 1 1 (3) 1.5 in. 2 2 1.5 0.25 1.25 in. = (1.5 1.25 ) 4.1172 in 2 2 (4.1172)(7 10 )= 19.21 10 lb in. 1.5 oc d c c t J c c JT c \uf070 \uf070 \uf074 \uf03d \uf03d \uf03d \uf03d \uf02d \uf03d \uf02d \uf03d \uf02d \uf03d \uf02d \uf03d \uf0b4\uf03d \uf03d \uf0b4 \uf0d7 19.21 kip in.T \uf03d \uf0d7 \uf074\uf020 (b) Analysis of solid spindle AB: 3 3 3 3 3 = 19.21 10 1.601 in 2 12 10 (2)(1.601) 1.006 in. 2s Tc J J Tc c c d c \uf074 \uf070 \uf074 \uf070 \uf0b4\uf03d \uf03d \uf03d \uf03d\uf0b4 \uf03d \uf03d \uf03d 2.01 in.\uf03dd \uf074 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. 268 4 in. 8 in. ds t 5 in.14 3 in. D C A B T PROBLEM 3.8 The solid spindle AB has a diameter ds \uf03d 1.5 in. and is made of a