BeerMOM_GE_C03-P001-P060BeerMOM_GE_C03-P061-P120BeerMOM_GE_C03-P121-P0161CCHHAAPPTTEERR 33PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 3.1 (a) Determine the maximum shearing stress caused by a 4.6-kN ⋅ m torque T in the 76-mm-diameter 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 (b) Solve part a, assuming that the solid shaft has been replaced by a hollow shaft of the same outer diameter and of 24-mm inner diameter. SOLUTION (a) Solid shaft: 44 6436638 mm 0.038 m2(0.038) 3.2753 10 m22(4.6 10 )(0.038)53.4 10 Pa3.2753 10dcJcTcJππτ−−== === = ××== = ×× 53.4 MPaτ= (b) Hollow shaft: ()2144 4 4 64213660.038 m2112 mm 0.012 m2(0.038 0.012 ) 3.2428 10 m22(4.6 10 )(0.038)53.9 10 Pa3.2428 10oidccdJccTcJππτ−−==== ==−= − = ××== = ×× 53.9 MPaτ= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 3.2 (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 JTJcTππττ−−−=−=−=× =×==××==×⋅× 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 TJcJcπππππτπτπ=−= −= ×======×==× 87.2 MPaτ= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 3.3 Knowing that the internal diameter of the hollow shaft shown is d = 22 mm, determine the maximum shearing stress caused by a torque of magnitude T = 900 N ⋅ m. SOLUTION ()221144 4 4 421max1211(40) 20 mm220.02 m11(22) 11 mm22(20 11 ) 228329 mm22(900)(0.02)78.8 MPa228329 10cdccdJccTcJππτ−== ==== ==−= −=== =× PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 3.4 Knowing that d = 30 mm, determine the torque T that causes a maximum shearing stress of 52 MPa in the hollow shaft shown. SOLUTION ()221144 4 4 421max12 6max11(40) 20 mm220.02 m11(30) 15 mm22(20 15 ) 171806 mm22(171806 10 )(52 10 )446.7 Nm0.02cdccdJccTcJJTcππττ−== ==== ==−= −==××== = ⋅ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 3.5 (a) For the 75 mm diameter solid cylinder and loading shown, determine the maximum shearing stress. (b) Determine the inner diameter of the hollow cylinder, of 100 mm outer diameter, for which the maximum stress is the same as in part a. SOLUTION (a) Solid shaft: 11(75) 37.5 mm22cd== = 6max332 (2)(4520)54.6 10 Pa (i.e.) 54.6 MPa(0.0375)Tc TJcτππ== = = × (b) Hollow shaft: 211(100) 50 mm22cd== = ()4421222max44 4 942126max2 (2)(4520)(0.050)0.050 3614 10 m(54.6 10 )ccJTccTcccπτπτπ−−===− = − = ×× 1110.0436 m 2 87.2 mmcdc=== 43.6 mm= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 3.6 (a) Determine the torque that can be applied to a solid shaft of 20-mm diameter without exceeding an allowable shearing stress of 80 MPa. (b) Solve Part a, assuming that the solid shaft has been replaced by a hollow shaft of the same cross-sectional area and with an inner diameter equal to half of its own outer diameter. SOLUTION (a) Solid shaft: 44 9411(0.020) 0.010 m22(0.10) 15.7080 10 m22cdJcππ−== ==− = × 96max(15.7080 10 )(80 10 )125.6640.010JTcτ−××== = 125.7 N mT =⋅  (b) Hollow shaft: Same area as solid shaft. ()()222 2 2 221 2 2 221244 4 4 9421132422(0.010) 0.0115470 m3310.0057735 m2(0.0115470 0.0057735 ) 26.180 10 m22Acc c c ccccccJccππ ππππ−=−=− = === ====−= − = × 69max2(8010)(26.18010)181.380.0115470JTcτ−××== = 181.4 N mT =⋅ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course