270_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

270_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.3.qxd 264 9/27/08 1:27 PM Page 264 CHAPTER 3 Torsion Solution 3.3-16 Td2 2 NUMERICAL DATA d2 104 mm d1 set tmax expression equal to 82 mm 32Td2 103 mm L 2.75 G 28 GPa Ip (p/32)( d 2 4 Ip 7.046 p a d24 d 14 ) d3 106 mm4 dreqd d24 d14 b p a d4 2 32 d4 b 1 then solve for d d14 d2 a d24 d14 d2 1 3 b dreqd 88.4 mm ; (c) RATIO OF WEIGHTS OF HOLLOW & SOLID SHAFTS (a) FIND ANGLE OF TWIST a f TL GIp f f (tmax) 48 MPa max WEIGHT IS PROPORTIONAL TO CROSS SECTIONAL AREA Td2 2L b 2Ip Gd2 2L Gd2 Ah As d12 B Ah As 0.524 ; So the weight of the tube is 52% of the solid shaft, but they resist the same torque. 0.091 radians 5.19° p2 Ad 42 p 2 d 4 reqd ; (b) REPLACE HOLLOW SHAFT WITH SOLID SHAFT - FIND DIAMETER d 2 p T tmax 32d 4 tmax 16T d3p ...
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