270_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

270_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online