296_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

296_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.5-3.7.qxd 290 9/27/08 1:10 PM Page 290 CHAPTER 3 Torsion Solution 3.5-4 Bar in a testing machine Strain gage at 45°: max 339 10 d tmax 50 mm T SHEAR STRESS (FROM EQ. 3-12) 6 500 N m 2 678 max pd 3 p(0.050 m)3 20.372 MPa SHEAR MODULUS SHEAR STRAIN (FROM EQ. 3-33) max 16(500 N # m) 16T 10 G 6 tmax gmax 20.372 MPa 678 * 10 6 30.0 GPa 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 When twisted by a torque T, the tube develops a maximum normal strain of 170 10 6. What is the magnitude of the applied torque T ? Problem 3.5-5 A steel tube (G Solution 3.5-5 G max IP 11.5 106 psi 170 10 d2 2.0 in. d1 1.5 in. p [(2.0 in.)4 32 Equate expressions: Td2 2IP 6 4 d1 2 p2 1d 32 2 1.07379 in. (1.5 in.)4] 340 max 10 6 SHEAR STRESS (FROM TORSION FORMULA) tmax Also, Tr IP max Td2 2IP G max Ggmax SOLVE FOR TORQUE T SHEAR STRAIN (FROM EQ. 3-33) 2 1.5 in. Steel tube 4 max ; 2GIPgmax d2 2(11.5 * 106 psi)(1.07379 in.4)(340 * 10 2.0 in. 4200 lb-in. ; 6 ) ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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