{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

296_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 296_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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

This is the end of the preview. Sign up to access the rest of the document.

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 ) ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online