298_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

298_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_3.5-3.7.qxd 9/27/08 292 1:10 PM Page 292 CHAPTER 3 Torsion MAXIMUM SHEAR STRESS INSIDE DIAMETER tmax Ggmax 2Gâmax Substitute numerical values: tmax T(d2/2) IP IP p4 (d 32 2 IP 4 d2 Td2 2tmax d4 2 Td2 4Gâmax 4 d1 ) 8Td2 pGâmax 4 d1 Td2 4Gâmax 4 d1 4 d2 (0.8 in.)4 8(750 lb-in.) (0.80 in.) p (6.2 * 106 psi) (880 * 10 0.4096 in.4 d1 0.60 in. 0.2800 in.4 ) 0.12956 in.4 ; 8Td2 pGâmax Problem 3.5-8 An aluminium tube has inside diameter d1 T 6 50 mm, shear modulus of elasticity G 27 GPa, and torque 4.0 kN m. The allowable shear stress in the aluminum is 50 MPa and the allowable normal strain is 900 10 6. Determine the required outside diameter d2. Solution 3.5-8 d1 T 50 mm Aluminum tube G 4.0 kN m NORMAL STRAIN GOVERNS 27 GPa allow 50 MPa allow 900 10 ALLOWABLE SHEAR STRESS allow)1 48.60 MPa allow REQUIRED DIAMETER Determine the required diameter d2. ( 6 t 50 MPa Tr IP (4000 N # m)(d2/2) p4 (0.050 m)4] [d2 32 48.6 MPa ALLOWABLE SHEAR STRESS BASED ON NORMAL STRAIN âmax ( allow)2 g 2 2G t 2G allow 48.6 MPa Rearrange and simplify: t 2Gâmax 2(27 GPa)(900 d4 2 6 10 ) (419.174 * 10 Solve numerically: d2 0.07927 m d2 79.3 mm ; 6 )d2 6.25 * 10 6 0 ...
View Full Document

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.

Ask a homework question - tutors are online