271_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

271_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.3.qxd 9/27/08 1:27 PM Page 265 265 SECTION 3.3 Circular Bars and Tubes Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is subjected to a torque produced by forces P 900 lb (see figure). The forces have their lines of action at a distance b 5.5 in. from the outside of the tube. P If the allowable shear stress in the tube is 6300 psi and the inner radius r1 1.2 in., what is the minimum permissible outer radius r2? P P r2 r1 P b b 2 r2 Solution 3.3-17 Circular tube in torsion SOLUTION OF EQUATION UNITS: Pounds, Inches Substitute numerical values: 6300 psi 4(900 lb)(5.5 in. + r2)(r2) p[(r4) 2 (1.2 in.)4] or P 900 lb b 5.5 in. allow r1 6300 psi 2.07360 r4 2 r2(r2 + 5.5) or 1.2 in. r4 2 0.181891 0.181891 r2 2 Find minimum permissible radius r2 r2 1.000402 r2 Solve numerically: TORSION FORMULA T 2P(b IP p4 (r 22 r2) r4) 1 2P(b + r2)r2 4P(b + r2)r2 p4 p (r4 r4) 2 1 (r2 r4) 1 2 All terms in this equation are known except r2. tmax Tr2 IP 1.3988 in. MINIMUM PERMISSIBLE RADIUS r2 1.40 in. ; 0 2.07360 0 ...
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