279_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

279_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.4.qxd 9/27/08 1:09 PM Page 273 SECTION 3.4 Nonuniform Torsion 273 Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T 36,000 lb-in. (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. The bar has length L 4.0 ft and is made of an aluminum alloy having shear modulus of elasticity G 3.9 106 psi. The allowable shear stress in the bar is 15,000 psi and the allowable angle of twist is 3.0°. If the diameter at end B is 1.5 times the diameter at end A, what is the minimum required diameter dA at end A? (Hint: Use the results of Example 3–5). Solution 3.4-9 dB Tapered bar MINIMUM DIAMETER BASED (From Eq. 3-27) 1.5 dA T 36,000 lb-in. L 4.0 ft G 3.9 b 48 in. f 106 psi allow 15,000 psi allow 3.0° dB/dA b2 + b + 1 TL a b G(IP)A 3b3 (36,000 lb-in.)(48 in.) p4 b dA 32 TL (0.469136) G(IP)A (0.469136) 2.11728 in.4 DIAMETER BASED UPON ALLOWABLE SHEAR d4 A STRESS tmax 1.5 (3.9 * 106 psi) a 0.0523599 rad MINIMUM UPON ALLOWABLE ANGLE OF TWIST 16T pd3 A d3 A 16 T ptallow 16(36,000 lb-in.) p(15,000 psi) d4 A 2.30 in. 2.11728 in.4 0.0523599 rad 40.4370 in.4 12.2231 in.3 dA 2.11728 in.4 fallow dA 2.52 in. ANGLE OF TWIST GOVERNS Min. dA 2.52 in. ; ...
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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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