295_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

295_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.5-3.7.qxd 9/27/08 1:10 PM Page 289 SECTION 3.5 Pure Shear Solution 3.5-3 d2 4.0 in. max Tubular bar T 70.0 k-in. 6400 psi max 70,000 lb-in. max 6400 psi From torsion formula, T Td2 2IP Tr IP Torsion formula: tmax 2IP tmax d2 2IPtmax L a b d2 GIP 2Ltmax Gd2 ‹f 2(48 in.)(6400 psi) (70.0 k-in.)(4.0 in.) 2(6400 psi) Td2 2tmax (4.0 * 106 psi)(4.0 in.) 21.875 in.4 Also, Ip TL GIp f (a) INSIDE DIAMETER d1 IP 289 f p4 (d 32 2 4 d1) p [(4.0 in.)4 32 4 d1 ] gmax 4 d1 ] Solve for d1: d1 21.875 in.4 2.40 in. ; 2.20° (c) MAXIMUM SHEAR STRAIN Equate formulas: p [256 in.4 32 0.03840 rad tmax G 6400 psi 4.0 * 106 psi 1600 * 10 6 rad ; ; (b) ANGLE OF TWIST L 48 in. G 4.0 106 psi Problem 3.5-4 A solid circular bar of diameter d 50 mm (see figure) is twisted in a testing machine until the applied torque reaches the value T 500 N m. At this value of torque, a strain gage oriented at 45° to the axis of the bar gives a reading P 339 10 6. What is the shear modulus G of the material? d = 50 mm Strain gage T 45° T = 500 N·m ...
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