03-02ChapGere.0008

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Solution 3.5-3 Tubular bar SECTION 3.5 Pure Shear 211 d 2 d 1 TT L d 2 5 4.0 in. T 5 70.0 k-in. 5 70,000 lb-in. s max 5 6400 psi t max 5 s max 5 6400 psi (a) I NSIDE DIAMETER d 1 Torsion formula: Also, Equate formulas: Solve for d 1 : d 1 5 2.40 in. (b) A NGLE OF TWIST f p 32 [256 in. 4 2 d 1 4 ] 5 21.875 in. 4 I p 5 p 32 ( d 2 4 2 d 1 4 ) 5 p 32 [(4.0 in.) 4 2 d 1 4 ] 5 21.875 in. 4 I P 5 Td 2 2 t max 5 (70.0 k-in.)(4.0 in.) 2(6400 psi) t max 5 Tr I P 5 Td 2 2 I P L 5 48 in. G 5 4.0 3 10 6 psi From torsion formula, (c) M AXIMUM SHEAR STRAIN 5 1600 3 10 2 6 rad g max 5 t max G 5 6400 psi 4.0 3 10 6 psi f 5 2.20 8 5 2(48 in.)(6400 psi) (4.0 3 10 6 psi)(4.0 in.) 5 0.03840 rad f 5 2 I P t max d 2 ¢ L GI P 5 2 L t max Gd 2 T 5 2 I P t max d 2 f 5 TL GI p Problem 3.5-4 A solid circular bar of diameter d 5 50 mm (see figure) is twisted in a testing machine until the applied torque reaches the value T 5 500 N ? m. At this value of torque, a strain gage oriented at 45° to the axis of the bar gives a reading
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