# HW5 - 6.1 A solid circular steel shaft having an outside...

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6.1 A solid circular steel shaft having an outside diameter of d = 0.75 in. is subjected to a pure torque of T = 650 lb-in. Determine the maximum shear stress in the shaft. Solution The polar moment of inertia for the shaft is 4 4 (0.75 in.) 0.031063 in. 32 J π = = The maximum shear stress in the steel shaft is found from the elastic torsion formula: max 4 (650 lb-in.)(0.75 in. / 2) 7,846.93 psi 0.0 7,850 ps 31063 in. i Tc J τ = = = = Ans. Comment: max = 7,850 psi

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6.7 A compound shaft (Fig. P6.7) consists of brass segment (1) and aluminum segment (2). Segment (1) is a solid brass shaft with an allowable shear stress of 60 MPa. Segment (2) is a solid aluminum shaft with an allowable shear stress of 90 MPa. If a torque of T C = 23,000 N-m is applied at C , determine the minimum required diameter of (a) the brass shaft and (b) the aluminum shaft. Fig. P6.7 Solution The polar moment of inertia for a solid shaft can be expressed as 4 32 J d π = Rearrange the elastic torsion formula to group terms with d on the left-hand side: 4 32 ( / 2) d T d τ = and simplify to 3 16 d T = From this equation, the unknown diameter of the solid shaft can be expressed as 3 16 T d πτ = Equilibrium: For this shaft, the internal torque magnitude in each segment equals the external torque; therefore, T 1 = T 2 = T C = 23,000 N-m . Minimum shaft diameters: (a) Brass shaft (1) 1 3 3 1 2 allow,1 16 16(23,000 N-m)(1,000 mm/m) (60 N/mm ) 125.0 mm T d = = Ans. (b) Aluminum shaft (2) 2 3 3 2 2 allow,2 16 16(23,000 N-m)(1,000 mm/m) (90 N/mm ) 109.2 mm T d = = Ans. Comment: (a) D min = 125.0 mm; (b) D min = 109.2 mm
6.9 A solid constant-diameter shaft is subjected to the torques shown in Fig. P6.9. The bearings shown allow the shaft to turn freely. (a) Plot a torque diagram showing the internal

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## This note was uploaded on 02/08/2011 for the course ME 650:291 taught by Professor Weng during the Fall '11 term at Rutgers.

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HW5 - 6.1 A solid circular steel shaft having an outside...

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