304_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

304_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_3.5-3.7.qxd 298 9/27/08 1:10 PM Page 298 CHAPTER 3 Torsion Solution 3.7-7 Splice in a propeller shaft EQUATE TORQUES SOLID SHAFT 16 T1 tmax pd3 For the same power, the torques must be the same. For the same material, both parts can be stressed to the same maximum stress. pd3tmax 16 T1 HOLLOW COLLAR p4 (d1 32 IP 4 d ) tmax 2tmax p 4 a b (d1 d1 32 2tmaxIP d1 T2 T2(d1/2) IP T2r IP ptmax 4 (d1 16 d1 ‹ T1 or a 4 d) T2 d1 4 b d pd3tmax 16 ptmax 4 (d 16d1 1 d1 d 0 1 d 4) (Eq. 1) MINIMUM OUTER DIAMETER Solve Eq. (1) numerically: d 4) Min. d1 1.221 d ; Problem 3.7-8 What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter 50 mm, inside diameter 40 mm, and shear modulus of elasticity 80 GPa) turning at 600 rpm if the allowable shear stress is 100 MPa and the allowable rate of twist is 3.0°/m? Solution 3.7-8 Hollow propeller shaft d2 50 mm d1 G 80 GPa n BASED UPON ALLOWABLE RATE OF TWIST T2 T2 GIP allow u GIP 40 mm 600 rpm allow 100 MPa p4 (d 32 2 IP allow d4) 1 3.0°/m 362.3 * 10 9 m4 T (80 GPa) (362.3 * 10 2 *a BASED UPON ALLOWABLE SHEAR STRESS tmax T1 T1(d2/2) IP T1 2t allowIP d2 2(100 MPa)(362.3 * 10 0.050 m 1449 N # m T2 9 m4) 94 m )(3.0°/m) p rad /degree b 180 1517 N m SHEAR STRESS GOVERNS Tallow T1 1449 N m MAXIMUM POWER 2p(600 rpm)(1449 N # m) 2pnT P 60 60 P Pmax 91,047 W 91.0 kW ; ...
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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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