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03-01ChapGere.0016

# 03-01ChapGere.0016 - 4 2 r 1 4 5 4 P b 1 r 2 r 2 p r 2 4 2...

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Problem 3.3-17 A circular tube of inner radius r 1 and outer radius r 2 is subjected to a torque produced by forces P 5 900 lb (see figure). The forces have their lines of action at a distance b 5 5.5 in. from the outside of the tube. If the allowable shear stress in the tube is 6300 psi and the inner radius r 1 5 1.2 in., what is the minimum permissible outer radius r 2 ? Solution 3.3-17 Circular tube in torsion 196 CHAPTER 3 Torsion P r 1 r 2 2 r 2 P P P b b P 5 900 lb b 5 5.5 in. t allow 5 6300 psi r 1 5 1.2 in. Find minimum permissible radius r 2 T ORSION FORMULA T 5 2 P ( b 1 r 2 ) All terms in this equation are known except r 2 . t max 5 Tr 2 I P 5 2 P ( b 1 r 2 ) r 2 p 2 ( r 2
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Unformatted text preview: 4 2 r 1 4 ) 5 4 P ( b 1 r 2 ) r 2 p ( r 2 4 2 r 1 4 ) I P 5 p 2 ( r 2 4 2 r 1 4 ) S OLUTION OF EQUATION U NITS : Pounds, Inches Substitute numerical values: or or 2 1.000402 r 2 2 2.07360 5 Solve numerically: r 2 5 1.3988 in. M INIMUM PERMISSIBLE RADIUS r 2 5 1.40 in. Ê r 2 4 2 0.181891 r 2 2 r 2 4 2 2.07360 r 2 ( r 2 1 5.5) 2 0.181891 5 6300 psi 5 4(900 lb)(5.5 in. 1 r 2 )( r 2 ) p [ ( r 2 4 ) 2 (1.2 in.) 4 ] P r 1 r 2 2 r 2 P b b A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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