Unformatted text preview: p 32 ≤ (25 mm) 4 ¢ b 2 1 b 1 1 3 b 3 ≤ (0.3 8 ) ¢ p 180 rad degrees ≤ T T A B L d B d A Problem 3.4-11 A uniformly tapered tube AB of hollow circular cross section is shown in the figure. The tube has constant wall thickness t and length L . The average diameters at the ends are d A and d B 5 2 d A . The polar moment of inertia may be represented by the approximate formula I P < p d 3 t /4 (see Eq. 3-18). Derive a formula for the angle of twist f of the tube when it is subjected to torques T acting at the ends. T T A B L t d B = 2 d A d A t A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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- Spring '08
- Moment Of Inertia, circular cross section, Nonuniform Torsion