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03-02ChapGere.0002

# 03-02ChapGere.0002 - p 32 ≤(25 mm 4 ¢ b 2 1 b 1 1 3 b 3...

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Problem 3.4-10 The bar shown in the figure is tapered linearly from end A to end B and has a solid circular cross section. The diameter at the smaller end of the bar is d A 25 mm and the length is L 300 mm. The bar is made of steel with shear modulus of elasticity G 82 GPa. If the torque T 180 N m and the allowable angle of twist is 0.3°, what is the minimum allowable diameter d B at the larger end of the bar? ( Hint: Use the results of Example 3-5.) Solution 3.4-10 Tapered bar SECTION 3.4 Nonuniform Torsion 205 d A 25 mm L 300 mm G 82 GPa T 180 N m allow 0.3 Find d B D IAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (From Eq. 3-27) f TL G ( I P ) A ¢ b 2 b 1 3 b 3 ( I P ) A 32 d A 4 b d B d A 0.914745 3 2 1 0 S OLVE NUMERICALLY : 1.94452 Min. d B b d A 48.6 mm 0.304915 b 2 b 1 3 b 3 (180 N # m)(0.3 m) (82
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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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