03-02ChapGere.0002

03-02ChapGere.0002 - p 32 (25 mm) 4 b 2 1 b 1 1 3 b 3 (0.3...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 5 25 mm and the length is L 5 300 mm. The bar is made of steel with shear modulus of elasticity G 5 82 GPa. If the torque T 5 180 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 5 25 mm L 5 300 mm G 5 82 GPa T 5 180 N ? m f allow 5 0.3 8 Find d B D IAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (From Eq. 3-27) f 5 TL G ( I P ) A ¢ b 2 1 b 1 1 3 b 3 Ê ( I P ) A 5 p 32 d A 4 b 5 d B d A 0.914745 b 3 2 b 2 2 1 5 0 S OLVE NUMERICALLY : b 5 1.94452 Min. d B 5 b d A 5 48.6 mm 0.304915 5 b 2 1 b 1 1 3 b 3 5 (180 N # m)(0.3 m) (82 GPa) ¢
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online