03-02ChapGere.0001

03-02ChapGere.0001 - SHEAR STRESS 5 12.2231 in. 3 d A 5...

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

View Full Document Right Arrow Icon
Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T 5 36,000 lb-in. (see figure). The diameter of the bar varies linearly from d A at the left-hand end to d B at the right-hand end. The bar has length L 5 4.0 ft and is made of an aluminum alloy having shear modulus of elasticity G 5 3.9 3 10 6 psi. The allowable shear stress in the bar is 15,000 psi and the allowable angle of twist is 3.0°. If the diameter at end B is 1.5 times the diameter at end A , what is the minimum required diameter d A at end A ? ( Hint: Use the results of Example 3-5). Solution 3.4-9 Tapered bar 204 CHAPTER 3 Torsion d B 5 1.5 d A T 5 36,000 lb-in. L 5 4.0 ft 5 48 in. G 5 3.9 3 10 6 psi t allow 5 15,000 psi f allow 5 3.0 8 5 0.0523599 rad M INIMUM DIAMETER BASED UPON ALLOWABLE
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SHEAR STRESS 5 12.2231 in. 3 d A 5 2.30 in. t max 5 16 T p d A 3 Ê d A 3 5 16 T p t allow 5 16(36,000 lb-in.) p (15,000 psi) M INIMUM DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (From Eq. 3-27) b 5 d B / d A 5 1.5 d A 5 2.52 in. A NGLE OF TWIST GOVERNS Min. d A 5 2.52 in. 5 40.4370 in. 4 d A 4 5 2.11728 in. 4 f allow 5 2.11728 in. 4 0.0523599 rad 5 2.11728 in. 4 d A 4 5 (36,000 lb-in.)(48 in.) (3.9 3 10 6 psi) ¢ p 32 ≤ d A 4 (0.469136) f 5 TL G ( I P ) A ¢ b 2 1 b 1 1 3 b 3 ≤ 5 TL G ( I P ) A (0.469136) T T A B L d B d A A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
View Full Document

This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

Ask a homework question - tutors are online