Mechanics_of_Materia - 204 CHAPTER 3 Torsion Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T 36,000

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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 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
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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 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) ¢ 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
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Solution 3.4-11 Tapered tube 206 CHAPTER 3 Torsion t 5 thickness (constant) d A , d B 5 average diameters at the ends d B 5 2 d A (approximate formula) I P 5 p d 3 t 4 T T A B L A NGLE OF TWIST Take the origin of coordinates at point O . I P ( x ) 5 p [ d ( x )] 3 t 4 5 p td A 3 4 L 3 x 3 d ( x ) 5 x 2 L ( d B ) 5 x L d A B L d B = 2 d A d A O X d (x) d x L For element of length dx : For entire bar: f 5 # 2 L L d f 5 4 TL 3 p Gtd A 3 # 2 L L dx x 3 5 3 TL 2 p Gtd A 3 d f 5 Tdx GI P ( x ) 5 Tdx G ¢ p td 3 A 4 L 3 x 3 5 4 TL 3 p Gtd A 3 # dx x 3 Problem 3.4-12
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This note was uploaded on 10/16/2011 for the course CIV 101 taught by Professor Prof.yoon during the Fall '11 term at Korea Advanced Institute of Science and Technology.

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Mechanics_of_Materia - 204 CHAPTER 3 Torsion Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T 36,000

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