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Mechanics_of_Materials_Chap_03-02 - 204 CHAPTER 3 Torsion...

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Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T 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 4.0 ft and is made of an aluminum alloy having shear modulus of elasticity G 3.9 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 1.5 d A T 36,000 lb-in. L 4.0 ft 48 in. G 3.9 10 6 psi allow 15,000 psi allow 3.0 0.0523599 rad M INIMUM DIAMETER BASED UPON ALLOWABLE SHEAR STRESS 12.2231 in. 3 d A 2.30 in. t max 16 T d A 3 d A 3 16 T t allow 16(36,000 lb-in.) (15,000 psi) M INIMUM DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (From Eq. 3-27) d B / d A 1.5 d A 2.52 in. A NGLE OF TWIST GOVERNS Min. d A 2.52 in. 40.4370 in. 4 d A 4 2.11728 in. 4 f allow 2.11728 in. 4 0.0523599 rad 2.11728 in. 4 d A 4 (36,000 lb-in.)(48 in.) (3.9 10 6 psi) ¢ 32 d A 4 (0.469136) f TL G ( I P ) A ¢ b 2 b 1 3 b 3 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 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 GPa) ¢ 32 (25 mm) 4 ¢ b 2 b 1 3 b 3 (0.3 ) ¢ 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 2 d A . The polar moment of inertia may be represented by the approximate formula I P d 3 t /4 (see Eq. 3-18). Derive a formula for the angle of twist 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 thickness (constant) d A , d B average diameters at the ends d B 2 d A (approximate formula) I 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 ) [ d ( x ) ] 3 t 4 td A 3 4 L 3 x 3 d ( x ) x 2 L ( d B ) 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 2 L L d f 4 TL 3 Gtd A 3 2 L L dx x 3 3 TL 2 Gtd A 3 d f Tdx GI P ( x ) Tdx G ¢ td 3 A 4 L 3 x 3 4 TL 3 Gtd A 3 # dx x 3 Problem 3.4-12 A prismatic bar AB of length L and solid circular cross section (diameter d ) is loaded by a distributed torque of constant intensity t per unit distance (see figure).
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