03-01 Chap Gere

# 03-01 Chap Gere - 3 Torsion Torsional Deformations Problem...

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181 Torsional Deformations Problem 3.2-1 A copper rod of length L 5 18.0 in. is to be twisted by torques T (see figure) until the angle of rotation between the ends of the rod is 3.0°. If the allowable shear strain in the copper is 0.0006 rad, what is the maximum permissible diameter of the rod? Solution 3.2-1 Copper rod in torsion 3 Torsion L 5 18.0 in. g allow 5 0.0006 rad Find d max 5 0.05236 rad f 5 3.0 85 (3.0) ¢ p 180 rad From Eq. (3-3): d max 5 0.413 in. d max 5 2 L g allow f 5 (2)(18.0 in.)(0.0006 rad) 0.05236 rad g max 5 r f L 5 d f 2 L L d TT L d Problem 3.2-2 A plastic bar of diameter d 5 50 mm is to be twisted by torques T (see figure) until the angle of rotation between the ends of the bar is 5.0°. If the allowable shear strain in the plastic is 0.012 rad, what is the minimum permissible length of the bar? Solution 3.2-2 Plastic bar in torsion d 5 50 mm 5 0.08727 rad g allow Find L min From Eq. (3-3): g max 5 r f L 5 d f 2 L 5 0.012 rad f 5 5.0 (5.0) ¢ p 180 rad L min 5 182 mm L min 5 d f 2 g allow 5 (50 mm)(0.08727 rad) (2)(0.012 rad) L d

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Problem 3.2-3 A circular aluminum tube subjected to pure torsion by torques T (see figure) has an outer radius r 2 equal to twice the inner radius r 1 . (a) If the maximum shear strain in the tube is measured as 400 3 10 2 6 rad, what is the shear strain g 1 at the inner surface? (b) If the maximum allowable rate of twist is 0.15 degrees per foot and the maximum shear strain is to be kept at 400 3 10 2 6 rad by adjusting the torque T , what is the minimum required outer radius ( r 2 ) min ? Solution 3.2-3 Circular aluminum tube 182 CHAPTER 3 Torsion L T r 1 r 2 T r 2 5 2 r 1 g max 5 400 3 10 2 6 rad (a) S HEAR STRAIN AT INNER SURFACE From Eq. (3-5b): g 1 5 200 3 10 2 6 rad g 1 5 1 2 g 2 5 1 2 (400 3 10 2 6 rad) 5 218.2 3 10 2 6 rad / in. 5 (0.15 8 / ft) ¢ p 180 rad degree ≤¢ 1 12 ft in. u allow 5 0.15 8 / ft (b) M INIMUM OUTER RADIUS From Eq. (3-5a): ( r 2 ) min 5 1.83 in. ( r 2 ) min 5 g max u allow 5 400 3 10 2 6 rad 218.2 3 10 2 6 rad / in. g max 5 r 2 f L 5 r 2 u r 1 r 2 Problem 3.2-4 A circular steel tube of length L 5 0.90 m is loaded in torsion by torques T (see figure). (a) If the inner radius of the tube is r 1 5 40 mm and the measured angle of twist between the ends is 0.5°, what is the shear strain g 1 (in radians) at the inner surface? (b) If the maximum allowable shear strain is 0.0005 rad and the angle of twist is to be kept at 0.5° by adjusting the torque T , what is the maximum permissible outer radius ( r 2 ) max ? Problems 3.2-3, 3.2-4, and 3.2-5
SECTION 3.2 Torsional Deformations 183 Solution 3.2-4 Circular steel tube L 5 0.90 m r 1 5 40 mm (a) S HEAR STRAIN AT INNER SURFACE From Eq. (3-5b): g 1 5 388 3 10 2 6 rad g min 5 g 1 5 r 1 f L 5 (40 mm)(0.008727 rad) 900 mm g max 5 0.0005 rad 5 0.008727 rad f 5 0.5 85 (0.5 8 ) ¢ p 180 rad degree (b) M AXIMUM OUTER RADIUS From Eq. (3-5a): ( r 2 ) max 5 51.6 mm ( r 2 ) max 5 (0.0005 rad)(900 mm) 0.008727 rad r 2 5 g max L f g max 5 g 2 5 r 2 f L ; r 1 r 2 Problem 3.2-5 Solve the preceding problem if the length L 5 50 in., the inner radius r 1 5 1.5 in., the angle of twist is 0.6°, and the allowable shear strain is 0.0004 rad.

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## This note was uploaded on 03/19/2008 for the course E M 316 taught by Professor Korkolis during the Spring '08 term at University of Texas.

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03-01 Chap Gere - 3 Torsion Torsional Deformations Problem...

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