03-01 Chap Gere

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

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Unformatted text preview: 3 Torsion Torsional Deformations Problem 3.2-1 A copper rod of length L 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 d T T L Copper rod in torsion d T T L L 18.0 in. f 3.0 ( 3.0) ¢ 0.0006 rad gmax dmax rad Find dmax rf L 2Lgallow f dmax 180 0.05236 rad allow ≤ From Eq. (3-3): df 2L 0.413 in. — ( 2)(18.0 in.)(0.0006 rad) 0.05236 rad Problem 3.2-2 A plastic bar of diameter d 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 d f Plastic bar in torsion 50 mm 5.0 allow ( 5.0) ¢ 180 0.012 rad ≤ d T rad L Lmin Find Lmin From Eq. (3-3): gmax T 0.08727 rad rf L df 2L df 2gallow Lmin 182 mm — ( 50 mm)(0.08727 rad) (2)(0.012 rad) 181 182 CHAPTER 3 Torsion Problem 3.2-3 A circular aluminum tube subjected to pure torsion by torques T (see figure) has an outer radius r2 equal to twice the inner radius r1. T (a) If the maximum shear strain in the tube is measured as 400 10 6 rad, what is the shear strain 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 10 6 rad by adjusting the torque T, what is the minimum required outer radius (r2)min? Solution 3.2-3 r2 max uallow r2 6 10 rad r1 0.15 ft 10 6 g1 200 Problems 3.2-3, 3.2-4, and 3.2-5 1 (400 2 10 6 10 6 rad) rad — Problem 3.2-4 A circular steel tube of length L torsion by torques T (see figure). (b) MINIMUM OUTER RADIUS From Eq. (3-5a): r2 f L r2u 400 10 6 rad 218.2 10 6 rad in. 0.90 m is loaded in (a) If the inner radius of the tube is r1 40 mm and the measured angle of twist between the ends is 0.5°, what is the shear strain 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? gmax uallow ( r2 ) min rad in. From Eq. (3-5b): 1 g 22 r1 ( r2 ) min (a) SHEAR STRAIN AT INNER SURFACE g1 r2 gmax rad 1 ft ( 0.15 ft) ¢ ≤¢ ≤ 180 degree 12 in. 218.2 L Circular aluminum tube 2r1 400 T 1.83 in. — Solution 3.2-4 Circular steel tube r2 L 0.90 m r1 40 mm f 0.5 ( 0.5 ) ¢ 0.008727 rad gmax (b) MAXIMUM OUTER RADIUS r1 rad ≤ 180 degree 0.0005 rad (a) SHEAR STRAIN AT INNER SURFACE From Eq. (3-5a): gmax g2 r2 f ; r2 L gmax L f ( r2 ) max ( 0.0005 rad)(900 mm) 0.008727 rad ( r2 ) max 51.6 mm — From Eq. (3-5b): gmin g1 g1 f L r1 388 ( 40 mm)(0.008727 rad) 900 mm 6 10 rad — Problem 3.2-5 Solve the preceding problem if the length L 50 in., the inner radius r1 1.5 in., the angle of twist is 0.6°, and the allowable shear strain is 0.0004 rad. Solution 3.2-5 Circular steel tube r2 L 50 in. r1 1.5 in. f 0.6 r1 ( 0.6 ) ¢ 0.010472 rad gmax rad ≤ 180 degree 0.0004 rad (a) SHEAR STRAIN AT INNER SURFACE From Eq. (3-5b): gmin g1 g1 r1 314 10 f L 6 ( 1.5 in.)(0.010472 rad) 50 in. rad — (b) MAXIMUM OUTER RADIUS From Eq. (3-5a): gmax g2 r2 f ;r L2 gmaxL f ( r2 ) max ( 0.0004 rad)(50 in.) 0.010472 rad ( r2 ) max 1.91 in. — 184 CHAPTER 3 Torsion Circular Bars and Tubes Problem 3.3-1 A prospector uses a hand-powered winch (see figure) to raise a bucket of ore in his mine shaft. The axle of the winch is a steel rod of diameter d 0.625 in. Also, the distance from the center of the axle to the center of the lifting rope is b 4.0 in. If the weight of the loaded bucket is W 100 lb, what is the maximum shear stress in the axle due to torsion? P W d b W Solution 3.3-1 Hand-powered winch Problem 3.3-2 When drilling a hole in a table leg, a furniture maker uses a hand-operated drill (see figure) with a bit of diameter d 4.0 mm. (a) If the resisting torque supplied by the table leg is equal to 0.3 N m, what is the maximum shear stress in the drill bit? (b) If the shear modulus of elasticity of the steel is G 75 GPa, what is the rate of twist of the drill bit (degrees per meter)? SECTION 3.3 Circular Bars and Tubes Problem 3.3-16 A hollow aluminum tube used in a roof structure has an outside diameter d2 100 mm and an inside diameter d1 80 mm (see figure). The tube is 2.5 m long, and the aluminum has shear modulus G 28 GPa. (a) If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist (in degrees) when the maximum shear stress is 50 MPa? (b) What diameter d is required for a solid shaft (see figure) to resist the same torque with the same maximum stress? (c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft? Solution 3.3-16 Hollow aluminum tube d2 80 mm L 2.5 m G d1 d2 100 mm d1 28 GPa max FOR THE SOLID SHAFT: tmax Td2 , 2Ip ¢ d2 d3 ( 100 mm) ( 80 mm) 4 100 mm d L ≤¢ ≤ GIp f 2tmaxL Gd2 f 2 (50 MPa)(2.5 m ) (28 GPa)(100 mm) f 5.12 83.9 mm 590,400 mm3 — (c) RATIO OF WEIGHTS 0.08929 rad Torque is the same. 2IPtmax d2 2tmax 4 ¢ ≤(d d2 32 2 ( 100 mm) 2 ( 80 mm) 2 (83.9 mm) 2 d2 0.51 — The weight of the tube is 51% of the weight of the solid shaft, but they resist the same torque. is the same as for tube. T 2 d1 Atube Asolid Wtube Wsolid — For the tube: T 2 d2 Wtube Wsolid (b) DIAMETER OF A SOLID SHAFT d 4 d1 d2 f max 4 d2 4 d1 ) 4 2Iptmax TL GIp d2 16 2tmax 4 ¢ ≤¢ ≤(d 32 2 d 3 d2 Solve for d 3: d 3 50 MPa T 2Iptmax 16T d3 tmax (a) ANGLE OF TWIST FOR THE TUBE Tr Ip d d1 d2 4 d1 ) 195 196 CHAPTER 3 Torsion Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is subjected to a torque produced by forces P 900 lb (see figure). The forces have their lines of action at a distance b 5.5 in. from the outside of the tube. If the allowable shear stress in the tube is 6300 psi and the inner radius r1 1.2 in., what is the minimum permissible outer radius r2? P P P r2 r1 P b Solution 3.3-17 Circular tube in torsion P SOLUTION OF EQUATION r2 r1 UNITS: Pounds, Inches Substitute numerical values: P b b 2 r2 2 r2 b 6300 psi 4 (900 lb)(5.5 in. r2 )( r2 ) [ ( r4 ) ( 1.2 in.) 4 ] 2 P 900 lb or b 5.5 in. r4 2.07360 2 r2 ( r2 5.5) or allow r1 6300 psi 1.2 in. Find minimum permissible radius r2 r4 2 0.181891 2 0.181891 r2 0 1.000402 r2 Solve numerically: TORSION FORMULA T r2 2P(b r2) 1.3988 in. MINIMUM PERMISSIBLE RADIUS IP tmax 2 ( r4 2 Tr2 IP r4 ) 1 2P ( b 4 2 ( r2 r2 r2 ) r2 r4 ) 1 4P ( b r2 ) r2 ( r4 r4 ) 2 1 All terms in this equation are known except r2. 1.40 in. — 2.07360 0 SECTION 3.4 Nonuniform Torsion 197 Nonuniform Torsion Problem 3.4-1 A stepped shaft ABC consisting of two solid circular segments is subjected to torques T1 and T2 acting in opposite directions, as shown in the figure. The larger segment of the shaft has diameter d1 2.25 in. and length L1 30 in.; the smaller segment has diameter d2 1.75 in. and length L2 20 in. The material is steel with shear modulus G 11 106 psi, and the torques are T1 20,000 lb-in. and T2 8,000 lb-in. Calculate the following quantities: (a) the maximum shear stress max in the shaft, and (b) the angle of twist C (in degrees) at end C. Solution 3.4-1 T1 d1 d2 B A C L1 L2 Stepped shaft T1 d1 L1 A 2.25 in. L1 1.75 in. L2 11 T1 20,000 lb-in. T2 8,000 lb-in. tBC 16 TBC 3 d2 TBC L2 G ( Ip ) BC 20 in. G C T2 fBC 30 in. d2 TBC L2 B d1 SEGMENT BC T2 d2 8,000 lb-in. 16(8,000 lb-in.) ( 1.75 in.) 3 0.015797 rad 106 psi 7602 psi ( 8,000 lb-in.)(20 in.) (11 106 psi) ¢ 32 ≤ (1.75 in.) 4 (a) MAXIMUM SHEAR STRESS Segment BC has the maximum stress tmax SEGMENT AB TAB T2 T1 12,000 lb-in. tAB 16 TAB ` ` d3 1 16(12,000 lb-in.) ( 2.25 in.) 3 fAB TABL1 G ( Ip ) AB ( 12,000 lb-in.)(30 in.) (11 0.013007 rad 106 psi) ¢ 7600 psi (b) ANGLE OF TWIST AT END C 5365 psi 32 ≤ (2.25 in.) 4 C fC AB BC 0.002790 ( 0.013007 rad 0.16 — T2 0.015797) rad 198 CHAPTER 3 Torsion Problem 3.4-2 A circular tube of outer diameter d3 70 mm and inner diameter d2 60 mm is welded at the right-hand end to a fixed plate and at the left-hand end to a rigid end plate (see figure). A solid circular bar of diameter d1 40 mm is inside of, and concentric with, the tube. The bar passes through a hole in the fixed plate and is welded to the rigid end plate. The bar is 1.0 m long and the tube is half as long as the bar. A torque T 1000 N m acts at end A of the bar. Also, both the bar and tube are made of an aluminum alloy with shear modulus of elasticity G 27 GPa. Tube Fixed plate End plate Bar T A Tube (a) Determine the maximum shear stresses in both the bar and tube. (b) Determine the angle of twist (in degrees) at end A of the bar. Bar d1 d2 d3 Solution 3.4-2 Bar and tube TORQUE Tube T 1000 N m (a) MAXIMUM SHEAR STRESSES Bar Bar: tbar T A Tube: ttube 70 mm Ltube d2 32 4 ( d3 60 mm G 0.5 m ( Ip ) tube 27 GPa Bar: fbar Tube: ftube 4 d2 ) A 106 mm4 BAR 40 mm ( Ip ) bar 4 d1 32 Lbar 251.3 1.0 m 103 mm4 G 27 GPa bar fA 1.0848 d1 79.6 MPa T ( d3 2 ) ( Ip ) tube — 32.3 MPa — (b) ANGLE OF TWIST AT END A TUBE d3 16T d3 1 9.43 TL bar G ( Ip ) bar 0.1474 rad TL tube G ( Ip ) tube tube — 0.1474 0.0171 rad 0.0171 0.1645 rad Problem 3.4-3 A stepped shaft ABCD consisting of solid circular segments is subjected to three torques, as shown in the figure. The torques have magnitudes 12.0 k-in., 9.0 k-in., and 9.0 k-in. The length of each segment is 24 in. and the diameters of the segments are 3.0 in., 2.5 in., and 2.0 in. The material is steel with shear modulus of elasticity G 11.6 103 ksi. 12.0 k-in. 3.0 in. Solution 3.4-3 24 in. 24 in. Stepped shaft 12.0 k-in. 9.0 k-in. 2.5 in. 9.0 k-in. 2.0 in. (a) SHEAR STRESSES tAB TAB rAB ( Ip ) AB 5660 psi tBC TBC rBC ( Ip ) BC 5870 psi tCD TCD rCD ( Ip ) CD 5730 psi tmax 5870 psi — 3.0 in. A G C B 103 11.6 rAB 1.25 in. LAB LBC ksi 1.5 in. rBC D rCD LCD 1.0 in. 24 in. (b) ANGLE OF TWIST AT END D TORQUES TAB 12.0 9.0 TBC 9.0 TCD fAB 18 k-in. POLAR MOMENTS OF INERTIA ( Ip ) AB ( Ip ) BC ( Ip ) CD 32 32 32 (3.0 in.) 4 7.952 in.4 4 4 0.007805 rad TBC LBC G ( Ip ) BC 0.009711 rad fCD 30 k-in. TAB LAB G ( Ip ) AB fBC 9.0 k-in. 9.0 9.0 TCD LCD G ( Ip ) CD 0.011853 rad (2.0 in.) 4 3.835 in. 1.571 in.4 AB fD (2.5 in.) D 1.68 BC — CD 9.0 k-in. 2.0 in. D C B A (a) Calculate the maximum shear stress max in the shaft. (b) Calculate the angle of twist D (in degrees) at end D. 9.0 k-in. 2.5 in. 0.02937 rad 24 in. 200 CHAPTER 3 Torsion Problem 3.4-4 A solid circular bar ABC consists of two segments, as shown in the figure. One segment has diameter d1 50 mm and length L1 1.25 m; the other segment has diameter d2 40 mm and length L2 1.0 m. What is the allowable torque Tallow if the shear stress is not to exceed 30 MPa and the angle of twist between the ends of the bar is not to exceed 1.5°? (Assume G 80 GPa.) Solutions 3.4-4 d1 = 50 mm A allow G L 1 = 1.25 m 30 MPa 1.5 C B L1 d2 = 40 mm B 0.02618 rad L 2 = 1.0 m TiLi a GI f 80 GPa Pi Segment BC has the smaller diameter and hence the larger stress. 16T d3 A T L2 Tallow T C ALLOWABLE TORQUE BASED UPON ANGLE OF TWIST ALLOWABLE TORQUE BASED UPON SHEAR STRESS tmax d2 Bar consisting of two segments T allow d1 T 3 d2tallow 16 f 32T L1 ¢4 G d1 Tallow 3.77 N # m TL1 GIP1 L2 4≤ d2 T L1 ¢ G IP1 TL2 GIP2 fallowG L4 L4 32 ¢ 1 2≤ d1 d2 L2 ≤ IP2 348 N m ANGLE OF TWIST GOVERNS Tallow Problem 3.4-5 A hollow tube ABCDE constructed of monel metal is subjected to five torques acting in the directions shown in the figure. The magnitudes of the torques are T1 1000 lb-in., T2 T4 500 lb-in., and T3 T5 800 lb-in. The tube has an outside diameter d2 1.0 in. The allowable shear stress is 12,000 psi and the allowable rate of twist is 2.0°/ft. Determine the maximum permissible inside diameter d1 of the tube. 348 N m — T2 = T1 = 1000 lb-in. 500 lb-in. A B T3 = T4 = 800 lb-in. 500 lb-in. C D d2 = 1.0 in. T5 = 800 lb-in. E SECTION 3.4 Solution 3.4-5 Nonuniform Torsion Hollow tube of monel metal REQUIRED POLAR MOMENT OF INERTIA BASED UPON d2 1.0 in. Tmaxr IP tmax 12,000 psi allow 2 /ft allow ALLOWABLE SHEAR STRESS d2 d1 0.16667 /in. IP Tmax ( d2 2 ) tallow 0.05417 in.4 REQUIRED POLAR MOMENT OF INERTIA BASED UPON 0.002909 rad/in. ALLOWABLE ANGLE OF TWIST From Table H-2, Appendix H: G 9500 ksi u TORQUES Tmax GIP Tmax Guallow IP 0.04704 in.4 SHEAR STRESS GOVERNS T2 T1 A T3 B T4 D C T5 E IP T1 1000 lb-in. T2 500 lb-in. T4 500 lb-in. 800 lb-in. T5 T3 0.05417 in.4 Required IP 800 lb-in. 4 d1 INTERNAL TORQUES 32 4 d2 4 ( d2 4 d1 ) 32IP ( 1.0 in.) 4 32(0.05417 in.4 ) 0.4482 in.4 TAB T1 1000 lb-in. TBC T1 T2 TCD T1 T2 T3 TDE T1 T2 T3 d1 500 lb-in. 0.818 in. — (Maximum permissible inside diameter) 1300 lb-in. T4 800 lb-in. Largest torque (absolute value only): Tmax 1300 lb-in. Problem 3.4-6 A shaft of solid circular cross section consisting of two segments is shown in the first part of the figure. The left-hand segment has diameter 80 mm and length 1.2 m; the right-hand segment has diameter 60 mm and length 0.9 m. Shown in the second part of the figure is a hollow shaft made of the same material and having the same length. The thickness t of the hollow shaft is d/10, where d is the outer diameter. Both shafts are subjected to the same torque. If the hollow shaft is to have the same torsional stiffness as the solid shaft, what should be its outer diameter d? 80 mm 1.2 m 60 mm 0.9 m d 2.1 m d t=— 10 201 202 CHAPTER 3 Solution 3.4-6 Torsion Solid and hollow shafts SOLID SHAFT CONSISTING OF TWO SEGMENTS 80 mm TORSIONAL STIFFNESS 60 mm T f kT Torque T is the same for both shafts. ‹ For equal stiffnesses, 1.2 m f1 © TLi GIPi G¢ 0.9 m T (1.2 m ) 32 32T ( 29,297 m G 3 ≤ (80 mm) 4 G¢ 98,741 m T (0.9 m ) 32 69,444 m 3 ) ≤ (60 3 3.5569 98,741 d mm) d4 0.0775 m 4 1 2 3.5569 m d4 36.023 10 6 m4 77.5 mm — 32T (98,741 m 3 ) G HOLLOW SHAFT d = outer diameter d t=— 10 2.1 m d0 inner diameter f2 TL GIp T (2.1 m ) ≤ [ d 4 ( 0.8d ) 4 ] 32 32T 2.1 m 32T 3.5569 m ¢ ≤ ¢ ≤ G 0.5904 d 4 G d4 UNITS: d G¢ 0.8d meters Problem 3.4-7 Four gears are attached to a circular shaft and transmit the torques shown in the figure. The allowable shear stress in the shaft is 10,000 psi. (a) What is the required diameter d of the shaft if it has a solid cross section? (b) What is the required outside diameter d if the shaft is hollow with an inside diameter of 1.0 in.? 8,000 lb-in. 19,000 lb-in. 4,000 lb-in. A 7,000 lb-in. B C D ...
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This note was uploaded on 10/01/2009 for the course MEGR 2144 taught by Professor Sharpe during the Fall '08 term at UNC Charlotte.

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