Torsion

TORSIONProblem 304 A steel shaft 3 ft long that has a diameter of 4 in is subjected to a torque of 15 kip·ft. Determine the maximum shearing stress and the angle of twist. Use G = 12 × 106 psi. Solution 304 max= D316T= (43)16(15)(1000)(12) max=14324psi max=14 3ksi answer =TLJG=15(3)(1000)(122)132 (44)(12 =0 0215rad [math] =1 23 [ math] answer 106) Problem 305 What is the minimum diameter of a solid steel shaft that will not twist through more than 3° in a 6-m length when subjected to a torque of 12 kN·m? What maximum shearing stress is developed? Use G = 83 GPa. Solution 305 =TLJG 3 180 =12(6)(10003)132 d4(83000) d=113 98mm answer max= d316T= (113 983)16(12)(10002) max=41 27MPa answer Problem 306 A steel marine propeller shaft 14 in. in diameter and 18 ft long is used to transmit 5000 hp at 189 rpm. If G = 12 × 106 psi, determine the maximum shearing stress. Solution 306 T=P2 f=2 (189)5000(396000) T=1667337 5lb in 5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1° in a length of 26 diameters. If the shearing stress is limited to 12 ksi. compute the shaft diameter. Solution 308 max= d316T 12(1000)=16T (23) T=18849 56lb in T=P2 f 18849 56=2 (240)P(396000 P=71 78hp answer Problem 309 A steel propeller shaft is to transmit 4. determine the maximum horsepower that can be transmitted. Compute the proper diameter if G = 83 GPa. What power can be transmitted by the shaft at 20 Hz? Solution 307 =TLJG 4 180 =T(5)(1000)132 d4(83000) T=0 1138d4 max= d316T 80= d316(0 11384) d=138mm answer T=P2 f 0 1138d4=P2 (20) P=14 3d4=14 3(1384) P=5186237285N mm/sec P=5186237 28W P=5 19MW answer Problem 308 A 2-in-diameter steel shaft rotates at 240 rpm. .max= d316T= (143)16(1667337 5) max=3094 6psi answer Problem 307 A solid steel shaft 5 m long is stressed at 80 MPa when twisted through 4°. Using G = 83 GPa. 000)}[/math] [math]d = 352.000\.41 \. d = 352 mm answer Problem 310 Show that the hollow circular shaft whose inner diameter is half the outer diameter has a torsional strength equal to 15/16 of that of a solid shaft of the same outside diameter. determine the relative angle of twist of gear D relative to gear A. .732.Solution 309 [math]T = \dfrac{P}{2\pi f} = \dfrac{4.000)}{2\pi(3)}[/math] [math]T = 238\. Using G = 28 GPa.732.08 \. \text{N}\cdot\text{m}[/math] Based on maximum allowable shearing stress: [math]\tau_{max} = \dfrac{16T}{\pi d^3}[/math] [math]50 = \dfrac{16(238\. P-311. \text{mm}[/math] Use the bigger diameter.41)(1000)}{\pi d^3}[/math] [math]d = 289.732.41(26d)(1000)}{\frac{1}{32}\pi d^4 (83\.71 \. \text{mm}[/math] Based on maximum allowable angle of twist: [math]\theta = \dfrac{TL}{JG}[/math] [math]1^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{238\.5(1\. Solution 310 Hollow circular shaft: max−hollow=16TD (D4−d4) max−hollow=16TD [D4−(21D)4] max−hollow=16TD (1615D4) max−hollow=162T15 D3 Solid circular shaft: max−solid= D316T max−solid=1615 162T15 D3 max−solid=1615 max−hollow ok! Problem 311 An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. Solution 312 max= d316T 20(1000)=16T (0 20)3 T=10 lb in L=T0 50lb in/in L=10 lb in0 50lb in/in L=20 in=62 83in . What will be the angular deformation of one end relative to the other end? G = 12 × 106 psi.Problem 311 =TLJG Rotation of D relative to A: D A=1JG TL D A=1132 (504)(28000)[800(2)–300(3)+600(2)](10002) D A=0 1106rad D A=6 34 answer Problem 312 A flexible shaft consists of a 0.50 lb⋅in/in.20-in-diameter steel wire encased in a stationary tube that fits closely enough to impose a frictional torque of 0. 28 N·m answer Problem 314 The steel shaft shown in Fig. 20 kW removed at B.5 deg/m.=TLJG If θ = dθ. T = 0. P-314 rotates at 4 Hz with 35 kW taken off at A. find the maximum shearing stress and the angle of rotation of gear A relative to gear C. and 55 kW applied at C. Using G = 83 GPa.5L and L = dL d =1JG 020 (0 5L)dL = 20 5L2 02 =1JG[0 25(20 )2−0 25(0)2] =100 2132 (0 204)(12 106) =0 5234rad=30 answer Problem 313 Determine the maximum torque that can be applied to a hollow circular steel shaft of 100-mm outside diameter and an 80-mm inside diameter without exceeding a shearing stress of 60 MPa or a twist of 0. . T = 4 198. Solution 313 Based on maximum allowable shearing stress: 14N mm T=6955 5N m max=16TD (D4−d4) 60=16T(100) (1004−804) T=6955486 Based on maximum allowable angle of twist: T=4198282 97N mm T=4198 28N m =TLJG 0 5 180 =T(1000)132 (1004−804)(83000) Use the smaller torque. Use G = 83 GPa. Solution 314 T=P2 TA=2 TB=2 TB=2 f (4)−35(1000)=–1392 6N m (4)−20(1000)=–795 8N m (4)55(1000)=2188 4N m Relative to C: max= d316T AB= (553)16(1392 6)(1000)=42 63MPa BC= (653)16(2188 4)(1000)=40 58MPa ∴ max= AB=42 63MPa answer =TLJG A C=1G JTL A C=183000 132 (554)1392 6(4)+132 (654)2188 4(2) (10002) A C=0 104796585rad A C=6 004 answer Problem 315 A 5-m steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from the left end where 20 kW are removed. Solution 315 T=P2 f TA=TC=2 (2)−20(1000)=–1591 55N m TB=2 (2)70(1000)=5570 42N m TD=2 (2)−30(1000)=–2387 32N m . determine the angle by which one end of the shaft lags behind the other end.5 m from the right end. (b) If a uniform shaft diameter of 100 mm is specified. Use G = 83 GPa. 30 kW are removed and another 20 kW leaves the shaft at 1. (a) Find the uniform shaft diameter so that the shearing stress will not exceed 60 MPa. At the right end. 6 mm answer Part (b) =TLJG D A=1JG TL D A=1132 (1004)(83000)[–1591 55(2)+3978 87(1 5)+2387 32(1 5)](10002) D A=0 007813rad D A=0 448 answer Problem 316 A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in Fig. G = 28 GPa. and the angle of rotation of the free end is limited to 6°. For steel. Determine the maximum permissible value of T subject to the following conditions: τst ≤ 83 MPa. τal ≤ 55 MPa. .Part (a) max= d316T For AB 60= d316(1591 55)(1000) d=51 3mm For BC 60= d316(3978 87)(1000) d=69 6mm For CD 60= d316(2387 32)(1000) d=58 7mm Use d = 69. P-316. G = 83 GPa and for aluminum. 04 N·m answer .Solution 316 Based on maximum shearing stress τmax = 16T / πd3: Steel st= (503)16(3T)=83 T=679042 16N mm T=679 04N m Aluminum al=16T (403)=55 T=691150 38N mm T=691 15N m Based on maximum angle of twist: = TLJG st+ TLJG al 6 180 =3T(900)132 (504)(83000)+T(600)132 (404)(28000) T=757316 32N mm T=757 32N m Use T = 679. 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