Torsion

TORSION
Problem 304 A steel shaft 3 ft long that has a diameter of 4 in is subjected to a torque of 15 kipft. 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 kNm? 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
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, compute the shaft diameter. 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. If the shearing stress is limited to 12 ksi, determine the maximum horsepower that can be transmitted.
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 4.5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1 in a length of 26 diameters. Compute the proper diameter if G = 83 GPa.
Solution 309 [math]T = \dfrac{P}{2\pi f} = \dfrac{4.5(1\,000\,000)}{2\pi(3)}[/math] [math]T = 238\,732.41 \, \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\,732.41)(1000)}{\pi d^3}[/math] [math]d = 289.71 \, \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\,732.41(26d)(1000)}{\frac{1}{32}\pi d^4 (83\,000)}[/math] [math]d = 352.08 \, \text{mm}[/math]
Use the bigger diameter, 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.
Solution 310 Hollow circular shaft: maxhollow=16TD (D4d4) maxhollow=16TD [D4(21D)4] maxhollow=16TD (1615D4) maxhollow=162T15 D3
Solid circular shaft: maxsolid= D316T
maxsolid=1615 162T15 D3 maxsolid=1615 maxhollow 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. P-311. Using G = 28 GPa, determine the relative angle of twist of gear D relative to gear A.
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 0.20-in-diameter steel wire encased in a stationary tube that fits closely enough to impose a frictional torque of 0.50 lbin/in. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. What will be the angular deformation of one end relative to the other end? G = 12 106 psi.
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
=TLJG If = d, T = 0.5L and L = dL
d =1JG
020 (0 5L)dL
= 20 5L2 02 =1JG[0 25(20 )20 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 0.5 deg/m. Use G = 83 GPa.
Solution 313 Based on maximum allowable shearing stress: 14N mm T=6955 5N m
max=16TD (D4d4) 60=16T(100) (1004804) T=6955486
Based on maximum allowable angle of twist: T=4198282 97N mm T=4198 28N m
=TLJG 0 5
180
=T(1000)132 (1004804)(83000)
Use the smaller torque, T = 4 198.28 Nm answer
Problem 314 The steel shaft shown in Fig. P-314 rotates at 4 Hz with 35 kW taken off at A, 20 kW removed at B, and 55 kW applied at C. Using G = 83 GPa, find the maximum shearing stress and the angle of rotation of gear A relative to gear C.
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. At the right end, 30 kW are removed and another 20 kW leaves the shaft at 1.5 m from the right end. (a) Find the uniform shaft diameter so that the shearing stress will not exceed 60 MPa. (b) If a uniform shaft diameter of 100 mm is specified, determine the angle by which one end of the shaft lags behind the other end. Use G = 83 GPa.
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
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 = 69.6 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. P-316. Determine the maximum permissible value of T subject to the following conditions: st 83 MPa, al 55 MPa, and the angle of rotation of the free end is limited to 6. For steel, G = 83 GPa and for aluminum, G = 28 GPa.
Solution 316
Based on maximum shearing stress max = 16T / d3: 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 = 679.04 Nm answer

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TORSION

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

Solution 305 =TLJG 3 180 =12(6)(10003)132 d4(83000) d=113 98mm answer

Solution 306 T=P2 f=2 (189)5000(396000) T=1667337 5lb in

Solution 307 =TLJG 4 180 =T(5)(1000)132 d4(83000) T=0 1138d4

max= d316T 80= d316(0 11384) d=138mm answer

Solution 308 max= d316T 12(1000)=16T (23) T=18849 56lb in

Solution 309 [math]T = \dfrac{P}{2\pi f} = \dfrac{4.5(1\,000\,000)}{2\pi(3)}[/math] [math]T = 238\,732.41 \, \text{N}\cdot\text{m}[/math]

Solid circular shaft: maxsolid= D316T

maxsolid=1615 162T15 D3 maxsolid=1615 maxhollow ok!

Solution 312 max= d316T 20(1000)=16T (0 20)3 T=10 lb in

=TLJG If = d, T = 0.5L and L = dL

Solution 313 Based on maximum allowable shearing stress: 14N mm T=6955 5N m

max=16TD (D4d4) 60=16T(100) (1004804) T=6955486

Based on maximum allowable angle of twist: T=4198282 97N mm T=4198 28N m

Use the smaller torque, T = 4 198.28 Nm answer

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

=TLJG A C=1G JTL

Use d = 69.6 mm answer

Based on maximum angle of twist:

6 180 =3T(900)132 (504)(83000)+T(600)132 (404)(28000) T=757316 32N mm T=757 32N m

Use T = 679.04 Nm answer

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