Homework#6-Solutions - Sec_3.5-3.7.qxd 9/27/08 1:10 PM Page...

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Unformatted text preview: Sec_3.5-3.7.qxd 9/27/08 1:10 PM Page 287 287 SECTION 3.5 Pure Shear Pure Shear Problem 3.5-1 A hollow aluminum shaft (see figure) has outside diameter d2 4.0 in. and inside diameter d1 2.0 in. When twisted by torques T, the shaft has an angle of twist per unit distance equal to 0.54°/ft. The shear modulus of elasticity of the aluminum is G 4.0 106 psi. d2 T T L (a) Determine the maximum tensile stress max in the shaft. (b) Determine the magnitude of the applied torques T. d1 d2 Probs. 3.5-1, 3.5-2, and 3.5-3 Solution 3.5-1 Hollow aluminum shaft d2 4.0 in. G 4.0 d1 2.0 in. (a) MAXIMUM TENSILE STRESS 0.54°/ft 106 psi max max MAXIMUM SHEAR STRESS max Gr (from Eq. 3-7a) r d2 /2 u max (4.0 max 6280 psi ; 2.0 in. 10 6 6283.2 psi Use the torsion formula tmax T rad/in. 106 psi)(2.0 in.)(785.40 max. (b) APPLIED TORQUE 1 ft prad (0.54°/ft) a ba b 12 in. 180 degree 785.40 occurs on a 45° plane and is equal to 10 6 tmaxIP r IP p [(4.0 in.)4 32 23.562 in.4 rad/in.) T (6283.2 psi) (23.562 in.4) 2.0 in. 74,000 lb-in. Tr IP ; (2.0 in.)4] Sec_3.5-3.7.qxd 290 9/27/08 1:10 PM Page 290 CHAPTER 3 Torsion Solution 3.5-4 Bar in a testing machine Strain gage at 45°: max 339 10 d tmax 50 mm T SHEAR STRESS (FROM EQ. 3-12) 6 500 N m 2 678 max pd 3 p(0.050 m)3 20.372 MPa SHEAR MODULUS SHEAR STRAIN (FROM EQ. 3-33) max 16(500 N # m) 16T 10 G 6 tmax gmax 20.372 MPa 678 * 10 6 30.0 GPa 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 When twisted by a torque T, the tube develops a maximum normal strain of 170 10 6. What is the magnitude of the applied torque T ? Problem 3.5-5 A steel tube (G Solution 3.5-5 G max IP 11.5 106 psi d2 170 10 p2 1d 32 2 2.0 in. d1 1.5 in. 4 d1 2 p [(2.0 in.)4 32 Equate expressions: Td2 2IP 6 1.07379 in. (1.5 in.)4] 340 max 10 6 SHEAR STRESS (FROM TORSION FORMULA) tmax Also, Tr IP max Td2 2IP G max Ggmax SOLVE FOR TORQUE T SHEAR STRAIN (FROM EQ. 3-33) 2 1.5 in. Steel tube 4 max ; 2GIPgmax d2 2(11.5 * 106 psi)(1.07379 in.4)(340 * 10 2.0 in. 4200 lb-in. ; 6 ) Sec_3.8.qxd 9/27/08 310 1:13 PM Page 310 CHAPTER 3 Torsion Solution 3.8-10 Bar enclosed in a tube TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) Bar: T1 Ta IP1 b IP1 + IP2 Ta Tube: T2 100.2783 N # m IP2 b IP1 + IP2 299.7217 N # m (a) MAXIMUM SHEAR STRESSES Bar: t1 Tube: t2 d1 25.0 mm G d2 30.0 mm d3 37.5 mm POLAR MOMENTS OF INERTIA Bar: IP1 Tube: IP2 p 1d 4 32 3 38.3495 * 10 4 d22 T2(d3/2) IP2 4 m 114.6229 * 10 T1L GIP1 T2L GIP2 1.03° f 9 32.7 MPa 49.0 MPa ; ; (b) ANGLE OF ROTATION OF END PLATE 80 GPa p4 d 32 1 T1(d1/2) IP1 ; 0.017977 rad (c) TORSIONAL STIFFNESS 9 m4 kT T f Problem 3.8-11 A solid steel bar of diameter d1 1.50 in. is enclosed by a steel tube of outer diameter d3 2.25 in. and inner diameter d2 1.75 in. (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has length L 30.0 in., is twisted by a torque T 5000 lb-in. acting on the end plate. (a) Determine the maximum shear stresses 1 and 2 in the bar and tube, respectively. (b) Determine the angle of rotation (in degrees) of the end plate, assuming that the shear modulus of the steel is G 11.6 106 psi. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.) 22.3 kN # m ; Tube A B T Bar End plate L d1 d2 d3 Sec_3.8.qxd 9/27/08 1:13 PM Page 311 SECTION 3.8 Statically Indeterminate Torsional Members Solution 3.8-11 311 Bar enclosed in a tube TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) Bar: T1 Ta IPI b IPI + IPI Ta Tube: T2 1187.68 lb-in. IP2 b IP1 + IP2 3812.32 lb-in. (a) MAXIMUM SHEAR STRESSES Bar: t1 T1(d1/2) IP1 T2(d3/2) IP2 Tube: t2 1790 psi 2690 psi ; ; (b) ANGLE OF ROTATION OF END PLATE d1 1.50 in. d2 1.75 in. d3 2.25 in. T1L GIP1 T2L GIP2 POLAR MOMENTS OF INERTIA 0.354° ; p4 d 32 1 (c) TORSIONAL STIFFNESS G 11.6 Bar: IP1 Tube: IP2 f 6 10 psi p4 1d 32 3 0.497010 in.4 4 d2 2 kT 4 1.595340 in. T f 809 k- in. T Problem 3.8-12 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 40 mm for the brass core and d2 50 mm for the steel sleeve. The shear moduli of elasticity are Gb 36 GPa for the brass and Gs 80 GPa for the steel. 0.006180015 rad ; Steel sleeve Brass core T Assuming that the allowable shear stresses in the brass and steel are 48 MPa and s 80 MPa, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.) b d1 d2 Probs. 3.8-12 and 3.8-13 Solution 3.8-12 Composite shaft shrink fit d1 d2 GB 40 mm 50 mm 36 GPa GS 80 GPa Allowable stresses: 48 MPa S 80 MPa B Sec_3.8.qxd 9/27/08 312 1:13 PM CHAPTER 3 Page 312 Torsion BRASS CORE (ONLY) Ta Eq. (3-44b): TS GSIPS b GBIPB + GSIPS 0.762082 T T TB TS (CHECK) ALLOWABLE TORQUE T BASED UPON BRASS CORE p4 d 32 1 IPB 251.327 * 10 9 TB(d1/2) 2tBIPB TB IPB d1 Substitute numerical values: m4 tB 9047.79 N m2 GBIPB TB STEEL SLEEVE (ONLY) 0.237918 T 2(48 MPa)(251.327 * 10 40 mm T 9 m4) 2535 N m ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE IPS GSIPS p4 (d 32 2 4 d1 ) 362.265 * 10 9 m4 2 28,981.2 N m tS Eq. (3-44a): TB TB Ta TS 2tSIPS d2 0.762082 T 2(80 MPa)(362.265 * 10 50 mm TS GBIPB b GBIPB + GS IPS TS SUBSTITUTE NUMERICAL VALUES: TORQUES Total torque: T TS(d2/2) IPS T 9 m4) 1521 N m STEEL SLEEVE GOVERNS Tmax 1520 N m ; 0.237918 T Problem 3.8-13 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 1.6 in. for the brass core and d2 2.0 in. for the steel sleeve. The shear moduli of elasticity are Gb 5400 ksi for the brass and Gs 12,000 ksi for the steel. Assuming that the allowable shear stresses in the brass and steel are b 4500 psi and s 7500 psi, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.) Sec_3.8.qxd 314 9/27/08 1:13 PM Page 314 CHAPTER 3 Torsion (a) Determine the allowable torque T1 that may be applied to the ends of the shaft if the angle of twist between the ends is limited to 8.0°. (b) Determine the allowable torque T2 if the shear stress in the brass is limited to b 70 MPa. (c) Determine the allowable torque T3 if the shear stress in the steel is limited to s 110 MPa. (d) What is the maximum allowable torque Tmax if all three of the preceding conditions must be satisfied? Brass sleeve Steel shaft d2 = 90 mm d1 = 70 mm T T A B 1.0 m L = 2.0 m 2 d1 C L = 2.0 m 2 d1 Brass sleeve d2 d1 Steel shaft d2 Solution 3.8-14 (a) ALLOWABLE TORQUE T1 BASED ON TWIST AT ENDS OF 8 DEGREES First find torques in steel (Ts) & brass (Tb) in segment in which they are joined - 1 degree stat-indet; use Ts as the internal redundant; see equ. 3-44a in text example Ts statics Gb IPb b Gs IPs + GbIPb now find twist of 3 segments: Tb T1 Ts Tb T1 a L L L Ts T1 4 4 2 + + Gb IPb Gs IPs Gs IPs T1 Gs IPs T1 a b Gs IPs + Gb IPb For middle term, brass sleeve & steel shaft twist the same so could use Tb(L/4)/(Gb IPb) instead Let a = allow; substitute expression for Ts then simplifiy; finally, solve for T1, allow Gs IPs L L L T1 a b T1 Gs IPs + Gb IPb 4 4 2 + + Gb IPb Gs IPs Gs IPs T1 fa L L L T1 T1 4 4 2 + + Gb IPb Gs IPs + Gb IPb Gs IPs T1 fa fa 1 L 1 2 + T1 a + b 4 Gb IPb GsIPs + Gb IPb Gs IPs T1, allow 4a Gb IPb(GsIPs + Gb IPb)Gs IPs c22 d 2 L Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb Sec_3.8.qxd 9/27/08 1:13 PM Page 315 SECTION 3.8 8a p b rad 180 NUMERICAL VALUES fa Gs 80 GPa Gb 40 GPa L d1 70 mm d2 p4 d1 32 IPs 2.357 315 13.69 kN m T2, allow 90 mm IPs Statically Indeterminate Torsional Members so T2 for hollow segment controls (c) ALLOWABLE p A d 4 d14 B 32 2 T1, allow 9.51 kN m TORQUE STRESS IN BRASS, T2 IPb 10 6 t d2 2 IPb 4.084 10 6 m4 ; T2, allow t Tb d2 2 IPb T2 a T2, allow t 2tbIPb d2 T3 a BASED ON ALLOWABLE SHEAR where from stat-indet analysis above GsIPS b Gs IPS + Gb IPb T3, allow 2ts (Gs IPs + Gb IPb) d1Gs T3, allow 13.83 kN m also check segment 3 with steel shaft alone ; d1 2 IPs T3 t T3, allow where from stat-indet analysis above d1 2 IPS Ts controls over T2 below also check segment 2 with brass sleeve over steel shaft Tb Ts BASED ON ALLOWABLE SHEAR b 6.35 kN m T2, allow T3 s 110 MPa First check segment 2 with brass sleeve over steel shaft m4 70 MPa b First check hollow segment 1 (brass sleeve only) T2 TORQUE STRESS IN STEEL, s IPb (b) ALLOWABLE 3.0 m T3, allow 7.41 kN m 2ts IPs d1 ; controls over T3 above (d) Tmax IF ALL PRECEDING CONDITIONS MUST BE CONSIDERED GbIPb b Gs IPS + Gb IPb 2tb(Gs IPS + Gb IPb) d2 Gb from (b) above Tmax 6.35 kN m ; max. shear stress in hollow brass sleeve in segment 1 controls overall ...
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This note was uploaded on 11/26/2011 for the course EM 319 taught by Professor Kennethm.liechti during the Spring '08 term at University of Texas at Austin.

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