MEE 320 Problems Chapter 3, Spring 2018

T 18 mm PROBLEM 3.2 For the cylindrical shaft shown, determine the maximum shearing stress caused by a torque of magnitude T  800 N  m. SOLUTION 4 max max 3 3 6 ; 2 2 2(800 N m) (0.018 m) 87.328 10 Pa             Tc J c J T c max 87.3 MPa    PROBLEM 3.1 The allowable shearing stress is 100 MPa in the 36-mm-diameter steel rod AB and 60 MPa in the 40-mm-diameter rod BC. Neglecting the effect of stress concentrations, determine the largest torque that can be applied at A. SOLUTION 4 3 max max , , 2 2 Tc J c T c J π π τ τ = = = Shaft AB: 6 max 6 3 100 MPa 100 10 Pa 1 1 (36) 18 mm 0.018 m 2 2 (100 10 )(0.018) 916 N m 2 AB AB c d T τ π = = × = = = = = × = ⋅ Shaft BC: 6 max 6 3 60 MPa 60 10 Pa 1 1 (40) 20 mm 0.020 m 2 2 (60 10 )(0.020) 1.754 N m 2 BC BC c d T τ π = = × = = = = = × = ⋅ The allowable torque is the smaller of TAB and TBC. 754 N m T = ⋅  5 750 mm 600 mm TB � 1200 N · m TC � 400 N · m dAB B A C dBC PROBLEM 3.17 The solid shaft shown is formed of a brass for which the allowable shearing stress is 55 MPa. Neglecting the effect of stress concentrations, determine the smallest diameters dAB and dBC for which the allowable shearing stress is not exceeded. SOLUTION 6 max 3 max 3 max 55 MPa 55 10 Pa 2 2           Tc T T c J c Shaft AB: 3 3 6 1200 400 800 N m (2)(800) 21.00 10 m 21.0 m (55 10 ) AB T c            minimum 2 42.0 mm dAB c    Shaft BC: 3 3 6 400 N m (2)(400) 16.667 10 m 16.67 mm (55 10 ) BC T c          minimum 2 33.3 mm dBC c    A 100 mm 40 mm C B D T � 1000 N · m PROBLEM 3.21 A torque of magnitude T 1000 N m  is applied at D as shown. Knowing that the allowable shearing stress is 60 MPa in each shaft, determine the required diameter of (a) shaft AB, (b) shaft CD. SOLUTION 1000 N m 100 (1000) 2500 N m 40 CD B AB CD C T r T T r       (a) Shaft AB: 6 all 3 6 3 3 6 60 10 Pa 2 2 (2)(2500) 26.526 10 m (60 10 )                Tc T T c J c 3 29.82 10 29.82 mm c     2 = 59.6 mm d  c  (b) Shaft CD: 6 all 3 6 3 3 6 60 10 Pa 2 2 (2)(1000) 10.610 10 m (60 10 )                Tc T T c J c 3 21.97 10 m 21.97 mm c     2 = 43.9 mm d  c  2.5 m 40 mm 50 mm A B T0 PROBLEM 3.34 (a) For the aluminum pipe shown (G  27 GPa), determine the torque T0 causing an angle of twist of 2°. (b) Determine the angle of twist if the same torque T0 is applied to a solid cylindrical shaft of the same length and cross-sectional area. SOLUTION (a) 4 4 4 4 6 4 3 9 50 mm 0.050 m, 40 mm 0.040 m ( ) (0.050 0.040 ) 2 2 5.7962 10 m 2 34.907 10 rad 2.5 m 27 10 Pa                       o i o i c c J c c L G TL   GJ 9 6 3 0 3 (27 10 )(5.7962 10 )(34.907 10 ) 2.5 2.1851 10 N m            GJ T L 0 T  2.19 kN m   Area of pipe: 2 2 2 2 2 ( ) (0.050 0.040 ) 2.8274 m        o i A c c (b) Radius of solid of same area: 4 4 6 2 3 0 9 6 0.030 m (0.030) 1.27235 10 m 2 2 (2.1851 10 )(2.5) 0.15902 rad (27 10 )(1.27235 10 )                   s s s s A c J c T L GJ 9.11 s     400 mm 375 mm 250 mm D 60 mm 36 mm TA � 800 N · m TB � 1600 N · m C B A PROBLEM 3.38 The aluminum rod AB ( G  27 GPa) is bonded to the brass rod BD ( G  39 GPa). Knowing that portion CD of the brass rod is hollow and has an inner diameter of 40 mm, determine the angle of twist at A. SOLUTION Rod AB: 9 4 4 9 3 / 9 9 27 10 Pa, 0.400 m 1 800N m 0.018 m 2 (0.018) 164.896 10 m 2 2 (800)(0.400) 71.875 10 rad (27 10 )(164.896 10 ) A B G L T c d J c TL GJ                        Part BC: 9 4 4 6 4 3 / 9 6 1 39 10 Pa 0.375 m, 0.030 m 2 800 1600 2400 N m, (0.030) 1.27234 10 m 2 2 (2400)(0.375) 18.137 10 rad (39 10 )(1.27234 10 ) B C G L c d T J c TL GJ                          Part CD:   1 1 2 2 4 4 4 4 6 4 2 1 3 / 9 6 1 0.020 m 2 1 0.030 m, 0.250 m 2 (0.030 0.020 ) 1.02102 10 m 2 2 (2400)(0.250) 15.068 10 rad (39 10 )(1.02102 10 ) C D c d c d L J c c TL GJ                        Angle of twist at A. / / / 3 105.080 10 rad A A B B C C D           6.02 A    PROBLEM 3.41 Two shafts, each of 22-mm diameter are connected by the gears shown. Knowing that G = 77 GPa and that the shaft at F is fixed, determine the angle through which end A rotates when a 130 N ⋅ m torque is applied at A. SOLUTION Calculation of torques. Circumferential contact force between gears B and E. AB EF E EF AB B E B T T r F T T r r r = = = 130 N m 150 (130) 177.3 N m 110 AB EF T T = ⋅ = = ⋅ Twist in shaft FE. 4 4 9 4 / 9 9 1 300 mm, 11 mm, 776 Pa 2 (0.011) 23 10 m 2 2 (177.3)(0.3) 0.03 rad (77 10 )(23 10 ) E F L c d G J c TL GJ π π ϕ − − = = = = = = = × = = = × × Rotation at E. / 0.03 rad E F ϕE = ϕ = Tangential displacement at gear circle E E B B r r δ ϕ ϕ = = Rotation at B. 15 (0.03) 0.0409 rad 11 E B E B r r ϕ ϕ = = = Twist in shaft BA. 9 4 / 9 9 0.35 m 23 10 m (130)(0.35) 0.0257 rad (77 10 )(23 10 ) A B L J TL GJ ϕ − − = = × = = = × × Rotation at A. / 0.0666 rad A B A B ϕ ϕ ϕ = + = 3.82 ϕA = °  PROBLEM 3.53 The composite shaft shown consists of a 5 mm thick brass jacket (G = 39 GPa) bonded to a 30 mm diameter steel core (G = 77 GPa). Knowing that the shaft is subjected to 565 N ⋅ m torques, determine (a) the maximum shearing stress in the brass jacket, (b) the maximum shearing stress in the steel core, (c) the angle of twist of end B relative to end A. SOLUTION Steel core: 1 4 4 9 4 1 1 9 9 2 1 1 1 15 mm 2 (0.015) 79.52 10 m 2 2 (77 10 )(79.52 10 ) 6123.04 N m c d J c G J π π − − = = = = = × = × × = ⋅ Torque carried by steel core 1 1 1 T G J L ϕ = Brass jacket: ( ) 2 1 4 4 4 4 6 4 2 2 1 9 6 2 2 2 15 mm 5 mm 20 mm (0.02 0.015 ) 0.1718 10 m 2 2 (39 10 )(0.1718 10 ) 6700.4 N m c c t J c c G J π π − − = + = + = = − = − = × = × × = ⋅ Torque carried by brass jacket 2 2 2 T G J L ϕ = Total torque 1 2 1 1 2 2 1 1 2 2 ( ) 565 0.044 rad/m 6123.04 6700.4 T T T G J G J L T L G J G J ϕ ϕ = + = + = = = + + (a) Maximum shearing stress in brass jacket 9 max 2 max 2 2 (39 10 )(0.02)(0.044) G G C L ϕ τ γ = = = × 6 2 34.37 10 N/m = × 34.37 MPa  (b) Maximum shearing stress in steel core 9 max 1 max 1 1 (77 10 )(0.015)(0.044) G G C L ϕ τ γ = = = × 6 2 50.82 10 N/m = × 50.82 MPa  (c) Angle of twist (L = 1.8 m) 3 (1.8)(0.044) 79.2 10 rad L L ϕ ϕ − = = = × = 4.53°  75 mm 30 mm PROBLEM 3.68 While a steel shaft of the cross section shown rotates at 120 rpm, a stroboscopic measurement indicates that the angle of twist is 2° in a 4-m length. Using G  77.2 GPa, determine the power being transmitted. SOLUTION Twist angle:   3 1 1 2 2 4 4 4 4 2 1 6 4 2 34.907 10 rad 1 0.015 m 2 1 0.0375 m 2 (0.0375 0.015 ) 2 2 3.0268 10 m , 4 m                     c d c d J c c J L 9 6 3 3 (77.2 10 )(3.0268 10 )(34.907 10 ) 4 2.0392 10 N m 120 120 rpm Hz 2 Hz 60 TL GJ T GJ L T f                 3 3 (2 ) 2 (2)(2.0392 10 ) 25.6 10 W 25.6 kW         P f T  60 mm 30 mm T T' PROBLEM 3.87 The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing that the radius of the fillet is r  8 mm and the allowable shearing stress is 45 MPa, determine the maximum power that can be transmitted. SOLUTION 3 3 3 2 2 1 30 mm 15 mm 15 10 m 2 60 mm, 8 mm 60 8 2, 0.26667 30 30 KTc KT c T J K c d c d D r D r d d                    From Fig. 3.32, K  1.18 Allowable torque. 3 3 6 (15 10 ) (45 10 ) 202.17 N m (2)(1.18) T        Maximum power. 3 2 (2 )(50)(202.17) 63.5 10 W P f T       P  63.5 kW 