ME 452 - Machine Design II Name of Student:_____________________________ Spring Semester 2015 Lecture Division Number:______________________ Homework No. 4 (50 points). Due at the beginning of lecture on Friday, February 13th. Solve the following problems from Chapter 6, Shigley’s Mechanical Engineering Design, Ninth Edition, R.G. Budynas and J.K. Nisbett. 1. (15 Points). Problem 6-15, page 349. 2. (15 Points). Problem 6-16, page 349. 3. (20 Points). Problem 6-17, page 349. 1 Solution to Homework Set 4. 1. Problem 6-15. (15 Points). The fully corrected endurance strength of the solid steel round bar from Eq. (6-18), see page 287, can be written as Se ka kb kc kd ke k f Se (1) The uncorrected endurance strength of the bar can be written from Eq. (6-8), see page 282, as Se 0.5 Sut (2a) The ultimate tensile strength of AISI 1095 HR steel from Table A-20, see page 1040, is Sut 120 kpsi (2b) Substituting Eq. (2b) into Eq. (2a), the uncorrected endurance strength of the solid steel round bar is Se 0.5 120 60 kpsi (3) The Marine factors. The surface modification factor can be written from Eq. (6-19), see page 287, as ka a Sutb (4) The coefficient and the exponent from Table 6-2, see page 288, respectively, are a 2.70 kpsi and b 0.265 (5) Substituting Eqs. (2b) and (5) into Eq. (4), the surface modification factor is ka 2.70 1200.265 0.759 (6) The size modification factor. Since the round bar (diameter d 1.8 inches) is not rotating then the equivalent diameter can be written from Eq. (6-24), see page 289, as d e 0.370 d 0.370 1.8 0.666 inches (7) Therefore, the size modification factor can be written from Eq. (6-20), see page 288, as kb 0.879 d e 0.107 (8) Substituting Eq. (7) into Eq. (8), the size modification factor is kb 0.879 0.6660.107 0.918 (9) The load modification factor for fully reversed bending, see Figure 6.23f, page 301, or repeated bending, see Figure 6.23e, page 301, from Eq. (6-26), see page 290, is kc 1 (10a) The temperature modification factor, see pages 290-292, is kd 1 2 (10b) The reliability modification factor, see page 292, is ke 1 (10c) The miscellaneous effects modification factor, see page 293, is kf 1 (11) Substituting Eqs. (3), (6), (9), (10), and (11) into Eq. (1), the fully corrected endurance strength is Se 0.759 0.918 1 60 41.806 kpsi (12) Stress Analysis. The fatigue stress concentration factor can be written from Eq. (6-32), see page 295, as K f 1 q( K t 1) (13) First, determine the theoretical stress concentration factor. The known geometric data for the bar is D 2 in, d 1.8 in and r 0.1 in (14a) Therefore, the geometric ratios are r 0.1 0.056 d 1.8 D 2 1.111 d 1.8 and (14b) The theoretical stress concentration factor from Table A-15-14, see page 1030, is K t 2.1 (14c) The notch sensitivity can be obtained from either Fig. 6-20, or from the curve-fit, see Eqs. (6-34) and (635a), page 296. Using the equations, the Neuber constant from Eq. (6-35a), page 296, is a 0.246 (3.08 103 ) Sut (1.51 105 ) Sut 2 (2.67 108 ) Sut 3 in (15a) where the ultimate tensile strength in Eq. (15a) must be in kpsi. Substituting Eq. (2) into Eq. (15a), the Neuber constant is a 0.246 (3.08 103 )120 (1.51105 )1202 (2.67 108 )1203 0.0477 in (15b) The notch sensitivity can be written from Eq. (6-34), page 296, as q 1 a 1 r (16a) Substituting Eqs. (14a) and (15b) into Eq. (16a), the notch sensitivity is q 1 0.869 0.0477 1 0.1 3 (16b) Substituting Eqs. (14c) and (16b) into Eq. (13), the fatigue stress concentration factor is K f 1 0.869(2.1 1) 1.956 (17) The maximum and minimum normal stresses due to the maximum and the minimum bending moments can be written, respectively, as max M max c I min and M min c I (18) The maximum and the minimum bending moments, respectively, are M max 25, 000 lb.in and M min 0 (19a) The area moment of inertia of a solid round bar is I d4 64 x 1.84 64 0.515 in 4 (19b) Substituting Eqs. (19a), (19b) into Eq. (18), the maximum normal stress and the minimum normal stress due to the maximum and minimum bending moments, respectively, are max 25, 000 (1.8 / 2) 43.689 kpsi 0.515 and min 0 (20a) (20b) This is commonly referred to as the repeated stress problem, see page 301. The amplitude ratio A = 1 and the stress ratio R = 0. The nominal values of the mean and alternating components of the normal stress from Eq. (6-36), see page 301, respectively, are m ,nom and a ,nom max min 2 max min 2 43.689 0 21.845 kpsi 2 (21a) 43.689 0 21.845 kpsi 2 (21b) The mean and alternating components of the normal stress, including the fatigue stress concentration factor, can be written as min m K f m,nom K f max (22a) 2 and min a K f a ,nom K f max (22b) 2 Substituting Eqs. (17) and (20) or (21) into Eqs. (22), the mean and alternating components of the stress components of the normal stress, respectively, are 4 43.689 0 42.728 kpsi 2 (23a) 43.689 0 42.728 kpsi 2 (23b) m 1.956 and a 1.956 For purposes of comparison two criteria of fatigue failure for infinite life will be presented: (i) the modified-Goodman line; and (ii) the Gerber parabola. (i) The modified-Goodman line, see Eq. (6-46), page 306, can be written as 1 (24) a m Nf Se Sut Substituting Eqs. (2), (12), (23a), and (23b) into Eq. (24), the factor of safety guarding against fatigue failure can be written as 1 42.728 42.728 (25a) N f 41.806 120 Rearranging this equation, the factor of safety guarding against fatigue failure is N f 0.73 (25b) Since the fatigue factor of safety is less than one then the solid steel round bar has a finite life. (ii) The Gerber parabola, see Eq. (6-47), page 306, or Table 6-7, page 307, can be written as 1/ 2 2 m Se 2 1 a Sut 2 N f ( )( ) 1 1 ( ) 2 Se m a Sut (26a) Substituting Eqs. (2), (12), (23a) and (23b) into Eq. (26a), the factor of safety guarding against fatigue failure can be written as 1/ 2 1 42.728 120 2 2 x 42.728 x 41.806 2 Nf ( )( ) 1 1 ( ) 2 41.806 42.728 42.728 x120 (26b) or as Nf 1/ 2 1 1202 41.806 2 ( ) 1 1 ( ) 2 41.806 x 42.728 60 (26c) Therefore, the factor of safety guarding against fatigue failure is N f 0.89 (26d) Note that the answer given by Eq. (26d) is more accurate but less conservative than the answer given by Eq. (25b). Since the fatigue factor of safety given by both Eqs. (25b) and (26d) is less than one then the solid steel round bar is predicted to have a finite life. Therefore, the number of cycles to failure must be determined. 5 The number of cycles to failure can be written from Eq. (6-16) see page 285, as K f rev N a 1/ b K fa a 1/ b (27a) Alternatively, if the stress concentration effects are ignored then the number of cycles to failure can be written as N a a 1/b (27b) The constant can be written from Eq. (6-14), see page 285, as ( f Sut ) 2 a Se (28a) The fatigue strength fraction from Fig. 6-18, see page 285, is f 0.82 (28b) Substituting Eqs. (2), (12), and (28b) into Eq. (28a), the coefficient is a (0.82 120) 2 231.607 kpsi 41.806 (28c) The exponent can be written from Eq. (6-15), see page 285, as f Sut 1 b log 3 Se (29a) Substituting Eqs. (2), (12), and (28b) into Eq. (29a), the exponent is 1 0.82 120 b log 0.124 3 41.806 (29b) Substituting Eqs. (17), (23b), (28c), and (29b) into Eq. (27a), the number of cycles to failure is 1 1.956 42.728 0.124 N 3728.733 cycles 3.729 103 cycles 231.607 (30a) Alternatively, substituting Eqs. (23b), (28c), and (29b) into Eq. (27b), the number of cycles to failure is 1 42.728 0.124 N 831108 cycles 8.311105 cycles 231.607 (30b) The factor of safety guarding against yielding. The stress concentration factors for static yielding of a ductile material will not be included here, see Section 3.13, page 110. The factor of safety guarding against yielding can be written from the Langer line, see Eq. (6-49), page 306, as 6 Ny Sy (31) a ,nom m ,nom The yield strength of AISI 1095 HR steel, from Table A-20, see page 1040, is S y 66 kpsi (32) Substituting Eqs. (21) and (32) into Eq. (31), the factor of safety guarding against yielding is Ny 66 1.51 21.845 21.845 (33) Since the factor of safety is greater than one then yielding is not predicted to have occurred. Conservative Approach. The factor of safety guarding against yielding can also be written as Ny Sy (34a) max Substituting Eqs. (20a) and (32) into Eq. (34a), the factor of safety guarding against yielding is Ny 66 1.51 43.689 (34b) Note that, in this case, the answer is the same as Eq. (33). 2. Problem 6-16. (15 Points). The corrected endurance strength of the shaft from Eq. (6-18), see page 287, can be written as Se ka kb kc kd ke k f Se (1) The uncorrected endurance strength of the shaft can be written from Eq. (6-8), see page 282, as Se 0.5Sut (2a) The ultimate tensile strength of AISI CD 1020 steel, from Table A-20, see page 1040, is Sut 470 MPa 470 kpsi 68.2 kpsi 6.89 (2b) Substituting Eq. (2b) into Eq. (2a), the uncorrected endurance strength of the shaft is Se 0.5 470 235 MPa (2c) The surface modification factor can be written from Eq. (6-19), see page 287, as ka a Sutb (3a) The coefficient and the exponent for cold drawn surface finish from Table 6-2, see page 288, are a 4.51 MPa and 7 b 0.265 (3b) Substituting Eq. (3b) into Eq. (3a), the surface modification factor is ka 4.51 4700.265 0.883 (3c) The size modification factor from Eq. (6-20), see page 288, is kb 1.24d 0.107 1.24 350.107 0.848 (4) The load modification factor from Eq. (6-26), see page 290, is kc 1 (5) The temperature modification factor, see pages 290-292, is kd 1 (6a) The reliability modification factor, see page 292, is ke 1 (6b) The miscellaneous effects modification factor, see page 293, is kf 1 (6c) Substituting Eqs. (2c), (3c)-(6c) into Eq. (1), the corrected endurance strength of the shaft is Se 0.883 0.848 1 235 175.964 MPa (7) Stress Analysis. The fatigue stress concentration factor can be written from Eq. (6-32), see page 295, as K f 1 q( K t 1) (8) First, determine the theoretical stress concentration factor. The known geometric data for the shaft is D 50 mm, d 35 mm and r 3 mm (9a) The geometric ratios are r 3 0.086 d 35 and D 50 1.43 d 35 (9b) Therefore, the theoretical stress concentration factor from Figure A-15-9, see page 1028, is K t 1.7 (10) The notch sensitivity can be obtained from either Fig. 6-20, or from the curve-fit, see Eqs. (6-34) and (635a), page 296. The equations will be used here. Therefore, the notch sensitivity can be written from Eq. (6-34), page 296, as 1 q (11) a 1 r 8 The Neuber constant can be written from Eq. (6-35a), page 296, is a 0.246 (3.08 103 ) Sut (1.51 105 ) Sut 2 (2.67 108 ) Sut 3 in (12a) Note that the units in Eq. (12a) are in because the ultimate tensile strength is in kpsi. Substituting Eq. (2b) into Eq. (12a), the Neuber constant is a 0.246 (3.08 103 )68.2 (1.51 105 )68.22 (2.67 108 )68.23 0.09771 in (12b) Then in SI units (see Example 6-6, page 296), the Neuber constant is a 0.09771 in 0.492 mm (12c) Substituting Eq. (12c) into Eq. (11), the notch sensitivity is q 1 0.779 0.492 1 3 (13) Then substituting Eqs. (10) and (13) into Eq. (8), the fatigue stress concentration factor is K f 1 0.779 (1.7 1) 1.545 (14) The free-body diagram of the shaft is shown in Figure 1. Figure 1. The free body diagram of the shaft. The sum of the reaction forces and the sum of the moments about the right-hand bearing can be written, respectively, as R1 R2 6 kN and 500 R1 175 x 6 0 (15a) Therefore, the reaction forces at the left bearing and the right bearing, respectively, are R1 = 2.1 kN and R2 = 3.9 kN (15b) The critical section of the shaft will be at the shoulder fillet between the 35 mm and 50 mm diameters where the bending moment is large, the diameter is smaller, and stress concentration exists. For the rotating shaft, the loading is completely reversed bending. The factor of safety guarding against yielding can be written as 9 Ny Sy rev (16a) The yield strength of AISI 1020 CD steel, from Table A-20, see page 1040, is S y 390 MPa (16b) The normal stress due to the completely reversed bending moment can be written as rev Mc I (17a) The completely reversed bending moment on the critical element at the critical section is M 200 2.1 420 Nm (17b) Substituting Eq. (17b) and the known geometry into Eq. (17a), the normal stress due to the completely reversed bending moment is rev 420 (35 / 2) 0.09978 kN/mm 2 99.78 MPa 4 ( / 64) 35 (18) Substituting Eqs. (16b) and (18) into Eq. (16a), the factor of safety guarding against yielding is Ny 390 3.91 99.78 (19) Note. The completely reversed bending stress is much less than the yield strength of the shaft material, therefore, yielding is not predicted to have occurred. Since this problem is a zero mean stress problem then the factor of safety guarding against fatigue failure can be written as S Se Nf e (20a) a K f rev Substituting Eqs. (7), (14), and (17) into Eq. (20a), the fatigue factor of safety is Nf 175.964 1.14 1.545 99.78 (20b) Since the factor of safety guarding against fatigue failure is greater than one then the shaft has an infinite life, that is, infinite life is predicted. 3. Problem 6-17. (20 Points). The corrected endurance strength of the shaft from Eq. (6-18), see page 287, can be written as Se ka kb kc kd ke k f Se (1) The ultimate tensile strength of AISI 1040 CD steel, from Table A-20, see page 1040, is Sut 85 kpsi 10 (2) The uncorrected endurance strength of the shaft can be written from Eq. (6-8), see page 282, as Se 0.5Sut 0.5 85 42.5 kpsi (3) The surface modification factor from Eq. (6-19), see page 287, can be written as ka aSutb (4a) The coefficient and the exponent from from Table 6-2, see page 288, are a 2.7 kpsi and b 0.265 (4b) Substituting Eqs. (4b) into Eq. (4a), the surface modification factor is ka 2.7 850.265 0.832 (5) The size modification factor from Eq. (6-20), see page 288, is kb 0.879d 0.107 0.879 1.6250.107 0.835 (6) The load modification factor from Eq. (6-26), see page 290, is kc 1 (7a) The temperature modification factor, see pages 290-292, is kd 1 (7b) The reliability modification factor, see page 292, is ke 1 (7c) The miscellaneous effects modification factor, see page 293, is kf 1 (7d) Substituting Eqs. (3) and (5)-(7d) into Eq. (1), the corrected endurance strength of the shaft is Se 0.832 0.835 1 42.5 29.526 kpsi (8) Stress Analysis. First, determine the reaction forces at bearings A and B. The free body diagram of the shaft is shown in Fig. 1a. 11 Figure 1a. The free body diagram of the shaft. The sum of the moments about the right-hand bearing B can be written as M B 0 (9a) that is 24 RA 16 F1 8 F2 0 (9b) Therefore, the reaction force at the left-hand bearing A is RA 2500(16) 1000(8) 2000 lb 24 (9c) The sum of the forces in the Y-direction can be written as F y 0 (10a) that is RA RB F1 F2 0 (10b) 2000 RB 2500 1000 0 (10c) or Therefore, the reaction force at bearing B is RB 1500 lb (10c) The shear force between the left-hand bearing A and the force F1 is VA RA 2000 lb (11a) The shear force between the forces F1 and F2 is V RA F1 2000 2500 500 lb (11b) The shear force between the force F2 and the right-hand bearing B is VB RB 1500 lb 12 (11c) The shear force diagram of the shaft is shown in Fig. 1b Figure 1b. The shear force diagram. Let the point of application of the force F1 be denoted as point C and the point of application of the force F2 be denoted as point D. Therefore, the bending moment at point C is M C 8 VA 8 x 2000 16000 lb.in (12a) Also, the bending moment at point D is M D 16VA 8 F1 16 x 2000 8 x 2500 12000 lb.in (12b) The bending-moment diagram of the shaft is shown in Fig. 1c. Figure 1c. The bending moment digram. The conclusion from Figs. 1a, 1b, and 1c is that the critical section of the shaft is at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters where: (i) the bending moment is large, (ii) the diameter is smaller, and (iii) stress concentration exists. Note that the fully reversed bending moment at the shoulder fillet is M 16000 2.5 x 500 14750 lb.in (13) The critical element is on the outer surface of the shaft at this critical section. The fatigue stress concentration factor can be written from Eq. (6-32), see page 295, as K f 1 q( K t 1) First, determine the theoretical stress concentration factor. The known data for the shaft is 13 (14) D 1.875 in, d 1.625 in and r 0.0625 in (15a) Therefore, the geometric ratios are r 0.0625 0.04 d 1.625 and D 1.875 1.15 d 1.625 (15b) Therefore, the theoretical stress concentration factor from Figure A-15-9, see page 1028, is K t 1.95 (16) The notch sensitivity can be obtained from either Fig. 6-20, or from the curve-fit, see Eqs. (6-34) and (6-35a), page 296. Here we will use the equations. Therefore, the notch sensitivity can be written from Eq. (6-34), page 296, as 1 (17) q a 1 r The Neuber constant from Eq. (6-35a), page 296, is a 0.246 (3.08 103 ) Sut (1.51 105 ) Sut 2 (2.67 108 ) Sut 3 in (18a) Note that the ultimate tensile strength in Eq. (17a) must be in kpsi. Substituting Eq. (2b) into Eq. (18a), the Neuber constant is a 0.246 (3.08 103 )85 (1.51 105 )852 (2.67 108 )853 0.07690 in (18b) Substituting Eq. (18b) into Eq. (17), the notch sensitivity is q 1 0.765 0.07690 1 0.0625 (19) Substituting Eqs. (16) and (19) into Eq. (14), the fatigue stress concentration factor is K f 1 0.765 (1.95 1) 1.727 (20) The normal stress due to the fully reversed bending moment on the critical element (this is the zero mean stress problem) can be written as Mc rev (21a) I Substituting Eq. (13) and the known geometry into Eq. (21a), the normal stress is rev 14750 (1.625 / 2) 35.0 kpsi ( / 64) 1.6254 14 (21b) Since this problem is a zero mean stress problem then the factor of safety guarding against fatigue failure can be written as S Se (22a) Nf e a K f rev Substituting Eqs. (8), (20), and (21b) into Eq. (22a), the fatigue factor of safety is Nf 29.526 0.488 1.727 35.0 (22b) Since the fatigue factor of safety is less than one then the shaft has a finite life. The S N diagram can be used to determine the life of the shaft. The number of cycles to failure can be written from Eq. (6-16) see page 285, as 1/ b 1/ b K f rev a (23) N a a The constant can be written from Eq. (6-14), see page 285, as a ( f Sut ) 2 Se (24) The fatigue strength fraction from Fig. 6-18, see page 285, is f 0.867 (25) Substituting Eqs. (2) and (25) into Eq. (24), the coefficient is a (0.867 85) 2 183.938 kpsi 29.526 (26) The exponent can be written from Eq. (6-15), see page 285, as f Sut 1 b log 3 Se (27) Substituting Eqs. (2), (8), and (26) into Eq. (27), the exponent is 1 0.876 85 b log 0.134 3 29.526 (28) Substituting Eqs. (19), (21b), (26), and (28) into Eq. (23), the number of cycles to failure is 1 1.727 35.0 0.134 N 4069.047 cycles 183.938 Therefore, the number of cycles to failure can be taken as 15 (29a) N 4.07 103 cycles (29b) Check for yielding. The factor of safety guarding against yielding (see Chapter 5 or use the Langer line) for the critical element on the outer surface of the shaft at the critical section can be written as Ny Sy a Sy rev (30) The normal stress due to the fully reversed bending moment on the critical element is given by Eq. (21a), that is Mc rev (31) I Since the fatigue stress concentration factor will not be included in this equation (see Section 3. 13, page 110) then the maximum bending moment, see Fig. 1b, is taken as M 16000 lb.in (32) Substituting Eq. (32) and the known geometry into Eq. (31), the normal stress is rev 16000 (1.625 / 2) 37.98 kpsi ( / 64) 1.6254 (33) The yield strength of AISI CD steel, see Table A-20, page 1040, is S y 71 kpsi (34) Substituting Eqs. (33) and (34) into Eq. (30), the factor of safety guarding against yielding for this element is 71 (35) Ny 1.87 37.98 Note that the completely reversed bending stress is much less than the yield strength of the shaft material, therefore, yielding is not predicted to have occurred. Aside: Consider the critical element that was investigated in the fatigue case. The maximum bending moment acting at this critical section is M 14750 lb.in (36) Substituting Eq. (36) and the known geometry into Eq. (31), the normal stress is rev 14750 (1.625 / 2) 35.0 kpsi ( / 64) 1.6254 (37) Substituting Eqs. (34) and (37) into Eq. (30), the factor of safety guarding against yielding for the critical element is 71 Ny 2.01 (38) 35.0 16

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