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Mechanical engineering Homework #8 Solutions Homework Problems from Text book: (Shigley’s Mechanical Engineering Design, Richard G Budynas and J Keith Nisbett, McGraw Hill, 10th Edition)

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Problem 6-5 230 kpsi, 150 000 cycles ut S N   Fig. 6-18, point is off the graph, so estimate: f = 0.77 Eq. (6-8): Sut > 200 kpsi, so 100 kpsi e e S S    Eq. (6-14):   2 2 ... ( ) 0.77(230) 313.6 kpsi 100 ut e f S a S    Eq. (6-15): 1 1 0.77(230) log log 0.08274 3 3 100 ut e f S b S                   Eq. (6-13): 0.08274 313.6(150 000) 117.0 kpsi . b f S aN Ans     ________________________________________________________________________________________________________________________________________________ Problem 6-9 Read from graph:   3 6 10 ,90 and (10 ,50). From b S aN  1 1 2 2 log log log log log log S a b N S a b N     From which 1 2 2 1 2 1 log log log log log log / S N S N a N N   6 3 6 3 log90log10 log50log10 log10 /10 2.2095    log 2.2095 0.0851 3 6 10 10 162.0 kpsi log50 / 90 0.0851 3 ( ) 162 10 10 in kpsi . a f ax a b S N N Ans           Check: 3 6 3 0.0851 10 6 0.0851 10 ( ) 162(10 ) 90 kpsi ( ) 162(10 ) 50 kpsi f ax f ax S S               The end points agree. ________________________________________________________________________________________________________________________________________________ https://www.coursehero.com/file/74784991/Homework-8-Solutionspdf/ This study resource was shared via CourseHero.com Problem 6-12 D = 1 in, d = 0.8 in, T = 1800 lbfin, f = 0.9, and from Table A-20 for AISI 1020 CD, Sut = 68 kpsi, and Sy = 57 kpsi. (a) 0.1 1 Fig. A-15-15: 0.125, 1.25, 1.40 0.8 0.8 ts r D K d d      Get the notch sensitivity either from Fig. 6-21, or from the curve-fit Eqs. (6-34) and (6-35b). Using the equations,          2 3 5 8 3 a 0.190 2.51 10 68 1.35 10 68 2.67 10 68 0.07335         1 1 0.812 0.07335 1 1 0.1 s q a r      Eq. (6-32): Kfs = 1 + qs (Kts  1) = 1 + 0.812(1.40  1) = 1.32 For a purely reversing torque of T = 1800 lbfin, 3 3 16 1.32(16)(1800) 23 635 psi 23.6 kpsi (0.8) fs a fs Tr K T K J d         Eq. (6-8): 0.5(68) 34 kpsi e S   Eq. (6-19): ka = 2.70(68)0.265 = 0.883 Eq. (6-20): kb = 0.879(0.8)0.107 = 0.900 Eq. (6-26): kc = 0.59 Eq. (6-18) (labeling for shear): Sse = 0.883(0.900)(0.59)(34) = 15.9 kpsi For purely reversing torsion, use Eq. (6-54) for the ultimate strength in shear. Eq. (6-54): Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi Adjusting the fatigue strength equations for shear, Eq. (6-14):     2 2 0.9(45.6) 105.9 kpsi 15.9 su se f S a S    Eq. (6-15): 1 1 0.9(45.6) log log 0.137 27 3 3 15.9 su se f S b S                   Eq. (6-16):   1 1 0.137 27 23.3 3 61.7 10 cycles . 105.9 b a N Ans a                  (b) For an operating temperature of 750 F, the temperature modification factor, from Table 6-4 is kd = 0.90. Sse = 0.883(0.900)(0.59)(0.9)(34) = 14.3 kpsi [Show More]

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