Chapter_7_solutions

Chapter_7_solutions - Chapter 7 7-1 (a) DE-Gerber, Eq....

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Chapter 7 7-1 (a) DE-Gerber, Eq. (7-10):   22 4 3 4 (2.2)(70) 3 (1.8)(45) 338.4 N m fa f s a AK M K T  4 3 4 (2.2)(55) 3 (1.8)(35) 265.5 N m fm f s m BK M K T  1/3 1/2 2 6 66 2(265.5) 210 10 8(2)(338.4) 11 210 10 338.4 700 10 d           d = 25.85 (10 3 ) m = 25.85 mm Ans. (b) DE-elliptic, Eq. (7-12) can be shown to be 1/3 1/3 338.4 265.5 16 16(2) 210 10 560 10 ey nA B d SS  d = 25.77 (10 3 ) m = 25.77 mm Ans. (c) DE-Soderberg, Eq. (7-14) can be shown to be 1/3 1/3 16 16(2) 338.4 265.5 210 10 560 10 d d = 27.70 (10 3 ) m = 27.70 mm Ans. (d) DE-Goodman: Eq. (7-8) can be shown to be 1/3 1/3 16 16(2) 338.4 265.5 210 10 700 10 eu t d d = 27.27 (10 3 ) m = 27.27 mm Ans. ________________________________________________________________________ Criterion d (mm) Compared to DE-Gerber DE-Gerber 25.85 DE-Elliptic 25.77 0.31% Lower Less conservative DE-Soderberg 27.70 7.2% Higher More conservative DE-Goodman 27.27 5.5% Higher More conservative ______________________________________________________________________________ 7-2 This problem has to be done by successive trials, since S e is a function of shaft size. The material is SAE 2340 for which S ut = 175 kpsi, S y = 160 kpsi, and H B 370. Chapter 7 - Rev. A, Page 1/45
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Eq. (6-19), p. 287: 0.265 2.70(175) 0.69 a k  Trial #1 : Choose d r = 0.75 in Eq. (6-20), p. 288: 0.107 0.879(0.75) 0.91 b k Eq. (6-8), p.282:  0.5 0.5 175 87.5 kpsi eu t SS Eq. (6-18), p. 287: S e = 0.69 (0.91)(87.5) = 54.9 kpsi 2 0.75 2 / 20 0.65 r ddr DD D   0.75 1.15 in 0.65 0.65 r d D  1.15 0.058 in 20 20 D r Fig. A-15-14: 2 0.75 2(0.058) 0.808 in r dd r  0.808 1.08 0.75 r d d 0.058 0.077 0.75 r r d K t = 1.9 Fig. 6-20, p. 295: r = 0.058 in, q = 0.90 Eq. (6-32), p. 295: K f = 1 + 0.90 (1.9 – 1) = 1.81 Fig. A-15-15: K ts = 1.5 Fig. 6-21, p. 296: r = 0.058 in, q s = 0.92 Eq. (6-32), p. 295: K fs = 1 + 0.92 (1.5 – 1) = 1.46 We select the DE-ASME Elliptic failure criteria, Eq. (7-12), with d as d r , and M m = T a = 0, 1/3 1/2 22 33 16(2.5) 1.81(600) 1.46(400) 43 54.9 10 160 10 r d            d r = 0.799 in Trial #2 : Choose d r = 0.799 in. 0.107 0.879(0.799) 0.90 b k S e = 0.69 (0.90)(0.5)(175) = 54.3 kpsi 0.799 1.23 in 0.65 0.65 r d D r = D / 20 = 1.23/20 = 0.062 in Chapter 7 - Rev. A, Page 2/45
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Figs. A-15-14 and A-15-15: 2 0.799 2(0.062) 0.923 in r dd r  0.923 1.16 0.799 r d d  0.062 0.078 0.799 r r d With these ratios only slightly different from the previous iteration, we are at the limit of readability of the figures. We will keep the same values as before. 1.9, 1.5, 0.90, 0.92 tt s s KK q q 1.81, 1.46 ff s Using Eq. (7-12) produces d r = 0.802 in. Further iteration produces no change. With d r = 0.802 in, 0.802 1.23 in 0.65 0.75(1.23) 0.92 in D d A look at a bearing catalog finds that the next available bore diameter is 0.9375 in. In nominal sizes, we select d = 0.94 in, D = 1.25 in, r = 0.0625 in Ans. ______________________________________________________________________________ 7-3 F cos 20 ( d / 2) = T A , F = 2 T A / ( d cos 20 ) = 2(340) / (0.150 cos 20 ) = 4824 N. The maximum bending moment will be at point C , with M C = 4824(0.100) = 482.4 N·m. Due to the rotation, the bending is completely reversed, while the torsion is constant. Thus, M a = 482.4 N·m, T m = 340 N·m, M m = T a = 0. For sharp fillet radii at the shoulders, from Table 7-1, K t = 2.7, and K ts = 2.2. Examining Figs. 6-20 and 6-21 (pp. 295 and 296 respectively) with 560 MPa, ut S conservatively estimate q = 0.8 and These estimates can be checked once a specific fillet radius is determined.
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Chapter_7_solutions - Chapter 7 7-1 (a) DE-Gerber, Eq....

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