18_ch 06 Mechanical Design budynas_SM_ch06

18_ch 06 Mechanical Design budynas_SM_ch06 - n f = 1 τ a S...

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164 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) Using σ max = 52 . 34(10 6 ) T from Prob. 6-24, n y = 420 52 . 34(3 . 86) = 2 . 08 Ans . 6-26 (a) τ max = 16 K fs T max π d 3 Fig. 6-21 for H B > 200, r = 3mm, q s . = 1 K fs = 1 + q s ( K ts 1) K fs = 1 + 1(1 . 6 1) = 1 . 6 T max = 2000(0 . 05) = 100 N · m, T min = 500 2000 (100) = 25 N · m τ max = 16(1 . 6)(100)(10 6 ) π (0 . 02) 3 = 101 . 9MPa τ min = 500 2000 (101 . 9) = 25 . 46 MPa τ m = 1 2 (101 . 9 + 25 . 46) = 63 . 68 MPa τ a = 1 2 (101 . 9 25 . 46) = 38 . 22 MPa S su = 0 . 67 S ut = 0 . 67(320) = 214 . 4MPa S sy = 0 . 577 S y = 0 . 577(180) = 103 . 9MPa S ± e = 0 . 5(320) = 160 MPa k a = 57 . 7(320) 0 . 718 = 0 . 917 d e = 0 . 370(20) = 7 . 4mm k b = ± 7 . 4 7 . 62 ² 0 . 107 = 1 . 003 k c = 0 . 59 S e = 0 . 917(1 . 003)(0 . 59)(160) = 86 . 8MPa Modified Goodman, Table 6-6
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Unformatted text preview: n f = 1 ( τ a / S e ) + ( τ m / S su ) = 1 (38 . 22 / 86 . 8) + (63 . 68 / 214 . 4) = 1 . 36 Ans . (b) Gerber, Table 6-7 n f = 1 2 ± S su τ m ² 2 τ a S e − 1 + ³ 1 + ± 2 τ m S e S su τ a ² 2 = 1 2 ± 214 . 4 63 . 68 ² 2 38 . 22 86 . 8 − 1 + ³ 1 + ´ 2(63 . 68)(86 . 8) 214 . 4(38 . 22) µ 2 = 1 . 70 Ans ....
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