12_ch 06 Mechanical Design budynas_SM_ch06

12_ch 06 Mechanical Design budynas_SM_ch06 - − 1 + ³ 1 +...

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158 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σ a = 32(7 . 5)(16) π (0 . 375 3 ) (10 3 ) = 23 . 18 kpsi r = 23 . 18 69 . 54 = 0 . 333 (a) Modified Goodman, Table 6-6 n f = 1 (23 . 18 / 34 . 22) + (69 . 54 / 188 . 1) = 0 . 955 Since finite failure is predicted, proceed to calculate N From Fig. 6-18, for S ut = 188 . 1 kpsi, f = 0 . 778 Eq. (6-14): a = [0 . 7781(188 . 1)] 2 34 . 22 = 625 . 8 kpsi Eq. (6-15): b =− 1 3 log 0 . 778(188 . 1) 34 . 22 =− 0 . 210 36 σ a S f + σ m S ut = 1 S f = σ a 1 ( σ m / S ut ) = 23 . 18 1 (69 . 54 / 188 . 1) = 36 . 78 kpsi Eq. (7-15) with σ a = S f N = ± 36 . 78 625 . 8 ² 1 / 0 . 210 36 = 710 000 cycles Ans . (b) Gerber, Table 6-7 n f = 1 2 ± 188 . 1 69 . 54 ² 2 ± 23 . 18 34 . 22 ²
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Unformatted text preview: − 1 + ³ 1 + ´ 2(69 . 54)(34 . 22) 188 . 1(23 . 18) µ 2 = 1 . 20 Thus, infinite life is predicted ( N ≥ 10 6 cycles). Ans . 6-21 (a) I = 1 12 (18)(3 3 ) = 40 . 5 mm 4 y = Fl 3 3 E I ⇒ F = 3 E I y l 3 F min = 3(207)(10 9 )(40 . 5)(10 − 12 )(2)(10 − 3 ) (100 3 )(10 − 9 ) = 50 . 3 N Ans . F max = 6 2 (50 . 3) = 150 . 9 N Ans . (b) M = . 1015 F N · m A = 3(18) = 54 mm 2 F F M 101.5 mm...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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