15_ch 06 Mechanical Design budynas_SM_ch06

# 15_ch 06 Mechanical Design budynas_SM_ch06 - r ± 0.375 in...

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Chapter 6 161 6-23 Preliminaries: Table A-20: S ut = 64 kpsi, S y = 54 kpsi S ± e = 0 . 5(64) = 32 kpsi k a = 2 . 70(64) 0 . 265 = 0 . 897 k b = 1 k c = 0 . 85 S e = 0 . 897(1)(0 . 85)(32) = 24 . 4 kpsi Fillet: Fig. A-15-5: D = 3 . 75 in, d = 2 . 5in, D / d = 3 . 75 / 2 . 5 = 1 . 5, and r / d = 0 . 25 / 2 . 5 = 0 . 10 K t = 2 . 1 . Fig. 6-20 with r ± 0.25 in, q ˙= 0 . 82 Eq. (6-32): K f = 1 + 0 . 82(2 . 1 1) = 1 . 90 σ max = 4 2 . 5(0 . 5) = 3 . 2 kpsi σ min = 16 2 . 5(0 . 5) =− 12 . 8 kpsi σ a = 1 . 90 ± ± ± ± 3 . 2 ( 12 . 8) 2 ± ± ± ± = 15 . 2 kpsi σ m = 1 . 90 ² 3 . 2 + ( 12 . 8) 2 ³ =− 9 . 12 kpsi n y = ± ± ± ± S y σ min ± ± ± ± = ± ± ± ± 54 12 . 8 ± ± ± ± = 4 . 22 Since the midrange stress is negative, S a = S e = 24 . 4 kpsi n f = S a σ a = 24 . 4 15 . 2 = 1 . 61 Hole: Fig. A-15-1: d /w = 0 . 75 / 3 . 75 = 0 . 20, K t = 2 . 5. Fig. 6-20, with
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Unformatted text preview: r ± 0.375 in, q ˙= . 85 Eq. (6-32): K f = 1 + . 85(2 . 5 − 1) = 2 . 28 σ max = 4 . 5(3 . 75 − . 75) = 2 . 67 kpsi σ min = − 16 . 5(3 . 75 − . 75) = − 10 . 67 kpsi σ a = 2 . 28 ± ± ± ± 2 . 67 − ( − 10 . 67) 2 ± ± ± ± = 15 . 2 kpsi σ m = 2 . 28 2 . 67 + ( − 10 . 67) 2 = − 9 . 12 kpsi...
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