22_ch 10 Mechanical Design budynas_SM_ch10

# 22_ch 10 Mechanical Design budynas_SM_ch10 - 2 ut 2 S e −...

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282 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design S yt = σ = F ± 16( K ) A D π d 3 + 4 π d 2 ² F = 155 . 3(10 3 ) [16(1 . 099)(1 . 338)] / [ π (0 . 162) 3 ] +{ 4 / [ π (0 . 162) 2 ] } = 85 . 8 lbf = min(110 . 8, 103 . 9, 85 . 8) = 85 . 8 lbf Ans . (e) Eq. (10-48): y = F F i k = 85 . 8 16 4 . 855 = 14 . 4in Ans . 10-30 F min = 9 lbf, F max = 18 lbf F a = 18 9 2 = 4 . 5 lbf, F m = 18 + 9 2 = 13 . 5 lbf A313 stainless: 0 . 013 d 0 . 1 A = 169 kpsi · in m , m = 0 . 146 0 . 1 d 0 . 2 A = 128 kpsi · in m , m = 0 . 263 E = 28 Mpsi, G = 10 Gpsi Try d = 0 . 081 in and refer to the discussion following Ex. 10-7 S ut = 169 (0 . 081) 0 . 146 = 243 . 9 kpsi S su = 0 . 67 S ut = 163 . 4 kpsi S sy = 0 . 35 S ut = 85 . 4 kpsi S y = 0 . 55 S ut = 134 . 2 kpsi Table 10-8: S r = 0 . 45 S ut = 109 . 8 kpsi S e = S r / 2 1 [ S r / (2 S ut )] 2 = 109 . 8 / 2 1 [(109 . 8 / 2) / 243 . 9] 2 = 57 . 8 kpsi r = F a / F m = 4 . 5 / 13 . 5 = 0 . 333 Table 6-7: S a = r 2 S
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Unformatted text preview: 2 ut 2 S e − 1 + ³ 1 + ´ 2 S e rS ut µ 2 S a = (0 . 333) 2 (243 . 9 2 ) 2(57 . 8) − 1 + ³ 1 + ± 2(57 . 8) . 333(243 . 9) ² 2 = 42 . 2 kpsi Hook bending ( σ a ) A = F a ± ( K ) A 16 C π d 2 + 4 π d 2 ² = S a ( n f ) A = S a 2 4 . 5 π d 2 ± (4 C 2 − C − 1)16 C 4 C ( C − 1) + 4 ² = S a 2 This equation reduces to a quadratic in C —see Prob. 10-28...
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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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