19_ch 10 Mechanical Design budynas_SM_ch10

19_ch 10 Mechanical - 6 505 lbf For simplicity we will round up to the next integer or half integer therefore use F i = 7 lbf k = 18 − 7 5 = 22

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Chapter 10 279 10-28 Use: E = 28 . 6 Mpsi, G = 11 . 5 Mpsi, A = 140 kpsi · in m , m = 0 . 190, rel cost = 1 . Try d = 0 . 067 in, S ut = 140 (0 . 067) 0 . 190 = 234 . 0 kpsi Table 10-6: S sy = 0 . 45 S ut = 105 . 3 kpsi Table 10-7: S y = 0 . 75 S ut = 175 . 5 kpsi Eq. (10-34) with D / d = C and C 1 = C σ A = F max π d 2 [( K ) A (16 C ) + 4] = S y n y 4 C 2 C 1 4 C ( C 1) (16 C ) + 4 = π d 2 S y n y F max 4 C 2 C 1 = ( C 1) ± π d 2 S y 4 n y F max 1 ² C 2 1 4 ± 1 + π d 2 S y 4 n y F max 1 ² C + 1 4 ± π d 2 S y 4 n y F max 2 ² = 0 C = 1 2 π d 2 S y 16 n y F max ± ³ ± π d 2 S y 16 n y F max ² 2 π d 2 S y 4 n y F max + 2 = 1 2 ´ π (0 . 067 2 )(175 . 5)(10 3 ) 16(1 . 5)(18) + ³ µ π (0 . 067) 2 (175 . 5)(10 3 ) 16(1 . 5)(18) 2 π (0 . 067) 2 (175 . 5)(10 3 ) 4(1 . 5)(18) + 2 = 4 . 590 D = Cd = 0 . 3075 in F i = π d 3 τ i 8 D = π d 3 8 D µ 33 500 exp(0 . 105 C ) ± 1000 ± 4 C 3 6 . 5 ²¶ Use the lowest F i in the preferred range. This results in the best fom. F i = π (0 . 067) 3 8(0 . 3075) · 33 500 exp[0 . 105(4 . 590)] 1000 ± 4 4 . 590 3 6 . 5 ²¸ =
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Unformatted text preview: 6 . 505 lbf For simplicity, we will round up to the next integer or half integer; therefore, use F i = 7 lbf k = 18 − 7 . 5 = 22 lbf/in N a = d 4 G 8 kD 3 = (0 . 067) 4 (11 . 5)(10 6 ) 8(22)(0 . 3075) 3 = 45 . 28 turns N b = N a − G E = 45 . 28 − 11 . 5 28 . 6 = 44 . 88 turns L = (2 C − 1 + N b ) d = [2(4 . 590) − 1 + 44 . 88](0 . 067) = 3 . 555 in L 18 lbf = 3 . 555 + . 5 = 4 . 055 in take positive root...
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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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