18_ch 14 Mechanical Design budynas_SM_ch14

# 18_ch 14 Mechanical Design budynas_SM_ch14 -...

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Unformatted text preview: budynas_SM_ch14.qxd 366 12/05/2006 17:39 Page 366 FIRST PAGES Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 14-21 Pt = Pn cos ψ = 6 cos 30° = 5.196 teeth/in 16 48 = 3.079 in, dG = (3.079) = 9.238 in 5.196 16 π (3.079)(300) = 241.8 ft/min V= 12 √ 33 000(5) 59.77 + 241.8 t W= = 682.3 lbf, K v = 241.8 59.77 dP = 0.8255 = 1.210 From Prob. 14-19: Y P = 0.296, ( K s ) P = 1.088, m G = 3, YG = 0.4056 ( K s ) G = 1.097, ( Y N ) P = 0.977, ( Y N ) G = 0.996, ( St ) P = ( St ) G = 28 260 psi, ( Z N ) P = 0.948, KB = 1 K R = 0.85 C H = 1, ( Z N ) G = 0.973, ( Sc ) P = ( Sc ) G = 93 500 psi √ C p = 2300 psi The pressure angle is: φt = tan−1 Eq. (13-19): tan 20° cos 30° = 22.80° 3.079 cos 22.8° = 1.419 in, 2 a = 1/ Pn = 1/6 = 0.167 in (rb ) P = (rb ) G = 3(rb ) P = 4.258 in Eq. (14-25): 3.079 + 0.167 2 Z= − 1/2 2 − 1.419 2 + 9.238 + 0.167 2 1/2 2 − 4.258 3.079 9.238 sin 22.8° + 2 2 = 0.9479 + 2.1852 − 2.3865 = 0.7466 Conditions O.K. for use π p N = pn cos φn = cos 20° = 0.4920 in 6 Eq. (14-21): mN = Eq. (14-23): I= Fig. 14-7: pN 0.492 = = 0.6937 0.95 Z 0.95(0.7466) sin 22.8° cos 22.8° 2(0.6937) J P = 0.45, ˙ 3 3+1 JG = 0.54 ˙ = 0.193 2 ...
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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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