4_ch 15 Mechanical Design budynas_SM_ch15

# 4_ch 15 Mechanical Design budynas_SM_ch15 - H = 4 54 hp Ans...

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382 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Our modeling is rough, but it convinces us that ( K v ) CI < ( K v ) steel , but we are not sure of the value of ( K v ) CI . We will use K v for steel as a basis for a conservative rating. Eq. (15-6): B = 0 . 25(12 6) 2 / 3 = 0 . 8255 A = 50 + 56(1 0 . 8255) = 59 . 77 Eq. (15-5): K v = ± 59 . 77 + 1178 59 . 77 ² 0 . 8255 = 1 . 454 Pinion bending ( σ all ) P = s w t = 2250 psi From Prob. 15-1, K x = 1, K m = 1 . 106, K s = 0 . 5222 Eq. (15-3): W t P = ( σ all ) P FK x J P P d K o K v K s K m = 2250(1 . 25)(1)(0 . 268) 6(1)(1 . 454)(0 . 5222)(1 . 106) = 149 . 6 lbf H 1 = 149 . 6(1178) 33 000 = 5 . 34 hp Gear bending W t G = W t P J G J P = 149 . 6 ³ 0 . 228 0 . 268 ´ = 127 . 3 lbf H 2 = 127 . 3(1178) 33 000 = 4 . 54 hp The gear controls in bending fatigue.
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Unformatted text preview: H = 4 . 54 hp Ans. 15-4 Continuing Prob. 15-3, Table 15-5: s ac = 50 000 psi s w t = σ c ,all = 50 000 √ 2 = 35 355 psi Eq. (15-1): W t = ³ σ c ,all C p ´ 2 Fd P I K o K v K m C s C xc Fig. 15-6: I = . 86 Eq. (15-9) C s = . 125(1 . 25) + . 4375 = . 593 75 Eq. (15-10) K s = . 4867 + . 2132 / 6 = . 5222 Eq. (15-11) K m = 1 . 10 + . 0036(1 . 25) 2 = 1 . 106 Eq. (15-12) C xc = 2 From Table 14-8: C p = 1960 µ psi...
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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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