12_ch 15 Mechanical Design budynas_SM_ch15

# 12_ch 15 Mechanical Design budynas_SM_ch15 - 2290 1 . 32(1)...

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390 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Gear: ( s ac ) G = 125 920 psi ( σ c ,all ) = 125 920(1 . 043)(1) 1(1)(1 . 118) = 117 473 psi W t = ± 117 473 2290 ² 2 ³ 0 . 71(2 . 000)(0 . 078) 1(1 . 412)(1 . 252)(0 . 526 25)(2) ´ = 156 . 6 lbf H 4 = 156 . 6(628 . 3) 33 000 = 3 . 0hp Rating: H = min(3 . 8, 3 . 3, 2 . 7, 3 . 0) = 2 . 7hp Pinion wear controls the power rating. While the basis of the catalog rating is unknown, it is overly optimistic (by a factor of 1.9). 15-11 From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So ( H B ) 11 and ( H B ) 21 are 180 Brinell and the bending stress numbers are: ( s at ) P = 44(180) + 2100 = 10 020 psi ( s at ) G = 10 020 psi The contact strength of the gear case, based upon the equation derived in Prob. 15-7, is ( s ac ) G = C p ( C L ) G C H µ S 2 H S F ± ( s at ) P ( K L ) P K x J P K T C s C xc N P IK s ² Substituting ( s at ) P from above and the values of the remaining terms from Ex. 15-1,
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Unformatted text preview: 2290 1 . 32(1) 1 . 5 2 1 . 5 10 020(1)(1)(0 . 216)(1)(0 . 575)(2) 25(0 . 065)(0 . 529) = 114 331 psi ( H B ) 22 = 114 331 23 620 341 = 266 Brinell The pinion contact strength is found using the relation from Prob. 15-7: ( s ac ) P = ( s ac ) G m . 0602 G C H = 114 331(1) . 0602 (1) = 114 331 psi ( H B ) 12 = 114 331 23 600 341 = 266 Brinell Core Case Pinion 180 266 Gear 180 266 Realization of hardnesses The response of students to this part of the question would be a function of the extent to which heat-treatment procedures were covered in their materials and manufacturing...
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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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