8_ch 15 Mechanical Design budynas_SM_ch15

# 8_ch 15 Mechanical Design budynas_SM_ch15 - = 17 023 psi (...

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386 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (c) ( S H ) P = ( S H ) G = ± σ c ,all σ c ² P = ± σ c ,all σ c ² G Substituting in the right-hand equality gives [ s ac C L / ( C R K T )] P ³ C p ´ W t K o K v K m C s C xc / ( Fd P I ) µ P = [ s ac C L C H / ( C R K T )] G ³ C p ´ W t K o K v K m C s C xc / ( Fd P I ) µ G Denominators cancel leaving ( s ac ) P ( C L ) P = ( s ac ) G ( C L ) G C H Solving for ( s ac ) P gives, ( s ac ) P = ( s ac ) G ( C L ) G ( C L ) P C H (1) From Eq. (15-14), ( C L ) P = 3 . 4822 N 0 . 0602 L ,( C L ) G = 3 . 4822( N L / m G ) 0 . 0602 . Thus, ( s ac ) P = ( s ac ) G (1 / m G ) 0 . 0602 C H = ( s ac ) G m 0 . 0602 G C H Ans. This equation is the transpose of Eq. (14-45). 15-8 Core Case Pinion ( H B ) 11 ( H B ) 12 Gear ( H B ) 21 ( H B ) 22 Given ( H B ) 11 = 300 Brinell Eq. (15-23): ( s at ) P = 44(300) + 2100 = 15 300 psi From Prob. 15-7, ( s at ) G = ( s at ) P J P J G m 0 . 0323 G = 15 300 ± 0 . 249 0 . 216 ² ( 3 0 . 0323 )
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Unformatted text preview: = 17 023 psi ( H B ) 21 = 17 023 − 2100 44 = 339 Brinell Ans. ( s ac ) G = 2290 1 . 0685(1) ¶ 15 300(0 . 862)(0 . 249)(1)(0 . 593 25)(2) 20(0 . 086)(0 . 5222) = 141 160 psi ( H B ) 22 = 141 160 − 23 600 341 = 345 Brinell Ans. ( s ac ) P = ( s ac ) G m . 0602 G C H . = 141 160(3 . 0602 )(1) = 150 811 psi ( H B ) 12 = 150 811 − 23 600 341 = 373 Brinell Ans. Care Case Pinion 300 373 Ans. Gear 339 345...
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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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