7_ch 15 Mechanical Design budynas_SM_ch15

7_ch 15 Mechanical Design budynas_SM_ch15 - that C R = µ K...

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Chapter 15 385 Rating of mesh is H rating = min(10 . 37, 9 . 29, 4 . 90, 4 . 93) = 4 . 90 kW Ans . with pinion wear controlling. 15-7 (a) ( S F ) P = ± σ all σ ² P = ( S F ) G = ± σ all σ ² G ( s at K L / K T K R ) P ( W t P d K o K v K s K m / FK x J ) P = ( s at K L / K T K R ) G ( W t P d K o K v K s K m / FK x J ) G All terms cancel except for s at , K L , and J , ( s at ) P ( K L ) P J P = ( s at ) G ( K L ) G J G From which ( s at ) G = ( s at ) P ( K L ) P J P ( K L ) G J G = ( s at ) P J P J G m β G Where β =− 0 . 0178 or β =− 0 . 0323 as appropriate. This equation is the same as Eq. (14-44). Ans. (b) In bending W t = ³ σ all S F FK x J P d K o K v K s K m ´ 11 = ³ s at S F K L K T K R FK x J P d K o K v K s K m ´ 11 (1) In wear ³ s ac C L C U S H K T C R ´ 22 = C p ³ W t K o K v K m C s C xc Fd P I ´ 1 / 2 22 Squaring and solving for W t gives W t = ³ s 2 ac C 2 L C 2 H S 2 H K 2 T C 2 R C 2 P ´ 22 ³ Fd P I K o K v K m C s C xc ´ 22 (2) Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing
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Unformatted text preview: that C R = µ K R and P d d P = N P , we obtain ( s ac ) 22 = C p ( C L ) 22 ¶ S 2 H S F ( s at ) 11 ( K L ) 11 K x J 11 K T C s C xc C 2 H N P K s I For equal W t in bending and wear S 2 H S F = ( √ S F ) 2 S F = 1 So we get ( s ac ) G = C p ( C L ) G C H ¶ ( s at ) P ( K L ) P J P K x K T C s C xc N P I K s Ans ....
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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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