29_ch 08 Mechanical Design budynas_SM_ch08

# 29_ch 08 Mechanical Design budynas_SM_ch08 - 15< 5 46 C...

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232 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Bearing on member: A b = td = 10(10) = 100 mm 2 σ b = 7(10 3 ) 100 =− 70 MPa n =− S y σ = 370 70 = 5 . 29 Strength of member At A , M = 1 . 4(200) = 280 N · m I A = 1 12 [10(50 3 ) 10(10 3 )] = 103 . 3(10 3 )mm 4 σ A = Mc I A = 280(25) 103 . 3(10 3 ) (10 3 ) = 67 . 76 MPa n = S y σ A = 370 67 . 76 = 5 . 46 At C , M = 1 . 4(350) = 490 N · m I C = 1 12 (10)(50 3 ) = 104 . 2(10 3 )mm 4 σ C = 490(25) 104 . 2(10 3 ) (10 3 ) = 117 . 56 MPa n = S y σ C = 370 117 . 56 = 3 .
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Unformatted text preview: 15 < 5 . 46 C more critical n = min(2 . 72, 5 . 29, 3 . 15) = 2 . 72 Ans. 8-45 F s = 3000 lbf P = 3000(3) 7 = 1286 lbf H = 7 16 in l = 1 2 + 1 2 + . 095 = 1 . 095 in L ≥ l + H = 1 . 095 + (7 / 16) = 1 . 532 in 1 2 " 1 2 " 1 3 4 " l 3000 lbf F s P O 3" 7" 3" Pivot about this point...
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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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