2_ch 05 Mechanical Design budynas_SM_ch05

2_ch 05 Mechanical Design budynas_SM_ch05 - A B + 2 B ) 1 /...

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116 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-2 S y = 50 kpsi MSS: σ 1 σ 3 = S y / n n = S y σ 1 σ 3 DE: ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 = S y / n n = S y / ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 (a) MSS: σ 1 = 12 kpsi, σ 3 = 0, n = 50 12 0 = 4 . 17 Ans. DE: n = 50 [12 2 (12)(12) + 12 2 ] 1 / 2 = 4 . 17 Ans. (b) MSS: σ 1 = 12 kpsi, σ 3 = 0, n = 50 12 = 4 . 17 Ans . DE: n = 50 [12 2 (12)(6) + 6 2 ] 1 / 2 = 4 . 81 Ans. (c) MSS: σ 1 = 12 kpsi, σ 3 =− 12 kpsi, n = 50 12 ( 12) = 2 . 08 Ans. DE: n = 50 [12 2 (12)( 12) + ( 12) 2 ] 1 / 3 = 2 . 41 Ans. (d) MSS: σ 1 = 0, σ 3 =− 12 kpsi, n = 50 ( 12) = 4 . 17 Ans. DE: n = 50 [( 6) 2 ( 6)( 12) + ( 12) 2 ] 1 / 2 = 4 . 81 5-3 S y = 390 MPa MSS: σ 1 σ 3 = S y / n n = S y σ 1 σ 3 DE: ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 = S y / n n = S y / ( σ 2 A
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Unformatted text preview: A B + 2 B ) 1 / 2 (a) MSS: 1 = 180 MPa, 3 = 0, n = 390 180 = 2 . 17 Ans. DE: n = 390 [180 2 180(100) + 100 2 ] 1 / 2 = 2 . 50 Ans. (b) A , B = 180 2 180 2 2 + 100 2 = 224 . 5, 44 . 5 MPa = 1 , 3 MSS: n = 390 224 . 5 ( 44 . 5) = 1 . 45 Ans. DE: n = 390 [180 2 + 3(100 2 )] 1 / 2 = 1 . 56 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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