3_ch 05 Mechanical Design budynas_SM_ch05

3_ch 05 Mechanical Design budynas_SM_ch05 -...

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Unformatted text preview: budynas_SM_ch05.qxd 11/29/2006 15:00 FIRST PAGES Page 117 Chapter 5 160 ± (c) σ A , σ B = − 2 160 − 2 2 + 1002 = 48.06, −208.06 MPa = σ1 , σ3 390 = 1.52 Ans. 48.06 − ( −208.06) 390 = 1.65 Ans. n= 2 + 3(1002 )]1/2 [−160 n= MSS: DE: (d) σ A , σ B = 150, −150 MPa = σ1 , σ3 390 = 1.30 Ans. n= MSS: 150 − ( −150) 390 = 1.50 Ans. n= DE: [3(150) 2 ]1/2 5-4 S y = 220 MPa (a) σ1 = 100, σ2 = 80, σ3 = 0 MPa MSS: DET: 220 = 2.20 Ans. 100 − 0 σ = [1002 − 100(80) + 802 ]1/2 = 91.65 MPa 220 n= = 2.40 Ans. 91.65 n= (b) σ1 = 100, σ2 = 10, σ3 = 0 MPa MSS: DET: n= 220 = 2.20 Ans. 100 σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa n= 220 = 2.31 Ans. 95.39 (c) σ1 = 100, σ2 = 0, σ3 = −80 MPa MSS: DE: n= 220 = 1.22 Ans. 100 − ( −80) σ = [1002 − 100( −80) + ( −80) 2 ]1/2 = 156.2 MPa 220 n= = 1.41 Ans. 156.2 (d) σ1 = 0, σ2 = −80, σ3 = −100 MPa 220 n= = 2.20 Ans. MSS: 0 − ( −100) σ = [( −80) 2 − ( −80)( −100) + ( −100) 2 ] = 91.65 MPa DE: n= 220 = 2.40 Ans. 91.65 117 ...
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