HW04 Solutions2011

HW04 Solutions2011 - ME 382 Fall 2011 HW04 Solutions...

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Unformatted text preview: ME 382 Fall 2011 HW04 Solutions Problem 1: Dowling 7.8 Given: = − 150 = 30 = − 45 SF=3.0 1 , 2 = + 2 ¡ ± ¢ £ − 2 ¤ 2 + 2 Plugging in given values gives: 1 = 165 , 2 = 15 , 3 = 0 (a) Tresca (maximum shear stress) criterion: ¥ = max(| σ 1 − σ 2 |, | σ 2 − σ 3 |, | σ 3 − σ 1 |) | σ 1 − σ 2 | = 150 MPa | σ 2 − σ 3 | = 15 MPa | σ 3 − σ 1 | = 165 MPa ¥ = 165 ¥ = = ¥ = (3.0)(165 ) = (b) von Mises (octahedral shear stress) criterion: ¦ ¦ = 1 √ 2 § ( σ 1 − σ 2 ) 2 + ( σ 2 − σ 3 ) 2 + ( σ 3 − σ 1 ) 2 ¦ ¦ = 1 √ 2 § (150) 2 + (15) 2 + (165) 2 = 158 ¦ ¦ = = ¦ ¦ = (3.0)(158 ) = ME 382 Fall 2011 HW04 Solutions Problem 2: Dowling 7.10 = 190 × 10 −6 = ¡ 760 × 10 −6 = 300 × 10 −6 = = = 0 From Table 4.2 and 5.2: σ Y = 260 MPa, E = 203 GPa and ν =0.293 . Use von Mises criterion. 3D Hooke’s Law (with = 0 ): = 1 ¡ ¢ = 1 ¡ ¢ Plugging in known values: 0.00019(203,000 ) = ¡ 0.293 ¡ 0.00076(203,000 ) = ¡ 0.293 Solving simultaneously (2 equations, 2 unknowns) gives: = ¡ 7.26 and = ¡ 156.4 For an isotropic material: = = 2(1 ¡ ) Plugging in known values: = 203 2(1 ¡ 0.293) = 78.5 = (78,500 )(0.0003) = 23.55 1 , 2 = £ + 2 ¤ ± ¥ ¦ ¡ 2 § 2 + 2 = ¡ 3.63 , ¡ 160 ¨ ¨ = 1 √ 2 © ( σ 1 ¡ σ 2 ) 2 + ( σ 2 ¡ σ 3 ) 2 + ( σ 3 ¡ σ 1 ) 2 = 158.2 = ¨¨¨ = 260 158.2 = . ME 382 Fall 2011 HW04 Solutions Problem 3: Dowling 7.19 Circular tube with axial load and torque: P=60kN T=1.0 kNm r i =23 mm From Table 4.2: σ Y = 469 MPa. Use von Mises criterion....
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This note was uploaded on 11/22/2011 for the course ME ME382 taught by Professor Dowling during the Fall '11 term at University of Michigan.

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HW04 Solutions2011 - ME 382 Fall 2011 HW04 Solutions...

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