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(| 1 2 |, | 2 3 |, | 3 1 |) | 1 2 | = 150 MPa | 2 3 | = 15 MPa | 3 1 | = 165 MPa = 165 = = = (3.0)(165 ) = (b) von Mises (octahedral shear stress) criterion: = 1 2 ( 1 2 ) 2 + ( 2 3 ) 2 + ( 3 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: Y = 260 MPa, E = 203 GPa and =0.293 . Use von Mises criterion. 3D Hookes 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 ( 1 2 ) 2 + ( 2 3 ) 2 + ( 3 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: Y = 469 MPa. Use von Mises criterion....
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HW04 Solutions2011 - ME 382 Fall 2011 HW04 Solutions...

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