Advanced strength and applied stress

This question is from Advanced Strength and Applied Stress Analysis, Second edition by Richard G Budynas.The partial answer for this question at the back of the book was:u = 2.87 kJ/m^3The correct solution should match the answers above.6.3For the stress matrix shown with E = 210 GPa and v = 0.3 show that thestenergy per unit volume using Eq. (6.2-11) gives the same result :using the principal stresses in Eq. (6.2-12).ult as that20 10 10[o] = 10 20 10MPa10 10 20- 2E lot + of + of – 2op, to.p; + op,) + 21 + v)i, +ig+ )[6.2-11]| * See Appendix ! for matrix nolation6.2STRAIN ENERGY413Equation (6.2-11) can be written in terms of the principal stresses of2, 02, and 03,which is basically the same element transformed to eliminate the shear stresses.Thus, using Eq. (6.2-5), the strain energy per unit volume for the general case canalso be expressed as” =35 [of + of + of – 21(0102 + 0203 + 0301)] [6.2-12]

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