a) Derive the following relations involving engineering stress (6), engineering strain (€), true stress () and true stra

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A Derive The Following Relations Involving Engineering Stress 6 Engineering Strain True Stress And True Stra 1 (34.98 KiB) Viewed 24 times

a) Derive the following relations involving engineering stress (6), engineering strain (€), true stress () and true strain (@). (i) § = ln(1 + E) (ii) ẽ = In 42 A (iii) Q = 0(1+E) (b) Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equation 6.5 for elastic deformation, that the modulus of elasticity is 103 GPa, and that elastic deformation terminates at a strain of 0.007. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 1520 MPa and 0.15, respectively. Furthermore, plastic deformation occurs between strain values of 0.007 and 0.60, at which point fracture occurs.