ME 4733: Deformation and Fracture of Engineering Materials Spring 2002 Problem Set 2 Solutions 1*) Prove that for the case of cubic crystals the modulus of elasticity in any given direction may be given by the following equation in terms of the three independent elastic constant s 11 , s 12 , s 44 , and the direction cosines of the crystallographic direction under study: 1 2 1 2 11 11 12 44 1 2 2 2 2 2 3 2 3 2 1 2 E s s s s ll ll ll = − − − + + ( ) ( ) where l 1 , l 2 , l 3 are direction cosines. Solution: As shown in the following figure, the axis direction is n, the three orthogonal crystal directions are e x = [ ] 100 , e y = [ ] 010 and e z = [ ] 001 . The three direction cosines are l i i = ⋅ ne , and n can be expressed in the crystal coordinates as n e = l i i . The stress state is σ = = σ σ nn ee ll i j i j , where σ is the applied tensile stress. The generated strain is ε = ε ij i j ee . The normal tensile strain in axis n direction is ε ε = ⋅ ⋅ = n n ε l l i ij j For cubic crystal, the elastic compliance matrix gives ε σ σ σ 11 11 11 12 22 12 33 = + + s s s
ME 4733: Deformation and Fracture of Engineering Materials
Jun 21, 2023
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