Isotropic linear elastic material Most general isotropic 4th order isotropic tensor Ck=A66k+(6k3+66) Replacing in: C (14) (15) Examples 1=A61(11+e+63)+p(e1+n)=(A+2)en+e22+s3(16) 012=2∈12 17 Practice problem: Write the matrix of coefficients C(elastic moduli) for an isotropic material( Voigt form) in Mathematica Compliance matrix for an isotropic elastic materi From experiments one finds 11=E[1 (18) E In these expressions, E is the Youngs Modulus, v the Poisson's ratio and G the shear modulus. They are referred to as the engineering constants 3� � � � � � � � � � � � � � �� � � �� � � �� Isotropic linear elastic materials Most general isotropic 4th order isotropic tensor: Cijkl = λδijδkl + µ δikδjl + δilδjk (12) Replacing in: σij = Cijkl�kl (13) gives: σij = λδij �kk + µ �ij + �ji (14) σij = λδij �kk + µ �ij + �ji (15) Examples σ11 = λδ11 �11 + �22 + �33 + µ �11 + �11 = λ + 2µ �11 + µ�22 + µ�33 (16) σ12 = 2µ�12 (17) Practice problem: Write the matrix of coefficients C (elastic moduli) for an isotropic material (Voigt form) in Mathematica. Compliance matrix for an isotropic elastic material From experiments one finds: σ11 − ν σ22 + σ33 1 E �11 = (18) σ22 − ν σ11 + σ33 1 E �22 = (19) σ33 − ν σ11 + σ22 1 E �33 = (20) σ23 G , 2�13 = σ13 G , 2�12 = σ12 (21) G 2�23 = In these expressions, E is the Young’s Modulus, ν the Poisson’s ratio and G the shear modulus. They are referred to as the engineering constants, 3