Show that for an isotropic homogeneous k the stiffness matri

Show that for an isotropic, homogeneous k, the stiffness matrix can be written K^e = k/2 J^e[(y_23^e)^2 + (x_32^e)^2 y_23^ey_31^2 + x_32^ex_13^e y_23^ey_12^e + x_32^ex_21^e y_23^ey_31^e + x_32^ex_13^e (y_31^e)^2 + (x_13^e)^2 y_31^e y_12^e + x_13^ex_21^e y_23^ey_12^e + x_32^ex_21^e y_31^ey_12^e + x_13^ex_21^e (y_12^e)^2 + 9x_21^e)^2] Since we know k^e will be symmetric, only calculate the upper right 6 entries.

Solution

Compliance matrix for an isotropic elastic material From experiments one finds: 11 = 1 E h 11 22 + 33 i 22 = 1 E h 22 11 + 33 i 33 = 1 E h 33 11 + 22 i 223 = 23 G , 213 = 13 G , 212 = 12 G (3.49) In these expressions, E is the Young’s Modulus, the Poisson’s ratio and G the shear modulus. They are referred to as the engineering constants, since they are obtained from experiments. The shear modulus G is related to the Young’s modulus E and Poisson ratio by the expression G = E 2(1+) . Equations (3.49) can be written in the following matrix form: 3.6. ISOTROPIC LINEAR ELASTIC MATERIALS 63 11 22 33 223 213 212 = 1 E 1 0 0 0 1 0 0 0 1 0 0 0 2(1 + ) 0 0 symm 2(1 + ) 0 2(1 + ) 11 22 33 23 13 12 (3.50) Invert and compare with: 11 22 33 23 13 12 = + 2µ 0 0 0 + 2µ 0 0 0 + 2µ 0 0 0 µ 0 0 symm µ 0 µ 11 22 33 223 213 212 (3.51) and conclude that: = E (1 + )(1 2) , µ = G = E 2(1 + ) (3.52)