1. (25 pts) In class, we have shown that the compliance matrix of a transversely isotropic material can be written as, U12 1 o 0 0 12 U23 1 trans,iso 23 0 0 12 12 where G23 2(+U23) If we invert this compliance matrix, [siyransis, and compare it with the elastic stiffness matrix, [ctrans/lso], we can express the elements in [soing the engineering moduli, E1, E2, v12, G12, and G23. Please express Cij in terms of the engineering moduli E1, E2, U12, G12, and G23. Note that the stiffness matrix of a transversely isotropic material is in a general form of trans,iso C C,2C,2 0 0 0 12 C12 C22 C23 0 0 0 [citransiso-102 23 22 0 0 00 C440 0 000 C66 0 0 0 0 0 C66