The constants  and  can be determined in terms of   by solving Equations (7.244), (7.247), the first of Equations(7.248) and (7.249). These are

(7.250)

At last, at the outer boundary the shear stress  must match the shear stress  in coordinate system. Thus, at , the shear stress then becomes

(7.251)

The right hand side of above equation can be written as

(7.252)

Thus, the equivalent axial shear modulus can be given combining Equation (7.251) and Equation (7.252). Then values of constants   and   are substituted from Equation (7.250). Thus, we get the result

(7.253)