Problem 10-53.  Solution 2.

In torsion, s_x = 0, s_y = 0, and s_z = 0.  By the stress-strain relations e_x = 0 and e_y = 0.  The strain transformation formulas with q = 30 deg. yield

e_x' = e_xcos²q + e_ysin²q + g_xysinqcosq  = g_xysqrt(3)/4

e_y' = e_xsin²q + e_ycos²q - g_xysinqcosq   = -g_xysqrt(3)/4

With e_x' = - e_y' = - 45(10^-6), the two equations are consistent, and they yield

 g_xy = - 45(10^-6)(4)/sqrt(3) = -1.0392(10^-4)

 t_xy = Gg_xy = -7.794 MPa

Because of the orientation of the (x, y) axes we have

t = t_rq = - t_xy = 7.794 MPa

This is the same value as for  t  in Solution 1.  The rest of the solution is the same as before.

Problem 10-53. Solution 3.

Since s_z = 0, we can use the stress-strain relations of Plane Stress:

s_x' = [E/(1-n²)](e_x' + e_y')

s_y' = [E/(1-n²)](e_x' + e_y')

With E = 200(10³) MPa, n = 0.32, and  e_x' = - e_y' = - 45(10^-6), obtain

s_x' = - s_y' = - 6.8182 MPa

With s_x = 0, s_y = 0, and q = 30 deg., the stress transformation formulas yield

s_x' = s_xcos²q + s_ysin²q + 2t_xysinqcosq  = t_xysqrt(3)/2

s_y' = s_xsin²q + s_ycos²q - 2t_xysinqcosq   = -t_xysqrt(3)/2

Thus

t_xy = 2s_x'/sqrt(3) = - 7.873 MPa

This differs slightly from the previous solution in which t_xy = - 7.794 MPa.  The reason for this is that the given values for E, n, and G are slightly inconsistent with a truly isotropic material. With E = 200 GPa, and n = 0.32, the isotropic G would be

G_iso = E/2(1 + n) = 75.76 GPa

The given G is 75 GPa.