Question

At a point in an elastic material there are two mutually perpendicular planes, one of which carries a direct tensile stress at 50 N/mm2 and a shear stress of 40 N/mm2, while the other plane is subjected to a direct compressive stress of 35 N/mm2 and a complementary shear stress of 40 N/mm2.

Determine the principal stresses at the point, the position of the planes on which they act and the position of the planes on which there is no normal stress.

Answer 1

$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\frac{\sigma_x-\sigma_y}{2}+{\tau_{xy}}^2}$

$\sigma_{1,2}=\frac{50+(-35)}{2}\pm\sqrt{\frac{50-(-35)}{2}+{40}^2}$

$\sigma_{1,2}=\frac{50-35}{2}\pm\sqrt{\frac{85}{2}+{40}^2}$

$\sigma_{1,2}=7.5\pm 58.363$

$\therefore \sigma_1=65.863MPa$

$\sigma_2=-50.863MPa$

$Tan\ 2\theta p=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}$

$2\theta p=Tan^{-1}\frac{2\tau_{xy}}{\sigma_x-\sigma_y}$

$\theta p=\frac{1}{2}Tan^{-1}\frac{2\tau_{xy}}{\sigma_x-\sigma_y}$

$\theta p=\frac{1}{2}Tan^{-1}\frac{2*40}{50-(-35)}$

$\theta p=\frac{1}{2}Tan^{-1}\frac{80}{85}$

=21.263° Which is position of principal plane

$\theta_s=\theta_p+45$

=21.263+45

=66.263° which is the position of shear plane