Question 4.5: Two mutually perpendicular planes of an element are subjecte...

Here $\sigma_x=10.5  MPa , \sigma_y=-3.5  MPa \text { and } \tau_{x y}=7.0  MPa$ . The principal stresses are given by

$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left\lgroup \frac{\sigma_x-\sigma_y}{2} \right\rgroup^2+\tau_{x y}^2}$

Substituting the given values, we get

$\begin{aligned} \sigma_{1,2} & =\frac{10.5-3.5}{2} \pm \sqrt{\left\lgroup \frac{10.5+3.5}{2}\right\rgroup^2+7.0^2} \\ & =3.5 \pm \sqrt{7^2+7^2}=3.5 \pm 9.9  MPa \end{aligned}$

Therefore, $\sigma_1=13.4  MPa \text { and } \sigma_2=-6.4  MPa$. Hence, the maximum principal stress is 13.4 MPa and minimum is 6.4 MPa (compressive).
The directions of the principal planes are given by

$\tan 2 \theta=\frac{2 \tau_{x y}}{\sigma_x-\sigma_y}$

or        $2 \theta=45^{\circ} \text { or } 225^{\circ}$

Therefore, $\theta_1=22^{\circ} 30^{\prime} \text { and } \theta_2=112^{\circ} 30^{\prime}$