Resultant stress on planes EF and GE, principal stresses

Assumption: the 30 MN/m² on plane GE is the normal (tensile) stress on it, as in the figure. Shear stresses on two perpendicular planes are equal in magnitude, so the shear on GE equals the shear on EF.

Components on plane EF

The 50 MN/m² resultant is at $30^\circ$ to the normal of EF:

$$\sigma_{EF} = 50\cos 30^\circ = 43.30\ \text{MN/m}^2, \qquad \tau_{EF} = 50\sin 30^\circ = 25.00\ \text{MN/m}^2$$

On the perpendicular plane GE: $\sigma_{GE} = 30$ and $\tau_{GE} = 25.00$ MN/m².

So $\sigma_x = 43.30$, $\sigma_y = 30$, $\tau_{xy} = 25.00$ (all in MN/m²).

(i) Resultant stress on plane GE

$$\sigma_R = \sqrt{30^2 + 25.00^2} = 39.05\ \text{MN/m}^2, \qquad \tan\alpha = \frac{25.00}{30} \Rightarrow \alpha = 39.81^\circ\ \text{to the normal of GE}$$

(ii) Principal stresses and planes

$$\sigma_{1,2} = \frac{43.30+30}{2} \pm \sqrt{\left(\frac{43.30-30}{2}\right)^2 + 25.00^2} = 36.65 \pm 25.87$$ $$\sigma_1 = 62.52\ \text{MN/m}^2, \qquad \sigma_2 = 10.78\ \text{MN/m}^2$$ $$\tan 2\theta_p = \frac{2(25)}{43.30-30} \Rightarrow 2\theta_p = 75.10^\circ,\ \theta_p = 37.55^\circ$$

$\sigma_1$ acts on the plane whose normal is at $37.55^\circ$ to the normal of EF; $\sigma_2$ acts on the plane at $127.55^\circ$ (both planes are at right angles).

(iii) Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = 25.87\ \text{MN/m}^2$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $82.55^\circ$ and $172.55^\circ$ from the normal of EF, with a normal stress of $36.65$ MN/m² on them.