Normal and resultant stress on 60 degree plane

Reading of the figure: $\sigma_x = 150$ MN/m² (left/right faces), $\sigma_y = 100$ MN/m² (top/bottom faces), $\tau_{xy} = 60$ MN/m². The plane makes $60^\circ$ with the horizontal, so its normal makes $60^\circ$ with the vertical. Measured anticlockwise from the $x$-axis, the plane is at $60^\circ$ and its normal is at $\theta = 150^\circ$ (equivalent to $-30^\circ$).

Normal stress

$$\begin{aligned} \sigma_n &= \frac{150+100}{2} + \frac{150-100}{2}\cos 300^\circ + 60\sin 300^\circ \\ &= 125 + 25(0.5) + 60(-0.866) = 85.54\ \text{MN/m}^2 \end{aligned}$$

Shear stress

$$\tau = -\frac{150-100}{2}\sin 300^\circ + 60\cos 300^\circ = -25(-0.866) + 60(0.5) = 51.65\ \text{MN/m}^2$$

Resultant stress and direction

$$\sigma_R = \sqrt{85.54^2 + 51.65^2} = 99.92\ \text{MN/m}^2$$ $$\tan\phi = \frac{\tau}{\sigma_n} = \frac{51.65}{85.54} \Rightarrow \phi = 31.12^\circ$$

The resultant makes $31.12^\circ$ with the normal to the plane (toward the shear direction), i.e. $58.88^\circ$ with the plane itself.

        resultant 99.92
         ^      
        /  \  
       /31.1\   normal sn = 85.54
      /  deg \ --->
   --/--------\--  plane (60 deg to horizontal)
   shear tau = 51.65 acts along the plane

Answer: $\sigma_n = 85.54$ MN/m², $\tau = 51.65$ MN/m², $\sigma_R = 99.92$ MN/m² at $31.12^\circ$ to the normal.