Express the stresses on plane P in terms of Eqs (2.1) and (2.2).

\sigma _{x^{\prime}}=\frac{\sigma _x+\sigma _y}{2}+\frac{\sigma _x-\sigma _y}{2}\cos 2\theta -\tau _{xy}\sin 2 \theta           (2.1b)

\tau _{x^{\prime}y^{\prime}}=\frac{\sigma_x- \sigma_y}{2}\sin 2\theta +\tau _{xy}\cos 2\theta                  (2.2)

67.2= \frac{-61.35+\sigma_y}{2}+\frac{-61.35-\sigma_y}{2} \cos 2 \theta

-68.57 \sin 2 \theta             (i)

 

84.64= \frac{-61.35-\sigma_y}{2} \sin 2 \theta+68.57 \cos 2 \theta            (ii)

where \sigma_y = normal stress on the plane perpendicular to Q,

\theta = angle between the planes P and Q.

From Eq. (i) and (ii),

\tan \theta=-\frac{128.55}{153.21}

or, \theta \approx-40^{\circ} ,  i.e. plane P is at 40^{\circ} clockwise to plane Q.

Substituting in Eq. (iii), \sigma_y=86.36  MPa.

Angle of inclination of plane R w.r.t. Q,

2 \theta=2 \times\left(-20^{\circ}\right)=-40^{\circ}

Normal stress on plane R,

\sigma_{\acute{x} }= \frac{-61.35+86.36}{2}-\frac{61.35+86.36}{2} \times \cos \left(-40^{\circ}\right)-68.57 \sin \left(-40^{\circ}\right) \simeq 0

 

\cos \left(-40^{\circ}\right)-68.57 \sin \left(-40^{\circ}\right) \simeq 0

 

\tau_{\acute{x} \acute{y}}= \frac{-61.35-86.36}{2} \sin \left(-40^{\circ}\right)+68.57 \cos \left(-40^{\circ}\right)=100  MPa