Question 4.3: Consider Figure 4.11(c) and take σx = 350 kg/cm² and σy = −7...
Consider Figure 4.11(c) and take \sigma_x=350 kg / cm ^2 \text { and } \sigma_y=-700 kg / cm ^2 (compressive). Find out the value of \phi defining the plane on which no normal stress acts. What is the magnitude of shear stress on that plane?
We know that
\sigma_{\text {average }}=\frac{1}{2}\left(\sigma_x+\sigma_y\right)
Here, \sigma_x=350 kg / cm ^2 \text { and } \sigma_y=-700 kg / cm ^2 . Therefore,
\begin{aligned} \sigma_{\text {avcrage }} & =\frac{1}{2}(350-700)=-175 kg / cm ^2 \\ \tau_{\max } & =\frac{1}{2}\left(\sigma_x-\sigma_y\right)=\frac{1}{2}(350+700)=525 kg / cm ^2 \end{aligned}
Now, \sigma_n is normal stress given by
\sigma_n=\sigma_{\text {average }}+\tau_{\max } \cos 2 \phi=-175+525 \cos 2 \phi
Equating \sigma_n \text { with } \theta=0^{\circ} \text {, we get }-175+525 \cos 2 \phi=0 \text {. Therefore, } \phi=35.26^{\circ} .
Thus, the plane on which normal stress does not act is making an angle \phi=35.26^{\circ} , shear stress on the same plane is given by
\tau_n=\tau_{\max } \sin 2 \phi
Putting \tau_{\max }=525 kg / cm ^2 \text { and } \phi=35.26^{\circ}, \text { we get }
\tau_n=525\left(\sin 2 \times 35.26^{\circ}\right)=494.9 kg / cm ^2
Therefore, value of \phi defining the angle is 35.26° and shear stress is 494.9 kg/cm².