Question 4.2: Construct Mohr’s circle for the case of (a) simple tension w...
Construct Mohr’s circle for the case of
(a) simple tension where \sigma_y=0 [Figure 4.10(a)];
(b) biaxial stress where \sigma_x is tension, \sigma_y compression and \left|\sigma_y\right|=2 \mid \sigma_x [Figure 4.10(b)];
(c) biaxial stress where \sigma_y=-\sigma_x [Figure 4.10(c)].
(a) Here \sigma_y=0 . We know that
\sigma_{\text {average }}=\frac{1}{2}\left(\sigma_x+\sigma_y\right)=\frac{\sigma_x}{2}
by putting \sigma_y=0 . Similarly,
\tau_{\max }=\frac{1}{2}\left(\sigma_x-\sigma_y\right)=\frac{\sigma_x}{2}
So for the Mohr’s circle, coordinates of centre are \left(\sigma_x / 2,0\right) \text { and radius is } \sigma_x / 2 . This can be drawn as shown in Figure 4.11(a).
(b) Here \left|\sigma_y\right|=2\left|\sigma_x\right| ; \sigma_y \text { is compressive, } \sigma_x is tensile. Therefore,
\begin{aligned} \sigma_{\text {averagc }} & =\frac{1}{2}\left(\sigma_x+\sigma_y\right)=\frac{1}{2}\left(\sigma_x-2 \sigma_x\right)=-\frac{\sigma_x}{2} \\ \tau_{\max } & =\frac{1}{2}\left(\sigma_x-\sigma_y\right)=\frac{1}{2}\left(\sigma_x+2 \sigma_x\right)=\frac{3 \sigma_x}{2} \end{aligned}
So the Mohr’s circle has its centre at \left(-\sigma_x / 2,0\right) \text { and radius }\left(3 \sigma_x / 2\right) . This is shown in Figure 4.11(b).
(c) Here \sigma_y=-\sigma_x , Therefore,
\begin{aligned} \sigma_{\text {average }} & =\frac{1}{2}\left(\sigma_x+\sigma_y\right)=0 \\ \tau_{\max } & =\frac{1}{2}\left(\sigma_x-\sigma_y\right)=\frac{1}{2}\left(2 \sigma_x\right)=\sigma_x \end{aligned}
So the derived Mohr’s circle has its centre at (0,0) and radius \sigma_x . This is shown in Figure 4.11(c).
