Question

At a point in a piece of stressed material the stresses are :
${\sigma }_x=\alpha KN/m^2$

${\tau }_{xy}={\tau }_{yx}=\beta KN/m^2$

Although the values of $\alpha$ and $\beta$ are not known yet the principal stresses are equal to each other being $\left(5kN/m^2\right).$ What is the radius of Mohr's circle?

• A. $2.5+(\alpha +\beta )$
• B. $2.5+\frac{(\alpha +\beta )}{2}$
• C. zero
• D. 2.5

Answer 1

The correct option is C zero

In general, the radius of Mohr's circle is given by

$r=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left({\tau }_{xy}\right)}^2}$
In this case principal stresses are equal i.e.${\sigma }_1={\sigma }_2=5kN/m^2.$ The shear stress on principal planes is also zero.${\tau }_{xy}=0$

$\therefore r=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+0}$

$\Rightarrow r=\frac{{\sigma }_x-{\sigma }_y}{2}=\frac{5-5}{2}$

$\Rightarrow r=0$

It means that when ${\sigma }_1={\sigma }_2,$ Mohr's circle degenerates into a point and no shear stresses at all develop in the xy plane.