Question

At a point in a piece of stressed material, the stresses are ${\sigma }_x=\alpha kN/m^2$ tensile (normal)

${\tau }_{xy}={\tau }_{yx}=\beta kB/m^2$(shear)

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

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

Answer 1

The correct option is C zero

The radius of Mohr's circle is given by = R = ${\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

In this case principal stresses are equal ${\sigma }_1={\sigma }_2=5kN/m^2$

Shear stress on principal planes is zero

$R=\frac{{\sigma }_1-{\sigma }_2}{2}=\frac{5-5}{2}=0$

$\therefore$ It means Mohr's circle becomes a point circle