Question

In a plane stress problem, there are normal tensile stresses ${\sigma }_{x}and{\sigma }_{y}with{\sigma }_x>{\sigma }_y,$ accompanied by shear stress${\tau }_{xy}$ at a point in the x-plane. If it is observed that the minimum principal stress on a certain section is zero, then

• A. ${\tau }_{xy}=\sqrt{{\sigma }_x\cdot {\sigma }_y}$
• B. ${\tau }_{xy}=\sqrt{\frac{{\sigma }_x}{{\sigma }_y}}$
• C. ${\tau }_{xy}=\sqrt{{\sigma }_x-{\sigma }_y}$
• D. ${\tau }_{xy}=\sqrt{{\sigma }_x+{\sigma }_y}$

Answer 1

The correct option is A

${\tau }_{xy}=\sqrt{{\sigma }_x\cdot {\sigma }_y}$

The magnitude of principal stress ${\sigma }_{1}and{\sigma }_2$ are given by

${\sigma }_{1,2}=\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$
It is given that minimum principal stress on a section is zero.
Therefore, ${\sigma }_2$ = 0

$\Rightarrow \left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2=0}$

$\Rightarrow \left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$
Squaring both sides

$\Rightarrow {\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)}^2={\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)}^2+{\left({\tau }_{xy}\right)}^2$

$\Rightarrow {\tau }_{xy}=\sqrt{{\sigma }_x.{\sigma }_y}$