Question

For a case of plane stress, ${\sigma }_x=40MN/m^2,{\sigma }_y=0,{\tau }_{xy}=80MN/m^2.$ What are the principal stresses$\left(inMN/m^2\right)$ and their orientation with x and y axes?

• A. ${\sigma }_1=80,{\sigma }_2=40,{\theta }_1=30{}^{\circ }$
• B. ${\sigma }_1=100,{\sigma }_2=-60,{\theta }_1=32{}^{\circ }$
• C. ${\sigma }_1=102.5,{\sigma }_2=-62.5,{\theta }_1=36{}^{\circ }$
• D. ${\sigma }_1=105,{\sigma }_2=62,{\theta }_1=36{}^{\circ }$

Answer 1

The correct option is C ${\sigma }_1=102.5,{\sigma }_2=-62.5,{\theta }_1=36{}^{\circ }$

Radius of Mohr's circle of stress,

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

$=\sqrt{{\left(20\right)}^2+{\left(80\right)}^2}$

$20\sqrt{17}MN/m^2$
The centre of Mohr's circle,

$C=\frac{{\sigma }_x+{\sigma }_y}{2}=\frac{40+0}{2}=20MN/m^2$

Major principal stress $=C+r=20(1+\sqrt{17})$

$=102.5MN/m^2$

Minor principal stress $=C-r=20(1-\sqrt{17})$

$=-62.5MN/m^2$

Orientation, $\theta =\frac{1}{2}{\tan }^{-1}\left(\frac{2{\tau }_{xy}}{{\sigma }_x-{\sigma }_y}\right)=38{}^{\circ }$
Although the orientaion given in the options is wrong, however the best choice will be (c).