Question

At a critical point in a component, the state of stress is given as ${\sigma }_{xx}=100MPa,{\sigma }_{yy}=220MPa,{\sigma }_{xy}={\sigma }_{yx}=80$ MPa and all other stress components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is
_____ (round off to one decimal place).

• A. 1.8

Answer 1

The correct option is A 1.8
As per given data

${\sigma }_{xx}=100MPa,{\sigma }_{yy}=220MPa$

${\sigma }_{xy}={\sigma }_{yx}=80MPa$

Principal stresses,
${\sigma }_{1,2}=\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\sigma }_{xy}^2}$

$=\left(\frac{100+220}{2}\right)\pm \sqrt{{\left(\frac{100-220}{2}\right)}^2+{80}^2}$

$=160\pm 100$

${\sigma }_1=160+100=260MPa$

${\sigma }_2=160-100=60MPa$

Both principal stresses are like in nature.

According to maximum shear stress theory.

Maximum shear stress

$=[max\left(\frac{{\sigma }_1-{\sigma }_2}{2}\right),\left(\frac{{\sigma }_1}{2}\right),\left(\frac{{\sigma }_2}{2}\right)]$

${\tau }_{max}=\frac{{\sigma }_1}{2}$

$\frac{{\sigma }_{yt}}{FOS\times 2}=\frac{{\sigma }_1}{2}$

$FOS=\frac{{\sigma }_{yt}}{{\sigma }_1}=\frac{468}{260}=1.8$