Question

A machine element is subjected to the following bi-axial state of stress; ${\sigma }_x=80MPa;{\sigma }_y=20MPa;{\tau }_{xy}=40MPa$. If the shear strength of the materials is 100 MPa, the factor of safety as per Tresca's maximum shear stress theory is

• A. 3.3
• B. 1.0
• C. 2.5
• D. 2.0

Answer 1

The correct option is D 2.0

${\sigma }_{1,2}=\frac{1}{2}[({\sigma }_x+{\sigma }_y)\pm \sqrt{{({\sigma }_x-{\sigma }_y)}^2+4{\tau }_{xy}^2}]$

$=\frac{1}{2}\left[\right(80+20)\pm \sqrt{(60{)}^2+4\times {40}^2}]$

${\sigma }_1=100MPa$

${\sigma }_2=0MPa$

$\left({\tau }_{max}\right)=\frac{{\sigma }_1}{2}=50MPa$

$F\left(Factorofsafety\right)=\frac{{\tau }_{ys}}{{\tau }_{max}}=\frac{100}{50}=2$