Question

Principal stresses at a point are 80 $N/m{m}^2$ and 40 $N/m{m}^2$, both tensile. The yield stess in simple tension for this material is 200 $N/m{m}^2$. The value of factors of safety according to maximum principal stress theory and maximum shear stress theory, respectively, are

• A. 2.5 and 2.5
• B. 2.5 and 5
• C. 5 and 5
• D. 5 and 1.67

Answer 1

The correct option is A 2.5 and 2.5

${\sigma }_1=80N/m{m}^2$

${\sigma }_2=40N/m{m}^2$

${\sigma }_3=0$
According to maximum principal stress theory,

${\sigma }_1\le \frac{{\sigma }_y}{FOs}$

$FOS=\frac{200}{80}=2.5$
According to maximum shear stress theory

${\tau }_{max,absolute}\le \frac{{\sigma }_y}{2\times FOS}$

$\therefore \frac{{\sigma }_1-{\sigma }_3}{2}\le \frac{{\sigma }_y}{2\times FOS}$

$\frac{80-0}{2}=\frac{200}{2\times FOS}$

$\therefore FOS=2.5$