Question

The principal stresses at a point are $2\sigma$ (tensile) and $\sigma$ (compressive), and the stress at elastic limit for the material in simple tension is 210 $N/m{m}^2$. According to maximum shear strain theory, the value of $\sigma$ at failure is

• A. 70 $N/m{m}^2$
• B. 105 $N/m{m}^2$
• C. 140 $N/m{m}^2$
• D. 210 $N/m{m}^2$

Answer 1

The correct option is A 70 $N/m{m}^2$

There is no theory like maximum shear strain theory. However if we use maximum shear stress theory.

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

Here, ${\sigma }_1=2\sigma$

${\sigma }_2=-\sigma$

${\sigma }_3=0$

$\therefore {\tau }_{max,absolute}=\frac{{\sigma }_1-{\sigma }_2}{2}=1.5\sigma$

$\therefore 1.5\sigma \le \frac{210}{2\times FOS}$

$\therefore \sigma =\frac{210}{1.5\times 2}[FOS=1]$

$=70N/m{m}^2$