Question

The strain measurement from a rectangular strain rosette were ${\epsilon }_0=900\times {10}^{-6},{\epsilon }_{45}=750\times {10}^{-6}$ and ${\epsilon }_{90}=300\times {10}^{-6}$. The magnitude of minimum and maximum principal strain respectively is:

• A. $264.59\times {10}^{-6},935.41\times {10}^{-6}$
• B. $935.41\times {10}^{-6},264.59\times {10}^{-6}$
• C. $623.61\times {10}^{-6},176.39\times {10}^{-6}$
• D. $176.39\times {10}^{-6},623.61\times {10}^{-6}$

Answer 1

The correct option is A $264.59\times {10}^{-6},935.41\times {10}^{-6}$


${\epsilon }_0=900\times {10}^{-6}$

${\epsilon }_{45}=750\times {10}^{-6}$

${\epsilon }_{90}=300\times {10}^{-6}$

We know that,

${\epsilon }_{45}=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\frac{{\epsilon }_x-{\epsilon }_y}{2}cos(2\times 45)+\frac{{\gamma }_{xy}}{2}sin(2\times 45)$

$\Rightarrow 750\times {10}^{-6}=\frac{900\times {10}^{-6}+300\times {10}^{-6}}{2}+0+\frac{{\gamma }_{xy}}{2}\times 1$

$\Rightarrow {\gamma }_{xy}=300\times {10}^{-6}$

$\Rightarrow {\epsilon }_{max/min}=\frac{{\epsilon }_x+{\epsilon }_y}{2}\pm \sqrt{{\left(\frac{{\epsilon }_x-{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$

$=[\frac{900+300}{2}\pm \sqrt{{\left(\frac{900-300}{2}\right)}^2+{\left(\frac{300}{2}\right)}^2}]\times {10}^{-6}$

${\epsilon }_{max}=935.41\times {10}^{-6}{\epsilon }_{min}=264.59\times {10}^{-6}$