Question

A delta rosette, as shown in figure, measures micro - strains as follows:
${ϵ}_A=-100\times {10}^{-6},{ϵ}_B=700\times {10}^{-6},{ϵ}_C=600\times {10}^{-6}$
What will be the principal strains, if modulus of elasticity is 200 GPa and Poisson's ratio is 0.3 ?

• A. $260\times {10}^{-6}$ and $540\times {10}^{-6}$
• B. $800\times {10}^{-6}$ and $-200\times {10}^{-6}$
• C. $1020\times {10}^{-6}$ and $420\times {10}^{-6}$
• D. None of these

Answer 1

The correct option is B $800\times {10}^{-6}$ and $-200\times {10}^{-6}$

${ϵ}_x={ϵ}_B=700\times {10}^{-6}$
${ϵ}_y={ϵ}_A=-100\times {10}^{-6}$
${ϵ}_x^{\prime }={ϵ}_C=600\times {10}^{-6}$
and $\theta ={45}^{\circ }$
${ϵ}_x^{\prime }=\frac{{ϵ}_x+{ϵ}_y}{2}+\frac{{ϵ}_x-{ϵ}_y}{2}\cos 2\theta +\frac{{\varphi }_{xy}}{2}\sin 2\theta$
$\Rightarrow 600\times {10}^{-6}=\left(\frac{700-100}{2}\right)\times {10}^{-6}+0+\frac{{\varphi }_{xy}}{2}$
$\Rightarrow {\varphi }_{xy}=600\times {10}^{-6}$
Principal strains, $\frac{{ϵ}_1}{{ϵ}_2}=\frac{{ϵ}_x+{ϵ}_y}{2}\pm \frac{1}{2}\sqrt{({ϵ}_x-{ϵ}_y{)}^2+({\varphi }_{xy}{)}^2}$
$=\frac{700-100}{2\times {10}^6}\pm \frac{\sqrt{(800{)}^2+(600{)}^2}}{2\times {10}^6}$
$\therefore {ϵ}_1=800\times {10}^{-6}$
and, ${ϵ}_2=-200\times {10}^{-6}$