A point on a machine component has the following state of plane strain:

ε_{xx }= (-) 720 μ, ε_{yy} = (-) 400 mu, ε_{xy} = 660 mu, (μ = 10^{-6})

If E = 70 GPa and G = 28 GPa, then the three principal stresses (σ_{1}, σ_{2}, σ_{3} in MPa) are

This question was previously asked in

HPPSC AE Civil 2016 (Deputy Manager) Official Paper

Option 3 : -42.18, -83.26, -31.36

**Calculation:**

Given,

εxx = -720 × 10^{-6} , εyy = -400 × 10-6, εxy = 660 × 10-6

E = 70 GPa, G = 28 GPa

∵ We know that relation between E and G,

E = 2G (1 + μ)

70 = 2 × 28 (1 + μ)

μ = 0.25

Principal strain is given by,

\({\varepsilon _1}/{\varepsilon _2} = \frac{{{\varepsilon _{xx}} + {\varepsilon _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\varepsilon _{xx}} - {\varepsilon _{yy}}}}{2}} \right)}^2} + {{\left( {\frac{{{\varepsilon _{xy}}}}{2}} \right)}^2}} \)

\({\varepsilon _1}/{\varepsilon _2} = \left[ {\frac{{ - 700 - 400}}{2} \pm \sqrt {{{\left( {\frac{{ - 700 + 400}}{2}} \right)}^2} + {{\left( {\frac{{660}}{2}} \right)}^2}} } \right] × {10^{ - 6}}\)

ϵ_{1} = -193.26 × 10^{-6}

ϵ_{2} = -926.74 × 10^{-6}

ϵ_{3} = 0

∵ We know that, relation between principal strain and principal stress

\({\varepsilon _1} = \frac{{{\sigma _1}}}{E} - \frac{{\mu \left( {{\sigma _2} + {\sigma _3}} \right)}}{E}\)

\({\varepsilon _2} = \frac{{{\sigma _2}}}{E} - \frac{{\mu \left( {{\sigma _1} + {\sigma _3}} \right)}}{E}\)

\({\varepsilon _3} = \frac{{{\sigma _3}}}{E} - \frac{{\mu \left( {{\sigma _1} + {\sigma _2}} \right)}}{E}\)

To save time, put the values in options in the above equations

So options 3 (-42.18, -83.26, -31.36) is correct.