When a body is subjected to biaxial stress i.e. direct stresses ( _x ) and ( _y ) in two mutually perpendicular planes accompanied by a simple shear stress ( _xy ), then minimum normal stress is

_x + _y2 - 12sqrt ( _x - _y )^2 + 4 _xy^2

This question belongs to Mechanical Engineering Strength Of Materials In ME

Strength of Materials in ME

When a body is subjected to biaxial stress i.e. direct stresses $$\left( {{\sigma _{\text{x}}}} \right)$$ and $$\left( {{\sigma _{\text{y}}}} \right)$$ in two mutually perpendicular planes accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ then minimum normal stress is

A) $$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$ B) $$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} - \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$ C) $$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$ D) $$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} - \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$

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Correct Answer: Option B

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When a body is subjected to biaxial stress i e direct...

When a body is subjected to biaxial stress i.e. direct stresses $$\left( {{\sigma _{\text{x}}}} \right)$$ and $$\left( {{\sigma _{\text{y}}}} \right)$$ in two mutually perpendicular planes accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ then minimum normal stress is

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