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 maximum normal stress is

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

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

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

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

Correct Answer: $$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$

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

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

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

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

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

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

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

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

$$\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$

$$\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$

When a body is subjected to direct tensile stresses ($${\sigma _{\text{x}}}$$ and $${\sigma _{\text{y}}}$$) in two mutually perpendicular directions, accompanied by a simple shear stress $${\tau _{{\text{xy}}}}{\text{,}}$$ then in Mohr's circle method, the circle radius is taken as

$$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} + \tau $$

$$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} + \tau $$

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

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

When a body is subjected to a direct tensile stress $$\left( {{\sigma _{\text{x}}}} \right)$$ in one plane accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ the maximum normal stress is

$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

$$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$

$$\frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

When a body is subjected to a direct tensile stress $$\left( {{\sigma _{\text{x}}}} \right)$$ in one plane accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ the minimum normal stress is

$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

$$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$

$$\frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

When a body is subjected to a direct tensile stress ($${\sigma _{\text{x}}}$$) in one plane accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), the maximum shear stress is

$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

$$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$

$$\frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm²Find the maximum amount of shear stress to which the body is subjected.

22.4mm

25mm

26.3mm

27.2mm

Tresca or maximum-shear stress criteria assumes that yielding occurs when the maximum shear stress reaches a value of the shear stress in the uniaxial tension test. Assume the principal stress being σ1, σ2, σ3 where σ1 is largest, and σ3 is the smallest principal stresses. Find the value of minimum shear stress to cause yielding, given that yield stress in tension is equal to σo?

τ = σo

τ = σo/2

τ = σo/3

τ = σo/4

If nominal shear stress $${\tau _{\text{v}}}$$ exceeds the design shear strength of concrete $${\tau _{\text{c}}}$$, the nominal shear reinforcement as per IS : 456-1978 shall be provided for carrying a shear stress equal to

$${\tau _{\text{v}}}$$

$${\tau _{\text{c}}}$$

$${\tau _{\text{v}}} - {\tau _{\text{c}}}$$

$${\tau _{\text{v}}} + {\tau _{\text{c}}}$$

A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be

400 MPa

500 MPa

900 MPa

1400 MPa