Correct Answer - Option 1 :
\(-\frac{{\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2v} \right)}}\)
Explanation:
For a solid cube, strain in x, y and z-axis are:
\(\begin{array}{l} {\epsilon_x} = \frac{{{\sigma _x}}}{E} - \frac{{v\left( {{\sigma _y} + {\sigma _z}} \right)}}{E}\\ {\epsilon_y} = \frac{{{\sigma _y}}}{E} - \frac{{v\left( {{\sigma _x} + {\sigma _z}} \right)}}{E}\\ {\epsilon_z} = \frac{{{\sigma _z}}}{E} - \frac{{v\left( {{\sigma _x} + {\sigma _y}} \right)}}{E} \end{array}\)
From the symmetry of cube, \(\epsilon_x =\epsilon_y=\epsilon_z=\epsilon\)
and \(\sigma_x=\sigma_y=\sigma_z=\sigma\)
Because of the rise in temperature, the cube will try to expand (tensile strain), and as the walls are constrained so a compressive force will develop on the walls of the cube.
So, \(\sigma_x=\sigma_y=\sigma_z=- \sigma\)
As the walls are constrained so total strain should be zero.
Tensile strain due to temperature rise + Compressive strain due to compressive force = 0
Volumetric strain of the cube,
\( \epsilon_v= \frac{{\left( {1 - 2v} \right)}}{E} \times \left(- \sigma_x - \sigma_y - \sigma_z\right) =- \frac{{\left( {1 - 2v} \right)}}{E} \times3\sigma \)
Volumetric thermal strain
\(\epsilon_v=3\alpha. \Delta T\) (Thermal tensile Strain)
\(- \frac{{\left( {1 - 2v} \right)}}{E} \times3\sigma = 3\alpha. \Delta T\)
∴ \(\sigma = \frac{{ - \alpha .{\rm{\Delta }}T.E}}{{\left( {1 - 2v} \right)}}\)