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A solid steel cube constrained on all six faces is heated so that the temperature rises uniformly by ΔT. If the thermal coefficient of the material is α, Young’s modulus is E and the Poisson’s ratio is v, the thermal stress developed in the cube due to heating is
1. \(-\frac{{\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2v} \right)}}\)
2. \(- \frac{{2\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2v} \right)}}\)
3. \(- \frac{{3\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2v} \right)}}\)
4. \(-\frac{{\alpha \left( {{\rm{\Delta }}T} \right)E}}{{3\left( {1 - 2v} \right)}}\)

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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)

\(- \frac{{\left( {1 - 2v} \right)}}{E} \times3\sigma = 3\alpha. \Delta T\)

∴ \(\sigma = \frac{{ - \alpha .{\rm{\Delta }}T.E}}{{\left( {1 - 2v} \right)}}\)

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