Question

A rectangular plate of uniform thickness having initial length \( a \) and width \( b \) is placed between two rigid immovable walls. The temperature of the plate is increased by \( \Delta T \). The plate is free to expand along the \( y \) and \( z \) directions. The mid-surface of the plate remains in the \( xy \)-plane. The Poisson’s ratio is \( \nu \) and the coefficient of thermal expansion is \( \alpha \). Assuming that the plate is initially free of stresses, the change in length of the plate after the increase in temperature is given by:

• A. ( a(1-\nu)\alpha\Delta T \)
• B. ( a(1+\nu)\alpha\Delta T \)
• C. ( a\alpha\Delta T \)
• D. ( 2a\alpha\Delta T \)

Hint:

When expansion is restrained in one direction, the Poisson effect from the free expansion in other directions adds extra strain in the constrained direction.

Answer 1

The Correct Option is B

The plate is restrained along the \( x \)-direction by rigid walls, so the free thermal expansion along \( x \) cannot occur. The free thermal strain in all directions would be:
\[ \varepsilon_{\text{free}} = \alpha \Delta T. \] Because expansion along \( x \) is prevented, stress develops along the restrained direction. The expansion in the \( y \) and \( z \) directions (which are free) contributes an additional strain in the \( x \)-direction due to Poisson’s effect:
\[ \varepsilon_{\text{poisson}} = \nu \alpha \Delta T. \] Thus, the total strain in the \( x \)-direction is the sum of the direct thermal strain and the Poisson strain:
\[ \varepsilon_x = \alpha\Delta T + \nu\alpha\Delta T = (1+\nu)\alpha\Delta T. \] Hence, the change in length of the plate is:
\[ \Delta a = a(1+\nu)\alpha\Delta T. \] Therefore, the correct option is (B).