Cylindrical bars P and Q have identical lengths and radii, but are composed of different linear elastic materials. The Young's modulus and coefficient of thermal expansion of Q are twice the corresponding values of P. Assume the bars to be perfectly bonded at the interface, and their weights to be negligible.

The bars are held between rigid supports as shown in the figure and the temperature is raised by AT. Assume that the stress in each bar is homogeneous and uniaxial. Denote the magnitudes of stress in P and Q by σ₁,and σ₂, respectively.

Which of the statement(s) given is/are CORRECT?

Correct Answer:
The correct statements are:

• The interface between P and Q moves to the left after heating
• ${\sigma }_1={\sigma }_2$

Step-by-Step Explanation:

1. Analysis of Stress in the Bars (σ₁ and σ₂):
The schematic shows two cylindrical bars, P (on the left) and Q (on the right), perfectly bonded at their interface and constrained between rigid, unyielding supports on both ends. When the temperature of the system is raised by ΔT, the bars attempt to expand thermally. Since they are confined, the rigid walls exert compressive reaction forces on the ends of the bars.

For static equilibrium of the combined assembly, the compressive force in bar P (denoted by R_P) and the compressive force in bar Q (denoted by R_Q) must be equal. Therefore, we have:
$R_P=R_Q=R$
Here, R is the axial internal force acting throughout the length of both bars.

The problem states that the cylindrical bars have identical lengths (L_P = L_Q = L) and identical radii. Since their radii are identical, their cross-sectional areas are also equal:
$A_P=A_Q=A$
Stress is defined as the internal normal force per unit cross-sectional area. The magnitude of stress in bar P (σ₁) and bar Q (σ₂) is:
${\sigma }_1=\frac{R}{A}$
and
${\sigma }_2=\frac{R}{A}$
Therefore, we directly find:
${\sigma }_1={\sigma }_2$

2. Movement of the Interface:
To determine how the interface moves, we calculate the individual changes in length of each bar. Let the material parameters of bar P be:
Young's modulus: $E_P=E$
Coefficient of thermal expansion: ${\alpha }_P=\alpha$
According to the problem, the parameters for bar Q are twice the corresponding values of P:
Young's modulus: $E_Q=2E$
Coefficient of thermal expansion: ${\alpha }_Q=2\alpha$

The total deformation of each bar consists of a thermal expansion component and a mechanical compression component.
For bar P:
$\Delta L_P=L\alpha \Delta T-\frac{RL}{AE}$
For bar Q:
$\Delta L_Q=L\left(2\alpha \right)\Delta T-\frac{RL}{A\left(2E\right)}=2L\alpha \Delta T-\frac{RL}{2AE}$

Since the ends of the combined bar are held between rigid supports, the total length cannot change, meaning the sum of the deformations must be zero:
$\Delta L_P+\Delta L_Q=0$
Substituting the expressions:
$\left(L\alpha \Delta T-\frac{RL}{AE}\right)+\left(2L\alpha \Delta T-\frac{RL}{2AE}\right)=0$
$3L\alpha \Delta T-\frac{3RL}{2AE}=0$
Solving for R:
$3L\alpha \Delta T=\frac{3RL}{2AE}\Rightarrow R=2AE\alpha \Delta T$

Now, we substitute this value of the axial reaction force R back into the deformation equations for P and Q:
$\Delta L_P=L\alpha \Delta T-\frac{\left(2AE\alpha \Delta T\right)L}{AE}=L\alpha \Delta T-2L\alpha \Delta T=-L\alpha \Delta T$
$\Delta L_Q=2L\alpha \Delta T-\frac{\left(2AE\alpha \Delta T\right)L}{2AE}=2L\alpha \Delta T-L\alpha \Delta T=L\alpha \Delta T$

Since $\Delta L_P$ is negative ($-L\alpha \Delta T$), bar P experiences net contraction. Since $\Delta L_Q$ is positive ($L\alpha \Delta T$), bar Q experiences net expansion. Because the left end of bar P and the right end of bar Q are held fixed by the rigid walls, the expansion of Q pushes the bonded interface to the left, compressing P. Thus, the interface between P and Q moves to the left after heating.