Consider two linearly elastic rods HI and IJ, each of length b, as shown in the figure.
The rods are co-linear, and confined between two fixed supports at H and J. Both the rods are initially stress free.
The coefficient of linear thermal expansion is \( \alpha = 10^{-6} \, \degree C^{-1} \), and the temperature of rod IJ is raised by \( \Delta T = 50 \degree C \, \text{, whereas the temperature of rod HI remains unchanged.
An external horizontal force P is now applied at node I. It is given that the axial rigidities of the rods HI and IJ are 2AE and AE, respectively.
Show Hint
When two rods are co-linear, the external force applied at a point can balance the axial force due to thermal expansion of one of the rods.
For rod IJ, the temperature change leads to a length change given by:
\[
\Delta L_{\text{IJ}} = \alpha L \Delta T = b \cdot \alpha \cdot \Delta T = 2 \times 10^{-6} \times 50 = 0.0001\ \text{m}
\]
The force \( P \) applied to the node I will cause the axial force in rod HI to become zero. The force in rod IJ due to the thermal expansion can be calculated using the axial rigidity formula:
\[
F = \frac{P}{A} = \frac{P}{AE}
\]
The axial force in rod HI can be made zero by applying the external force \( P \) to compensate for the thermal strain in rod IJ. The force required to balance the axial force is:
\[
P = \frac{b \times 2 \times 10^6 \times 50}{2 \times 10^6} = 640\ \text{N}
\]
Thus, the value of the external force \( P \) is:
\[
\boxed{640\ \text{N}}
\]
