Question

A brass rod is fixed rigidly at two ends at 27\(^\circ\)C. If it is cooled to temperature -43\(^\circ\)C, tension in rod becomes T\(_{0}\). Find temperature (in \(^\circ\)C) at which tension will be 1.4 T\(_{0}\) :

• A. -71\(^\circ\)C
• B. -65\(^\circ\)C
• C. -50\(^\circ\)C
• D. -82\(^\circ\)C

Hint:

In thermal stress problems, always identify the reference temperature where the stress is zero. The change in temperature (\(\Delta T\)) is always measured from this reference temperature. Tension arises from cooling (rod wants to contract but can't), and compression arises from heating (rod wants to expand but can't).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A rod is fixed between two supports, preventing its thermal expansion or contraction. When the temperature changes, a thermal stress is induced, resulting in tension (on cooling) or compression (on heating). We are given a relationship between tensions at two different temperatures and need to find one of the temperatures.
Step 2: Key Formula or Approach:
The thermal stress (\(\sigma\)) developed in a rod due to a temperature change \(\Delta T\) is given by \(\sigma = Y \alpha \Delta T\), where Y is Young's modulus and \(\alpha\) is the coefficient of linear expansion.
The tension (force) in the rod is \(T = \sigma \cdot A = (Y A \alpha) \Delta T\), where A is the cross-sectional area.
Since Y, A, and \(\alpha\) are constants for the rod, the tension is directly proportional to the change in temperature from the stress-free state.
\[ T \propto \Delta T \] The stress-free temperature is the temperature at which the rod was fixed, which is 27\(^\circ\)C.
Step 3: Detailed Explanation:
Let the initial stress-free temperature be \(T_{ref} = 27^\circ\)C.
Case 1: Temperature is cooled to \(T_1 = -43^\circ\)C.
The change in temperature is \(\Delta T_1 = T_{ref} - T_1 = 27 - (-43) = 70^\circ\)C.
The tension developed is \(T_0\). So, we can write:
\[ T_0 = C \cdot \Delta T_1 = C \cdot 70 \quad \text{...(i)} \] where \(C = Y A \alpha\) is a constant.
Case 2: Temperature is cooled to an unknown temperature \(T_2 = t\).
The tension becomes \(T' = 1.4 T_0\).
The change in temperature is \(\Delta T_2 = T_{ref} - T_2 = 27 - t\).
The new tension is:
\[ 1.4 T_0 = C \cdot \Delta T_2 = C \cdot (27 - t) \quad \text{...(ii)} \] Now, we divide equation (ii) by equation (i):
\[ \frac{1.4 T_0}{T_0} = \frac{C \cdot (27 - t)}{C \cdot 70} \] \[ 1.4 = \frac{27 - t}{70} \] \[ 1.4 \times 70 = 27 - t \] \[ 98 = 27 - t \] \[ t = 27 - 98 = -71^\circ\text{C} \] Step 4: Final Answer:
The required temperature is -71\(^\circ\)C.