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\(_{0}\) :
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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).
To solve for the temperature at which the tension in a brass rod becomes 1.4 times its original tension (\(T_0\)), we apply the concept of thermal stress due to temperature change.
The tension in the rod due to temperature change is given by:
T = \alpha EA \Delta T, where:
\alpha is the coefficient of linear expansion.
E is the Young's modulus of brass.
A is the cross-sectional area of the rod.
\Delta T is the change in temperature.
Initially, when the rod is cooled from \(27^\circ\)C to \(-43^\circ\)C:
\Delta T_0 = (27 - (-43)) = 70\ ^\circ C
Thus, the initial tension \(T_0\) is T_0 = \alpha EA (70).
For the tension to be 1.4 times the original tension T = 1.4 T_0:
Using T = \alpha EA \Delta T, we set:
\alpha EA \Delta T = 1.4 \alpha EA \cdot 70
Cancel out common terms, giving:
\Delta T = 1.4 \times 70 = 98\ ^\circ C
This change in temperature, \Delta T\), is from the initial temperature:
New temperature = \(27\ ^\circ C - 98\ ^\circ C = -71\ ^\circ C\)
Therefore, the temperature at which the tension in the rod will be 1.4 times \(T_0\) is \(-71\ ^\circ C\).
