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What will happen if a rod is tied with fixed supports rigidly at both ends and temperature is increased ?
Solution
If a rod is tied with fixed supports rigidly at both ends, then it will have expansion with increase in temperature.
At both ends, forces are applied due to rigid supports hence compressive strain is produced. And corresponding thermal stress is produced in rod. As result rod can be banded.
E.g. length of track of steel is $5 \mathrm{~m}$ and cross sectional area is $40 \mathrm{~cm}^{2}$. Thermal expansion is prevented by decrease of $10^{\circ} \mathrm{C}$ in temp.
Coefficient of linear expansion is $\alpha_{l(\text { steel })}=1.2 \times 10^{-5} \mathrm{~K}^{-1}$
There will be slight change in length due to temperature change (it is compressive strain).
$\therefore$ Compressive strain $=\frac{\Delta l}{l}$
$\therefore$ Compressive strain $=\frac{\alpha_{l} l_{\mathrm{T}}}{l}$
$\therefore \frac{\Delta l}{l}=\alpha_{l(\text { steel })} \Delta \mathrm{T}$
$=1.2 \times 10^{-5} \times 10$
$=1.2 \times 10^{-4}$
Now, Young's modulus $Y=\frac{\text { Compressive stress }}{\text { Compressive strain }}$
$\therefore \quad$ Compressive stress $=\mathrm{Y} \times$ Compressive strain
$\frac{\Delta \mathrm{F}}{\mathrm{A}}=\mathrm{Y} \times \frac{\Delta l}{l}$
$\therefore \Delta \mathrm{F}=\mathrm{AY} \times 1.2 \times 10^{-4}$
$=40 \times 10^{-4} \times 2 \times 10^{11} \times 1.2 \times 10^{-4}$
$=96 \times 10^{3} \mathrm{~N}$
$\approx 10^{5} \mathrm{~N}$
Thus, by this value of force with two rigid supports, tracks can be banded.
