Question

A rod of length \( 5 m \) is prevented from thermal expansion by fixing its ends rigidly. Its cross-sectional area is \( 40 cm ^{2} \). Calculate thermal stress developed on a temperature rise of \( 20^{\circ} C \), if \( Y=2 \times 10^{11} \) \( Nm ^{-2} \) and \( \alpha=1.2 \times 10^{-5} K ^{-1} \).

Answer 1

Y = 2 x 1011 Nm^-2

A = 40 cm²

ΔT = 20°C

α = 1.2 x 10^-5 k^-1

\(Y= \frac{\text{Stress}}{\text{Strain}}\)

\(Y = \frac{\frac FA}{\frac{\Delta l}l}\)

\(Y = \frac F{A\frac{l(1+ α\Delta T)}l}\)

\(Y = \frac F{A{(1+ α\Delta T)}}\) 

\(F = Y. A(1 + α \Delta T)\)

\(= 2 \times 10^{11} \times 40 \times 10^{-4}\times(1 + 1.2 \times 10^{-5}\times 20)\)

\(= 80\times 10^7 (1 + 24 \times 10^{-5})\)

\(= 80\times 10^7 + 1920 \times 10^2\)

\(= 8000\times 10^5 + 1.920 \times 10^5\)

\(= 8000.192 \times 10^5 N\)