Question

PART ONE

Answer 1

Answer:$\Delta L=0.0101\ m$

Explanation:

Given

Length of track $L_o=28\ m$ when

$T_o=2^{\circ}C$

Coefficient of linear expansion $\alpha =11\times 10^{-6}\ ^{\circ}C^{-1}$

When Temperature rises to $T=35^{\circ}C$

$\Delta T=35-2=33^{\circ}C$

and we know length expand on increasing temperature

$L=L_o[1+\alpha \Delta T]$

$L-L_o=L_o\alpha \Delta T$

$\Delta L=28\times 11\times 10^{-6}\times (33)$

$\Delta L=0.0101=10.164\ mm$

(b)When rails are clamped thermal stress induced

we know $E=\frac{stress}{strain}$

$Stress=E\times strain$

$Stress=20\times 10^{10}\times \frac{\Delta L}{L_o}$

$Stress=20\times 10^{10}\times \frac{0.0101}{28}$

$Stress=72.14\ MPa$

$Stress=72.14\times 10^{6}\ N/m^2$