Question

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at ${250}^{o}C$, if the original lengths are at ${40.0}^{o}C$? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass $=2.0\times {10}^{-5}{K}^{-1}$, steel $=1.2\times {10}^{-5}{K}^{-1})$

Answer 1

Initial temperature, $T_1={40}^{o}C$
Final temperature, $T_2={250}^{o}C$
Change in temperature, $△T=T_2-T_1={210}^{o}C$
Length of the brass rod at $T_1,l_1=50cm$
Diameter of the brass rod at $T_1,d_1=3.0cm$
Length of the steel rod at $T_2,l_2=50cm$
Diameter of the steel rod at $T_2,d_2=3.0mm$
Coefficient of linear expansion of brass, ${\alpha }_1=2.0\times {10}^{-5}{K}^{-1}$
Coefficient of linear expansion of steel, ${\alpha }_2=1.2\times {10}^{-5}{K}^{-1}$
For the expansion in the brass rod, we have:
Change in length $\left(△l_1\right)$ / Original length $\left(l_1\right)={\alpha }_1△T$

$\therefore △l_1=50\times (2.1\times {10}^{-5})\times 210$
$=0.2205cm$
For the expansion in the steel rod, we have:
Change in length $\left(△l_2\right)$/ Original length $\left(l_2\right)={\alpha }_2△T$

$\therefore △l_1=50\times (1.2\times {10}^{-5})\times 210$
$=0.126cm$
Total change in the lengths of brass and steel,
$△l=△l_1+△l_2$
$=0.2205+0.126$
$=0.346cm$
Total change in the length of the combined rod $=0.346cm$
Since the rod expands freely from both ends, no thermal stress is developed at the junction.