Two steel rods and an aluminium rod of equal length `l_0` and equal cross section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at `0^0 C`. Find the length of the system when the temperature is raised to `theta`. Coefficient of linear expansion of aluminium and steel are `alpha_a and alpha_s` respectively. Young's modulus of aluminium is `Y_a` and of steel is `Y_s`.
steel
Aluminium
Steel
`

Explanation

Let the initial length of the system at `0^@C`
` =l_0`
When temperature changes by `theta`
Final length of the steel at system temperature.
So, strain of the system`=(l-(l_0))/(l_0)`
` Steel`
` Aluminium`
` Steel`
But the total strain of the system
= total stress of system/ total Young's modulus of system
Now, total stress due to two steel rod
+ stress due to aluminium
`= (gamma _s alpha_s theta) +( gamma_s alpha _s theta )+ (( gamma_(Al)) (alpha_(Al)) theta)`
`= (2 gamma_s alpha_s theta) +( gamma_(Al) alpha_(Al) theta)`
Now Young's modulus system
`= gamma_s + gamma_s+ gamma_(Al) = (2 gamma_s + gamma_(Al)`
`:. Strain of system = (2 gamma_s alpha_s theta )+ (gamma_(Al) alpha_(Al) theta)/ (2 gamma_s + gamma_(Al))`
`(l-l)/ (l_0)= ((2gamma_s alpha_s theta) + (gamma_(Al) alpha_(Al) theta)) / (2 gamma_s+ gamma_(Al))`
l= (l_0)[ 1+((2gamma_s alpha_s theta)+(2gamma_(Al) alpha_(Al) theta) / (2 gamma_s+ gamma_(Al))`