Two same length rods of brass and steel of equal cross-sectional area are joined end to end as shown in figure and supported between two rigid vertical walls. Initially the rods are unstrained. If the temperature of system is raised by `Deltat`. Find the displacement of the junction of two rods. Given that the coefficient of linear expansion and young's modulus of brass and steel are `apha_(b).alpha_(s)(alpha_(b)gtalpha_(s)),Y_(b)` and `Y_(s)` respectively.

Explanation

Here the initial lengths of the two rods are `l_(0)` and as `alpha_(b)gtalpha_(s)` , we can directly state that the junction of the two rods is displacedf toward right. Due to this brass rod is some what expanded but less as compared to free expansion and steel is overall compressed due to the stress developed between the two rods.
Figure shows the final situation of rods at higher temperature `Deltat`. in figure A is the initial position of junction and B is its final position.

If only brass rod is present, expansion in it due to increase in temperature by `DeltaT` is
`Deltal_(b)=alpha_(b)l_(0)DeltaT`
Which is equal to AC in figure. But finally the rod is expanded by x, thus the elastic compression in the rod due to stress between the two rods is BC, which is given as
`BC=alpha_(b)l_(0)DeltaT-x`
Thus elastic strain in brass rod is
`("Strain")_("brass")=(alpha_(b)l_(0)Deltat-x)/(l_(0))`[ As `alpha_(b)l_(0)DeltaT` is small]
Similary if steel rod alone is there, it would have been expanded by
`Deltal_(s)=alpha_(s)l_(0)DeltaT`
Which is equal to AD in figure . But finally steel rod is compressed by x, thus the elastic compression in steel rod due to stress between the two rods is BD, which is given as
`BD=alpha_(s)=l_(0)DeltaT+x`
Thus elastic strain in steel rod is
`("Strain)_("steel")=(alpha_(s)l_(0)DeltaT-x)/(l_(0))` [As `alpha_(s)l_(0)DeltaT` is small]
As thetwo rods are in contact, stress develoed in the two rod must be equal thus we have
`("Stress")_("brass")=("Stress")_("steel")`
or `Y_(b)xx("Strain")_("brass")=Y_(S)xx("Strain")_("steel")`
or `Y_(b)[(alpha_(b)l_(b)DeltaT-x)/(l_(@)0]=Y_(s)[(alpha_(s)l_(@)DeltaT+x)/(l_(0))]`
or `x=((Y_(b)alpha_(b)-Y_(s)alpha_(s))l_(@)DeltaT)/((Y_(b)+Y_(s)))`