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Two rods of same length made of brass and steel of equal cross-sectional area are joined end to end as shown in figure (a) 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 the rods. Given that the coefficients of linear expansion and Young.s modulus of brass and steel are alpha_(b), alpha_(s) (alpha_(b) gt alpha_(s)), y_(b) and y_(s) respectively. |
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Answer» SOLUTION :Here the initial lengths of the two rods are `l_(0)`and as `alpha_(b) gt alpha_(s)`,we can directly state that the junction of the two rods is displaced 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 (b) 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 b. 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_(0)l_(0)DeltaT` is small] Similarly if steel rod along is there, it would have been expanded by `Deltal_(s)=alpha_(s)l_(0)DeltaT` which is equal to AD in figure b. But finally steel rod is compressed by x, thus the elastic compression in steelrod due to stress between the two rods in 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 the two rods are in contact, stress developed in the two rods must be equal so, we have `("Stress")_("brass") =("Stress")_("steel")` (or) `y_(b) xx ("Strain")_("brass") =y_(s) xx ("Strain")_("steel") or y_(b)[(alpha_(0)l_(0)DeltaT-x)/(l_(0))]=y_(s)[(alpha_(s) l_(0) DeltaT+x)/(l_(0))]` (or) `x=((y_(b)alpha_(b)-y_(s) alpha_(s)) l_(0) DeltaT)/((y_(b)+y_(s)))` |
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