# The linear expansion of rod having variable coefficient of linear expansion across its length

It is known that solids expand on heating. Consider a case of rod of negligible area of cross-section such that it expands linearly upon heating.Now the increment in the length of the rod is proportional to the change in temperature as well as its length... Thus: $$dl=l*\alpha*\Delta T———(a)$$ ...where $$\alpha$$ = coefficient of linear expansion of the material with which the rod is made with.

Consider a rod (of length $$l_o$$)with a variable $$\alpha$$ such that $$\alpha = Kx$$...where x is the distance of a point on the rod from one of its end.calculate the the final length of rod when the temperature of the rod is increased by $$\Delta$$T Now using (a) $$dl=dx*kx*\Delta T———(b)$$ next step is to integrate (b).. $$\int dl =\int kxdx*\Delta T$$

My doubt is regarding the limits to be taken for x.Is it from o to $$l_o$$?If so what about extra length which is generated slowly by the change in temperature...shouldn’t we include the expansion of that part in to consideration?

• If the total length of the rod is being heated evenly, then all parts of the rod expand. – user207455 May 3 '19 at 6:28
• @SolarMike yeah they do....hence the doubt is arises.. – Crypton May 3 '19 at 6:56

Let x be the location along the rod (after heating has occurred) of the cross section that was at location $$x_0$$ before heating, and let the coefficient of linear expansion be a linear function of the initial (material) location $$a(x_0)=kx_0$$. Then for the segment of the rod between $$x_0$$ and $$x_0+dx_0$$, we have after heating that $$dx=(1+\alpha\Delta T)dx_0=(1+kx_0\Delta T)dx_0$$If we integrate this equation from $$x_0=0$$ to $$x_0=l$$ we obtain: $$x(l)=l+k\frac{l^2}{2}\Delta T$$This is the final length of the rod.