An unconstrained rail 100feet long and cross section of 13 i

Solution

Let L be the lenght of the rail. Due to the cooling the lenght would be reduced to L_f so L_f-L=L*a*(T_f-T), where a is the linear expansion coefficient of the material.

Therefore, the axial strain is e_axial=(L_f-L)/L_f

Now, we know that nu=-e_trans/e_axial, where nu is the Poisson\'s coefficient of the material an e_trans= is the cross secction strain. Thus e_trans=-nu*e_axial. But e_trans=(A_f-A)/A

If the material of the rail is steel a=6.11x10^-6 ºF^-1 and nu=0.3 so L_f=100+100*(6.11x10^-6*(-25-95))=99.92 ft

e_axial=(99.92-100)/100=-7.33x10^-4, e_trans=-0.3*(-7.33x10^-4)=2.2x10^-4 and A_f=13+13*(2.2x10^-4)=13.003 in²