Heat it!

We know that heating the steel will change its length:-

$\Delta L \longrightarrow$ change in length.

$L \longrightarrow$ original length.

Change in length is given by:- $\Delta L = l\alpha\Delta T$ Plugging in values we get $\Delta L=2\times10^{-4} m$

Also we know that:- $E=\dfrac{FL}{A\Delta L}$ Rearranging:- $F=\dfrac{EA\Delta L}{L}$ Plugging in values we get $\boxed{F=600.00N}$

First, assume that the fixed support at one end of the rod is removed. The rod is now free to expand due to the increase in temperature. The total length of expansion of the rod is given by the expression $\alpha L\Delta T$ . But because the rod is in reality fixed at both ends, The actual elongation should be zero. And so, from this, it is clear that that fixed support is exerting a force that is required to compress the rod by the previously computed deformation. From solid mechanics, the normal deformation due to a normal force is given by: $\delta =\frac { FL }{ AE }$ . Equating, $\alpha L\Delta T=\frac { FL }{ AE }$ and so, $\alpha \Delta TAE=F$

E= stress/strain

strain= δL/L = α δT = 0.000001 x 20

stress = Ex Strain = 2000000 x 20 x 0.000001 = 40 N/m²

we know Stress= F/A force F = stress x A = 40 x 15 = 600.00