From strain diagram

By similar triangular

$\frac{X_u}{0.0035}=\frac{d-X_u}{\frac{0.87f_y}{ES}+0.002}\\ \frac{d-X_u}{X_u}=\frac{4.35{\times}10^{-6}f_y+0.002}{0.0035}\Big[2{\times}10^5=ES \ N/mm^2\Big] \\ \frac{d}{X_u}-1=\frac{4.35{\times}10^6f_y+0.002}{0.0035} \\ \frac{d}{x_u}=\frac{4.35{\times}10^6f_y+0.002}{0.0035}+1 \\ X_u=\frac{d}{\frac{4.35{\times}10^6f_y+0.002}{0.0035}+1}$.

For $F_{e250},f_y=250 \\ X_u=0.53d$

For $F_{e415},f_y=250 \\ X_u=0.48 d$

For $F_{e500},f_y=500 \\ X_u=0.46 d$