$$P = 1200 kN \space M20\space \&\space Fe415$$

$P_u=1.5 ×1200=1800 kN \\ \text { Assume 1% steel of gross c/s }\\ \text {99% concrete of gross c/s } \\ Asc =0.01 A_g \\ Ac = 0.99 A_g \\ Now \space P_{uc}=0.4 f_{ck} A_c+0.67 f_y A_{sc} \\ 1800 ×10^3=1.05[0.4 ×20×0.99 A_g+0.67 ×415×0.01 A_g]\\ A_g=160206.13 mm^2 \\ A_g=\pi/4∅^2=451.64 ≈460 mm \\ (A_g)_{provided}=\pi/4×460 ×460 =166190 mm^2 \\ Asc = 0.01×166190=1661.9 mm^2 \\ \text { Provide 6 – 20 mm∅ } (Asc)p = 6×\pi/4×20^2=1885 mm^2 \\ (Ac)p=(Ag)p-(Asc)p=166190-1885=164305 mm^2 \\ \dfrac {\pi×D_k×a_{sh}}{Pitch ×A_k }\space\space\&\space\space\dfrac {0.36f_{ck}}{f_y} ×(\dfrac {A_g}{A_K} -1) \\ D_K=D-(2×\text { clear cover})=450-(2×50)=350mm \\ A_k=\pi/4×350^2=96211 mm^2 \\ \dfrac {\pi×350×50}{50 ×96211}\space \space \& \space\space \dfrac {0.36×20}{415} ×(\dfrac {166190}{96211}-1) \\ 11.42 ×10^{-3} \space\space\&\space\space 12.31×10^{-3} \\ R.H.S \gt L.H.S \\ P_u=0.4×f_{ck}×A_c+0.67×f_y×Asc=0.4×20×164305+0.67×415×1885 \\ P_u=1838.5 kN \gt1800 kN \hspace {0.5 cm}Safe $