The beam is subjected to a maximum factored shear force of $400$ KN. The materials are the concrete of grade M20 and TOR steel. Refer table for the values of allowable shear stress(Zc)

$$b = 350 mm $$ $ d = 500 mm \\ V_{UD} = 500 KN \\ F_ck=20N/mm^2\\ f_y=415 N/mm^2\\ Ast=4-25 mm \phi=1964mm^2\\ Z_uv=\dfrac {V_{UD}}{b\times d}=\dfrac {400 \times 10^3}{350\times 500}=2.28 N/mm^2 \lt 2.8 N/mm^2\\ pt=\dfrac {100\times Ast}{b\times d}=\dfrac {100\times 1964}{350\times 500}=1.12\%$

shear stress in concrete (Zuc):-

By interpolation

$$\dfrac {0.67-0.62}{1.25-1}=\dfrac {(Zuc-0.62)}{1.12-1}$$ $ \therefore Z_{uc}=0.64 N/mm^2\\ V_{uc}=Z_{uc}bd=0.64 \times 350\times500=112 KN \\ V_{u\space min}=0.4bd=0.4\times350\times500=70 KN \\ Assume \space \phi_s=10mm \\ a_{sv}=2\times\dfrac \pi4 \times10^2=157 mm^2\\ V_{uD}= 400 KN\\ V_{uc}+V_{u\space min}=112+70=182 KN\\ here \space V_{uD} \gt V_{uc}+ V_{u\space min}$

Design and provide shear reinforcement

$$V_{us}=V_{uD}-V_{uc}=400-112=288KN $$

Spacing :

$S_1=\dfrac {0.87f_ya_{sv}d}{V_{us}}\\ S_1=\dfrac {0.87\times415\times157\times500}{288\times10^3}=98.4\\ S_2=0.75d=0.75\times500=375 mm \\ S_3=300 mm$

Hence $10$ mm $\phi2$ strirups at $90$ mm c/c