A rectangular beam of width $(b)$ $230$ mm and effective depth $(d)$ $450$ mm is reinforced with four bars of $12$ mm diameter. The grade of concrete is $M20$ and grade of steel is $Fe500$. Given that for $M20$ grade of concrete the ultimate shear strength, $\tau_{IX} = 0.36 \: N/mm^2$ for steel percentage $p=0.25$, and $\tau_{IX} =0.48 \: N/mm^2$ for $p=0.50$. For a factored shear force of $45$ kN, the diameter (in mm) of $Fe500$ steel two legged stirrups to be used at spacing of $375$ mm, should be
- $8$
- $10$
- $12$
- $16$