Find the area of steel required for an ultimate moment of $600KNm.$ Use $M20/F_{e415}$ .

Data: $b_f =1100 mm \\ D_f =120mm \\ b_w =300mm \\ d = 600 mm \\ M_d =600KN/m \\ f_ck = 20N/mm^2 \\ f_y =415 N/mm^2 $

Assume, $X_u =D_f =120$ mm

$M_uf=C_u\times L_a \\ M_uf=0.36f_ckb_fX_u\times (d-0.42X_u)\\ M_uf=0.36\times20\times 1100\times 120\times (600-0.42\times 120)\\ M_uf=522.33KNm\\ \therefore M_d \gt M_uf \hspace {1cm} \text { N.A lies in web}\\ Assume, \dfrac 37 X_u=D_f\\ \therefore X_u=\dfrac 73D_f=\dfrac 73\times 120=280 mm \\ M_uw=[0.36f_ckb_wX_u(d-0.42X_u)] +[0.446f_ck(b_f-b_w)D_f\times (d-\dfrac {d_f}2)]\\ M_uw=[0.36\times 20\times 300\times 280\times (600-0.42\times 280)]+[0.446\times 20\times (1100-300)\times 120\times (600-\dfrac {120}2)]\\ M_uw=754.16KNm\\ M_d \lt M_uw ..... \text{ Follow case 2B}\\ y_f=0.15X_u+(0.65D_f)\\ y_f=0.15 X_u+ 0.65\times 120 \\ y_f=0.15X_u+78 \\ M_d=(C_{u1}\times L_{a1})+(C_{u2}\times L_{a2})\\ M_d=[0.36f_ckb_wX_0\times(d-0.42X_u)]+[0.446f_ck\times(b_f-b_w)\times y_f\times(d-\dfrac {y_f}2)] \\ 600\times10^6=0.36\times20\times300\times X_u\times(600-0.42X_u)+[0.446\times20\times(1100-300)\times(0.15X_u+78)\times(600-\dfrac {0.15X_u+78}2)]\\ X_u=170.64mm\\ y_f=0.15X_u +78=0.15\times 170.64+78\\ =103.59 mm \\ Now, \\ C_{u1}=T_{u1}\\ [0.36\times f_ck\times b_w\times X_u]=[0.87f_yAst]\\ [0.36\times20\times300\times170.64]=[0.87\times415\times Ast_1]\\ Ast_1=1020.866mm^2\\ Also, \\ C_{u2}=T_{u2}\\ [0.446\times f_ck\times(b_f-b_w)\times y_f]=[0.87f_yAst_2]\\ [0.446\times20\times(1100-300)\times103.59]=[0.87\times415\times Ast_2]\\ Ast_1=2048 mm^2\\ Ast=Ast_1+Ast_2\\ Ast=1020.86+2048\\ Ast=3069mm^2\\ Provide 4-32mm \phi\\ \therefore Ast_p=4\times\dfrac \pi4 \times32^2=3216 mm^2$