Assuming concrete of $M20$ grade and HYSD steel of grade $F_{e415}$ . Calculate ultimate moment of resistance of the section.

Data: $b_f =1250 mm\\ D_f = 110 mm \\ b_w =320 mm\\ d = 420 mm\\ Ast = 7-28 mm\\ =7\times\dfrac \pi4 \times 28^2=4310.26 mm^2\\ Assume \space X_u \lt D_f [\text{ N.A lies in the flange }]\\ C_u=T_u\\ 0.36 f_ckb_fX_u=0.87f_yAst \\ 0.36\times20\times1250\times X_u=0.87\times415\times4310.26\\ X_u=172.91 mm \\ X_u \gt D_f.... \text{ Assumption incorrect }\\ Case 2A :- \dfrac 37 X_u \geq D_f \to \text{ N.A lies in the web }\\ C_{u_1}+ C_{u2}=T_{u1 } + T_{u2}\\ (0.36f_ckb_wX_u)+[0.446f_ck\times(b_f-b_w)\times D_f] =[0.87 f_y Ast]\\ 0.36\times20\times320 \times X_u+[0.446\times20\times(1250-320)\times110]= [0.87\times415\times4310.26] \\ X_u=279.38 mm \\ \dfrac 37\times X_u=\dfrac 37\times 279.38=119.73 \gt D_f\\ \text{ Hence Assumption is correct }\\ X_{u\space max}=0.48\times420=201.6 mm\\ X_u\gt X_{u\space max}\to \text{ Over-reinforced section }\\ \dfrac 37 \times X_{u\space max}=86.4 mm \lt D_f\\ Since \dfrac 37 \lt D_f...... \text { Assumption incorrect for X_{u max}} \\ Follow Case 2B :- \\ y_f=(0.15X_{u\space max})+(0.65D_f)\\ y_f=(0.15\times201.6)+ (0.65\times110)=01.74 mm \\ Restrict X_u=X_{u\space max} \\ M_u= (C_{u1}\times L_{a1}) +(C_{u2}\times L_{a2}) \\ M_u =[(0.36f_ckb_wX_{u\space max}) \times (d-0.42 X_{u\space max})]+[0.446f_ck\times(b_f-b_w)\times y_f\times (d-\frac {y_f}2)]\\ M_{u\space max}=[(0.36\times20\times320\times201.6)\times(420- 0.42 \times201.6)]+[0.446\times20\times(1250-320)\times 101.74 \times(420-\dfrac {101.74}2)] \\ M_{u\space max}=467.29 KNm$