Design the reinforcement for a R.C beam of size 230mm x 500mm overall depth supported between an effective span of 6 m. It is subjected to udl D.L + L.L. (service load) of 35 KN/m. Use $M_{20}$ Concrete and $F_{e415}$ steel. - | $\frac{dc}{d}$ | 0.05 | 0.1 | 0.15 | 0.2 | |----------------|-------|-------|-------|-------| | $f_{sc}$ | 355.1 | 351.9 | 342.4 | 329.2 |

$$b = 230 \ mm,\ D=500 \ mm,\ D_c=50 \ mm \ (assumed),\ l=6m,\ D.L+L.L=35KN/m,\ M_{20} \ \ \& \ \ F_{e415}$$

$$f_{cc}=0.446f_ck=8.92N/mm^2$$ ![enter image description here][1] $$BM_{\max}=\frac{wl^2}{8}=\frac{35{\times}6^2}{8}=157.5KNm$$

$M_d$ $=1.5 {\times}BM \\ =1.5 {\times}157.5 \\ =236.5KNm$

$M_{u \max}$ $= 0.138 f_ckbd^2 \\ =0.138{\times}20{\times}230{\times}450^2 \\ =128.54 KNm$

$\therefore M_d \gt M_{u \max}$

Design a doubly reinforced section.

$M_{u1}=128.54KNm$

$M_{u2}$ $=M_d-M_{u1} \\ =236.25-128.54 \\ =107.71 KNm.........(1)$

$M_{u1}=T_{u1}{\times}L_{u1}=0.87{\times}f_yAst_1{\times}(d-0.42M_{u \max}) \\ 128.54{\times}10^6=[(0.87{\times}415{\times}{\times}Ast_1){\times}(450-0.42{\times}0.48{\times}450)] \\ \therefore Ast_1=991 \ mm^2$

$M_{u2}$ $=T_{u}{\times}L_{a2}..........(2) \\ =0.87{\times}f_yAst_2{\times}(d-d_c) \\ 107.71{\times}10^6=[(0.87{\times}415{\times}Ast_2){\times}(450-50)] \\ \therefore Ast_1=745.81 \ mm^2$

$Ast$ $=Ast_1+Ast_2 \\ =1736.81 \ mm^2$

Provide4 - 25 mm steel

$Ast=1964 mm^2$

$M_{u2}$ $=C_{u2}{\times}L_{a2} \\ =(f_{sc}-f_{cc}){\times}Asc{\times}(d-d_c) \\ 107.71{\times}10^6=(350-8.92){\times}(450-50) $

$Asc=789.47 \ mm^2$

Provide 4 - 16 mm

$Asc=804 \ mm^2$