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reinforced concrete beam

A reinforced concrete beam 230 mm wide is to carry load 40 KN/m. The beam is simply supported on a span of 8 m. Design a section when: - 1. Depth is not restricted. - 2. Effective depth is restricted to 500 mm -

Use $M_{20}$ grade of concrete $F_{e415}$ grade of steel

numericals • 972  views
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## Case 1: Depth is not restricted

$B.M=\frac{wl^2}{8}=\frac{40{\times}8^2}{8}=320KNm \\ M_u=1.5{\times}320=480KNm \\ B=230 \ mm \\ M_{u \max}=0.138 f_ckbd^2 \\ 480{\times}10^2=0.138{\times}20{\times}230{\times}d^2 \\ d=869.56 \ mm \\ Ast_x=\frac{0.5f_ckbd}{f_y}{\times}\bigg(1-\sqrt{1-\frac{4.6M_u}{f_ckbd^2}}\bigg) \\ \frac{0.5{\times}20{\times}230{\times}869.5}{415}{\times}\bigg(1-\sqrt{1-\frac{4.6{\times}480{\times}10^6}{20{\times}230{\times}869.5}}\bigg) \\ Ast_x=1907.13 \ mm^2 \\ Provide \ 4-25 \ mm \ {\phi} \\ Ast_p=1963 \ mm^2$

## Case 2:-Effective depth is restricted to 500 mm

$d=500 \ mm \\ M_d=480 KNm$

$M_{u \max}$ $=0.138 \ f_ckbd^2 \\ =0.138{\times}20{\times}230{\times}500^2 \\ =158.7KNm$

$M_d \gt M_{u \max} \ singly \\ M_{u1}=158.7 \ KNm \\ M_{u2}=M_dM_d-M_{u1}=480-158.7=321.3KNm \\ M_{u1}=T_{u1}{\times}L_{a1} \\ 158.7{\times}10^6=0.87f_yAst{\times}(d-0.42X_{u \max}) \\ 158.7{\times}10^6=0.87{\times}415{\times}Ast_1{\times}(500-0.42{\times}0.48{\times}500) \\ Ast_1=1101.08 \\ M_{u2}=T_{u2}{\times}L_{a2} \\ 321.3{\times}10^6=0.87f_yAst{\times}(d-d_c) \\ 321.3{\times}10^6=0.87{\times}415{\times}Ast_1{\times}(500-50) \\ Ast_2=1977.56 \ mm^2 \\ Ast=1101.8+1977.56=3078.64$

Provide 4 of 32 mm ${\phi}$

$F_{sc}=350 N/mm^2 \\ F_{cc}=8.92N/mm^2$

$M_{u2}=C_{u2}{\times}L_{a2} \\ 321.3{\times}10^6=(f_{u2}-f_{cc}){\times}Asc{\times}(d-d_c) \\ 321.3{\times}10^6=(350-8.92){\times}Asc{\times}(500-50) \\ Asc=2093.35 mm^2$

Provide 3 – 32 mm ${\phi}$ steel bars

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