written 8.4 years ago by
teamques10
★ 68k
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modified 8.4 years ago
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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