written 8.8 years ago by
teamques10
★ 69k
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modified 8.8 years ago
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Data : - b=200 mmd=450 mmM20,Fe415Mumax=70KNm
Find Ast = ?
Mumax=0.138fckbd2
70×106 =0.138×20×200×d2d=356.1d=360 mm
But d=450 mm given (whichever big)
∴ Take d=450∴D=d+dc=450+50=500
$Ast \ x\frac{0.5f_ckbd}{f_y}{\times}\Bigg(1-\sqrt{1-\frac{4.6M_u}{f_ckbd^2}}\Bigg) \\
Ast \ x=\frac{0.5{\times}20{\times}200{\times}450}{415}\Bigg(1-\sqrt{1-\frac{4.6{\times}70{\times}10^6}{20{\times}200{\times}450^2}}\Bigg) …
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