Question

The rectangular beam of width, 300 mm is having effective depth of 365 mm. the concrete grade is m20 and the grade of reinforcing steel is fe415. as per limit state method, the area of steel in a balanced section is :

Answer 1

For Fe415 steel, $(\frac{x_u}{d})_{balanced} = 0.479$

$(\frac{x_u}{d})_{balanced} =\frac{0.87\ f_y\ A_{st}}{0.36\ f_{ck}bd} \\ \\ \Rightarrow 0.479= \frac{0.87\ \times 415\ \times A_{st}}{0.36\ \times 20 \times 300 \times 365} \\ \\ \Rightarrow A_{st}=\frac{0.479 \times 0.36 \times 20 \times 300 \times 365}{0.87 \times 415} \\ \\ \Rightarrow A_{st}= 1045.96\ mm^2$

$\text{Area\ of\ tension\ steel\ in\ balanced\ section\ is\ 1045.96\ } mm^2$

Answer 2

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For Fe415 steel, $(\frac{x_u}{d})_{balanced} = 0.479$

$(\frac{x_u}{d})_{balanced} =\frac{0.87\ f_y\ A_{st}}{0.36\ f_{ck}bd} \\ \\ \Rightarrow 0.479= \frac{0.87\ \times 415\ \times A_{st}}{0.36\ \times 20 \times 300 \times 365} \\ \\ \Rightarrow A_{st}=\frac{0.479 \times 0.36 \times 20 \times 300 \times 365}{0.87 \times 415} \\ \\ \Rightarrow A_{st}= 1045.96\ mm^2$

$\text{Area\ of\ tension\ steel\ in\ balanced\ section\ is\ 1045.96\ } mm^2$

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