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Moment of Force

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Question

A beam of rectangular section 250 m wide and 500 mm effective depth is provided with 4 bars of 18 mm diameter as tensile steel of grade Fe250. The depth of neutral axis from the extreme compression fibre is _____ cm. [Use M20 concrete]
12.3

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Solution

The correct option is A 12.3
$A_{s t} = 4 \times \frac{π}{4} \times 18^{2} = 1018 m m^{2}$
$b = 250 m m, d = 500 m m, f_{c k} = 20 M P a$
Equating compressive and tensile forces
$C = T$
$0.36 f_{c k} x_{u} b = 0.87 A_{s t} f_{y}$
$x_{u} = \frac{0.87 \times 1018 \times 250}{0.36 \times 20 \times 250} = 122.99 m m ≃ 123 m m$
Also, $x_{u, l i m} = 0.53 \times 500$
$= 265 m m > 123 m m (x_{u} < x_{u, l i m} so under reinforced section)$
$x_{u} = 123 m m = 12.3 c m$

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