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Question

The amount of steel required such that there is not flexure collapse given combination of loads is equal to ___ $m m^{2}$ .

$M_{u} = 80 k N m; T_{u} = 40 k N m; V_{u} = 60 k N$

The dimension of the beam are 300 mm x 500 mm and effective cover = 40 mm
Take M25 concrete and Fe415 steel.
974

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Solution

The correct option is A 974
Given, Effective cover = 40 mm
d = D - 40 = 500 - 40 = 460 mm

$M_{L} = M_{u} + \frac{T_{u}}{1.7} (1 + \frac{D}{b})$

$= 80 + \frac{40 \times (1 + \frac{500}{300})}{1.7} = 142.75 k N m$

$A_{s t} = \frac{0.5 f_{c k}}{f_{y}} [1 - \sqrt{1 - \frac{4.6 M_{c}}{f_{c k} b d^{2}}}] b d$

$= \frac{0.5 \times 25}{415} [1 - \sqrt{1 - \frac{4.6 \times 142.75 \times 10^{6}}{25 \times 300 \times 460^{2}}}] \times 300 \times 460$

$= 974 m m^{2}$

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