The amount of steel required such that there is not flexure collapse given combination of loads is equal to ___ mm2. Mu=80kNm;Tu=40kNm;Vu=60kNThe dimension of the beam are 300 mm x 500 mm and effective cover = 40 mm Take M25 concrete and Fe415 steel.974

The amount of steel required such that there is not flexure collapse g

1.	The amount of steel required such that there is not flexure collapse given combination of loads is equal to ___ mm2. Mu=80kNm;Tu=40kNm;Vu=60kNThe dimension of the beam are 300 mm x 500 mm and effective cover = 40 mm Take M25 concrete and Fe415 steel.974
Answer» 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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The amount of steel required such that there is not flexure collapse given combination of loads is equal to ___ mm2. Mu=80kNm;Tu=40kNm;Vu=60kNThe 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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