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Find the design strength corresponding to limiting condition of 'no tension' in the column section. Assume eccentricity of loading with respect to major axis only. Use M25/Fe415.
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Ixx=bD312=550×350312=1965.1×106mm4

Iyy=bD312=350×550312=4852.6×106mm4Here Iyy>Ixx

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PG  126  table

$P_{uc}=0.36 f_{ck} bD=0.36×25×350×550 \\ P_{uc}=1732.5 kN \\ P_u=P_{uc}+∑P_{usi}=1732.5+722.5=2455kN \\ Now M_u=M_{uc} + ∑ M_{usi} \\ M_{uc}=P_{uc}×(0.5D-0.42D)=1732.5×0.08×550=76.23kNm \\ Now, \\ M_{us1}= P_{us1}×e_1=262.27×0.235=01.63kNm \\ M_{us2}= P_{us2}×e_1=248.78×0.085=21.4kNm \\ M_{us3}= P_{us3}×e_1=175.34×0.085=14.9kNm \\ M_{us4}= P_{us4}×e_1=36.1×0235=-8.48kNm …

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