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 $M25/Fe415.$

Welcome back.

and 2 others joined a min ago.

1.6kviews

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

$M25/Fe415.$

written 8.8 years ago by

teamques10 ★ 69k

• modified 4.3 years ago

reinforced concrete structures theory

ADD COMMENT EDIT

1 Answer

7views

written 8.8 years ago by

teamques10 ★ 69k

• modified 8.8 years ago

$I_{xx}=\dfrac {bD^3}{12}=\dfrac {550×350^3}{12}=1965.1×10^6 mm^4$

$I_{yy}=\dfrac {bD^3}{12}=\dfrac {350×550^3}{12}=4852.6×10^6 mm^4 \\ Here \space I_{yy} \gt I_{xx}$

enter image description here

$PG\space \space 126\space \space table$

$P_{uc}=0.36 f_{ck} bD=0.36×25×350×550 \\ P_{uc}=1732.5 kN \\ P_u=P_{uc}+∑P_{usi}=1732.5+722.5=2455kN \\ Now M_u=M_{uc} + ∑ M_{usi} \\ M_{uc}=P_{uc}×(0.5D-0.42D)=1732.5×0.08×550=76.23kNm \\ Now, \\ M_{us1}= P_{us1}×e_1=262.27×0.235=01.63kNm \\ M_{us2}= P_{us2}×e_1=248.78×0.085=21.4kNm \\ M_{us3}= P_{us3}×e_1=175.34×0.085=14.9kNm \\ M_{us4}= P_{us4}×e_1=36.1×0235=-8.48kNm \\ ∑M_{us}=59.65\\ M_u=76.23+59.65 .\\ M_u=135.88kNm$

ADD COMMENT EDIT

Community

Content

Company