and 4 others joined a min ago.
0
1.5kviews
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 $M25/Fe415.$
1 Answer
0
4views
written 7.8 years ago by
teamques10
★ 64k
|
• modified 7.8 years ago |
$$I_{xx}=\dfrac {bD^3}{12}=\dfrac {550×350^3}{12}=1965.1×10^6 mm^4$$
$I_{yy}=\dfrac {bD^3}{12}=\dfrac {350×550^3}{12}=4852.6×10^6 mm^4 \\ Here \space I_{yy} \gt I_{xx}$
$$PG\space \space 126\space \space table $$
$P_{uc}=0.36 f_{ck} bD=0.36×25×350×550 \\ P_{uc}=1732.5 kN \\ P_u=P_{uc}+∑P_{usi}=1732.5+722.5=2455kN \\ Now M_u=M_{uc} + ∑ M_{usi} \\ M_{uc}=P_{uc}×(0.5D-0.42D)=1732.5×0.08×550=76.23kNm \\ Now, \\ M_{us1}= P_{us1}×e_1=262.27×0.235=01.63kNm \\ M_{us2}= P_{us2}×e_1=248.78×0.085=21.4kNm \\ M_{us3}= P_{us3}×e_1=175.34×0.085=14.9kNm \\ M_{us4}= P_{us4}×e_1=36.1×0235=-8.48kNm \\ ∑M_{us}=59.65\\ M_u=76.23+59.65 .\\ M_u=135.88kNm$
ADD COMMENT
EDIT
Please log in to add an answer.