0
1.4kviews
Design a short column square is c/s to carry an axial load of $2200$ kN using $M20/Fe415.$ Adopt LSM. Also design lateral ties.
1 Answer
0
5views

$$P=220 kN\hspace {0.5 cm} M20 \space \& \space Fe415$$

$ P_u=1.5×2200=3300 kN \\ \text { Assume 1% steel of gross c/s } \\ \text { 99% concrete of gross c/s } \\ A_{sc}=0.01 A_g \\ A_c=0.99 A_g \\ P_{uc}=0.4 f_{ck} A_c+0.67 f_y A_{sc} \\ 3300 ×10^3=0.4 ×20×0.99 A_g+0.67 ×415×0.01 A_g \\ A_g=308396.8 mm^2 \\ A_g=side^2 \\ side =\sqrt{308396.8 } \\ =555.33 \\ side ≈560 mm (A_g)_{provided}=560 ×560 =313600mm^2 \\ A_{sc}=0.01×313600=3136 mm^2 \\ \text {Provide 8-25 mm∅ } \\ A_{sc} \text { provided }=8×\pi/4×25^2=3928mm^2 \\ A_s provided=A_{qp}-A_{scp}=313600-3928=309672 mm^2 \\ Check \\ P_u=[(0.4f_{ck}(Ac)_p )+(0.67f_y (Asc)_p)] \\ P_u=[(0.4×20×309672)+(0.67×415×3928)] \\ P_u=35569.55 kN \gt 3300kN \space \text { Therefore Safe. }\\ \text { Lateral Ties }\\ ∅_{ties} ≥ \dfrac {∅_{main}}4=\dfrac {25}4=6.25mm ≅8mm \\ Spacing \\ S_1=\text { least lateral Dimension }=560mm \\ S_2=16 ×∅_{main}=16×25=400mm \\ S_3=48×∅_{ties}=48×8=384mm \\ S_4=300mm \\ \text { Provided 8mm ∅ ties @ 300 mm c/c }$

enter image description here

Please log in to add an answer.