04-21-2010, 11:13 AM
If we adopt the load of 39390.1KN/column (as estimated in the latest review on the design load), and assume that the circular or the polygonal core structure was to support the transverse load due to horizontal forces (wind or seismic), then we could design the column as a laterally braced column (i.e. it is designed to support mainly vertical load). In that case, we cannot afford not to recognize the gravity of the vertical load that a typical column has to support, as such we have to try to minimize on the section dimension by employing high grade concrete and or steel. If we are to employ grade 75 concrete and 460 steel, then the combined strength of the concrete and the steel should be such that they will be able to equal or surpass that due to the external load. Let’s assume that the section will be made of only concrete (unreinforced), then the cross section of concrete required should be equal or greater than 39390.1 x 10^3/0.45 x 75 = 1167114.0mm^2!!. If we decided to use 800mm x 800mm reinforced column for the say the first ¼ total height of the building from the foundation, then the cross sectional area of concrete provided = 800 x 800 = 640000 mm^2, which implies that steel is to provide 1167114.0 - 640000 = 527114mm^2. If we use the steel/concrete modular ratio of 15, then area of steel reinforcement = 527114/15 = 35140.9mm^2. For 50mm diameter steel, number of steel strands = 35140.9/0.25pi (50)^2 = 18. Note that we did not include the effect of the reduction of the area of concrete due the presence of steel. This will not have much influence on the area of steel to be provided in as much as we used a modification factor of 0.45 instead of anything in the range of 0.5 to 0.6. The ratio of the area of steel to concrete = 35140.9/640000 = 0.055 = 5.5%. Though this is within the limits allowed in the codes, I would prefer to upgrade it (putting into consideration the fact that these columns will in actuality participate in the carrying the lateral load and also the fact that the structure will be subjected to twist, the grave consequences if something should go wrong, as such has to be well reinforced-that is, to 18 + 14 = 32 strands) to make 32/4 = 8 numbers on each face of the column. I do not know if 50mm diameter steel does exist, so let’s scale the dimension down to 32mm diameter. In this case, lets adopt 8% of the cross sectional area of concrete as the area of steel to be provided. This implies that the strands of steel to be provided = (8% of 640000)/0.25pi (32)^2 =64 strands. This could be provided in 2 rings of 8/6/per face and central core in cross of2 x 4 strands. (4 x 8 + 4 x 6 + 2 x 4 = 64)(Refer to the screen shot above).
Regards
Teddy