BIAXIAL COLUMN DESIGN
- 1. AHSANULLAH UNIVERSITY OF SCIENCE & TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING NAME: IFAT HASAN SHAWON ID: 10.01.03.107
- 2. ON BIAXIAL COLUMN DESIGN
- 3. Generally a column is something which carries load from beam and slab. In other words columns are defined as the members that carries load mainly in compression. columns carry bending moment as well, about one or both axes of the cross section.
- 4. In beam column slab, normally slab transfer loads to beam and beam transfer loads to column and finally column transfer loads to footing. In slab column, slab transfer loads directly to column and column transfer loads to footing.
- 5. Short column: In short column the strength is governed by strength of the materials and the geometry of the cross section. Slender column: A column is said to be slender if its cross-sectional dimensions are small compared with its length.
- 6. The column having axial load acting in such a way that the load is eccentric about both the axes in the plane of the column then it is called biaxially loaded column.
- 7. There are situations in which for rectangular and square columns axial compression is accompanied by simultaneous bending about both principal axes of the section.
- 8. Figure shows inter action diagram for compression plus biaxial bending: a) uniaxial bending about Y axis; b) uniaxial bending about X axis; c) biaxial bending about diagonal axis; d) interaction surface
- 9. An analysis of the damage observed in 24 reinforced concrete (RC) columns tested under uniaxial and biaxial horizontal loading . The test results show that for biaxial loading conditions specific damage occurs for lower drift demands when compared with the corresponding uniaxial demand .
- 10. Load contour method Reciprocal load method Strain compatibility method Equivalent eccentricity method 45 slice through interaction surface
- 11. The 12 20 in column shown in fig is reinforced with eight no.9 bars arranged around the column perimeter, providing an area Ast = 8 in². A factored load Pu of 255 kips is to be applied with eccentricities eʏ =3in,ex= 6in. material strengths are ƒ’c = 4 ksi and ƒʏ= 60 ksi. Check adequacy by trial design.
- 13. In a typical design situation given the size and reinforcement of the trial column and the load eccentricities ex and ey following steps should be followed At first we need to calculate ratio γ then we shall calculate e/h. After that we shall calculate nominal loads Pnxo and Pnyo for uniaxial bending around the X and Y axes respectively, and the nominal load Po for concentric loading. Then 1/Pn is computed from equation, 1/ Pn =1/ Pnyo+1/ Pnxo -1/Po
- 14. From the eqaution Pn is calculated, where Pn =approximate value of nominal load in biaxial bending with eccentricities ex and ey. The design requirement is that the factored load Pu must not exceed ɸPn . ɸ = 0.65 for tied column and 0.70 for spiral column according to ACI code.
- 15. By the reciprocal load method first considering bending about the Y axis, γ = 15/20 = 0.75 and e/h = 6/20 = 0.3 With the reinforcement ratio of Ast/bh= 8/240 = 0.033, using the avg graps A.6 γ =0.7 and A.6 γ =0.8 Pnyo/ ƒ’cAg (avg) =0.62+0.66/2= 0.64 Pnyo= 0.64*4*240= 614kips Po/ ƒ’cAg =1.31 Po=1.31*4*240=1258kips
- 16. Then for the bending about the X axis, γ = 7/12 = 0.6 and e/h = 3/12 = 0.25 Graph A.5 Appendix A gives Pnxo/ ƒ’cAg (avg) =0.65 Pnxo= 0.65*4*240= 624kips Po/ ƒ’cAg =1.31 Po=1.31*4*240=1258kips
- 17. Now substituting these values in the equation 1/ Pn =1/ Pnyo+1/ Pnxo -1/Po =1/624+1/614-1/1258 =0.00244 From which Pn =410 kips. Thus according to Bresler method the design load, Pu =0.65*4109 = 267 kips can be applied safely.