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Column Design : Biaxial

Course Title : Prestressed Concrete SessionalCourse No : CE 416

Presented By Tanmay Biswas

ID : 10.01.03.161

WHAT IS COLUMN?

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.

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.

FUNCTION OF COLUMN:

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.

TYPES OF COLUMN

Columns are defined as members that carry loads chiefly in compression the design of column then requires the computation of the cross sections failure surface.

The biaxial bending method discussed rectangular column to be designed if bending is present about only one of the two principal axis of section.

Biaxial Column

Biaxially Loaded Column

Intraction Curve

Interaction Diagram

Figure : 1

Figure 1:a) Uniaxial bending about y axisb) Biaxial bending about x axisc) Biaxial bending about diagonal axisd) Interaction surface

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Approximation of section through intersection surface

Method 1 : Strain compatibility methodThis is the most nearly theoretically correct method of solving biaxially loaded column.

Method 2 : Equivalent Eccentricity methodAn approximate method. Limited to columns that are symmetrical about two axes with a ratio of side lengths between 0.5 and 2.0.

Analysis And Design

Method 3 : Reciprocal load method

Method 4 : Contour load method

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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.

EXAMPLE: DESIGN OF COLUMN FOR BIAXIAL BENDING

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

Design procedures:

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.

By the reciprocal load method first considering bending aboutthe Y axis, γ = 15/20 = 0.75 and e/h = 6/20 = 0.3With the reinforcement ratio of Ast/bh= 8/240 = 0.033, using the

avg graps A.6 γ =0.7 and A.6 γ =0.8Pnyo/ 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

Solution:

Then for the bending about the X axis, γ = 7/12 = 0.6 and e/h = 3/12 = 0.25Graph A.5 Appendix A givesPnxo/ ƒ’cAg (avg) =0.65Pnxo= 0.65*4*240= 624kips Po/ ƒ’cAg =1.31 Po=1.31*4*240=1258kips

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.

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