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 .
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.
12.
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.
