Determine size of footing
Column subjected to axial column load
and uplift
Column subjected to both axial column load and moment
or eccentric loading
Since factor of safety is included in determining allowable
soil bearing capacity, there is no need to add addition factor of safety in
determine the footing sizes. But,
since the bottom of footing is at a depth below ground surface, the weight of
soil and footing above the bearing area should be subtracted from the allowable
soil capacity. The required footing
area is column load divided by the net allowable soil bearing capacity.
A = P / Qa-net
[2.1]
Where
A: required footing area.
P: Axial column load
Qa net = net allowable soil bearing capacity.
The weigh of footing and the soil above should be heavy
enough to offset the uplift forces from wind or seismic.
Wt ³
U * F.S.
[2.2]
Where:
Wt: Total weight of footing
U: uplift force
F.S.: factor of safety.
This situation usually occurs at column at building bracing
location. The factor of safety for
uplift force in most of building codes is 1.5.
Example 1: Determine footing sizes for axial loads and
uplift.
Given:
-
Column loads:
-
Live load: 25 kips
-
Dead load: 25 kips
-
Uplift = 20 kips
-
Factor of safety for uplift = 1.5
-
Footing information:
-
Top of footing at 1 ft below ground surface, unit weigh
weight of soil: 100 lbs/ft3.
-
Allowable soil bearing capacity = 3000 psf
-
Unit weight of concrete: 150 lbs/ft3.
Requirement: Determine footing sizes for axial loads
and uplift.
Solution:
-
Total column service load = 25+25=50 kips
-
Assume a footing depth of 1 ft,
-
Net allowable soil bearing capacity =
3000-150*1-100*1=2750 psf
-
Required footing area = 50*1000/2750=18.2 ft2.
-
Try 4’6”x4’6” footing, footing area = 20.2 ft2.
-
Required weight of footing to offset uplift = 20*1.5=30
kips
-
Weight of footing above footing = 100*4.5*4.5/1000=2.0
kips
-
Required weight of footing = 30-25-2=3 kips
-
Required volume of footing = 3/0.15=20 ft3.
-
Required depth of footing = 20/20.2 =1 ft
-
Use 4’6”x4’6”x1’ footing.
Columns at the base of a moment revisiting frame are often
subjected to moment in addition to axial load.
Columns that at edge of buildings often have to be designed with
eccentricity due to limitation of property line. The bearing pressure at the bottom of footing will distribute
in trapezoidal or triangular shape. The
footing has to be sized so that maximum footing pressure does not exceed
allowable soil bearing capacity.
Figure 2.1 Footing pressures with eccentricity not more
than 1/6 footing width
When eccentricity is less than 1/6 width of footing,
footing pressure under the footing is distributed in trapezoidal shape. When
eccentricity equals to 1/6 width of footing, footing pressure distributes
triangularly with zero pressure at one end of the footing.
The soil bearing capacity can be calculated as
Q = P / A ±
M / S
[2.3]
P: Axial column Load
A: footing area
M = P*e, column moment in the x direction, e is
eccentricity in x direction.
S = LB2/6
section modulus of footing area in x direction
For a rectangular footing, the equation can be written as
Q= P / A ±
M / S
= P/(BL) ±
P*e/(LB2/6)
= (P/A) [1±e*B/6]
[2.4]
L, B are length and width of footing.
When footing is subjected to moments or eccentricities in
both direction, the equations become
Q = P / A ±
Mx / Sx ±
My / Sy
[2.5]
Or
Q = (P/A) [1±ex*B/6±ey*L/6]
[2.6]
Example 2: Determine maximum and minimum footing
pressure for footing with eccentricity < B/6.
Given:
-
Column loads:
-
Live load: 25 kips
-
Dead load: 25 kips
-
Live load moment = 20 ft-kips
-
Dead load moment = 20 kips
-
Footing information:
-
Footing sizes = 6 ft x 6 ft
Requirement:
Determine maximum and minimum footing
pressure.
Solution:
-
Total axial load = 25+25=50 kips
-
Total column moment = 20+20=40 ft-kips
-
Eccentricity = 40/50=0.8 ft
< B/6 = 1 ft
-
Maximum footing pressure = [50,000/(6x6)][1+0.8*6/6] =
2500 psf
-
Minimum footing pressure = [50,000/(6x6)][1-0.8*6/6] =
277 psf
When eccentricity exceeds 1/6 width of footing, soil
pressure under pressure distributes in a triangular shape with a portion of the
footing have zero pressure. The
resultant of footing pressure, R coincides with column load, P as shown below.
Since the center of the resultant is at 1/3 length of the triangle, the
length of the bearing area is three times of the distance from the center of the
column load to the edge of footing.
Figure 2.2 Footing pressure with eccentricity greater than 1/6 footing width
Therefore,
P = Qmax [3(B/2-2)L/2]
Then,
Qmax = 2P/[3(B/2-e)L]
[2.7]
Example 3: Determine maximum footing pressure for
footing with eccentricity > B/6
Given:
-
Column loads:
-
Live load: 25 kips
-
Dead load: 25 kips
-
Live load moment = 30 ft-kips
-
Dead load moment = 30 kips
-
Footing information:
-
Footing sizes = 6 ft x 6 ft
Requirement: Determine maximum and minimum footing
pressure.
Solution:
-
Total axial load = 25+25=50 kips
-
Total column moment = 30+30=60 ft-kips
-
Eccentricity = 60/50=1.2 ft
> B/6 = 1 ft
-
Maximum footing pressure = 2*50,000/[3*(6/2-1.2)*6] =
3086 psf
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