Combined footings
Contents:
Combined footings and strap footings are normal used when
one of columns is subjected to large eccentric loadings.
When two columns are reasonably close, a combined footing is designed for
both columns as shown in Figure 3.1. When
two columns are far apart, a strap is designed to transfer eccentric moment
between two columns as shown in Figure 3.1.
The goal is to have uniform bearing pressure and to minimize differential
settlement between columns.
Figure 3.1
Combined footing and strap footing
Design procedure:
Service load design:
- Determine
the size of combined footing.
Structural analysis:
- Perform
structural analysis to determine moment and shear in various section of
the footing.
Reinforced concrete design:
- Check
punching shear & direct shear
- Design
longitudinal reinforcements.
- Design
transverse reinforcements.
- Design
column dowels.
The size of the footing shall be determined to have uniform
bearing pressure under the footing so that differential settlement is minimized.
The resultant of bearing pressures needs to coincide with the resultant of
column loads. The procedures are as
follows:
- Determine
the location of the resultant of column loads.
- Calculated
the required length of footing. The length of the footing is twice the
distance from the edge footing of the exterior column to the resultant of
column loads.
- Calculate
the width of footing. The
required area of footing is the total column load divided by allowable net
soil bearing pressure. The width of footing is the required footing area
divided by the length of footing.
Example 3.1: Determine size of a combined footing
Given:
-
Column information:
-
Column A: Live load = 40 kips, Dead load = 50 kips
-
Column B: Live load = 80 kips, Dead load = 100 kips.
-
Distance between two columns: 15 ft.
-
Footing information:
-
Allowable soil bearing capacity; 3000 psf
-
Distance from column A to edge of footing: 1 ft.
-
Allowable soil bearing capacity = 3000 psf
-
Weight of soil above footing = 120 psf
-
Depth of footing= 24”
-
Depth of soil above footing = 12”
Requirements: Determine the size of a combined
footing.
Solution:
Total column load of A = 40+50=90 kips
Total column load of B = 80+100 = 180 kips
Take moment about A,
Location of resultant from A= 180*15/(90+180) = 10 ft.
The length of footing = 2*(10+1) = 22 ft
Use 22 ft
Net soil bearing capacity = 3000-2*150-120=2580 psf
Required footing area = (90+180)/2.58=104.7 ft2.
Required width of footing = 104.7/22=4.8’
Use 5 ft
Structural analysis of a combined footing is the same as
analyzing an invert simply support beam supported by two columns with factored
soil pressure as loading. The
procedures are as follows:
- Calculate
factored footing pressure.
- Calculate
maximum shear at an effective
depth from the face of column
- Calculate
maximum positive and negative moment in the footing.
Maximum positive moment occurs at face of column.
Maximum negative moment occurs between two columns at zero-shear.
It is worth to mention that because of load factors, the
centroid of factored column loads does not necessary located at the center of
the footing. It means that the
factored footing pressure is no longer uniform.
The correct way to solve the problem to analyze the footing with
trapezoid shape of factored footing pressure.
Example 3.2. Determine maximum shear and moment
of a combined footing
Given:
Design code: ACI 318-05
Requirements: Determine maximum shear and moment in
longitudinal direction.
Solution:
Factored column loads:
Column A: Pua = 1.2*50+1.6*40=124 kips
Column B: Pub = 1.2*100+1.6*80=248 kips
Location of resultant from column A= 248*15/(124+248)=10
ft.
Since the location of resultant is at center of footing,
factored footing pressure is uniform.
Factored footing pressure per linear foot of footing, Qu =
(124+248)/22=16.9 k/ft
Shear diagram:
At point 1: Vu = 16.9*1.5-124= -98.7 kips
At point 3: Vu = 16.9*(1.5+14)-124=138 kips
At point 4: Vu = 16.9*(1.5+14+1)-124-248= -93.2 kips
Moment diagram:
At point 1: Mu = 16.9*1.52/2-124*0.5=
-43 ft-kips
At point 2:
Location of point 2: from triangular relation between
point 1 and point 3 in shear diagram
X = 14*98.7/(98.7+122.9)=6.24’
from inside face of column A
Mu = 16.9*(1.5+6.24)2/2-124*(0.5+6.24)=-329.5
ft-kips
At point 3: Mu = 16.9*(1.5+14)2/2-124*(0.5+14)=232.1 ft-kips
At point 4: Mu = 16.9*5.52/2=255.6
ft-kips
Design procedure:
- Check
both punching shear and direct shear. The
critical section of punching shear is at ½ effective depth from face of
column. The critical section of
direct shear is at one effective depth of column.
For column at the edge of footing the critical section of punching
shear only has three sides along the column.
Critical sections of punching shear and direct shear are shown below.
- Design
longitudinal reinforcements. Longitudinal reinforcements are design based on the
maximum moments from structural analysis.
Reinforcement for negative moment should be placed near top face of
the footing. Positive
reinforcement should be placed near bottom face of the footing.
- Design
transverse reinforcements. Transverse
reinforcements are designed based on moment in the transverse direction at
face of column. They should be
placed near bottom face of the footing.
- Design
column dowels.
Example 3.3: Reinforced concrete design of a
combined footing
Given:
-
A combined footing with loading, shear, and moment as
shown in example 3.1 & 3.2
-
Compressive strength of concrete for footing at 28
days: 4000 psi
-
Yield strength of rebar: 60 ksi
Design code; ACI 318-05
Requirement: Check shear stresses and design
flexural reinforcements.
Solution:
a. Check punching shear for column A
Assume the reinforcements are #6 bars, the effective depth
d = 24" - 3" (cover) - 0.75" (one bar size)
= 20.3 " = 1.7'
Factored footing pressure = (124+248)/(22*5)=3.38 kips/ft2.
The perimeter of punching shear is
P = 2*(6”+12”+20.3”/2)+(12”+20.3”)=88.6”
The punch shear stress can be calculated as
vu =
[124-(3.38)(1+1.7)(0.5+1+1.7/2)](1000)/(20.3*88.6) =57 psi
The shear strength of concrete is
f
vc = 0.75 x 4 x
4000 = 189.7 psi
O.K.
Check punching shear for column B
The perimeter of punching shear is
P = 4*(12+20.3)=129.2”
The punch shear stress can be calculated as
vu = [248-(3.38)(1+1.7)2](1000)/(20.3*129.2)
= 85.2 psi < 189.7 psi O.K.
b. Check direct shear:
The critical section of direct shear is at one effective
depth from the face of column. From
Example 3.2, the maximum direction shear is 138 kips at inside face of column
B. The distance from zero shear (point 2) to the maximum direct shear (point 3)
is 14-6.24= 7.76’. From triangular relationship, the direct shear at critical
section is
Vu
= 138*(7.76-1.2)/7.76= 116.7 kips
The shear strength of concrete for footing section,
f
Vc = f
vc*b*d
= (0.75 x 2 x
4000)*60*20.3/1000=115.6 kips »
116.7 kips (less than 1% difference) O.K.
c. Determine Maximum positive reinforcement in
longitudinal direction
The maximum positive moment at the face of the column B is
Mu
= 255.6 k-ft. for 5’ width
of footing
Use trail method for reinforcement design
Assume depth of stress block, a = 0.9".
T = Mu/[f(d-a/2)]
= [(255.6)(12)]/[(0.9)(20.4-0.9/2)] = 170.8 kips
Calculate new a,
a = T/[(0.85)(fc')(b)] = 170.8/[(0.85)(4)(60)] =
0.84 »
0.9”
As = T/fy = 170.8 / 60 = 2.84 in2
.
The reinforcement ratio is
r
= As/bd = 2.84/(60)(20.4) = 0.0023
Minimum reinforcement ratio,
r
min= 0.0033 > rmin
=( (4/3)*0.0023=0.0031
Use rmin
= 0.0031,
As
= (0.0031 )(60)(20.4) = 3.8 in2.
Use 9-#6 bars in both directions, As = 3.96 in2.
d. Maximum negative reinforcement in longitudinal
direction
The maximum negative moment between column is
Mu
= 329.5 k-ft. for 5’ width
of footing
Use trail method for reinforcement design
Assume a = 1.2".
T =
(329.5)(12)/[(0.9)(20.4-1.2/2)] = 221.8 kip
Calculate new a,
a = 221.8/[(0.85)(4)(60)] = 1.1"
»
1.2”
As = 221.8 /60 = 3.7 in2
.
The reinforcement ratio is
r
= As/bd = 3.7/(60)(20.4) = 0.0031
Minimum reinforcement ratio,
r
min= 0.0033 < rmin
=( (4/3)*0.0031=0.0041
Use r
= 0.0033,
As
= (0.0033 )(60)(20.4) = 4.0 in2.
Use 10-#6 bars in both directions, As = 4.4 in2.
e. Determine reinforcement in transverse direction
The distance from face of column to the edge of the footing
is
l = (5– 1)/2
=2'
The factored moment at the face of the column is
Mu
= (3.38)(2)2/2 = 6.76 k-ft. per foot
width of footing
Use trail method for reinforcement design
Assume a = 0.1".
T =
(6.76)(12)/[(0.9)(20.4-0.1/2)] = 4.4 kip
Calculate new a,
a = 4.4/[(0.85)(4)(12)] = 0.11
»
0.1” assumed
As = 4.4/60 = 0.073
for one foot section.
The reinforcement ratio is
r
= As/bd = 0.073/(12)(20.4) = 0.0003
Minimum reinforcement ratio,
r
min= 0.0033 < rmin
=( (4/3)*0.0003=0.0004
Use rmin
=0.0004
As
= (0.0004 )(22)(12)(20.4) = 2.2 in2.
Use 16 #4 bars, As = 0.2*15= 3 in2.
Maximum spacing = (22*12-3-3)/15= 17.2”
<Maximum spacing, 18” O.K.
e. Designing column dowels.
The bearing capacity of concrete at column base is
Pc
= (0.7)(0.85)(4)(12)(12) = 342.7 kips
Which is greater than factored column loads of both A and
B.
The minimum dowel area is
As,min
= (0.0005)(12)(12) = 0.72 in2
Use 4 - #4 dowels As
= 0.8 in2
The footing is shown in below
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