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Example 3.2. Determine maximum shear and moment of a combined footing
Example 3.3: Reinforced concrete design of a combined footing
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:
Structural analysis:
Reinforced concrete design:
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:
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:
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.
Given:
A combined footing as shown in Example 3.1
Column size: 1 ft by 1ft for both A & B
Design code: ACI 318-05
Requirements: Determine maximum shear and moment in longitudinal direction.
Solution:
Factored column loads:
Column A: P_{ua} = 1.2*50+1.6*40=124 kips
Column B: P_{ub} = 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, Q_{u} = (124+248)/22=16.9 k/ft
Shear diagram:
At point 1: V_{u} = 16.9*1.5-124= -98.7 kips
At point 3: V_{u} = 16.9*(1.5+14)-124=138 kips
At point 4: V_{u} = 16.9*(1.5+14+1)-124-248= -93.2 kips
Moment diagram:
At point 1: M_{u} = 16.9*1.5^{2}/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
M_{u} = 16.9*(1.5+6.24)^{2}/2-124*(0.5+6.24)=-329.5 ft-kips
At point 3: M_{u} = 16.9*(1.5+14)^{2}/2-124*(0.5+14)=232.1 ft-kips
At point 4: M_{u} = 16.9*5.5^{2}/2=255.6 ft-kips
Design procedure:
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
v_{u} = [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 v_{c} = 0.75 x 4 x 4000 = 189.7 psi O.K.
The perimeter of punching shear is
P = 4*(12+20.3)=129.2”
The punch shear stress can be calculated as
v_{u} = [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
V_{u} =
138*(7.76-1.2)/7.76= 116.7 kips
The shear strength of concrete for footing section,
f V_{c} = f v_{c}*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
M_{u} = 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)(f_{c}')(b)] = 170.8/[(0.85)(4)(60)] = 0.84 » 0.9”
As = T/f_{y} = 170.8 / 60 = 2.84 in^{2} .
The reinforcement ratio is
r = As/bd = 2.84/(60)(20.4) = 0.0023
Minimum reinforcement ratio,
r _{min}= 0.0033 > r_{min} =( (4/3)*0.0023=0.0031
Use r_{min} = 0.0031,
A_{s} = (0.0031 )(60)(20.4) = 3.8 in^{2}.
Use 9-#6 bars in both directions, A_{s} = 3.96 in^{2}.
d. Maximum negative reinforcement in longitudinal direction
The maximum negative moment between column is
M_{u} = 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 in^{2} .
The reinforcement ratio is
r = A_{s}/bd = 3.7/(60)(20.4) = 0.0031
Minimum reinforcement ratio,
r _{min}= 0.0033 < r_{min} =( (4/3)*0.0031=0.0041
Use r = 0.0033,
A_{s} = (0.0033 )(60)(20.4) = 4.0 in2.
Use 10-#6 bars in both directions, A_{s} = 4.4 in2.
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
M_{u} = (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
A_{s} = 4.4/60 = 0.073 for one foot section.
The reinforcement ratio is
r = A_{s}/bd = 0.073/(12)(20.4) = 0.0003
Minimum reinforcement ratio,
r _{min}= 0.0033 < r_{min} =( (4/3)*0.0003=0.0004
Use r_{min} =0.0004
A_{s} = (0.0004 )(22)(12)(20.4) = 2.2 in2.
Use 16 #4 bars, A_{s} = 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
P_{c} = (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
A_{s,min} = (0.0005)(12)(12) = 0.72 in^{2}
Use 4 - #4 dowels A_{s} = 0.8 in^{2}
The footing is shown in below