CE-REF.COM
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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
Service load design:
Structural analysis
Reinforced concrete design
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: 22 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 footing A & B.
Solution:
Assume a footing width of 6 ft, the eccentricity of footing A is e = 6/2-1=2’.
The distance between footing reaction, L = 22-2=20’
The eccentric moment is M = (40+50)*2=180 ft-kips
The shear produced by M is, V = 180/20=9 kips
Reaction at footing A = 40+50+9 =99 kips
Net soil bearing capacity = 3000-2*150-120=2580 psf
Required footing area of A = 99/2.58=38.4 ft2.
Use 6’ by 6.5’ footing, area = 39 ft2.
Reaction at footing B = 100+80-9=171 kips
Required footing area = 171/2.58=66.3 ft2.
Use 8’ by 8.5’ footing, A = 68 ft2.
Given: The strap footing in example 3.4
Design code: ACI 318-05
Requirement: Determine maximum factored shears and moment in the footings and strap.
Solution:
Factored column load of A = 1.2*50+1.6*40=124 kips
Factored column load of B = 1.2*100+1.6*80=248 kips
Factored eccentric moment, M_{u} = 124*2=248 ft-kips
Factored shear, V_{u} = 248/20=12.4 kips
Factored footing reaction at A = 124+12.4= 136.4 kips
Factored footing pressure per linear foot of A = 136.4/6=22.7 k/ft
Factored footing reaction at B = 248-12.4= 235.6 kips
Factored footing pressure per linear foot at B = 235.6/8=29.5 k/ft.
Shear diagram:
At point 1: V_{u} = 22.7*1.5-124= -90 kips
At point 2: V_{u} = 22.7*6-124= 12.2 kips
At point 3: V_{u} = 22.7*6-124=12.2 kips
At point 4: V_{u} = 12.4+29.5*3.5= 115.7 kips
At point 5: V_{u} = 29.5*-3.5=-103.5 kips
Moment diagram:
At point 1: M_{u} = 22.7*1.5^{2}/2-124*0.5= -36.5 ft-kips
At point 2: M_{u} = 22.7*6^{2}/2-124*5= -211.4 ft-kips
At point 3: M_{u} = 22.7*6*(6/2+13)-124*(5+13)=-52.8 ft-kips
At point 4: M_{u} = 22.7*6*(6/2+13+3.5)-124*(5+13+3.5)+29.5*3.5^{2}/2=183 ft-kips
At point 5: M_{u} = 29.5*3.5^{2}/2= 180.7 ft-kips
Design procedure:
Given:
A strap footing with loading, shear, and moment as shown in example 3.4 & 3.5
Compressive strength of concrete for footing at 28 days: 3000 psi
Yield strength of rebars: 60 ksi
Design code: ACI 318-05
Requirement: design footing depth and flexural reinforcements.
Solution:
1. Design footing strap
Assume a 2’-6” by 2’ footing strap and the reinforcement is #8 bars, with 2” top cover the effective depth, d = 24-2-1=21”
a. Check direct shear
From Example 3.5, the factored shear force on footing strap, Vu = 12.4 kips
Factored shear strength of concrete,
f V_{c} = f v_{c}*b*d = (0.75 x 2 x 3000)*30*21/1000=51.7 kips
Minimum shear strength of concrete without shear reinforcement.
1/2f V_{c} =0.5*51.7=25.8 kips > 12.4 kips no shear reinforcement is required
b. Design flexural reinforcement
From Example 3.5, the maximum factored moment at point 2, M_{u} =211.4 ft-kips
Use trail method for reinforcement design
Assume a = 1.8 ".
T = M_{u}/f(d-a/2)
= (211.4)(12)/[(0.9)(21-1.8/2)]=140.2 kip
Calculate new a,
a = T/0.85f_{c}'b = 140.2/[(0.85)(3)(30)] = 1.83 in » 1.8” assumed
A_{s} = T/f_{y} = 140.2/60 = 2.33 in^{2}.
The reinforcement ratio is
r = A_{s}/bd = 2.33/[(21)(30)] = 0.0037
Minimum reinforcement ratio,
r_{min} = 0.0033
Use 4#7 bars, As = 0.6*4= 2.4 in^{2}.
a. Check punching shear
Assume a 16” depth of footing and #6 bars, the effective depth
d = 16" - 3" (bottom cover) – 0.75 (one bar size) = 12.3 " = 1.02’
Factored footing pressure = 22.7/6.5=3.49 kips/ft^{2}.
The perimeter of punching shear is at one half effective depth from face of column.
Since the distance from exterior face of column to exterior face of footing, 6", is less than effective depth, d,
the perimeter of punching shear is only 3 sides. At the interior face of column the length is column width, 12" plus d.
At the other two faces is column width + d + 6".
P = 2*(6”+12”+12.3”/2)+(12”+12.3”)=72.6”
The punch shear stress can be calculated as
v_{u} = [124-(3.49)(1+1.02)(0.5+1+1.02/2](1000)/[(12.3)(72.6)]= 123 psi
The shear strength of concrete is
f v_{c} = 0.75 x 4 x Ö3000 = 164.3 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 3.4, the maximum direction shear is –90 kips at inside face of column.
The factored shear at one effective depth, d, from the face of the column is
V_{u} = 90*(4.5-1.02)/4.5=69.6 kips
Factored shear strength of concrete,
f V_{c} = f v_{c}*b*d = (0.75 x 2 x Ö3000)*6.5*12*12.3/1000 = 78.8 kips > 69.6 kips O.K.
c. Determine maximum negative reinforcement in longitudinal
direction
The location of zero shear is at
X = 4.5*90/(90+12.2) = 3.96’ from inside face of the column
The maximum negative moment is at zero shear, at 3.96’ from inside face of column, or 5.46’ from exterior end of footing.
M_{u} = 22.7*5.46^{2}/2-124*(0.5+3.96)=-214.7 ft-kips
Use trail method for reinforcement design
Assume a = 1.3".
T = M_{u}/f(d-a/2)
= (214.7)(12)/[(0.9)(12.3-1.3/2)]= 245.7 kip
Calculate new a,
a = T/0.85f_{c}'b = 245.7/[(0.85)(3)(6.5*12)] = 1.24 in » 1.3” assumed
A_{s} = T/f_{y} = 245.7/60 = 4.1 in^{2}.
The reinforcement ratio is
r = A_{s}/bd = 4.1/[(78)(12.3)] = 0.0043
Minimum reinforcement ratio,
r_{min} = 0.0033
Use 8-#7 bars, 4#7 extended from footing strap, 2 #7 in each side of footing,
As = 0.6*8=4.8 in^{2}. Place reinforcement at top face of footing.
d. Determine reinforcement in transverse direction
The distance from face of column to the edge of the footing is
l = (6.5– 1)/2 =2.75'
The factored moment at the face of the column is
M_{u} = (3.49)(2.75)^{2}/2 = 13.2 k-ft. per foot width of footing
Use trail method for reinforcement design
Assume a = 0.5".
T = M_{u}/f(d-a/2)
= (14.7)(12)/[(0.9)(12.3-0.5/2)]= 16.3 kip
Calculate new a,
a = T/0.85f_{c}'b = 16.3/[(0.85)(3)(12)] = 0.53 in » 0.5” assumed
A_{s} = T/f_{y} = 16.3/60 = 0.27 in^{2}. per one foot section.
The reinforcement ratio is
r = A_{s}/bd = 0.27/[(12)(12.3)] = 0.0018
Minimum reinforcement ratio,
rmin = 0.0033 > rmin =(4/3)*0.0018=0.0024
Use r_{min} =0.0024
A_{s} = (0.0024 )(6)(12)(12.3) = 2.1 in2.
Use 5 #6 bars, A_{s} = 0.44*5= 2.2 in2.
Place reinforcement at bottom face of footing.
Assume a footing depth of 20”
and #8 bars, the effective depth =20-3-1=16”
The factored footing pressure = 29.5/8.5= 3.47 ksf
a. Check punching shear
The perimeter of punching shear is at d/2 beyond faces of column
P = 4*(12+16)=112”
The punch shear stress can be calculated as
v_{u} = [248-(3.47)(1+1.33)^{2}](1000)/(16)(112) = 127.9 psi <186 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 3.4, the maximum direct shear is 115.7 kips at inside face of column.
The factored shear at one effective depth from the face of the column is
V_{u} = 12.2+(115.7-12.2)*(3.5-1.33)/3.5= 76.3 kips
Factored shear strength of concrete,
f V_{c} = f v_{c}*b*d = (0.75 x 2 x Ö3000)*8.5*12*16/1000= 134.1 kips > 85 kips O.K.
c. Determine maximum positive reinforcement in longitudinal
direction
M_{u} = 180.7 ft-kips
Use trail method for reinforcement design
Assume a = 0.6".
T = M_{u}/f(d-a/2)
= (180.7)(12)/[(0.9)(16-0.6/2)]= 153.4 kip
Calculate new a,
a = T/0.85f_{c}'b = 153.4/[(0.85)(3)(8.5*12)] = 0.59 in » 0.6” assumed
A_{s} = T/f_{y} = 153.4/60 = 2.6 in^{2}.
The reinforcement ratio is
r = A_{s}/bd = 2.6/[(8.5*12)(16)] = 0.0016
Minimum reinforcement ratio,
r_{min} = 0.0033 >r_{min} =(4/3)*0.0016=0.0021
Use r_{min} =0.0021
A_{s} = (0.0021 )(8.5)(12)(16) = 3.5 in^{2}.
Use 6-#7 bars, As = 0.6*6= 3.6 in^{2}.
d. Determine reinforcement in transverse direction
The distance from face of column to the edge of the footing is
l = (8.5– 1)/2 =3.75'
The factored moment at the face of the column is
M_{u} = (3.47)(3.75)^{2}/2 = 24.4 k-ft. per foot width of footing
Use trail method for reinforcement design
Assume a = 0.7".
T = M_{u}/f(d-a/2)
= (24.4)(12)/[(0.9)(16-0.7/2)]= 20.8 kip
Calculate new a,
a = T/0.85f_{c}'b = 20.8/[(0.85)(3)(12)] = 0.68 in » 0.7” assumed
A_{s} = T/f_{y} = 20.8/60 = 0.35 in^{2} per one foot section.
The reinforcement ratio is
r = A_{s}/bd = 0.35/[(12)(16)] = 0.0018
Minimum reinforcement ratio,
r_{min} = 0.0033 > r_{min} =(4/3)*0.0018=0.0024
Use r_{min} =0.0024
A_{s} = (0.0024)(8)(12)(16) = 3.7 in2.
Use 7 #7 bars, A_{s} = 0.6*7= 4.2 in2.
4. 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 in2
Use 4 - #4 dowels A_{s} = 0.8 in2
The footing is shown in below