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This portion of reinforced concrete design of spreading footing follows the requirement of ACI code 318-99. Factored loads should be used instead of service load. Factored footing pressure is used to determine footing reinforcement.
The topics include:
Example 6: Determine footing reinforcement for footing subjected to axial column load
Example 7: Design of column dowel
The footing needs to be reinforced for the bending moment producing from upward footing pressure. According to ACI code, the critical section is at the face of column. The factored moment at the critical section can be calculated as
M_{u} = Q_{u} * l^{2}/2 [2.11]
Where
Q_{u} is factored footing pressure
l is the distance from the face of column to the edge of footing.
Figure 2.5 Critical moment section of a footing
The footing reinforcements are designed based on ACI strength design method. At ultimate stress situation, the concrete at top portion is subjected to compression. The compressive stresses distribute uniformly over a depth a. The resultant of compressive stress, C is located at a distance, a/2, from the top surface. Tensile force is taken by rebar at an effective distance, d, from the top surface.
Figure 2.6 Tensile and compressive forces and stresses on a footing secti0n
According to ACI code, the ultimate compressive is 0.85f’_{c}, where f’_{c} is compressive strength of concrete. Therefore, the compressive stress, C = 0.85 f’_{c} a b, where b is the width of the footing in calculation. By equilibrium, the tensile force is equal to the compression resultant,
T = C = 0.85f’_{c} a b [2.12]
Since T = A_{s}/f_{y}, the depth of stress block,
a = T/(0.85f_{c}'b) = A_{s}f_{y}/(0.85f’_{c} b) = A_{s}f_{y}d/(0.85f’_{c} bd),
Let the reinforcement ratio, r = A_{s}/bd, then
a = rf_{y}d/0.85f’_{c}
Let m = f_{y}/0.85f’_{c} [2.13]
then, a = rdm
The nominal moment strength of the section,
M_{n} = T (d-a/2) = A_{s}f_{y} (d-a/2) = A_{s}f_{y} (d-rdm/2) = A_{s}f_{y} d- A_{s}f_{y} drm/2
ACI code requires that the factored moment,
M_{u} £ f M_{n}
Where, f = 0.9, is the strength reduction factor for beam design. Let M_{u} = f M_{n} , We have M_{u} = f (A_{s}f_{y} d- A_{s}f_{y} drm/2)
Divide both side by bd^{2}, we have M_{u}/fbd = (A_{s}/bd)f_{y} -(A_{s}/bd) f_{y} rm/2) = rf_{y} - f_{y} r^{2}m/2)
Let R_{n} = M_{u}/fbd^{2}, [2.14]
we can rewrite the equation as
r^{2}(m/2) - r - R_{n}/f_{y} = 0
Solving the equation, the reinforcement ratio,
r = (1/m)[1-Ö(1-2mR_{n}/f_{y})] [2.15]
The area of reinforcement is A_{s} = rbd
Steps for calculating reinforcement.
The procedure for calculating reinforcement is as follows
1. Calculate factored moment, Mu.
2. Calculated factor m = f_{y}/0.85f’_{c}
3. Calculate factor R_{n} = M_{u}/fbd^{2}
4. Calculate reinforcement ratio, r = (1/m)[1-Ö(1-2mR_{n}/f_{y})]
5. Calculate area of reinforcement is A_{s} = rbd
Trial and error method
There are many ways to determine reinforcements. One simple method is using a trial and error method by assuming the depth of compression block, a. The steps are as follows
ACI code requires the minimum reinforcement ratio,
r _{min}= 3 Öf_{c}' / f_{y} but not less than 200/f_{y}. [2.17]
In addition, it also said that the minimum reinforcement does not need to be more than 4/3 of the calculated value,
r _{min}= (4/3)r [2.18]
Given:
Column loads:
Live load: 25 kips
Dead load: 25 kips
Footing and column information:
Footing sizes = 4 ft 6 in. x 4 ft 6 in. x 1ft
Column size: 1 ft x 1 ft
Concrete strength at 28 day = 3000 psi
Yield strength of rebars = 60 ksi
Requirement: Calculate footing reinforcements
Solution:
Calculate factored column load,
P_{u} = 1.2*25+1.6*25=70 kips
Factored footing pressure = 70/(4.5*4.5) =3.45 ksf
Distance from critical section to edge of footing = (4.5-1)/2=1.75’
Factored moment at critical section M_{u} = (3.45)*1.75^{2}/2=5.29 ft-kips/ft
Effective depth = 12”-3” (cover)-0.5” (rebar size) = 8.5”
Factor R_{n} = (5.29)(1000)(12)/[(0.9)(12)(8.5^{2})]= 81.4 psi
Factor m = 60000/[(0.85)(3000)] = 23.5
The reinforcement ratio is r = (1/23.5){1-Ö[1-(2)(23.5)(81.4)/60000]} = 0.0014
If use trial and error method, assume a = 1”
T = (5.29*12)/[0.85*(8.5-1/2)]=9.33 kips/ft
Check a = 9.33/[0.85*3*12] =0.31”
Assume a = 0.31”
T = (5.29*12)/[0.85*(8.5-0.31/2)]= 9 kips
Check a = 9/[0.85*3*12]=0.3” (close enough)
As =9/60=0.15 in^{2}/ft
The reinforcement ratio, r = 0.15/[8.5*12]=0.0015
Less than r_{min} = 200/ 60=0.0033 or r_{min} = (4/3) 0.00162=0.00216
For a footing width of 4’6”, As = 0.216*8.5*4.5*12= 0.99in^{2}.
Use 5#4 in both direction, As = 5*0.2=1.0 in^{2}.
Reinforcements should be placed at the tension side at the bottom of the footing.
For a square footing, rebars are placed uniformly in both directions. ACI code requires that the rebars be placed not more than 18 inch apart.
For a rectangular footing, rebars in the long direction are placed uniformly but not the short direction. ACI code (15.4.4.2) requires a certain portion of reinforcements in short direction to be placed within a band equal to the width of footing in the short direction. The distribution ratio is calculated based on the aspect ratio of footing as
[2.19]
where is the ratio of
length to short side.
Dower rebars that go from the bottom of footing into the footing need the meet the following requirements:
Transfer vertical column forces when column load exceeds the compressive strength of concrete.
Transfer moment at column base
Meet minimum reinforcement in ACI code
Meet splice requirement for column reinforcement.
The bearing strength of column at the column base
fP_{c} = 0.65*0.85f'_{c}A_{g}
Where A_{g} is the gross section area of column
The bearing strength of footing at the column base is
fP_{c} = 0.65*0.85f'_{c}aA_{g} a = ÖA_{2}/A_{1} £ 2.
Edge length of A1, A2 are shown in the figure below.
Where there is no moment at the column base, the area reinforcement through column base can be calculated as
A_{s} = (P_{u} - fP_{c}) / f_{y} [2.20]
P_{u} is factored column load. When P_{u} < fP_{c}, ACI code requires that the minimum reinforcement for dowel through column base is 0.005A_{g}. A_{g} is the cross section area of column. The diameter of the dowel should not exceeds the longitudinal reinforcement of column by 0.15 in.
When the column base is subjected to both axial loads and moments, the column dowel needs to be designed to resist column moment. The design procedure is the same as design of beam-columns.
The length of dowel that below the column base need to meet minimum development length of ACI 318-05 code.
Basic development length for compression member is the larger of
L_{db} = 0.02 (f_{y}d_{b}/Öf_{c}’) [2.21]
L_{db} =0.0003 f_{y}d_{b} [2.22]
or 8 in
Where d_{b} is the diameter of rebar
The length of dowel is modified by the area of reinforcement as
L_{d} = (A_{s} required / A_{s} provided)( L_{db}) [2.23]
The length of dowel that projects above the footing needs to meet the compression splice requirement of column reinforcement.
When, f_{y} £ 60,000 psi, L_{ab} = 0.0005 f_{y}d_{b}, > L_{d} or 12” [2.24]
When, f_{y} > 60,000 psi, L_{ab} = (0.0009 f_{y} – 24)d_{b} > L_{d} or 12” [2.25]
When column base is subject to moment and the rebars are in tension, length of splice and anchor should be designed based on tension requirement.
Given:
Column loads:
Live load: 20 kips
Dead load: 40 kips
Footing and column information:
Footing sizes = 4 ft x 4 ft x 1ft
Column size: 1 ft x 1 ft
Concrete strength at 28 day for footing = 3000 psi
Concrete strength at 28 day for column = 4000 psi
Yield strength of rebars = 60 ksi
Column reinforcement: 4#6
Footing reinforcement: 4#4 each way.
Design code: ACI 318-05
Requirement: Design column dowels including sizes and length
Solution:
Determine number and size of rebar
Factored column load = 1.2*40+1.6*20=80 kips
Bearing strength of column = 0.65*0.85*4*12*12=317.5 kips >80 kips
A_{1} = 12*12=144 in^{2}.
Effective depth = 12-3-1=8 in.
Edge length of A_{2} = 12+8*2*2 = 44” < 48”
Area of A_{2} = 44*44=1936 in^{2}.
The ratio, a = ÖA_{2}/A_{1}=Ö1936/144=3.7 >2 Use 2.
The bearing strength of footing at column base = 0.65*0.85*2*3*144=477 kips >80 kips
Use minimum reinforcement, A_{s} = 0.005*144=0.72 in^{2}
Use 4#4, A_{s} = 0.8 in^{2}
Length of dowel in footing:
Basic development for #4 bars:
L_{db} = 0.02 (f_{y}d_{b}/Öf'_{c}) = 0.02(60,000*0.5/Ö3000) = 11”
L_{db} =0.0003 f_{y}d_{b}=0.0003*60000*0.5= 9 in.
Required development length, L_{d} = 11”*(0.72/0.8) = 9.9”
Since the distance from the top of footing to the #4 bottom reinforcement is 7.5”, use dowel with 90 degree hook, 7.5 in vertical and 2.5 “ turn.
Splice length in column:
For #6 bars in column,
L_{ap} = 0.0005 f_{y}d_{b} = 0.0005*60000*0.75=22.5 in
L_{d} = 0.02*60000*0.75/Ö40000=14 in.
L_{d} = 0.0003*60000*0.75=13.5 in.
Use L_{ap} = 22.5 in.