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Design reinforcement for spread footings

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:

Design footing reinforcements

Moment calculation

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

Mu = Qu * l2/2                                                                                                            [2.11]


Qu 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

Calculating Reinforcement

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 = As/fy, the depth of stress block,

a = T/(0.85fc'b) = Asfy/(0.85f’c b) = Asfyd/(0.85f’c bd),

Let the reinforcement ratio, r = As/bd, then

a = rfyd/0.85f’c                                                                                   

Let m = fy/0.85f’c                      [2.13]

then, a = rdm

The nominal moment strength of the section,

Mn = T (d-a/2) = Asfy (d-a/2) = Asfy (d-rdm/2) = Asfy d- Asfy drm/2

ACI code requires that the factored moment,

Mu £ f Mn

Where, f = 0.9, is the strength reduction factor for beam design. Let Mu = f Mn , We have Mu = f (Asfy d- Asfy drm/2)

Divide both side by bd2, we have Mu/fbd =  (As/bd)fy -(As/bd) fy rm/2) = rfy - fy r2m/2)

Let Rn = Mu/fbd2,                    [2.14]

we can rewrite the equation as

r2(m/2) - r - Rn/fy = 0

Solving the equation, the reinforcement ratio,

r = (1/m)[1-Ö(1-2mRn/fy)]                    [2.15]

The area of reinforcement is As = rbd


Steps for calculating reinforcement.


The procedure for calculating reinforcement is as follows

1. Calculate factored moment, Mu.

2. Calculated factor m = fy/0.85f’c  

3. Calculate factor Rn = Mu/fbd2

4. Calculate reinforcement ratio, r = (1/m)[1-Ö(1-2mRn/fy)]

5. Calculate area of reinforcement is As = 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

  1. Assume a depth of the stress block, a.
  2. Calculate tensile force, T using equation [2.16]
  3. Calculate new depth of the stress block, a, using equation [2.13]
  4. If the new depth, a, is not close to the assumed, a, in step 1, repeat step 2 and 3 with new depth, a.
  5. If the new depth, a, is close to the assume, a, calculate area of reinforcement using equation [2.16]

Minimum and maximum reinforcements

ACI code requires the minimum reinforcement ratio, 


r min= 3 Öfc' / fy but not less than 200/fy.                  [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]


Example 6: Determine footing reinforcement for footing subjected to axial column load


  • 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


  1. Calculate factored column load,

  2. Pu = 1.2*25+1.6*25=70 kips

  3. Factored footing pressure = 70/(4.5*4.5) =3.45 ksf

  4. Distance from critical section to edge of footing = (4.5-1)/2=1.75’

  5. Factored moment at critical section Mu = (3.45)*1.752/2=5.29 ft-kips/ft

  6. Effective depth = 12”-3” (cover)-0.5” (rebar size) = 8.5”

  7. Factor Rn = (5.29)(1000)(12)/[(0.9)(12)(8.52)]= 81.4 psi

  8. Factor m = 60000/[(0.85)(3000)] = 23.5

  9. The reinforcement ratio is  r = (1/23.5){1-Ö[1-(2)(23.5)(81.4)/60000]} = 0.0014

  10. If use trial and error method, assume a = 1”

  11. T = (5.29*12)/[0.85*(8.5-1/2)]=9.33 kips/ft

  12. Check a = 9.33/[0.85*3*12] =0.31”

  13. Assume a = 0.31”

  14. T = (5.29*12)/[0.85*(8.5-0.31/2)]= 9 kips

  15. Check a = 9/[0.85*3*12]=0.3”                   (close enough)

  16. As =9/60=0.15 in2/ft

  17. The reinforcement ratio, r = 0.15/[8.5*12]=0.0015  

  18. Less than rmin = 200/ 60=0.0033 or rmin = (4/3) 0.00162=0.00216

  19. For a footing width of 4’6”, As = 0.216*8.5*4.5*12= 0.99in2.

  20. Use 5#4 in both direction, As = 5*0.2=1.0 in2.

Placing reinforcements.

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 ( 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


where is the ratio of length to short side.

Design column dowels

Dower rebars that go from the bottom of footing into the footing need the meet the following requirements:

  1. Transfer vertical column forces when column load exceeds the compressive strength of concrete.

  2. Transfer moment at column base

  3. Meet minimum reinforcement in ACI code

  4. Meet splice requirement for column reinforcement.

Bearing strength of concrete at base of column

The bearing strength of column at the column base

fPc = 0.65*0.85f'cAg

Where Ag is the gross section area of column


The bearing strength of footing at the column base is

fPc = 0.65*0.85f'caAg            a = ÖA2/A1    £ 2.

Edge length of A1, A2 are shown in the figure below.

Reinforcement required at the base of column

Where there is no moment at the column base, the area reinforcement through column base can be calculated as

As = (Pu - fPc) / fy                                                                                         [2.20]

Pu is factored column load.  When Pu < fPc, ACI code requires that the minimum reinforcement for dowel through column base is 0.005Ag.  Ag 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.

Length of dowel for compression

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

Ldb = 0.02 (fydb/Öfc’)                                                                                       [2.21]

Ldb =0.0003 fydb                                                                                              [2.22]

or  8 in

Where db is the diameter of rebar


The length of dowel is modified by the area of reinforcement as

Ld = (As required / As provided)( Ldb)                                                              [2.23]

The length of dowel that projects above the footing needs to meet the compression splice requirement of column reinforcement. 

When, fy £ 60,000 psi, Lab = 0.0005 fydb                 > Ld or 12”                         [2.24]

When, fy > 60,000 psi, Lab = (0.0009 fy – 24)db         > Ld 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.

Example 7: Design of column dowel


  • 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


Determine number and size of rebar

  1. Factored column load = 1.2*40+1.6*20=80 kips

  2. Bearing strength of column = 0.65*0.85*4*12*12=317.5 kips         >80 kips

  3. A1 = 12*12=144 in2.

  4. Effective depth = 12-3-1=8 in.

  5. Edge length of A2 = 12+8*2*2 = 44”               < 48”

  6. Area of A2 = 44*44=1936 in2.

  7. The ratio, a = ÖA2/A1=Ö1936/144=3.7          >2                    Use 2.

  8. The bearing strength of footing at column base = 0.65*0.85*2*3*144=477 kips >80 kips

  9. Use minimum reinforcement, As = 0.005*144=0.72 in2

  10. Use 4#4, As = 0.8 in2

Length of dowel in footing:

  1. Basic development for #4 bars:

  2. Ldb = 0.02 (fydb/Öf'c) = 0.02(60,000*0.5/Ö3000) = 11”

  3. Ldb =0.0003 fydb=0.0003*60000*0.5= 9 in.

  4. Required development length, Ld = 11”*(0.72/0.8) = 9.9”

  5. 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:

  1. For #6 bars in column,

  2. Lap = 0.0005 fydb =    0.0005*60000*0.75=22.5 in  

  3. Ld = 0.02*60000*0.75/Ö40000=14 in.

  4. Ld = 0.0003*60000*0.75=13.5 in.

  5. Use Lap = 22.5 in.