Design of a square footing
The procedure for designing a square footing is as follows:
Service load design:

Determine size of footing.
Reinforced concreter design:

Determine depth of footing for punching shear and direct
shear

Determine footing reinforcement for bending moment.

Determine column dowel to transfer column load.
Example:
Design of a square footing
Given:

Column loads:

Live load: 80 kips

Dead load: 120 kips

Footing uplift: 0 kips

Column size: 1 ft. x 1 ft.

Soil information:

Allowable soil bearing capacity:
4000 psf

Soil cover above footing: 1 ft

Unit weight of soil: 120 pcf

Materials used:

Concrete strength at 28 day = 3000
psi

Yield strength of rebars = 60 ksi
Design code: ACI 31805
Requirement: Determine
size, depth, and reinforcement for a square footing.
Solution:
Service load design:
1. Determine
footing sizes:

Assume a footing depth of 18”,net soil bearing capacity
,

Q_{net} =
4000 – 150*18/12120*1 = 3655 psf

Required footing are, A = (100+100) (1000) / 3655 = 54.7
ft^{2}

Use 7'6" by 7'6" square footing. The footing area is
56.3 ft^{2}.
Reinforced concrete design:
2. Determine footing
depth
The factored footing pressure can be calculated as
Q_{u} =
(1.2 x 120 + 1.6 x 80) / 56.3 = 4.83 psf
a. Check punching shear

Assume the reinforcements are #6 bars, the effective
depth

d = 18"  3" (cover)  0.75" (one bar size) = 14.3 " =
1.2'

The punch shear stress can be calculated as

v_{u} =
4.83 [7.5^{2}(1+1.2)^{2}]*1000/[4 x 1.2
x (1+1.2) x 144] = 163.3 psi

The shear strength of concrete is fv_{c} =
0.75 x 4 x (3000)1/2 = 164 psi O.K.
b. Check direct shear:

The distance from the critical section of direct shear
to the edge of the footing,

l =
7.5/2  1/2 – 1.2 = 2.05'

The direct shear stress is

v_{u} = (4.83)(1000)(2.05)
/ (12)(14.3) = 57.7 psi per
foot width of footing.

The shear strength of concrete for direct shear is

f v_{c} =
0.75 x 2 x (3000)^{1/2}
= 82 psi >
57.7 psi O.K.
3. Determine
footing reinforcement

The distance from face of column to the edge of the
footing is

l =
7.5/2 – 1/2 = 3.25'

The factored moment at the face of the column is

M_{u} =
(4.83)(3.25)^{2}/2 = 25.5 kft. per
foot width of footing

Factor R_{n} =
(25.5)(1000)(12)/[(0.9)(12)(14.3^{2})]=139.5 psi

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

The reinforcement ratio is r =
(1/23.5){1Ö[1(2)(23.5)(139.5)/60000]}
= 0.0024

Minimum reinforcement ratio, r =
0.0033 or r =
(4/3)*0.0024 =0.0032

A_{s} = (0.0032
)(7.5)(12)(14.3) = 4.1 in^{2}.

Use 10  #6 bars in both directions, A_{s }= 4.4
in^{2}.
4. Designing column
dowels:

The bearing capacity of concrete at column base is

P_{c} =
(0.65)(0.85)(3)(12)(12) = 238.7 kips

The factor column load is

P_{u} =
(1.2)(120)+ (1.6)(80) = 272 kips

The required area of column dowels is

A_{s} = (272
 238.7) / 60 = 0.56 in^{2}

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

