Design of rectangular footings

Designing procedure:

Service load design:

1. Determine required footing size from required footing area and limitation of footing width.

Reinforced concreter design:

2. Determine footing depth for punching shear and direct shear in the longitudinal direction.

3. Determine footing reinforcement for longitudinal direction.

4. Determine footing reinforcement for transverse direction.  Distribute reinforcement based on length to width aspect ratio.

5. Determine column dowel to transfer column load.

Example 9: Design of a rectangular footing

Given:

• Column loads:

• Live load: 60 kips

• Dead load: 100 kips

• Footing uplift: 0 kips

• Column size: 1 ft 6 in. x 1 ft.

• Footing information:

• One side of footing is limited to 5’ due to property line

• Soil information:

• Allowable soil bearing capacity: 3500 psf

• Soil cover above footing: 1 ft

• Unit weight of soil: 100 pcf

• Materials used:

• Concrete strength at 28 day = 3000 psi

• Yield strength of rebars = 60 ksi

Design code: ACI-318-05

Requirement:  Determine size, depth, and reinforcement for a square footing.

Solution:

Service load design

1. Determine footing sizes

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

2. Q_net = 3500 – 150*18/12-100*1 = 3175 psf

3. Required footing are, A = (60+100) (1000) / 3175 = 50.4 ft²

4. Since one side of the footing is limited to 5', the length of footing is

5. L = 50.4/5 = 10.1'        Use 10’, the footing area is 50 ft².

Reinforced concrete design:

2. Determine footing depth

The factored footing pressure can be calculated as

Q_u = (1.2 x 100 + 1.6x 60) / 50 = 4.32 psf

a. Check punching shear

Assume the reinforcements are #8 bars, the effective depth

d = 18" - 3" (cover) - 1" (one bar size) = 14" = 1.16'

The punch shear stress can be calculated as

v_u = (4.32)[50-(1.5+1.6)(1+1.16)](1000)/[(2)(1.16)(1.5+1.16+1.16)(144)]= 118 psi

The shear strength of concrete is f v_c = 0.75 x 4 x ( 3000)^1/2 = 164 psi            O.K.

b. Check direct shear:

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

2. l = (10– 1.5)/2 – 1.16 = 3.09'

3. The direct shear stress is v_u = (4.32)(1000)(3.09) / (12)(14) = 79.4 psi    per foot width of footing.

4. The shear strength of concrete for direct shear is f v_c = 0.75 x 2 x ( 3000)^1/2 = 82 psi    > 79.4 psi     O.K.                                                   

3. Determine footing reinforcement.  

Longitudinal direction

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

l = (10– 1.5)/2 =4.25 '

The factored moment at the face of the column is

M_u = (4.32)(4.25)²/2 = 39 k-ft.        per foot width of footing

Factor R_n = (39)(1000)(12)/[(0.9)(12)(14²)]= 221.1 psi

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

The reinforcement ratio is  r = (1/23.5){1-Ö[1-(2)(23.5)(221.1)/60000]} = 0.00386

Minimum reinforcement ratio,

r = 0.0033  < r = 0.00386

Use calculated reinforcement

A_s = (0.00386)(5)(12)(14) = 3.24 in².

Use 5#8, A_s = 0.79*5=3.95 in².

Transverse direction

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 = (4.32)(2)²/2 = 8.6 k-ft. per foot width of footing

Factor R_n = (39)(1000)(12)/[(0.9)(12)(14²)]= 48.8 psi

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

The reinforcement ratio is  r = (1/23.5){1-Ö[1-(2)(23.5)(48.8)/60000]} = 0.00082

Minimum reinforcement ratio,

r = 0.0033 or r_min =(4/3)*0.00082=0.001

Use r_min =0.0011

A_s = (0.0011 )(10)(12)(14) = 1.9 in².

Use 11 #4 bars, A_s = 0.2*11=2.2 in².

Distribute reinforcements

1. The aspect ratio, b = 10/5=2

2. The distribution ratio, g = 2/(2+1) = 0.67

3. The reinforcement in the 5’ width center band is

4. N=11*0.67=7.4

5. Use 7 #4 in the center 5’ band, spacing = 5*12/7 = 8.6 in.                O.K.

6. Use 2#4 each side

7. Maximum spacing = [(10*12-5*12)/2 –3 (cover)]/2=13.5 in.               O.K.

4. Designing column dowels. 

1. The bearing capacity of concrete at column base is

2. P_c = (0.65)(0.85)(3)(18)(12) = 359.1 kips

3. The factor column load is

4. P_u = (1.2)(100)+ (1.6)(60) = 216 kips             < 359.1 kips

5. Use minimum dowel area ,

6. A_s,min = (0.0005)(18)(12) = 1.08 in²

7. Use 4 - #5 dowels As = 1.2 in²

8. The footing is shown in below