﻿ Rectangular footing design

CE-REF.COM

Custom Search

## Design of rectangular footings

### Designing procedure:

Reinforced concreter design:

#### Example 9: Design of a rectangular footing

Given:

• 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, 08, 11

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

Solution:

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

2. Qnet = 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:

The factored footing pressure can be calculated as

Qu = (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

vu = (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 vc = 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 vu = (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 vc = 0.75 x 2 x ( 3000)1/2 = 82 psi    > 79.4 psi     O.K.

#### 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

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

Factor Rn = (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

As = (0.00386)(5)(12)(14) = 3.24 in2.

Use 5#8, As = 0.79*5=3.95 in2.

#### 4. 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

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

Factor Rn = (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 rmin =(4/3)*0.00082=0.001

Use rmin =0.0011

As = (0.0011 )(10)(12)(14) = 1.9 in2.

Use 11 #4 bars, As = 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.

1. The bearing capacity of concrete at column base is

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

3. The factor column load is

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

5. Use minimum dowel area ,

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

7. Use 4 - #5 dowels As = 1.2 in2

8. The footing is shown in below

### Related Topics

Reinforced concrete design

Bearing capacity