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## Design of a square footing

The procedure for designing a square footing is as follows:

Reinforced concreter design:

#### Example: Design of a square footing

Given:

• Live load: 80 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 318-05

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

Solution:

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

2. Qnet = 4000 – 150*18/12-120*1 = 3655 psf

3. Required footing are, A = (100+100) (1000) / 3655 = 54.7 ft2

4. Use 7'-6" by 7'-6" square footing. The footing area is 56.3 ft2.

Reinforced concrete design:

The factored footing pressure can be calculated as

Qu = (1.2 x 120 + 1.6 x 80) / 56.3 = 4.83 psf

a. Check punching shear

1. Assume the reinforcements are #6 bars, the effective depth

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

3. The punch shear stress can be calculated as

4. vu = 4.83 [7.52-(1+1.2)2]*1000/[4 x 1.2 x (1+1.2) x 144] = 163.3 psi

5. The shear strength of concrete is fvc = 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 = 7.5/2 - 1/2 – 1.2 = 2.05'

3. The direct shear stress is

4. vu = (4.83)(1000)(2.05) / (12)(14.3) = 57.7 psi            per foot width of footing.

5. The shear strength of concrete for direct shear is

6. f vc = 0.75 x 2 x (3000)1/2  = 82 psi         > 57.7 psi        O.K.

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

2. l = 7.5/2 – 1/2 = 3.25'

3. The factored moment at the face of the column is

4. Mu = (4.83)(3.25)2/2 = 25.5 k-ft. per foot width of footing

5. Factor Rn = (25.5)(1000)(12)/[(0.9)(12)(14.32)]=139.5 psi

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

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

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

9. As = (0.0032 )(7.5)(12)(14.3) = 4.1 in2.

10. Use 10 - #6 bars in both directions, As = 4.4 in2.

1. The bearing capacity of concrete at column base is

2. Pc = (0.65)(0.85)(3)(12)(12) = 238.7 kips

3. The factor column load is

4. Pu = (1.2)(120)+ (1.6)(80) = 272 kips

5. The required area of column dowels is

6. As = (272 - 238.7) / 60 = 0.56 in2

7. The minimum dowel area is

8. As,min = (0.0005)(12)(12) = 0.72 in2

9. Use 4 - #4 dowels As = 0.8 in2

10. The footing is shown in below ### Related Topics

Reinforced concrete design

Bearing capacity