The Pad foundations are a type of shallow foundation made of reinforced concrete. Pad Foundations are simple, cost-effective, and typically used for light to moderate loads in stable soil conditions near the surface. Still, they are not suitable for unstable ground or heavy loads.
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Table of Contents
Design of a Pad Foundation
Please note: This design is based on BS 6399 Part 1:1984 Code of Practice of Dead Load and Imposed Load, BS 8110 Part 1:1985 Code of Practice for design and Construction, and BS 8110 Part 1: 1997 Code of Practice for design and construction.
Let’s start with making assumptions:
- Ultimate designed axial load = 2014.1KN
- Serviceability axial load = 1438.64KN
- Serviceability moment = 16.81kNm
- Fcu = 30N/mm2
- Fy = 460N/mm2
- Soil bearing capacity = 200kN/m2
For the serviceability limit state
- Assumed footing weight = 120KN
- Total serviceability load = 1438.64 + 120 = 1558.64KN
Therefore, from above:
- P = N/BD + 6M/BD2
- Where P is the safe bearing capacity
- Assume square footing, i.e., B = D
- P = N/B2 + 6M/B3 = 1558.64/ B2 + 100.86/ B3
Therefore:
- 200 kN/mm2 = 1558.64/ B2 + 100.86/ B3
- Solve for B, which will be equal to 2.8m
- Provide a base 2.8 m square, area = 7.8m2
Eccentricity, e:
- Eccentricity, e = M/N = 16.81 kNm/1558.64kN
- e = 0.01m
- D/6 = 2.8/6 = 0.47
- Therefore, e = 0.01m < D/6 = 0.47
For the ultimate limit state
Assume a 625mm thick footing, and with the footing constructed on a blinding layer of concrete, the minimum cover is taken as 50mm.
Therefore:
- self-weight of footing = Area * h * γconcrete
- 7.8 * 0.625 * 24 = 117KN < assumed weight, 120KN
Bending Reinforcements
At the column, which is the critical section
- M = earth pressure * pad size * (projection/2)2
From above;
- Provide: Y25 – at 125mm centres
- As prov = 3927mm2
Maximum spacing = 750mm. Therefore, the reinforcement provided meets the requirements specified by the code for minimum and maximum bar spacing in a slab.
- Provide: Y25 – at 200mm centres
- As prov = 2454mm2
Punching shear
- Critical perimeter = column perimeter + 8 *1.5d
- Critical perimeter = 4 * 300 + 12 * 565= 7980mm
- Area within perimeter = (300 + 3d)2 = (300 + 3 * 565)2
- Area within perimeter = 3.98 * 106
Therefore:
- Punching shear force, V = 257kN/m2 (7.8m2 – 3.98m2)V = 982KN
- Punching shear stress, v = 0.22N/mm2
- From table 3.9 BS 8110 -1- 1985, vc = 0.4N/mm2
Therefore:
- Modified vc = [30/25]⅓ * 0.4 = 0.43 N/mm2
- 0.5vc < v < (vc + 0.4)
- 0.5*0.43 < 0.22 < (0.43 + 0.4) 0.215 < 0.22 < 0.83 OK
Shear stress
At the critical section for shear, 1.0d from the column face
- V= 257kN/m2 * 2.8 * 0.625 = 449.75kN
- V = V/bd = 449.75 * 103/2800*565 = 0.28 N/mm2
- 0.28 N/mm2 < 0.43 N/mm2
Therefore, the section is adequate in shear.
Conclusion on Pad Foundation
As defined above, the Pad foundations are a type of shallow foundation made of reinforced concrete. Pad Foundations are simple, cost-effective, and typically used for light to moderate loads in stable soil conditions near the surface. Still, they are not suitable for unstable ground or heavy loads.
Today, we designed a pad foundation in Accordance with BS 8110. I know you will have learned something, and you need to learn and read more because this profession of structures is broader than what you learned today.
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