Calculations › Pad Foundation
Pad Foundation
EN 1997-1
Checks a reinforced concrete pad foundation for bearing pressure, punching shear and sliding to EN 1997-1.
Diagram
Example inputs
Length Lx1500 mm
Width Ly1500 mm
Foundation depth500 mm
Cover50 mm
Column width (x)300 mm
Column depth (y)300 mm
Axial load (ULS)500 kN
Axial load (SLS/char.)350 kN
Moment at base (ULS)0 kNm
Allowable bearing (qa)200 kN/m²
Concrete gradeC25/30
National annexUK
Results
Pad Foundation — 1500×1500mm
C25/30 · depth 500mm · qa=200 kN/m² · EN 1997-1
Results
✓ Bearing PASS
util 0.84168.06EN 1997-1 §6.5
| Pad plan area = Lx × Ly | 2.25 | m² | |
| Self-weight = A × d × γc | 28.13 | kN | |
| Total SLS load = NEdSLS + Wsw | 378.13 | kN | |
| Eccentricity e = MEd / NEdSLS | 0 | m | |
| Effective area Aeff = (Lx − 2e) × Ly | 2.25 | m² | |
| σmax bearing pressure = N_total / Aeff | 168.06 | kN/m² | EN 1997-1 §6.5 |
| qa allowable bearing | 200 | kN/m² |
✓ Overturning PASS
util 0.000.00EN 1997-1 §6.5.4
| MEd moment at foundation | 0 | kNm | |
| NEdSLS characteristic load | 350 | kN | |
| Eccentricity e = MEd / NEdSLS | 0 | m | |
| e_limit (middle third) = Lx / 6 | 0.25 | m | EN 1997-1 §6.5.4 |
✓ Punching Shear PASS
util -0.25-41.15EN 1992-1-1 §6.4
| d effective depth = h − cover − φ/2 | 442 | mm | |
| u1 control perimeter (2d) = 2(cw+cd) + 2π×2d | 6754.34 | mm | EN 1992-1-1 §6.4.2 |
| pNet net upward pressure (ULS) = (NEd,ULS − 1.35Wsw) / A | 0.205 | N/mm² | |
| vEd punching shear stress = pNet × (A − Acontrol) / u1 | -41.15 | N/mm | EN 1992-1-1 §6.4.3 |
| k size effect factor = 1 + √(200/d) | 1.673 | — | |
| vRd,c concrete shear capacity = 0.18/γC × k × (100ρfck)^(1/3) | 0.379 | N/mm² | EN 1992-1-1 §6.4.3 |
| VRd,c total capacity = vRd,c × u1 × d | 1131.47 | kN |
✓ Base Bending PASS
util 0.000.04EN 1992-1-1 §6.1
| Cantilever length lx = (Lx − cw) / 2 | 600 | mm | |
| pULS upward pressure (ULS) = NEd,ULS / A | 0.222 | N/mm² | |
| mEd base moment (per m) = pULS × lx² / 2 | 0.04 | kNm/m | EN 1992-1-1 §6.1 |
| K bending parameter = mEd / (fcd × b × d²) | 0 | — | |
| K' limiting parameter | 0.168 | — | |
| z lever arm = d[0.5 + √(0.25 − K/1.134)] | 419.9 | mm | |
| As,req reinforcement (per m) = mEd / (fyd × z) | 0.22 | mm²/m | EN 1992-1-1 §6.1 |
WARNINGS
Punching shear assumes ρ = 0.1% minimum — verify actual reinforcement ratio after design.
Try it yourself
Run this calc with your own inputs