Pile group settlement
Capacity is usually not what governs a pile design — settlement is. The equivalent raft method replaces the group with a single footing at the depth where the load has spread out, and computes the elastic settlement of that footing in the soil beneath. The load spreads at 1 in 4 from two-thirds of the pile length.
| Quantity | Value | Note |
|---|---|---|
| Group load | 6,000 kN | 9 piles |
| Group footprint | 4.0 × 4.0 m | as entered |
| Pile length | 15.0 m | as entered |
| Raft depth | 10.0 m | two-thirds of the pile length |
| Raft size | 6.5 × 6.5 m | load spread at 1 in 4 |
| Raft pressure | 142.0 kPa | load over the raft area |
| Soil modulus | 20 MPa | beneath the raft |
| Soil settlement | 47.0 mm | elastic, flexible area |
| Pile shortening | 1.33 mm | QL/(2AE) |
| Total | 48.4 mm | the number that is checked |
| Limit | 25 mm | typical for a framed structure |
Total settlement is 48 mm, above the 25 mm usually allowed for a framed structure — the group is capacity-adequate but serviceability-limited, which is the normal outcome and the reason this check exists.
(2/3) L · Where the load has spread.
B + 2(L/3)(1/4) · 1-in-4 spread.
q B(1−ν²) I / E_s · I = 1.12 for a square.
Q L/(2 A E_p) · Uniform shaft friction.
A clear path from inputs to results.
Capacity is usually not what governs a pile design — settlement is. The equivalent raft method replaces the group with a single footing at the depth where the load has spread out, and computes the elastic settlement of that footing in the soil beneath.
How to use
- Enter the group load and footprint.
- Set the pile length and diameter.
- Add the soil modulus and Poisson’s ratio.
- Read the soil settlement, the pile shortening and the total.
The formula
Reference: Equivalent raft method (Tomlinson), with the load spreading at 1 in 4 from two-thirds of the pile length.
Worked example
6,000 kN on a 3 × 3 group of 15 m piles
- 6,000 kN, 4 × 4 m footprint, nine 400 mm piles 15 m long.
- Soil modulus 20 MPa, Poisson 0.3, pile modulus 30 GPa.
- 1
The equivalent raft sits at two-thirds of 15 m, or 10 m.
- 2
The load spreads at 1 in 4 over the remaining 5 m, so the raft is 6.5 m square.
- 3
The raft pressure is 6,000/42.25 = 142 kPa.
- 4
Elastic settlement is 142 × 6.5 × 0.91 × 1.12 / 20,000 = 47 mm.
- 5
The piles shorten by 1.3 mm, giving 48 mm total — above the 25 mm limit.
47 mm soil · 1.3 mm pile · 48 mm total.
Interpreting the result
The soil beneath the raft does almost all of it, and the pile compression is a rounding error. That is why the soil modulus is the input that matters, and why a settlement estimate is really an estimate of the modulus rather than of the geometry.
Assumptions & limitations
- A rigid group load spread uniformly.
- Homogeneous elastic soil beneath the raft.
- No consolidation settlement in clay.
- No pile-to-pile interaction beyond the raft idealisation.
Frequently asked questions
Why two-thirds of the pile length?
Because that is where the load has transferred to the soil in a friction pile. An end-bearing pile on rock would use the full length instead.
Does this work in clay?
The elastic part does, but clay also consolidates, and the consolidation settlement is usually larger and slower. A one-dimensional consolidation calculation is needed for that.
How does a pile group differ from a single pile?
The group settles more than a single pile at the same load per pile, because the stress bulbs overlap. That is what the equivalent raft captures.
Why is 25 mm the limit?
It is a common serviceability limit for framed structures. Masonry and cladding are more sensitive, and some codes use 50 mm for isolated foundations.
What if the modulus is unknown?
Use a correlation from SPT or CPT and treat the answer as a range. The modulus is often uncertain by a factor of two, and the settlement scales directly with it.
Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.