Pile capacity
A pile carries load two ways: at the tip, where the soil fails in bearing, and along the shaft, where it fails in friction. The two are computed separately and added, and in sand the tip term depends on the length-to-diameter ratio — which is why a longer pile in sand is disproportionately stronger, and why a pile in clay behaves quite differently.
| Quantity | Value | Basis |
|---|---|---|
| Diameter | 400 mm | as entered |
| Embedded length | 15.0 m | as entered |
| L/D | 37.5 | slenderness |
| Cross-section | 0.1257 m² | πD²/4 |
| Perimeter | 1.257 m | πD |
| End-bearing pressure qp | 8000 kPa | Meyerhof, capped at 4 pa N60 |
| Unit shaft friction fs | 40.0 kPa | 2 N60, capped at 100 |
| Tip capacity | 1005 kN | qp × Ap |
| Shaft capacity | 754 kN | fs × perimeter × L |
| Ultimate | 1759 kN | tip + shaft |
| Allowable | 704 kN | ÷ 2.5 |
In sand the tip term is capped at 4 pa N60, so lengthening the pile stops helping once L/D passes about 10 — at this slenderness of 37.5 the cap is governing.
qp = 0.4 pa N60 (L/D) ≤ 4 pa N60 · Meyerhof.
fs = 2 N60 ≤ 100 kPa · Driven pile.
qp = 9 cu · Undrained.
(Qp + Qs) / FS · FS 2.5 to 3.
A clear path from inputs to results.
A pile carries load two ways: at the tip, where the soil fails in bearing, and along the shaft, where it fails in friction. The two are computed separately and added, and in sand the tip term depends on the length-to-diameter ratio.
How to use
- Choose the soil type and the installation method.
- Enter the diameter and embedded length.
- Set the SPT blow count for sand or the undrained strength for clay.
- Read the tip and shaft capacity and the allowable load.
The formula
Reference: Meyerhof (1976) as summarised in FHWA deep-foundation design guidance.
Worked example
A 400 mm driven pile, 15 m into sand
- 400 mm diameter, 15 m embedded.
- N60 = 20, driven.
- Factor of safety 2.5.
- 1
The slenderness is 15/0.4 = 37.5.
- 2
The tip pressure is 0.4 × 100 × 20 × 37.5 = 30,000 kPa, capped at 4 × 100 × 20 = 8,000 kPa.
- 3
Tip capacity is 8,000 × 0.1257 = 1,005 kN.
- 4
Unit shaft friction is 2 × 20 = 40 kPa, so the shaft carries 40 × 1.257 × 15 = 754 kN.
- 5
The ultimate capacity is 1,759 kN and the allowable is 704 kN.
8,000 kPa tip · 1,759 kN ultimate · 704 kN allowable.
Interpreting the result
The cap at 4 pa N60 is what stops a long pile in sand from developing an implausible tip resistance. Because it binds at an L/D of about 10, extra length in sand buys shaft friction only — which is linear, and much less than the tip would have been.
Assumptions & limitations
- A single pile in uniform soil with no group effect.
- Driven piles; bored piles take two-thirds of the shaft friction in sand.
- No settlement, downdrag, scour or uplift.
- The factor of safety is applied to the ultimate capacity.
Frequently asked questions
Why is the tip pressure capped?
Because a pile in sand cannot mobilise more than about 4 pa N60 at the base, however long it is. Beyond that the soil fails locally rather than in a general bearing failure.
What factor of safety should I use?
2.5 to 3 for a single pile without a load test, and 2.0 if a static load test confirms the capacity.
Does a pile group carry more than the sum?
Usually less per pile, because the zones of influence overlap. A group efficiency factor or a block failure check is needed.
Why does clay behave differently?
Because undrained clay has no friction angle: the tip is 9 cu and the shaft is α cu, both linear in strength and length. There is no slenderness benefit.
Can I add the tip and shaft directly?
Meyerhof does, but they mobilise at different settlements — the shaft peaks early and the tip later. For a settlement-limited design the sum overstates the capacity at working load.
Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.