Design of spread foundations
A spread footing (shallow foundation) is a wide concrete pad under a column or a long strip under a wall. It spreads the load over more soil.
The bearing pressure is q = Q / A (load ÷ footing area). The soil has a safe allowable bearing capacity qa (for example 150–300 kPa for firm soil). Design rule: q ≤ qa. So the needed area is A = Q ÷ qa; for a square footing, B = √A.
Also check: settlement (the footing must not sink unevenly), the depth (below soft top soil, frost and moving soil), and the concrete thickness against punching and bending. Footings are made of reinforced concrete with bars at the bottom.
Design of pile foundations
When the top soil is soft, we use piles: long columns of concrete or steel driven or bored deep into the ground. A pile cap on top joins the piles and takes the column.
A pile carries load in two ways. Skin friction is the grip of soil along the sides: Qs = π D L fs (D diameter, L length, fs friction per area). End bearing is the push of strong soil under the tip: Qb = qb × π D² / 4.
Ultimate capacity = Qs + Qb. Allowable capacity = ultimate ÷ safety factor (about 2.5). For a group of piles, add the single-pile loads, then check that the group as a whole does not fail or settle.
Design of retaining structures
A retaining wall holds back soil that is higher on one side. The soil pushes sideways with an earth pressure. For dry sand-like soil (Rankine): Ka = (1 − sin φ) / (1 + sin φ), where φ is the friction angle (30° gives 0.33). Pressure at depth z is Ka γ z, a triangle like water. Total push per metre of wall: P = ½ Ka γ H², acting at H/3 above the base.
Three stability checks: (1) Overturning: FS = resisting moment ÷ overturning moment, aim for at least 2. (2) Sliding: friction under the base ÷ P, aim for at least 1.5. (3) Bearing: the pressure under the base must stay below qa. Put drainage (weep holes and gravel) behind the wall, because trapped water adds large pressure.
Try it
At home: fill a tray with dry sand and stand a book upright against a side. Pile more sand behind it: the book tips. Now put a wide flat board under a heavy stone on soft sand: it sinks less than a narrow stick does.
In the 3D: predict, then check. If H goes from 3 m to 6 m, how many times does P grow?
Key formulas and definitions
- q = Q / A; q ≤ q_a; B = √(Q / q_a) (square footing)
- Pile: Q_s = π D L f_s; Q_b = q_b π D² / 4
- Allowable pile load = (Q_s + Q_b) ÷ 2.5
- K_a = (1 − sin φ) / (1 + sin φ)
- P = ½ K_a γ H², acting at H/3
- FS overturning = resisting moment ÷ overturning moment ≥ 2
Worked examples
1. Find K_a for a soil with φ = 30°.
sin 30° = 0.5. K_a = (1 − 0.5) / (1 + 0.5) = 0.5 / 1.5 = 0.333.
2. A wall holds H = 3 m of soil, γ = 18 kN/m³, K_a = 0.333. Find P and where it acts.
P = ½ × 0.333 × 18 × 9 = 27 kN per metre. It acts at H/3 = 1 m above the base.
3. The wall weighs 60 kN per metre, with its weight 1.0 m from the toe (tipping point). P = 27 kN acts 1 m above the base. Find the overturning FS.
Resisting moment = 60 × 1.0 = 60 kNm/m. Overturning moment = 27 × 1 = 27 kNm/m. FS = 60 / 27 = 2.22, which is above 2. OK.
4. A column load of 600 kN sits on a 2 m × 2 m footing. Find q. What footing is needed if q_a = 200 kPa?
q = 600 / 4 = 150 kPa, which is below 200. OK. Needed area = 600 / 200 = 3 m², so B = √3 = 1.73 m.
5. A pile has D = 0.4 m, L = 10 m, f_s = 30 kPa and q_b = 2000 kPa. Find Q_s and Q_b.
Q_s = π × 0.4 × 10 × 30 = 377 kN. Tip area = π × 0.16 / 4 = 0.1257 m². Q_b = 2000 × 0.1257 = 251 kN.
6. For that pile find the allowable load (FS = 2.5). How many piles carry a 1000 kN column?
Ultimate = 377 + 251 = 628 kN. Allowable = 628 / 2.5 = 251 kN. Piles = 1000 / 251 = 3.98, so use 4 piles.
Common mistakes
- Using the earth pressure as a rectangle. It is a triangle: zero at the top.
- Forgetting water behind the wall. Wet soil and trapped water push much harder.
- Using the footing width instead of the area in q = Q / A.
- Counting only skin friction or only end bearing for a pile. Add both.