How to design a pad footing

Two loads, two limit states, four checks · worked example

A pad footing is a block of concrete under a single column, wide enough that the soil can carry the load and deep enough that the block does not break doing it. It is the cheapest foundation there is, and the one most buildings in Nigeria sit on.

This page carries on from the load takedown: the same three-storey building, the same interior column, and the number that arrived at the bottom of it.

The one thing that trips everybody up

A footing is designed at two different limit states in the same calculation, and using the wrong load in the wrong place is the single most common mistake on a foundation sheet.

What you are doingLoad to useWhy
Sizing the plan area Service
Gk + Qk
Allowable bearing pressure already has a factor of safety of 2 to 3 inside it. Factoring the load as well would apply the safety factor twice.
Bending, shear, steel Ultimate
1.4Gk + 1.6Qk
This is reinforced concrete design, exactly like a beam or a slab, and BS 8110 designs concrete at ultimate.

Size the pad on the service load. Reinforce it on the ultimate load. If a footing comes out enormous, check this first — sizing on the factored load makes every pad about 45% too big.

Step 1 — The two loads

From the takedown, the ground-floor interior column carries three storeys. Adding the same items up unfactored gives the service load alongside the ultimate one:

Per storey, service
Floor Gk 6.04 x 16 = 96.6 kN, Qk 1.5 x 16 = 24.0 kN
Beams 8 m x 1.57 = 12.6 kN
Column 0.225² x 24 x 3.0 = 3.6 kN
= 136.8 kN

N (service) = 3 x 136.8 = 410 kN
N (ultimate) = 589 kN (from the takedown)

Materials: fcu 25, fy 460, column 225 × 225, allowable bearing pressure 150 kN/m², founding 1.0 m below ground.

That 150 kN/m² is a presumed value, not a measured one — fine for sizing a bungalow, not fine for a block of flats. Where the number comes from is a page of its own.

Step 2 — Size the pad

The soil carries the column load plus the footing itself and the earth backfilled on top of it. You do not know those until you have a size, so start with an allowance of about 15% and check it afterwards.

Areq = 1.15 N / qallow = 1.15 x 410 / 150 = 3.14 m²
side = sqrt(3.14) = 1.77 m -> use 1.8 m x 1.8 m (A = 3.24 m²)

Now check the allowance instead of trusting it. Take the pad 400 mm thick, so 600 mm of backfill sits on it:

Pad 1.8 x 1.8 x 0.4 x 24 = 31.1 kN
Backfill (3.24 - 0.05) x 0.6 x 18 = 34.5 kN
-------
65.6 kN (16% of N)

q = (410 + 65.6) / 3.24 = 146.8 kN/m² <= 150 PASS

The takedown page estimated 2.7 m² and a 1.65 m square by dividing the column load straight by the bearing pressure. That is the right first move, and it is why the real pad is a size bigger: the footing has to carry itself too.

Step 3 — The pressure that causes bending

Here is the second thing worth getting straight. The footing's own weight is carried by the ground directly underneath it — it goes straight down, bends nothing, and must be left out of the design pressure. Only the column load bends the pad.

qult = Nult / A = 589 / 3.24 = 181.8 kN/m²

Step 4 — Bending at the column face

The pad is a cantilever sticking out of the column on all four sides, loaded upwards by the soil. The critical section is the face of the column.

cantilever l = (1800 - 225) / 2 = 787.5 mm

M = q l² / 2 per metre width
= 181.8 x 0.7875² / 2 = 56.4 kNm/m

Steel is needed both ways, so there are two layers. Design both on the upper layer's effective depth — it is the shallower of the two, and using the deeper one for both overstates half the footing.

d = 400 - 50 cover - 12 (lower bar) - 12/2 = 332 mm

K = M / (b d² fcu) = 56.4 x 10&sup6; / (1000 x 332² x 25) = 0.020
0.020 < 0.156, so no compression steel

z = d[0.5 + sqrt(0.25 - K/0.9)] = 0.98d -> capped at 0.95d = 315 mm

As = M / (0.87 fy z) = 56.4 x 10&sup6; / (0.87 x 460 x 315) = 447 mm²/m

As,min = 0.13% bh = 0.0013 x 1000 x 400 = 520 mm²/m governs

Provide T12 @ 200 c/c each way = 565 mm²/m

Minimum steel beating the calculated steel is normal in footings and is not a sign of an error — a pad is thick because of shear, not because of bending, and 0.13% of a thick section is a lot of steel. The same thing happens with minimum links in beams.

Step 5 — Vertical shear

The first shear check is the ordinary beam one, taken on a plane a distance d out from the column face.

outstand = 787.5 - 332 = 455 mm
V = 181.8 x 0.455 = 82.8 kN/m
v = V / (b d) = 82.8 x 10³ / (1000 x 332) = 0.249 N/mm²

100As/bd = 100 x 565 / (1000 x 332) = 0.170
vc = (0.79/1.25)(0.170)1/3(400/332)1/4 = 0.367 N/mm²

0.249 < 0.367 PASS

Step 6 — Punching shear

The second is the one that decides the thickness: the column trying to punch a plug straight through the pad. BS 8110 checks it on a square perimeter 1.5d out from the column face, against the load outside that perimeter.

perimeter side = 225 + 2(1.5 x 332) = 1221 mm (< 1800, so it is
inside the pad and applies)
u = 4 x 1221 = 4884 mm

V = q (A - area inside u) = 181.8 (3.24 - 1.49) = 318 kN
v = V / (u d) = 318 x 10³ / (4884 x 332) = 0.196 N/mm²

0.196 < vc = 0.367 PASS

And the absolute limit, at the column face itself, where the concrete would simply crush:

u0 = 4 x 225 = 900 mm
v = 589 x 10³ / (900 x 332) = 1.97 N/mm²
limit = 0.8 sqrt(fcu) = 0.8 sqrt(25) = 4.0 N/mm² PASS

If punching fails, thicken the pad — do not add more steel. Links in a footing are awkward to fix, easy to get wrong on site, and cost more than the extra 50 mm of concrete that would have solved it.

Step 7 — Anchorage, and what to buy

The bars have to be developed past the section where they are needed, or the steel is there and the bond is not.

available beyond the column face = 787.5 - 50 cover = 737 mm
required anchorage, T12 in fcu 25 = 40 x 12 = 480 mm
737 > 480 straight bars, no bobs needed

The pad in full: 1800 × 1800 × 400 deep, 9 no. T12 × 1700 long each way in the bottom, on 50 mm blinding, founded 1.0 m down.

ItemWorkingQuantity
Excavation1.8 × 1.8 × 1.0, plus working space3.24 m³
Blinding50 mm under, 1.9 m square0.18 m³
Concrete1.8 × 1.8 × 0.41.30 m³
Formwork4 × 1.8 × 0.42.88 m²
Reinforcement18 × 1.70 m × 0.889 kg/m27.2 kg

Sense check: 27.2 kg in 1.30 m³ is 21 kg/m³. Pads normally land between 20 and 40 kg/m³ — an order of magnitude less than a beam, because a footing is mostly concrete doing nothing but spreading load. A pad that comes out at 90 kg/m³ has been designed as if it were a beam. See the rebar weight chart for the 0.889 kg/m and what a tonne of T12 actually is.

When a pad will not do

Pads stop working for two reasons, and they look completely different on a drawing:

Structura designs pads, strips and rafts, and stops at piles deliberately — see what it will not do. Settlement is not calculated for any foundation type; bearing pressure is a strength check, and a footing can pass it while still settling more than the building can tolerate.

Run this footing yourself

Structura designs pad, strip and raft foundations and shows every step above — service sizing, ultimate bending, both shear checks, the bar bending schedule and the quantities — with a PASS or FAIL on each. Single members, footings included, are free to run, as many as you like.

Run the whole building instead and the takedown feeds the footing for you, so the 410 kN never gets copied across by hand.

Open Structura in your browser

On Android, get it on Google Play; on iPhone and iPad, get it on the App Store — same account as the browser.

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