What size should a beam be?

Depth comes from the span; width comes from what has to fit inside it

A beam size is two separate questions that get asked as one. The depth is a structural answer — span, load, and whether the beam is continuous. The width is mostly a detailing answer — how many bars have to sit side by side, how thick the wall above it is, and a fire minimum.

Unlike a column, a beam does not care how many storeys the building has. A column accumulates every floor above it; a beam carries the floor it is in and nothing else. The beam in a bungalow and the beam on the third floor of a duplex are the same beam if the span and the bay are the same.

Estimating the load in one line

A load takedown gives the real figure. For a trial section this is close enough:

w = 12 x (width of floor carried, m) kN/m, ultimate

Beam at 3.5 m centres w = 12 x 3.5 = 42 kN/m
Worked example w = 40.2 kN/m (slab 38.0 + own web 2.2)

The 12 kN/m² is the ultimate load of an ordinary reinforced concrete residential floor — a 150 to 160 mm slab, finishes, partitions and 1.5 kN/m² imposed — and it is where the worked beam on this site lands when the arithmetic is done properly.

Add anything standing on the beam separately. A 225 blockwork wall 3 m high, rendered both faces, is about 11.7 kN/m unfactored — 16 kN/m at ultimate, which is 40% again on top of the floor load. A beam under a wall is a different beam from the one next to it, and this is the single commonest reason a beam sized off a table is too small.

Where span over twelve comes from

Everyone quotes a fraction of the span, and the fractions are roughly right, but they are usually presented as folklore. They are not. They fall out of the point where the two governing checks meet.

Work a section to the singly reinforced limit, K = K' = 0.156 — the most moment the concrete will take before you have to add compression steel. At that point the moment per unit of section is fixed:

M/bd² = K' fcu = 0.156 x 25 = 3.9 N/mm² (grade 25)

fs = (2/3)(460) = 307 N/mm² (steel provided = steel required)
MF = 0.55 + (477 - 307) / [120(0.9 + 3.9)] = 0.846

Allowable span/d simply supported 20 x 0.846 = 16.9
continuous 26 x 0.846 = 22.0

So a beam worked to its flexural limit runs out of stiffness at a span/depth of about 17, or 22 continuous. Add cover, a link and half a bar — about 43 mm — and turn it into an overall depth:

Simply supported h >= span/16.9 + 43 ~= span / 15
Continuous h >= span/22.0 + 43 ~= span / 18

Which is why span/12 is a safe rule and not a precise one. It is deeper than the check demands, and that margin is why beams in Nigerian residential practice almost never fail deflection — the worked beam uses a span/depth of 9.8 against an allowable 22.3. Slabs are the opposite: there, deflection is what decides the thickness, every time.

Trial depths, by span and load

Overall depth h in mm, for a 225 mm wide beam in grade 25 concrete, 25 mm cover, T8 links and T20 main bars, rounded up to the next 25 mm. Each cell is the deeper of what strength needs and what deflection needs.

Simply supported

Spanw = 20
kN/m
30405060
3.0 m225250275300325
4.0 m300325350400425
5.0 m350375425475525
6.0 m400450500550600

Continuous over its supports

Spanw = 20
kN/m
30405060
3.0 m200250275300325
4.0 m250300350375400
5.0 m300350400450500
6.0 m350425475525575

Strength is taken as M = wL²/8 simply supported and M = 0.11wL² continuous, which is BS 8110 Table 3.5's worst coefficient. The continuous column is only valid if Table 3.5 is: three or more spans, spans within 15% of each other, and Qk <= Gk. Outside those limits you analyse the frame and the moment can be a good deal larger.

These are trial sections, not designs. Depth halves the steel far faster than it costs concrete, so where you have the headroom, take the next size up and let the design confirm it — a beam that arrives at the singly reinforced limit is one revision away from needing compression steel.

The width, and how many bars fit across it

Bars need a clear gap between them of at least the bar diameter and at least the aggregate size plus 5 mm — 25 mm for the usual 20 mm granite — or the concrete cannot get down between them and the steel is not doing what the calculation says it is.

Usable width = b - 2(cover + link) = b - 66 mm (25 cover, T8 link)
Fits if n x (bar) + (n - 1) x (gap) <= usable width
Beam widthT12T16T20T25
150 mm2222
225 mm4443
300 mm6655
375 mm9876

Bars in one layer. Where a row is within a few millimetres of fitting, treat it as not fitting — bars are not fixed to the millimetre on site, and 20 mm cover instead of 25 buys back a bar.

A second layer is not free. It moves the centroid of the steel up, which cuts the effective depth by 30 to 40 mm, and d is squared in everything. Where the steel will not fit in one layer, a deeper or wider beam is nearly always cheaper than the layer.

What overrides the table

What a 225 × 450 actually reaches

The default Nigerian beam — nine by eighteen inches — worked out properly. Effective depth 407 mm, grade 25 concrete:

Mcap = 0.156 b d² fcu = 0.156 x 225 x 407² x 25
= 145 kNm singly reinforced
As at that moment = 1153 mm² -> 4T20 (1257), which fits the width
Ultimate loadMax span
simply supported
Max span
continuous
20 kN/m6.85 m8.10 m
30 kN/m6.20 m6.65 m
40 kN/m5.40 m5.75 m
50 kN/m4.80 m5.15 m
60 kN/m4.40 m4.70 m

The 20 kN/m row is capped by deflection, not by strength; the rest are capped by the singly reinforced limit. Shear is not in this table and should be checked — at 60 kN/m over 4.4 m the end reaction is 132 kN, which is real link design rather than the minimum provision.

This is the honest version of "nine by eighteen carries anything". It carries an ordinary residential bay to about 5 m and then stops, and the thing that most often stops it early is not a longer span — it is a wall standing on it.

Size it, then check it

Structura designs the beam from the section you pick: moment, K against K', lever arm, steel area, shear and links, and the deflection check with both numbers shown — so when it fails you can see whether it was strength or stiffness that ran out. It checks Table 3.5's own conditions before using its coefficients, and the bar bending schedule comes with the answer.

Beams are free to run, as many as you like.

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