Designing a beam to BS 8110 is four questions in order: how much moment, how much steel to resist it, does it need links, and will it sag too much. This page works one beam through all four.
The beam carries the 160 mm slab from the slab example, over a 4.0 m simply supported span. Section 225 × 450 overall, fcu 25, fy 460, cover 25 mm, T8 links, T20 main bars.
Step 1 — Load and moment
The beam picks up the slab either side, halfway to the next beam — 3.5 m of floor in total here — plus its own web below the slab.
Self weight 1.4 x 0.225 x (0.450 - 0.160) x 24 = 2.2 kN/m
w = 40.2 kN/m
M = wL²/8 = 40.2 x 4.0² / 8 = 80.4 kNm
V = wL/2 = 40.2 x 4.0 / 2 = 80.4 kN
Only the web counts in the self weight. The 160 mm above it is already in the slab load — weigh it twice and every member below inherits the error.
Step 2 — Effective depth
= 450 - 25 - 8 - 10 = 407 mm
d is measured to the centroid of the tension steel, which is why the link diameter and half the bar diameter come off. With two layers of bars it comes off again — a detail worth getting right, because d is squared in everything that follows.
Step 3 — Flexural steel
= 80.4 x 10&sup6; / (225 x 407² x 25) = 0.086
K <= K' = 0.156 PASS — singly reinforced
z = d[0.5 + sqrt(0.25 - K/0.9)]
= 407[0.5 + sqrt(0.25 - 0.096)] = 363 mm (<= 0.95d = 387)
As = M / (0.87 fy z)
= 80.4 x 10&sup6; / (0.87 x 460 x 363) = 553 mm²
Provide 2T20 = 628 mm²
What K' is really telling you
K' = 0.156 is the point at which the concrete in compression is working as hard as the code will allow with the neutral axis at x = 0.5d. Past it you cannot make the section stronger by adding tension steel alone — the concrete crushes first — so you either deepen the beam or add compression steel in the top face to help. K > K' is not a failure; it is a fork in the method.
Step 4 — Shear and links
100As/bd = 100 x 628 / (225 x 407) = 0.686
vc = (0.79/1.25)(0.686)1/3(400/d)1/4 = 0.557 N/mm²
(400/d = 0.98, taken as 1.0)
v > vc, so links are needed
v < vc + 0.4 = 0.957, so minimum links govern
Asv/sv = 0.4b / (0.87 fyv) = 0.4 x 225 / (0.87 x 250) = 0.414
sv = 100.5 / 0.414 = 243 mm (max 0.75d = 305 mm)
Provide T8 links @ 225 c/c
Note which rule won. The shear the beam actually has to carry needed less than the code's minimum link provision, so the minimum set the spacing. Students often size links for the shear and never check the minimum — and then wonder why the answer at the back of the book is closer together.
Step 5 — Deflection
fs = (2/3)(460)(553/628) = 270 N/mm²
M/bd² = 2.16
MF = 0.55 + (477 - 270) / [120(0.9 + 2.16)] = 1.11
allowable = 20 x 1.11 = 22.3
9.83 <= 22.3 PASS, with a great deal to spare
Beams are deep relative to their span, so deflection rarely governs — the opposite of slabs, where it usually does.
The answer
| Item | Provided |
|---|---|
| Section | 225 × 450 mm |
| Bottom steel | 2T20 (628 mm²) |
| Links | T8 @ 225 c/c |
| Top steel | 2T12 hangers, to carry the links |
Continuous beams add one step before all of this: finding the moments. BS 8110 Table 3.5 gives coefficients that save you an analysis — but only if Qk <= Gk, the spans are roughly equal and there are three or more of them. Outside those limits the coefficients are not valid and you analyse the frame properly.
Run your own beam
Structura designs singly and doubly reinforced beams, sizes and spaces the links, runs the deflection check, and checks Table 3.5's own validity before using it. Every line above appears on the sheet with its clause, and the bar bending schedule comes with it.
Beams are free to run, as many as you like.
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.