Flexural verification to CIRSOC 201, step by step

Everyone remembers φMn ≥ Mu. That is the last line of the check, not the check. The steps that decide the outcome are others, and one is why φ is not 0.9.

Checking a beam in bending reduces, in most people's memory, to one inequality: φMn ≥ Mu. It is true, and it is the last line. What decides the outcome happens earlier, in steps that are rarely read closely because they look like paperwork.

This is what Stabileo's CIRSOC 201 module returns for a 20×40 cm beam of H-25 concrete, fy 420 steel, 25 mm cover and 8 mm stirrups, under a moment of 60 kN·m:

d = 35.9 cm, d' = 4.1 cm
Mu = 60.00 kN·m
As,min = 2.39 cm²
As,max (singly reinforced) = 11.58 cm²
As,req (tension) = 4.73 cm²
Tension reinforcement: 2 Ø20 (6.28 cm²)
a = 6.21 cm, c = 7.31 cm
εt = 11.74‰
εt ≥ 5‰ → tension-controlled → φ = 0.90
φMn = 77.90 kN·m

Ten steps. The tenth is the one everyone remembers. The ones that decide are the third and the ninth.

Step 3: the minimum governs more often than it seems

Article 9.6.1.2 asks for a minimum flexural reinforcement equal to the greater of 0.25·√f'c/fy·bw·d and 1.4/fy·bw·d. It tends to be read as a floor that is almost never reached. In housing beams it is reached constantly.

Mu [kN·m]	As by strength [cm²]	As,min [cm²]	Governs
15	1.12	2.39	the minimum
30	2.28	2.39	the minimum
60	4.73	2.39	strength
90	7.38	2.39	strength

The same 20×40 H-25 beam under increasing moments: what decides the reinforcement.

The crossover is at Mu = 31.5 kN·m. Below that moment, sizing for strength achieves nothing: the section carries 2.39 cm² either way. It is worth knowing which side of the crossover a beam is on before arguing about its bars.

And the minimum rarely depends on the concrete

Of the article's two branches, the √f'c one only overtakes the other once the concrete is quite good.

Concrete	0.25·√f'c/fy	1.4/fy	Governs
H-20	2.6620	3.3333	1.4/fy
H-25	2.9762	3.3333	1.4/fy
H-30	3.2603	3.3333	1.4/fy
H-35	3.5215	3.3333	√f'c

The two branches of the minimum, for fy 420, per mille.

The crossover is at f'c = 31.36 MPa. Across the whole range used in housing — H-20, H-25, H-30 — the minimum reinforcement does not depend on the concrete: it comes from 1.4/fy and nothing else. Going from H-20 to H-30 does not lower the minimum by a single square centimetre.

Step 9: φ is not 0.9 by decree

The strength reduction factor is learnt as 0.90 for bending and used as a constant. It is not one. It comes from the strain in the tension steel at the ultimate state, and the module says so in step 9: εt ≥ 5‰ → tension-controlled → φ = 0.90. Put more steel in the section and εt falls, taking φ with it.

Mu [kN·m]	Reinforcement	εt [‰]	φ	φMn [kN·m]
30	2 Ø16	20.03	0.90	51.6
60	2 Ø20	11.74	0.90	77.9
90	2 Ø25	6.44	0.90	115.2
120	2 Ø32	2.76	0.707	133.5

The same beam, loaded further each time. εt and φ come from steps 8 and 9.

In the last row the section entered the transition zone and φ dropped to 0.707. Capacity did not grow the way the steel did: between the last two rows the bar area goes from 9.82 to 16.08 cm² nominal and φMn only moves from 115.2 to 133.5 kN·m. A good part of what is added is lost in the factor.

An over-reinforced section does not fail the check. It fails in φ, and the result still says it passes.

Why compression steel goes in

The usual answer is "so the bars fit". That is the consequence, not the reason. At Mu = 200 kN·m on this same beam, the module finds εt = 1.51‰, below the limit, and adds compression reinforcement. With it the neutral axis rises, εt returns to 7.46‰ and φ returns to 0.90.

So compression steel is not there to contribute capacity. It is there to give the section its ductility back and recover the factor. The capacity follows, as a result.

Try it on a real beam

Below is the model: a simply supported 5 m beam, 20×40, H-25, with 12 kN/m of dead load and 8 kN/m of live load, plus its own weight. It opens in PRO's design workflow, with CIRSOC 201 as the code in force.

The beam in Stabileo, in PRO's design workflow. Three buttons, in this order: "Compute demands", "Run code check" and "Design all" — only the third one picks the bars, and until there are bars there is nothing to verify: the table says "no reinforcement". The 5 m span is modelled as two elements; the first closes at D/C = 0.89 governed by positive bending in the span, the second at 0.86 governed by shear, both under the 1.2D+1.6L combination. Change the load and run it again. PRO is in development; the concrete module is the part that is implemented and tested. Open full size

In short

Before sizing, check whether the moment is above or below the crossover with the minimum. Below it, the reinforcement is already decided.
From H-20 to H-30, the minimum does not depend on the concrete. It comes from 1.4/fy.
φ is a result, not an input. Read it together with εt.
If compression steel appears, it is because the section lost ductility, not because it lacks capacity.