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Flexural verification to CIRSOC 201, step by step

Almost everyone remembers φMn ≥ Mu. That is the last line of the check, not the check. The outcome is decided earlier, and one of those steps can pull φ below 0.9.

Most people remember a flexural check as one inequality: φMn ≥ Mu. That is true, and it is the last line. What decides the outcome happens earlier, in steps that rarely get a close look because they seem 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:

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

Ten steps. The tenth is the one everyone remembers. The ones that decide the result 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 is usually treated as a floor that is almost never reached. In residential beams it is reached all the time.

Mu [kN·m]As by strength [cm²]As,min [cm²]Governs
151.122.39the minimum
302.282.39the minimum
604.732.39strength
907.382.39strength
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 discussing its reinforcement.

And the minimum rarely depends on the concrete

Of the article's two expressions, the √f'c one only exceeds the other for fairly strong concrete.

Concrete0.25·√f'c/fy1.4/fyGoverns
H-202.66203.33331.4/fy
H-252.97623.33331.4/fy
H-303.26033.33331.4/fy
H-353.52153.3333√f'c
The two expressions for the minimum, for fy 420, per mille.

The crossover is at f'c = 31.36 MPa. Across the whole range used in residential work (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 automatically 0.9

The strength reduction factor is taught as 0.90 for bending and then treated as a constant. It is not. 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]Reinforcementεt [‰]φφMn [kN·m]
302 Ø1620.030.9051.6
602 Ø2011.740.9077.9
902 Ø256.440.90115.2
1202 Ø322.760.707133.5
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 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 design 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 and the second at 0.87, both governed by the anchorage of the bars (Table 25.4.2.3): bending and shear stay below. 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

  1. Before sizing, check whether the moment is above or below the crossover with the minimum. Below it, the reinforcement is already decided.
  2. From H-20 to H-30, the minimum does not depend on the concrete. It comes from 1.4/fy.
  3. φ depends on εt: always read the two together.
  4. If compression steel appears, it is because the section lost ductility, not because it lacks capacity.
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