I have a question regarding a simplification/derivation of the (ULS) bending moment capacity of a given T-cross-section, with compression-zone on the flanged side. The problem is that the rebar-strain $\varepsilon_s$ (here) reaches $\varepsilon_{ud}$ (which cannot be exceeded) before the concrete strain $\varepsilon_c = \varepsilon_{cu}$. This means that a rectangular stress block approximation cannot be used (ref. Eurocode 2, point 3.1.7(2))

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The simplification is from a Norwegian book, hence I can't upload the relevant pages. The translation reads approximately:

Usually, the flange is so large that the compression zone (heightwise) is small. This leads to a large rebar strain $\varepsilon_s$ before the ultimate compressive strain $\varepsilon_{cu}$ is reached.

If $\varepsilon_s$ reaches $\varepsilon_{ud}$ [ultimate rebar-strain] before $\varepsilon_{c}$ reaches $\varepsilon_{cu}$ [the rebar fractures before the concrete] one usually assumes constant compressive concrete stress acting only in the flange, such that the concrete force resultant acts in the middle of the flange.

Moment equilibrium (rebar force) about the flange's middle yields design moment resistance $M_{Rd}$

$$M_{Rd}=f_{yd}\cdot A_s \cdot \Big(d-\frac{t}{2}\Big),$$

where $f_{yd}$ is rebar design yield stress.

One must check with this assumption, that the stress in the flange does not exceed the design concrete compressive stress $f_{cd}$.

Since flange moment capacity is assumed to be

$$M_{Rd}=\sigma_{cd}\cdot b_{eff}\cdot t\cdot \Big(d-\frac{t}{2}\Big),$$

we must have

$$\sigma_{cd}=\frac{M_{Rd}}{t\cdot b_{eff}\Big(d-\frac{t}{2}\Big)}\leq f_{cd}$$

If this is the case, the capacity-estimate is conservative.

So the 'math' is clear, but I can't see the justification behind the assumption. Has the author used the lower bound theorem (theory of plasticity), or is there anything else I can't spot?


  • $\begingroup$ It's an interesting question. But could you define the terms that appear in the equations and figures? I didn't understand a few of them. $\endgroup$ Apr 10, 2023 at 16:03

1 Answer 1


I agree with you.

we should establish the neutral axis and apply the $\varepsilon_{cu} \ ,$ The you find out the $\varepsilon_s \ ,$ And these are your strains. and the stress, S, of the steel bars is just $S_{steel}= E_{steel}*\varepsilon_s $

From here you proceed as normal and calculate the C=T and the moment arm d


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