I'm looking at the usual UK/EU standards formula for the reinforcing steel area percentage, in a retaining wall. Essentially in "usual" cases it comes down to 0.13% of the cross-sectional area of Tw (wall thickness) in the upstand of the vertical wall itself, and Tb (base thickness [depth]) for the outer and inner base. If Fy is below a certain amount then the percentage changes but it's still calculated as a simple percentage.

Example scenario

I'm imagining trying to apply this to the retaining wall shown in the diagrams below. In this situation, the retaining wall is built onto one edge of a large and deep foundation. The foundation itself is very deep and very wide compared to the retaining wall built on it. Essentially when the foundation is considered as the base of the retaining wall, it continues on to a distance where the retaining wall is no longer affecting or affected by it.

The kind of dimensions I have in mind are - foundation/base is indefinitely wide (for these purposes), and say 900mm deep. The retaining wall upstand is say 200mm wide and 600mm high, and built at one edge of it. The foundations are deemed not to need reinforcement steel (perhaps because it's basically a large monolithic concrete block or it's on on sufficiently stable ground/rock beneath it). So the only reinforcing steel is the steel required for the retaining wall aspect.

Apparently, the inner and outer steel area is calculated as a percentage of the base's cross-section area. But I'm looking at the three diagrams below, which are roughly to scale.

  • First (left) diagram - the outer steel is run down to the bottom of the foundations, so essentially the foundations act like a base that's 900mm thick and very wide.
  • Second diagram - outer steel is only run down to 350mm deep; a more typical depth if it were on a slab not a foundation.
  • Third diagram - the 2nd layout is re-visualised as equivalent to two parts - essentially a 400mm base with rebar along the bottom, resting on a non-reinforced 500mm slab.

enter image description here

My confusion

What's puzzling me is why are the steel area requirements, which are based on Tb, different in these three versions? Intuitively, I expected that for any given lateral pressure, the inner/outer steel would reach a sort of limit point where there just wasn't any need to add more, because in the first two cases it's embedded with right angle bends to a point where the limiting failure mode in the base slab is pullout failure/concrete failure rather than steel elastic limitations (ie not governed by the cross section area of steel), and if the foundation itself is considered structurally adequate, then I'd have thought the second and third would be structurally equivalent and therefore need equal areas of steel too, as they are essentially the same layout.

(Alternatively, the 1st layout seems stronger than the 2nd, as there's even more grip and harder pullout, and the entire foundation slab is engaged rather than just the top half of it, so why does it seem to need more rather than less steel cross-sectional area, to achieve equal wall stability?)

But instead, the standard equation is simply 0.13% of Tb, so if the base is made thicker, which should add not remove strength, then the area of inner/outer steel required seems to have to go up, not down. That feels counter-intuitive.

So I'm left with a mess of confusion: Why does the steel area percentage differ in the three cases shown? Why does adding strength and solidity to the bottom of the base (i.e., as extra depth) cause an increase to the required steel area? Why isn't there some kind of limit based on lateral pressure rather than base dimensions beyond which more isn't needed? Why isn't a further limit reached where it doesn't matter how much deeper the base and the outer steels go, because the extra depth doesn't affect the retaining wall stability? What is the logic behind the cross section area of steel being an unqualified percentage of Tb?

What am I missing, and what should actually happen here?

  • $\begingroup$ You can do whatever you want as long as you are willing to submit proof that your design is safe. Intuition is not enough. Often this proof is more costly than just doing what they say you should do. $\endgroup$
    – joojaa
    Commented Oct 26, 2018 at 5:56
  • $\begingroup$ @joojaa - the question is about the logic behind a formula in an engineering EU standard, which confuses me because it looks illogical on several accounts, and a concrete example (pun unintended!) to show why it doesn't seem sensible or even logical to me. Your comment ignores that and suggests getting "proof" for some calculations for this hypothetical example..... maybe you're misunderstanding and believe I'm asking about a specific work rather than the rationale and basis for the EU standard formula? $\endgroup$
    – Stilez
    Commented Oct 26, 2018 at 9:52

1 Answer 1


I am not sure which guidance manual you are looking at, but it must in turn be referring to EN1992-1-1 section This section defines the minimum amount of reinforcement in a beam or slab in order for the beam or slab to be considered reinforced concrete. And it does indeed use the value of 0.13% (with adjustments for concrete and steel grade). So it is a much more generic requirement than just something for retaining walls. It is relatively easy to see why it must be a percentage of the concrete area, if you consider the different behaviour of plain concrete and reinforced concrete:

  • Plain concrete at low bending moments will have tension and compression in the concrete cross section. If the bending moment increases, the tension stress will reach the capacity, a crack will form and the structure will fail suddenly. This is a brittle failure.

  • Reinforced concrete at low bending moments will also tension and compression in the concrete. If the load increases, the tension stress in the concrete will reach the capacity and a crack will form. When the crack forms, the reinforcement will take over and transfer the tension that was previously transferred by the concrete. If there is enough reinforcement for this to be possible, the crack will be small, and the structure will not fail until the load increases enough for the reinforcement to yield, resulting in a ductile failure. (Meaning that we will get large deformations before the structure fails complete. A "hey, that looks unsafe — we better fix that before someone gets hurt"—thing.) The ductile failure makes the structure much safer even if it doesn't have a larger failure load.

The important bit in this context is as follows: If the amount of reinforcement is not enough to absorb the tension force that occurs in the concrete just before the first crack forms, the reinforcement will yield immediately at the creation of the first crack, and the crack will therefore become very large very fast, resulting in almost the same behaviour of plain concrete. (A brittle failure.) At a given concrete type and steel grade you will therefore want concrete area times concrete tension capacity to be less than steel area times yield stress in order to ensure a ductile failure. Which can be reformulated as steel area larger than x percent of the concrete area. (I believe the value of 0.13% also includes an empirical adjustment, so I am not going to recalculate the exact value, but hopefully the principle should be clear.)

It is worth noting that none of this has anything to do with the actual load on the retaining wall, so you will have to check that as well. And it is of course also perfectly possible to construct an unreinforced retaining wall which is completely safe for a given load — it just won't give a ductile failure.

And addition to all that, I feel I have to note that the reinforcement arrangement you have drawn on the sketches, is not a good arrangement and you should not construct it like that: When the rebar on the inner side is bend around the inner corner like that, there is a great risk of concrete cover breaking off at the bend long before the bar reaches its yield capacity. I can add a few sketches to explain this better, if you are interested, but I hesitate to add even more stuff on something tangential in case it was just a hastily drawn couple of lines? (You can also take a look at for example EN1992-1-1 annex J.2 for a couple of examples of how the reinforcement can be arranged at a corner in bending.)

  • $\begingroup$ +1, good answer. It's also worth remembering that such large slabs also generate significant heat during the curing process, and the resulting temperature differential between the core and the material in contact with the ground can lead to early internal stresses, which are themselves managed to some extent by this minimum steel reinforcement. $\endgroup$
    – Wasabi
    Commented Oct 26, 2018 at 19:05

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