What is the effective depth of a beam with two layers of tension reinforcement?

In EC2 for calculating crack width and deflection, does the effective depth d refers to the depth from compression face to centroid of the tension reinforcement or is that calculated to the centroid of outermost tension reinforcement? It seems from output file that software OASYS Adsec 8.4 uses the latter definition. So I need to confirm which is correct.

Looking at section "7.3.3 - Control of cracking without direct calculation", we have equations (7.6) and (7.7) which define the maximum bar diameter for crack control:

\begin{align} \phi_s &= \phi_s^*\cdot\dfrac{f_{ct,eff}}{2.9}\cdot\dfrac{k_c h_{cr}}{2(h-d)} &&\text{Bending (part of section in compression)}\tag{7.6} \\ \phi_s &= \phi_s^*\cdot\dfrac{f_{ct,eff}}{2.9}\cdot\dfrac{k_c h_{cr}}{(h-d)}&&\text{Tension (entire section under tension)}\tag{7.7} \end{align}

The variables are all defined immediately after the equations are declared. For us, the only variable that matters is

$d$ is the effective depth to the centroid of the outer layer of reinforcement

So if this is what OASYS is calculating, then it is certainly correct. Nowhere else in EC2 is $d$ given that definition (it's usually the depth to the centroid of the entire reinforcement). That is not to say, however, that the use of this definition to calculate "crack width and deflection" (as per your question) is incorrect, however.

Section "7.3.4 - Calculation of crack widths" has equations (7.8) to (7.10):

\begin{align} w_k &= s_{r,max}\left(\epsilon_{sm}-\epsilon_{cm}\right) \tag{7.8}\\ \epsilon_{sm}-\epsilon_{cm} &= \dfrac{\sigma_s-k_t\dfrac{f_{ct,eff}}{\rho_{p,eff}}\left(1+\alpha_e\rho_{p,eff}\right)}{E_s} \geq 0.6\dfrac{\sigma_s}{E_s} \tag{7.9} \\ \rho_{p,eff} &= \dfrac{A_s + \xi_1^2 A_p}{A_{c,eff}} \tag{7.10} \end{align}

where

$A_{c,eff}$ is the effective tension area. $A_{c,eff}$ is the area of concrete surrounding the tension reinforcement of depth, $h_{c,ef}$, where $h_{c,ef}$ is the lesser of $2.5(h-d)$, $(h-x)/3$ or $h/2$ (see Figure 7.1)

That possible value $2.5(h-d)$ does not define which $d$ should be used, however. Looking at Figure 7.1(a) (shown below), it is not entirely clear which $d$ is being described: if it is the standard $d$, then the centroid would be closer to the lower layer; if it is the "outer layer" $d$, then it would be equally spaced between both layers. If you squint, it looks like the shown centroid is indeed a bit closer to the lower layer, but you never know if that's just an imprecision in how the figure was drawn. The fact that it is labelled as "level of steel centroid" isn't very helpful either: is it the centroid of all the steel or just the steel relevant to cracking, which is what's being discussed in this section? Gun to my head, I'd choose the former (all steel), but I can't guarantee that's correct.

What bothers me most is that the theory behind sections 7.3.3 and 7.3.4 is almost certainly the same, so I find it odd that $(h-d)$ would have different definitions in these related topics. It's also been a long time since I've looked at cracking theory, so I can't even look at this from that perspective.

• Thanks for digging up the definition of d in the code. It is pretty confusing for anyone. It is clear from Figure 7.1(a) that d is calculated based on all two layers of reinforcement, else the line labelled "A" would be drawn through the bottom row of rebars. – Jennifer Feb 27 '18 at 5:55
• @Jennifer: my understanding of the "outermost layer" is that it is the ones closest to the lateral sides, not the bottom. By this interpretation, the "outermost layer"'s centroid would be halfway between the two layers. – Wasabi Feb 27 '18 at 16:53
• I don't think your interpretation of outermost layer makes any sense. Outermost layer should be the bottom-most layer or top-most layer. – Jennifer Feb 28 '18 at 4:19