D4. Normative verifications: the behavior of reinforced concrete elements
D4. Normative verifications: the behavior of reinforced concrete elements
D.4.1 Reinforcement mapping – reinforced concrete normative verification / connecting struts
Calculations of reinforcement mapping are carried out by the current software generally by using the method of Capra Maury (Annals of ITCBTP  Institut des Techniques de la Construction du bâtiment et des Travaux Publics  of December 1978) or the method of Wood and Armer (“The reinforcement of slabs according to a predetermined field of moments" Concrete February 1968, August 1968).
These methods render it possible to determine the necessity of reinforcement in the 4 directions Axi, Ayi, Axs, Ays on the lower and upper faces of an element according to the 2 directions of reinforcement considered orthogonal, X and Y.
A good approach to the calculation of these mappings can be made from Wood's method by simplifying:
Either an element subjected to the following stress components:

Nxx, Nyy, Nxy membrane stresses (positive if traction)

Mxx, Myy, Mxy bending stresses
The first step is to calculate the following intermediate stresses:

Nwx = Nxx + Nxy

Nwy = Nyy + Nxy

Mwx = Mxx + Mxy if Mxx is positive, Mwx = Mxx  Mxy otherwise

Mwy = Myy+ Mxy if Myy is positive, Mwy = Myy Mxy otherwise
D.4.2 Bending elements: slabs
The slabs work mainly subjected to the bending moments Mxx, Myy, Mxy.
The membrane efforts Nxx, Nyy, Nxy are often negligible.
The Mxy bending moments can be important, they are null in the zones where the principal bending moments are aligned with the principal axes (often confused with the directions of the reinforcement) and therefore in the middle of the span and on continuous supports.
They should not be neglected especially in the corners of the slabs and in the case of concentrated loads.
The calculations of passive reinforcement carried out in the determination of mappings are according to the rules brought by the Eurocode, because they are adapted to the study of a rectangular section subjected to combined loading (N, M), they can thus be carried out in SLS and the ULS.
The calculation of the reinforcement in SLS considering the crack openings is more delicate and requires the use of welltested software. Indeed, the presence of MXY bending moments requires steel calculations on several facets because cracking does not necessarily occur according to the direction of the reinforcement.
As the behavior of slabs in bending is similar to that of beams, it is necessary to redistribute the bending moment diagrams to have a trusslike behavior.
The reinforcement areas Ax are calculated directly from the efforts (Nwx, Mwx) and the reinforcements Ay from (Nwy, Mwy).
Example: 5m square slab
The slab is hinged on 3 sides and clamped on the 4th, it is subjected to a uniform permanent load of 50kN/m², a thickness of 20cm, and a concrete cover of 3cm.
Horizontal bending moments M_{YY}  Bending moments M_{XY}
Ay reinforcement is calculated manually with fyd=200MPa (SLS).
Ay manual = 10 x (M_{YY} + M_{XY})/(0.9 x 0.17 x 200) cm²
Values that can be compared with the reinforcement areas calculated by the software:
Reinforcement Ays software / Reinforcement Ayi software
There is a good agreement between the reinforcement calculated manually and the one determined by the software (CAPRA MAURY method).
It can be deduced that the Mxy bending moments are cumulated with the M_{xx} and M_{yy }bending moments.
D.4.3 Elements submitted to shear forces in their plane: concrete walls
The bracing walls of a building are subjected to normal stresses and shear forces in their plane.
We have the examples of "large dimension walls of lightly reinforced concrete" studied in EC81 chapter 5.4.3.5. or transfer slabs of buildings subjected to horizontal forces (inclined columns).
For these elements, the components Mxx, Myy, Mxy, Vxz, and Vyz are very low or even null.
They are only subjected to membrane efforts Nxx, Nyy, Nxy:

Nxx and Nyy being the tensile/compression stresses along x and y axes

Nxy the shear in the plane of the wall.
Taking into consideration the previous calculations of reinforcement distribution, this gives:

Nwx = Nxx + Nxy

Nwy= Nyy +  Nxy
Hence the reinforcement:

Ax = (Nxx + Nxy)/fyd

Ay = (Nyy + Nxy)/fyd

Ax and Ay being the sum of the reinforcement in X and Y (2 faces included)

and fyd is the design stress of the rebars
These formulas lead to the observation that if the normal stresses are zero (Nxx=Nyy=0), then the reinforcements Ax and Ay are shear reinforcements and their values are equal to Ax= Nxy/fyd and Ay= Nxy/fyd.
Hence, shear efforts require reinforcement in both directions, as opposed to the classical shear verifications performed according to EC2. This is because the strength of the concrete is not accounted for.
Therefore, the shear reinforcements calculated by the mappings are "higher" than those usually calculated using EC2.
Article 5.4.3.5.2 of EC81 concerning the study of "Poorly reinforced large dimension concrete walls" was consulted so that no shear reinforcement is required Ved is lower than Vrdc.
Thus, it is recommended in these cases to use reinforcement maps only to consider local effects, and to perform calculations of the main reinforcement by taking crosssections at the base of the walls and to determine the reinforcement from the resulting torsors at the level of these crosssections.
Example of a sail:
Let us study an isolated concrete wall, 5m high, 4m wide, and 20cm thick.
It is clamped at the bottom and subjected to a seismic horizontal load of 2000kN at the top.
To avoid peak efforts, the horizontal load is linearized over the width of the concrete wall (500kN/m).
The resulting membrane forces Nxx, Nyy, and Nxy are illustrated below.
Stresses Nxx (horizontal) (kN/m) – Stresses Nyy (vertical) (kN/m)
Stresses Nxy (shear)  Reinforcement mapping (cm²/m²) for one layer
The steel sections are calculated manually and compared with the map values.
This table shows that, on one hand, the manual calculation provides a good approximation of the required reinforcement section and, on the other hand, that the shear efforts Nxy are added to the two membrane efforts Nxx and Nyy, which does not reflect the real behavior of reinforced concrete walls.
Normative verifications of a reinforced concrete wall:
The reinforced concrete wall is recalculated according to EC2.
At a crosssection at the base of the wall, the resulting efforts are equal to:

Mflection = 2000 x 5 = 10 000kN.m

Vu = 2000kN (ALS)
Hence the bending reinforcement: Atension = 10 000 /(0.9 x 3.9 x 50) = 57 cm^{2} or 29cm^{2} per layer.
Vrdc = 997kN is lower than Vu = 2000kN so, 11cm²/m of shear reinforcement must be distributed (6cm²/m per layer) considering cot(θ)=1 or 5 cm²/m with cot(θ)=2.5.
Comparing the 2 methods:
The bending reinforcement is more important using reinforcement mappings, (36+23+15) cm²/m x 0.5 m= 37cm² compared to 29cm² using the normative verifications (30% greater).
For the shear efforts, 7cm²/m must be distributed using the reinforcement mappings whereas the normative verifications allow distributing only 2.5cm²/m considering cot(θ) = 2.5.
Summary of the comparison between the calculation of a reinforced concrete wall either by the reinforcement mappings or by the normative verifications for reinforced concrete (EC2).
D.4.4 Crosssection method
Most software allow to “cut” the elements to calculate the resulting torsors in their centers.
Let us do a study case on the base of a reinforced concrete wall:
The horizontal crosssection at point 0 at the base of the wall allows obtaining the resulting torsor consisting of the normal efforts, the shear efforts, and the bending moments (inplane or outofplane), by the integration of the stresses.
The reinforcement can then be determined by using a beam type calculation, for which one must be sure to stay within the definition range of a beam element.
The use of crosssections is above all very useful for the engineer to quantify the stress paths in a structure.
D.4.5 Scope of validity of steel mapping
D.4.5.1 Mapping and cuts
Reinforcement mapping is the result of a numerical calculation carried out for each element independently, therefore without considering the overall reinforced concrete behavior of the structure.
Note: The crosssection method is the only one that respects the behavior of the reinforced concrete walls and is considered normative. The usual additional verifications must be applied: redistribution of bending moment diagram, verification of the connecting struts, lapping, and minimal reinforcement ... is still to be done.
Therefore, the engineer needs to validate the obtained results with other normative methods.
Example of a wallbeam calculated with the reinforcement mapping:
The example of a beam on two supports is studied here. The beam is modeled as a tall wallbeam, to show on a simple case the inconsistencies of the reinforcement mappings.
Let the isostatic beam with a 10 m span, 3 m high, and a uniform load of 200 kN/m². This beam is modeled in plate elements working in their plane with a mesh size of 0.5 x 0.5 m².
Visualization of horizontal efforts Fxx
Calculation of reinforcement with reinforcement mappings:
The bending capacity is equal to F0 x d0 = 2 layers x 0.5 m x (24 x 2.75 + 16 x 2.25 + 8 x 1.75) = 116cm².m.
The resulting reinforcement is equal to Ax = 116 cm².m / 2.9 m = 40 cm².
Calculation of the reinforcement with the crosssection method:
This second method consists of “cutting” the beam in the middle (AA).
The software integrates the efforts Fxx along the height of the crosssection to deduce the resulting torsor in the middle of the crosssection.
The bending moment at the center is equal to 3358kN.m, (we then find the classical bending moment using Strength of Materials, Mu=1.35ql2/8 = 3375 kN.m), hence the reinforcement calculated according to the rules of reinforced concrete: A = 3358/(0.9 x 2.9 x 43) = 30 cm².
Conclusion and comparison between the two methods:
Reinforcement resulting from the reinforcement mappings / Reinforcement resulting from the crosssection method (Reinforced concrete calculation).
Conclusion: This example shows the limitations of steel mapping since it does not respect the deformation rule for reinforced concrete. The cutoff method makes it possible to optimize the reinforcement.
D.4.5.2 Strutandtie method: Finite element contribution
Case of a wallbeam
The Eurocode 2 strongly encourages the use of the strutandtie method, in this case, finite elements can help the engineer to define the operating scheme of the connecting struts as well as their inclinations. Let us consider the example of the following wallbeam:
As the span of 3m50 is smaller than 3 times the height of the beam, the classical rules for beams do not apply (EC25.3.1), this wallbeam is calculated using the strutandtie method.
As this example is quite simple, we can manually define the strut behavior:
The usual rules estimate the height of the connecting strut Z in 1.90m, we obtain tg(Θ) = 2.18, i.e. a tie with H=515KN, or a theoretical reinforcement area A=11.8cm² (ULS calculation).
In more complex cases, the engineer will have to define a strutandtie behavior which may be difficult. The finite elements then bring precious help for the engineer, we propose to follow the following method:
Representative isolated model:
It consists of creating an isolated representative model of the problem.
Isolated representative model
The principal stresses:
We will also refer to part 1, chapter E.3.3.
What to remember: There are 2 principal stresses, the min S1, the max S2, they are represented as perpendicular arrows, the length of each arrow depends on the intensity of the stress. The S1 stresses show the negative compressions and S2 positive tensions.
Principal stresses S2 (compressions)
Principal stresses S1 (tractions)
In our case, the long blue arrows show compressed zones, the red ones show tensioned zones.
Nota bene: when the 2 arrows S1 and S2 are almost equal, i.e. when the representation is a cross, then the zone is in pure shear.
Definition of strutandtie behavior:
The visualization of the compressive stresses at the base allows to visualize the direction of the connecting struts, its angle is of the order tg(Θ)= 2. Hence, a tie with 562kN from which a reinforcement area A = 13cm² with some uncertainty due to the graphical method.
Reinforcement mapping:
The reinforcement mapping directly reinforces the lower tie:
The reinforcement area along 1m summed results in 5.07cm² per layer, i.e. 10cm² in total. This more precise value can be retained.
Balance sheet
This approach makes it possible to identify the stress path and to set up a model of a suitable connecting strut (compatible with the stress path).
Load path in a wallbeam with multiple openings, S1 stresses
Load path in a wallbeam with multiple openings, S2 stresses
The reader might refer to §8 "Strutandtie modeling" of the FIB Bulletin nº45 for details on modeling this type of approach and to SETRA Bulletin Ouvrage d'Art nº14, pages 23 to 32.
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