## Exw

where

a's - fyd likewise

s Es where

s Es

The design bending moment about the centroidal axis is expressed as

MRd - Nc(h-P2x) + A's as'(-2-d) + Asas(-2-d) . In case both the reinforcing bars are yielded (as = a's = fyd ),: NRd = -b-pi-x-fcd - A's'fyd + As'fyd

pi and p2 are factors given in Tables 6.i or 6.2.

6.1.3. Design of reinforcing bars in case of bending without axial force and in case of bending with great eccentricity axial force

Let's take a transversal section with axis of symmetry y (Fig. 6.3) and load effects in the plane of symmetry. Given the design load effects at ultimate limit state MEd and NEd, the bending moment about the tension reinforcement is calculated:

MEsd = MEd - NEd ■ ys and the reinforcement area As required by MEsd is calculated like in the case of simple bending moment. The axial force NEd is taken into account subsequently, by correcting of an equivalent quantity the resistance of the tensioned reinforcement steel. In such cases, the applied axial force must be taken into account with its sign: if NEd is a compression force, it will reduce the area of tensioned steel.

A general rule is adopted: only the tensioned steel (As ) is provided, and only in case the tensioned steel is not sufficient, some compressed steel (A's ) is added. In order to ensure that the structure has a ductile behaviour, the strain £s of the tensioned steel must be greater that the strain corresponding to the limit of elasticity, that is £s ^ £yd = fyd/Es . This implies that the neutral axis does not exceed the depth where £cu is the strain at the compressed end. This limitations is valid for isostatic members; for other cases, other limitations apply (see note).

The value of £cu depends exclusively on the concrete class. The £cu2 (for parabola and exponential -rectangle diagram) and £cu3 (for bilateral and uniform) values are identical [Table 3.i-EC2]. The £yd value depends on the steel design stress fyd = fyk/ys.

If MEsd is greater than the moment Mlim, that corresponds to xlim in presence of tensioned reinforcement only, a certain amount of compressed steel has to be put in place. The difference

AMEsd = MEsd - Mlim is to be absorbed by two sets of reinforcement, one compressed and one tensioned, of area As = AM|Esd' , which both work at the design limit of elasticity.

The area of reinforcement steel, A's , has to be added to the section As corresponding to the limit bending moment.

Note

The procedure exposed in the general guidelines follows from the principle of committing, as far as possible, compression to concrete and traction to steel.

More severe limitations of x than those above- are required in order to meet ductility requirements in the case of indeterminate structures in bending, where redistribution of moments may take place.

Moreover, both in statically indeterminate and determinate cases, the verification of serviceability tensional stresses [7.2(2)-EC2] implicitly requires that the depth of neutral axis at ultimate limit state is limited. In the design process point [9.2.1.1(1-EC2] must also be taken into account requiring a minimal quantity of tensioned reinforcement steel to avoid that fragility situations arise side steel. In other words, it's necessary that the resisting bending moment of the reinforced section is greater than the moment that causes cracking.

6.1.4 Rectangular section

6.1.4.1 Use of parabola-rectangle and exponential-rectangle stress-strain relations

Design is simple if the fc and fc values (respectively the resultant and its distance from the edge, for an element of unitary width and depth), given at point 7.2, are used.

Given the section dimensions (b, h, d, effective depth defined as distance of the tensioned reinforcement steel centroid from the compressed edge), materials and action effects, the bending moment (Fig. 6.3) about the tensioned reinforcement elements is calculated: MEsd = MEd - NEd .ys .

In order to determine if As is sufficient, or if also A's is necessary, the following procedure is followed: xlim is determined the limit bending moment MRd,lim with tensioned reinforcement only is calculated

MRd,lim = Fc'Zlim where Fc = fc-b-xym-fcd is the resultant of compression stresses and Zlim = (d- fc -xym) is the inner lever arm.

a) If MEsd is smaller than Mrd,lim, As alone is needed. In order to determine it the value of x corresponding to MEsd must be defined. It results:

fc-b-x -fcd- (d- fc -x) = MEsd which, developed, becomes:

0 0