## Fdb

cd y

Once determined x e as, the moment resistance results:

MRd - A'sfyd(h-d') + Asas(-2-d") + P1xbfcd(^2-P2x)(

d) In the fourth field (NRd3 ^ NEd ^ NRd4 ) the moment resistance can be determined, with a good approximation, by the relation of proportionality indicated in fig. 6.8, which shows the final end of the interaction diagram M-N.

NRd3 Nr.J NRJ4

Figure 6.8. Terminal end of the interaction diagram M-N

The moment resistance reaches a maximum for x = x2 where the analytic function that expresses it has an edge point due to the discontinuity between (6.4) and (6.6). The derivative for x = x2 is positive if (6.4) is used and is negative with (6.6).

### 6.1.6 Interaction diagram MRd-NRd

In the case of sections subjected to bending with axial force with small eccentricity, such as those of columns, the most logical solution is the one with double symmetric reinforcement. Such sections can also withstand simple bending and, if it's the case, composed bending in relation if the dimensions and placement of reinforcement.

In order to have an overview on the problem, let's consider the load capacity of a rectangular section (Fig. 6.9) with dimension h = 600 mm, b = 400 mm of fck 30 concrete, in the four conditions:

- no reinforcement

- symmetric reinforcement (fyk = 450 N/mm2 ) at the edges in percentage 0,5 - 1,0 - 1,5 on each edge, distanced at d' =50 mm.

The calculation of the resistance of this section at the ultimate limit state is developed, for each of the four reinforcement conditions, by associating the parabola-rectangle diagrams for concrete and the bilinear diagram or steel, with 7 configurations characterized by strain values of plane section (reference to Fig. 1) given in Table 6.7. Steel B450 (fyk = 450 N/mm2).

£c = - 0,0035 at upper end |
£s = + 0,05000 (bottom reinf.) |

£c= - 0,0035 at upper end |
£s = + 0,02500 (bottom reinf.) |

£c= - 0,0035 at upper end |
£s= + 0,01000 (bottom reinf.) |

£c= - 0,0035 at upper end |
£s= + 0,00196 (bottom reinf.) |

£c= - 0,0035 at upper end |
£s= 0,00000 (bottom reinf.) |

£c= - 0,0035 at upper end |
£c= 0,00000 at lower end |

£c= - 0,0020 everywhere |
£c = - 0,0020 everywhere |

Development of the calculation in relation with the third strain condition for the reinforced section with 1% bilateral reinforcement (As = A's = 2400 mm2 )

Upper reinforcement: £'s = - 0,0023 and therefore a's = -391 N/mm2; F's = -939 kN Lower reinforcement: £s = + 0,010 and therefore as = + 391 N/mm2 , Fs = +939 kN Neutral axis: x/d = 3,5/(3,5+10) , and as d = 550 mm, x = 142,6 mm

Distance of Fc from the compressed edge: fcx = 59 mm

Moment about the concrete section centroid

MRd = Fc (d/2- p2x ) + Fs (d - d')= 189,1 + 469,5 = 658,6 kNm

The eccentricity of NRd is MRd / NRd = 839 mm.

Fig. 6.9 graphically shows the 29 pairs of results obtained. For each value of the percentage of reinforcement the points are joined by a straight line. The result is convex polygons.

The following observations arise from the observation of Fig. 6.9:

3. the polygons include the domains of resistance: the points of co-ordinates Med, Ned placed inside the polygon are in a safe zone; the points on the polygon strictly verify the ultimate limit state; external point do not meet resistance conditions at ultimate limit state

4. a straight line parallel to the N axis intersects the polygon in two points. This means that with a given reinforcement, a given bending moment can be withstood by two different values of the axial force. The limit case of a single N value happens when the straight line passes by the highest point of the polygon.

5. Only one value of M can be associated to a given value of N, as it's possible to verify by tracing a line parallel to the M axis.

6. The polygon related to the non-reinforced section denotes the possibility to withstand bending moments only if they come with an adequate axial force (provided by self weight or by prestressing).

7. The branches on the left of the M axis denote resistance to bending with positive axial force

8. The straight line of inclination M/N = h/30 = 20 mm defines the field of use of the polygons: in fact, according to [6.1(4)-EC2] a minimal eccentricity must always be taken into account, adding up to the bigger value between h/30 and 20 mm.

Such polygons as those traced below are called "M-N interaction diagrams".

Axial Force NRl)<RN) Figure 6.9. Interaction diagram for rectangular section

### 6.1.7 Biaxial bending and bending with axial force

Biaxial bending may be separated into separate uniaxial bending components under circumstances laid down in Eurocode 2 5.8.9 . For pure biaxial bending, where the bending components lay on the two centroid axis of inertia, the problem solution has computational difficulties. If a design software is not used, and calculation developed by hand, it should be processed by iterations; in such case it's convenient to adopt the rectangular stress diagram for concrete.

Given a section subjected to an axial force Ned applied on the centre of gravity, and to two bending components Meyd e Mezd expressed by two vectors orientated along a couple of orthogonal axis y e z with origin in the centre of gravity. In general, the existence of a stress distribution that gives place to resistance greater or equal to the action effects must be demonstrated, as well as the fact that the straight line that connects the centre of gravity of the compressed zone with the centre of gravity of the tensioned reinforcement is perpendicular to the resultant bending vector Med. This implies that the eccentric axial force must also lay on that line.

Axial Force NRl)<RN) Figure 6.9. Interaction diagram for rectangular section

6.2 Shear C6.2.2 Shear capacity of members without shear reinforcement

6.2.2 Members not requiring 6.2.2.1. Shear flexure capacity design shear reinforcement Most shear failures occur in the region of the member cracked in flexure. It is necessary to make a distinction between shear flexure and shear tension. In this chapter only shear flexure is regarded, which can be considered as the general case.

In ENV 1992-1-1 the equation for the shear capacity of members without shear reinforcement was VRdi = [TRd k (1.2 + 40 pl) + 0.15acp]bwd (6.7)

Where

TRd basic shear strength, which follows from TRd = 0.25fctk,o.o5/yc. k factor allowing for the size effect, equal to k = 1.6 - d (m) > 1 pl flexural tensile reinforcement ratio, As/bwd < 0.02 CTcp design axial stress (if any) = Nsd/Ac bw minimum web section

The are two shortcomings with regard to the use of this equation. At first the role of the concrete strength is not correct, as was demonstrated in [Walraven, 1987, pp. 68 - 71.] For lower strength concrete classes the deviations were not yet very large, but if the strength increases the deviations soon reach an unacceptable level.

The second problem is that the equation has principally been derived for beams, failing in shear flexure and is not valid for members which typically fail in shear tension. Such members are for instance prestressed hollow core slabs, which nearly always fail in shear tension, in the area where the member is not cracked in flexure. Applied to such members Eq. 6.7 would give unnecessary conservative results.

The recommendations for the determination of the shear flexure capacity of members not reinforced in shear are given in chapter 6.2.2 of prEN 1992-1-1:2001. The basic formula is given as Eq. 6.2.a in this document. This equation has been derived in the following way. The basic equation adopted, which was believed to take appropriate account of the most important influencing factors like concrete strength, longitudinal reinforcement ratio and cross-sectional height was

where k = size factor = 1 + (200/d)1/2 pl = longitudinal reinforcement ratio fc = concrete cylinder strength (N/mm2) C = coefficient to be determined A selection was made of a representative number of shear tests, considering a parameter variation as wide as possible and as well as possible distributed within practical limits. This was already done by König and Fischer (1995). An overview of the test parameters is given in Fig. 6.10. Then for every test result the optimum value C was determined. If the distribution is normal, Fig. 6.11 a lower bound value for C was determined according to the level 2 method described in [Taerwe, 1993 ] with the equation:

where a sensitivity factor, equal to 0.8 for the case of one dominating variable (concrete strength) ß reliability index, taken equal to 3.8 according to [Eurocode, Basis of Structural design, Draft version 2001 ] v standard deviation If the distribution turns out to be log-normal, Fig. 6.11, the equation is

In these equations a reliability index ß = 3.8 means a probability of occurrence of 0.0072%. König and Fisher (1995) carried out this procedure for 176 shear tests.

ratio of reinforcement relative frequencies tor NSC and HSC ret frequencies ratio of reinforcement relative frequencies tor NSC and HSC ret frequencies

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