## Dog

Remembering that MEsd = As fyd z, with z = (d- fc -x), finally

As Fc must be equal and contrary to Ft , the resultant of traction of the reinforcement steel As , it can also be determined by:

- If MEsd is greater than MRd,lim , some reinforcement steel A's in compression is needed. To calculate it, AMEsd = MEsd - MRd,lim. from which:

The tensioned reinforcement is:

In such cases AMEsd must be sensibly smaller than MRd,lim , viz. it must not distort the problem.

### 6.1.4.2 Rectangular diagram of concrete stresses

With reference to Fig. 6.4, y = hx is the depth of the compressed zone and nfcd is the design tensile stress. Values of the X and ^ factors are given in Table 6.4. It results:

Figure 6.4. Rectangular section with rectangular concrete stress diagram

Rearranging:

Figure 6.4. Rectangular section with rectangular concrete stress diagram

Also in this case, the considerations and developments of the previous point apply. 6.1.4.3 T-sections

Two situations can arise in T-sections:

- the neutral axis is in the flange: no difference as if the section was rectangular;

- the neutral axis crosses the web: its determination and the subsequent developments are simple if the rectangular stress diagram provided for at paragraph [3.1.7(3)-EC2] is adopted. An approximate method is also presented.

### 6.1.4.3.1 General method Introduction

- As very high values of resisting moments can be reached with T-sections, especially if medium/high strength concrete is used, the tensioned reinforcement steel should be laid on two layers, each one of area As. The upper layer will have distance d from the compressed edge so that, as this reinforcement layer will have at least strain £yd , the lower layer is also surely yielded. In some cases, a lower bulb for the placement of tensioned reinforcement steel could be needed.

- The rectangular stress diagram is adopted for calculation. The block of compressive stresses is defined by a uniform value nfcd and by the extension y = Xx where X and n are two factors lower than 1 and function of fck according to the formulae [3.19 to 3.22-EC2]. Values of and n are given in Table 2.3.

- In case of simple bending moment, the equilibrium to rotation between external and internal moment is written with reference to the layer that corresponds to half way between the reinforcement layers As (steel reinforcement centroid). In this way the contributions of the two tensioned reinforcement layers do not appear in the equation. The position of the neutral axis is determined through this equation. Then, the reinforcement elements are determined by equilibrium to shifting.

- In case of bending with axial force the reference layer of bending moments M* (sum of the given moment and of the one deriving from the shifting of the force NEd) will still be the above-indicated one. In this case the equilibrium to shifting, that is used to determine As, also contain NEd.

### Procedure

With reference to Fig. 6.5 y is assumed as the basic parameter. M* is the given bending moment about the As reinforcement centroid. s is the distance of this centroid from the concrete compressed edge. The bending moment resistance is expressed as the sum of the moments of blocks of compression, about the same layer. The equilibrium gives:

By developing, it results:

E c rifcd

E c rifcd

Taking:

with k function of the geometrical data only, y is given by:

If M* is lower than the limit moment, no compressed reinforcement bars are needed. The tensioned reinforcement area is given by:

2As = (Fc +NEd )/fyd and the resultant of compression Fc is: Fc = n fcd-[bw-y + (b - bw)-c]

If it is not necessary to put the tensioned reinforcement bars on two layers, in the above-shown formulae is is sufficient to identify s as d, while the first member of (7.13) gives the necessary area of reinforcement.

### 6.1.4.3.2 Approximate design method

The method applies to those T-beams where the flange is able to withstand all compressive forces deriving from bending moment and axial force, without addition of compressed reinforcement bars. It's assumed that tensile stresses are uniformly distributed.

With reference to Fig. 6.6, the total bending moment M* about the centroid of tensioned reinforcement bars is calculated: M* = MEd - y* NEd (NEd positive if traction).

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