## Info

B - effective tension area for upper surface C - effective tension area for lower surface

Figure 7.9 Definition of the effective tension area

### Figure 7.9 Definition of the effective tension area

In situations, where bonded reinforcement is fixed at reasonably close spacing within the tension zone (spacing < 5(c+^/2), the maximum final crack spacing can be calculated from the expression:

s=3,4c+0,425kik2^

where *

c ki k2

bar diameter.

cover to the reinforcement coefficient which takes account of the bond properties of the bonded reinforcement;

ki = 0.8 for high bond bars ki= 1.6 for bars with an effectively plain surface coefficient which takes account of the distribution of strain;

k2 = 0.5 for bending k2= 1.0 for pure tension

For cases of eccentric tension or for local areas, intermediate values of k2 should be used which can be calculated from the relation:

Where £1 is the greater and £2 is the lesser tensile strain at the boundaries of the section considered, assessed on the basis of a cracked section. As can be seen, with regard to crack spacing, the cover, c, is introduced explicitly into the expression of the crack width as suggested by A. Beeby. This suggestion is backed up by the following figure, taken from reference [1], which clearly shows the dependence of the crack spacing on this parameter. In EC2, c is implicitly taken as 25 mm.

0 50 100 150 200 250

transfer length I, [mm]

Fig.7.10 Influence of cover on the transfer length acccording to Beeby [1]

Also, on a formal level, the formula of prEN gives Srmax instead of Srm. Srmax is obtained as 1.7 times Srm. Thus,

Pp,eff of EC2 becomes

Pp,eff of prEN.

In the case in which 2 different bar diameters are used, an equivalent diameter ^eg has to be determined in order to apply the above formulation. In EC2 [4] it is recommended to use the average diameter

MC90[3] and prEN [10] suggest for this case the equivalent diameter ^eg, although MC90 provides no definition of ^eg. The definition of ^eg depends on the definition of pf. To show the difference, first ^m will be applied to the equilibrium equation (7.33) to derive ^eg and then ^eg will be derived by considering 2 diameters in the determination of the steel area (7.35).

(7.33) changes in case of the use of 2 different diameters into (7.34).

Hence:

ctm ctm

since Srmax = 2lt and Srmax = ^/3.6p and, according to MC90[5]

2f A

ctm ct,eff

ct,eff req

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