## Info

0,0026

4,06

0,57

Line 2: it is defined by the strain £s = £syd of the lower layer of reinforcement. In this case the depth of the neutral axis, deduced by (6.2) results:

For positions of the neutral axis xi < x < X2, both reinforcement layers are subjected to a stress Os = fyd (compression for the upper one, traction for the lower one), with strain £s ^ £syd.

Configuration 3 is characterized by strain value £s = 0 (and therefore Os = 0) for the lower reinforcement (x = d). Therefore in the 3-4 range stress is Os = fyd for the upper reinforcement and Os = £s-Es for the lower.

Line 3' is defined by £c = 0 at the lower edge of section.

In configuration 4: £s = £s' = £c2 (in absolute value), which is always greater than £syd. It results then Os = Os' = fyd . In the transition from 3 to 4 the upper reinforcement is compressed at stress fyd, the lower reinforcement at stress increasing from 0 to fyd.

The following values of the axial force resistance NRd correspond to configurations 1,2,3,4: NRdi = pibxifcd NRd2 = pibx2fcd

In these two cases there are no contributions from the reinforcement bars because these are subjected to ±fyd, and generate two equal and opposite forces in equilibrium.

pi values are given in Table 6.i from the parabola-exponential rectangle model and in Table 6.2 for the rectangle model.

Calculation of the moment resistance in the four above-defined sectors. a) Ned < NRdi , that is x < xi

The position x of the neutral axis must be preliminary determined by the equation of equilibrium to shifting. Keeping in mind that the upper reinforcement A's is compressed in elastic field and that the lower reinforcement As is tensioned at stress fyd, the equilibrium is written as:

Taken o's =E£'s where £'s is given by (6.i), it results:

Taken o's =E£'s where £'s is given by (6.i), it results:

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