2.3.3 Two-Directional Reinforcement Meshes with k < 0

Two-Directional Reinforcement Meshes with k < 0

If the main axial forces N₁ and N₂ have different signs in a two-directional reinforcement mesh, a tension force is respectively obtained for the equilibrium of forces in the two reinforcement directions, as well as a compression force in the selected direction of the compressive strut.

Rows 5 and 6 of Table IV (Figure 2.15) provide examples for this possible state of equilibrium.

However, for a wall subjected to both tension and compression, a compression strut may expectedly result in the direction γ and another one in the direction β for the selected direction of the concrete compression strut (arithmetic mean between the two reinforcement directions). This is exactly the case when the arithmetic mean in the diagram above is to the left of the zero crossing of the force distribution of Z_y. However, this kind of equilibrium is not possible. The reinforcement of the conjugated direction is determined, that is, the value γ_0y is used for the compression strut direction γ.

$\tan {\gamma }_{0y }= -k · cot \alpha$

This means that no force occurs in the second reinforcement direction y under the angle β. Row 7 in Table IV (Figure 2.15) shows an example of this equilibrium of forces. In the add-on module RF-CONCRETE Surfaces, such a state of equilibrium is reached when a compression force in the reinforcement direction y results for the routinely assumed direction of the compression strut (arithmetic mean between the directions of both reinforcement sets).

We have thus described all possible states of equilibrium for two-directional reinforcements.