RF-CONCRETE Surfaces Version 5

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RF-CONCRETE Surfaces Version 5

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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, there will be a tension force for the equilibrium of forces in the two reinforcement directions respectively, and a compression force in the selected direction of the compressive strut.

Figure 2.16 Two-directional reinforcement in tension and compression

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

For a wall subjected to both tension and compression, however, a compression strut may expectedly result in direction γ and another one in direction β for the selected direction of the concrete compression strut (arithmetic mean between the two directions of reinforcement). This is the case when the arithmetic mean in the diagram above is to the left of the zero-crossing of the force distribution of Zy. However, this kind of equilibrium is not possible. We determine the reinforcement of the conjugated direction, that is, the value γ0y is used for the compression strut direction γ.

tan γ0y = -k · cot α 

Equation 2.12 (2.12)

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 RF-CONCRETE Surfaces add-on module, such a state of equilibrium is reached when a compression force in the direction of the reinforcement direction y results for the routinely assumed direction of the compression strut (arithmetic mean between the directions of both reinforcement directions).

Thus, we have described all possible states of equilibrium for two-directional reinforcements.

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