3.6.11: Optimize Reinforcement 🔷
Fig. 188.8.131.52: Calculation of reinforcement quantities for a 1mx1m plate and an in-plane tensile 50kN load
Fig. 184.108.40.206 shows a rectangular plate of size
with a uniform tensile line-load of
which translates to two point-loads of
each. The first step consists of defining a reinforced concrete cross section via a “Cross Section”-component (see section 3.3.2). The definition of reinforcement layers does not impact the displacements or cross section forces of the model. It merely forms the basis for calculating reinforcement quantities using linear elastic cross section forces. For reinforcement design it is assumed that the material in layer zero (usually concrete) has no tensile and infinite compressive strength. Since the latter is a simplification, one should assure a sufficient height of the concrete cross section by using the “Optimize Cross Section”-component first.
The input-plugs of the “Optimize Reinforcement”-component are similar to those of the “Optimize Cross Section”-component:
Under the submenu “Settings” reside two plugs which let you specify the partial safety factors for concrete (“gammaMc”) and steel (“gammaMs”). The former exists for future use, since currently concrete is assumed to be infinitely strong in compression. The latter is set to 1.15 which constitutes the standard value according to Eurocode 2 (see ). The output of “Optimize Reinforcement” comprises these plugs:
The “ShellView”-component lets one display the thickness of the reinforcement layers and the stresses there (see fig. 220.127.116.11). The input-plug “LayerInd” sets the index of the layer to be displayed. The concrete cross section has index zero, index one corresponds to the top-, index four to the bottom-most reinforcement layer. The Z-direction of the local coordinate system points to the top of the cross section. The entry “Local layer axes” under “Display Scales” in the component “ModelView” lets one enable, disable and scale the arrows of the local coordinate system.
In the example of fig. 18.104.22.168 the default reinforcement material BSt500 with a characteristic yield strength of
leads to a necessary amount of reinforcement of
. This is equivalent to a layer thickness of 0.00575 cm. The “2” in the denominator results from the fact that there are two reinforcement layers (of index one and four) which point in the direction of the tensile force.