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188.8.131.52 Example shear design
Example shear design
The shear design of a plate according to EN 1992-1-1 is presented by means of the design details (see example for statically required reinforcement, Figure 2.39).
In the detailed results, the shear forces determined in RFEM are shown first.
The required longitudinal reinforcement is determined from these internal forces.
The analysis of the shear resistance is shown further below in the details. It starts with determining the allowed tensile reinforcement in the direction of the principal shear force.
The second reinforcement direction at the bottom surface of the plate and the first reinforcement direction at the top surface of the plate are the only reinforcement directions to which tension is assigned and which run approximately parallel to the direction of the principal shear force.
These yield an Applied Longitudinal Reinforcement asl of 0.61 cm2/m.
The shear resistance VRd,c of the plate without shear reinforcement is determined with the following parameters:
- d = 0.160 m
- bw = 1.00 m
- fck = 20.0 N/mm2 for concrete C20/25
- k1 = 0.15
- σcp = 0.00 N/mm2 for concrete C20/25
The same result can be found in the design details:
The shear resistance VRd,c of the plate without shear reinforcement is compared to the acting shear force VEd.
- VRd,c = 35.142 kN/m ≥ VEd = 29.56 kN/m
It has therefore been determined that the shear resistance of the plate without shear reinforcement is sufficient and no further checks are necessary.