2.4.4.4 Example of shear design

Example of 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 applied and which run approximately parallel to the direction of the principal shear force.

These yield the Applied Longitudinal Reinforcement a_sl of 0.61 cm²/m.

The shear resistance V_Rd,c of the plate without shear reinforcement is determined with the following parameters:

$C_{Rd,c} =\hspace{0.17em}\frac{0.18}{{\gamma }_c} =\hspace{0.17em}\frac{0.18}{0.15} =\hspace{0.17em}0.12$

$k =\hspace{0.17em}1 + \sqrt{\left(\frac{200}{d}\right)} = 1 + \sqrt{\left(\frac{200}{160}\right)} =\hspace{0.17em}2.11 \le 2.00 \to k =\hspace{0.17em}2.00 d \text{in [mm]}$

• d = 0.160 m

${\rho }_I =\hspace{0.17em}\frac{a_{sl}}{\left(b_w · d\right)} =\hspace{0.17em}\frac{0.613}{\left(100 · 16\right)} =\hspace{0.17em}0.000383 \le 0.02$

• b_w = 1.00 m
• f_ck = 20.0 N/mm²   for concrete C20/25
• k₁ = 0.15
• σ_cp = 0.00 N/mm²   for concrete C20/25

The same result can be found in the design details:

The shear resistance V_Rd,c of the plate without shear reinforcement is compared to the acting shear force V_Ed.

• V_Rd,c = 35.142 kN/m ≥ V_Ed = 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.