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Overview

1 Introduction

2 Theoretical Background

3 Input Data

4 Calculation

5 Results

6 Result Evaluation

7 Printout

8 General Functions

9 Literature

2.5.6 Shear Design

Shear Design

In the shear design, the applied tensile reinforcement is determined first.

Image 2.74 Applied tensile reinforcement

From all reinforcement layers and directions, a total of 1.54 cm²/m of tension reinforcement can be applied. With it, the shear force V_Rd,c that can be resisted without shear reinforcement is determined.

Image 2.75 Design shear resistance without shear reinforcement

With the applied tensile reinforcement, the longitudinal reinforcement ratio ρ_l is determined:

$ρ_{l} = \frac{A_{s l}}{(b_{w} \cdot d)} = \frac{1.54}{(100 \cdot 125.5)} = 0.00012 \leq 0.02$

In a 3D model type (in contrast to a plate), an additional axial force can occur. It must be considered via the corresponding concrete longitudinal stress.

$σ_{c p} = \frac{n_{β}}{(b_{w} \cdot h)} = \frac{- 425.783}{(100 \cdot 129)} = - 0.33 {N/m}^{2}$

The factor k for considering the plate thickness is calculated as follows:

$k = 1 + \sqrt{\frac{200}{d}} = 1 + \sqrt{\frac{200}{1255}} = 1.399 \leq 2.0 d in [mm]$

The following factors are also included in the design:

Table 2.2
Factor of concrete longitudinal stress	k₁ = 0.15
Concrete compressive strength for C30/37	f_ck = 30.0 N/mm²
Safety factor	$C_{r d, c} = \frac{0.18}{γ_{c}} = \frac{0.18}{1.5} = 0.12$

Thus, the design shear resistance V_Rd,c without shear reinforcement can be determined according to Equation (6.2a):

$V_{R d, c} = [C_{r d, c} \cdot k {(100 \cdot ρ_{l} \cdot f_{c k})}^{\frac{1}{3}} + k_{1} \cdot σ_{c p}] \cdot b_{w} \cdot d = = [0.12 \cdot 1.399 {(100 \cdot 0.00012 \cdot 30.0)}^{\frac{1}{3}} + 0.15 \cdot 0.33] \cdot 1000 \cdot 1255 = 212.00 kN/m$

According to Equation (6.2b), the minimum value of the design shear resistance V_Rd,c without shear reinforcement is determined from the minimum reinforcement ratio ν_min:

$ν_{min} = 0.035 \cdot k^{\frac{3}{2}} \cdot {f_{c k}}^{\frac{1}{2}} = 0.035 \cdot 1 . 399^{\frac{3}{2}} \cdot 30 . 0^{\frac{1}{2}} = 0.317$

$V_{R d, c} = (0.317 \cdot 0.15 \cdot 0.33) \cdot 1000 \cdot 1255 = 459.96 kN/m$

Because the plate's design shear resistance V_Rd,c = 459.96 kN/m is greater than the applied shear force V_Ed = 259.726 kN/m, no shear reinforcement is required in the example.

Should the plate's shear resistance be insufficient, the program first checks if the maximum shear resistance of the concrete compression strut V_Rd,max is sufficient. V_Rd,max is determined with the minimum inclination of the compression strut θ. When the design shear resistance of the concrete compression strut is greater than the applied shear force V_Ed, the statically required shear reinforcement req a_sw can be determined. Then the design for the shear reinforcement V_Rd,sy is carried out.

Parent Chapter

2.5 Shells