104x

2023-05-24

VE 1023 | Design of a Flat Slab According to DIN EN 1992-1-1 with NA: Punching Shear Design

Description

The model is based on the example 4 of [1]: Point-supported slab. The internal forces and the required longitudinal reinforcement can be found the in verification example 1022. In this example, punching is examined in the axis B/2.

The required punching shear reinforcement is determined with manual calculation. Then, the results are compared with the RFEM results.

Verification Example 1022 | Slab Designed According to DIN EN 1992-1-1 with NA

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Verification Example 1023 | Flat Slab Designed According to DIN EN 1992-1-1 with NA: Punching Shear Design

Analytical Solution

The middle effective depth d_eff results from the concrete cover and the diameter of the longitudinal reinforcement:

d_{eff} = (d_{y} + d_{z}) / 2 = (18.0 + 20.0) / 2 = 19.0 cm

Effective Depth of Slab According to DIN EN 1992-1-1, 6.4.2 (1)

The basic control perimeter u₁ is then calculated:

u_{1} = 4 \cdot a + 2 \cdot 2.0 d \cdot π = 4 \cdot 45 + 2 \cdot 2.0 \cdot 19 \cdot π = 418.76 cm

Basic Control Perimeter in Case of Rectangluar Column According to DIN EN 1992-1-1, 6.4.2, Figure 6.13

With the coefficient ß=1.10, a shear force to be supported in the control perimeter is calculated:

v_{Ed} = \frac{β \cdot V_{Ed}}{u_{1} \cdot d} = \frac{1.10 \cdot 0.7873 MN}{4.19 m \cdot 0.19 m} = 1.088 MN / m

Maximum Shear Stress According to DIN EN 1992-1-1, 6.4.3 (3)

The punching shear resistance of the slab without shear reinforcement:

V_{Rd, c} = C_{Rd, c} \cdot k \cdot {(100 \cdot ρ_{l} \cdot f_{ck})}^{1 / 3} + k_{1} \cdot σ_{cp} \geq (v_{\min} + k_{1} \cdot σ_{cp}) = 0.12 \cdot 2.0 \cdot {(100 \cdot 0.0166 \cdot 35)}^{1 / 3} + 0.10 \cdot 0 = 0.929 MN / m^{2} \geq 0.586 MN / m^{2}

where

v_{\min} = (0.0525 / γ_{c}) \cdot k^{3 / 2} \cdot {f_{ck}}^{1 / 2} = (0.0525 / 1.5) \cdot 2 . 0^{3 / 2} \cdot 35^{1 / 2} = 0.586 MN / m^{2}

ρ_{l} = \sqrt{ρ_{ly} \cdot ρ_{lz}} = \sqrt{\frac{31.42 {cm}^{2} / m \cdot 10^{- 4}}{0.18 m} \cdot \frac{31.42 {cm}^{2} / m \cdot 10^{- 4}}{0.20 m}} = 0.0166 \leq 0.02

C_{Rd, c} = 0.18 / γ_{c} = 0.18 / 1.5 = 0.12

k_{1} = 1 + \sqrt{\frac{200 mm}{190 mm}} = 2.02 \leq 2.0

Punching Shear Resistance of Slab Without Shear Reinforcement According to DIN EN 1992-1-1, 6.4.4 (1)

Punching shear reinforcement is therefore necessary.

Maximum value of punching shear resistance:

v_{Rd, \max} = 1.4 \cdot v_{Rd, c} = 1.4 \cdot 0.929 = 1.300 MN / m^{2} \geq v_{Ed} = 1.088 MN / m^{2}

Maximum Value of Punching Shear Rresistance According to DIN EN 1992-1-1/NA, 6.4.5 (3)

Determination of the punching shear reinforced area:

V_{Rd, c, out} = C_{Rd, c} \cdot k \cdot {(100 \cdot ρ_{l} \cdot f_{ck})}^{1 / 3} + k_{1} \cdot σ_{cp} \geq (v_{\min} + k_{1} \cdot σ_{cp}) V_{Rd, c} = 0.10 \cdot 2.0 \cdot {(100 \cdot 0.0166 \cdot 35)}^{1 / 3} + 0.10 \cdot 0 = 0.774 MN / m^{2} \geq 0.586 MN / m^{2}

where

C_{Rd, c} = 0.15 / γ_{c} = 0.15 / 1.5 = 0.10

Design Value of Shear Resistance According to DIN EN 1992-1-1, 6.2.2 (1)

u_{out} = \frac{β \cdot V_{Ed}}{v_{Rd, c, out} \cdot d} = \frac{1.10 \cdot 0.787}{0.774 \cdot 0.19} = 5.889 m

Outer Perimeter According to DIN EN 1992-1-1, 6.4.5 (4)

distance from the outer perimeter to the edge of the load surface:

a_{out} = (u_{out} - u_{0}) / 2 π = (5.889 - 4 \cdot 0.45) / 2 π = 0.651 m

Distance from Outer Perimeter to Load Surface Edge

The required number of shear reinforcement rows:

n = \frac{a_{out} - 0.5 d - 1.5 d}{0.75 d} + 1 = \frac{0.651 - 0.5 \cdot 0.19 - 1.5 \cdot 0.19}{0.75 \cdot 0.19} + 1 = 2.90 \to 3 rows

Required Number of Shear Reinforcement Rows

With 3 rows of reinforcement, the radial spacing s_r is:

s_{r} = \frac{a_{out} - 0.5 d - 1.5 d}{n - 1} = \frac{0.651 - 0.5 \cdot 0.19 - 1.5 \cdot 0.19}{3 - 1} = 0.135 m

Radial Spacing of Perimeters of Shear Reinforcement

Reinforcement per row:

A_{sw} = \frac{v_{Ed} - 0.75 \cdot v_{Rd, c}}{1.5 \cdot (\frac{d}{s_{r}}) \cdot f_{ywd, ef} \cdot (\frac{1}{u_{1} \cdot d}) \cdot \sin (α)} = \frac{1.088 - 0.75 \cdot 0.929}{1.5 \cdot (\frac{0.190}{0.135}) \cdot 297.5 \cdot (\frac{1}{4.188 \cdot 0.190}) \cdot \sin (90^{\circ})} \cdot 10^{4} = 4.98 {cm}^{2}

Punching Shear Resistance with Shear Reinforcement According to DIN EN 1992-1-1, 6.4.5 (1)

This results in the following punching shear reinforcement in the individual reinforcement rows:

Reinforcement row	Distance from column edge	Factor k_sw	Punching shear reinforcement
-	-	-	A_sw*k_sw
1	0.5d=0.095 m	2.5	2.5*4.98cm²=12.45cm²
2	0.085+0.99=0.194 m	1.4	1.4*4.98cm²=6.97cm²
3	0.085+2*0.99=0.293 m	1.0	1.0*4.98cm²=4.98cm²

Results

In the following tables, the results of RFEM 6 are compared with the analytical solution.

Punching shear design
Designation	Variable	Unit	RFEM result	Analytical solution	Ratio
Design shear force	V_Ed	[kN]	787.3	787.3	1.0
Basic control perimeter	u₁	[m]	4.188	4.188	1.0
Maximum shear stress	v_Ed	[MN/m²]	1.088	1.088	1.0
Punching shear resistance without shear reinforcement	v_Rd,c	[MN/m²]	0.929	0.929	1.0
Maximum value of punching shear resistance	v_Rd,max	[MN/m²]	1.300	1.300	1.0
Outer perimeter	u_out	[m]	5.889	5.889	1.0
Distance from the outer perimeter to the edge of the load surface	a_out	[m]	0.651	0.651	1.0
Required number of shear reinforcement rows	n	[-]	3	3	1.0
Punching shear resistance with shear reinforcement	A_sw	[cm₂]	4.98	4.98	1.0
First reinforcement row	A_sw,1	[cm₂]	12.46	12.45	1.0
Distance from the outer perimeter to the edge of the load surface	l_sw,1	[m]	0.095	0.095	1.0
Second reinforcement row	A_sw,2	[cm₂]	6.98	6.97	1.0
Distance from the outer perimeter to the edge of the load surface	l_sw,2	[m]	0.194	0.194	1.0
Third reinforcement row	A_sw,3	[cm₂]	4.98	4.98	1.0
Distance from the outer perimeter to the edge of the load surface	l_sw,3	[m]	0.293	0.293	1.0

01 Verification Example 1023 | Design of a Flat Slab According to DIN EN 1992-1-1 with NA: Punching Shear Reinforcement

References

German Concrete and Structural Engineering Association E. V, examples for design according to Eurocode 2. Volume 1: Building construction, Berlin: Ernst & Sohn 2012, 1st prize corrected reprint of 1. Edition, 978-3-433-01877-4

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The punching shear design, in line with EN 1992-1-1, should be performed for slabs with a concentrated load or reaction. The node where the design of punching shear resistance is performed (that is, where there is a punching problem) is called a node of punching shear. The concentrated load at these nodes can be introduced by columns, concentrated force, or nodal supports. The end of the linear load introduction on slabs is also regarded as a concentrated load and therefore, the shear resistance at wall ends, wall corners, and ends or corners of line loads and line supports should be controlled as well.

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Design Level 2: Design of damage equivalent stress acc. to 6.8.5 and 6.8.7(1) (simplified fatigue design): The design using damage equivalent stress ranges is performed for the fatigue combination according to EN 1992‑1‑1, Chapter 6.8.3, Eq. (6.69) with the specifically defined cyclic action Q_fat.

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How can I use graphic templates for multi printing of a reinforcement?

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Although all the member design checks have been fulfilled, I get the "Not Covered Reinforcement" result. What is the reason?

Where can I find the graphical results of punching shear in the concrete design? Especially the distribution of shear forces at the critical perimeter or the punching loads.

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Can I define a different bar size in the RFEM 6 Concrete Design add-on other than the default bar sizes available in the drop-down?

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