Column and Foundation Properties & Structural Loads
Foundation Plate
- Dimension in the x-direction: wx = 2.70 m
- Dimension in the y-direction: wy = 1.80 m
- Dimension in the z-direction: t = 1.00 m
- Self-weight: Gp,k = 121.5 kN with γc = 25 kN/m³
Column
- Dimension in the x-direction: cx = 0.40 m
- Dimension in the y-direction: cy = 0.40 m
- Dimension in the z-direction: h = 3.00 m
- Self-weight: Gc,k = 12 kN with γc = 25 kN/m³
Load Case 1 – Permanent Loads in z-Direction
Vertical: VG,k = 366.5 kN
Including the self-weight of the column Gc,k = 12 kN and the foundation Gp,k = 121.5 kN, the sum of the permanent vertical loads is:
VG,k,+add = 121.5 kN + 12 kN + 366.5 kN = 500 kN
The self-weight of the foundation is automatically taken into account with the self-weight of the structure as long as the "Active self-weight" is selected. If you want to enter the self-weight manually, it is necessary to define additional loads for the foundation.
The "Support Forces" as well as the "Support Forces with Foundation Load" are displayed under the results of the static analysis. For the design shear force Vz, the foundation self-weight and the additional foundation loads are not included, because they are not part of the structural system. These loads are only considered in Vz,+add of the design shear force with the additional foundation loads.
Load Case 2 – Permanent Loads in x-Direction
Horizontal: HG,k = 50 kN
Load Case 3 – Variable Loads in x-Direction
Horizontal: HQ,k = 26 kN
Load Case 4 – Variable Loads in z-Direction
Vertical: VQ,k = 110 kN
This information is crucial for performing the geotechnical design check and the reliable design of the foundation and the column. The utilization of RFEM 6 facilitates precise modeling and calculation, thereby enabling the comprehensive examination of intricate interactions between structural components.
Global Stability (Equilibrium of Structure) | (EQU) DIN EN 1997-1, 6.5.1 and 2.4.7.2
The design of the overall stability is a decisive aspect in the building design. In this design, the load combination CO10 – 0.9 ⋅ LC1 + 1.10 ⋅ LC2 + 1.50 ⋅ LC3 is considered
Md,dst ≤ Md,stb
Partial Safety Factors
| γG,stb = 0.9 | favorable permanent loads |
| γG,dst = 1.1 | unfavorable permanent loads |
| γQ,stb = 0 | favorable variable loads |
| γQ,dst = 1.5 | unfavorable variable loads |
Calculation of Destabilizing Moment
Md,dst = MG,k,dst ⋅ γG,dst + MQ,k,dst ⋅ γQ,dst = 50 kN ⋅ 4 m ⋅ 1.1 + 26 kN ⋅ 4 m ⋅ 1.5 = 376 kNm
Calculation of Stabilizing Moment
Md,stb = MG,k,stb ⋅ γG,stb + MQ,k,stb ⋅ γQ,stb = 500 kN ⋅ 2.70 m ⋅ 1/2 ⋅ 0.9 + 0 kN = 607.5 kNm
Design
η = 376 kNm / 607.5 kNm = 0.619 ≤ 1.0
Ground Failure Resistance | (GEO) DIN EN 1997-1, 6.5.2
A ground failure analysis is an essential part of geotechnical planning, as it assesses the bearing capacity of the subsoil under maximum loads. The ground failure resistance design is performed using the "Concrete Foundations" add-on in RFEM 6 and clarified in a manual calculation. The calculation steps required for the design can also be viewed directly in RFEM 6, so that you can comprehensively understand the results.
For this design, the load combination LC2 – 1.35 ⋅ LC1 + 1.35 ⋅ LC2+ 1.50 ⋅ LC3+ 1.50 ⋅ LC4 is considered with Vz,d = VG,k ⋅ γG + VQ,k ⋅ γQ = 500 kN ⋅ 1.35 + 110 kN ⋅ 1.50 = 840 kN.
According to DIN EN 1997-1/NA, it is necessary to use the modified Design Approach 2—hereinafter Approach 2*—for the ground failure resistance design. In order to determine the characteristic resistances Rk,i of the soil, it is necessary to apply the characteristic values of the load. The design approach to be used can be defined in the global settings of the add-on. Approach 2* is preset for DIN EN 1997‑1/NA.
Eccentricity ex of Effective Vertical Load in x-Direction
The characteristic values of the loading of the vertical load with additional foundation loads Vz,+add and the characteristic value of the resulting design bending moment My,+add in the foundation base center are required to determine the eccentricity of the effective vertical loads.
Vz+add = VG,k,tot + VQ,k = 500 kN + 110 kN = 610 kN
My,+add = (50 kN + 26 kN) ⋅ 4 m = 304 kNm
ex = -My,+add/Vz,+add = -304 kNm / 610 kN = -0.498 m
Effective Foundation Length, Width, and Base
The eccentric loading reduces the allowable foundation base.
wx - 2 ⋅ | ex | = 2.70 m - 2 ⋅ 0.498 m = 1.704 m
wy - 2 ⋅ | ey | = 1.80 m - 2 ⋅ 0.000 m = 1.800 m
Effective length: L' = max(wx - 2 ⋅ | ex |; wy - 2 ⋅ | ey |) = 1.800 m
Effective width: B' = min(wx - 2 ⋅ | ex |; wy - 2 ⋅ | ey |) = 1.704 m
A' = L' ⋅ B' = 1.800 m ⋅ 1.704 m = 3.0672 m²
Bearing Resistance
The bearing resistance is determined by a user-defined entry. In this example, the characteristic value of the allowable soil stress is σRk = 420 kN/m².
σRd = σRk / γR,v = 420 kN/m² / 1.4 = 300 kN/m²
σEd = Vz,d / A' = 840 kN / 3.0672 m² = 273.87 kN/m²
Design
η = σEd / σRd = 273.87 kN/m² / 300 kN/m² = 0.913 ≤ 1.0
Sliding Resistance (GEO) DIN EN 1997-1, 6.5.3
The sliding resistance design plays an important role in the evaluation of the stability against lateral sliding. This design is also performed using the add-on in RFEM 6 and illustrated by a manual calculation for the load combination LC2 – 1.35 ⋅ LC1 + 1.35 ⋅ LC2+ 1.50 ⋅ LC3+ 1.50 ⋅ LC4. To fulfill this design, the resulting horizontal force must be smaller than the sliding resistance.
Hd ≤ Rd + Rp,d
Action Hd
Hx,d = HG,k ⋅ γG + HQ,k ⋅ γQ = 50 kN ⋅ 1.35 + 26 kN ⋅ 1.50 = 106.5 kN
Sliding Resistance Rs,d
In order to calculate the characteristic effective vertical force in the soil joint, it is necessary to define the corresponding characteristic action Ech = CO6 – LC1 + LC2 + LC3 + LC4 for the design load. When using the load combination wizard, the table of the corresponding characteristic action is entered automatically. For all load combinations within the ULS, the corresponding characteristic action must be assigned at this point in order to perform the sliding design.
V'k = 500 kN + 100 kN = 610 kN
Rs,k = V'k ⋅ tan(δs,k) = 610 kN ⋅ tan(25°) = 284.4487 kN
Rs,d = Rs,k / γR,h = 284.4487 kN / 1.1 = 258.589 kN
The passive earth pressures (earth resistances Rp,d) can be defined in the geotechnical configurations. They are deactivated in the default settings and also neglected in this example, as they have a positive effect on the sliding resistance design.
Design
η = Hx,d / Rs,d = 106.5 kN / 258.589 kN = 0.412 ≤ 1
Highly Eccentric Loads in Core (Limitation of Gaping Joint) | DIN EN 1997‑1, A 6.6.5
The highly eccentric loading in core is a decisive factor in the foundation design. This section presents the corresponding design and provides an explanation of the calculation with RFEM 6.
The eccentricity of the soil pressure resultants under permanent and variable actions may be so large that the foundation base remains to be subjected to compression up to its centroid. This means that the second core width must be observed. This is limited by an ellipse so that the condition (ey/L)² + (ex/B)² ≤ 1/9 = elim applies.
In the case of foundations on non-cohesive and cohesive soils, no gaping joint may arise in the foundation base as a result of characteristic permanent actions. Accordingly, the resultant must be within the first core width. For a biaxial eccentricity to lie within the rhomboid boundary, the condition (|ey|)/L + (|ex|)/B ≤ 1/6 = elim must be met.
Mark 1: The limitation of the 1st core width is represented by a rhombus.
Mark 2: The limitation of the second core width is represented by an ellipse.
Mark 3: The resultant's point of application is the point at which the entire resulting load acts on the foundation base.
To ensure that the various conditions are managed with the correct load combinations, it is necessary to defined them as follows. First, you need to create a new design situation in "SLS – Characteristic | Permanent”.
In the next step, copy the characteristic load combinations, comprising solely permanent loads, and assign them to the new design situation.
In the table entry for the "Concrete Foundations" add-on, the Design Situations tab, it is necessary to select the assignment to the design situation type "EN 1997 | DIN | 2022-10” to “SLS – Characteristic | Permanent”. This adjustment ensures that the relevant standards and guidelines are taken into account correctly.
Permanent Actions (1st Core Width)
In this example, the design only has to be performed with the load combination CO11 – LC1 + LC2.
Design Bending Moment with Additional Foundation Loads in Middle of Foundation Base
The calculated design bending moment at the center of the foundation base, taking into account additional loads, results as follows:
My,+add = 500 kN ⋅ 0 m + 50 kN ⋅ 4 m = 200 kNm
Design Shear Force with Additional Foundation Loads
Vz,+add = 500 kN
Eccentricity of Effective Vertical Load in x-Direction
ex = -My,+add/Vz,+add = -200 kNm / 500 kN = -0.400 m
This value indicates the extent to which the vertical load deviates from the geometric center. This is crucial for the stability analysis.
Allowable Relative Load Eccentricity
In the case of rectangular solid cross-sections, the first core width is closed by a rhombus and limited by elim = 1/6.
Design
η = (|ex| / wx + |ey| / wy) / elim = (0.400 m/2.70 m + 0/1.80 m) / (1/6) = 0.889 ≤ 1
Permanent and Variable Actions (2nd Core Width)
In this example, the design only has to be performed with the load combination CO6 – LC1 + LC2 + LC3 + LC4.
Design Bending Moment with Additional Foundation Loads in Middle of Foundation Base
The calculated design bending moment at the center of the foundation base, taking into account additional loads, results as follows:
My,+add = 610 kN ⋅ 0 m + (50 kN + 26 kN) ⋅ 4 m = 304 kNm
Design Shear Force with Additional Foundation Loads
Vz,+add = 500 kN + 110 kN = 610 kN
Eccentricity of Effective Vertical Load in x-Direction
ex = -My,+add/Vz,+add = -304 kNm / 610 kN = -0.498 m
This value indicates the extent to which the vertical load deviates from the geometric center. This is crucial for the stability analysis.
Allowable Relative Load Eccentricity
For rectangular solid cross-sections, the second core width is closed by an ellipse and limited by elim = 1/9.
Design
η = ((ex / wx)² + (ey / wy)²) / elim = ((-0.498 m/2.70 m)²+(0/1.80 m)²) / (1/9) = 0.307 ≤ 1
Load with Large Eccentricities | DIN EN 1997-1, 6.5.4
If the eccentricity of the load resultants exceeds 1/3 of the side length for rectangular foundations or 0.6 of the radius for circular foundations, it is necessary to implement specific measures. If the eccentricity exceeds the above-mentioned limit value of 1/3, special measures must be taken according to 2.4.2. The add-on provides you with precise and comprehensive results that enhance the quality and safety of the foundation design. In this example, the design only has to be performed with the load combination CO6 – LC1 + LC2 + LC3 + LC4.
Internal Forces to Apply
My,+add = (50 kN + 26 kN) ⋅ 4 m = 304 kNm
Vz,+add = 500 kN + 110 kN = 610 kN
Resulting Eccentricity
ex = -My,+add / Vz,+add = -304 kNm / 610 kN = -0.498 m
Allowable Eccentricity
ex,max = wx / 3 = 2.70 m / 3 = 0.90 m
Design
η = |ex | / ex,max = |-0.498 m| / 0.90 m = 0.554 ≤ 1
Since Mz = 0 in this example, the other direction can be neglected here.
In our example, there is no groundwater. Therefore, it is not necessary to perform the uplift design.