# Design of Concrete Columns Subjected to Axial Compression with RF-CONCRETE Members

### Technical Article

This article deals with rectilinear elements of which the cross-section is subjected to axial compressive force. The purpose of this article is to show how very many parameters defined in the Eurocodes for concrete column calculation are considered in the RFEM structural analysis software.

#### What is axial compression?

A section of a structural element is stressed by axial compression when the forces acting on one side of the section are reduced at the section's center of gravity to a single force N. Thus, the normal force N is perpendicular to the cross-section and directed towards the cross-section. In contrast to combined bending, this stress is never encountered in practice, because a real column is always subjected to either the asymmetry of the loading or to imperfections in the column construction, as can be seen in this technical article.

#### Slenderness Criterion for Isolated Elements

It is assumed that second-order effects (imperfections, asymmetry, etc.) can be neglected if the element is only loaded by an axial compressive force N_{Ed} and if the slenderness criterion is met.

Slenderness Criterion

λ < λ_{lim}

λ ... slenderness coefficient

λ_{lim} ... limiting slenderness

#### Slenderness and Effective Length According to EN 1992-1-1

$\mathrm{\lambda}=\frac{{\mathrm{l}}_{0}}{\mathrm{i}}$

λ |
slenderness coefficient |

l_{0} |
effective length = k_{cr} ⋅ l |

i |
radius of gyration of uncracked concrete section |

k_{cr} |
effective length factor = 0.5 ⋅ √[(1 + k_{1} / (0.45 + k_{1})) ⋅ (1 + k_{2} / (0.45 + k_{2}))] according to 5.8.3.2(3) formula (5.15) |

l |
free length |

k_{1}, k_{2} |
flexibility coefficients at both ends of element |

#### Limiting Slenderness According to EN 1992-1-1

Limiting slenderness

λ_{lim} = (20 ⋅ A ⋅ B ⋅ C) / √n according to 5.8.3.1(1) formula (5.13N)

A = 1 / (1 + 0.2 φ_{ef}) = 0.7 if φ_{ef} is unknown

B = √(1 + 2 ⋅ ω) = 1.1 if ω is unknown

C = 1.7 - r_{m} = 0.7 if r_{m} is unknown

n = N_{Ed} / (A_{c} ⋅ f_{cd}) ... relative normal force

φ_{ef} ... effective creep coefficient

ω ... mechanical reinforcement ratio

r_{m} ... moment ratio

N_{Ed} ... design value of acting axial force

A_{c} ... total area of pure concrete section

f_{cd} ... Design value of the compressive strength of concrete

#### Compressive Stress in Steel

The concrete shrinkage under axial compression is limited to ε_{c2} in the case of the σ-ε parabola-rectangle diagram. By static friction of concrete and steel, the shortening is identical for the reinforcement, and we can deduce its stress.

${\mathrm{\sigma}}_{\mathrm{s}}=\left\{\begin{array}{l}{\mathrm{f}}_{\mathrm{yd}}\mathrm{if}{\mathrm{\epsilon}}_{\mathrm{c}2}{\mathrm{\epsilon}}_{\mathrm{ud}}\\ {\mathrm{E}}_{\mathrm{s}}\xb7{\mathrm{\epsilon}}_{\mathrm{c}2}\mathrm{otherwise}\end{array}\right.$

σ_{s} |
stress in reinforcement |

f_{yd} |
design yield strength of reinforcing steel = f_{yk} / γ_{s} |

ε_{c2} |
relative compression strain for maximum stress |

E_{s} |
modulus of elasticity |

f_{yk} |
characteristic yield strength |

γ_{s} |
partial safety factor of steel |

ε_{ud} |
design value of limit deformation = f_{yd} / E_{s} |

#### Compressive Stress in Concrete

Stress in concrete

f_{cd} = α_{cc} ⋅ f_{ck} / γ_{c}

α_{cc} ... factor considering long-term actions on compressive strength

f_{ck} ... characteristic compressive strength of concrete

γ_{c} ... partial safety coefficient relating to concrete

#### Dimensions of the Concrete Cross-Section

The force that can be balanced by the concrete cross-section corresponds to its maximum load-bearing capacity for compression, which depends directly on its section and its design resistance.

Equilibrium force of concrete

F_{c} = A_{c} ⋅ f_{cd}

The reinforcement will balance the rest of the axial compressive load.

Equilibrium force of reinforcement

F_{s} = N_{Ed} - F_{c}

From these two equilibrium equations, it is possible to deduce the concrete cross-section to be designed, then that of the reinforcing steel.

Area of concrete cross-section

A_{c} ≥ N_{Ed} / (f_{cd} + A_{s} / A_{c} ⋅ σ_{s})

A_{s} = F_{s} / σ_{s} ... reinforcement area

#### Application of the Theory Using RF-CONCRETE Members

In this article, we will analyze the results obtained automatically for the reinforcement calculation. As the objective is also to determine the concrete cross-section to be designed, the RFEM basis model will have a specified width and an unknown height greater than or equal to the width.

We will consider the following parameters:

- Permanent loads: N
_{g}= 1 390 kN - Variable loads: N
_{q}= 1,000 kN - Column length: l = 2.1 m
- Rectangular cross-section to be determined: width b = 40 cm / unknown height ≥ 40 cm
- The column's self-weight can be ignored.
- Column not integrated into bracing.
- Concrete strength class: C25/30
- Steel: S 500 A for inclined graph
- Diameter of longitudinal reinforcement: ϕ = 20 mm
- Diameter of transverse reinforcement: ϕt = 8 mm
- Concrete cover: 3 cm

#### Material Properties

Design value for the compressive strength of concrete

f_{cd} = 1 ⋅ 25 / 1.5 = 16.7 MPa

Relative compression strain for maximum stress

ε_{c2}= 2‰

Design yield strength of reinforcing steel

f_{yd} = 500 / 1.15 = 435 MPa

Limit strain in reinforcement

ε_{ud} = f_{yd} / E_{s} = 435 / (2 ⋅ 10^{5}) = 2.17‰

Stress in reinforcement

σ_{s} = 2 ⋅ 10^{5} ⋅ 0.002 = 400 MPa as ε_{c2} < ε_{ud}

In order to check the material settings in RF-CONCRETE Members, Image 02 shows the provided stresses and strains for concrete and the required reinforcement.

#### Ultimate Limit State

Ultimate limit state design loads

N_{Ed} = 1.35 ⋅ N_{g} + 1.5 ⋅ N_{q}

N_{Ed} = 1.35 ⋅ 1390 + 1.5 ⋅ 1000 = 3.38 MN

#### Second Order Effects not Taken into Account in ULS

In order to apply the load at the head of the column correctly, we have modeled a member that is restrained only at the base and is free at its head. However, we want to consider the column being fixed at the head to some beams by assuming that the column is less stiff than the beams. We can then consider that the member is fixed at both ends. Thus, in theory, the flexibility coefficients should be zero for a perfect restraint, but the perfect restraint does not exist in practice; the minimum value to be considered for the flexibility coefficients is: k_{1} ou k_{2} = 0.1.

Effective length factor

k_{cr} = 0.5 ⋅ (1 + 0.1 / (0.45 + 0.1)) = 0.59

Image 04 shows the possibility of setting the effective length factor for a member type element in RFEM.

Since the height of the cross-section has to be determined, it is assumed that h > b, and thus, that the radius of gyration of a rectangular cross-section is more governing for the small width.

Governing radius of gyration in plane parallel to width b = 40 cm

i_{z} = b / √12

Slenderness

λ_{z} = (0.59 ⋅ 2.1 ⋅ √12) / 0.40 = 10.73 m

Image 05 shows the slenderness values determined for the member after calculation in RFEM Table 4.10.

To verify our slenderness, we manually determine the limiting slenderness by assuming h = b.

Limiting slenderness

n = 3.38 / (0.40² ⋅ 16.7) = 1.26

λ_{lim} = 20 ⋅ 0.7 ⋅ 1.1 ⋅ 0.7 / √1.26 = 9.6 m

λ_{z} > λ_{lim} → The condition is not fulfilled.

However, we are still going to calculate the column with regard to axial compression because, the difference being small, we note below that with the determination of the section's real height, the condition will be respected.

#### Real Height to Be Calculated

In order to determine the real height h of the cross-section, the following hypothesis can be adopted for the reinforcement ratio to be considered: A_{s} / A_{c} = 1%. We can then deduce the real cross-section to be designed and its height as a function of the stress in the reinforcement and the width of the cross-section b.

Area of concrete cross-section

A_{c} ≥ 3.38 / (16.7 + 400 / 100) = 0.163 m²

Cross-section height

A_{c} = b ⋅ h → h ≥ 0.163 / 0.4 = 0.41 m

The assumption h > b made for the slenderness calculation is correct, and we can retain a section height by choosing a multiple of 5 cm; that is to say, h = 45 cm.

Image 06 shows the steps to automatically determine the height of the rectangular cross-section in RF-CONCRETE Members, using the "Optimize" function.

#### Load-Bearing Cross-Section

Equilibrium force of concrete

F_{c} = 0.40 ⋅ 0.45 ⋅ 16.7 = 3 MN

Equilibrium force of reinforcement

F_{s}= 3.376 - 3 = 0.38 MN

We deduce the corresponding reinforcement area:

Reinforcement area

A_{s} = 0.38 / 400 ⋅ 10^{4} = 9.5 cm²

Having set the reinforcing steels for a diameter of 20 mm in RF-CONCRETE Members, the provided reinforcement determined automatically by the add-on module is 4 members, with distribution in the corners as requested, i.e. 1 HA 20 per corner, which results in the following reinforcement area:

A_{s} = 4 ⋅ 3.142 = 12.57 cm²

Mechanical reinforcement ratio

ω = (A_{s} ⋅ f_{yd}) / (A_{c} ⋅ f_{cd}) = 12.57 ⋅ 435 / (40 ⋅ 45 ⋅ 16.7) = 0.182

Final check of limiting slenderness as h > b

n = 3.38 / (0.40 ⋅ 0.45 ⋅ 16.7) = 1.125

B = √(1 + 2 ⋅ ω) = 1.17

λ_{lim} = 20 ⋅ 0.7 ⋅ 1.17 ⋅ 0.7 / √1.125 = 10.81 m

λ_{z} < λ_{lim} → The slenderness criterion is fulfilled.

#### Application in Other Add-On Modules

The RF-CONCRETE Columns add-on module also allows for determining the reinforcement of a structural element subjected to axial compression. A technical article detailing the differences between RF-CONCRETE Members and RF-CONCRETE Columns can be found here.

#### Author

#### M.Eng. Milan Gérard

Sales & Technical Support

Milan Gérard works at the Paris site. He is responsible for sales and provides technical support to our French-speaking customers.

#### Keywords

Eurocodes Compression Reinforcement Slenderness

#### Reference

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RF-CONCRETE Members also includes the design of a shear joint. In order to perform this design, you should select the "Shear joint available" check box in window 1.6, Shear Joint tab.

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- Why do I get a discontinuous area in the distribution of internal forces? In the area of the supported line, the shear force VEd shows a jump, which does not seems to be plausible.
- Is it possible to perform design without an additional reinforcement in RF‑CONCRETE Surfaces?
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- For the design, I can only select the surrounding reinforcement for a rectangular cross-section. Why?
- What does the "Crack formation in the first 28 days" option means?
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