Calculation of Tie Reinforcement with RF-CONCRETE Members
This article deals with determining the concrete reinforcement for a beam stressed by tension only according to EN 1992-1-1. The aim is to show the tensile load of a member-type element (without imposed deformations) and to define the concrete reinforcement in accordance with the standard's construction rules and provisions using the RFEM structural analysis software.
What does tension for a concrete element mean?
A section of a structural element is stressed by simple tension when the forces acting on one side of the section are reduced at the section's center of gravity to a single force N. This normal force N is then perpendicular to the section and directed towards the side where the forces act. The self-weight is neglected in the concrete and the section is uniformly subjected to tension.
Tensile Stress in Steel
For steel with a σ - ε diagram showing an inclined graph, the equation to the right of the graph corresponding to the tensile behavior of steel is written according to the steel's characteristic values set forth in §3.2. 7 (2) of EN 1992-1-1.
|σs||Stress in reinforcement|
|fyd||Design value of yield strength = fyk / γs|
|k||Characteristic limits ratio = ftk / fyk|
|Es||Modulus of elasticity|
|εs||Deformation in reinforcement = εud = 0.9 ⋅ εuk|
|fyk||Characteristic yield strength|
|γs||Partial safety factor of steel|
|ftk||Characteristic value of tensile strength|
|εud||Design value of limit deformation|
Please note: the concrete under tension is neglected for pure tension. In this case, only steels fully balance the tensile force NEd . Thus, the necessary reinforcement area is determined according to the tensile force and the provided stress.
As = NEd / σs
As ... Reinforcement area
NEd ... Ultimate normal force
Application of the Theory Using RF-CONCRETE Members
We look at an example of an element subjected to simple tension by analyzing the results obtained for the longitudinal reinforcements. You will find the input data below:
- Permanent loads: Ng = 100 kN
- Variable loads: Nq = 40 kN
- Square section: 20/20 cm
- Concrete strength class: C25/30
- Steel: S 500 A for inclined graph
- Diameter of longitudinal reinforcement: ϕl = 12 mm
- Diameter of transverse reinforcement: ϕt = 6 mm
- Concrete cover: 3 cm
- Control of cracking not required.
In order to verify the material settings in RF-CONCRETE Members, Image 02 describes the materials used for concrete and reinforcement.
Ultimate Limit State
Design loading in the ultimate limit state:
NEd = 1.35 ⋅ 100 + 1.5 ⋅ 40 = 195.00 kN
Provided Tensile Stress
Ultimate limit state for a durable, transient design situation:
fyd = 500 / 1.15 = 435 MPa
k = 525 / 500 = 1.05 according to Table C.1 of EN 1992-1-1
εuk = 25‰
εud = 0.9 ⋅ 25 = 22.5‰
σs = 435 + (1.05 ⋅ 435 - 435) / (2.5 - 435 / (200,000)) ⋅ [2.25 - 435 / (200,000)] = 454 MPa
Required Longitudinal Reinforcement
Longitudinal reinforcements for the ultimate limit state:
As = 0.195 / 454 ⋅ 104 = 4.30 cm²
Provided Longitudinal Reinforcement
Having configured the reinforcing steel with a diameter of 12 mm in RF-CONCRETE Members, the provided reinforcement determined automatically by the add-on module is 4 bars, with symmetrical distribution on the lower and upper parts of the section, i.e. 2 x 2 HA12, which results in the following reinforcement area:
As = 4 ⋅ 1.13 = 4.52 cm²
With the transverse reinforcement also having been defined by the user, RF-CONCRETE Members can automatically determine the spacings according to the standard, and check whether their arrangement conforms.
In our case, by imposing stirrups with a diameter of 6 mm, the program gives us a spacing of 0.122 m, but also displays warning message No. 155) in the Notes column that can be seen in Image 07.
The formula referring to §9.2.2 (8) of EN 1992-1-1 is defined below.
Sl,max = 0.75 ⋅ d
Sl,max ... Maximum transverse spacing of stirrups
d ... Effective height
d = h - e - ∅t - ∅l/2
h ... Cross-section height
e ... Concrete Cover
The previous formulas give us the following results:
d = 0.200 - 0.03 - 0.006 - 0.012 / 2 = 0.158 m
Sl,max = 0.75 ⋅ 0.158 = 0.12 m
Therefore, warning message 155 appears because the spacing between the stirrup legs of a beam in the transverse direction exceeds the limit value given by the standard. The problem can be solved by increasing the number of stirrup legs in the settings for the stirrup reinforcement, as detailed in this FAQ.
Having set the parameters beforehand, RF-CONCRETE Members gives the number of rebars required according to the defined arrangement, in order to verify the tensile load according to the internal forces coming from RFEM. Depending on the warning messages displayed, it is also possible for the user to modify the reinforcement and its arrangement after the calculation.
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.
Do you have further questions or need advice? Contact us via phone, email, or chat or find suggested solutions and useful tips on our FAQ page available 24/7.
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.
- Why does the note 155) appear as a result of the reinforced concrete design of members for the existing stirrup reinforcement?
- What is the maximum number of reinforcement groups that can be created in a design case in RF‑CONCRETE Surfaces?
- In RF‑CONCRETE Surfaces, I obtain a high amount of reinforcement in relation to a lever arm that is almost zero. How is such a small lever arm of internal forces created?
- When converting from the manual definition of reinforcement areas to the automatic arrangement of reinforcement according to Window 1.4, the result of the deformation calculation differs, although the basic reinforcement has not been modified. What is the reason for this change?
- 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?
- I obtain different results when comparing the deformation analysis in the RF‑CONCRETE add-on modules and another calculation program. What could be the reason for this?
- 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?
- I have recently purchased RSTAB with the CONCRETE add-on module. Can I use it to perform the stability analysis of reinforced concrete columns?
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