1. Home
  2. Diamonds
  3. Validation
  4. Design
  5. DC EC 05: longitudonal reinforcement in column under compression and bending

DC EC 05: longitudonal reinforcement in column under compression and bending


Geometry Cross-section: b=250mm
Material: C25/30
Concrete cover: d=40mm
Reinforcement distribution: Identical
Only upper and lower
Load Design forces: N_{Ed.ULS}=300kN
Standard EN 1992-1-1 [- -]
How to model a column in Diamonds that can be compared with handcalculations
  • Diamonds always looks for the most economical reinforcement ratio. This is not always neat according to the local axes of the rod, but often according to the main axes of inertia. While in hand calculations the load is placed along a local axis and the calculation is also performed in that direction. To impose that behavior in Diamonds, you must choose only top and bottom reinforcement.
    By choosing an ‘identical’ reinforcement distribution, you ask him to choose the top and bottom reinforcement equally. This is a reinforcement distribution for which graphs/tables are suitable for manual calculations.
  • Eurocode 2 offers simplified methods (EN 1992-1-1 §5.8.7 and §5.8.8) to include 2nd order effects and imperfections into the results. The method based on nominal curvature (EN 1992-1-1 §5.8.8) is implemented in Diamonds. This method results in an increased bending moment.
    In Diamonds the method for nominal curvature is selected by default. Because we neglected these effects in this example, the method must be deselected.
  • According to Eurocode 2, reinforcement is required for the forces in ULS, possibly increased with additional reinforcement to limit the stresses in SLS. It is not known in advance whether additional reinforcement will be required for SLS. Since we are only doing a ULS calculation in this example, we turn off the stress limits in SLS in Diamonds (by making a copy of C25/30 and editing the copy).



Method 1 : using diagrams


    \[v_d=\frac{N_{Ed.ULS}}{b\cdot d\cdot f_{cd}}=\frac{300kN}{250mm\cdot 360mm\cdot 16,6MPa}=0,2\]

    \[\mu_d=\frac{M_{Ed.ULS}}{b\cdot d^2 \cdot f_{cd}}=\frac{120kNm}{250mm\cdot (360mm)^2\cdot 16,6MPa}=0,222 \]

Graph 10.2a from Gewapend beton: numeri (click here to open the graph) shows \omega _1=\omega _2 as a function of \mu_d, v_d and \delta_1. Using this graph and the results above, we can deduct:

    \[\omega _1=\omega _2=0.15\]


    \[A_{s1}=A_{s2}==\omega_1 \cdot b \cdot d\cdot \frac{f_{cd}}{f_{yd}}= 0,15 \cdot 250mm \cdot 360mm \cdot \frac{16.6MPa}{434,8MPa} = 518mm^2  \]

Longitudonal reinforcement calculated by Diamonds (EN 1992-1-1 [- -])

Results Independent reference Diamonds Difference
Longitudonal reinforcement 518mm² 514mm² ≈ 0%


  • EN 1992-1-1: 2005 + AC: 2010
  • Van Hooymissen, L., Spegelaere, M., Van Gysel, A., & De Vylder, W. (2002). Gewapend beton. Academia Press, example 1 on page 10.54.
    Keep in mind that the calculations in this book are made done using the NBN B15-002, which is comparable to ENV 1992-1. We used EN 1992-1 [- -] in our verification example.
    But it still remains a good reference if you want to understand how reinforcement calculations work.
  • Gruyaert, E., & Minne, P. (2019). Gewapend beton: numeri.
    This reference is a summary of Gewapend beton (2002) but with updated formula and principles according to EN 1992-1-1. This document contains multiple graphs and tables helping the design of reinforced concrete.
  • Tested in Diamonds 2023r01.

Article Attachments

Was this article helpful?

Related Articles

Need Support?
Can't find the answer you're looking for? Don't worry we're here to help!