Eurocode 2 Interaction Diagrams: ULS Section Envelopes

What are ULS section envelopes and N–M interaction diagrams?

An N–M interaction diagram is the boundary of the axial-force and bending-moment combinations a section can resist at the ultimate limit state. Calculate it by choosing limiting strain profiles, converting strains into concrete and steel stresses, and integrating those stresses into forces and moments. Repeating the calculation produces the curve.

This guide develops the calculation for a rectangular reinforced-concrete section, then extends the same formulation to biaxial N–My–Mz envelopes. It uses EN 1992-1-1:2004, the parabola–rectangle concrete diagram and the horizontal-top steel diagram. Check the adopted edition, amendments and National Annex for your project.

1. Set up the section in kN and m

Use metres for dimensions and coordinates, square metres for areas, kN for forces and kN·m for moments. Store stresses and elastic moduli in kN/m². A stress of 1 MPa is 1,000 kN/m².

Input	Worked example
Section width b	0.300 m
Section depth h	0.500 m
Concrete	C30/37; fck = 30,000 kN/m²
Steel	fyk = 500,000 kN/m²; Es = 200,000,000 kN/m²
Example design factors	αcc = 1.00; γc = 1.50; γs = 1.15
Reinforcement	Two 20 mm bars 0.050 m from each face
Area of each two-bar layer	0.0006283185 m²

Take r downwards from the compressed top face. The moment reference is the section centre, with z = h/2 − r. Use tension-positive strain, stress and axial force. With this convention, top-face compression produces a negative My. Keep this convention throughout the calculation.

f_{cd} = \frac{α_{cc} f_{ck}}{γ_{c}} = 20,000 kN/m²; f_{yd} = \frac{f_{yk}}{γ_{s}} \approx 434,782.6 kN/m²

A_{s} = 2 \frac{π ϕ^{2}}{4} = 0.0006283185 m^{2}

The example uses αcc = 1.00 to make the arithmetic explicit. Use your actual project value; for αcc = 0.85 the concrete design strength becomes 17,000 kN/m².

2. Use the EC2 material laws and limiting strains

For concrete up to C50/60, EC2 Table 3.1 gives the following values for the parabola–rectangle diagram. Strain is dimensionless: 2‰ means 0.002, not 0.02.

Parameter	C12/15 to C50/60	Meaning
εc2	0.002 = 2.0‰	Strain at the start of the constant-stress plateau
εcu2	0.0035 = 3.5‰	Ultimate compressive strain for the bending branch
n	2	Exponent of the parabolic branch
Uniform compression limit	0.002 = 2.0‰	Limit for this concrete diagram in pure compression

See EN 1992-1-1 Table 3.1 and §3.1.7; Sétra’s Eurocode 2 guide, pp. 65–69 explains the material diagrams and strain pivots.

Let e be the positive magnitude of concrete compression strain. For compression, the signed stress is negative. Ignore concrete tension in this ULS section calculation.

σ_{c} = {\begin{matrix} 0 & for ε ≥ 0 \\ - f_{cd} [1 - {(1 - \frac{e}{ε_{c2}})}^{2}] & for 0 < e ≤ 0.002 \\ - f_{cd} & for 0.002 < e ≤ 0.0035 \end{matrix}

An over-limit strain is an inadmissible state, not a value to clip back into the concrete stress law. The pure-compression and intermediate full-compression limits also need the profile checks in step 3.

For higher-strength concrete, do not reuse the preceding values. EC2 Table 3.1 changes the parabola–rectangle parameters. In the following expressions only, insert the numerical value of fck in MPa, between 50 and 90:

ε_{c2} = (2 + 0.085 {(f_{ck} - 50)}^{0.53}) \times 10^{−3}

ε_{cu2} = (2.6 + 35 {(\frac{90 - f_{ck}}{100})}^{4}) \times 10^{−3}

n = 1.4 + 23.4 {(\frac{90 - f_{ck}}{100})}^{4}

For reinforcing steel, use the EC2 horizontal-top design diagram in this example: linear elasticity followed by a plateau in tension or compression.

σ_{s} = \max (- f_{yd}, \min (E_{s} ε_{s}, f_{yd}))

ε_{yd} = \frac{f_{yd}}{E_{s}} \approx 0.002174 = 2.174 ‰

EN 1992-1-1 §3.2.7 permits this horizontal-top model without an imposed steel strain limit. That modelling permission does not mean the physical steel can stretch indefinitely. If you choose the inclined-top diagram instead, apply the design strain limit εud; its recommended value is 0.9εuk, subject to the National Annex. The minimum characteristic strains εuk for ductility classes A, B and C are 2.5%, 5.0% and 7.5%, respectively, corresponding to recommended design limits of 2.25%, 4.50% and 6.75% when those minima are used. Do not add an arbitrary 10‰ limit to this horizontal-top example.

3. Generate admissible strain profiles

Let x be the neutral-axis depth measured from the compressed face. Plane sections remain plane, so each concrete strip and bar takes its strain from the same linear profile.

Partly compressed section, 0 < x ≤ h. Fix the top-face strain at −0.0035 and calculate the strain at depth r:

ε (r) = ε_{cu2} (\frac{r}{x} - 1)

Fully compressed section, x ≥ h. Rotate the profile about the EC2 compression pivot. For this concrete diagram, the pivot is 3h/7 below the compressed face and has strain −0.002.

a = h (1 - \frac{ε_{c2}}{ε_{cu2}}) = \frac{3 h}{7}

e_{top} = \frac{ε_{c2}}{1 - \frac{a}{x}}; ε (r) = e_{top} (\frac{r}{x} - 1)

At x = h the two branches meet at the same profile. As x tends to infinity, the strain tends to −0.002 throughout the section: pure compression. Add this uniform-strain state explicitly. Also add pure tension, with all bars at +εyd and zero concrete stress. For this symmetric example those two states have zero moment.

4. Integrate the stresses: a worked calculation

Take a neutral-axis depth of 0.200 m. The top reinforcement layer has strain −0.002625 and the bottom layer +0.004375. Both layers reach the yield plateau for this strain profile; their signed stresses are −434,782.6 and +434,782.6 kN/m², respectively.

Divide the concrete depth into 100 equal strips. Each strip is 0.005 m deep, with an area of 0.0015 m². Evaluate the strain and stress at each strip’s midpoint. The following expressions define the depth, area, force and moment contribution of strip j:

Δr = \frac{h}{n}; r_{j} = (j - \frac{1}{2}) Δr; A_{j} = b Δr

F_{j} = σ_{c,j} A_{j}; M_{y,j} = F_{j} (\frac{h}{2} - r_{j})

Calculate each reinforcement layer separately using its actual coordinate and area. Because the strips contain the gross concrete area, subtract the concrete displaced by the reinforcement before adding its steel force:

F_{s,net} = A_{s} (σ_{s} - σ_{c,bar})

Evaluate the replacement concrete stress at the bar coordinate—the usual point-bar approximation. The resulting top- and bottom-layer net forces are −260.616 and +273.182 kN. Multiply each by its lever arm about the section centre, then add all concrete and reinforcement contributions.

Concrete integration	N (kN)	My (kN·m)
100 midpoint strips	−958.972	−268.800
400 midpoint strips	−958.869	−268.800

The values are independently calculated from the stated section and material assumptions. Refining the concrete integration from 100 to 400 strips changes the axial force by approximately 0.103 kN. Check both the force and moment convergence rather than assuming a fixed strip count is sufficient.

5. Trace the complete N–M interaction diagram

Repeat the integration for a sequence of neutral-axis depths. Begin with depths between the compressed face and the far face of the section, then continue beyond the section depth using the full-compression pivot. Increase the sampling density where the curve changes rapidly.

Choose a neutral-axis depth and the corresponding admissible strain profile.
Evaluate the concrete and reinforcement strains at their coordinates.
Apply the selected material laws to obtain signed stresses.
Sum the force and moment contributions about the same reference point.
Store the N–My pair and repeat for the next profile.
Add the uniform-tension and uniform-compression endpoints explicitly.
Calculate the opposite bending branch with the other face compressed.

For the symmetric example, the two moment branches mirror each other. For asymmetric geometry or reinforcement, compute each branch independently. Plot My horizontally and N vertically, keeping the calculated point order rather than sorting points by only one coordinate.

Endpoint	N (kN)	My (kN·m)
Uniform tension	+546.364	0
Uniform compression	−3,477.522	0

At uniform compression, the steel strain is −0.002 and its stress is −400,000 kN/m²: it has not reached the yield plateau. Assuming yielded steel here overestimates the resistance.

The finite-depth sweep and uniform endpoints cover the relevant uniaxial limiting profiles for the chosen horizontal-top steel diagram. An inclined-top steel model also requires steel-controlled profiles and its design strain cap. The two modelling choices must remain consistent.

6. Extend the calculation to an N–My–Mz envelope

For biaxial bending, replace concrete strips with small area cells and rotate the strain gradient through the full range of section directions. One strain-plane representation is:

ε (y, z) = ε_{0} + κ_{y} z - κ_{z} y

Use the same material laws at every cell and bar. Integrate their signed forces using their y and z coordinates:

N = \sum_{j} F_{j}; M_{y} = \sum_{j} z_{j} F_{j}; M_{z} = - \sum_{j} y_{j} F_{j}

For each orientation, measure the projected concrete depth between the extreme faces and use that depth in the strain-pivot construction. Traverse its admissible profiles, calculate the resulting N–My–Mz points, and repeat at progressively finer angles. Retain all bending directions and apply the displaced-concrete correction at the bars.

A My–Mz envelope at fixed N is a slice of the surface. For each orientation, bracket and solve for strain states that produce the target axial force. For an N–My curve at Mz = 0, enforce the zero-Mz condition. A projection of the entire surface is not the same calculation and can overestimate the resistance of the intended slice.

The biaxial formulation adds several numerical tasks: cell integration, angle and strain-profile sampling, equilibrium solving and convergence all need checking. Validate symmetric sections and selected uniaxial cases before relying on a biaxial surface.

7. Check design actions and common calculation errors

Compare the factored design action triplet with the resistance domain. Member imperfections and any required second-order effects must already be included in the design moments. The envelope covers section resistance under axial force and bending; shear, torsion, anchorage, detailing and serviceability require their own checks.

Convert MPa to kN/m² before multiplying by areas in m².
Use 0.002 and 0.0035 for the stated concrete strains, not 2 and 3.5.
Use the −0.002 pure-compression limit for this parabola–rectangle model.
Calculate each bar stress; not every bar yields in every profile.
Correct displaced concrete if using gross concrete cells or strips.
Refine the mesh and the curve sampling separately.
Keep load and resistance sign conventions and reference axes identical.

Calculate interaction envelopes with COSA

Working through the formulation builds an understanding of strain compatibility and section resistance. For repeated project calculations, COSA generates section interaction envelopes directly, with 2D slices and a 3D view, alongside the section-analysis and reporting workflow.

Interaction envelopes are available with a paid COSA licence. Explore the product to choose the licence for your workflow.

Calculate interaction envelopes with COSA →

For an overview of the available tools, see COSA’s parametric sections, batch analysis and multi-section reports.

References

EN 1992-1-1:2004: Table 3.1; §3.1.7 concrete design diagrams; §3.2.7 and Annex C reinforcing steel; §6.1 and Figure 6.1 strain domains and section resistance. Apply adopted amendments and the project National Annex.
Sétra: Methodological guide to Eurocode 2, pp. 65–69 (material laws and strain pivots)
European Commission JRC: EN 1992-1-1 training material
European Commission JRC: Eurocode 2 worked examples
fib StructuralCodes: EC2 concrete material parameters and reinforcement material parameters