Short Column Design per Eurocode 2 (EN 1992-1-1) — Worked Example

Eurocode 2 (EN 1992-1-1:2004) is the governing structural concrete code for the European Union and is increasingly adopted in the Gulf Cooperation Council states. This article covers a braced, short column design — the most common case in low-to-mid-rise RC frames.

Problem Statement

Parameter	Value
Column cross-section	300 mm × 300 mm
Concrete class	C25/30 (fck = 25 MPa, fctm = 2.6 MPa)
Reinforcement	B500B (fyk = 500 MPa)
Column height	3.0 m (clear height between floors)
Axial load (design)	NEd = 800 kN
Bending moments	MEd,top = 30 kN·m; MEd,bot = 20 kN·m
Frame type	Braced (non-sway)

Step 1 — Material Design Strengths

EC2 uses design values (characteristic divided by partial factor):

fcd = αcc × fck / γC = 1.0 × 25 / 1.5 = 16.67 MPa
fyd = fyk / γS       = 500 / 1.15       = 434.8 MPa

Where γC = 1.5 (concrete) and γS = 1.15 (steel) per EC2 Table 2.1N.
αcc = 1.0 is the coefficient for sustained loading effects on compressive strength (national annex — using 1.0 here).

Step 2 — Slenderness Classification (§5.8.3)

Effective length for a braced column with both ends nominally fixed (partial fixity from beams):

l₀ = 0.7 × l = 0.7 × 3.0 = 2.1 m

Radius of gyration for a 300 × 300 square section:

i = b/√12 = 300/√12 = 300/3.464 = 86.6 mm

Slenderness ratio:

λ = l₀ / i = 2,100 / 86.6 = 24.2

Limiting slenderness (EC2 §5.8.3.1):

λ_lim = 20 × A × B × C / √n

n = NEd / (Ac × fcd) = 800,000 / (300² × 16.67) = 800,000 / 1,500,300 = 0.533

A = 0.7  (simplified, when φef unknown)
B = 1.1  (simplified, when ω unknown)
C = 1.7 - rm = 1.7 - (MEd,bot/MEd,top) = 1.7 - (20/30) = 1.7 - 0.667 = 1.033

λ_lim = 20 × 0.7 × 1.1 × 1.033 / √0.533
       = 20 × 0.7 × 1.1 × 1.033 / 0.730
       = 15.91 / 0.730
       = 21.8

Check: λ = 24.2 > λ_lim = 21.8 — second-order effects must be considered.

However, since λ = 24.2 is only slightly above the limit, we can use the nominal stiffness method (§5.8.7) or nominal curvature method (§5.8.8). For a short column close to the limit, the moment amplification is small. We proceed with the nominal curvature method for this example.

Step 3 — Second-Order Moment (Nominal Curvature Method, §5.8.8)

First-order equivalent moment (§5.8.8.2):

M₀Ed = max(0.6 × MEd,max + 0.4 × MEd,min; 0.4 × MEd,max)
     = max(0.6 × 30 + 0.4 × 20; 0.4 × 30)
     = max(18 + 8; 12)
     = max(26; 12)
     = 26 kN·m

Curvature (§5.8.8.3):

1/r = Kr × Kφ × (fyd / (Es × 0.45 × d))

Assuming: d ≈ 0.9 × h = 0.9 × 300 = 270 mm

fyd/Es = 434.8 / 200,000 = 0.002174 (yield strain)
1/r₀ = fyd / (Es × 0.45 × d)
      = 434.8 / (200,000 × 0.45 × 270)
      = 434.8 / 24,300,000
      = 1.789 × 10⁻⁵ /mm

For Kr (accounting for axial load, §5.8.8.3(3)):

nu = 1 + ω  (≈ 1.35 for initial estimate, refine iteratively)
n = 0.533
nbal = 0.4

Kr = (nu - n)/(nu - nbal) = (1.35 - 0.533)/(1.35 - 0.40) = 0.817/0.95 = 0.860

Taking Kφ = 1.0 (no creep simplification):

1/r = 0.860 × 1.0 × 1.789×10⁻⁵ = 1.539×10⁻⁵ /mm

Nominal second-order moment:

e₂ = (1/r) × l₀² / c  (c = 10 for parabolic distribution)
   = 1.539×10⁻⁵ × 2,100² / 10
   = 1.539×10⁻⁵ × 4,410,000 / 10
   = 67.9 mm

M₂ = NEd × e₂ = 800 × 0.0679 = 54.3 kN·m

Total design moment:

MEd,tot = M₀Ed + M₂ = 26.0 + 54.3 = 80.3 kN·m

Also check minimum eccentricity (§6.1(4)):

e₀ = max(h/30; 20 mm) = max(300/30; 20) = max(10; 20) = 20 mm
M₀_min = NEd × e₀ = 800 × 0.020 = 16 kN·m  < 80.3 kN·m ✓

Step 4 — Interaction Check

Non-dimensional design parameters:

ν = NEd / (Ac × fcd) = 800,000 / (300² × 16.67) = 0.533
μ = MEd / (Ac × fcd × h) = 80.3×10⁶ / (300² × 16.67 × 300) = 80.3×10⁶ / 450,090,000 = 0.178

From a standard EC2 interaction diagram for a symmetric section (d’/h = 30/300 = 0.10):

At ν = 0.533 and μ = 0.178, the required mechanical reinforcement ratio ω ≈ 0.30.

Required As (total, symmetric):

ω = As,total × fyd / (Ac × fcd)
0.30 = As,total × 434.8 / (300² × 16.67)
As,total = 0.30 × 90,000 × 16.67 / 434.8
         = 0.30 × 1,500,300 / 434.8
         = 450,090 / 434.8
         = 1,035 mm²

Minimum reinforcement (EC2 §9.5.2):

As,min = max(0.1 × NEd/fyd; 0.002 × Ac)
       = max(0.1 × 800,000/434.8; 0.002 × 90,000)
       = max(184; 180)
       = 184 mm²  <<  1,035 mm²  ✓ (flexure + axial governs)

Maximum reinforcement (§9.5.2(3)):

As,max = 0.04 × Ac = 0.04 × 90,000 = 3,600 mm²  > 1,035 ✓

Step 5 — Bar Selection

Provide 4 × 20 mm dia bars (one at each corner, symmetric):

As,provided = 4 × π/4 × 20² = 4 × 314.2 = 1,257 mm² > 1,035 mm² ✓

Links (EC2 §9.5.3):

Minimum link diameter = max(6 mm; φ_long/4) = max(6; 20/4) = max(6; 5) = 6 mm → use 8 mm dia
Maximum link spacing = min(20 × φ_long; b; 400 mm) = min(400; 300; 400) = 300 mm → use 250 mm

Provide 8 mm dia links at 250 mm c/c.

Summary

Item	Value
Section	300 × 300 mm
NEd	800 kN
MEd (first-order)	26.0 kN·m
M₂ (second-order)	54.3 kN·m
MEd,total	80.3 kN·m
ν	0.533
μ	0.178
Required As	1,035 mm²
Provided As	4–20ϕ = 1,257 mm²
Links	8ϕ @ 250 mm
Slenderness λ	24.2 (second-order effects included)

Important Note on National Annexes

EC2 is a framework code — each EU member state publishes a National Annex that may modify partial factors, αcc values, or limiting slenderness formulas. The values used here follow the recommended Eurocode values; always verify against the applicable national annex for your project jurisdiction.

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