Buckling Load Calculator (Euler)

Critical Load · Buckling Stress · Slenderness Ratio · Buckling Check

Buckling Load Calculator


Knicklast
Case 1
pin–pin
Knicklast
Case 2
pin–fixed
Knicklast
Case 3
fixed–fixed
Knicklast
Case 4
Cantilever
Or enter a manual β value:
β = effective length factor
Always I_min (weak axis)!
IPE 240: I_z = 284 cm⁴ | IPE 300: I_z = 604 cm⁴

Euler Cases & Formulas

Euler Buckling Load (general):
F_crit = π² · E · I / (β · L)²  [N]
β = effective length factor | L = system length | L_eff = β·L
Effective length factors β (Euler Cases):
CaseEnd ConditionsβTypical Use
Euler 1pin – pin1.0Truss members
Euler 2pin – fixed0.7Portal columns
Euler 3fixed – fixed0.5Frame columns
Euler 4fixed – free (cantilever)2.0Masts, chimneys
Buckling Stress:
σ_crit = π² · E / λ²  [N/mm²]
λ = L_eff / i = β · L / i  (slenderness ratio)
i = √(I / A)  [mm] radius of gyration
Slenderness limits (guidelines):
λ ≤ 100 Steel compression members (EC3, typical)
λ ≤ 150 Timber columns (EC5)
λ ≤ 200 Angle sections, secondary members
Euler validity: σ_crit ≤ proportionality limit (approx. 2/3 f_y)
Buckling Check (simplified):
F_Ed ≤ F_crit / γ
γ = 1.5 (EC3 safety format) or γ = 3.0 (classical Euler)
Utilisation ratio η = F_Ed / (F_crit / γ) ≤ 1.0

Symbol Reference
F_critCritical Euler buckling load [N]
EYoung's modulus [N/mm²]
I_minMin. second moment of area [cm⁴]
βEffective length factor [–]
LSystem length (column height) [m]
L_eff = β·LEffective (buckling) length [m]
λSlenderness ratio [–]
i = √(I/A)Radius of gyration [cm]
σ_critEuler buckling stress [N/mm²]


Column Buckling – Stability Theory Fundamentals

What is Column Buckling?

Buckling is a stability failure of slender compression members: once the critical load is exceeded, the column deflects sideways — without the material's compressive strength being reached. Buckling is the most common failure mode for columns, struts, frame rafters and compression chords in trusses. The critical load depends on the bending stiffness EI, the effective length β·L and the end conditions — not on material compressive strength!

Euler Cases Compared (L = 4 m, IPE 240, Steel)
CaseβL_eff [m]F_crit [kN]
Euler 11.04.0371
Euler 20.72.8757
Euler 30.52.01,484
Euler 42.08.093
I_z(IPE 240) = 284 cm⁴, E = 210,000 N/mm²
Radius of Gyration i_z – Typical Sections
Sectioni_z [cm]A [cm²]
IPE 2401.9739.1
IPE 2702.2145.9
HEB 2005.0778.1
HEB 3007.58149.1
CHS 100×53.4215.1

Detailed Formula Derivation

1. Euler's Equation – Derived from the Differential Equation of the Deflected Column
The differential equation of the loaded compression member:
E·I·y'' + F·y = 0
Solution (trial function y = A·sin(kx), k² = F/EI) gives k·L = n·π.
For the lowest critical load (n = 1):
F_crit = π² · E · I / L_eff²  with  L_eff = β · L
Valid only in the elastic range (Euler hyperbola stays below the proportionality limit).
2. Slenderness Ratio and Buckling Stress
λ = L_eff / i_min = β · L / √(I_min / A)
σ_crit = F_crit / A = π² · E / λ²  [N/mm²]
Euler validity: σ_crit ≤ σ_P (proportionality limit, approx. 0.67 · f_y)
Steel S235: σ_P ≈ 157 N/mm² → λ_min ≈ π·√(E/σ_P) ≈ 115
For λ < 115, Tetmajer/Johnson parabola or EC3 buckling curves (a–d) apply.
3. Effect of Euler Case on Buckling Load
Effective length L_eff = β · L – β depends on boundary conditions:
  • β = 1.0 (Euler 1): Both ends pinned – corresponds to half a sine wave
  • β = 0.7 (Euler 2): One end pinned, one fixed – asymmetric, L_eff = 0.7·L
  • β = 0.5 (Euler 3): Both ends fixed – L_eff = 0.5·L, F_crit 4× higher than Case 1
  • β = 2.0 (Euler 4): Cantilever (fixed base, free top) – L_eff = 2·L, F_crit 4× lower than Case 1

Practical Example: Portal Frame Column HEB 200, Euler 2

Problem:

Steel column HEB 200, length L = 6.0 m, fixed at base, pinned at top (Euler 2, β = 0.7). Design compressive force F_Ed = 1,200 kN. Check against buckling (γ = 1.5).

Section data HEB 200:

I_z = 2,000 cm⁴, A = 78.1 cm², i_z = 5.07 cm, E = 210,000 N/mm²

Solution:
  • L_eff = β · L = 0.7 × 6.0 = 4.2 m
  • F_crit = π² × 210,000 × 2,000×10⁴ / (4,200)² = 23,550 kN
  • F_Rd = F_crit / γ = 23,550 / 1.5 = 15,700 kN
  • F_Ed = 1,200 kN ≤ 15,700 kN → ✓ Check satisfied
  • λ = 4,200 / (5.07×10) = 82.8 → Euler range valid

Frequently Asked Questions

The Euler formula is valid only in the elastic range, i.e. as long as σ_crit ≤ the proportionality limit (Steel S235: approx. 157 N/mm², limiting slenderness λ_G ≈ 115). For more slender members (λ > 115 for S235) the Euler formula is conservative. For stockier members (λ < 115), subdivision formulas per Tetmajer, Johnson parabola, or the EC3 buckling curves a–d (which account for geometric imperfections and residual stresses) must be used.

A compression member always buckles in the direction of least bending resistance — i.e. about the axis with the smallest second moment of area I_min. For an IPE section this is I_z (weak axis), because the flanges are narrow. This is why lateral restraints (purlins, bracing) are so effective: they reduce the effective length in the weak axis, which greatly increases the buckling load.

In the classical Euler approach γ = 3.0 is recommended (old German DIN guideline), since imperfections, initial bow, and residual stresses must also be covered. Under Eurocode EC3, the check is performed with partial factor γ_M1 = 1.0 (steel) and a reduction factor χ (buckling curve a–d): N_Ed ≤ χ · A · f_y / γ_M1. For preliminary sizing, γ = 1.5 is a common indicative value.

Most effective measures in descending order of impact:
  1. Reduce the effective length – add lateral restraints; changing β from 1.0 to 0.5 multiplies F_crit by 4.
  2. Increase cross-section height/width – I grows with h³, width only linearly.
  3. Use hollow sections instead of IPE – RHS/CHS have balanced I in all axes.
  4. Improve base fixity – a rigid base reduces β from 2.0 to 0.7 (factor 8 on F_crit!).

Column buckling describes the lateral deflection of a member under axial compression. Plate buckling (local buckling) describes the stability failure of thin-walled plates or shells under compression, shear or combined loading (e.g. web buckling, shell buckling of silos/tanks). Both are stability problems (equilibrium in the deformed position), but with different formulas and design procedures.

Summary

Critical Buckling Load

F_crit = π²·E·I / (β·L)²
Unit: N → kN

Slenderness Ratio

λ = β·L / i
i = √(I/A)

Buckling Check

F_Ed ≤ F_crit / γ
η = F_Ed·γ / F_crit ≤ 1

Typical Applications
  • Portal frame columns: Steel IPE/HEB, Euler case 2 or 3
  • Truss members: Compression chords and verticals, Euler case 1 (pinned)
  • Timber columns: Slenderness λ ≤ 150 (EC5)
  • Masts and chimneys: Cantilever columns, Euler case 4 (β = 2.0)
  • Mechanical engineering: Screw bolts, piston rods, struts

Is this page helpful?            
Thank you for your feedback!

Sorry about that

How can we improve it?