Learn: Euler critical buckling load
A slender compressed column may become laterally unstable before its material yields. Euler’s load predicts the ideal elastic bifurcation load for a column represented by an effective buckling length.
What the formula is saying
The restoring bending rigidity is EI. Instability resistance varies inversely with effective length squared; end restraint is included through that effective length, not by changing π².
Read the symbols in plain language
- E
- Young’s modulus
Elastic stiffness: the stress change needed for a unit strain in the stated material model. It is not a strength limit.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- I
- Second moment of area
The area-weighted square of distance from the stated axis. It measures the spread of the cross-section, not its area or mass.
m⁴The fourth power of metres is used for a second moment of area; 1 m⁴ = 10¹² mm⁴.
- Lcr
- Effective buckling length
Length of the equivalent pin-ended buckling half-wave, including the effect of end restraint; it is not automatically the physical member length.
mMetres measure length; 1 m = 1000 mm.
- Ncr
- Result to find
Euler critical buckling load. The Euler coefficient combines bending rigidity and inverse squared effective length.
N
Sort out the units first
E is Pa, I is m⁴ and L is the effective length in m. The answer is N. If the physical column length is Lphys and the length factor is K, enter K × Lphys as L.
Assumptions before calculating
Assume an initially straight, prismatic, elastic column under concentric axial compression, with idealized end restraints. Use the weak-axis I for the buckling plane being checked.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Ncr and explain the result in the stated output unit.
- E · Young’s modulus
- 200000000000 Pa
- I · Second moment of area
- 0.000008 m⁴
- Lcr · Effective buckling length
- 3 m
Find bending rigidity
Elastic modulus and section inertia jointly resist lateral curvature.
(200000000000) × (0.000008) = 1600000 N·m²Square the effective length
Use the effective buckling length for the actual idealized end-restraint condition.
(3)^2 = 9 m²Calculate ideal elastic buckling load
The Euler coefficient combines bending rigidity and inverse squared effective length.
π^2 × (1600000) ÷ (9) ≈ 1754596.338 N
Does this worked answer make sense?
Doubling effective length reduces Ncr to one quarter. Doubling I doubles Ncr, which explains why section geometry matters for slender columns.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- E · Young’s modulus
- 210000000000 Pa
- I · Second moment of area
- 0.000012 m⁴
- Lcr · Effective buckling length
- 4 m
Find bending rigidity
Elastic modulus and section inertia jointly resist lateral curvature.
(210000000000) × (0.000012) = 2520000 N·m²Square the effective length
Use the effective buckling length for the actual idealized end-restraint condition.
(4)^2 = 16 m²Calculate ideal elastic buckling load
The Euler coefficient combines bending rigidity and inverse squared effective length.
π^2 × (2520000) ÷ (16) ≈ 1554462.693 N
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- E · Young’s modulus
- 200000000000 Pa
- I · Second moment of area
- 0.00001 m⁴
- Lcr · Effective buckling length
- 5 m
Find: Learn: Euler critical buckling load
A hint, not the answer
The restoring bending rigidity is EI. Instability resistance varies inversely with effective length squared; end restraint is included through that effective length, not by changing π².
E is Pa, I is m⁴ and L is the effective length in m. The answer is N. If the physical column length is Lphys and the length factor is K, enter K × Lphys as L.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find bending rigidity
Elastic modulus and section inertia jointly resist lateral curvature.
(200000000000) × (0.00001) = 2000000 N·m²Square the effective length
Use the effective buckling length for the actual idealized end-restraint condition.
(5)^2 = 25 m²Calculate ideal elastic buckling load
The Euler coefficient combines bending rigidity and inverse squared effective length.
π^2 × (2000000) ÷ (25) ≈ 789568.3521 N
Avoid the common trap
Do not enter physical length unless it equals the effective length. Do not use the strong-axis inertia automatically or interpret the ideal load as permission to apply that load in practice.
When this method applies — and when it does not
Euler load is not a design resistance. Initial crookedness, residual stress, local buckling, yielding, connection flexibility and safety factors are omitted; stocky columns may not remain elastic up to this load.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Euler critical buckling load. See the Stresses in Beams and Beam Displacements modules; match the load and support conditions, not just the equation’s appearance.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.