Learn: Euler critical stress
Euler buckling can also be expressed as an average axial stress instead of a total load. This form uses the geometric slenderness L/i to express how vulnerable a column is to elastic buckling.
What the formula is saying
Dividing Euler load by area and using i² = I/A gives σcr = π²E/λ². The area is absorbed into the radius of gyration, so it must not be divided out a second time.
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.
- λ
- Slenderness ratio
Slenderness ratio. The inverse-square length effect is retained after dividing Euler load by area.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- σcr
- Result to find
Euler critical stress. Combine the elastic modulus and slenderness to obtain an average axial stress.
Pa
Sort out the units first
E and the result are in Pa. λ is the geometric slenderness ratio using the same length units in L and i; it is not the Eurocode nondimensional slenderness λ-bar.
Assumptions before calculating
Apply the same ideal straight, prismatic, concentric and linearly elastic column assumptions as Euler load. The effective length and radius of gyration refer to the same buckling plane.
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 σcr and explain the result in the stated output unit.
- E · Young’s modulus
- 200000000000 Pa
- λ · Slenderness ratio
- 120
Square geometric slenderness
The inverse-square length effect is retained after dividing Euler load by area.
(120)^2 = 14400Find ideal critical stress
Combine the elastic modulus and slenderness to obtain an average axial stress.
π^2 × (200000000000) ÷ (14400) ≈ 137077838.9 Pa
Does this worked answer make sense?
At fixed E, doubling slenderness quarters the critical stress. The result should be compared with material yield only as an applicability check, not as a complete design.
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
- λ · Slenderness ratio
- 150
Square geometric slenderness
The inverse-square length effect is retained after dividing Euler load by area.
(150)^2 = 22500Find ideal critical stress
Combine the elastic modulus and slenderness to obtain an average axial stress.
π^2 × (210000000000) ÷ (22500) ≈ 92116307.74 Pa
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
- λ · Slenderness ratio
- 100
Find: Learn: Euler critical stress
A hint, not the answer
Dividing Euler load by area and using i² = I/A gives σcr = π²E/λ². The area is absorbed into the radius of gyration, so it must not be divided out a second time.
E and the result are in Pa. λ is the geometric slenderness ratio using the same length units in L and i; it is not the Eurocode nondimensional slenderness λ-bar.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Square geometric slenderness
The inverse-square length effect is retained after dividing Euler load by area.
(100)^2 = 10000Find ideal critical stress
Combine the elastic modulus and slenderness to obtain an average axial stress.
π^2 × (200000000000) ÷ (10000) ≈ 197392088 Pa
Avoid the common trap
Do not use λ-bar in place of L/i. Keep λ squared in the denominator and keep E in stress units, not force units.
When this method applies — and when it does not
If the predicted stress exceeds yield strength, the elastic model is outside its useful range. This formula omits imperfections and partial factors and is not a code resistance check.
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 stress. 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.