Euler Buckling Formula Explained

Why a perfectly straight column suddenly snaps sideways — and why that has nothing to do with strength.

Mechanics of MaterialsMechanical Engineering Year 2Free preview
⏱️ About 18 min

Stand a plastic ruler on end and press down on it from above. At first it carries the load perfectly, staying dead straight. Push a little harder and nothing changes — still straight. Push just a fraction harder and, with no warning, it whips sideways into a bent shape and collapses. The ruler did not break, crush, or yield; its material is just as strong at the instant it buckles as it was a moment before. What failed was its <em>stability</em>. This is column buckling, and it is governed by a completely different law from the stress and strain equations you have used so far. Leonhard Euler, in 1757, found the load at which a slender column loses stability — and his answer contains no strength term at all, only stiffness and geometry.

💡
The big idea: A slender column under axial compression stays straight until the load reaches a <strong>critical value P_cr</strong>, at which a bent configuration becomes energetically favourable and the column buckles sideways. For the ideal pinned-pinned column Euler's formula gives P_cr = π²EI/L², depending on flexural rigidity (EI) and length (L) but not on material strength. Dividing by the area gives the <strong>critical stress</strong> σ_cr = π²E/(L/r)², where r = √(I/A) is the radius of gyration and L/r the slenderness ratio. Because buckling is a stability phenomenon, a slender column can buckle at a stress far below its yield stress.
🎯 By the end, you'll be able to
  • Describe the bifurcation of equilibrium at the critical load
  • Apply Euler's formula P_cr = π²EI/L² for a pinned-pinned column
  • Compute the slenderness ratio L/r and the radius of gyration r = √(I/A)
  • Explain why a slender column can buckle at a stress well below yield
📎 Helpful to know first

The bifurcation idea

Imagine loading an ideally straight, ideally centred column, increasing the axial force P slowly from zero. For every value of P below some threshold the only possible equilibrium shape is the straight one: push the column sideways a little and it springs back. The straight shape is stable. At a single, sharply defined load — the critical load P_cr — that changes. The straight shape becomes indifferent, and a whole family of slightly-bent shapes suddenly becomes possible as well. The column has reached a bifurcation point: the load can no longer increase, and the smallest imperfection tips it sideways.

This is why buckling feels sudden. There is no gradual warning, no creeping deflection in a perfect column — up to P_cr it is straight, and at P_cr it buckles. Real columns have tiny imperfections, so in practice you see a rapid but not instantaneous sideways growth as the load approaches P_cr; we treat that real behaviour in the lesson on eccentric loading.

🔑 Buckling is a stability failure, not a stress failure

This is the single most important idea in the module, so fix it firmly: a column does not buckle because its stress got too high. It buckles because, above P_cr, the straight shape is no longer a stable equilibrium. The material may be nowhere near its yield stress when this happens.

Contrast the two failure modes a compression member can suffer:

  • Material failure (yielding/crushing): the stress reaches the yield strength σ_y and the material gives out. This governs short, stocky columns.
  • Stability failure (buckling): the member becomes unstable and buckles sideways at P_cr, even though the stress is below yield. This governs long, slender columns.

A slender column can buckle at a stress a fraction of its yield stress. Strength is simply the wrong yardstick for a slender column — stiffness and geometry are what matter.

Euler's critical load

For a slender column with both ends free to rotate but held in position (the pinned-pinned condition), Euler's analysis gives the critical load as:

\[ P_{cr} = \frac{\pi^{2}\,E\,I}{L^{2}} \]
Euler buckling load for an ideal pinned-pinned column. E is Young's modulus, I the second moment of area of the cross-section about the bending axis, and L the column length. Note the absence of any strength term.

Critical stress and the slenderness ratio

It is often more useful to work in stress than in load. Divide P_cr by the cross-sectional area A, and use the radius of gyration r, defined by r² = I/A (so r bundles the section's geometry into a single length):

\[ \sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^{2}\,E}{(L/r)^{2}}, \qquad r = \sqrt{\dfrac{I}{A}} \]
Critical (buckling) stress. The load has vanished: σ_cr depends only on the material (through E) and on the slenderness ratio L/r. A larger r for the same area means a less slender, more buckle-resistant column.
✨ The radius of gyration rewards shape, not material

The radius of gyration r = √(I/A) measures how far a section's area is spread from its centroid. For a solid square of side b, I = b⁴/12 and A = b², so r = b/√12 — about 0.29 b. Two sections with the same area can have very different radii of gyration: a thin tube spreads its material far from the centre and has a large r, making it far more buckle-resistant than a compact square of the same area. This is why columns are hollow tubes, angles, or I-sections rather than solid blocks — for a buckling problem you want material pushed outward, where it raises I the most.

Critical stress versus slenderness ratio. The Euler hyperbola sigma_cr = pi squared E over (KL/r) squared falls as slenderness grows. A horizontal yield cutoff at sigma_y = 250 MPa caps the usable stress. The bold governing curve is the smaller of the two: flat at the yield stress for stocky columns, then following the hyperbola down for slender columns. The two meet at the transition slenderness KL/r of about 89. 0 130 250 critical stress (MPa) 50 89 150 200 slenderness ratio KL/r sigma_y = 250 transition (KL/r)_c = 89 stocky yielding governs slender Euler buckling governs
Critical stress versus slenderness. The bold governing curve is the smaller of the Euler hyperbola and the yield cutoff — stocky columns yield, slender columns buckle, and the two meet at the transition slenderness of about 89.
📝 Worked example: A steel column (E = 200 GPa) has pinned-pinned ends, length L = 2.0 m, and a solid square cross-section of side b = 50 mm. Find (a) the Euler critical load P_cr, (b) the critical stress, and (c) the slenderness ratio, and confirm the column buckles well below its yield stress of 250 MPa.
  1. Second moment of area: I = b⁴/12 = 50⁴/12 = 5.208 × 10⁵ mm⁴.
  2. Use N and mm: E = 200 000 N/mm², L = 2000 mm.
  3. (a) P_cr = π²EI/L² = π² × 200 000 × 5.208 × 10⁵ / 2000².
  4. Numerator = 9.8696 × 200 000 × 5.208 × 10⁵ = 1.028 × 10¹². Denominator = 4.000 × 10⁶.
  5. P_cr = 2.57 × 10⁵ N = 257 kN.
  6. Area A = b² = 2500 mm². (b) σ_cr = P_cr/A = 257 000/2500 = 103 MPa.
  7. Radius of gyration r = b/√12 = 50/3.464 = 14.43 mm. (c) Slenderness L/r = 2000/14.43 = 139.
  8. Check: σ_cr = 103 MPa is well below σ_y = 250 MPa — the column buckles at less than half of its yield stress (slenderness 139 is well into the slender/Euler regime).
✓ P_cr = 257 kN; σ_cr = 103 MPa; slenderness L/r = 139. The column buckles at a stress less than half of yield — a stability failure.
🎮 Interactive: column buckling visualizer LIVE
Predict first: Start on 'pinned-pinned'. Set L = 2.0 m and section b = 50 mm. The banner should read 'stability failure — buckling governs' because P_cr (about 257 kN) is far below the yield load P_y (about 625 kN). Now drag b up toward 100 mm: P_cr rises fast, and at some point the banner flips to 'material failure — yielding governs'. That flip is the whole lesson — the failure mode itself changes with slenderness.

An interactive column buckling visualizer: an end-condition selector, length and section-size sliders, a canvas drawing the buckled mode shape for the chosen end fixity, and a banner stating whether stability or material failure governs.

Pick an end condition and vary length L and section size b. The readout reports slenderness KL/r, the Euler load P_cr, the yield load P_y, and which one governs — with an explicit banner that flips as you cross the buckling-to-yielding transition.
✏️ Practice: A pinned-pinned steel column (E = 200 GPa) has length L = 1.5 m and a solid square cross-section of side b = 40 mm. Using P_cr = π²EI/L² with I = b⁴/12, find the Euler critical load, in kN.
kN
Solution
  1. I = b⁴/12 = 40⁴/12 = 2.133 × 10⁵ mm⁴.
  2. Use N and mm: E = 200 000 N/mm², L = 1500 mm.
  3. P_cr = π²EI/L² = 9.8696 × 200 000 × 2.133 × 10⁵ / 1500².
  4. Numerator = 4.211 × 10¹¹. Denominator = 2.250 × 10⁶.
  5. P_cr = 1.87 × 10⁵ N = 187 kN.
✏️ Practice: For a steel column (E = 200 GPa) the Euler critical stress is σ_cr = π²E/(L/r)², which depends only on material and slenderness. If the slenderness ratio L/r = 120, find σ_cr, in MPa.
MPa
Solution
  1. σ_cr = π²E/(L/r)² = 9.8696 × 200 000 / 120².
  2. Numerator = 1.974 × 10⁶. Denominator = 14 400.
  3. σ_cr = 137 MPa.

Check your understanding

1. The Euler critical load of an ideal pinned-pinned column is:
Euler's formula is P_cr = π²EI/L² — it depends on flexural rigidity and length, with no material-strength term.
2. A long, slender column buckles. Compared with its yield stress, the stress at which it buckles is:
Buckling is a stability failure: a slender column becomes unstable at P_cr, by which point the axial stress can be far below yield.
3. The radius of gyration r is defined as:
r = √(I/A) bundles the section geometry into a single length; the slenderness ratio is L/r.
🔎 Related lessons
Effective Length & End Conditions Design of Columns — Safety Factors · soon
✅ Key takeaways
  • A column loses stability at a critical load P_cr and buckles sideways — a stability failure, not a stress failure.
  • Pinned-pinned Euler load: P_cr = π²EI/L² (no strength term).
  • Critical stress σ_cr = π²E/(L/r)² depends only on E and the slenderness ratio L/r.
  • Radius of gyration r = √(I/A); spreading material outward raises r and resists buckling.
  • A slender column can buckle at a stress far below yield — strength is the wrong yardstick for slender columns.
➡️ Euler's formula was derived for pinned-pinned ends. Real columns are built into foundations, cantilevered, or braced in many other ways — and each end condition shifts P_cr dramatically. The next lesson generalises Euler with a single factor, K, that folds in how the ends are held.
Want to test yourself on this? Try the Mechanical Aptitude test →
🎓 Go deeper: external courses & trusted references

External sites are listed for reference only. This course is independent and has no affiliation with, or endorsement from, the institutions named.