Mechanics of Materials · Chapter 10 of 10 · Advanced
Columns and Buckling
A slender column can fail long before it crushes, by suddenly bowing sideways. Buckling is an instability, not a strength limit, and Euler's load tells you when it strikes.
Readiness check
This closing chapter blends bending stiffness with stability. Tick only what you can do closed-notes.
- Recall the flexural rigidity EI.
- Compute the radius of gyration r = √(I/A).
- Recall the yield or compressive strength.
- Work with π² and squares.
- Distinguish strength failure from instability.
The core idea
Above a critical load a slender column stops being straight and bows out sideways; that load is Pcr = π²EI/Le², set by stiffness and length, not strength.
Pcr = π²EI/Le²Le = KLσcr = π²E/(L/r)²Compression can fail a column two ways: a short, stout one crushes or yields, but a long, slender one buckles, deflecting sideways suddenly at the Euler load Pcr. Buckling depends on bending stiffness EI and the effective length Le = KL, where K reflects the end conditions, not on the material's strength. Expressed as a stress, σcr = π²E/(L/r)² falls with the slenderness ratio L/r. Euler's formula governs only while σcr stays below the yield stress; below that slenderness, the column yields instead.
The skills, taught in order
Buckling is a stiffness-and-stability problem. Five skills cover instability, Euler's load, end conditions, slenderness, and the formula's limits.
10.1 Stability and buckling
Buckling is a qualitatively different failure: the straight column becomes unstable and deflects sideways, even though the stress is still below yield. It is sudden and often catastrophic, so it must be checked separately from strength.
10.2 Euler's critical load
For a slender, pinned column the critical load is Pcr = π²EI/L². Note it uses E and I (stiffness), not the strength: a stronger steel of the same stiffness buckles at the same load. The smallest I of the section governs, so columns buckle about their weak axis.
10.3 Effective length and end conditions
End restraints change the buckled shape, captured by the effective length Le = KL.
| End conditions | K | Pcr relative to pinned |
|---|---|---|
| Fixed-free (cantilever) | 2.0 | 0.25× |
| Pinned-pinned | 1.0 | 1× |
| Fixed-pinned | 0.7 | ≈ 2× |
| Fixed-fixed | 0.5 | 4× |
10.4 Slenderness and critical stress
Dividing Pcr by area gives σcr = π²E/(Le/r)², where r = √(I/A) is the radius of gyration and Le/r the slenderness ratio. The more slender the column, the lower the stress at which it buckles.
10.5 Limits of Euler's formula
Euler's curve only applies where σcr is below the yield stress, that is, for slender columns above a critical slenderness. Stockier columns fail by yielding or by inelastic buckling, handled by empirical curves (such as the Johnson formula). Always check which regime a column is in.
Engineering connection: struts, building columns, jack screws, landing-gear legs, and any slender compression member are buckling-governed; bracing (lowering Le) is the cheapest fix.
Worked example 1: Euler load and validity
A pinned steel column is 4 m long with cross-sectional area 2000 mm² and least moment of inertia 2.2×10⁶ mm⁴ (E = 200 GPa, yield 250 MPa). Find the critical load and confirm Euler's formula applies.
- ProblemFind Pcr for the column in Figure 1 and check Euler validity.
- Given / findL = 4 m, A = 2000 mm², I = 2.2×10⁶ mm⁴, E = 200 GPa, σy = 250 MPa, pinned (K = 1). Find Pcr and σcr.
- AssumptionsSlender, ideal, pinned column; smallest I governs.
- ModelEuler's load, then σcr = Pcr/A compared with yield; check slenderness.
- EquationsPcr = π²EI/L² r = √(I/A), L/r σcr = Pcr/A
- SolvePcr = π²(200 000)(2.2×10⁶)/(4000)² = 271 kN. r = √(2.2×10⁶/2000) = 33.2 mm, so L/r = 121. σcr = 271 000/2000 = 136 MPa, below the 250 MPa yield, so Euler applies.
- CheckThe Euler limit for steel is L/r ≈ √(π²E/σy) = 89; at L/r = 121 the column is comfortably slender, so it buckles before it yields. Good.
- ConclusionThe column buckles at 271 kN, well below the 500 kN it would carry if strength alone (250 MPa × 2000 mm²) governed. For slender members, stability, not stress, sets the capacity.
Worked example 2: the effect of end conditions
Take the same column (pinned Pcr = 271 kN) and find its critical load if instead it were fixed-free, fixed-pinned, or fixed-fixed.
- ProblemFind Pcr for the other end conditions of the column in Figure 2.
- Given / findPinned Pcr = 271 kN. Find Pcr for fixed-free (K = 2), fixed-pinned (K = 0.7), fixed-fixed (K = 0.5).
- AssumptionsSame section and material; only the end restraint changes.
- ModelReplace L with Le = KL, so Pcr scales as 1/K² relative to the pinned case.
- EquationsPcr = π²EI/(KL)² Pcr/Ppinned = 1/K²
- SolveFixed-free: 271/2² = 68 kN. Fixed-pinned: 271/0.7² = 554 kN. Fixed-fixed: 271/0.5² = 1086 kN.
- CheckThe ratios are exactly 0.25, 2.0, and 4.0, as 1/K² predicts. Restraint helps dramatically: fixing both ends quadruples the buckling load for free.
- ConclusionEnd conditions matter as much as the section. Bracing or fixing ends to lower Le is the cheapest way to raise buckling capacity, the practical lever for column design and the close of the course.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Checking only crushing stress | Slender column fails far below P/A = σy | "Is it slender enough to buckle?" | Compare P with the Euler load Pcr, not just strength. |
| Using strength in Pcr | Stronger steel assumed to buckle later | "Does Pcr contain σy?" | Pcr depends on E and I, not the strength. |
| Ignoring end conditions | Capacity off by up to 16× | "What is the effective length Le = KL?" | Use the K for the actual restraints. |
| Wrong axis (using large I) | Buckling load overestimated | "Which is the smallest I?" | Columns buckle about the weak axis; use the least I. |
Practice ladder
A pinned column has EI = 200 GPa × 1.0×10⁶ mm⁴ and length 2.5 m. Find its Euler load.
Show answer
Pcr = π²EI/L² = π²(200 000)(1.0×10⁶)/(2500²) = 1.974×10¹¹/6.25×10⁶ = 31.6 kN.
A fixed-free strut (cantilever column) is 1.5 m long with EI = 200 GPa × 0.5×10⁶ mm⁴. Find its buckling load.
Show answer
Le = KL = 2 × 1.5 = 3 m. Pcr = π²EI/Le² = π²(200 000)(0.5×10⁶)/(3000²) = 9.87×10¹⁰/9×10⁶ = 11.0 kN. The free top makes it weak.
A solid steel rod 20 mm in diameter, pinned, 1.2 m long, carries axial compression. Find Pcr and check Euler validity (E = 200 GPa, yield 250 MPa).
Show answer
I = πd⁴/64 = π(20)⁴/64 = 7854 mm⁴; A = 314 mm²; r = √(I/A) = 5 mm; L/r = 240 (very slender). Pcr = π²(200 000)(7854)/(1200²) = 1.55×10¹⁰/1.44×10⁶ = 10.8 kN. σcr = 10 800/314 = 34 MPa ≪ 250, so Euler is valid. The thin rod buckles at a tiny load.
Find a real slender compression member (a tent pole, a scaffold tube, a table leg). Estimate its Euler load and the end conditions, and judge whether buckling or crushing governs.
What good work looks like
The least I and effective length identified, Pcr computed, σcr compared with yield to confirm the regime, and a comment on bracing.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one compression member: compute its Euler load with the correct effective length, confirm the regime against yield, and suggest how bracing would change it.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write Euler's critical load.
Pcr = π²EI/Le², with Le = KL.
2. What does buckling depend on, and not depend on?
It depends on stiffness EI and length; it does not depend on the material's strength.
3. Give the effective-length factor K for the four standard end conditions.
Fixed-free 2.0, pinned-pinned 1.0, fixed-pinned 0.7, fixed-fixed 0.5.
4. Write the critical stress in terms of slenderness.
σcr = π²E/(Le/r)², with r = √(I/A).
5. When is Euler's formula invalid?
When σcr exceeds the yield stress (stocky columns); they yield or buckle inelastically instead.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 10 (Columns) |
| Cross-reference | Hibbeler, Ch. 13 · Gere and Goodno, Ch. 11 |
| Core topics | 10.1 Stability · 10.2 Euler's load · 10.3 Effective length · 10.4 Slenderness · 10.5 Limits of Euler |
| Engineering connection | Struts, building columns, jack screws, and any slender compression member. |
| Read next | You have completed the Mechanics of Materials course. Return to the course hub or continue to Machine Elements. |