Mechanics of materials / Sheet 03
Buckling
Formula reference for mechanical engineering interviews. Buckling is a stability failure, so yield strength does not appear in the Euler equation at all. Check slenderness before choosing a formula.
Scope: Columns, plates, shellsSections: 10Units: SI, N and mmRevision: A
34 formulas
Euler buckling
Pcr = π² E I / Le²
No Sy anywhere. Only E, I and length.
Effective length
Le = K L
K comes from the end conditions, section 02
Critical stress
σcr = π² E / λ²
Simply Pcr divided by A
Radius of gyration
r = √(I / A)
Always use Imin, the weak axis
Slenderness ratio
λ = Le / r
Compute this first, every time
Transition slenderness
λc = √(2 π² E / Sy)
About 120 to 130 for structural steel
Length sensitivity
Pcr ∝ 1 / L²
Halving the unbraced length quadruples capacity
Effective length factors
| End conditions | Theoretical K | Design K | Capacity vs pinned |
|---|---|---|---|
| Pinned to pinned | 1.0 | 1.0 | ×1 |
| Fixed to fixed | 0.5 | 0.65 | ×4 |
| Fixed to pinned | 0.7 | 0.80 | ×2 |
| Fixed to free | 2.0 | 2.1 | ×0.25 |
Theoretical K assumes a joint with zero rotation, which no real bolted or welded connection delivers. Quote the design value, and say why it differs. The fixed to free case is the dangerous one: a cantilevered strut carries a quarter of the pinned capacity.
Short and intermediate columns
Long column, λ > λc
σcr = π² E / λ²
Euler; elastic instability
Intermediate, λ < λc
σcr = Sy - (1/E)(Sy λ / 2π)²
Johnson parabolic formula
Short column
σ = P / A ≥ Sy
Pure crushing, no instability
Rankine-Gordon
1/PR = 1/Pc + 1/Pe
Blends crushing and Euler across the whole range
Secant, eccentric load
σmax = (P/A)[1 + (e c / r²) sec((Le/2r)√(P/AE))]
e is load eccentricity; e c / r² is the eccentricity ratio
If Euler returns a critical stress above yield, you used the wrong formula. That result is the signal to switch to Johnson or to a crushing check.
Radius of gyration by section
Solid round
r = d / 4
Same about every axis
Hollow round
r = √(do² + di²) / 4
Highest r per unit area, which is why tubes win
Rectangle, weak axis
r = b / √12 = 0.289 b
b is the smaller dimension
Square, side a
r = 0.289 a
Equal about both axes
Thin-walled tube
r ≈ Rmean / √2
Quick estimate for t much less than R
Plates and shells
Thin plate
σcr = k π² E / [12(1 - ν²)] · (t/b)²
k depends on edge support and aspect ratio
Thin cylinder, axial
σcr ≈ 0.6 E t / R
Theoretical only; real shells fail far below this
Post-buckling reserve
plates yes, columns no, shells worst
Aircraft skins are allowed to buckle in service
Local buckling
governed by D/t or b/t
Sets the floor on wall thickness for a tube
Design practice
Factor of safety
3 to 5 on Pcr
Higher than for yielding, because of imperfection sensitivity
Cheapest fix
add a mid-span brace
Halves Le, quadruples Pcr, adds no weight to the member
Second fix
move material outward
Raise I at constant A, so switch to a tube
Non-fix
stronger alloy
E barely changes within a material family, so Pcr barely changes
Two-axis check
λx = Lex/rx, λy = Ley/ry
Bracing can differ per axis; the larger λ governs
Worked chain: sizing a strut
Step 1
Imin, A, r = √(Imin/A)
Identify the weak axis before anything else
Step 2
Le = K L, then λ = Le/r
Per axis if the bracing differs
Step 3
compare λ with λc
Chooses Euler or Johnson
Step 4
Pcr = σcr A, then n = Pcr / P
Target 3 to 5, not 1.5
Where it bites in practice
Actuators
lead screws, push rods
Long, slender, in compression, often near cantilevered
Structures
uprights, frames, legs
Check both axes; bracing usually differs
Thermal restraint
σ = E α ΔT with no room to expand
A slender restrained member buckles rather than yields
Sheet metal
webs, brackets, thin walls
Local buckling and crippling before global instability
Numbers worth memorizing
| Material | E (GPa) | Sy (MPa) | λc |
|---|---|---|---|
| Mild steel A36 | 200 | 250 | 126 |
| Alloy steel 4140 | 200 | 655 | 78 |
| Aluminium 6061-T6 | 69 | 276 | 70 |
| Aluminium 7075-T6 | 71 | 503 | 53 |
Note what the table shows: the two steels have identical E, so identical Euler capacity at the same geometry, despite a 2.6 times difference in yield. Strength only changes where the transition sits, not the buckling load itself.
Concepts that decide the interview
- Buckling is a stability failure, not a strength failure. Sy does not appear in the Euler equation, so a stronger alloy within the same family does essentially nothing. Fix it with geometry, bracing, or end fixity.
- A column buckles about its weakest axis. Use Imin, and check both axes separately, because effective lengths often differ per axis.
- Capacity goes as 1 over length squared, so a single mid-span brace quadruples it at no weight cost. This is almost always the best answer to "it is buckling, what do you do."
- End fixity is worth up to a factor of four, but real joints never reach the theoretical value, which is why design K values are higher than textbook ones.
- Compute slenderness before choosing a formula. Euler applied to a stubby column returns a critical stress above yield, which is nonsense.
- Buckling is imperfection-sensitive: crookedness, eccentricity and residual stress all reduce the real failure load, so safety factors run 3 to 5.
- Columns have no post-buckling reserve, plates do, and thin shells are the worst case of all. That difference explains why aircraft skins are permitted to buckle but a strut is not.
- Tubes are the efficient section because they maximise r for a given area, but thin walls invite local buckling, so there is an optimum D/t rather than "thinner is always better."
Sheet 03 of 07 / Rev AIdeal straight concentrically loaded members unless noted