Buckling — Euler column buckling load derivation: Worked examples

The scenario matrix

Before working problems, let's enumerate what kinds of question exist. Each row is a "case class". Every worked example below is tagged with the cell it fills, so by the end no cell is empty.

#	Case class	What's unknown / what's weird	Example
A	Forward — everything given, find $P_{cr}$	plug in	Ex 1
B	Inverse (design) — given required load, find $I$ or radius	rearrange for geometry	Ex 2
C	End-condition swap — same column, different $K$	$L_{e} = K L$ changes	Ex 3
D	Regime boundary — does it buckle or yield first?	compare $σ_{cr}$ vs $σ_{y}$	Ex 4
E	Degenerate / limiting input — $L \to 0$ , $L \to \infty$ , $I \to 0$ , hollow→thin wall	check the trend, not a number	Ex 5
F	Cross-section geometry — solid vs hollow tube of equal area	which has bigger $I$ ?	Ex 6
G	Real-world word problem — spacecraft with a safety factor	translate words → symbols	Ex 7
H	Exam twist — column can buckle about two different axes	smallest $I$ governs	Ex 8

Recall Why these cells and no others?

Every buckling problem is: (1) which direction is the arrow of unknowns — forward or inverse (A vs B)? (2) what's the support, i.e. $K$ (C)? (3) is Euler even valid, or does yield cap it (D, E)? (4) what's the cross-section giving $I$ (F, H)? Cell G is just "all of the above, dressed in a story with a safety factor." Cover these and you've covered buckling.

Case A — Forward: just plug in

Worked example Ex 1 (Cell A) — plain pin-pin strut

Aluminium rod: $E = 70 GPa$ , solid circular radius $r_{0} = 5 mm$ , length $L = 1.0 m$ , pinned both ends. Find $P_{cr}$ .

Forecast: guess the order of magnitude — hundreds of newtons? thousands? (A 1 m aluminium pencil-thick rod is very floppy in compression.)

Second moment of area of a solid circle: $I = \frac{π r _{0}^{4}}{4} = \frac{π ( 0.005 ) ^{4}}{4} = 4.909 \times 1 0^{- 10} m^{4}$ Why this step? $I$ is the only geometry the formula needs — it encodes how far material sits from the bending axis (see Second Moment of Area). The $r_{0}^{4}$ tells you thickness matters enormously.
Pin-pin means $K = 1$ , so $L_{e} = 1.0 m$ . Why? Pins fix position but allow rotation → the buckled shape is a full half sine wave over $L$ (see Column End Conditions and Effective Length).
Substitute: $P_{cr} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 4.909 \times 1 0 ^{- 10} )}{1. 0 ^{2}} \approx 339 N$

Verify: Units: $Pa \cdot m^{4} / m^{2} = \frac{N}{m ^{2}} \cdot m^{2} = N$ ✓. Order-of-magnitude: our forecast of "hundreds of N" holds. A person could snap this by leaning on it — correct for a slender rod.

Case B — Inverse (design): find the geometry

Worked example Ex 2 (Cell B) — size a lander leg for a target load

A lander leg (pin-pin) must survive $P = 2000 N$ without buckling. Steel, $E = 200 GPa$ , $L = 0.8 m$ . Find the minimum $I$ , then the radius if it's a solid rod.

Forecast: we're running the formula backwards. Will $I$ be bigger or smaller than Ex 1's $4.9 \times 1 0^{- 10}$ ? (Higher load and stiffer steel pull in opposite directions — guess before computing.)

Set $P_{cr} = P$ and solve Euler for $I$ : $I \geq \frac{P L _{e}^{2}}{π ^{2} E} = \frac{2000 ( 0.8 ) ^{2}}{π ^{2} ( 200 \times 1 0 ^{9} )} = 6.485 \times 1 0^{- 10} m^{4}$ Why this step? In design the unknown moves to the left. We're not asking "what load?" but "what shape resists this load?" — this is failure-mode-driven sizing.
Back out the radius from $I = π r_{0}^{4} /4$ : $r_{0} = (\frac{4 I}{π})^{1/4} = (\frac{4 ( 6.485 \times 1 0 ^{- 10} )}{π})^{1/4} \approx 5.38 \times 1 0^{- 3} m = 5.38 mm$ Why? We invert the circle formula to get a manufacturable dimension.

Verify: Plug $r_{0} = 5.38 mm$ forward: $I = π (0.00538)^{4} /4 = 6.49 \times 1 0^{- 10}$ ✓, and $P_{cr} = π^{2} (200 \times 1 0^{9}) (6.49 \times 1 0^{- 10}) /0.64 \approx 2001 N \approx P$ ✓. (Real design would multiply $P$ by a safety factor first — done in Ex 7.)

Case C — End-condition swap

Worked example Ex 3 (Cell C) — same rod, four different supports

Take the Ex 1 aluminium rod ( $E I$ fixed, pin-pin gives $339 N$ ). Recompute $P_{cr}$ for fixed–fixed ( $K = 0.5$ ), fixed–pin ( $K \approx 0.7$ ), and fixed–free ( $K = 2$ ).

Forecast: rank them strongest → weakest before computing. Clamping should help; a free top should hurt.

Because $P_{cr} \propto 1/ K^{2}$ at fixed $E I, L$ , write everything relative to the pin-pin value $P_{1} = 339 N$ (where $K = 1$ ): $P_{cr} (K) = \frac{P _{1}}{K ^{2}}$ Why this step? $E, I, L$ don't change — only the wave that fits between the supports changes, and that's entirely captured by $K$ (see Column End Conditions and Effective Length).
Evaluate each:
- Fixed–fixed, $K = 0.5$ : $P_{cr} = 339/0.25 = 1356 N$
- Fixed–pin, $K = 0.7$ : $P_{cr} = 339/0.49 \approx 692 N$
- Fixed–free, $K = 2$ : $P_{cr} = 339/4 \approx 84.8 N$
Why this step? We just apply the $1/ K^{2}$ law to each support. Notice the huge spread — from $84.8 N$ to $1356 N$ , a factor of 16 — for the identical rod: how you clamp the ends matters as much as the rod itself.

Figure 1 — Four buckled columns side by side. The vertical dotted line is the undeflected axis; each curve is the shape the column bows into. Fixed–fixed (leftish) packs the tightest curve → shortest effective wave → strongest; the red fixed–free curve stretches into a lazy quarter wave → longest wave → weakest. The arrow labelled $P$ on top of each is the same axial load direction.

Verify: Ranking fixed-fixed > fixed-pin > pin-pin > fixed-free holds: $1356 > 692 > 339 > 84.8$ ✓. Products $P_{cr} \cdot K^{2}$ all equal $339$ ✓ (constant, as predicted).

Case D — Regime boundary: buckle or yield?

Worked example Ex 4 (Cell D) — the transition slenderness

For a steel column ( $E = 200 GPa$ , yield $σ_{y} = 250 MPa$ ), find the slenderness ratio $λ = L_{e} / r$ (here $r$ is the gyration radius $I / A$ ) at which Euler buckling stress exactly equals yield stress. Below it the column crushes; above it, it buckles.

Forecast: will the transition $λ$ be around 10, 100, or 1000? (Think: steel is stiff, so it stays "long-column" until quite slender.)

Set critical stress equal to yield: $σ_{cr} = \frac{π ^{2} E}{λ ^{2}} = σ_{y}$ Why this step? This is the exact frontier of Yield vs Buckling — Failure Mode Selection. The material's own strength $σ_{y}$ finally enters — but only as a cap, never inside the Euler formula itself.
Solve for $λ$ : $λ = π \frac{E}{σ _{y}} = π \frac{200 \times 1 0 ^{9}}{250 \times 1 0 ^{6}} = π 800 \approx 88.9$ Why this step? We isolate $λ$ by multiplying both sides by $λ^{2} / σ_{y}$ and taking the square root — this turns "when are the two stresses equal?" into a single dividing number on the slenderness axis. Any real column can now be classified instantly: compute its own $λ$ and compare to $88.9$ .

Verify: Slenderness above $\approx 89$ → Euler governs; below → yield governs. Check: at $λ = 88.9$ , $σ_{cr} = π^{2} (200 \times 1 0^{9}) /88. 9^{2} \approx 2.50 \times 1 0^{8} = 250 MPa = σ_{y}$ ✓.

Case E — Degenerate / limiting inputs

Worked example Ex 5 (Cell E) — what happens at the extremes?

No arithmetic answer here — we test the behaviour of $P_{cr} = π^{2} E I / L_{e}^{2}$ (and its cross-section $I$ ) at edge cases and confirm each makes physical sense.

Forecast: which of these blow up, and which vanish?

$L \to 0$ (very short column): $P_{cr} = π^{2} E I / L^{2} \to \infty$ . Why this step? The formula says an infinitely short column needs infinite load to buckle. That's why short columns never buckle — they yield first (Ex 4). Euler correctly "hands off" to yield here.
$L \to \infty$ (very long column): $P_{cr} \to 0$ . Why? A very long rod buckles under a feather. Matches everyday experience (a long wire won't stand up).
$I \to 0$ (vanishingly thin cross-section): $P_{cr} \to 0$ . Why? No material away from the axis → no bending stiffness → collapses instantly.
Hollow tube, wall $t \to 0$ at fixed mean radius $r_{m}$ (thin-wall limit): the exact ring moment $I = \frac{π}{4} (r_{o u t}^{4} - r_{in}^{4})$ collapses to the thin-wall formula $I \to π r_{m}^{3} t$ . Why this step? As the wall gets thin, $r_{o u t} \approx r_{in} \approx r_{m}$ and the fourth-power difference linearises into a single $t$ term. This is the exact formula we use in Ex 6 — Case E shows it is a limit of the general ring, not a separate rule. Note $I$ stays finite and non-zero even as $t \to 0$ relative to $r_{m}$ , because material sits far out; only if $r_{m}$ also shrinks does $I \to 0$ .
Doubling $L$ : $P_{cr} \to P_{cr} /4$ . Why? $P_{cr} \propto 1/ L^{2}$ : this is the "longer is weaker" trap, now made exact.

Verify: Numeric spot-check of the doubling law using Ex 1: $L = 2 m$ gives $P_{cr} = π^{2} (70 \times 1 0^{9}) (4.909 \times 1 0^{- 10}) /4 \approx 84.8 N = 339/4$ ✓. Thin-wall check: with $r_{m} = 15 mm$ , $t = 0.833 mm$ , exact ring $I = \frac{π}{4} ((r_{m} + t /2)^{4} - (r_{m} - t /2)^{4})$ matches $π r_{m}^{3} t$ to better than $1%$ ✓.

Case F — Cross-section geometry

Worked example Ex 6 (Cell F) — solid rod vs hollow tube, same area

Two aluminium columns of equal cross-sectional area $A = π (0.005)^{2} = 7.854 \times 1 0^{- 5} m^{2}$ . One is a solid rod ( $r_{0} = 5 mm$ ); the other a thin tube with mean radius $r_{m} = 15 mm$ . Compare their $I$ , and hence $P_{cr}$ .

Forecast: equal material — so equal buckling strength? (Trap: guess before reading step 2.)

Solid rod: $I_{solid} = π r_{0}^{4} /4 = 4.909 \times 1 0^{- 10} m^{4}$ (from Ex 1). Why this step? We need a baseline $I$ for the "all material bunched near the axis" case, so we can measure how much the tube gains by spreading the same material outward.
Thin tube of mean radius $r_{m}$ and wall thickness $t$ : since area $A = 2 π r_{m} t$ , we get $t = A / (2 π r_{m}) = 7.854 \times 1 0^{- 5} / (2 π \cdot 0.015) = 8.33 \times 1 0^{- 4} m$ . Its second moment is $I_{tube} = π r_{m}^{3} t$ : $I_{tube} = π (0.015)^{3} (8.33 \times 1 0^{- 4}) = 8.836 \times 1 0^{- 9} m^{4}$ Why this step? $I = π r_{m}^{3} t$ pushes the same material further from the axis, and $I$ grows with distance squared — this is the entire point of Second Moment of Area.
Ratio: $I_{tube} / I_{solid} = 8.836 \times 1 0^{- 9} /4.909 \times 1 0^{- 10} = 18.0$ . Since $P_{cr} \propto I$ , the tube is 18× stronger against buckling with identical mass.

Figure 2 — Two cross-sections with the same shaded area. Left: solid black disc of radius $r_{0}$ , material crowded near the centre. Right: the red ring (tube) of mean radius $r_{m}$ , the same amount of metal flung out to the rim where it resists bending hardest. The black arrow in each marks the radius used in its $I$ formula.

Verify: $P_{cr}$ ratio equals $I$ ratio $= 18.0$ ✓ (both share $E, L, K$ ). This is why spacecraft struts are tubes, never solid rods.

Case G — Real-world word problem with safety factor

Worked example Ex 7 (Cell G) — spacecraft strut sizing with a factor of safety

A launch strut carries a peak compressive load of $P_{applied} = 8000 N$ during liftoff (Spacecraft Load Paths & Launch Loads). Design rule: it must not buckle until $1.5 \times$ that load. Fixed–pin ends ( $K = 0.7$ ), $L = 1.2 m$ , titanium $E = 110 GPa$ . Find the minimum required $I$ .

Forecast: with a 1.5 safety factor and a long strut, expect $I$ noticeably larger than the bare-minimum case.

Required critical load includes the safety factor $n_{s} = 1.5$ : $P_{cr} \geq n_{s} P_{applied} = 1.5 \times 8000 = 12000 N$ Why this step? Real loads have uncertainty; we design so the collapse load sits comfortably above the worst expected load.
Effective length: $L_{e} = K L = 0.7 \times 1.2 = 0.84 m$ . Why this step? Fixed–pin ends are neither fully clamped nor freely pinned, so the wavelength that fits is shorter than the physical length — we bake that into $L_{e}$ before substituting, because Euler only ever "sees" $L_{e}$ , never $L$ directly (see Column End Conditions and Effective Length).
Rearrange Euler for $I$ : $I \geq \frac{P _{cr} L _{e}^{2}}{π ^{2} E} = \frac{12000 ( 0.84 ) ^{2}}{π ^{2} ( 110 \times 1 0 ^{9} )} = 7.797 \times 1 0^{- 9} m^{4}$ Why? Same inverse move as Ex 2, now stacked with the $K$ and safety factor of the real story.

Verify: Forward check: $P_{cr} = π^{2} (110 \times 1 0^{9}) (7.797 \times 1 0^{- 9}) /0.8 4^{2} \approx 12000 N = 1.5 \times 8000$ ✓. Units of $I$ : $N \cdot m^{2} / Pa = N \cdot m^{2} / (N / m^{2}) = m^{4}$ ✓.

Case H — Exam twist: two axes to buckle about

Worked example Ex 8 (Cell H) — rectangular strut buckles the easy way

A pin-pin strut has a rectangular cross-section $b \times h = 20 mm \times 10 mm$ , $L = 0.6 m$ , $E = 70 GPa$ . About which axis does it buckle, and what is $P_{cr}$ ?

Forecast: a rectangle has two different $I$ values. Which one does the column choose?

Two second moments (bending about each in-plane axis): $I_{1} = \frac{b h ^{3}}{12} = \frac{0.020 ( 0.010 ) ^{3}}{12} = 1.667 \times 1 0^{- 9} m^{4}$ $I_{2} = \frac{h b ^{3}}{12} = \frac{0.010 ( 0.020 ) ^{3}}{12} = 6.667 \times 1 0^{- 9} m^{4}$ Why this step? Bending about the axis where material is close gives a small $I$ ; about the axis where it's far, a big $I$ (see Second Moment of Area).
A column buckles at the first opportunity — the lowest load — so it picks the smaller $I$ : $I_{m i n} = I_{1} = 1.667 \times 1 0^{- 9} m^{4}$ . Why? Nature reaches the smaller critical load first; the strong direction never gets a say.
$P_{cr} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 1.667 \times 1 0 ^{- 9} )}{0. 6 ^{2}} \approx 3199 N$ Why this step? We substitute $I_{m i n}$ (never $I_{2}$ ) into Euler, because the physical column has already chosen to fold about the weak axis — using the larger $I$ would overpredict strength and be unsafe.

Figure 3 — The $20 \times 10 mm$ rectangle end-on. The black dashed line is the strong axis (large $I_{2}$ , wide spread of material). The red double-arrow marks the thin direction — the column bows out of the wide face along this weak axis, because that's where $I$ is smallest.

Verify: Confirm $I_{1} < I_{2}$ : $1.667 \times 1 0^{- 9} < 6.667 \times 1 0^{- 9}$ ✓ (ratio exactly $(b / h)^{2} = 4$ ). $P_{cr} = 3199 N$ using $I_{m i n}$ ✓.

Self-check

Which

I

governs buckling of a rectangular strut?

the smallest one — the column folds about its weakest axis

For fixed–free vs pin-pin, how does

P_{cr}

change?

drops by factor

K^{2} = 4

(weaker)

Above what does Euler govern, below what does yield?

above the transition slenderness

λ = π E / σ_{y}

Euler governs; below it, yield

Difference between

r_{0}

and

r

r_{0}

is the physical rod radius (enters

I = π r_{0}^{4} /4

);

r = I / A

is the radius of gyration (enters slenderness

λ = L_{e} / r

)

Why is a tube stronger than a solid rod of equal area?

I = π r_{m}^{3} t

places material far from the axis, giving far larger

I

(and

P_{cr} \propto I

)

In design (inverse) problems, what do you solve Euler for?

the geometry —

I

or radius — after setting

P_{cr} = n_{s} P_{applied}