Buckling — Euler column buckling load derivation: Exercises

Definition The symbols we use on this page (all built in the parent note)

$P_{cr}$ — critical buckling load: the smallest axial squeeze at which a slightly bent shape can balance instead of the column staying straight (in newtons, N).
$E$ — Young's modulus: how stiff the material is (how hard it resists being stretched or compressed), in pascals (Pa $=$ N/m $^{2}$ ). Steel $\approx 200 GPa$ , aluminium $\approx 70 GPa$ .
$I$ — second moment of area: how the material of the cross-section is distributed about the bending axis; larger when material sits farther out. Units m $^{4}$ . See Second Moment of Area.
$L$ — the real length of the column; $L_{e} = K L$ — the effective length, the length of an "equivalent pin–pin column" that buckles the same way, with $K$ set by the end supports. See Column End Conditions and Effective Length.
$A$ — cross-sectional area (m $^{2}$ ); $σ_{cr} = P_{cr} / A$ — the critical stress, i.e. force per unit area at buckling (Pa).
$r = I / A$ — radius of gyration: the single distance at which you could concentrate all the area and keep the same $I$ ; $λ = L_{e} / r$ is the slenderness ratio (a pure number).

Everything below uses only: $P_{cr} = \frac{π ^{2} E I}{L _{e}^{2}}, L_{e} = K L, σ_{cr} = \frac{π ^{2} E}{( L _{e} / r ) ^{2}}, r = \frac{I}{A}$

Reference figures you'll reuse

Figure — Buckling — Euler column buckling load derivation

Figure s01 — the four end conditions and their $K$ . Read it left to right: the shape of the buckled wave that each support allows fixes the effective-length factor $K$ (from Column End Conditions and Effective Length). Pin–pin fits one half sine wave ( $K = 1$ ); a clamped base with free top fits only a quarter wave, so it acts like a pin–pin column of double length ( $K = 2$ ); clamping both ends squeezes in a shorter wave ( $K = 0.5$ ). You will use this picture directly in Exercises 2.1 and 5.1 — each time, find the matching column in s01 first, read off $K$ , then compute.

Figure s02 — the column-strength curve. Critical stress vs slenderness $λ = L_{e} / r$ . Left of the crossover (stubby columns) the material yields first — the flat cap at $σ_{y}$ ; right of it (slender columns) Euler's $1/ λ^{2}$ curve governs; and the dashed middle band is the real-world inelastic / intermediate region (Exercises 3.2 and 4.2 place their columns on this map). This is the theme of Yield vs Buckling — Failure Mode Selection.

Level 1 — Recognition

Recall Solution 1.1

WHAT: pin–pin means the effective-length factor $K = 1$ , so $L_{e} = K L = 1.2 m$ . WHY that $K$ : pinned ends can rotate but not move sideways, so exactly one half sine wave fits over the full length (see figure s01, leftmost case). Plug in: $P_{cr} = \frac{π ^{2} E I}{L _{e}^{2}} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 5.0 \times 1 0 ^{- 10} )}{( 1.2 ) ^{2}}$ $P_{cr} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 5.0 \times 1 0 ^{- 10} )}{1.44} \approx 240 N$

Recall Solution 1.2

Answer: (b) yield stress $σ_{y}$ . WHY: Euler buckling is a stiffness-and-geometry event, not a strength event. The formula $P_{cr} = π^{2} E I / L_{e}^{2}$ contains only $E$ (stiffness), $I$ (geometry of cross-section), and $L_{e}$ (geometry of length). A column can buckle while every fibre is still perfectly elastic — nothing has "broken" in the material sense.

Level 2 — Application

Recall Solution 2.1

WHAT: fixed–free has $K = 2$ — read it straight off the second column of figure s01 — so $L_{e} = 2 L = 2.4 m$ . WHY that $K$ : a clamped-base/free-top column can only fit a quarter sine wave; mirror it and you see it behaves like half of a pin–pin column of length $2 L$ (figure s01, second case). Longer effective length $\Rightarrow$ weaker. $P_{cr} = \frac{π ^{2} E I}{( 2 L ) ^{2}} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 5.0 \times 1 0 ^{- 10} )}{( 2.4 ) ^{2}} \approx 60 N$ Since $P_{cr} \propto 1/ L_{e}^{2}$ and $L_{e}$ doubled, the load dropped by $2^{2} = 4$ : $240/4 = 60 N$ . ✓

Recall Solution 2.2

Step 1 — geometry $I$ : $I = \frac{π r _{0}^{4}}{4} = \frac{π ( 0.006 ) ^{4}}{4} = 1.018 \times 1 0^{- 9} m^{4}$ Step 2 — Euler load ( $K = 1$ , $L_{e} = 0.9 m$ ): $P_{cr} = \frac{π ^{2} ( 200 \times 1 0 ^{9} ) ( 1.018 \times 1 0 ^{- 9} )}{( 0.9 ) ^{2}} \approx 2481 N$

Level 3 — Analysis

Recall Solution 3.1

First: WHY $I_{tube} \approx π R^{3} t$ (the thin-wall approximation). The exact second moment of a circular tube with outer radius $R_{o}$ and inner radius $R_{i}$ is $I = \frac{π}{4} (R_{o}^{4} - R_{i}^{4})$ . Write the wall in terms of a mean radius $R = \frac{1}{2} (R_{o} + R_{i})$ and thickness $t = R_{o} - R_{i}$ , so $R_{o} = R + \frac{t}{2}$ , $R_{i} = R - \frac{t}{2}$ . Then $R_{o}^{4} - R_{i}^{4} = (R_{o}^{2} - R_{i}^{2}) (R_{o}^{2} + R_{i}^{2}) = [(R_{o} - R_{i}) (R_{o} + R_{i})] (R_{o}^{2} + R_{i}^{2}) = (t) (2 R) (2 R^{2} + \frac{t ^{2}}{2}) .$ WHY drop the $t^{2}$ term: for a thin wall $t ≪ R$ , so $t^{2} /2$ is negligible beside $2 R^{2}$ . Keeping only the leading term: $I \approx \frac{π}{4} \cdot t \cdot 2 R \cdot 2 R^{2} = π R^{3} t .$ This is valid only because the wall is thin — it is the reason a tube beats a rod: nearly all its area sits at the far radius $R$ .

Solid: $r_{0} = A / π = 2.83 \times 1 0^{- 4} / π = 9.49 \times 1 0^{- 3} m$ . $I_{solid} = \frac{π r _{0}^{4}}{4} = \frac{π ( 9.49 \times 1 0 ^{- 3} ) ^{4}}{4} = 6.37 \times 1 0^{- 9} m^{4}$ Tube: wall $t = A / (2 π R) = 2.83 \times 1 0^{- 4} / (2 π \cdot 0.04) = 1.126 \times 1 0^{- 3} m$ (check $t / R \approx 0.028 ≪ 1$ , so thin-wall is safe). $I_{tube} \approx π R^{3} t = π (0.04)^{3} (1.126 \times 1 0^{- 3}) = 2.264 \times 1 0^{- 7} m^{4}$ Ratio (same $E$ , same $L$ , so $P_{cr}$ ratio $=$ $I$ ratio): $\frac{P _{cr, tube}}{P _{cr, solid}} = \frac{2.264 \times 1 0 ^{- 7}}{6.37 \times 1 0 ^{- 9}} \approx 35.5$ Interpretation: identical mass, but the tube resists buckling ~35× better because $I$ scales with the square of how far material sits from the axis. This is exactly why launch vehicles and lander legs are tubes, not rods.

Recall Solution 3.2

Radius of gyration for a solid circle: $I = π r_{0}^{4} /4$ , $A = π r_{0}^{2}$ , so $r = \frac{I}{A} = \frac{π r _{0}^{4} /4}{π r _{0}^{2}} = \frac{r _{0}}{2} = 4.0 \times 1 0^{- 3} m$ Crossover: set $σ_{cr} = σ_{y}$ : $\frac{π ^{2} E}{λ ^{2}} = σ_{y} \Rightarrow λ = π \frac{E}{σ _{y}} = π \frac{70 \times 1 0 ^{9}}{270 \times 1 0 ^{6}} = 50.6$ Length ( $K = 1$ so $L_{e} = L^{*}$ ): $L^{*} = λ r = 50.6 \times 4.0 \times 1 0^{- 3} = 0.202 m$ Meaning: locate $λ = 50.6$ on figure s02 — it sits right at the crossover, on the boundary of the dashed inelastic band. Shorter than ~20 cm this rod yields first (left region); longer than that Euler buckling governs. A real rod of this exact length would in fact fail slightly below both curves, in the inelastic regime.

Level 4 — Synthesis

Recall Solution 4.1

(a) Required load: $P_{cr} = 1.5 \times 2500 = 3750 N$ . Rearrange Euler for $I$ ( $L_{e} = L = 0.75 m$ ): $I \geq \frac{P _{cr} L _{e}^{2}}{π ^{2} E} = \frac{3750 ( 0.75 ) ^{2}}{π ^{2} ( 200 \times 1 0 ^{9} )} = 1.069 \times 1 0^{- 9} m^{4}$ (b) Wall thickness (using $I \approx π R^{3} t$ from Exercise 3.1): $t \geq \frac{I}{π R ^{3}} = \frac{1.069 \times 1 0 ^{- 9}}{π ( 0.025 ) ^{3}} = 2.177 \times 1 0^{- 5} m \approx 0.022 mm$ Reality check: such a thin wall would fail by local wall wrinkling long before global Euler buckling — a reminder that Euler is a necessary but not sufficient check. Still, it correctly sizes the global-buckling requirement.

Recall Solution 4.2

Euler load: $I = π r_{0}^{4} /4 = π (0.005)^{4} /4 = 4.909 \times 1 0^{- 11} m^{4}$ (note the exponent: $(0.005)^{4} = 6.25 \times 1 0^{- 10}$ , times $π /4$ gives $4.909 \times 1 0^{- 11}$ ). $P_{cr} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 4.909 \times 1 0 ^{- 11} )}{( 0.30 ) ^{2}} \approx 377 N$ Yield load: $A = π r_{0}^{2} = 7.854 \times 1 0^{- 5} m^{2}$ , $P_{y} = σ_{y} A = (270 \times 1 0^{6}) (7.854 \times 1 0^{- 5}) \approx 21206 N$ Compare: $P_{cr} = 377 N ≪ P_{y} = 21206 N$ , so Euler buckling governs; the strut fails at $\approx 377 N$ . Its slenderness $λ = L_{e} / r = 0.30/ (0.005/2) = 120$ places it far out on the right (slender) branch of figure s02, well clear of the inelastic band — so pure Euler is trustworthy here.

Level 5 — Mastery

Recall Solution 5.1

(a) Fixed–fixed has $K = 0.5$ (third column of figure s01), so $L_{e} = 0.5 L$ . Since $P_{cr} \propto 1/ L_{e}^{2}$ : $P_{cr, new} = \frac{π ^{2} E I}{( 0.5 L ) ^{2}} = \frac{1}{0.25} \cdot \frac{π ^{2} E I}{L ^{2}} = 4 \times 800 = 3200 N$ (b) Factor $= 1/ K^{2} = 1/0.25 = 4$ . Clamping quadruples the buckling load — free strength from geometry. (c) Same load by shortening a pin–pin column: need $L_{e}^{'2} = L^{2} /4$ , i.e. $L^{'} = L /2$ . So halving the length gives the identical $4 \times$ gain — the two routes ( $K$ and $L$ ) enter the formula the same way, through $L_{e}^{2}$ .

Recall Solution 5.2

Effective length: $L_{e} = 0.7 \times 1.1 = 0.77 m$ . Required $P_{cr} = 2.0 \times 1800 = 3600 N$ . (a) $I \geq \frac{P _{cr} L _{e}^{2}}{π ^{2} E} = \frac{3600 ( 0.77 ) ^{2}}{π ^{2} ( 70 \times 1 0 ^{9} )} = 3.090 \times 1 0^{- 9} m^{4}$ (b) $t \geq \frac{I}{π R ^{3}} = \frac{3.090 \times 1 0 ^{- 9}}{π ( 0.02 ) ^{3}} = 1.229 \times 1 0^{- 4} m \approx 0.123 mm$ (c) WHY $A \approx 2 π R t$ : for a thin ring, the wall is a strip of length equal to the circumference $2 π R$ (measured at the mean radius) and width $t$ ; since $t ≪ R$ the inner and outer radii are both $\approx R$ , so the exact area $π (R_{o}^{2} - R_{i}^{2}) = π (R_{o} + R_{i}) (R_{o} - R_{i}) = π (2 R) (t) = 2 π R t$ . Then: $A \approx 2 π R t = 2 π (0.02) (1.229 \times 1 0^{- 4}) = 1.545 \times 1 0^{- 5} m^{2}$ $P_{y} = σ_{y} A = (270 \times 1 0^{6}) (1.545 \times 1 0^{- 5}) \approx 4171 N$ Since $P_{y} \approx 4171 N > P_{cr} = 3600 N$ , the tube reaches its Euler limit before yielding — buckling correctly governs, exactly as intended for a slender strut. ✓ (In practice thin-wall local buckling and the inelastic band of figure s02 would also be checked.)