Question bank — Buckling — Euler column buckling load derivation

False — $P_{cr}\propto 1/L^2$, so doubling $L$ quarters $P_{cr}$, not halves it. The length enters squared, a fact people persistently underweight.

True — $P_{cr}={\pi }^{2}EI/L_e^2$ is linear in $E$, so a stiffer material buckles at proportionally higher load. But note $E$ (stiffness), not ${\sigma }_y$ (strength), is what counts.

False — yield stress ${\sigma }_y$ appears nowhere in Euler's formula. Same $E$, $I$, $L$ means same $P_{cr}$; the alloy only raises the yield cap, not the buckling load.

False — buckling depends on $I$, not $A$. A thin-walled tube pushes the same area outward, giving far larger $I$ and buckling load than a solid rod of equal area.

False — a nontrivial bent shape requires $\sin (kL)=0$ (with $k=\sqrt{P/EI}$), i.e. $kL=n\pi$ for integer $n$. Only the discrete eigenvalue loads $P=n^2{\pi }^{2}EI/L^2$ permit it; buckling is an eigenvalue problem.

False (within linear theory) — small-deflection theory leaves $A$ completely undetermined. You need nonlinear (elastica) post-buckling theory to relate amplitude to load.

False — at exactly $P_{cr}$ the straight and bent shapes are both equilibria (neutral stability). Real columns bow because they have tiny imperfections; a mathematically ideal one could sit straight.

True — clamping both ends forces a shorter effective wave ($K=0.5$, $L_e=L/2$), so $P_{cr}={\pi }^{2}EI/(0.5L{)}^2=4\times$ the pin–pin value.

True — only a quarter sine wave fits, giving $K=2$, $L_e=2L$, so $P_{cr}$ drops to $1/4$ of the pin–pin case for the same $L$.

True — its effective length factor is $K\approx 0.7$ (exactly, the root of $\tan (kL)=kL$), so $L_e\approx 0.7L$. That gives $P_{cr}\approx {\pi }^{2}EI/(0.7L{)}^2\approx 2\times$ the pin–pin load — stronger than pin–pin ($K=1$) but weaker than fixed–fixed ($K=0.5$).

False — short columns have small slenderness $\lambda$, so predicted ${\sigma }_{cr}={\pi }^{2}E/{\lambda }^2$ exceeds ${\sigma }_y$; they yield/crush before Euler buckling can occur. Euler governs only the slender regime.

Error: the formula uses $I$, not $A$. Bending resistance comes from how far material sits from the neutral axis ($I$), a completely different quantity from area.

Error: you must use the effective length $L_e=KL$ with $K=2$, giving $P_{cr}={\pi }^{2}EI/(2L{)}^2$ — one quarter of what you computed.

Error: it's $M=-Py$. The sign encodes positive feedback — the offset load deepens the bend rather than restoring it, which is precisely why the straight state goes unstable.

Error: one boundary condition only gives $B=0$. You must also apply $y(L)=0$, which forces $\sin (kL)=0$ — that condition, not the ODE alone, selects the buckling loads.

Error: $n=0$ gives $k=0$, i.e. the trivial straight ($y\equiv 0$) solution — no buckling at all. The first physical buckling mode is $n=1$, giving $P_{cr}={\pi }^{2}EI/L^2$.

Error: higher $n$ requires higher load ($P\propto n^2$). A load rising from zero reaches $n=1$ first, so nature always buckles in the lowest mode (a single half sine wave).

Error: curvature (small-slope) is the second derivative $d^{2}y/d{x}^2$, the rate of change of slope. Using the first derivative confuses tilt with bending and destroys the whole beam relation.

Because buckling is a bending phenomenon; resistance to bending depends on how far material lies from the neutral axis, which is exactly what the second moment of area $I$ measures.

It means the moment produced by the offset load acts to increase the deflection — positive feedback. Any tiny bend feeds a moment that bends it further, so above $P_{cr}$ the straight state can't survive a nudge.

Because combining $M=-Py$ with $EI{y}^{\prime \prime }=M$ gives a restoring-type equation in the space variable $x$ (with $k^2=P/EI$); its sine/cosine solutions mean the buckled shape is literally a spatial standing wave.

Because loading increases from zero and the column collapses at the first load that permits a bent shape. Higher modes need larger loads that the column never reaches intact.

Because ${\sigma }_{cr}={\pi }^{2}E/{\lambda }^2$ falls with ${\lambda }^2$. Large $\lambda$ (long, thin) means buckling stress far below ${\sigma }_y$ → Euler wins; small $\lambda$ means it exceeds ${\sigma }_y$ → the material yields first.

Because linear theory homogeneously scales any solution — if $A\sin (\pi x/L)$ solves it, so does any multiple. Only nonlinear (large-deflection) terms, dropped by the small-slope assumption, pin the amplitude down.

Clamped ends prevent rotation, forcing the bent shape into a shorter effective wavelength ($L_e=0.5L$). A shorter effective length means larger $P_{cr}={\pi }^{2}EI/L_e^2$.

Because peak axial compression during launch can drive slender struts and lander legs past $P_{cr}$ in an instant; buckling is sudden and catastrophic, so it must be designed against with a safety factor regardless of duration.

Using the general form $P_{cr}={\pi }^{2}EI/L_e^2$ with $L_e=KL$, as $L\to \infty$ we get $P_{cr}\to 0$ for any fixed $K$ — an infinitely long column buckles under any positive load. Slenderness is the enemy; length dominates because it enters squared.

The general form $P_{cr}={\pi }^{2}EI/(KL{)}^2\to \infty$ as $L\to 0$, which is unphysical — well before that the compressive stress reaches ${\sigma }_y$ and the column yields/crushes. The Euler curve is capped by the yield line for short columns.

Then there's a bending moment from the very start, so the column bends immediately at any load — there's no clean sudden bifurcation. Deflection grows continuously; this is the beam-column regime, not pure Euler buckling.

It has no sharp critical load — deflection grows smoothly and blows up as $P$ approaches $P_{cr}$. The ideal $P_{cr}$ acts as an asymptote the real column approaches but formally never sits at.

The smallest $I$. The column buckles about its weakest bending axis first, so design must use $I_{\min }$, not the average or the larger value.

The trivial straight equilibrium $y\equiv 0$ — no bending at all. It's a valid solution of the ODE but represents "not buckled," so the first buckling mode is $n=1$.