Visual walkthrough — Shell buckling — thin-walled cylinder under axial load
Step 0 — The objects and symbols we are allowed to use
Before any physics, let us name the picture. A rocket tank is a hollow tube. Look at the labelled tube below.
The word thin means (read " much smaller than "). Typically . This "thinness" is the whole reason buckling — not crushing — wins. See Yield vs stability failure modes.
Step 1 — What "buckling" looks like: the dimple
WHAT. Push down on the top of the can. The wall stays flat and straight… until, at one special load, it pops inward and outward into a pattern of little dents.
WHY we draw this first. Every symbol later (, wavelength, wavenumber) describes this shape. You cannot understand the algebra until you have watched the shape appear.
PICTURE.
Notice in the figure the dents repeat: several waves go around the circumference and several go up the length. That "several" is what the next steps must count and then optimise away.
Step 2 — Restoring effect #1: bending stiffness
WHAT. To fold the wall into a dimple you must curve it. Metal hates being curved — it pushes back. That push-back is called bending (flexural) rigidity .
WHY this tool, not simple ? For a narrow beam you would use . But a cylinder wall is a wide plate: when you bend it one way, Poisson coupling makes it want to curl the other way too (anticlastic curvature), and suppressing that stiffens it. That is exactly the extra factor . This is the plate result — see Plate bending and flexural rigidity D.
PICTURE.
Step 3 — Restoring effect #2: membrane hoop stiffness
WHAT. Here is the effect a flat plate does not have. When you dent the wall inward, you have shortened the circumference at that ring — you have stretched or compressed the hoop. The material resists that stretch too.
WHY this exists only for a cylinder. A flat sheet can dimple with no in-plane stretching. A curved sheet cannot: pushing radially by changes the circumference. That change is proportional to curvature . This is the same circumferential stretch behind Hoop stress in pressurised cylinders.
PICTURE.
Step 4 — The tug-of-war sets a favourite wavelength
WHAT. We now have two restoring springs with opposite tastes:
- hates short waves.
- hates long waves.
The wall will buckle at whatever wavelength makes the total restoring energy smallest (weakest spot = easiest place to fold). Somewhere in between short and long, the two effects trade off and the wall is at its most vulnerable.
WHY minimise? Buckling starts at the easiest mode. So the physical critical stress is the minimum over all possible wavelengths — the valley of the curve below.
PICTURE.
The restoring energy per unit area then has the shape
- The first term climbs as grows (short waves are heavily curved → big).
- The second term climbs as shrinks (long waves stretch lots of hoop → big divided by tiny ).
- Between them the sum bottoms out. That valley is the buckle the cylinder actually picks.
Step 5 — The geometric-mean magic ()
WHAT. Find the valley bottom of .
WHY this specific algebra tool? There is a famous inequality: for any two positive numbers , the sum is smallest — relative to their product — exactly when , and then . Our two terms are a perfect match: their product is independent of : So the minimum of the sum is fixed no matter what is, and it equals , reached when the two springs contribute equally.
PICTURE.
Step 6 — The energy balance: why equals the valley depth
This is the step the parent note glossed over. We now show carefully why the critical stress pairs with the valley and why and do not survive.
WHAT. Energy method: the buckled shape is stable only if bending it costs more energy than the axial squeeze gives back. At the exact critical load, the two are equal — that is the tipping point.
WHY the energy method (not force balance)? Buckling is a stability question: "does a tiny dent grow or die?" The clean test is energy — a dent grows precisely when it releases net energy. This is the same logic as Euler column buckling.
PICTURE.
The two energies, per unit surface area, for a dent of amplitude and wavenumber :
\qquad \underbrace{W_{\text{squeeze}} = \tfrac{1}{2}\,\sigma\,t\,k^{2}\,w_0^{2}}_{\text{work the axial stress gives up}} $$ - $w_0^2$ — appears in **both** terms with the same power. When we set them equal, $w_0^2$ **cancels**. That is *why* the answer never contains the dent's size — the critical stress is an onset condition, independent of how deep the eventual dent goes. - $\sigma\,t$ — axial force per unit circumferential length (stress × wall thickness). - $k^{2}$ on the work side — as the axially-squeezed wall bows sideways it shortens along the axis by an amount $\propto k^{2}w_0^2$; that shortening is how the squeeze does work. The **same** $k^{2}$ sits on the bending term. **Set $W_{\text{squeeze}} = U_{\text{restore}}$** (the tipping point) and cancel $\tfrac12 w_0^2$: $$ \sigma\,t\,k^{2} = D\,k^{2}+\frac{K}{k^{2}} \;\;\Longrightarrow\;\; \sigma\,t = D + \frac{K}{k^{4}}. $$ Buckling happens at the **smallest** such $\sigma$, so minimise the right side over $k$. Its minimum (Step 5, with $k^2$ playing the role of the variable) is the geometric mean, giving $$ \boxed{\;\sigma_{cr}\,t = 2\sqrt{D\,K} = 2\sqrt{D\,\frac{Et}{R^{2}}}\;}. $$ - The **length $L$ never entered** the energy per unit area, so for a long cylinder it drops out entirely: a short can and a tall can (same $R,t$) buckle at the same *stress*. --- ## Step 7 — Finite length & the discrete-mode caveat **WHAT.** Steps 4–6 let $k$ vary *continuously* — legitimate only for a **long** cylinder, where whatever optimal wavelength we want fits comfortably many times. **WHY flag it.** A real tank has *ends* (bulkheads, rings). The axial half-waves must fit a whole number of times into the length, so the axial wavenumber is **discrete**: $$ k_{n} = \frac{n\pi}{L},\qquad n = 1,2,3,\dots $$ For a **long** cylinder the spacing between allowed $k_n$ is tiny, so one of them lands essentially on the continuous optimum → our formula holds. For a **short** cylinder ($L$ not many times $R$), the optimum wavelength may **not fit**; the cylinder is forced onto a stiffer discrete mode and buckles at a **higher** $\sigma_{cr}$ than $0.605\,Et/R$. The classical formula is therefore a *long-cylinder* result. **PICTURE.** ![[deepdives/dd-physics-3.6.07-d2-s08.png]] > [!mistake] Out-of-scope regimes for this formula > - **Very short cylinders** ($L\lesssim R$): behave more like a flat plate strip; discrete axial modes raise $\sigma_{cr}$. Use plate/short-shell theory, not $0.605\,Et/R$. > - **End supports matter**: clamped vs simply-supported ends shift the allowed $k_n$ and hence the load — see [[NASA SP-8007 buckling of thin-walled cylinders]]. > - This page's clean geometric-mean result is the **moderate-to-long** cylinder ($L$ several $R$ or more), which covers most rocket tanks and interstages in [[Rocket tank and interstage structural design]]. --- ## Step 8 — Plug in $D$, simplify to the famous number, and check the edges **WHAT.** Substitute $D=\dfrac{Et^{3}}{12(1-\nu^{2})}$ into $\sigma_{cr}\,t = 2\sqrt{D\cdot\dfrac{Et}{R^{2}}}$ and clean up. **WHY.** We kept $D$ symbolic to keep Steps 5–6 readable. Now we cash it in. **PICTURE.** ![[deepdives/dd-physics-3.6.07-d2-s09.png]] $$ \sigma_{cr}\,t = 2\sqrt{\frac{Et^{3}}{12(1-\nu^{2})}\cdot\frac{Et}{R^{2}}} = 2\sqrt{\frac{E^{2}t^{4}}{12(1-\nu^{2})R^{2}}} = \frac{2E t^{2}}{R}\sqrt{\frac{1}{12(1-\nu^{2})}}. $$ Divide both sides by $t$, and use $\sqrt{1/12}=1/(2\sqrt3)$ so the $2$ cancels: > [!formula] Classical axial buckling stress — the destination > $$ \boxed{\;\sigma_{cr} = \frac{E}{\sqrt{3(1-\nu^{2})}}\,\frac{t}{R}\;} $$ > - $E$ — stiffer metal, stronger against buckling (top, so it *helps*). > - $t/R$ — the thinness ratio. Thin wall (small $t$) or fat can (big $R$) ⇒ weaker. This is the single geometry knob. > - $\sqrt{3(1-\nu^{2})}$ — a pure number. For $\nu=0.3$: $\sqrt{3(1-0.09)}=\sqrt{2.73}=1.652$. > > Therefore for metals: > $$ \sigma_{cr}\approx 0.605\,E\,\frac{t}{R}. $$ **The degenerate & edge cases — never skip these:** > [!example] Limit 1 — Flat plate: $R\to\infty$ > As $R\to\infty$, $t/R\to0$, so $\sigma_{cr}\to0$. A flat plate has **no membrane hoop term** (Step 3's $K=Et/R^2\to0$), so this *particular* mechanism disappears — flat plates buckle by a different rule. Consistent: our formula switches off exactly where its physics does. > [!example] Limit 2 — Very thin wall: $t\to0$ > $\sigma_{cr}\propto t\to0$. An infinitely thin foil carries no axial buckling stress at all. Matches intuition — you cannot stand on a sheet of cling film. > [!example] Limit 3 — Incompressible material: $\nu\to0.5$ > $\sqrt{3(1-0.25)}=\sqrt{2.25}=1.5$, coefficient $=1/1.5=0.667$. The plate stiffening rises slightly. Isotropic elasticity only admits $-1<\nu<\tfrac12$, and real engineering metals have $\nu\le0.5$ (rubber approaches the $\tfrac12$ limit). Across that **whole physical range** the denominator $3(1-\nu^2)$ stays positive and finite, so the formula never blows up. Good — it is well-behaved for every real material. > [!mistake] The one edge case the formula does NOT tell you about > **All of the above is for a *perfect* cylinder.** Real cans have dents ~$t$. Because Step 5's valley is *broad and flat* (many wavelengths have nearly equal energy), the real cylinder sits on a knife-edge and buckles at only **20–70\%** of $\sigma_{cr}$. You must multiply by a knockdown factor $\gamma$ (see [[Imperfection sensitivity and knockdown factors]] and [[NASA SP-8007 buckling of thin-walled cylinders]]): > $$ \gamma = 1 - 0.901\left(1-e^{-\phi}\right),\qquad \phi=\tfrac{1}{16}\sqrt{R/t}. $$ --- ## The one-picture summary ![[deepdives/dd-physics-3.6.07-d2-s10.png]] This final figure compresses the whole walkthrough: the dented shell → two springs ($D$ bending vs $K$ membrane) → energy balance $\sigma t\,k^2 = Dk^2 + K/k^2$ → geometric-mean valley $2\sqrt{DK}$ → the boxed result → knockdown to reality. Everything used in [[Rocket tank and interstage structural design]] hangs on this one chain. > [!recall]- Feynman retelling — the whole derivation in plain words (click) > Imagine squeezing an empty soda can from the top. To crumple it, the wall has to do two hard things at once. **First**, it must *bend* into little dents — and metal resists bending, especially into tight tiny folds; call that resistance $D$, and it's huge for tight dents because bending strength grows like thickness cubed. **Second**, when a dent pushes inward it has to *shrink the ring* around the can — stretching or squeezing the metal sideways — and that resistance, call it $K$, is worst for big broad bulges. So the two resistances have opposite tastes: one hates tiny dents, the other hates big ones. Now the squeeze from above *gives back* energy when the wall bows — and a dent actually grows only when that giveaway beats the two resistances. At the tipping point they're exactly equal, and because the dent's depth appears on both sides it cancels out — that's why the answer doesn't care how deep the eventual dent is. The can folds at the dent-size *in between* where the two resistances are as easy as possible. There's a classic bit of maths — the smallest a "spring one plus spring two" can be is twice the square root of their product, when they're equal — so the weakest spot has strength $2\sqrt{DK}$. Cash in what $D$ and $K$ are, tidy the numbers, and out pops one clean rule: the can can take about **0.6 times its metal-stiffness times its thinness ($t/R$)** before folding — as long as it's a good few times taller than it is wide, so the ideal dent-size fits. Then reality bites: because lots of dent-sizes are almost equally weak, a real dented can folds at only a third to a half of that. --- ## Active recall What are the two competing restoring effects in cylinder buckling? ::: Bending stiffness $D$ (hates short-wavelength dents) and membrane hoop stiffness $K=Et/R^2$ (hates long-wavelength dents). In the energy balance, why does the dent amplitude $w_0$ drop out? ::: Both the restoring energy and the squeeze work scale as $w_0^2$, so setting them equal cancels $w_0^2$ — the critical stress is an onset condition independent of dent depth. Precisely, what is the wavenumber $k$? ::: $k=2\pi/\lambda$, radians of pattern per metre; big $k$ = short waves, small $k$ = long waves. A dent $\sin(kx)$ has curvature $\propto k^2$. Why does the length $L$ drop out of $\sigma_{cr}$ for a long cylinder? ::: The energy per unit area doesn't contain $L$, and the discrete axial modes $k_n=n\pi/L$ are so closely spaced that one lands on the continuous optimum. What happens for a very short cylinder? ::: The optimal wavelength may not fit; discrete modes force a stiffer shape, so $\sigma_{cr}$ is *higher* than $0.605\,Et/R$ — the classical formula is a long-cylinder result. Why is the critical value a geometric mean $2\sqrt{DK}$? ::: The product of the two spring contributions is independent of wavenumber, so the sum's minimum is $2\sqrt{DK}$ (from $a+b\ge2\sqrt{ab}$), reached when they contribute equally. As $R\to\infty$ what happens to $\sigma_{cr}$ and why? ::: It $\to 0$; the membrane term $K=Et/R^2\to0$ (a flat plate has no hoop restoring effect), so this mechanism vanishes. What is the physical range of $\nu$, and does the formula stay finite there? ::: Isotropic elasticity requires $-1<\nu<\tfrac12$; for all such $\nu$ the denominator $3(1-\nu^2)$ stays positive, so $\sigma_{cr}$ never blows up. For $\nu=0.3$ what is the classical coefficient? ::: $1/\sqrt{3(1-0.09)}=1/1.652\approx0.605$.