Shell buckling — thin-walled cylinder under axial load: Worked examples

This page is the drill hall for shell buckling. The parent note built the classical formula $σ_{cr} = \frac{E}{3 ( 1 - ν ^{2} )} \frac{t}{R} \approx 0.605 E \frac{t}{R}, γ = 1 - 0.901 (1 - e^{- ϕ}), ϕ = \frac{1}{16} R / t .$ Here we hammer it against every kind of input a problem can throw at you — normal numbers, tiny and huge geometry ratios, degenerate limits, a word problem, and an exam-style trap. Before we start, let us name what "every kind" means.

The scenario matrix

Every symbol below was earned in the parent note. Just so we never trip: $E$ = Young's modulus (stiffness of the material), $t$ = wall thickness, $R$ = cylinder radius, $ν$ = Poisson ratio, $σ_{cr}$ = perfect-cylinder buckling stress, $P_{cr}$ = the axial load (force, in newtons) that stress corresponds to via $P_{cr} = σ_{cr} \cdot A$ where $A = 2 π R t$ is the wall cross-section, $γ$ = knockdown factor (the fraction of the perfect value a real dented cylinder actually reaches), $σ_{a l l o w} = γ σ_{cr}$ = realistic allowable stress, $P_{a l l o w} = σ_{a l l o w} \cdot A$ = realistic allowable load, and $σ_{Y}$ = yield stress (where the material itself gives).

Cell	What the case stresses	Covered by
A. Standard forward	plug numbers in, get $σ_{cr}$ , $P_{cr}$	Ex 1
B. Scaling law	how the answer responds to a change	Ex 2
C. Buckling vs yield	which failure mode actually wins	Ex 3
D. Thick-wall limit ( $R / t$ small)	formula meaning at the edge of validity	Ex 4
E. Thin-wall / large- $R / t$ limit	knockdown $γ \to 0$ danger zone	Ex 5
F. Degenerate input ( $t = 0$ , $ν \to 1$ )	what breaks and why	Ex 6
G. Inverse / design problem	solve for $t$ given a required load	Ex 7
H. Real-world word problem	translate a mission into numbers	Ex 8
I. Exam twist	a hidden trap (mixing $γ$ , area, units)	Ex 9

We now walk one worked example per cell.

Worked example Example 1 — Cell A: Standard forward calculation

Aluminium interstage: $E = 70$ GPa, $ν = 0.3$ , $R = 1.8$ m, $t = 3$ mm. Find classical $σ_{cr}$ , then $P_{cr}$ , then the realistic allowable using the knockdown.

Forecast: Will the realistic load be closer to the classical value, or a small fraction of it? Guess a fraction.

Step 1 — Geometry ratio $t / R = 0.003/1.8 = 1.667 \times 1 0^{- 3}$ . Why this step? The classical stress depends on geometry only through $t / R$ , so compute it once.

Step 2 — $σ_{cr} = 0.605 E (t / R) = 0.605 \times 70 \times 1 0^{9} \times 1.667 \times 1 0^{- 3} = 70.6$ MPa. Why this step? $ν = 0.3$ collapses the coefficient to $0.605$ , so we can use it directly.

Step 3 — Wall area $A = 2 π R t = 2 π (1.8) (0.003) = 0.03393 m^{2}$ . Why this step? Load = stress × the area that actually carries axial force, which is the thin ring of wall.

Step 4 — $P_{cr} = σ_{cr} A = 70.6 \times 1 0^{6} \times 0.03393 = 2.40$ MN (perfect cylinder).

Step 5 — Knockdown: $ϕ = \frac{1}{16} R / t = \frac{1}{16} 600 = 1.531$ , so $γ = 1 - 0.901 (1 - e^{- 1.531}) = 0.298$ . Why this step? Real cylinders are imperfection-sensitive — see Imperfection sensitivity and knockdown factors. We must scale down.

Step 6 — $σ_{a l l o w} = 0.298 \times 70.6 = 21.0$ MPa, $P_{a l l o w} \approx 0.71$ MN.

Verify: $P_{a l l o w} / P_{cr} = 0.71/2.40 = 0.298 = γ$ . ✓ Units: (Pa)(m²) = N. ✓ The realistic load is under a third of the classical — the knockdown dominates.

Figure s01 below draws this result as two bars — the tall teal bar is the perfect-cylinder load $P_{cr}$ , the short orange bar is the honest $γ P_{cr}$ ; the plum arrow is the knockdown chopping the capacity down. Read it as "the number the formula promises vs the number nature delivers."

Figure — Shell buckling — thin-walled cylinder under axial load

Worked example Example 2 — Cell B: Scaling law (forecast then verify)

Take Example 1's tank. What happens to $σ_{cr}$ and $P_{cr}$ if we double the thickness to $t = 6$ mm (all else equal)?

Forecast: Both double? Both quadruple? One of each? Commit before reading.

Step 1 — $σ_{cr} \propto t / R$ . Doubling $t$ doubles the stress: $70.6 \to 141.2$ MPa. Why this step? $R$ , $E$ , $ν$ are fixed, so only the linear-in- $t$ factor moves.

Step 2 — Area $A = 2 π R t \propto t$ , so it also doubles: $0.03393 \to 0.06786 m^{2}$ . Why this step? $P_{cr} = σ_{cr} A$ has two factors of $t$ hidden in it.

Step 3 — $P_{cr} \propto σ_{cr} \cdot t \propto t^{2}$ : the load quadruples, $2.40 \to 9.59$ MN. Why this step? We multiply the two independent effects — stress up ×2 and area up ×2 — to get the load's response; combining them is the whole point of a scaling argument.

Verify: Direct compute $σ_{cr} = 0.605 (70 \times 1 0^{9}) (0.006/1.8) = 141.2$ MPa ✓ and $P = 141.2 \times 1 0^{6} \times 0.06786 = 9.58$ MN ✓ (rounding). Stress doubles, load quadruples — the $t^{2}$ law.

Worked example Example 3 — Cell C: Does buckling or yield win?

Steel shell: $E = 200$ GPa, yield $σ_{Y} = 250$ MPa, $R / t = 400$ , $ν = 0.3$ . Which fails first?

Forecast: Steel is strong — will it reach 250 MPa yield, or buckle first?

Step 1 — Classical $σ_{cr} = 0.605 E / (R / t) = 0.605 \times 200000/400 = 302.5$ MPa. Why this step? $σ_{cr} = 0.605 E (t / R)$ and $t / R = 1/400$ ; this is the perfect buckling stress.

Step 2 — $302.5 > 250$ : naively you might say "yield wins." But apply the knockdown first. Why this step? The perfect value is a fantasy; the honest comparison uses $σ_{a l l o w}$ . This is the Yield vs stability failure modes decision.

Step 3 — $ϕ = \frac{1}{16} 400 = \frac{1}{16} (20) = 1.25$ , $γ = 1 - 0.901 (1 - e^{- 1.25}) = 1 - 0.901 (0.7135) = 0.357$ . Why this step? $ϕ$ measures how imperfection-prone this geometry is; feeding it through the SP-8007 formula turns the perfect stress into a realistic one — without $γ$ we would compare a fantasy against yield.

Step 4 — $σ_{a l l o w} = 0.357 \times 302.5 = 108$ MPa. Why this step? This is the actual stress the shell can hold; only this number is fair to compare against the 250 MPa yield.

Verify: $108 MPa ≪ 250 MPa$ yield ✓ — the shell buckles at ~108 MPa, far below yield. Using $σ_{Y}$ would over-predict strength by more than 2×. Stability governs.

Worked example Example 4 — Cell D: Thick-wall edge of validity (

$R / t$ small) A stubby titanium bottle: $E = 110$ GPa, yield $σ_{Y} = 830$ MPa (typical Ti-6Al-4V), $ν = 0.3$ , $R / t = 25$ . Compute $σ_{cr}$ and $γ$ , and comment on whether the formula is even trustworthy.

Forecast: With such a thick wall, is the buckling stress huge or tiny? And is $γ$ close to 1?

Step 1 — $σ_{cr} = 0.605 \times 110000/25 = 2662$ MPa. Why this step? Small $R / t$ makes $t / R$ large, so the classical stress rockets up.

Step 2 — $ϕ = \frac{1}{16} 25 = \frac{1}{16} (5) = 0.3125$ , $γ = 1 - 0.901 (1 - e^{- 0.3125}) = 0.758$ . Why this step? Fewer near-degenerate modes when the wall is thick, so imperfections hurt less — $γ$ climbs toward 1.

Step 3 — Sanity flag: even $σ_{a l l o w} = γ σ_{cr} = 0.758 \times 2662 = 2018$ MPa is more than twice the titanium yield of $830$ MPa. So the wall would yield at 830 MPa long before it reaches 2018 MPa buckling — the elastic formula's assumption (buckling in the elastic range) is violated. Why this step? The parent defined "thin" as $R / t ≳ 30$ . Below that, buckling is not the governing mode; comparing concrete numbers ( $2018 > 830$ ) proves it rather than hand-waving.

Verify: $σ_{a l l o w} = 2018$ MPa $> σ_{Y} = 830$ MPa ✓ — buckling does not govern; yield does. At $R / t = 25 < 30$ the formula returns a stress no real metal survives elastically → correctly signals stability is not the failure mode here. The math is self-consistent; the physics tells us to switch modes. ✓

Worked example Example 5 — Cell E: Thin-wall / huge-

$R / t$ danger zone A giant lightweight tank: $R / t = 1600$ , $E = 70$ GPa (aluminium), $ν = 0.3$ . Find $γ$ and $σ_{a l l o w}$ , and compare $γ$ against the $R / t = 600$ case of Example 1.

Forecast: Bigger $R / t$ (thinner) — does $γ$ go up or crash down?

Step 1 — $σ_{cr} = 0.605 \times 70000/1600 = 26.5$ MPa. Why this step? Thinner wall ⇒ smaller $t / R$ ⇒ smaller classical stress.

Step 2 — $ϕ = \frac{1}{16} 1600 = \frac{1}{16} (40) = 2.5$ , $γ = 1 - 0.901 (1 - e^{- 2.5}) = 0.173$ . Why this step? Larger $R / t$ pushes $ϕ$ up, so $e^{- ϕ} \to 0$ and $γ \to 1 - 0.901 = 0.099$ (the asymptotic floor).

Step 3 — $σ_{a l l o w} = 0.173 \times 26.5 = 4.58$ MPa. Why this step? The classical stress is already small; multiplying by the tiny $γ$ shows how brutally thin shells lose capacity — this is the number an engineer must actually respect.

Verify: Compare with Ex 1 ( $R / t = 600$ ): $γ$ fell from $0.298$ to $0.173$ as the wall got thinner ✓ — thinner shells are more imperfection-sensitive. As $R / t \to \infty$ , $γ \to 0.099$ , never zero: the SP-8007 curve has a floor. ✓ See NASA SP-8007 buckling of thin-walled cylinders.

Figure s02 below plots $γ$ against $R / t$ : the teal curve sags downward as the wall thins, the plum dashed line is the $0.099$ floor it never crosses, and the two dots mark Example 1 ( $R / t = 600$ ) and Example 5 ( $R / t = 1600$ ) — read the falling curve as "the thinner you build, the less of the promised strength you get to keep."

Worked example Example 6 — Cell F: Degenerate inputs (what breaks?)

Two "trick" inputs. (a) Set $t = 0$ . (b) Let $ν \to 1$ . What does the formula say, and does it make physical sense?

Forecast: Will $σ_{cr}$ go to zero, infinity, or blow up?

Step 1 (a) — With $t = 0$ : $σ_{cr} = 0.605 E (0/ R) = 0$ . A wall of zero thickness carries zero buckling stress. Why this step? No material means no restoring stiffness at all — the trivial correct limit.

Step 2 (a) — But $ϕ = \frac{1}{16} R /0 \to \infty$ and $γ \to 0.099$ . Product $σ_{a l l o w} = γ \cdot 0 = 0$ . Consistent: no wall, no strength. Why this step? Check both factors so no accidental $0 \times \infty$ slips through; here it cleanly gives $0$ .

Step 3 (b) — With $ν \to 1$ : denominator $3 (1 - ν^{2}) = 3 (1 - 1) = 0$ , so $σ_{cr} \to \infty$ . Why this step? $ν = 0.5$ is the incompressible limit for real solids; $ν \to 1$ is unphysical. The blow-up is the formula telling you the input is illegal, not a real infinite strength.

Verify: (a) $t = 0 \Rightarrow σ_{cr} = 0$ ✓, physically sensible. (b) $ν = 1 \Rightarrow$ division by zero ✓, flags an impossible material. Both degenerate cases behave: one gives a true limit, one signals a forbidden input.

Worked example Example 7 — Cell G: Inverse design problem (solve for

$t$ ) An aluminium interstage ( $E = 70$ GPa, $ν = 0.3$ , $R = 1.5$ m) must safely carry an axial load of $P = 1.2$ MN. Using the realistic allowable (with knockdown), find the minimum wall thickness $t$ . (Iterate, since $γ$ depends on $t$ .)

Forecast: Roughly millimetres or centimetres?

Step 1 — Required allowable stress if wall area is $A = 2 π R t$ : $σ_{a l l o w} = \frac{P}{2 π R t} .$ And $σ_{a l l o w} = γ (t) \cdot 0.605 E \frac{t}{R}$ . Why this step? We equate demand and capacity; both depend on $t$ , so we iterate.

Step 2 — First guess ignore knockdown drift, take $γ \approx 0.3$ . Then $\frac{P}{2 π R t} = 0.3 \cdot 0.605 E \frac{t}{R} \Rightarrow t^{2} = \frac{P}{2 π ( 0.3 ) ( 0.605 ) E} .$ $t^{2} = \frac{1.2 \times 1 0 ^{6}}{2 π ( 0.3 ) ( 0.605 ) ( 70 \times 1 0 ^{9} )} = 1.503 \times 1 0^{- 5} m^{2}$ , so $t = 3.88 \times 1 0^{- 3}$ m $= 3.88$ mm. Why this step? $R$ cancels (both sides carry $1/ R$ ), giving a clean $t^{2}$ equation. Keep everything in SI (metres) so units stay consistent.

Step 3 — Refine $γ$ at $t = 3.88 \times 1 0^{- 3}$ m. With $R = 1.5$ m, the ratio $R / t = 1.5/0.00388 = 386.6$ (both in metres — dimensionless). Then $ϕ = \frac{1}{16} 386.6 = 1.229$ , $γ = 1 - 0.901 (1 - e^{- 1.229}) = 0.363$ . Why this step? Our first $γ$ was a guess; recompute at the found thickness, with $R$ and $t$ in the same unit so $R / t$ is a pure number.

Step 4 — Re-solve with $γ = 0.363$ : $t^{2} = \frac{1.2 \times 1 0 ^{6}}{2 π ( 0.363 ) ( 0.605 ) ( 70 \times 1 0 ^{9} )} = 1.242 \times 1 0^{- 5} m^{2}$ , $t = 3.52 \times 1 0^{- 3}$ m $= 3.52$ mm. One more pass ( $R / t = 1.5/0.00352 = 426$ , $γ \approx 0.371$ ) shifts $t$ by under $0.02$ mm → converged near $t \approx 3.5$ mm. Why this step? Each iteration changes $γ$ less than the last (the SP-8007 curve is flat here), so the correction shrinks geometrically — stopping once the change is far below manufacturing tolerance is justified.

Verify: At $t = 3.52 \times 1 0^{- 3}$ m: $σ_{cr} = 0.605 (70 \times 1 0^{9}) (0.00352/1.5) = 99.4$ MPa, $R / t = 426$ , $γ = 0.371$ , $σ_{a l l o w} = 36.9$ MPa. Capacity $= 36.9 \times 1 0^{6} \times 2 π (1.5) (0.00352) = 1.22$ MN $≳ 1.2$ MN ✓. Millimetres, as forecast. See Rocket tank and interstage structural design.

Worked example Example 8 — Cell H: Real-world word problem

A launch team stacks a 40-tonne upper stage on top of an aluminium interstage during vertical integration on the pad (no thrust yet, just gravity). The interstage has $R = 1.6$ m, $t = 2.5$ mm, $E = 70$ GPa, $ν = 0.3$ . Take $g = 9.81 m/s^{2}$ . Does it survive the stack-up, with a safety factor of 1.5?

Forecast: 40 tonnes sounds huge — will a 2.5 mm wall hold it?

Step 1 — Applied axial load: $P = m g = 40000 \times 9.81 = 3.92 \times 1 0^{5}$ N $= 0.392$ MN. Why this step? On the pad the only axial load is the weight above; convert tonnes to newtons.

Step 2 — Classical $σ_{cr} = 0.605 (70 \times 1 0^{9}) (0.0025/1.6) = 66.2$ MPa. Area $A = 2 π (1.6) (0.0025) = 0.02513 m^{2}$ . Why this step? Build the capacity side.

Step 3 — Knockdown: $R / t = 1.6/0.0025 = 640$ , $ϕ = \frac{1}{16} 640 = 1.581$ , $γ = 1 - 0.901 (1 - e^{- 1.581}) = 0.284$ . So $σ_{a l l o w} = 0.284 \times 66.2 = 18.8$ MPa. Why this step? The bare classical stress is optimistic; the knockdown converts it into the stress a real, slightly-dented interstage can actually hold, which is what safety must be judged on.

Step 4 — Allowable load $P_{a l l o w} = σ_{a l l o w} A = 18.8 \times 1 0^{6} \times 0.02513 = 0.472$ MN. Why this step? Turn the allowable stress back into a force so it can be compared directly with the applied weight $P$ .

Step 5 — Safety factor $= P_{a l l o w} / P = 0.472/0.392 = 1.20$ . Why this step? Compare capacity to demand as a ratio.

Verify: Required SF is $1.5$ , achieved SF is $1.20 < 1.5$ ✗ — it does not meet the margin. Physically believable: a 2.5 mm can holding 40 t sits on a knife-edge. The team must thicken the wall or add stiffeners. ✓

Worked example Example 9 — Cell I: Exam twist (the hidden trap)

Exam question: "A cylinder has classical $σ_{cr} = 150$ MPa and knockdown $γ = 0.4$ . A student computes the allowable load as $P = σ_{cr} \times A$ using the classical stress, then separately says 'the stress is safe because $150 < 250$ MPa yield.' Find both of the student's mistakes and the correct allowable stress." (Given $A = 0.02 m^{2}$ , $σ_{Y} = 250$ MPa.)

Forecast: Spot the two traps before computing.

Step 1 — Trap 1: The allowable stress must include the knockdown: $σ_{a l l o w} = γ σ_{cr} = 0.4 \times 150 = 60$ MPa, not 150 MPa. Why this step? Using the perfect-cylinder value ignores imperfection sensitivity — the whole point of $γ$ .

Step 2 — Trap 2: Comparing to yield is the wrong test. The governing limit is the lower of buckling-allowable and yield: $min (60, 250) = 60$ MPa. Buckling governs. Why this step? A structure fails at whichever mode triggers first; $60 < 250$ so stability wins.

Step 3 — Correct allowable load: $P_{a l l o w} = σ_{a l l o w} A = 60 \times 1 0^{6} \times 0.02 = 1.2$ MN. The student's $150 \times 1 0^{6} \times 0.02 = 3.0$ MN over-predicts by $2.5 \times$ . Why this step? The load is the quantity the launch vehicle actually experiences; converting the correct stress into a force (and contrasting with the student's inflated force) exposes the size of the error.

Verify: Correct $σ_{a l l o w} = 60$ MPa ✓; correct $P_{a l l o w} = 1.2$ MN vs student's $3.0$ MN ✓; the $2.5 \times$ error is exactly $1/ γ = 1/0.4$ ✓. Two traps: (1) forgot $γ$ , (2) compared against yield instead of taking the min. ✓

Recall Which cell did each example hit? (click to reveal)

Ex1 A standard · Ex2 B scaling · Ex3 C buckle-vs-yield · Ex4 D thick limit · Ex5 E thin/danger · Ex6 F degenerate · Ex7 G inverse design · Ex8 H word problem · Ex9 I exam trap. Every cell covered.

Active recall

In an inverse (

t

-solving) problem, why must you iterate?

Because

γ

depends on

R / t

, which depends on the unknown

t

— one pass gives a guess, then you refine

γ

and re-solve.

R / t \to \infty

, what value does

γ

approach?

The floor

1 - 0.901 = 0.099

; thinner shells never quite reach zero but get very fragile.

In Example 3, why isn't "302.5 MPa > 250 MPa yield" the right comparison?

You must apply the knockdown first:

σ_{a l l o w} = 0.357 \times 302.5 = 108

MPa, which is well below yield — buckling governs.

Two mistakes in the Example 9 exam trap?

(1) Forgetting the knockdown

γ

on the stress; (2) comparing to yield instead of taking the minimum of buckling-allowable and yield.

What end conditions and length range does the classical

0.605 E t / R

assume?

Simply-supported ends and a moderately-long cylinder; very short shells buckle higher, very long ones become Euler columns.

Links: Shell buckling — thin-walled cylinder under axial load · Imperfection sensitivity and knockdown factors · NASA SP-8007 buckling of thin-walled cylinders · Yield vs stability failure modes · Rocket tank and interstage structural design · Euler column buckling · Hoop stress in pressurised cylinders · Plate bending and flexural rigidity D