Shell buckling — thin-walled cylinder under axial load: Exercises

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Throughout we use the two headline results from the parent note: $σ_{cr} = \frac{E}{3 ( 1 - ν ^{2} )} \frac{t}{R} \approx 0.605 E \frac{t}{R} (ν = 0.3), P_{cr} = σ_{cr} (2 π R t) .$ And the knockdown (from Imperfection sensitivity and knockdown factors, NASA SP-8007 buckling of thin-walled cylinders): $γ = 1 - 0.901 (1 - e^{- ϕ}), ϕ = \frac{1}{16} R / t, σ_{a l l o w} = γ σ_{cr} .$

Level 1 — Recognition

Recall Solution 1.1

It is a stiffness (stability) failure — the straight shape stops being a stable equilibrium at $P_{cr}$ , and the shell jumps to a buckled shape. The material is nowhere near yield; see Yield vs stability failure modes. The deciding ratio is the ==radius-to-thickness ratio $R / t$ == (equivalently $t / R$ ). Larger $R / t$ means a thinner, "floppier" wall and a lower buckling stress, since $σ_{cr} \propto t / R$ .

Recall Solution 1.2

(a) Raising Young's modulus $E$ (stiffer material) raises $σ_{cr}$ linearly — a stiffer wall resists dimpling harder. (b) Raising radius $R$ lowers $σ_{cr}$ (it's in the denominator): a wider can is flatter locally, so its membrane hoop stiffness $\propto E t / R^{2}$ falls. (c) No $L$ appears. In the classical minimisation the optimum buckle wavelength is short compared to the tube, so the length cancels out — this is the "many small diamonds" regime. (d) $0.605 = \frac{1}{3 ( 1 - ν ^{2} )}$ evaluated at $ν = 0.3$ : $3 (1 - 0.09) = 2.73 = 1.652$ , and $1/1.652 = 0.605$ .

Level 2 — Application

Recall Solution 2.1

Step 1 — ratio. $t / R = 0.0025/1.0 = 2.5 \times 1 0^{- 3}$ . Why: the formula needs only this ratio. Step 2 — stress. $σ_{cr} = 0.605 \times 200 \times 1 0^{9} \times 2.5 \times 1 0^{- 3} = 3.025 \times 1 0^{8}$ Pa $= 302.5$ MPa. Why: $ν = 0.3$ gives the $0.605$ coefficient. Step 3 — bearing area. $A = 2 π R t = 2 π (1.0) (0.0025) = 0.01571$ m². Why: load = stress × the thin ring of wall that carries it. Step 4 — load. $P_{cr} = σ_{cr} A = 3.025 \times 1 0^{8} \times 0.01571 = 4.75 \times 1 0^{6}$ N $\approx 4.75$ MN.

Recall Solution 2.2

Step 1 — $R / t$ . $R / t = 1.0/0.0025 = 400$ . Why: $ϕ$ depends on $R / t$ . Step 2 — $ϕ$ . $ϕ = \frac{1}{16} 400 = \frac{1}{16} (20) = 1.25$ . Why: SP-8007 measures "how imperfection-prone" via $R / t$ . Step 3 — $γ$ . $γ = 1 - 0.901 (1 - e^{- 1.25}) = 1 - 0.901 (1 - 0.2865) = 1 - 0.901 (0.7135) = 0.357$ . Step 4 — allowable. $σ_{a l l o w} = 0.357 \times 302.5 = 108.0$ MPa. Barely a third of the perfect-shell value — a knife-edge structure.

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

Level 3 — Analysis

Recall Solution 3.1

(a) $σ_{cr} \propto t / R$ , so doubling $t$ doubles $σ_{cr}$ (factor $2$ ). (b) $P_{cr} = σ_{cr} (2 π R t) \propto (t / R) \cdot (R t) = t^{2}$ . So doubling $t$ makes $P_{cr}$ rise by $2^{2} =$ factor 4. Why: both the stress capacity and the bearing area grow linearly in $t$ . (c) $σ_{cr} \propto 1/ R$ , so doubling $R$ halves $σ_{cr}$ (factor $\frac{1}{2}$ ). A bigger, flatter can is easier to dimple.

Recall Solution 3.2

Step 1 — the coefficient is $1/ 3 (1 - ν^{2})$ . Why: $0.605$ was only the $ν = 0.3$ special case. Step 2 — plug $ν = 0.34$ . $1 - ν^{2} = 1 - 0.1156 = 0.8844$ . Then $3 \times 0.8844 = 2.653$ , $2.653 = 1.629$ . Step 3 — invert. coefficient $= 1/1.629 = 0.614$ . So titanium gives $σ_{cr} \approx 0.614 E t / R$ — slightly higher coefficient than aluminium/steel, because a larger Poisson ratio stiffens the plate rigidity $D$ a touch (recall $D \propto 1/ (1 - ν^{2})$ ; see Plate bending and flexural rigidity D).

Recall Solution 3.3

Shell A: $ϕ = \frac{1}{16} 200 = \frac{1}{16} (14.14) = 0.884$ . $γ_{A} = 1 - 0.901 (1 - e^{- 0.884}) = 1 - 0.901 (1 - 0.4131) = 1 - 0.901 (0.5869) = 0.471$ . Shell B: $ϕ = \frac{1}{16} 800 = \frac{1}{16} (28.28) = 1.768$ . $γ_{B} = 1 - 0.901 (1 - e^{- 1.768}) = 1 - 0.901 (1 - 0.1708) = 1 - 0.901 (0.8292) = 0.253$ . Shell B ( $R / t = 800$ ) has the smaller $γ$ . Why: thinner walls have imperfection amplitude comparable to $t$ relative to the wall, and the buckle modes crowd together (near-degenerate), so a real dent knocks the load down harder. Larger $R / t \Rightarrow$ larger $ϕ \Rightarrow$ smaller $γ$ .

Level 4 — Synthesis

Recall Solution 4.1

Step 1 — classical buckling. $σ_{cr} = 0.605 \times 70000 / 300 = 141.2$ MPa (using $E$ in MPa: $70000$ ). Why: $σ_{cr} = 0.605 E / (R / t)$ . Step 2 — knockdown. $ϕ = \frac{1}{16} 300 = \frac{1}{16} (17.32) = 1.083$ . $γ = 1 - 0.901 (1 - e^{- 1.083}) = 1 - 0.901 (1 - 0.3386) = 1 - 0.901 (0.6614) = 0.404$ . Step 3 — allowable buckling stress. $σ_{a l l o w} = 0.404 \times 141.2 = 57.0$ MPa. Step 4 — compare to yield. $57.0$ MPa $≪ σ_{Y} = 270$ MPa. The shell buckles first, at ~57 MPa. Stability governs, not strength — see Yield vs stability failure modes. You cannot design to $σ_{Y}$ here.

Recall Solution 4.2

Qualitative: an inward radial dimple is opposed by internal pressure trying to push the wall back outward (pressure resists inward deflection). This adds a stabilising restoring term, so pressurised cylinders buckle at a higher axial load than empty ones — the "pressure stabilisation" used in real rocket tanks (a can is far stronger when its lid is on and it's slightly pressurised). Hoop stress: $σ_{θ} = pR / t = (0.3 \times 1 0^{6}) (1.8) / (0.003) = 1.8 \times 1 0^{8}$ Pa $= 180$ MPa. Why: this is the membrane tension the pressure creates around the circumference. Comment: $180$ MPa is a substantial membrane tension — the same order as classical buckling stresses — so this pressure meaningfully stiffens the wall against axial buckling and must be part of a real design margin (though we must also check that $σ_{θ}$ itself stays below yield).

Level 5 — Mastery

Recall Solution 5.1

Design requirement: allowable load $\geq SF \times P = 1.5 \times 2.0 = 3.0$ MN. The allowable load is $P_{a l l o w} = γ σ_{cr} (2 π R t) = γ (0.605 E t / R) (2 π R t) = γ \cdot 0.605 E \cdot 2 π t^{2}$ . Note $R$ cancels in the load except through $γ (R / t)$ . So $P_{a l l o w} = 1.209 π E γ t^{2}$ .

Iteration 1 — guess $t = 4$ mm.

$R / t = 1.5/0.004 = 375$ ; $ϕ = \frac{1}{16} 375 = \frac{1}{16} (19.36) = 1.210$ ; $γ = 1 - 0.901 (1 - e^{- 1.210}) = 1 - 0.901 (1 - 0.2982) = 1 - 0.901 (0.7018) = 0.368$ .
$P_{a l l o w} = 1.209 π (70 \times 1 0^{9}) (0.368) (0.004)^{2} = 1.209 π (70 \times 1 0^{9}) (0.368) (1.6 \times 1 0^{- 5})$ .
Compute: $1.209 π = 3.798$ ; $\times 70 \times 1 0^{9} = 2.659 \times 1 0^{11}$ ; $\times 0.368 = 9.78 \times 1 0^{10}$ ; $\times 1.6 \times 1 0^{- 5} = 1.566 \times 1 0^{6}$ N $= 1.57$ MN. Too small (need 3.0 MN).

Iteration 2 — guess $t = 6$ mm.

$R / t = 250$ ; $ϕ = \frac{1}{16} 250 = \frac{1}{16} (15.81) = 0.988$ ; $γ = 1 - 0.901 (1 - e^{- 0.988}) = 1 - 0.901 (1 - 0.3724) = 1 - 0.901 (0.6276) = 0.434$ .
$P_{a l l o w} = 1.209 π \times 70 \times 1 0^{9} \times 0.434 \times (0.006)^{2} = 3.798 \times 70 \times 1 0^{9} \times 0.434 \times 3.6 \times 1 0^{- 5}$ .
$= 2.659 \times 1 0^{11} \times 0.434 = 1.154 \times 1 0^{11}$ ; $\times 3.6 \times 1 0^{- 5} = 4.15 \times 1 0^{6}$ N $= 4.15$ MN. More than enough.

Iteration 3 — guess $t = 5$ mm.

$R / t = 300$ ; $ϕ = \frac{1}{16} 300 = 1.083$ ; $γ = 0.404$ (from Ex 4.1).
$P_{a l l o w} = 2.659 \times 1 0^{11} \times 0.404 \times (0.005)^{2} = 1.074 \times 1 0^{11} \times 2.5 \times 1 0^{- 5} = 2.69 \times 1 0^{6}$ N $= 2.69$ MN. Just short of 3.0 MN.

Iteration 4 — guess $t = 5.4$ mm.

$R / t = 277.8$ ; $ϕ = \frac{1}{16} 277.8 = \frac{1}{16} (16.67) = 1.042$ ; $γ = 1 - 0.901 (1 - e^{- 1.042}) = 1 - 0.901 (1 - 0.3528) = 1 - 0.901 (0.6472) = 0.417$ .
$P_{a l l o w} = 2.659 \times 1 0^{11} \times 0.417 \times (0.0054)^{2} = 1.109 \times 1 0^{11} \times 2.916 \times 1 0^{- 5} = 3.23 \times 1 0^{6}$ N $= 3.23$ MN. OK.

Answer: minimum thickness $\approx 5.4 mm$ (round up to a standard gauge, e.g. 5.5–6 mm, for margin). Then $R / t = 1.5/0.0054 \approx 278$ , which is $≫ 30$ , so the thin-shell buckling formula is fully valid. Notice: had we naively used the classical stress (no $γ$ ), we'd have picked a wall about $0.417 \approx 0.65$ times as thick — dangerously thin.

Recall Solution 5.2

(a) $σ_{θ} = pR / t = (0.2 \times 1 0^{6}) (1.5) / (0.0054) = 5.556 \times 1 0^{7}$ Pa $= 55.6$ MPa. This is well below $σ_{Y} = 270$ MPa (margin $\approx 4.9 \times$ ), so hoop yield is not a concern at this pressure. (b) The final wall thickness is set by the most demanding of: (i) axial buckling with knockdown $γ σ_{cr}$ (the governing case here), while (ii) internal pressure only helps buckling (stabilisation) and (iii) hoop yield $pR / t < σ_{Y}$ is a separate strength check that must also pass — you size for buckling, then verify pressure doesn't yield the wall.

Recall Quick self-audit (click to reveal)

If you can do these from a blank page, you've mastered D4: Which ratio sets $σ_{cr}$ , and does $L$ appear? ::: $t / R$ sets it; length $L$ does not appear in axial shell buckling. Load-bearing area of the wall? ::: $A = 2 π R t$ (the thin annulus), never $π R^{2}$ . Why must knockdown be recomputed per shell? ::: $γ = 1 - 0.901 (1 - e^{- ϕ})$ with $ϕ = \frac{1}{16} R / t$ — it depends on $R / t$ . Why is the L5 thickness solve iterative? ::: $γ$ depends on $t$ through $R / t$ , so $P_{a l l o w} (t)$ is implicit in $t$ .