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

3.6.7 · D4 · Physics › Spacecraft Structures & Systems Engineering › Shell buckling — thin-walled cylinder under axial load

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Poore note mein hum parent note ke do headline results use karte hain: $σ_{cr} = \frac{E}{3 ( 1 - ν ^{2} )} \frac{t}{R} \approx 0.605 E \frac{t}{R} (ν = 0.3), P_{cr} = σ_{cr} (2 π R t) .$ Aur 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

Yeh ek stiffness (stability) failure hai — straight shape $P_{cr}$ par ek stable equilibrium nahi rehti, aur shell ek buckled shape mein jump kar jaati hai. Material yield ke paas kahin nahi hai; dekho Yield vs stability failure modes. Deciding ratio hai ==radius-to-thickness ratio $R / t$ == (equivalently $t / R$ ). Bada $R / t$ matlab thinner, "floppier" wall aur lower buckling stress, kyunki $σ_{cr} \propto t / R$ .

Recall Solution 1.2

(a) Young's modulus $E$ (stiffer material) badhane se $σ_{cr}$ linearly badhta hai — ek stiffer wall dimpling ko zyada strongly resist karti hai. (b) Radius $R$ badhane se $σ_{cr}$ ghatta hai (yeh denominator mein hai): ek wider can locally flatter hoti hai, toh uska membrane hoop stiffness $\propto E t / R^{2}$ girata hai. (c) Nahi, $L$ appear nahi karta. Classical minimisation mein optimum buckle wavelength tube se chhoti hoti hai, toh length cancel ho jaati hai — yeh "many small diamonds" regime hai. (d) $0.605 = \frac{1}{3 ( 1 - ν ^{2} )}$ evaluate kiya $ν = 0.3$ par: $3 (1 - 0.09) = 2.73 = 1.652$ , aur $1/1.652 = 0.605$ .

Level 2 — Application

Recall Solution 2.1

Step 1 — ratio. $t / R = 0.0025/1.0 = 2.5 \times 1 0^{- 3}$ . Kyun: formula ko sirf yahi ratio chahiye. 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. Kyun: $ν = 0.3$ deta hai $0.605$ coefficient. Step 3 — bearing area. $A = 2 π R t = 2 π (1.0) (0.0025) = 0.01571$ m². Kyun: load = stress × wall ka thin ring jo ise carry karta hai. 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 — $R / t$ . $R / t = 1.0/0.0025 = 400$ . Kyun: $ϕ$ depend karta hai $R / t$ par. Step 2 — $ϕ$ . $ϕ = \frac{1}{16} 400 = \frac{1}{16} (20) = 1.25$ . Kyun: SP-8007 "imperfection-prone" measure karta hai $R / t$ ke through. 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. Perfect-shell value ka barely ek third — ek knife-edge structure.

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

Level 3 — Analysis

Recall Solution 3.1

(a) $σ_{cr} \propto t / R$ , toh $t$ double karne se $σ_{cr}$ double hota hai (factor $2$ ). (b) $P_{cr} = σ_{cr} (2 π R t) \propto (t / R) \cdot (R t) = t^{2}$ . Toh $t$ double karne se $P_{cr}$ $2^{2} =$ factor 4 se badhta hai. Kyun: stress capacity aur bearing area dono linearly $t$ mein badhte hain. (c) $σ_{cr} \propto 1/ R$ , toh $R$ double karne se $σ_{cr}$ half ho jaata hai (factor $\frac{1}{2}$ ). Ek bada, flatter can dimple karna asaan hota hai.

Recall Solution 3.2

Step 1 — coefficient $1/ 3 (1 - ν^{2})$ hai. Kyun: $0.605$ sirf $ν = 0.3$ ka special case tha. Step 2 — $ν = 0.34$ plug karo. $1 - ν^{2} = 1 - 0.1156 = 0.8844$ . Phir $3 \times 0.8844 = 2.653$ , $2.653 = 1.629$ . Step 3 — invert karo. coefficient $= 1/1.629 = 0.614$ . Toh titanium deta hai $σ_{cr} \approx 0.614 E t / R$ — aluminium/steel se thoda zyada coefficient, kyunki bada Poisson ratio plate rigidity $D$ ko thoda stiffen karta hai (yaad karo $D \propto 1/ (1 - ν^{2})$ ; dekho 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 ( $R / t = 800$ ) ki $γ$ chhoti hai. Kyun: thinner walls mein imperfection amplitude $t$ ke relative comparable hota hai wall ke respect mein, aur buckle modes crowd together hote hain (near-degenerate), toh ek real dent load ko zyada neeche girata hai. Bada $R / t \Rightarrow$ bada $ϕ \Rightarrow$ chhota $γ$ .

Level 4 — Synthesis

Recall Solution 4.1

Step 1 — classical buckling. $σ_{cr} = 0.605 \times 70000 / 300 = 141.2$ MPa ( $E$ MPa mein use karke: $70000$ ). Kyun: $σ_{cr} = 0.605 E / (R / t)$ . 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 — yield se compare karo. $57.0$ MPa $≪ σ_{Y} = 270$ MPa. Shell pehle buckle karti hai, ~57 MPa par. Stability govern karta hai, strength nahi — dekho Yield vs stability failure modes. Tum yahan $σ_{Y}$ ke liye design nahi kar sakte.

Recall Solution 4.2

Qualitative: ek inward radial dimple ka internal pressure opposition karta hai jo wall ko wapas outward push karne ki koshish karta hai (pressure inward deflection ko resist karta hai). Yeh ek stabilising restoring term add karta hai, toh pressurised cylinders ek higher axial load par buckle hoti hain empty ones se — "pressure stabilisation" jo real rocket tanks mein use hoti hai (ek can bahut stronger hoti hai jab uska lid on ho aur woh thodi pressurised ho). 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. Kyun: yeh woh membrane tension hai jo pressure circumference ke around create karta hai. Comment: $180$ MPa ek substantial membrane tension hai — classical buckling stresses ke same order mein — toh yeh pressure wall ko axial buckling ke against meaningfully stiffen karta hai aur ek real design margin ka hissa hona chahiye (halanki hamen yeh bhi check karna hoga ki $σ_{θ}$ khud yield se neeche rahe).

Level 5 — Mastery

Recall Solution 5.1

Design requirement: allowable load $\geq SF \times P = 1.5 \times 2.0 = 3.0$ MN. Allowable load hai $P_{a l l o w} = γ σ_{cr} (2 π R t) = γ (0.605 E t / R) (2 π R t) = γ \cdot 0.605 E \cdot 2 π t^{2}$ . Note karo ki $R$ load mein cancel ho jaata hai siwa $γ (R / t)$ ke through. Toh $P_{a l l o w} = 1.209 π E γ t^{2}$ .

Iteration 1 — $t = 4$ mm guess karo.

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

Iteration 2 — $t = 6$ mm guess karo.

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

Iteration 3 — $t = 5$ mm guess karo.

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

Iteration 4 — $t = 5.4$ mm guess karo.

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

Answer: minimum thickness $\approx 5.4 mm$ (standard gauge tak round up karo, jaise 5.5–6 mm, margin ke liye). Tab $R / t = 1.5/0.0054 \approx 278$ , jo $≫ 30$ hai, toh thin-shell buckling formula fully valid hai. Notice karo: agar hum naively classical stress use karte (koi $γ$ nahi), toh humne $0.417 \approx 0.65$ times chhoti wall pick ki hoti — 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. Yeh $σ_{Y} = 270$ MPa se kaafi neeche hai (margin $\approx 4.9 \times$ ), toh is pressure par hoop yield concern nahi hai. (b) Final wall thickness sabse demanding se set hoti hai: (i) axial buckling with knockdown $γ σ_{cr}$ (yahan governing case), jabki (ii) internal pressure sirf buckling mein help karta hai (stabilisation) aur (iii) hoop yield $pR / t < σ_{Y}$ ek alag strength check hai jo bhi pass hona chahiye — tum buckling ke liye size karo, phir verify karo ki pressure wall ko yield nahi karta.

Recall Quick self-audit (click to reveal)

Agar tum yeh ek blank page se kar sako, tumne D4 master kar liya hai: Kaun sa ratio $σ_{cr}$ set karta hai, aur kya $L$ appear karta hai? ::: $t / R$ set karta hai; length $L$ axial shell buckling mein appear nahi karta. Wall ka load-bearing area? ::: $A = 2 π R t$ (thin annulus), kabhi $π R^{2}$ nahi. Knockdown har shell ke liye recompute kyun karna padta hai? ::: $γ = 1 - 0.901 (1 - e^{- ϕ})$ with $ϕ = \frac{1}{16} R / t$ — yeh $R / t$ par depend karta hai. L5 thickness solve iterative kyun hai? ::: $γ$ depend karta hai $t$ par $R / t$ ke through, toh $P_{a l l o w} (t)$ $t$ mein implicit hai.