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

3.6.7 · D1 · Physics › Spacecraft Structures & Systems Engineering › Shell buckling — thin-walled cylinder under axial load

Is page par assume kiya gaya hai ki parent page ke notation ke baare mein tumhe kuch nahi pata. Hum har letter, har ratio aur har picture ko zero se build karte hain, ek aisi order mein jahan har idea apne pehle wale idea par lean karta hai.

1 · Cheez khud: thin-walled cylinder

Kisi bhi formula se pehle, cheez ko picture karo. s01 mein, magenta arrow follow karo (woh $R$ hai, round top ki spoke), left edge par orange band (woh wall thickness $t$ hai), aur right par navy double-arrow (woh height $L$ hai).

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

2 · Usse squeeze karna: force, stress aur area

Tank ko upar stack kiye weight aur thrust ki wajah se apni axis ke along squeeze kiya jaata hai.

3 · Material: $E$ aur $ν$

Do numbers metal khud ko describe karte hain, shape se independent.

4 · Do failure modes: yield vs stability

Parent page ka poora point yahi hai ki yeh cylinder obvious tarike se fail nahi karta.

5 · Do restoring effects (tug-of-war)

Jab wall thodi radial amount $w$ andar ya bahar dimple karti hai, do "springs" wapas fight karti hain.

s04 mein, top strip (small $k$ ) mein kuch long humps hain; bottom strip (large $k$ ) mein kai tight humps hain. Same wall, do possible fold patterns.

Definition Membrane (hoop) spring constant

$K = \frac{E t}{R ^{2}}$ Wall ko $w$ radially push karna circumference (hoop) ko stretch ya squeeze karta hai. Stored stretch-energy dimple se ladti hai, aur uski stiffness ek spring constant hai $K = \frac{E t}{R ^{2}} .$ Notice karo: yahan koi $(1 - ν^{2})$ nahi. Hoop stretch circumferential fibres ka uniaxial stretch hai — ek single direction ka lengthen hona — toh yeh plain axial stiffness $E t$ use karta hai, two-directional plate stiffness nahi. Yahi honest reason hai ki $ν$ $D$ mein appear karta hai lekin $K$ mein nahi: $D$ ek bending (two-way plate) effect hai, $K$ ek stretch (one-way membrane) effect hai. (Fuller treatments mein ek mild $ν$ correction rehta hai, lekin is level par $K = E t / R^{2}$ woh clean number hai jo $2 D K$ mein enter karta hai.)

Picture: round hoop smaller ya bigger banane se resist karna — yeh sirf isliye exist karta hai kyunki wall curved hai ( $1/ R$ ). Ek flat plate mein koi hoop nahi, toh aisa koi term nahi. Yeh wahi circumferential stretch idea hai jaise Hoop stress in pressurised cylinders.

Intuition Hoop energy

$K / k^{4}$ kyun scale karti hai Yahan subtle part hai. Jab wall ek jagah dimple karti hai, material ko kahin jaana hota hai. Ek long gentle fold (small $k$ ) ke liye, wall sach mein radially in aur out push karti hai, toh poori circumference stretch honi chahiye — badi hoop energy. Ek tight short fold (bada $k$ ) ke liye, metal hoop stretch karne ki jagah sideways wall ke along neighbouring dimple mein slosh kar sakta hai, toh almost koi circumference change nahi — tiny hoop energy.

Buckle pattern ki geometry ke through ise work karne par (radial motion jo folds ko compatible rakhne ke liye chahiye woh tighter hone par fall off hoti hai) hoop energy milti hai $U_{hoop} \propto \frac{K}{k ^{4}} w_{0}^{2} .$ Words mein: long loose folds hoop ke liye expensive hain (woh real circumference stretch force karte hain); tight folds material ko sideways shuffle karke cheat karte hain. Yahi hai $1/ k^{4}$ — bending term ka exact mirror image.

Intuition Do springs kyun, optimum

$k^{*}$ , aur unka geometric mean Dono costs side by side rakho (per unit dimple depth): $U (k) = D k^{4} + \frac{K}{k ^{4}} .$ Bending tight folds ke liye worst hai (bada $k$ ); hoop loose folds ke liye worst hai (chhota $k$ ). Wall woh $k$ choose karti hai jo total ko minimum banaye. Derivative zero set karo: $\frac{d U}{d k} = 4 D k^{3} - \frac{4 K}{k ^{5}} = 0 ⟹ k^{4} \cdot k^{4} = \frac{K}{D} ⟹ k^{*} = (\frac{K}{D})^{1/8} .$ $K = E t / R^{2}$ aur $D = E t^{3} / [12 (1 - ν^{2})]$ substitute karne par natural fold size milti hai $k^{*} \sim [12 (1 - ν^{2}) / (R^{2} t^{2})]^{1/8} \sim 1/ R t$ , yani wavelength $λ^{*} = 2 π / k^{*} \sim R t$ — dimples lagbhag $R t$ wide hote hain, $R$ ya $L$ dono se chhote.

$k^{*}$ par dono costs equal hote hain ( $D k^{4} = K / k^{4}$ ), aur $a + b \geq 2 ab$ se (equality jab $a = b$ ) minimum value geometric mean hai $min_{k} (D k^{4} + \frac{K}{k ^{4}}) = 2 D K .$ Isliye parent ka clean result padhta hai $σ_{cr} t = 2 D \cdot \frac{E t}{R ^{2}}$ — do springs exactly balance, aur buckle length $L$ vanish ho gayi (wall ne apna natural fold size $\sim R t$ find kar liya, tube height se independent — provided tube bahut short na ho, dekho §6).

Formula

$2 D K$ se boxed answer tak — step by step algebra Balanced result se shuru karo aur $K = E t / R^{2}$ aur $D = \frac{E t ^{3}}{12 ( 1 - ν ^{2} )}$ substitute karo: $σ_{cr} t = 2 \frac{E t ^{3}}{12 ( 1 - ν ^{2} )} \cdot \frac{E t}{R ^{2}} .$ Root ke andar multiply karo ( $t^{3} \cdot t = t^{4}$ , $E \cdot E = E^{2}$ ): $σ_{cr} t = 2 \frac{E ^{2} t ^{4}}{12 ( 1 - ν ^{2} ) R ^{2}} .$ Perfect squares ko root se bahar nikalo ( $E^{2} = E$ , $t^{4} = t^{2}$ , $R^{2} = R$ ): $σ_{cr} t = 2 \cdot \frac{E t ^{2}}{R 12 ( 1 - ν ^{2} )} .$ Dono sides ko $t$ se divide karo (yahi reason hai hum divide karte hain — hume stress chahiye tha, stress×thickness nahi): $σ_{cr} = \frac{2 E t}{R 12 ( 1 - ν ^{2} )} .$ Constant simplify karo: $12 = 23$ , toh upar ka $2$ $12 = 23$ ke andar ke $2$ ko cancel karta hai, $3$ bachta hai: $σ_{cr} = \frac{E}{3 ( 1 - ν ^{2} )} \frac{t}{R} \approx 0.605 E \frac{t}{R} (ν = 0.3) .$ Pehle mila har $t$ , $R$ , $E$ , $ν$ exactly ek jagah land karta hai — yahi tug-of-war ka poora payoff hai.

6 · Length $L$ aur ends: local vs global buckling

Humne $L$ define kiya tha lekin woh $σ_{cr}$ se drop out ho gaya. Yeh sirf middle regime mein true hai — ends aur aspect ratio $L / R$ decide karte hain ki kis tarah ki buckling hoti hai.

Definition Aspect ratio

$L / R$ aur do buckling families

Local shell buckling (short/medium tube): wall chhote diamond dimples mein fold hoti hai size $\sim R t$ ke, aur $σ_{cr} = 0.605 E t / R$ $L$ ko ignore karta hai, kyunki natural fold tube se bahut chhoti hai. Yahi woh regime hai jo parent page solve karta hai. s05 mein, left panel.
Global (Euler) column buckling (bahut long, slender tube): poora cylinder ek bent straw ki tarah sideways bow karta hai, exactly Euler column buckling mode, aur ab $L$ strongly matter karta hai — critical load $L^{2}$ badhne par fall karta hai. s05 mein, right panel.

Quantitative switch-over: local shell stress $0.605 E t / R$ ko ek tube ke global Euler stress se compare karo ( $σ_{Euler} = π^{2} E (R^{2} + ...) /...$ ; ek thin ring ke liye Euler stress $\approx \frac{π ^{2}}{2} E (R / L)^{2}$ hai pinned ends ke liye). Dono equal set karne par, crossover roughly yahan hota hai $\frac{L}{R} \approx \frac{π ^{2} /2}{0.605} \cdot \frac{R}{t} \approx 2.85 R / t .$ Concrete rule of thumb: $L / R ≲ 2.85 R / t$ ke liye (yani zyaatar tanks) local dimpling govern karta hai aur tum $0.605 E t / R$ use karte ho; sirf is se aage bahut slender tubes ke liye Euler formula par switch karo. Example: $R / t = 600$ crossover $L / R \approx 2.85 600 \approx 70$ deta hai — ek tube ko global bowing jeeetne se pehle ~70 radii long hona padega.

7 · Reality gap: knockdown $γ$ aur $ϕ$

Prerequisite map

Equipment checklist

Right side cover karo aur dekho kya tum har ek memory se bata sakte ho.

R / t

ratio kya measure karta hai, aur ratio kyun raw sizes nahi?

Wall kitni "thin/flimsy" hai; proportion, metres nahi, buckling decide karta hai — same

R / t

utna hi tinny lagta hai.

Stress

σ

ki definition aur uske units?

Force per unit area,

σ = P / A

; pascals (

N / m^{2}

) mein measure hota hai, usually MPa.

Axial load kaunsa area carry karta hai, aur uska formula?

Sirf wall ring:

A = 2 π R t

Young's modulus

E

kya describe karta hai?

Material stiffness in stretch — unit fractional stretch per stress kitna chahiye.

Poisson's ratio

ν

kya describe karta hai, aur topic ko yeh kyun chahiye?

Lengthwise stretch per sideways thinning; yeh

D

mein

(1 - ν^{2})

plate-stiffening factor produce karta hai.

Yield aur buckling failure mein difference?

Yield = material crush (strength); buckling = shape unstable ho jaati hai (stiffness) — buckling pehle strike karta hai yahan.

Radial displacement

w

kya hai?

Wall perfectly round se in/out kitna poke karti hai — ek dimple ki depth.

Buckle wavenumber

k

kya count karta hai?

Folds kitne tightly packed hain — bada

k

kai short folds, chhota

k

few long folds; wavelength

λ = 2 π / k

Flexural rigidity

D

ka formula aur meaning?

D = E t^{3} / [12 (1 - ν^{2})]

— wall ko curve karne ka resistance; tight folds ke liye

D k^{4}

cost.

Bending energy

D k^{4}

kyun scale karti hai?

Curvature fold ki 2nd derivative hai,

\propto k^{2}

; bending energy

\propto

curvature

^{2} \propto (k^{2})^{2} = k^{4}

Membrane spring constant

K

(exact) kya hai, aur uski cost kya hai?

K = E t / R^{2}

(koi

ν

nahi — yeh ek one-way hoop stretch hai, two-way plate bending nahi);

K / k^{4}

cost, loose folds ke liye worst.

Hoop energy

1/ k^{4}

kyun scale karti hai?

Loose folds real circumference stretch force karte hain; tight folds material ko sideways shuffle karne dete hain, toh hoop cost folds ke tighter hone par fall off hoti hai.

ν

do springs mein kahan enter karta hai, aur sirf wahan kyun?

Sirf

D

mein (two-directional plate bending, Poisson coupling);

K

mein nahi (single-direction hoop stretch).

Optimal wavenumber

k^{*}

aur fold size kya hai?

k^{*} = (K / D)^{1/8}

; wavelength

λ^{*} \sim R t

— natural dimple size,

L

se independent.

σ_{cr}

geometric mean

2 D K

kyun hai?

D k^{4} + K / k^{4}

minimize karna

k^{*}

par do springs ko equally balance karta hai;

a + b \geq 2 ab

geometric mean deta hai.

2 D K

se boxed formula tak algebra dikhao.

D, K

sub karo,

E^{2} t^{4} / [12 (1 - ν^{2}) R^{2}]

milega,

E t^{2} / R

bahar nikalo,

t

se divide karo,

12 = 23

2 cancel karta hai →

E t / [R 3 (1 - ν^{2})]

Buckling local se global kab switch hoti hai

L / R

par?

Roughly

L / R \approx 2.85 R / t

; iske neeche local dimpling govern karta hai (

0.605 E t / R

use karo), iske upar Euler bowing jeetta hai.

Clamped vs free ends

σ_{cr}

ko kaise affect karte hain?

Clamped = stiffest, highest

σ_{cr}

; free end = weakest (global Euler load ~4× drop kar sakta hai). Hamesha boundary condition check karo.

γ

aur

ϕ

kya karte hain?

γ

perfect stress ko real dents ke liye knock down karta hai;

ϕ = \frac{1}{16} R / t

empirical

γ = 1 - 0.901 (1 - e^{- ϕ})

mein feed karta hai.