Visual walkthrough — Shell buckling — thin-walled cylinder under axial load

3.6.7 · D2 · Physics › Spacecraft Structures & Systems Engineering › Shell buckling — thin-walled cylinder under axial load

---

Step 0 — Jinhe use karne ki permission hai: woh objects aur symbols

Kisi bhi physics se pehle, picture ko naam dete hain. Ek rocket tank ek hollow tube hai. Neeche labelled tube dekho.

Thin ka matlab hai $t\ll R$ (padho "$t$ much smaller than $R$"). Typically $R/t\gtrsim 30$. Yahi "thinness" ki wajah se buckling — crushing se pehle — jeet jaata hai. Dekho Yield vs stability failure modes.

---

Step 1 — "Buckling" kaisa dikhta hai: the dimple

KYA. Can ke top ko neeche push karo. Wall flat aur seedha rehta hai… jab tak, ek special load par, woh andar-bahar ek pattern of little dents mein pop nahi kar jaata.

Pehle yeh kyun draw karte hain. Baad ke har symbol ($w$, wavelength, wavenumber) isi shape ko describe karta hai. Jab tak aap shape ko appear hote nahi dekha, algebra samajh nahi aayega.

PICTURE.

Figure mein notice karo ki dents repeat hote hain: kai waves circumference ke around jaate hain aur kai length ke upar jaate hain. Yahi "kai" hai jo agli steps ko count aur phir optimise away karna padega.

---

Step 2 — Restoring effect #1: bending stiffness $D$

KYA. Wall ko dimple mein fold karne ke liye use curve karna padta hai. Metal ko curve hona pasand nahi — woh push back karta hai. Usi push-back ko bending (flexural) rigidity $D$ kehte hain.

Simple $EI$ nahi, yeh tool kyun? Ek narrow beam ke liye aap $EI$ use karte. Lekin cylinder wall ek wide plate hai: jab aap ise ek taraf bend karte ho, Poisson coupling ise doosri taraf curl karna chahta hai (anticlastic curvature), aur use suppress karna ise stiffen karta hai. Yahi exactly extra factor $\frac{1}{1-{\nu }^2}$ hai. Yeh plate result hai — dekho Plate bending and flexural rigidity D.

PICTURE.

---

Step 3 — Restoring effect #2: membrane hoop stiffness $Et/R^2$

KYA. Yahan woh effect hai jo flat plate ke paas nahi hoti. Jab aap wall ko andar dent karte ho, aapne us ring par circumference chhota kar diya hai — aapne hoop ko stretch ya compress kar diya hai. Material us stretch ko bhi resist karta hai.

Yeh sirf cylinder ke liye kyun exist karta hai. Ek flat sheet bina in-plane stretching ke dimple kar sakti hai. Ek curved sheet nahi kar sakti: radially $w$ se push karne par circumference badal jaati hai. Woh change curvature $1/R$ ke proportional hai. Yahi circumferential stretch Hoop stress in pressurised cylinders ke peeche bhi hai.

PICTURE.

---

Step 4 — Tug-of-war ek favourite wavelength set karta hai

KYA. Ab hamare paas opposite tastes wale do restoring springs hain:

• $D$ ko short waves se nafrat hai.
• $K$ ko long waves se nafrat hai.

Wall woh wavelength par buckle karegi jahan total restoring energy sabse chhoti hoti hai (weakest spot = fold karne ki sabse aasaan jagah). Short aur long ke beech mein kahin, dono effects trade off karte hain aur wall sabse zyada vulnerable hoti hai.

Minimise kyun karte hain? Buckling easiest mode par start hoti hai. Isliye physical critical stress sabse possible wavelengths mein minimum hoti hai — neeche curve ki valley.

PICTURE.

Restoring energy per unit area ka shape phir yeh hota hai: $U(k)\sim \underset{\text{bending: short waves ke liye grow karta hai}}{\underbrace{D{k}^2}}+\underset{\text{membrane: long waves ke liye grow karta hai}}{\underbrace{\frac{K}{k^2}}}.$

• Pehla term $k$ badhne par badhta hai (short waves highly curved hain → $D$ bada).
• Doosra term $k$ ghatne par badhta hai (long waves bahut saara hoop stretch karti hain → tiny $k^2$ se divide hota bada $K$).
• Beech mein sum bottom out karta hai. Woh valley hi woh buckle hai jise cylinder actually pick karta hai.

---

Step 5 — Geometric-mean magic ($a+b\ge 2\sqrt{ab}$)

KYA. $U(k)=D{k}^2+\frac{K}{k^2}$ ka valley bottom dhundho.

Yeh specific algebra tool kyun? Ek famous inequality hai: kisi bhi do positive numbers $a,b$ ke liye, sum $a+b$ unke product ke relative sabse chhota hota hai — exactly jab $a=b$, aur tab $a+b=2\sqrt{ab}$. Hamare dono terms perfect match hain: unka product $k$ se independent hai: $\left(D{k}^2\right)\cdot \left(\frac{K}{k^2}\right)=DK(\text{the }k\text{ cancels!}).$ Isliye sum ka minimum fix hai chahe $k$ kuch bhi ho, aur woh $2\sqrt{DK}$ ke barabar hai, tab pahuncha jab dono springs equally contribute karte hain.

PICTURE.

---

Step 6 — Energy balance: ${\sigma }_{cr}t$ valley depth ke barabar kyun hota hai

Yeh woh step hai jise parent note ne gloss over kar diya tha. Ab hum carefully dikhate hain ki critical stress valley $U_{\min }$ ke saath kyun pair karti hai aur $w^2$ aur $L$ kyun survive nahi karte.

KYA. Energy method: buckled shape tabhi stable hai jab use bend karne mein zyada energy lage jitni axial squeeze wapas deta hai. Exact critical load par, dono equal hote hain — woh tipping point hai.

Energy method kyun (force balance nahi)? Buckling ek stability question hai: "kya ek tiny dent grow karegi ya maregi?" Clean test energy hai — ek dent exactly tab grow hoti hai jab woh net energy release kare. Yahi logic Euler column buckling mein bhi hai.

PICTURE.

Dono energies, per unit surface area, amplitude $w_0$ aur wavenumber $k$ ki dent ke liye:

\qquad \underbrace{W_{\text{squeeze}} = \tfrac{1}{2}\,\sigma\,t\,k^{2}\,w_0^{2}}_{\text{axial stress jo kaam deta hai}} $$ - $w_0^2$ — **dono** terms mein same power ke saath appear hota hai. Jab hum inhe equal set karte hain, $w_0^2$ **cancel** ho jaata hai. Isliye answer mein kabhi dent ka size nahi hota — critical stress ek onset condition hai, eventual dent kitni deep jaaye usse independent. - $\sigma\,t$ — axial force per unit circumferential length (stress × wall thickness). - $k^{2}$ work side par — jab axially-squeezed wall sideways bow karti hai toh woh axis ke along $\propto k^{2}w_0^2$ amount se shorten hoti hai; woh shortening hi squeeze ka kaam karne ka tarika hai. **Wahi** $k^{2}$ bending term par bhi baiṭha hai. **$W_{\text{squeeze}} = U_{\text{restore}}$ set karo** (tipping point) aur $\tfrac12 w_0^2$ cancel karo: $$ \sigma\,t\,k^{2} = D\,k^{2}+\frac{K}{k^{2}} \;\;\Longrightarrow\;\; \sigma\,t = D + \frac{K}{k^{4}}. $$ Buckling **sabse chhote** aisi $\sigma$ par hoti hai, isliye right side ko $k$ ke upar minimise karo. Iska minimum (Step 5, jahan $k^2$ variable ka role play karta hai) geometric mean hai, jo deta hai: $$ \boxed{\;\sigma_{cr}\,t = 2\sqrt{D\,K} = 2\sqrt{D\,\frac{Et}{R^{2}}}\;}. $$ - **Length $L$ kabhi enter nahi ki** energy per unit area mein, isliye ek long cylinder ke liye woh completely drop out ho jaati hai: ek chhota can aur ek lamba can (same $R,t$) same *stress* par buckle karte hain. --- ## Step 7 — Finite length aur discrete-mode caveat **KYA.** Steps 4–6 ne $k$ ko *continuously* vary karne diya — legitimate sirf ek **long** cylinder ke liye, jahan jo bhi optimal wavelength chahiye woh comfortably kai baar fit hoti hai. **Ise flag kyun karte hain.** Ek real tank ke *ends* hote hain (bulkheads, rings). Axial half-waves ek whole number of times length mein fit hone chahiye, isliye axial wavenumber **discrete** hai: $$ k_{n} = \frac{n\pi}{L},\qquad n = 1,2,3,\dots $$ Ek **long** cylinder ke liye allowed $k_n$ ke beech spacing tiny hai, isliye unka ek essentially continuous optimum par land karta hai → hamaara formula hold karta hai. Ek **short** cylinder ke liye ($L$ bahut zyada $R$ nahi), optimal wavelength **fit nahi ho sakti**; cylinder ek stiffer discrete mode par force hoti hai aur $0.605\,Et/R$ se **zyada** $\sigma_{cr}$ par buckle karti hai. Classical formula isliye ek *long-cylinder* result hai. **PICTURE.** ![[deepdives/dd-physics-3.6.07-d2-s08.png]] > [!mistake] Is formula ke out-of-scope regimes > - **Bahut short cylinders** ($L\lesssim R$): flat plate strip jaisi behave karti hain; discrete axial modes $\sigma_{cr}$ badhate hain. $0.605\,Et/R$ nahi, plate/short-shell theory use karo. > - **End supports matter karte hain**: clamped vs simply-supported ends allowed $k_n$ aur hence load shift karte hain — dekho [[NASA SP-8007 buckling of thin-walled cylinders]]. > - Is page ka clean geometric-mean result **moderate-to-long** cylinder ke liye hai ($L$ kaafi $R$ ya zyada), jo [[Rocket tank and interstage structural design]] mein zyaadatar rocket tanks aur interstages cover karta hai. --- ## Step 8 — $D$ plug in karo, famous number tak simplify karo, aur edges check karo **KYA.** $D=\dfrac{Et^{3}}{12(1-\nu^{2})}$ ko $\sigma_{cr}\,t = 2\sqrt{D\cdot\dfrac{Et}{R^{2}}}$ mein substitute karo aur clean up karo. **Kyun.** Humne $D$ symbolic rakha tha Steps 5–6 ko readable rakhne ke liye. Ab use cash in karte hain. **PICTURE.** ![[deepdives/dd-physics-3.6.07-d2-s09.png]] $$ \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})}}. $$ Dono sides ko $t$ se divide karo, aur $\sqrt{1/12}=1/(2\sqrt3)$ use karo taki $2$ cancel ho jaye: > [!formula] Classical axial buckling stress — the destination > $$ \boxed{\;\sigma_{cr} = \frac{E}{\sqrt{3(1-\nu^{2})}}\,\frac{t}{R}\;} $$ > - $E$ — stiffer metal, buckling ke against zyada strong (upar hai, isliye yeh *help* karta hai). > - $t/R$ — thinness ratio. Thin wall (chhota $t$) ya fat can (bada $R$) ⇒ weaker. Yeh ek geometric knob hai. > - $\sqrt{3(1-\nu^{2})}$ — ek pure number. $\nu=0.3$ ke liye: $\sqrt{3(1-0.09)}=\sqrt{2.73}=1.652$. > > Isliye metals ke liye: > $$ \sigma_{cr}\approx 0.605\,E\,\frac{t}{R}. $$ **Degenerate aur edge cases — inhe kabhi skip mat karo:** > [!example] Limit 1 — Flat plate: $R\to\infty$ > Jab $R\to\infty$, $t/R\to0$, isliye $\sigma_{cr}\to0$. Flat plate ke paas **koi membrane hoop term nahi** hota (Step 3 ka $K=Et/R^2\to0$), isliye yeh *particular* mechanism disappear ho jaata hai — flat plates alag rule se buckle karti hain. Consistent: hamaara formula exactly wahan switch off ho jaata hai jahan iska physics karta hai. > [!example] Limit 2 — Bahut thin wall: $t\to0$ > $\sigma_{cr}\propto t\to0$. Infinitely thin foil bilkul bhi axial buckling stress carry nahi karta. Intuition se match karta hai — aap cling film ki sheet par khade nahi ho sakte. > [!example] Limit 3 — Incompressible material: $\nu\to0.5$ > $\sqrt{3(1-0.25)}=\sqrt{2.25}=1.5$, coefficient $=1/1.5=0.667$. Plate stiffening thoda badhta hai. Isotropic elasticity sirf $-1<\nu<\tfrac12$ admit karta hai, aur real engineering metals mein $\nu\le0.5$ hota hai (rubber $\tfrac12$ limit approach karta hai). Us **poore physical range** mein denominator $3(1-\nu^2)$ positive aur finite rehta hai, isliye formula kabhi blow up nahi karta. Achha — yeh har real material ke liye well-behaved hai. > [!mistake] Woh ek edge case jo formula nahi batata > **Yeh sab ek *perfect* cylinder ke liye hai.** Real cans mein ~$t$ ke dents hote hain. Kyunki Step 5 ki valley *broad aur flat* hai (kai wavelengths ki energy almost equal hoti hai), real cylinder ek knife-edge par baith kar sirf $\sigma_{cr}$ ke **20–70%** par buckle karta hai. Ek knockdown factor $\gamma$ se multiply karna padta hai (dekho [[Imperfection sensitivity and knockdown factors]] aur [[NASA SP-8007 buckling of thin-walled cylinders]]): > $$ \gamma = 1 - 0.901\left(1-e^{-\phi}\right),\qquad \phi=\tfrac{1}{16}\sqrt{R/t}. $$ --- ## Ek picture ka summary ![[deepdives/dd-physics-3.6.07-d2-s10.png]] Yeh final figure poori walkthrough compress karti hai: dented shell → do springs ($D$ bending vs $K$ membrane) → energy balance $\sigma t\,k^2 = Dk^2 + K/k^2$ → geometric-mean valley $2\sqrt{DK}$ → boxed result → knockdown to reality. [[Rocket tank and interstage structural design]] mein use hone wali har cheez isi ek chain par hang karti hai. > [!recall]- Feynman retelling — poori derivation plain words mein (click) > Socho tum ek empty soda can ko upar se squeeze kar rahe ho. Use crumple karne ke liye wall ko ek saath do mushkil kaam karne hote hain. **Pehla**, use chhote dents mein *bend* karna padta hai — aur metal bending resist karta hai, especially tight tiny folds mein; us resistance ko $D$ kaho, aur woh tight dents ke liye huge hoti hai kyunki bending strength thickness cubed jaisi grow karti hai. **Doosra**, jab ek dent andar push karta hai toh use can ke around ring *shrink* karni padti hai — metal ko sideways stretch ya squeeze karna padta hai — aur woh resistance, ise $K$ kaho, bade broad bulges ke liye sabse zyada hoti hai. Isliye dono resistances ke opposite tastes hain: ek ko tiny dents se nafrat hai, doosre ko bade ones se. Ab upar se squeeze *wapas* energy deta hai jab wall bow karti hai — aur ek dent tabhi actually grow hota hai jab woh giveaway dono resistances ko beat kar le. Exact tipping point par woh equal hote hain, aur kyunki dent ki depth dono sides par appear hoti hai woh cancel out ho jaati hai — isliye answer care nahi karta ki eventual dent kitni deep jaati hai. Can woh dent-size par fold hota hai *beech mein* jahan dono resistances as easy as possible hain. Maths ka ek classic bit hai — ek "spring one plus spring two" sabse chhota ho sakta hai twice the square root of their product, jab woh equal hon — isliye weakest spot ki strength $2\sqrt{DK}$ hai. $D$ aur $K$ kya hain cash in karo, numbers tidy karo, aur ek clean rule nikal aata hai: can approximately **0.6 times its metal-stiffness times its thinness ($t/R$)** tak le sakta hai fold hone se pehle — jab tak woh kaafi zyada baar jitna chaura hai utna lamba ho, taaki ideal dent-size fit ho. Phir reality bolta hai: kyunki kaafi saari dent-sizes almost equally weak hoti hain, ek real dented can iske sirf ek tihaayi se aadhe par fold ho jaata hai. --- ## Active recall Cylinder buckling mein do competing restoring effects kya hain? ::: Bending stiffness $D$ (short-wavelength dents se nafrat) aur membrane hoop stiffness $K=Et/R^2$ (long-wavelength dents se nafrat). Energy balance mein dent amplitude $w_0$ kyun drop out ho jaata hai? ::: Restoring energy aur squeeze work dono $w_0^2$ ke saath scale karte hain, isliye inhe equal set karne par $w_0^2$ cancel ho jaata hai — critical stress ek onset condition hai jo dent depth se independent hai. Precisely, wavenumber $k$ kya hai? ::: $k=2\pi/\lambda$, radians of pattern per metre; bada $k$ = short waves, chhota $k$ = long waves. Ek dent $\sin(kx)$ ki curvature $\propto k^2$ hoti hai. Ek long cylinder ke liye length $L$, $\sigma_{cr}$ se kyun drop out ho jaata hai? ::: Energy per unit area mein $L$ contain nahi hota, aur discrete axial modes $k_n=n\pi/L$ itne closely spaced hote hain ki ek continuous optimum par land kar jaata hai. Bahut short cylinder ke liye kya hota hai? ::: Optimal wavelength fit nahi ho sakti; discrete modes ek stiffer shape force karte hain, isliye $\sigma_{cr}$ $0.605\,Et/R$ se *zyada* hota hai — classical formula ek long-cylinder result hai. Critical value geometric mean $2\sqrt{DK}$ kyun hai? ::: Dono spring contributions ka product wavenumber se independent hai, isliye sum ka minimum $2\sqrt{DK}$ hai ($a+b\ge2\sqrt{ab}$ se), tab pahuncha jab woh equally contribute karte hain. $R\to\infty$ par $\sigma_{cr}$ ka kya hota hai aur kyun? ::: Woh $\to 0$ ho jaata hai; membrane term $K=Et/R^2\to0$ (flat plate ke paas koi hoop restoring effect nahi), isliye yeh mechanism vanish ho jaata hai. $\nu$ ki physical range kya hai, aur kya formula wahin finite rehta hai? ::: Isotropic elasticity require karta hai $-1<\nu<\tfrac12$; aise sabhi $\nu$ ke liye denominator $3(1-\nu^2)$ positive rehta hai, isliye $\sigma_{cr}$ kabhi blow up nahi karta. $\nu=0.3$ ke liye classical coefficient kya hai? ::: $1/\sqrt{3(1-0.09)}=1/1.652\approx0.605$.