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

3.6.7 · D3 · Physics › Spacecraft Structures & Systems Engineering › Shell buckling — thin-walled cylinder under axial load

Yeh page shell buckling ka drill hall hai. Parent note ne classical formula build kiya tha: $σ_{cr} = \frac{E}{3 ( 1 - ν ^{2} )} \frac{t}{R} \approx 0.605 E \frac{t}{R}, γ = 1 - 0.901 (1 - e^{- ϕ}), ϕ = \frac{1}{16} R / t .$ Yahan hum isse har tarah ke inputs ke against hammer karte hain — normal numbers, tiny aur huge geometry ratios, degenerate limits, ek word problem, aur ek exam-style trap. Shuru karne se pehle, seedha baat karte hain ki "har tarah" ka matlab kya hai.

The scenario matrix

Neeche har symbol parent note mein earn kiya gaya hai. Bas clear kar lete hain: $E$ = Young's modulus (material ki stiffness), $t$ = wall thickness, $R$ = cylinder radius, $ν$ = Poisson ratio, $σ_{cr}$ = perfect-cylinder buckling stress, $P_{cr}$ = axial load (force, newtons mein) jo us stress se correspond karta hai via $P_{cr} = σ_{cr} \cdot A$ jahan $A = 2 π R t$ wall cross-section hai, $γ$ = knockdown factor (perfect value ka woh fraction jo ek real dented cylinder actually reach karta hai), $σ_{a l l o w} = γ σ_{cr}$ = realistic allowable stress, $P_{a l l o w} = σ_{a l l o w} \cdot A$ = realistic allowable load, aur $σ_{Y}$ = yield stress (jahan material khud deta hai).

Definition Classical formula ke peeche boundary conditions aur length assumptions

Classical $σ_{cr} = 0.605 E t / R$ ek aise cylinder ke liye derive kiya gaya hai jo dono ends par simply supported ho (ends round hold hote hain lekin rotate karne ke liye free) aur "moderately long" ho — roughly $1 ≲ L / R$ aur itna lamba nahi ki woh Euler column jaisa behave kare. Us middle band mein buckle ek short-wavelength diamond pattern choose karta hai aur length $L$ answer se bahar ho jaata hai. Do edge cases is band se bahar hain: bahut chhote cylinders ( $L / R ≪ 1$ ) zyada stress par buckle karte hain (ends unhe stiffen karte hain), aur bahut lamba / slender wale ( $L / R$ huge) shell dimpling ki jagah Euler column buckling se govern hone wale whole-body column bending pe switch ho jaate hain. Neeche har example assume karta hai ki hum moderately-long, simply-supported band ke andar hain jab tak stated na ho.

Cell	Yeh case kya stress karta hai	Kis example mein covered
A. Standard forward	numbers plug in karo, $σ_{cr}$ , $P_{cr}$ nikalo	Ex 1
B. Scaling law	kisi change pe answer kaise respond karta hai	Ex 2
C. Buckling vs yield	kaun sa failure mode actually jeetta hai	Ex 3
D. Thick-wall limit ( $R / t$ small)	validity ke edge par formula ka matlab	Ex 4
E. Thin-wall / large- $R / t$ limit	knockdown $γ \to 0$ danger zone	Ex 5
F. Degenerate input ( $t = 0$ , $ν \to 1$ )	kya toot ta hai aur kyun	Ex 6
G. Inverse / design problem	required load diya, $t$ ke liye solve karo	Ex 7
H. Real-world word problem	ek mission ko numbers mein translate karo	Ex 8
I. Exam twist	ek hidden trap ( $γ$ , area, units mix karna)	Ex 9

Ab hum har cell ka ek worked example chalate hain.

Worked example Example 1 — Cell A: Standard forward calculation

Aluminium interstage: $E = 70$ GPa, $ν = 0.3$ , $R = 1.8$ m, $t = 3$ mm. Classical $σ_{cr}$ nikalo, phir $P_{cr}$ , phir knockdown use karke realistic allowable nikalo.

Forecast: Kya realistic load classical value ke karib hogi, ya uski ek choti fraction? Pehle fraction guess karo.

Step 1 — Geometry ratio $t / R = 0.003/1.8 = 1.667 \times 1 0^{- 3}$ . Yeh step kyun? Classical stress geometry par sirf $t / R$ ke through depend karta hai, isliye ise ek baar compute kar lo.

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. Yeh step kyun? $ν = 0.3$ coefficient ko $0.605$ pe collapse kar deta hai, isliye hum ise directly use kar sakte hain.

Step 3 — Wall area $A = 2 π R t = 2 π (1.8) (0.003) = 0.03393 m^{2}$ . Yeh step kyun? Load = stress × woh area jo actually axial force carry karti hai, jo ki wall ka thin ring hai.

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 $γ = 1 - 0.901 (1 - e^{- 1.531}) = 0.298$ . Yeh step kyun? Real cylinders imperfection-sensitive hote hain — dekho Imperfection sensitivity and knockdown factors. Hume scale down karna hoga.

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. ✓ Realistic load classical ka ek third se bhi kam hai — knockdown dominate karta hai.

Figure s01 neeche yeh result do bars ki tarah draw karta hai — tall teal bar perfect-cylinder load $P_{cr}$ hai, short orange bar honest $γ P_{cr}$ hai; plum arrow knockdown dikhata hai jo capacity ko kaat ta hai. Ise padho jaise "formula jo number promise karta hai vs nature jo number actually deliver karta hai."

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

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

Example 1 ka tank lo. Agar hum thickness double kar dein $t = 6$ mm par (baaki sab same), $σ_{cr}$ aur $P_{cr}$ ka kya hoga?

Forecast: Dono double honge? Dono quadruple? Ek-ek? Padhne se pehle commit karo.

Step 1 — $σ_{cr} \propto t / R$ . $t$ double karne se stress double hoga: $70.6 \to 141.2$ MPa. Yeh step kyun? $R$ , $E$ , $ν$ fixed hain, to sirf linear-in- $t$ factor move karta hai.

Step 2 — Area $A = 2 π R t \propto t$ , to yeh bhi double hogi: $0.03393 \to 0.06786 m^{2}$ . Yeh step kyun? $P_{cr} = σ_{cr} A$ mein $t$ ke do factors chhupe hue hain.

Step 3 — $P_{cr} \propto σ_{cr} \cdot t \propto t^{2}$ : load quadruple ho jaata hai, $2.40 \to 9.59$ MN. Yeh step kyun? Hum do independent effects multiply karte hain — stress ×2 upar aur area ×2 upar — load ka response paane ke liye; inhe combine karna ek scaling argument ka pura point hai.

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

Worked example Example 3 — Cell C: Kya buckle hoga ya yield karega?

Steel shell: $E = 200$ GPa, yield $σ_{Y} = 250$ MPa, $R / t = 400$ , $ν = 0.3$ . Pehle kaun fail hoga?

Forecast: Steel strong hai — kya yeh 250 MPa yield reach karega, ya pehle buckle ho jayega?

Step 1 — Classical $σ_{cr} = 0.605 E / (R / t) = 0.605 \times 200000/400 = 302.5$ MPa. Yeh step kyun? $σ_{cr} = 0.605 E (t / R)$ aur $t / R = 1/400$ ; yeh perfect buckling stress hai.

Step 2 — $302.5 > 250$ : naively tum keh sakte ho "yield jeetta hai." Lekin pehle knockdown apply karo. Yeh step kyun? Perfect value ek fantasy hai; honest comparison $σ_{a l l o w}$ use karta hai. Yeh Yield vs stability failure modes decision hai.

Step 3 — $ϕ = \frac{1}{16} 400 = \frac{1}{16} (20) = 1.25$ , $γ = 1 - 0.901 (1 - e^{- 1.25}) = 1 - 0.901 (0.7135) = 0.357$ . Yeh step kyun? $ϕ$ measure karta hai ki yeh geometry kitni imperfection-prone hai; ise SP-8007 formula ke through feed karne se perfect stress ek realistic ek mein convert hota hai — $γ$ ke bina hum ek fantasy ko yield se compare kar rahe hote.

Step 4 — $σ_{a l l o w} = 0.357 \times 302.5 = 108$ MPa. Yeh step kyun? Yeh woh actual stress hai jise shell hold kar sakta hai; sirf yahi number 250 MPa yield ke against fairly compare karne ke liye sahi hai.

Verify: $108 MPa ≪ 250 MPa$ yield ✓ — shell ~108 MPa par buckle karta hai, yield se kaafi neeche. $σ_{Y}$ use karna strength 2× se zyada over-predict karta. Stability governs.

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

$R / t$ small) Ek stubby titanium bottle: $E = 110$ GPa, yield $σ_{Y} = 830$ MPa (typical Ti-6Al-4V), $ν = 0.3$ , $R / t = 25$ . $σ_{cr}$ aur $γ$ compute karo, aur comment karo ki formula actually trustworthy bhi hai ya nahi.

Forecast: Itni thick wall ke saath, kya buckling stress huge hoga ya tiny? Aur kya $γ$ 1 ke karib hoga?

Step 1 — $σ_{cr} = 0.605 \times 110000/25 = 2662$ MPa. Yeh step kyun? Chhota $R / t$ matlab $t / R$ bada hai, to classical stress rocket ki tarah upar jaata hai.

Step 2 — $ϕ = \frac{1}{16} 25 = \frac{1}{16} (5) = 0.3125$ , $γ = 1 - 0.901 (1 - e^{- 0.3125}) = 0.758$ . Yeh step kyun? Jab wall thick hoti hai to near-degenerate modes kam hote hain, isliye imperfections kam hurt karte hain — $γ$ 1 ki taraf chadhta hai.

Step 3 — Sanity flag: $σ_{a l l o w} = γ σ_{cr} = 0.758 \times 2662 = 2018$ MPa bhi titanium yield $830$ MPa se double se zyada hai. To wall 830 MPa par yield karegi aur 2018 MPa buckling reach karne se pehle — elastic formula ka assumption (elastic range mein buckling) violated ho jaata hai. Yeh step kyun? Parent ne "thin" ko $R / t ≳ 30$ define kiya tha. Usse neeche, buckling governing mode nahi hai; concrete numbers compare karna ( $2018 > 830$ ) ise hand-waving ki jagah prove karta hai.

Verify: $σ_{a l l o w} = 2018$ MPa $> σ_{Y} = 830$ MPa ✓ — buckling govern nahi karta; yield karta hai. $R / t = 25 < 30$ par formula ek aisa stress return karta hai jo koi real metal elastically survive nahi kar sakta → correctly signal karta hai ki stability yahan failure mode nahi hai. Math self-consistent hai; physics humhe mode switch karne ko kehti hai. ✓

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

$R / t$ danger zone Ek giant lightweight tank: $R / t = 1600$ , $E = 70$ GPa (aluminium), $ν = 0.3$ . $γ$ aur $σ_{a l l o w}$ nikalo, aur $γ$ ko Example 1 ke $R / t = 600$ case se compare karo.

Forecast: Bada $R / t$ (thinner) — kya $γ$ upar jayega ya crash karega?

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

Step 2 — $ϕ = \frac{1}{16} 1600 = \frac{1}{16} (40) = 2.5$ , $γ = 1 - 0.901 (1 - e^{- 2.5}) = 0.173$ . Yeh step kyun? Bada $R / t$ matlab $ϕ$ bada, to $e^{- ϕ} \to 0$ aur $γ \to 1 - 0.901 = 0.099$ (asymptotic floor).

Step 3 — $σ_{a l l o w} = 0.173 \times 26.5 = 4.58$ MPa. Yeh step kyun? Classical stress already chhota hai; tiny $γ$ se multiply karne par dikha deta hai ki kitni brutally thin shells capacity kho deti hain — engineer ko actually yahi number respect karni chahiye.

Verify: Ex 1 ( $R / t = 600$ ) se compare karo: $γ$ $0.298$ se $0.173$ tak gira jab wall thinner hui ✓ — thinner shells zyada imperfection-sensitive hoti hain. Jab $R / t \to \infty$ , $γ \to 0.099$ , kabhi zero nahi: SP-8007 curve ka ek floor hai. ✓ Dekho NASA SP-8007 buckling of thin-walled cylinders.

Figure s02 neeche $γ$ ko $R / t$ ke against plot karta hai: teal curve neeche ki taraf sagging hai jaise wall thin hoti hai, plum dashed line woh $0.099$ floor hai jise yeh kabhi cross nahi karta, aur do dots Example 1 ( $R / t = 600$ ) aur Example 5 ( $R / t = 1600$ ) mark karte hain — girte curve ko padho jaise "jitna thinner banao, promised strength ka utna hi kam hissa rakh sakte ho."

Worked example Example 6 — Cell F: Degenerate inputs (kya toot ta hai?)

Do "trick" inputs. (a) $t = 0$ set karo. (b) $ν \to 1$ le jao. Formula kya kehta hai, aur kya yeh physically sensible hai?

Forecast: Kya $σ_{cr}$ zero jayega, infinity, ya blow up karega?

Step 1 (a) — $t = 0$ ke saath: $σ_{cr} = 0.605 E (0/ R) = 0$ . Zero thickness ki wall zero buckling stress carry karti hai. Yeh step kyun? Koi material nahi matlab bilkul koi restoring stiffness nahi — trivial correct limit.

Step 2 (a) — Lekin $ϕ = \frac{1}{16} R /0 \to \infty$ aur $γ \to 0.099$ . Product $σ_{a l l o w} = γ \cdot 0 = 0$ . Consistent: koi wall nahi, koi strength nahi. Yeh step kyun? Dono factors check karo taaki koi accidental $0 \times \infty$ slip na kare; yahan yeh cleanly $0$ deta hai.

Step 3 (b) — $ν \to 1$ ke saath: denominator $3 (1 - ν^{2}) = 3 (1 - 1) = 0$ , to $σ_{cr} \to \infty$ . Yeh step kyun? $ν = 0.5$ real solids ke liye incompressible limit hai; $ν \to 1$ unphysical hai. Blow-up formula ka signal hai ki input illegal hai, real infinite strength nahi.

Verify: (a) $t = 0 \Rightarrow σ_{cr} = 0$ ✓, physically sensible. (b) $ν = 1 \Rightarrow$ division by zero ✓, ek impossible material flag karta hai. Dono degenerate cases sahi behave karte hain: ek true limit deta hai, ek forbidden input signal karta hai.

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

$t$ ke liye solve karo) Ek aluminium interstage ( $E = 70$ GPa, $ν = 0.3$ , $R = 1.5$ m) ko safely ek axial load $P = 1.2$ MN carry karna hai. Realistic allowable use karte hue (knockdown ke saath), minimum wall thickness $t$ nikalo. (Iterate karo, kyunki $γ$ $t$ par depend karta hai.)

Forecast: Roughly millimetres mein hoga ya centimetres mein?

Step 1 — Required allowable stress agar wall area $A = 2 π R t$ hai: $σ_{a l l o w} = \frac{P}{2 π R t} .$ Aur $σ_{a l l o w} = γ (t) \cdot 0.605 E \frac{t}{R}$ . Yeh step kyun? Hum demand aur capacity barabar karte hain; dono $t$ par depend karte hain, isliye hum iterate karte hain.

Step 2 — Pehla guess: knockdown drift ignore karo, $γ \approx 0.3$ lo. Phir $\frac{P}{2 π R t} = 0.3 \cdot 0.605 E \frac{t}{R} \Rightarrow t^{2} = \frac{P}{2 π ( 0.3 ) ( 0.605 ) E} .$ $t^{2} = \frac{1.2 \times 1 0 ^{6}}{2 π ( 0.3 ) ( 0.605 ) ( 70 \times 1 0 ^{9} )} = 1.503 \times 1 0^{- 5} m^{2}$ , to $t = 3.88 \times 1 0^{- 3}$ m $= 3.88$ mm. Yeh step kyun? $R$ cancel ho jaata hai (dono sides par $1/ R$ hai), ek clean $t^{2}$ equation deta hai. Sab kuch SI (metres) mein rakho taaki units consistent rahein.

Step 3 — $t = 3.88 \times 1 0^{- 3}$ m par $γ$ refine karo. $R = 1.5$ m ke saath, ratio $R / t = 1.5/0.00388 = 386.6$ (dono metres mein — dimensionless). Phir $ϕ = \frac{1}{16} 386.6 = 1.229$ , $γ = 1 - 0.901 (1 - e^{- 1.229}) = 0.363$ . Yeh step kyun? Hamara pehla $γ$ ek guess tha; found thickness par recompute karo, $R$ aur $t$ ko same unit mein rakho taaki $R / t$ ek pure number ho.

Step 4 — $γ = 0.363$ ke saath re-solve karo: $t^{2} = \frac{1.2 \times 1 0 ^{6}}{2 π ( 0.363 ) ( 0.605 ) ( 70 \times 1 0 ^{9} )} = 1.242 \times 1 0^{- 5} m^{2}$ , $t = 3.52 \times 1 0^{- 3}$ m $= 3.52$ mm. Ek aur pass ( $R / t = 1.5/0.00352 = 426$ , $γ \approx 0.371$ ) $t$ ko $0.02$ mm se kam shift karta hai → $t \approx 3.5$ mm ke paas converge. Yeh step kyun? Har iteration mein $γ$ pehle se kam change hota hai (SP-8007 curve yahan flat hai), to correction geometrically shrink hota hai — rukna justified hai jab change manufacturing tolerance se kaafi neeche ho.

Verify: $t = 3.52 \times 1 0^{- 3}$ m par: $σ_{cr} = 0.605 (70 \times 1 0^{9}) (0.00352/1.5) = 99.4$ 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 mein, jaise forecast kiya. Dekho Rocket tank and interstage structural design.

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

Ek launch team vertical integration ke dauran pad par ek 40-tonne upper stage ko ek aluminium interstage ke upar stack karta hai (abhi koi thrust nahi, sirf gravity). Interstage ka $R = 1.6$ m, $t = 2.5$ mm, $E = 70$ GPa, $ν = 0.3$ hai. $g = 9.81 m/s^{2}$ lo. Kya yeh stack-up survive karta hai, 1.5 ke safety factor ke saath?

Forecast: 40 tonnes bahut bada lagta hai — kya 2.5 mm ki wall ise hold kar payegi?

Step 1 — Applied axial load: $P = m g = 40000 \times 9.81 = 3.92 \times 1 0^{5}$ N $= 0.392$ MN. Yeh step kyun? Pad par sirf axial load upar ka weight hai; tonnes ko newtons mein convert karo.

Step 2 — Classical $σ_{cr} = 0.605 (70 \times 1 0^{9}) (0.0025/1.6) = 66.2$ MPa. Area $A = 2 π (1.6) (0.0025) = 0.02513 m^{2}$ . Yeh step kyun? Capacity side build karo.

Step 3 — Knockdown: $R / t = 1.6/0.0025 = 640$ , $ϕ = \frac{1}{16} 640 = 1.581$ , $γ = 1 - 0.901 (1 - e^{- 1.581}) = 0.284$ . To $σ_{a l l o w} = 0.284 \times 66.2 = 18.8$ MPa. Yeh step kyun? Bare classical stress optimistic hai; knockdown ise woh stress mein convert karta hai jo ek real, thodi-si dented interstage actually hold kar sakti hai, aur safety isi par judge honi chahiye.

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. Yeh step kyun? Allowable stress ko force mein wapas convert karo taaki directly applied weight $P$ se compare ho sake.

Step 5 — Safety factor $= P_{a l l o w} / P = 0.472/0.392 = 1.20$ . Yeh step kyun? Capacity ko demand se ratio ke roop mein compare karo.

Verify: Required SF 1.5 hai, achieved SF 1.20 $< 1.5$ hai ✗ — yeh margin meet nahi karta. Physically believable: ek 2.5 mm ka can 40 t hold karte hue knife-edge par baitha hai. Team ko wall thicken karni hogi ya stiffeners add karne honge. ✓

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

Exam question: "Ek cylinder ka classical $σ_{cr} = 150$ MPa aur knockdown $γ = 0.4$ hai. Ek student allowable load $P = σ_{cr} \times A$ compute karta hai classical stress use karke, phir alag se kehta hai 'stress safe hai kyunki $150 < 250$ MPa yield hai.' Student ki dono galtiyan nikalo aur sahi allowable stress batao." (Diya gaya: $A = 0.02 m^{2}$ , $σ_{Y} = 250$ MPa.)

Forecast: Do traps computing se pehle dhundho.

Step 1 — Trap 1: Allowable stress mein knockdown shamil hona chahiye: $σ_{a l l o w} = γ σ_{cr} = 0.4 \times 150 = 60$ MPa, na ki 150 MPa. Yeh step kyun? Perfect-cylinder value use karna imperfection sensitivity ignore karta hai — $γ$ ka pura point yahi hai.

Step 2 — Trap 2: Yield se compare karna galat test hai. Governing limit buckling-allowable aur yield ka minimum hota hai: $min (60, 250) = 60$ MPa. Buckling governs. Yeh step kyun? Structure tab fail hota hai jab koi bhi mode pehle trigger ho; $60 < 250$ to stability jeetti hai.

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. Student ka $150 \times 1 0^{6} \times 0.02 = 3.0$ MN $2.5 \times$ over-predict karta hai. Yeh step kyun? Load woh quantity hai jise launch vehicle actually experience karta hai; sahi stress ko force mein convert karna (aur student ke inflated force se contrast karna) error ki magnitude expose karta hai.

Verify: Correct $σ_{a l l o w} = 60$ MPa ✓; correct $P_{a l l o w} = 1.2$ MN vs student ka $3.0$ MN ✓; $2.5 \times$ error exactly $1/ γ = 1/0.4$ hai ✓. Do traps: (1) $γ$ bhool gaya, (2) minimum lene ki jagah yield se compare kiya. ✓

Recall Har example ne kaun sa cell hit kiya? (reveal karne ke liye click karo)

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. Har cell covered.

Active recall

Ek inverse (

t

-solving) problem mein iterate kyun karna padta hai?

Kyunki

γ

R / t

par depend karta hai, jo ki unknown

t

par depend karta hai — ek pass ek guess deta hai, phir tum

γ

refine karke re-solve karte ho.

Jab

R / t \to \infty

γ

kis value pe jaata hai?

Floor

1 - 0.901 = 0.099

par; thinner shells kabhi zero tak nahi pahunchte lekin bahut fragile ho jaate hain.

Example 3 mein, "302.5 MPa > 250 MPa yield" sahi comparison kyun nahi hai?

Pehle knockdown apply karna hoga:

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

MPa, jo yield se kaafi neeche hai — buckling governs.

Example 9 exam trap mein do galtiyan kaun si hain?

(1) Stress par knockdown

γ

bhoolna; (2) buckling-allowable aur yield ka minimum lene ki jagah yield se compare karna.

Classical

0.605 E t / R

kaun si end conditions aur length range assume karta hai?

Simply-supported ends aur moderately-long cylinder; bahut chhote shells zyada stress par buckle karte hain, bahut lamba wale Euler columns ban jaate hain.

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