Buckling — Euler column buckling load derivation: Worked examples

3.6.6 · D3 · Physics › Spacecraft Structures & Systems Engineering › Buckling — Euler column buckling load derivation

Scenario matrix

Problems karne se pehle, chaliye kaunse kism ke questions hote hain yeh enumerate karte hain. Har row ek "case class" hai. Neeche har worked example us cell ke saath tagged hai jise woh fill karta hai, toh end mein koi cell khaali nahi bachegi.

#	Case class	Kya unknown hai / kya ajeeb hai	Example
A	Forward — sab diya hai, $P_{cr}$ nikaalein	plug in karein	Ex 1
B	Inverse (design) — required load diya hai, $I$ ya radius nikaalein	geometry ke liye rearrange karein	Ex 2
C	End-condition swap — same column, alag $K$	$L_{e} = K L$ badalta hai	Ex 3
D	Regime boundary — pehle buckle hoga ya yield?	$σ_{cr}$ vs $σ_{y}$ compare karein	Ex 4
E	Degenerate / limiting input — $L \to 0$ , $L \to \infty$ , $I \to 0$ , hollow→thin wall	number nahi, trend check karein	Ex 5
F	Cross-section geometry — equal area ka solid vs hollow tube	kisका bada $I$ hai?	Ex 6
G	Real-world word problem — safety factor wala spacecraft	words → symbols translate karein	Ex 7
H	Exam twist — column do alag axes ke baare mein buckle ho sakta hai	sabse chota $I$ govern karta hai	Ex 8

Recall Yeh cells kyon hain, aur doosri kyon nahi?

Har buckling problem mein yeh hai: (1) unknowns ka arrow kis direction mein hai — forward ya inverse (A vs B)? (2) support kya hai, yaani $K$ (C)? (3) kya Euler valid bhi hai, ya yield ne cap laga di (D, E)? (4) $I$ dene wala cross-section kya hai (F, H)? Cell G bas "yeh sab, ek story mein safety factor ke saath" hai. Inhe cover karo toh buckling cover ho jaati hai.

Case A — Forward: bas plug in karo

Worked example Ex 1 (Cell A) — plain pin-pin strut

Aluminium rod: $E = 70 GPa$ , solid circular radius $r_{0} = 5 mm$ , length $L = 1.0 m$ , dono ends pinned. $P_{cr}$ nikaalein.

Forecast: order of magnitude guess karo — hundreds of newtons? thousands? (1 m ka aluminium pencil-thick rod compression mein bahut floppy hota hai.)

Solid circle ka second moment of area: $I = \frac{π r _{0}^{4}}{4} = \frac{π ( 0.005 ) ^{4}}{4} = 4.909 \times 1 0^{- 10} m^{4}$ Yeh step kyon? $I$ hi wo akela geometry hai jo formula ko chahiye — yeh encode karta hai ki material bending axis se kitni door hai (dekhein Second Moment of Area). $r_{0}^{4}$ batata hai ki thickness kitna zyada matter karti hai.
Pin-pin matlab $K = 1$ , toh $L_{e} = 1.0 m$ . Kyon? Pins position fix karte hain par rotation allow karte hain → buckled shape $L$ pe ek poora half sine wave hota hai (dekhein Column End Conditions and Effective Length).
Substitute karein: $P_{cr} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 4.909 \times 1 0 ^{- 10} )}{1. 0 ^{2}} \approx 339 N$

Verify: Units: $Pa \cdot m^{4} / m^{2} = \frac{N}{m ^{2}} \cdot m^{2} = N$ ✓. Order-of-magnitude: humara "hundreds of N" ka forecast sahi nikla. Ek insaan isme lean karke ise tod sakta hai — slender rod ke liye correct hai.

Case B — Inverse (design): geometry nikaalein

Worked example Ex 2 (Cell B) — target load ke liye lander leg size karo

Ek lander leg (pin-pin) ko $P = 2000 N$ bina buckle hue survive karna hai. Steel, $E = 200 GPa$ , $L = 0.8 m$ . Minimum $I$ nikaalein, phir radius agar yeh solid rod ho.

Forecast: hum formula ulta chalaa rahe hain. Kya $I$ Ex 1 ke $4.9 \times 1 0^{- 10}$ se bada hoga ya chota? (Zyada load aur stiff steel opposite directions mein pull karte hain — compute karne se pehle guess karo.)

$P_{cr} = P$ set karo aur Euler mein $I$ ke liye solve karo: $I \geq \frac{P L _{e}^{2}}{π ^{2} E} = \frac{2000 ( 0.8 ) ^{2}}{π ^{2} ( 200 \times 1 0 ^{9} )} = 6.485 \times 1 0^{- 10} m^{4}$ Yeh step kyon? Design mein unknown left side pe jaata hai. Hum "kya load hai?" nahi pooch rahe balki "kaun sa shape yeh load resist karega?" — yeh hai failure-mode-driven sizing.
$I = π r_{0}^{4} /4$ se radius back out karo: $r_{0} = (\frac{4 I}{π})^{1/4} = (\frac{4 ( 6.485 \times 1 0 ^{- 10} )}{π})^{1/4} \approx 5.38 \times 1 0^{- 3} m = 5.38 mm$ Kyon? Hum circle formula ko invert karte hain taaki ek manufacturable dimension mile.

Verify: $r_{0} = 5.38 mm$ forward plug karo: $I = π (0.00538)^{4} /4 = 6.49 \times 1 0^{- 10}$ ✓, aur $P_{cr} = π^{2} (200 \times 1 0^{9}) (6.49 \times 1 0^{- 10}) /0.64 \approx 2001 N \approx P$ ✓. (Real design mein pehle $P$ ko safety factor se multiply karte — Ex 7 mein kiya hai.)

Case C — End-condition swap

Worked example Ex 3 (Cell C) — same rod, chaar alag supports

Ex 1 wala aluminium rod lo ( $E I$ fixed, pin-pin se $339 N$ milta hai). Fixed–fixed ( $K = 0.5$ ), fixed–pin ( $K \approx 0.7$ ), aur fixed–free ( $K = 2$ ) ke liye $P_{cr}$ recompute karo.

Forecast: compute karne se pehle strongest → weakest rank karo. Clamping se help milni chahiye; free top se hurt hona chahiye.

Kyunki $P_{cr} \propto 1/ K^{2}$ fixed $E I, L$ pe, sab kuch pin-pin value $P_{1} = 339 N$ (jahan $K = 1$ ) ke relative likhein: $P_{cr} (K) = \frac{P _{1}}{K ^{2}}$ Yeh step kyon? $E, I, L$ nahi badlte — sirf wave jo supports ke beech fit hoti hai badlti hai, aur woh poori tarah $K$ se capture hoti hai (dekhein Column End Conditions and Effective Length).
Har ek evaluate karo:
- Fixed–fixed, $K = 0.5$ : $P_{cr} = 339/0.25 = 1356 N$
- Fixed–pin, $K = 0.7$ : $P_{cr} = 339/0.49 \approx 692 N$
- Fixed–free, $K = 2$ : $P_{cr} = 339/4 \approx 84.8 N$
Yeh step kyon? Hum bas $1/ K^{2}$ law ko har support pe apply kar rahe hain. Bahut bada spread notice karo — $84.8 N$ se $1356 N$ tak, 16 ka factor — bilkul same rod ke liye: aap ends ko kaise clamp karte hain, yeh rod ke material jaisa hi matter karta hai.

Figure 1 — Chaar buckled columns side by side. Vertical dotted line undeflected axis hai; har curve woh shape hai jisme column bow karta hai. Fixed–fixed (leftish) sabse tight curve pack karta hai → sabse chota effective wave → strongest; red fixed–free curve ek lazy quarter wave mein stretch hoti hai → sabse lamba wave → weakest. Har ek ke upar $P$ ka arrow same axial load direction hai.

Verify: Ranking fixed-fixed > fixed-pin > pin-pin > fixed-free holds: $1356 > 692 > 339 > 84.8$ ✓. Products $P_{cr} \cdot K^{2}$ sab $339$ ke barabar hain ✓ (constant, jaise predict kiya tha).

Case D — Regime boundary: buckle hoga ya yield?

Worked example Ex 4 (Cell D) — transition slenderness

Ek steel column ke liye ( $E = 200 GPa$ , yield $σ_{y} = 250 MPa$ ), wo slenderness ratio $λ = L_{e} / r$ dhundho (yahan $r$ gyration radius $I / A$ hai) jis par Euler buckling stress exactly yield stress ke barabar ho. Usse neeche column crush hoga; usse upar, buckle hoga.

Forecast: kya transition $λ$ 10 ke aaspaas hogi, 100, ya 1000? (Sochein: steel stiff hai, toh "long-column" tab tak rahega jab tak bahut slender na ho jaaye.)

Critical stress ko yield ke barabar set karo: $σ_{cr} = \frac{π ^{2} E}{λ ^{2}} = σ_{y}$ Yeh step kyon? Yeh Yield vs Buckling — Failure Mode Selection ki exact frontier hai. Material ki apni strength $σ_{y}$ finally enter hoti hai — lekin sirf ek cap ke roop mein, Euler formula ke andar kabhi nahi.
$λ$ ke liye solve karo: $λ = π \frac{E}{σ _{y}} = π \frac{200 \times 1 0 ^{9}}{250 \times 1 0 ^{6}} = π 800 \approx 88.9$ Yeh step kyon? Hum $λ^{2} / σ_{y}$ se dono sides multiply karte hain aur square root lete hain taaki "dono stresses kab equal hain?" ka jawaab slenderness axis par ek single dividing number ban jaaye. Ab kisi bhi real column ko instantly classify kiya ja sakta hai: apna $λ$ compute karo aur $88.9$ se compare karo.

Verify: Slenderness $\approx 89$ se upar → Euler govern karta hai; neeche → yield govern karta hai. Check: $λ = 88.9$ par, $σ_{cr} = π^{2} (200 \times 1 0^{9}) /88. 9^{2} \approx 2.50 \times 1 0^{8} = 250 MPa = σ_{y}$ ✓.

Case E — Degenerate / limiting inputs

Worked example Ex 5 (Cell E) — extremes par kya hota hai?

Yahan koi arithmetic answer nahi — hum $P_{cr} = π^{2} E I / L_{e}^{2}$ (aur uske cross-section $I$ ) ka behaviour edge cases par test karte hain aur confirm karte hain ki har ek physically sense karta hai.

Forecast: inme se kaunse blow up hote hain, aur kaunse vanish?

$L \to 0$ (bahut chota column): $P_{cr} = π^{2} E I / L^{2} \to \infty$ . Yeh step kyon? Formula kehta hai infinitely short column ko buckle karne ke liye infinite load chahiye. Isliye hi chote columns kabhi buckle nahi hote — pehle yield ho jaate hain (Ex 4). Euler yahan yield ko correctly "hand off" karta hai.
$L \to \infty$ (bahut lamba column): $P_{cr} \to 0$ . Kyon? Bahut lamba rod ek feather ke neeche buckle ho jaata hai. Everyday experience se match karta hai (ek lamba wire khada nahi rahega).
$I \to 0$ (bahut patla cross-section): $P_{cr} \to 0$ . Kyon? Axis se door koi material nahi → bending stiffness nahi → instantly collapse.
Hollow tube, wall $t \to 0$ fixed mean radius $r_{m}$ par (thin-wall limit): exact ring moment $I = \frac{π}{4} (r_{o u t}^{4} - r_{in}^{4})$ thin-wall formula $I \to π r_{m}^{3} t$ mein collapse ho jaata hai. Yeh step kyon? Jab wall thin hoti jaati hai, $r_{o u t} \approx r_{in} \approx r_{m}$ aur fourth-power difference ek single $t$ term mein linearise ho jaata hai. Yahi exact formula hai jo hum Ex 6 mein use karte hain — Case E dikhata hai ki yeh general ring ka ek limit hai, koi alag rule nahi. Note karein ki $I$ finite aur non-zero rehta hai even as $t \to 0$ $r_{m}$ ke relative, kyunki material bahar door hai; sirf agar $r_{m}$ bhi shrink ho tab $I \to 0$ .
$L$ double karna: $P_{cr} \to P_{cr} /4$ . Kyon? $P_{cr} \propto 1/ L^{2}$ : yeh "lamba matlab kamzor" trap hai, ab exact ban gaya.

Verify: Doubling law ka numeric spot-check Ex 1 se: $L = 2 m$ deta hai $P_{cr} = π^{2} (70 \times 1 0^{9}) (4.909 \times 1 0^{- 10}) /4 \approx 84.8 N = 339/4$ ✓. Thin-wall check: $r_{m} = 15 mm$ , $t = 0.833 mm$ ke saath, exact ring $I = \frac{π}{4} ((r_{m} + t /2)^{4} - (r_{m} - t /2)^{4})$ $π r_{m}^{3} t$ se $1%$ se behtar match karta hai ✓.

Case F — Cross-section geometry

Worked example Ex 6 (Cell F) — same area ka solid rod vs hollow tube

Do aluminium columns jinki equal cross-sectional area $A = π (0.005)^{2} = 7.854 \times 1 0^{- 5} m^{2}$ hai. Ek solid rod hai ( $r_{0} = 5 mm$ ); doosra ek thin tube hai mean radius $r_{m} = 15 mm$ ke saath. Unke $I$ aur hence $P_{cr}$ compare karo.

Forecast: equal material — toh equal buckling strength? (Trap: step 2 padhne se pehle guess karo.)

Solid rod: $I_{solid} = π r_{0}^{4} /4 = 4.909 \times 1 0^{- 10} m^{4}$ (Ex 1 se). Yeh step kyon? Hume ek baseline $I$ chahiye "sab material axis ke paas bunched" case ke liye, taaki measure kar sakein ki tube kitna gain karti hai same material ko bahar spread karke.
Mean radius $r_{m}$ aur wall thickness $t$ wala thin tube: kyunki area $A = 2 π r_{m} t$ , hume milta hai $t = A / (2 π r_{m}) = 7.854 \times 1 0^{- 5} / (2 π \cdot 0.015) = 8.33 \times 1 0^{- 4} m$ . Uska second moment hai $I_{tube} = π r_{m}^{3} t$ : $I_{tube} = π (0.015)^{3} (8.33 \times 1 0^{- 4}) = 8.836 \times 1 0^{- 9} m^{4}$ Yeh step kyon? $I = π r_{m}^{3} t$ same material ko axis se aur door push karta hai, aur $I$ distance ke square ke saath badhta hai — yahi pura point hai Second Moment of Area ka.
Ratio: $I_{tube} / I_{solid} = 8.836 \times 1 0^{- 9} /4.909 \times 1 0^{- 10} = 18.0$ . Kyunki $P_{cr} \propto I$ , tube 18× zyada strong hai buckling ke against identical mass ke saath.

Figure 2 — Do cross-sections jinki same shaded area hai. Left: solid black disc radius $r_{0}$ ka, material centre ke paas crowded. Right: red ring (tube) mean radius $r_{m}$ ki, same amount ka metal bahar rim pe flung jahan yeh bending ko sabse zyada resist karta hai. Har ek mein black arrow us radius ko mark karta hai jo uske $I$ formula mein use hota hai.

Verify: $P_{cr}$ ratio $I$ ratio ke barabar hai $= 18.0$ ✓ (dono share karte hain $E, L, K$ ). Isliye spacecraft struts tubes hoti hain, solid rods kabhi nahi.

Case G — Safety factor wala real-world word problem

Worked example Ex 7 (Cell G) — safety factor ke saath spacecraft strut sizing

Ek launch strut liftoff ke dauran peak compressive load $P_{applied} = 8000 N$ carry karta hai (Spacecraft Load Paths & Launch Loads). Design rule: yeh $1.5 \times$ us load tak buckle nahi hona chahiye. Fixed–pin ends ( $K = 0.7$ ), $L = 1.2 m$ , titanium $E = 110 GPa$ . Minimum required $I$ nikaalein.

Forecast: 1.5 safety factor aur lamba strut ke saath, $I$ bare-minimum case se noticeably bada expect karo.

Required critical load mein safety factor $n_{s} = 1.5$ include hai: $P_{cr} \geq n_{s} P_{applied} = 1.5 \times 8000 = 12000 N$ Yeh step kyon? Real loads mein uncertainty hoti hai; hum design karte hain taaki collapse load worst expected load se comfortably upar rahe.
Effective length: $L_{e} = K L = 0.7 \times 1.2 = 0.84 m$ . Yeh step kyon? Fixed–pin ends na fully clamped hain na freely pinned, toh jo wavelength fit hoti hai woh physical length se choti hai — hum woh $L_{e}$ mein substitute karne se pehle bake in karte hain, kyunki Euler sirf $L_{e}$ "dekhta" hai, $L$ directly kabhi nahi (dekhein Column End Conditions and Effective Length).
Euler ko $I$ ke liye rearrange karo: $I \geq \frac{P _{cr} L _{e}^{2}}{π ^{2} E} = \frac{12000 ( 0.84 ) ^{2}}{π ^{2} ( 110 \times 1 0 ^{9} )} = 7.797 \times 1 0^{- 9} m^{4}$ Kyon? Same inverse move jaise Ex 2 mein, ab real story ke $K$ aur safety factor ke saath stacked.

Verify: Forward check: $P_{cr} = π^{2} (110 \times 1 0^{9}) (7.797 \times 1 0^{- 9}) /0.8 4^{2} \approx 12000 N = 1.5 \times 8000$ ✓. $I$ ki units: $N \cdot m^{2} / Pa = N \cdot m^{2} / (N / m^{2}) = m^{4}$ ✓.

Case H — Exam twist: buckling ke liye do axes

Worked example Ex 8 (Cell H) — rectangular strut easy way buckle karta hai

Ek pin-pin strut ka rectangular cross-section hai $b \times h = 20 mm \times 10 mm$ , $L = 0.6 m$ , $E = 70 GPa$ . Yeh kis axis ke baare mein buckle karta hai, aur $P_{cr}$ kya hai?

Forecast: rectangle ke do alag $I$ values hote hain. Column kaunsa choose karta hai?

Do second moments (har in-plane axis ke baare mein bending): $I_{1} = \frac{b h ^{3}}{12} = \frac{0.020 ( 0.010 ) ^{3}}{12} = 1.667 \times 1 0^{- 9} m^{4}$ $I_{2} = \frac{h b ^{3}}{12} = \frac{0.010 ( 0.020 ) ^{3}}{12} = 6.667 \times 1 0^{- 9} m^{4}$ Yeh step kyon? Jis axis ke baare mein material paas hai wahan bending se chota $I$ milta hai; jahan door hai, bada $I$ (dekhein Second Moment of Area).
Column pehle mauke par — sabse kam load par — buckle karta hai, toh woh chota $I$ choose karta hai: $I_{m i n} = I_{1} = 1.667 \times 1 0^{- 9} m^{4}$ . Kyon? Nature pehle chote critical load tak pahunchti hai; strong direction ko kabhi mauqa nahi milta.
$P_{cr} = \frac{π ^{2} ( 70 \times 1 0 ^{9} ) ( 1.667 \times 1 0 ^{- 9} )}{0. 6 ^{2}} \approx 3199 N$ Yeh step kyon? Hum Euler mein $I_{m i n}$ substitute karte hain ( $I_{2}$ kabhi nahi), kyunki physical column ne already weak axis ke baare mein fold karne ka choice kar liya hai — bade $I$ ka use strength overpredict karta aur unsafe hota.

Figure 3 — $20 \times 10 mm$ rectangle end-on. Black dashed line strong axis hai (bada $I_{2}$ , material ka wide spread). Red double-arrow thin direction mark karta hai — column is weak axis ke along wide face se bahar bow karta hai, kyunki wahan $I$ sabse chota hai.

Verify: Confirm karo $I_{1} < I_{2}$ : $1.667 \times 1 0^{- 9} < 6.667 \times 1 0^{- 9}$ ✓ (ratio exactly $(b / h)^{2} = 4$ ). $P_{cr} = 3199 N$ $I_{m i n}$ use karke ✓.

Self-check

Which

I

governs buckling of a rectangular strut?

the smallest one — the column folds about its weakest axis

For fixed–free vs pin-pin, how does

P_{cr}

change?

drops by factor

K^{2} = 4

(weaker)

Above what does Euler govern, below what does yield?

above the transition slenderness

λ = π E / σ_{y}

Euler governs; below it, yield

Difference between

r_{0}

and

r

r_{0}

is the physical rod radius (enters

I = π r_{0}^{4} /4

);

r = I / A

is the radius of gyration (enters slenderness

λ = L_{e} / r

)

Why is a tube stronger than a solid rod of equal area?

I = π r_{m}^{3} t

places material far from the axis, giving far larger

I

(and

P_{cr} \propto I

)

In design (inverse) problems, what do you solve Euler for?

the geometry —

I

or radius — after setting

P_{cr} = n_{s} P_{applied}