Finite element method — nodes, elements, stiffness matrix: Visual walkthrough

3.6.18 · D2 · Physics › Spacecraft Structures & Systems Engineering › Finite element method — nodes, elements, stiffness matrix

Yahan sirf itna assume kiya gaya hai ki tum jaante ho length, force, aur ratio kya hote hain. Har nayi idea — stretch, strain, stiffness — pehle wali picture ke upar build hoti hai. Agar ek stretched rubber band aur spring tum samajhte ho, tum ready ho.

Step 1 — Rod draw karo aur uske do ends ko naam do

KYA. Hum ek seedhi rod lete hain. Hum uske do ends par dots lagate hain aur unhe node 1 (baayein) aur node 2 (daayein) kehte hain. Node sirf ek tracked point hai — ek jagah jahan hum agree karte hain ki material kitna move hua, ye measure karenge.

KYUN. Hum rod ke har ek atom ko track nahi kar sakte; wo infinitely many numbers hote. Finite element method ki poori trick yahi hai: sirf ends ko track karo, phir ek rule se beech ko bharo. Toh sabse pehla kaam in do ends ko choose karna hai.

PICTURE. Rod flat padi hai. Uski natural (unloaded) length $L$ hai. Baayein dot node 1 hai, daayein dot node 2 hai.

Figure — Finite element method — nodes, elements, stiffness matrix

Teen symbols ab meaning rakhte hain, picture se anchored:

$L$ — rod ki rest length (dots ke beech ki doori, kuch hone se pehle).
$A$ — rod ki cross-section area (kitni moti hai, end se dekhi jaaye).
$E$ — Young's modulus, material kitna stiff hai uske liye ek number (steel mein bada, rubber mein chhota). Hum isse Step 4 mein properly milenge; dekho 3.6.1-Stress-and-strain-in-structural-members.

Step 2 — Do end values se andar ke har point ko describe karo

KYA. Rod ke andar koi bhi point lo, node 1 se $x$ doori par. Uska displacement $u (x)$ do end displacements ka ek weighted blend hai: $u (x) = N_{1} (x) (1 - \frac{x}{L}) d_{1} + N_{2} (x) \frac{x}{L} d_{2}$

Term by term padhte hain:

$d_{1}$ — baayein end ki movement, woh raw quantity jise hum blend kar rahe hain.
$(1 - \frac{x}{L})$ — node 1 ki movement point $x$ tak kitni pahunchti hai. Baayein end par poori ( $= 1$ ), daayein jaate-jaate $0$ ho jaati hai.
$d_{2}$ — daayein end ki movement.
$\frac{x}{L}$ — node 2 ki movement $x$ tak kitni pahunchti hai. Baayein sar zero, daayein sar poori.

Ye do weights $N_{1}, N_{2}$ shape functions hain.

KYUN. Hume sirf ends se middle guess karne ka koi rule chahiye. Sabse simple aur honest guess hai do end movements ke beech ek straight line. Ek straight line exactly wohi linear blend hai jo upar diya gaya hai — aur isme woh ek property hai jo hum insist karte hain: har node par, blend uss node ki apni value wapas deta hai ( $N_{1} = 1$ node 1 par, aur node 2 par $= 0$ ). Ek chhoti rod ke liye kuch zyada fancy zaroori nahi hai.

PICTURE. Do shape functions seedhi ramps hain jo X ki tarah cross karti hain: $N_{1}$ $1$ se $0$ tak slide karti hai, $N_{2}$ $0$ se $1$ tak chadhti hai. Har $x$ par ye $1$ mein add hoti hain (kuch lost nahi hota) — ise partition of unity kehte hain.

Recall

$N_{1} + N_{2} = 1$ har jagah kyun hona chahiye? Agar poori rod daayein ek hi amount $c$ se slide kare (toh $d_{1} = d_{2} = c$ ), koi bhi point alag nahi hila hona chahiye: $u (x) = N_{1} c + N_{2} c = (N_{1} + N_{2}) c$ . Har $x$ par ye $c$ ke barabar ho, iske liye hume $N_{1} + N_{2} = 1$ chahiye. ::: Weights ka sum ek hona chahiye taaki pure sliding se pure sliding mile — koi fake stretching nahi.

Step 3 — Stretch measure karo: strain

KYA. Stretch ka matlab "kitna move hua" nahi, balki "per length kitna move hua" hai. Ye per-length stretch strain hai, jise $ϵ$ likhte hain. Hume ye milti hai ye poochhne se ki $u$ kitni tezi se badlti hai jab hum $x$ ke saath chalte hain — yaani $u$ -versus- $x$ line ka slope: $ϵ = \frac{d u}{d x} = - \frac{1}{L} d_{1} + \frac{1}{L} d_{2} = \frac{d _{2} - d _{1}}{L}$

Term by term:

$\frac{d u}{d x}$ — derivative: displacement line ka slope, yaani movement mein change divided by position mein change.
$- \frac{1}{L} d_{1}$ — node 1 ka daayein move karna rod ko un-stretch karta hai (baayein end ko daayein end ki taraf push karta hai), isliye minus.
$+ \frac{1}{L} d_{2}$ — node 2 ka daayein move karna rod ko stretch karta hai.
$\frac{d _{2} - d _{1}}{L}$ — gap change $(d_{2} - d_{1})$ ko rest length $L$ par spread kiya.

KYUN derivative aur kuch nahi? Strain ye jawaab deta hai "original length ke fraction ke roop mein kitna lamba hua?" Ye rod ke saath position mein change ke against position ki rate of change hai — aur woh tool jo ek quantity ki doosri ke against rate of change measure karta hai woh derivative hai. Koi doosra operation tumhe "per unit length" nahi deta.

Gaur karo: kyunki $u$ ek straight line hai (Step 2), uska slope har jagah same hai. Toh is element mein ek constant strain hai poori length mein — theek hai, jab tak element chhota ho.

PICTURE. Seedhi $u (x)$ line ka ek hi slope hai; woh slope $ϵ$ hai. Stretch $=$ (nayi length $-$ purani length) $\div$ purani length.

Step 4 — Stretch ko internal push/pull mein badlo: Hooke's law

KYA. Ek stretched rod wapas kheenchti hai. Internal pull-per-area stress $σ$ hai, jo strain se Hooke's law ke zariye judi hai: $σ = E ϵ = \frac{E}{L} (d_{2} - d_{1})$

$σ$ — cross-section par spread internal force (force ÷ area).
$E$ — Young's modulus: material ka stiffness constant. Bada $E$ ⇒ same stretch ke liye zyada stress.
$ϵ$ — Step 3 se strain.

Phir total internal force hai stress times kitna area kheench raha hai: $F_{internal} = σ A = \frac{E A}{L} (d_{2} - d_{1})$

$A$ — cross-section area; moти rod same stress par zyada total force carry karti hai.
Bundle $\frac{E A}{L}$ rod ka spring constant hai — stiffness $=$ (material stiffness $\times$ thickness) ÷ length.

KYUN. Chhoti stretches ke liye, real materials ek linear spring ki tarah behave karte hain: stretch double karo, pull double ho jaata hai. Hooke's law usi straight-line behaviour ka mathematical statement hai, aur yehi ek puri continuous rod ko ek single number $E A / L$ — ek plain spring — se replace karne deta hai. Compare karo 3.6.1-Stress-and-strain-in-structural-members se.

PICTURE. Rod ko stiffness $k = E A / L$ ki ek coil spring ke roop mein redraw kiya gaya hai. Lambi rod = softer spring; moти ya stiffer material = stronger spring.

Step 5 — Har end par forces balance karo (Newton's third law)

KYA. Element apne nodes par kheenchta hai; reaction mein nodes forces $F_{1}, F_{2}$ feel karte hain. Likho ki har end kya feel karta hai: $F_{1} = - F_{internal} = \frac{E A}{L} (d_{1} - d_{2}), F_{2} = + F_{internal} = \frac{E A}{L} (d_{2} - d_{1})$

$F_{2} = + k (d_{2} - d_{1})$ — agar node 2 node 1 se aage hai, rod stretched hai aur node 2 ko wapas baayein kheenchti hai... ruko: $F_{2}$ yahan woh force hai jo node ko us stretched state ko hold karne ke liye supply karni padti hai, isliye ye outward point karta hai. Sign convention: ek stretched rod ko stretch mein rakhne ke liye dono ends par outward forces chahiye.
$F_{1} = - F_{2}$ — equal aur opposite. Jo bhi hum ek end par push karte hain, doosra end mirror image feel karta hai.

OPPOSITE SIGNS KYUN? Newton's third law. Rod force invent nahi kar sakti; node 1 par har pull node 2 par opposite direction mein equal pull se matched hai. Isliye $F_{1} = - F_{2}$ hamesha hota hai, aur ye woh deep reason hai ki matrix ki rows mein se har ek ka sum zero hoga.

PICTURE. Spring ke ends par do arrows: stretched rod ke liye ye ek doosre se dur point karte hain; squashed rod ke liye ye ek doosre ki taraf point karte hain. Same size, opposite directions.

Step 6 — Do equations ko ek matrix mein pack karo

KYA. Do force equations ko expand karo: $F_{1} = \frac{E A}{L} d_{1} - \frac{E A}{L} d_{2}, F_{2} = - \frac{E A}{L} d_{1} + \frac{E A}{L} d_{2}$ Inhe stack karo aur $\frac{E A}{L}$ factor out karo: $[F_{1} F_{2}] = \frac{E A}{L} [1 - 1 - 1 1] [d_{1} d_{2}]$

Matrix ko influences ke grid ke roop mein padhte hain, $K_{ij}$ = node $j$ ki unit movement se node $i$ par force:

top-left $1$ : node 1 ko ek unit daayein move karo ⇒ node 1 par force $+ k$ (ye resist karta hai).
top-right $- 1$ : node 2 ko ek unit daayein move karo ⇒ node 1 par force $- k$ (ye saath kheench jaata hai).
bottom row: mirror image, Newton's third law se.

MATRIX KYUN? Kyunki do forces do displacements par depend karte hain, aur har force dono displacements par depend karta hai. Matrix exactly wohi bookkeeping device hai "har output har input se bana." $\frac{E A}{L}$ ek constant number mein aur pure pattern $[1 - 1 - 1 1]$ mein clean split geometry of connection (pattern) ko material size (number) se alag dikhata hai.

PICTURE. Ek 2×2 grid jiske cells sign se coloured hain — orange $+$ ke liye, teal $-$ ke liye — har column (ek displacement) se har row (usse produce hone wali force) tak arrow ke saath.

Step 7 — Degenerate case: $K^{e}$ invert kyun nahi ho sakti

KYA. Poori rod ko same amount daayein slide karo: $d_{1} = d_{2} = c$ . Plug in karo: $K^{e} [c c] = \frac{E A}{L} [c - c - c + c] = [00]$ Nonzero motion se zero force. Determinant hai: $det K^{e} = (\frac{E A}{L})^{2} (1 \cdot 1 - (- 1) (- 1)) = 0.$

YE KYUN MATTER KARTA HAI. Ek rigid slide kuch stretch nahi karta, isliye koi energy store nahi hoti aur koi force nahi chahiye — ye ek rigid body mode hai. Mathematically iska matlab hai ki kaafi saare displacement states same (zero) force dete hain, toh hum nahi kar sakte $K^{e}$ ko undo karke unique $d$ recover karein. Ilaaj, jo parent note ki assembly mein apply hota hai, ek boundary condition hai: kam se kam ek node ko pin karo taaki sliding forbidden ho. Tabhi reduced matrix invertible banti hai.

PICTURE. Rod space mein float kar rahi hai dono arrows equal ke saath — ye drift karta hai, koi spring stretch nahi, koi restoring force nahi. Ek ghost copy dikhata hai isse node 1 par pinned, ab well-behaved.

Ek-picture summary

KYA. Ek diagram poori chain chalata hai: do dots → ek straight blend → ek slope (strain) → ek spring pull (stress×area) → equal-and-opposite end forces → 2×2 matrix, saath mein zero-determinant rigid slide side mein laga hua.

Yehi recipe, fancier shape functions aur zyada DOFs ke saath, beam, shell, aur solid matrices produce karta hai jo 3.6.12-Modal-analysis-and-natural-frequencies, 3.6.5-Buckling-of-columns-and-shells, aur 3.7.4-Thermal-analysis-of-spacecraft-components mein use hoti hain.

Recall Feynman retelling — plain words mein wapas bol do

Humare paas ek rod thi aur humne refuse kiya ki har point track karein, toh humne sirf uske do ends aur unki movements $d_{1}, d_{2}$ track kiye. Beech bharne ke liye humne do end movements ke beech ek straight line kheenchi — wohi shape-function blend hai. Us line ka slope batata hai rod per unit length kitni stretch hui; wohi strain hai. Strain ko material stiffness $E$ se multiply karo to internal stress mile, aur area $A$ se multiply karo to total pull mile; wohi ek plain spring hai stiffness $k = E A / L$ ke saath. Ek stretched spring apne dono ends ko outward equal aur opposite force se dhakelta hai, jo simply Newton's third law hai. "Har end par force, har end ki movement ke terms mein" likhne ke liye ek chhota 2×2 grid chahiye — aur nikalta hai $\frac{E A}{L} [1 - 1 - 1 1]$ . $+ 1$ 's woh node hai jo apni movement resist karta hai; $- 1$ 's woh node hai jo apne neighbour dwara drag hota hai. Finally, agar tum poori rod ko bina stretch kiye slide karo, tum bilkul koi force feel nahi karte — toh matrix ka determinant zero hai aur invert nahi ho sakti jab tak rod ko pin na karo.

Prerequisites revisited: 3.6.1-Stress-and-strain-in-structural-members · 2.4.8-Numerical-methods-for-differential-equations Where this leads: 3.6.12-Modal-analysis-and-natural-frequencies · 3.6.5-Buckling-of-columns-and-shells · 3.7.4-Thermal-analysis-of-spacecraft-components