Finite element method — nodes, elements, stiffness matrix

3.6.18 · Physics › Spacecraft Structures & Systems Engineering

Why FEM Exists

Real spacecraft structures ki geometries complex hoti hain: curved panels, ribs, equipment ke liye cutouts. In shapes ke liye elasticity ke partial differential equations ko analytically solve karna textbook examples se zyada kisi bhi cheez ke liye impossible hai.

FEM ki philosophy: discretization. Continuous structure (infinite degrees of freedom) ko simple shapes ke finite collection se replace karo jinka behavior hum exactly calculate kar sakte hain.

The Three Pillars

1. Nodes

Nodes kyun? Hum har point par displacement track nahi kar sakte — woh infinite unknowns hain. Sirf nodes par track karke aur unke beech interpolate karke, hum problem ko equations ke finite set tak reduce kar dete hain.

Physical meaning: Har node ek physical location represent karta hai. Applied forces/moments nodes par jaati hain. Boundary conditions (fixed supports) node displacements ko constrain karti hain.

2. Elements

Common spacecraft elements:

Beam elements: 2 nodes, truss members, longerons model karta hai
Shell elements: 3-4 nodes, thin panels, skin model karta hai
Solid elements: 4-8 nodes, thick blocks, joints model karta hai

Shape Functions — The Interpolation Magic

Nodes 1 aur 2 wale ek 2-node beam element ke liye, element ke kisi bhi point $x$ par displacement $u (x)$ hai:

$u (x) = N_{1} (x) u_{1} + N_{2} (x) u_{2}$

jahan shape functions $N_{1}, N_{2}$ is tarah choose ki gayi hain:

$N_{1} = 1$ node 1 par, $N_{1} = 0$ node 2 par
$N_{2} = 0$ node 1 par, $N_{2} = 1$ node 2 par
$N_{1} + N_{2} = 1$ har jagah (partition of unity)

Length $L$ par linear interpolation ke liye:

$N_{1} (x) = 1 - \frac{x}{L}, N_{2} (x) = \frac{x}{L}$

Yeh kyun kaam karta hai: Nodes par, hume exact values milti hain. Nodes ke beech, hume ek smooth blend milta hai. Yeh ek continuous field ko sirf node values ka function bana deta hai.

3. Stiffness Matrix

Derivation from First Principles (1D Bar Element)

Ek uniform bar consider karo, length $L$ , cross-section $A$ , Young's modulus $E$ . Do nodes, har ek mein 1 DOF (axial displacement).

Step 1: Shape functions use karke displacement field

$u (x) = N_{1} (x) d_{1} + N_{2} (x) d_{2} = (1 - \frac{x}{L}) d_{1} + \frac{x}{L} d_{2}$

Step 2: Strain-displacement relation

$ϵ (x) = \frac{d u}{d x} = - \frac{1}{L} d_{1} + \frac{1}{L} d_{2} = \frac{1}{L} (d_{2} - d_{1})$

Yeh step kyun? Strain displacement ka gradient hai. Hamare linear shape functions element mein constant strain produce karte hain (chhote elements ke liye acceptable).

Step 3: Hooke's law se stress

$σ = E ϵ = \frac{E}{L} (d_{2} - d_{1})$

Step 4: Internal force

$F_{in t er na l} = σ \cdot A = \frac{E A}{L} (d_{2} - d_{1})$

Step 5: Nodes par equilibrium

Node 1 ko element se tension milta hai: $F_{1} = - F_{in t er na l} = - \frac{E A}{L} (d_{2} - d_{1}) = \frac{E A}{L} (d_{1} - d_{2})$

Node 2 ko compression milta hai: $F_{2} = + F_{in t er na l} = \frac{E A}{L} (d_{2} - d_{1})$

Opposite signs kyun? Newton's third law — element node 1 ko pull karta hai jaise node 1 element ko pull karta hai.

Step 6: Matrix form

$[F_{1} F_{2}] = \frac{E A}{L} [1 - 1 - 1 1] [d_{1} d_{2}]$

Element stiffness matrix hai:

Global Assembly

Har element un nodes mein contribute karta hai jinhein woh connect karta hai. $N$ nodes aur multiple elements wali structure ke liye:

Global stiffness matrix $K_{g l o ba l}$ banao (size $N \times N$ 1D ke liye, $6 N \times 6 N$ 3D ke liye)
Har element ke liye, element DOFs ko global DOFs se map karo
$K^{e}$ entries ko corresponding global positions mein add karo

Example: Global nodes 2 aur 5 ko connect karne wala element. Uska $K_{11}^{e}$ $K_{g l o ba l, 22}$ mein add hoga, $K_{12}^{e}$ $K_{g l o ba l, 25}$ mein add hoga, wagera.

The FEM Workflow

Preprocessing: Geometry define karo, mesh create karo (nodes place karo, elements define karo), material properties assign karo, loads/BCs apply karo
Element calculations: Har element ke liye $K^{e}$ compute karo (analytical formulas ya complex shapes ke liye numerical integration use karo)
Assembly: Global $K$ build karo
Boundary conditions: Fixed DOFs ke rows/columns eliminate karo (ya penalty method use karo)
Solve: $Kd = F$ (typically sparse matrix solver, kyunki zyaatar nodes zyaatar doosre nodes se connect nahi hote)
Postprocessing: $d$ se strains, stresses calculate karo; Von Mises stress check karo, deformed shape plot karo

FEM ek approximation hai. Accuracy improve hoti hai:

Zyada elements (h-refinement): Chhote elements → gradients ko better capture karna
Higher-order shape functions (p-refinement): Linear ki jagah quadratic/cubic interpolation

Rule of thumb: Jahan stress gradients high hon wahan zyada elements rakho (holes, filets, joints ke paas).

Common mistake Common Misconception: "Zyada nodes matlab accurate"

Kyun sahi lagta hai: Zyada data points reality ko better capture karenge.

Sach yeh hai:

Element quality count se zyada matter karti hai. Ek badly shaped element (bahut thin triangle, sharp angle) dense mesh mein bhi nonsense produce karta hai.
Garbage in, garbage out: Wrong material properties ya boundary conditions → mesh density chahे jo bhi ho, galat answer.
Numerical ill-conditioning: Extreme aspect ratios (1:1000 length:width) $K$ mein rounding errors cause karte hain.

Fix yeh hai:

Element quality metrics check karo (aspect ratio < 10, skewness < 0.5)
Mesh convergence study run karo: 1000, 2000, 4000 elements se solve karo. Agar solution <2% change ho, toh converge ho gaya.
Adaptive meshing use karo: software automatically wahan refine karta hai jahan error high ho.

Energy Interpretation

Stiffness matrix potential energy minimize karne se nikalta hai. Ek structure ke liye:

$Π = U_{s t r ain} - W_{e x t er na l}$

jahan strain energy:

$U = \frac{1}{2} d^{T} Kd$

Equilibrium par, $\frac{\partial Π}{\partial d _{i}} = 0$ saare $i$ ke liye → $Kd = F$ .

Yeh kyun matter karta hai: FEM guarantee karta hai ki hum woh displacement field dhundhte hain jisme BCs ke consistent sabse kam total energy ho. Yeh nature ka actual behavior hai (minimum energy principle).

Recall Ek 12-saal ke bacche ko explain karo

Ek spacecraft panel ko ek trampoline socho. Jab tum uس par khade hote ho, woh ek complicated curved shape mein jhuk jaata hai. Hum us curve ko kaise calculate karein?

Hum poore trampoline ke liye "bending equation" ek saath solve nahi kar sakte — woh bahut mushkil hai. Toh hum shortcut use karte hain:

Trampoline ko chhote squares mein kaat do (elements). Har square itna chhota hai ki hum pretend kar sakte hain woh simple tarike se jhukta hai — jaise corners ko seedhi lines se connect karna.
Corners mark karo (nodes). Hum calculate karenge ki har corner kitna neeche move karta hai.
Har chhota square ek spring ki tarah kaam karta hai. Agar tum ek corner neeche push karo, square push back karta hai — woh uski "stiffness" hai. Zyada stiff squares (mota material) zyada push back karte hain.
Saare springs connect karo. Corners jahan squares milte hain unse multiple springs attached hote hain — unki stiffnesses add up ho jaati hain.
Spring system solve karo. Agar main ek corner par 100 N apply karun aur doosra corner fixed rakhuun, toh har corner kitna move karta hai? Woh toh sirf spring equations hain — easy!
Squares chhote banao. Jitne chhote squares, utna haari seedhi-line approximation real smooth curve se milti hai.

Yahi FEM hai: Ek mushkil "continuous" problem ko bahut saare easy "discrete" spring problems mein badlo. Spacecraft engineers iska use karte hain predict karne ke liye ki satellite panel launch vibrations ke under crack karega ya nahi — 100 prototypes banaye aur test kiye bina.

Connections

3.6.1-Stress-and-strain-in-structural-members — FEM inheen quantities ke liye solve karta hai
3.6.5-Buckling-of-columns-and-shells — Eigenvalue problem: $(K - λ K_{G}) ϕ = 0$
3.6.12-Modal-analysis-and-natural-frequencies — $(K - ω^{2} M) ϕ = 0$ , same $K$ + mass matrix
2.4.8-Numerical-methods-for-differential-equations — FEM PDEs ke liye spatial discretization hai
3.7.4-Thermal-analysis-of-spacecraft-components — Same FEM framework, stiffness ki jagah conductance

#flashcards/physics

Question: Finite element method ke peeche fundamental idea kya hai? :: Ek continuous structure (infinite unknowns) ko nodes par connected simple elements ke finite mesh se replace karo (finite unknowns), jahan displacement har element ke andar shape functions use karke interpolate kiya jaata hai.

Question: Boundary conditions apply karne se pehle element stiffness matrix singular kyun hoti hai?

Kyunki structure rigid body motion (translation/rotation) kar sakta hai zero strain energy ke saath. Constraints ke bina, infinite solutions hain — system underconstrained hai.

Question: 1D bar element stiffness matrix

K^{e} = \frac{E A}{L} [1 - 1 - 1 1]

first principles se derive karo.

(1) Displacement

u (x) = N_{1} d_{1} + N_{2} d_{2}

linear shape functions ke saath. (2) Strain

ϵ = d u / d x = (d_{2} - d_{1}) / L

. (3) Stress

σ = E ϵ

. (4) Force

F = σ A = E A (d_{2} - d_{1}) / L

. (5) Equilibrium: node 1 ko

- F

milta hai, node 2 ko

+ F

milta hai. (6)

[F_{1}, F_{2}]^{T}

[d_{1}, d_{2}]^{T}

se relate karne wale matrix ke roop mein likho.

Question: Global assembly mein, agar node 5 ko 3 elements share karte hain, toh

K_{g l o ba l}

mein uski row kaise populate hoti hai?

Teeno element stiffness matrices se contributions add karo — node 5 se connected springs ki stiffnesses add up ho jaati hain. Isliye shared nodes mein aksar

K

mein bade diagonal entries hote hain.

Question: FEM accuracy improve karne ke kaunse do types ke mesh refinement hain?

h-refinement (zyada elements, chhota size) steep gradients ko better capture karta hai. p-refinement (higher-order shape functions jaise quadratic) nodes add kiye bina smooth variations ko better approximate karta hai.

Question: Sirf element count nahi, element quality kyun check karni chahiye?

Badly shaped elements (extreme aspect ratio, sharp angles) dense mesh mein bhi numerical errors aur nonsense results cause karte hain. Aspect ratio < 10 aur skewness < 0.5 jaisi quality metrics critical hain.

Question:

Kd = F

ki energy interpretation kya hai?

Yeh minimum total potential energy

Π = \frac{1}{2} d^{T} Kd - F^{T} d

ki condition hai.

\frac{\partial Π}{\partial d _{i}} = 0

set karne se equilibrium equation milti hai. FEM woh displacement field dhundhta hai jo energy minimize kare.

Question:

N

nodes aur 6 DOF per node wali 3D structure ke liye, BCs apply karne se pehle

K_{g l o ba l}

ki size kya hai?

6 N \times 6 N

. Har node mein translations

(u_{x}, u_{y}, u_{z})

aur rotations

(θ_{x}, θ_{y}, θ_{z})

hote hain, toh har node ke liye 6 equations.