Finite element method — nodes, elements, stiffness matrix: Worked examples

3.6.18 · D3 · Physics › Spacecraft Structures & Systems Engineering › Finite element method — nodes, elements, stiffness matrix

Yeh deep dive assume karta hai ki tumne parent note padh li hai aur tumhe pata hai ki node, element, shape function, aur 1D bar stiffness matrix $K^{e} = \frac{E A}{L} [1 - 1 - 1 1]$ kya hote hain. Yahan theory mein kuch naya nahi hai — hum sirf un tools ko har tarah ki situation par use karte hain jo is topic mein aa sakti hain, taaki koi bhi exam ya real design task tumhe surprise na kare.

Shuru karne se pehle, teen simple reminders taaki neeche har symbol ka matlab samajh aaye:

Recall Har symbol ka matlab (tap karke kholein)

$E$ ::: Young's modulus — material kitna stretching resist karta hai (stiff = bada $E$ ), units pascals ( $Pa = N / m^{2}$ ). $A$ ::: bar ki cross-section area — kitni moti hai, units $m^{2}$ . $L$ ::: element ki length, units metres. $k = E A / L$ ::: ek bar ka "spring constant" — ise force $k$ se kheencho aur yeh 1 metre stretch hogi. Units $N / m$ . $d_{i}$ ::: node $i$ ka displacement (kitna hila), units metres. $F_{i}$ ::: node $i$ par lagaya gaya force, units newtons.

The scenario matrix

Is topic mein jo bhi problem ho sakti hai woh in case classes mein se ek hogi. Neeche diye worked examples mein label hai ki woh kaun sa cell cover karte hain, taaki milke poori table cover ho jaye.

#	Case class	Kya cheez ise alag banati hai	Covered by
C1	Single element, tension	positive displacement, force kheenchta hai	Ex 1
C2	Single element, compression	negative displacement, sign flip	Ex 2
C3	Two elements in series, assembly	global $K$ ke beech mein "2"	Ex 3
C4	Different stiffnesses $k_{1} \neq = k_{2}$	non-symmetric assembly, weakest link	Ex 4
C5	Degenerate / singular $K$	koi boundary condition nahi → rigid body motion	Ex 5
C6	Zero / limiting input	$A \to 0$ , $L \to \infty$ , $E \to \infty$	Ex 6
C7	Real-world word problem	spacecraft strut sizing, units har jagah	Ex 7
C8	Exam twist — load middle node par, dono ends fixed	force stiffness ratio se split hoti hai	Ex 8

Example 1 — Single bar in tension (C1)

Forecast: pehle andaza lagao — node 2 node 1 ki taraf jayega ya door, aur roughly kitne microns?

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

Spring constant $k = E A / L$ compute karo. Yeh step kyun? Ek bar bilkul ek spring ki tarah behave karta hai; $k$ mein saari geometry aur material aa jaati hai. $k = \frac{E A}{L} = \frac{( 70 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 4} )}{0.5} = 2.8 \times 1 0^{7} N/m$
$F = Kd$ likho aur fixed row delete karo. Yeh step kyun? Node 1 hil nahi sakta, isliye uski row mein koi unknown nahi; hum sirf node 2 ki equation rakhte hain. $k d_{2} = F_{2}$
Displacement ke liye solve karo. Yeh step kyun? Yahi FEM ka poora point hai — displacement hamaara primary unknown hai. $d_{2} = \frac{F _{2}}{k} = \frac{1400}{2.8 \times 1 0 ^{7}} = 5.0 \times 1 0^{- 5} m = 50 μ m$
Strain se phir Hooke's law ke zariye stress recover karo (parent ke Steps 2–3). Yeh step kyun? Displacement akela kisi structure ko nahi todta — stress todta hai. $ε = \frac{d _{2} - d _{1}}{L} = \frac{5.0 \times 1 0 ^{- 5}}{0.5} = 1 0^{- 4}, σ = E ε = 70 \times 1 0^{9} \times 1 0^{- 4} = 7 MPa$

Verify: $d_{2} > 0$ → node node 1 se door jaata hai (bahar ka pull → tension → elongation). ✔ Units: $N \div (N/m) = m$ . ✔ 7 MPa aluminium yield (~270 MPa) se kaafi neeche hai, isliye linear elastic FEM valid hai. ✔ Dekho 3.6.1-Stress-and-strain-in-structural-members.

Example 2 — Single bar in compression (C2)

Forecast: node 2 kis taraf jayega, aur kya hamare equations mein kuch badlega?

$K$ mein kuch nahi badlega — sirf $F_{2}$ ka sign badlega. Yeh step kyun? Stiffness geometry/material ki property hai, load direction ki nahi. Sirf right-hand side flip hoti hai. $d_{2} = \frac{- 1400}{2.8 \times 1 0 ^{7}} = - 5.0 \times 1 0^{- 5} m$
Negative sign interpret karo. Yeh step kyun? FEM mein minus sign error nahi hai — yeh direction ki information hai. Negative $d_{2}$ matlab node 2 node 1 ki taraf jaata hai: bar chhoti ho jaati hai → compression.
Stress. $σ = E \frac{d _{2} - d _{1}}{L} = 70 \times 1 0^{9} \cdot \frac{- 5.0 \times 1 0 ^{- 5}}{0.5} = - 7 MPa$ Negative sign batata hai yeh compressive hai.

Verify: magnitude Ex 1 jaisi, sign ulta — kyunki hamaari equation linear hai ( $d_{2} \propto F_{2}$ ). ✔ Real strut jo compression mein ho usse 3.6.5-Buckling-of-columns-and-shells ke against bhi check karna chahiye; FEM axial stress akela buckling nahi pakadta. ✔

Example 3 — Two bars in series, assembly (C3)

Forecast: kaun sa node zyada hil ta hai, aur kis ratio mein?

Har element ko global $3 \times 3$ matrix mein assemble karo. Yeh step kyun? Node 2 shared hai, isliye dono elements position $(2, 2)$ mein stiffness dalte hain — woh add hoti hain. $K_{global} = k 1 - 1 0 - 1 2 - 1 0 - 1 1$
Boundary condition $d_{1} = 0$ lagao: row/column 1 hatao. Yeh step kyun? Rigid-body freedom hata deta hai (Ex 5 dekho) aur solvable $2 \times 2$ mein shrink karta hai. $k [2 - 1 - 1 1] [d_{2} d_{3}] = [0 P]$
$2 \times 2$ system solve karo. Row 1: $2 d_{2} - d_{3} = 0 \Rightarrow d_{3} = 2 d_{2}$ . Row 2: $- d_{2} + d_{3} = P / k$ . Substitute karo: $d_{2} = P / k$ . $d_{2} = \frac{P}{k}, d_{3} = \frac{2 P}{k}$

Verify: $d_{3} / d_{2} = 2$ — node 3 wall se do springs door hai, isliye double stretch accumulate karta hai. ✔ Har bar same internal force $P$ carry karta hai (series springs load share karte hain): bar 1 stretch $= d_{2} - 0 = P / k$ , bar 2 stretch $= d_{3} - d_{2} = P / k$ . Equal. ✔

Example 4 — Unequal stiffnesses, weakest link (C4)

Forecast: kaun sa bar zyada stretch hoga, aur kya node 3 ki total motion soft bar ya stiff bar jaise dikhegi?

Do alag $k$ 's ke saath assemble karo. Yeh step kyun? Position $(2, 2)$ ab $k_{1} + k_{2}$ milega, $2 k$ nahi. $K_{global} = k_{1} - k_{1} 0 - k_{1} k_{1} + k_{2} - k_{2} 0 - k_{2} k_{2}$
Row/column 1 delete karo (node 1 fixed). $[k_{1} + k_{2} - k_{2} - k_{2} k_{2}] [d_{2} d_{3}] = [0 P]$
Solve karo. Row 1 se: $(k_{1} + k_{2}) d_{2} = k_{2} d_{3}$ . $k_{1} = 3 k, k_{2} = k$ ke saath: $4 k d_{2} = k d_{3} \Rightarrow d_{3} = 4 d_{2}$ . Row 2: $- k d_{2} + k d_{3} = P \Rightarrow d_{2} (- 1 + 4) = P / k$ . $d_{2} = \frac{P}{3 k}, d_{3} = \frac{4 P}{3 k}$
Bar stretches compare karo. Stiff bar 1: $d_{2} - 0 = P /3 k$ . Soft bar 2: $d_{3} - d_{2} = 4 P /3 k - P /3 k = P / k$ .

Verify: har bar phir se force $P$ carry karta hai (series). Bar 1 stretch $= P / k_{1} = P /3 k$ ✔, bar 2 stretch $= P / k_{2} = P / k$ ✔ — soft bar 3× zyada stretch hoti hai. Structure ki compliance sabse weak link se dominate hoti hai. ✔

Example 5 — The degenerate, singular case (C5)

Forecast: andaza lagao ki kya unique displacement exist karta hai.

Full $3 \times 3$ $K$ ka determinant compute karo. Yeh step kyun? Unique solution ke liye $det \neq = 0$ chahiye. Yeh solvable system ka litmus test hai. $det k 1 - 1 0 - 1 2 - 1 0 - 1 1 = 0$
Rigid-body mode identify karo. Yeh step kyun? Zero determinant matlab ek nonzero $d$ exist karta hai jisme $Kd = 0$ . Usse dhundho. $d = (1, 1, 1)^{⊤}$ se $Kd = 0$ : har node same distance slide karta hai → koi bar stretch nahi → zero internal force. Yeh structure space mein float kar raha hai.
Fix karo. Yeh step kyun? System solvable hone se pehle hume free-float freedom hatani hogi. Kam se kam ek DOF fix karo (1D mein). 2D mein 3 non-colinear DOFs fix karo; 3D mein 6. Isse rigid-body columns delete ho jaate hain aur $det \neq = 0$ aa jaata hai.

Verify: $K$ ki row sums hain $1 - 1 + 0 = 0$ , $- 1 + 2 - 1 = 0$ , $0 - 1 + 1 = 0$ — har row zero sum karta hai, yeh rigid-body (constant-displacement) null vector ka algebraic fingerprint hai. ✔ Yeh parent ki warning se match karta hai ki unconstrained $K$ singular hoti hai.

Example 6 — Zero and limiting inputs (C6)

Forecast: kaun si limits strut ko floppy banati hain (bada displacement) aur kaun si rigid (zero displacement)?

(a) Vanishing area $A \to 0$ . Yeh step kyun? Ek degenerate geometry test karta hai — bina material wala bar. $d_{2} = \frac{F _{2} L}{E A} A \to 0 + \infty$ Koi cross-section nahi matlab koi resistance nahi: infinitely thin strut bina limit ke stretch hogi. Physically, woh toot chuki hoti — yeh signal hai ki model apni valid range se bahar chali gayi.
(b) Infinite length $L \to \infty$ . $d_{2} = \frac{F _{2} L}{E A} L \to \infty + \infty$ Lambi springs softer hoti hain ( $k = E A / L \to 0$ ): series mein zyada material zyada total stretch accumulate karta hai. Sensible hai.
(c) Infinite stiffness $E \to \infty$ . $d_{2} = \frac{F _{2} L}{E A} E \to \infty 0$ Perfectly rigid material bilkul deform nahi karta — FEM answer rigid-body idealisation par collapse karta hai.

Verify: teen limits ( $\infty, \infty, 0$ ) intuition se match karte hain: softness displacement badhati hai, stiffness khatam karti hai. ✔ Dhyan raho ki jab $k \to 0$ hota hai toh element stiffness matrix $0$ ki taraf jaati hai, jo phir se singular hai — Ex 5 ka limiting form. ✔

Example 7 — Real-world word problem: spacecraft strut sizing (C7)

Forecast: area ka andaza $mm^{2}$ mein lagao — kuch, kuch sau, ya kuch hazaar?

$d = F L / (E A)$ se stiffness requirement likho. Yeh step kyun? Deflection yahan binding design limit hai, isliye $A$ ke liye displacement formula invert karte hain. $A \geq \frac{F L}{E d _{m a x}}$
Numbers substitute karo (SI throughout). Yeh step kyun? Mixed units orders-of-magnitude errors ka #1 source hain — millimetres ko pehle metres mein convert karo: $d_{m a x} = 0.2 \times 1 0^{- 3} m$ . $A \geq \frac{5000 \times 0.8}{110 \times 1 0 ^{9} \times 0.2 \times 1 0 ^{- 3}} = \frac{4000}{2.2 \times 1 0 ^{7}} = 1.818 \times 1 0^{- 4} m^{2} = 181.8 mm^{2}$
Resulting stress ko titanium yield (~830 MPa) ke against check karo. Yeh step kyun? Stiffness meet karna strength guarantee nahi karta — dono check karo hamesha. $σ = \frac{F}{A} = \frac{5000}{1.818 \times 1 0 ^{- 4}} = 2.75 \times 1 0^{7} Pa = 27.5 MPa$ Safety factor $= 830/27.5 \approx 30$ — is design mein stiffness govern karti hai, strength nahi.

Verify: $A = 1.818 \times 1 0^{- 4}$ back-substitute karo: $d = \frac{5000 \times 0.8}{110 \times 1 0 ^{9} \times 1.818 \times 1 0 ^{- 4}} = 2.0 \times 1 0^{- 4} m = 0.2 mm$ — exactly limit par. ✔ $A$ ki units: $\frac{N \cdot m}{Pa \cdot m} = \frac{N \cdot m}{( N/m ^{2} ) \cdot m} = m^{2}$ . ✔ Thermal load isme add ho sakti hai — dekho 3.7.4-Thermal-analysis-of-spacecraft-components.

Example 8 — Exam twist: middle node par load, dono ends fixed (C8)

Forecast: agar $k_{1} = k_{2}$ ho toh kya node 2 bilkul hilta hai? Agar bar 1 bahut zyada stiff ho toh kaun sa bar $P$ ka zyada hissa leta hai?

Assemble karo, phir dono boundary conditions lagao. Yeh step kyun? Node 1 aur node 3 dono fixed hone se sirf node 2 ki row bachti hai — ek single equation. Global $(2, 2)$ entry hai $k_{1} + k_{2}$ ; nodes 1 aur 3 ke saath coupling terms zero displacements multiply karte hain aur vanish ho jaate hain. $(k_{1} + k_{2}) d_{2} = P$
Solve karo. $d_{2} = \frac{P}{k _{1} + k _{2}}$ Dono bars yahan parallel springs ki tarah act karte hain (dono simultaneously node 2 ki motion resist karte hain), isliye unki stiffnesses add hoti hain.
Har bar mein force. Yeh step kyun? Har fixed end jo reaction feel karta hai wahi exam usually poochhta hai. Bar 1 $d_{2}$ se stretch hoti hai: $F_{1} = k_{1} d_{2} = \frac{k _{1}}{k _{1} + k _{2}} P$ . Bar 2 $d_{2}$ se compress hoti hai: $F_{2} = k_{2} d_{2} = \frac{k _{2}}{k _{1} + k _{2}} P$ .

Verify: $F_{1} + F_{2} = \frac{( k _{1} + k _{2} )}{k _{1} + k _{2}} P = P$ ✔ (load fully reacted). $k_{1} = k_{2} = k$ ke saath: $d_{2} = P /2 k$ aur har bar $P /2$ leta hai — symmetric, as expected. ✔ Stiffer bar bada share leta hai — series case (Ex 4) ke opposite, jahan softer bar zyada stretch hoti thi. Jaano kis topology mein ho tum!

Recall Self-test

Ex 4 mein kaun sa bar zyada stretch hua aur kyun? ::: Soft bar ( $k_{2} = k$ ), kyunki series mein dono bars same force $P$ carry karte hain aur stretch $= P / k$ hoti hai, isliye chota $k$ → badi stretch. Ex 8 mein $k_{1} ≫ k_{2}$ ke saath, kaun sa bar $P$ ka zyada hissa carry karta hai? ::: Bar 1 (stiff wala), fraction $k_{1} / (k_{1} + k_{2}) \to 1$ . Unconstrained $K$ singular kyun hoti hai? ::: Ek rigid-body mode $(1, 1, 1)$ zero force deta hai → non-unique solutions → $det = 0$ .

Dynamic loads aur in stiffness matrices ka natural-frequency calculations mein kaise use hota hai, yeh jaanne ke liye 3.6.12-Modal-analysis-and-natural-frequencies par jao; underlying solver mathematics ke liye dekho 2.4.8-Numerical-methods-for-differential-equations.