Worked examples — FEM for structures — assembling global stiffness

3.6.19 · D3 · Physics › Spacecraft Structures & Systems Engineering › FEM for structures — assembling global stiffness

Yeh page parent topic on assembling global stiffness ki "battle-tested" companion hai. Parent ne dikhaya tha ki scatter-add kaise kaam karta hai. Yahan hum har situation drill karenge jo aap ko mil sakti hai: shared nodes, disconnected pieces, parallel bars, rotated bars, degenerate (rigid-body) matrices, aur wo exam twists jo logon ko trip karte hain.

Shuru karne se pehle, ek vaada: koi bhi symbol use nahi hoga jab tak usse draw na kar diya jaaye. Agar aapne kabhi stiffness matrix nahi dekha, toh pehle parent ka "Core Idea" padho — lekin neeche sab kuch apni notation khud earn karta hai.

Recall Teen words jinka hum baar baar use karenge

Node ::: mesh mein ek point jahan elements milte hain; displacements yahan measure hote hain. DOF (degree of freedom) ::: ek independent tarika jisme ek node move kar sakta hai (jaise horizontal, ya vertical). Hum inhe $1,2,3,\dots$ se number karte hain poore mesh mein. Scatter-add ::: chhote element matrix ko lo aur uski har entry ko bade global matrix $\mathbf{K}$ mein add karo, un rows/columns par jo us element ke global DOF numbers se milti hain.

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Scenario Matrix

Har assembly problem neeche diye cells ka mixture hota hai. Jo worked examples follow karte hain unhe un cell(s) ke label ke saath likha gaya hai jo wo cover karte hain, isliye mil ke ye poora table cover karte hain.

Cell	Kya tricky hai	Covered by
A. Series chain (shared interior node)	overlapping entries ko sum karna padta hai	Ex 1
B. Disconnected / no shared node	zero coupling entry; block-diagonal $\mathbf{K}$	Ex 2
C. Sign bookkeeping	off-diagonals negative kyun, diagonals positive kyun	Ex 3
D. Zero / degenerate (rigid-body) matrix	$\det \mathbf{K}=0$ boundary conditions se pehle	Ex 4
E. Rotated element (transformation)	local $\to$ global, angle $\ne 0$; aur $\theta =0^{\circ }$ limit	Ex 5
F. Vertical bar (limiting angle $90^{\circ }$)	$\cos \theta =0$: kaun si entries survive karti hain?	Ex 6
G. Parallel bars (same node pair)	do elements dono nodes share karte hain $\Rightarrow$ stiffnesses add hoti hain	Ex 7
H. Real-world word problem	spacecraft bracket, DOFs khud choose karo	Ex 8
I. Exam twist (DOFs mis-numbered)	non-adjacent nodes $\Rightarrow$ scattered off-diagonals	Ex 9

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Cell A — Series Chain (Shared Interior Node)

Forecast: padhne se pehle guess karo ki kaun si ek entry $k$ ki jagah $2k$ hogi.

Step 1 — Bar I ko DOFs $[1,2]$ par place karo. $\mathbf{K}\leftarrow k\left[\begin{array}{ccc}1& -1& 0\\ -1& 1& 0\\ 0& 0& 0\end{array}\right]$ Yeh step kyun? Bar I sirf DOFs 1 aur 2 ko touch karta hai, isliye uska $2\times 2$ block us corner mein land karta hai; row/col 3 zero rehti hai kyunki Bar I ko node 3 ka koi pata nahi hai.

Step 2 — Bar II ko DOFs $[2,3]$ par place karo. $\mathbf{K}\leftarrow \mathbf{K}+k\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& -1\\ 0& -1& 1\end{array}\right]$ Yeh step kyun? Bar II DOFs 2 aur 3 ko touch karta hai. Uska top-left $1$ upar land karta hai us $1$ ke jo pehle se $K_{22}$ par hai — yehi overlap hai.

Step 3 — overlap sum karo (yehi toh poora point hai). $\mathbf{K}=k\left[\begin{array}{ccc}1& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right]$ Yeh step kyun? Node 2 dono bars se belong karta hai, isliye uski self-stiffness $K_{22}=1+1=2k$ hai. Figure mein middle node dekho: do springs usse khainchte hain, isliye wo double resistant hai.

Verify: har row $0$ sum karta hai ($1-1+0=0$, $-1+2-1=0$, $0-1+1=0$). Poori chain ka rigid slide kuch bhi stretch nahi karta, isliye zero net force — correct hai. $K_{13}=0$ kyunki nodes 1 aur 3 koi bhi element share nahi karte. ✓

Sparse Matrix Storage se compare karo: dekho $\mathbf{K}$ tridiagonal hai — zyaadatar entries already zero hain.

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Cell B — Disconnected Pieces (No Shared Node)

Forecast: kya $K_{23}$ (do pieces ka coupling) zero hoga ya nonzero?

Step 1 — dono blocks scatter karo. $\mathbf{K}=k\left[\begin{array}{cccc}1& -1& 0& 0\\ -1& 1& 0& 0\\ 0& 0& 1& -1\\ 0& 0& -1& 1\end{array}\right]$ Yeh step kyun? Bar I top-left $2\times 2$ fill karta hai; Bar II bottom-right $2\times 2$. Kuch bhi overlap nahi karta.

Step 2 — coupling entry padho. $K_{23}=0$. Yeh step kyun? $K_{ij}\ne 0$ tabhi hoga jab DOFs $i,j$ ek element share karein. DOFs 2 aur 3 alag, unconnected bars se belong karte hain.

Verify: $\mathbf{K}$ block-diagonal hai — do independent springs, bilkul waise jaisa physics demand karta hai. Har block singular hai (akela bar freely slide kar sakta hai), jo Cell D ko foreshadow karta hai. ✓

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Cell C — Sign Bookkeeping

Forecast: us force ka sign predict karo jo node $i$ par lagta hai jab node $j$ forward move karta hai.

Step 1 — $u_i=0,u_j=+1$ impose karo. Force vector $=\mathbf{k}\left[\begin{array}{c}0\\ 1\end{array}\right]=k\left[\begin{array}{c}-1\\ +1\end{array}\right]$. Yeh step kyun? $\mathbf{k}$ ka column $j$ hi force answer hai "unit push at $j$" ke liye.

Step 2 — dono numbers interpret karo. $F_j=+k$ (stretched bar $j$ ko wapas pull karta hai — restoring, positive diagonal). $F_i=-k$ (wahi bar $i$ ko forward drag karta hai — negative off-diagonal). Yeh step kyun? Newton's third law: bar apne dono ends ko ek doosre ki taraf pull karta hai.

Verify: $F_i+F_j=-k+k=0$: koi external agent nahi, internal forces cancel hote hain. Isliye har off-diagonal coupling negative hoti hai aur diagonals positive spring-like elements ke liye. ✓

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Cell D — Zero / Degenerate Matrix (Rigid-Body Mode)

Forecast: compute karne se pehle determinant guess karo.

Step 1 — rigid-body vector $\mathbf{u}=(1,1,1{)}^T$ test karo. $\mathbf{K}\mathbf{u}=k\left[\begin{array}{c}1-1+0\\ -1+2-1\\ 0-1+1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right].$ Yeh step kyun? Agar poori structure slide karne par zero force produce ho, toh wo displacement ek zero-energy (rigid-body) mode hai, isliye $\det \mathbf{K}=0$.

Step 2 — confirm karo $\det \mathbf{K}=0$. Expand karo: $1(2\cdot 1-(-1)(-1))-(-1)((-1)(1)-0)=1(2-1)+1(-1)=1-1=0$. Yeh step kyun? Zero eigenvalue $\Rightarrow$ zero determinant $\Rightarrow$ koi unique solution nahi — structure space mein float kar jaayega.

Step 3 — boundary condition lagao (node 1 fix karo). Row/col 1 delete karo: ${\mathbf{K}}_{\text{red}}=k\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]$, $\det =k^2(2\cdot 1-1)=k^2\ne 0$. Yeh step kyun? Rigid-body freedom hataana (ek node nail karna) $\mathbf{K}$ ko positive-definite aur invertible banata hai — dekho Boundary Conditions in FEM.

Verify: original determinant $0$ (singular ✓), reduced determinant $k^2>0$ (solvable ✓). Isliye FEM code hamesha Direct vs Iterative Solvers mein solve karne se pehle BCs apply karta hai. ✓

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Cell E — Rotated Element (Transformation), $\theta =0^{\circ }$ Limit ke Saath

Forecast: kya $45^{\circ }$ par horizontal-only entry $K_{u_x{u}_x}$ poora $k$ hoga, ya uska ek fraction?

Step 1 — transformation kyun zaroori hai. Ek bar sirf apni khud ki length ke saath motion resist karta hai. Apne local axis $\xi$ mein, stiffness do axial displacements $(u_{\xi ,A},u_{\xi ,B})$ relate karta hai: ${\mathbf{k}}_{\text{loc}}=k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right].$ Yeh step kyun? Global $x,y$ directions bar ke saath aligned nahi hain, isliye ek global push isse sirf partly stretch karta hai. Hume project karna padega.

Step 2 — transformation matrix $\mathbf{T}$ explicitly likho. $c=\cos \theta ,s=\sin \theta$ lo. Ek node ka axial displacement uska $(x,y)$ motion bar direction par project karta hai: $u_{\xi }=c{u}_x+s{u}_y$. Dono nodes A aur B ke liye yeh projection stack karo (global DOF order $u_{xA},u_{yA},u_{xB},u_{yB}$): $\underset{\text{2 local}}{\underbrace{\left[\begin{array}{c}{u}_{\xi ,A}\\ u_{\xi ,B}\end{array}\right]}}=\underset{\mathbf{T}(2\times 4)}{\underbrace{\left[\begin{array}{cccc}c& s& 0& 0\\ 0& 0& c& s\end{array}\right]}}\underset{\text{4 global}}{\underbrace{\left[\begin{array}{c}{u}_{xA}\\ u_{yA}\\ u_{xB}\\ u_{yB}\end{array}\right]}}.$ Yeh step kyun? $\mathbf{T}$ ek rectangular dictionary hai jo 4 global movements ko 2 axial wale mein convert karta hai jo bar actually feel karta hai. Uski top row node A project karta hai, bottom row node B.

Step 3 — global stiffness $\mathbf{K}={\mathbf{T}}^T{\mathbf{k}}_{\text{loc}}\mathbf{T}$ paane ke liye sandwich karo. $\mathbf{K}=k\left[\begin{array}{cccc}{c}^2& cs& -c^2& -cs\\ cs& s^2& -cs& -s^2\\ -c^2& -cs& c^2& cs\\ -cs& -s^2& cs& s^2\end{array}\right].$ Yeh step kyun? ${\mathbf{T}}^T$ (ek $4\times 2$ matrix) $2\times 2$ axial stiffness ko wapas 4 global DOFs mein push karta hai — har entry hai "$k$ us hisaab se scale hua ki wo do directions bar ke saath kitni align hain." Yeh Coordinate Transformations in FEM ka standard bar result hai.

Step 4 — $\theta =45^{\circ }$ plug karo: $c=s=\frac{1}{\sqrt{2}}$, isliye $c^2=s^2=cs=\frac{1}{2}$. ${\mathbf{K}}_{45}=\frac{k}{2}\left[\begin{array}{cccc}1& 1& -1& -1\\ 1& 1& -1& -1\\ -1& -1& 1& 1\\ -1& -1& 1& 1\end{array}\right].$ Yeh step kyun? $45^{\circ }$ par ek horizontal push aur ek vertical push dono bar ko equally stretch karte hain, isliye $x$ aur $y$ stiffness fifty-fifty share karte hain.

Step 5 — $\theta =0^{\circ }$ limit (horizontal bar): $c=1,s=0$. $c^2=1,s^2=0,cs=0$, isliye ${\mathbf{K}}_0=k\left[\begin{array}{cccc}1& 0& -1& 0\\ 0& 0& 0& 0\\ -1& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right].$ Yeh step kyun? Ek horizontal bar ki koi vertical stiffness nahi hoti (uske $y$-rows/cols vanish ho jaate hain); sirf $x$-DOFs 1-D pattern $k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ carry karte hain — bilkul Cell A wala plain bar horizontally embedded.

Verify: $45^{\circ }$ par, $K_{u_x{u}_x}=\frac{k}{2}$ — $k$ ka ek fraction, forecast confirm karta hai. $0^{\circ }$ par, $K_{u_x{u}_x}=k$ (full) aur vertical stiffness $=0$. Dono matrices ke zero row sums hain (rigid slide kuch stretch nahi karta). Neeche Cell F ($\theta =90^{\circ }$) ke saath, teen angles $0^{\circ },45^{\circ },90^{\circ }$ poori range bracket karte hain. ✓

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Cell F — Vertical Bar (Limiting Angle $90^{\circ }$)

Forecast: guess karo ki horizontal stiffness $K_{u_x{u}_x}$ zero hai ya nonzero.

Step 1 — $c=0,s=1$ substitute karo. $c^2=0,s^2=1,cs=0$, isliye ${\mathbf{K}}_{90}=k\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& -1\\ 0& 0& 0& 0\\ 0& -1& 0& 1\end{array}\right].$ Yeh step kyun? Ek vertical bar horizontal motion resist nahi kar sakta — vertical rod ko sideways push karo toh wo swing karta hai, stretch nahi. Isliye saare $x$-rows/cols zero hain.

Step 2 — surviving block interpret karo. Sirf $y$-DOFs (rows/cols 2 aur 4) jaana pehchana $k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ carry karte hain. Yeh step kyun? Yeh Cell C ke 1-D bar se match karta hai, bas vertical direction mein embedded hai.

Verify: Ex 5 se compare karo jab $\theta$ sweep karta hai $0^{\circ }\to 45^{\circ }\to 90^{\circ }$: $x$-stiffness slide karta hai $k\to \frac{k}{2}\to 0$ aur $y$-stiffness slide karta hai $0\to \frac{k}{2}\to k$ — perfectly complementary, koi jumps nahi. Zero $x$-stiffness physically correct hai, aur warn karta hai ki ek unbraced vertical bar ek singular sub-system hai jise bracing/BCs ki zaroorat hai. ✓

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Cell G — Parallel Bars (Same Node Pair)

Forecast: kya do stiffnesses add honge ya average honge?

Step 1 — Bar I scatter karo (DOFs $[1,2]$). $\mathbf{K}\leftarrow k_1\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right].$ Yeh step kyun? Bar I dono DOFs touch karta hai, poora $2\times 2$ fill karta hai.

Step 2 — Bar II ko same DOFs $[1,2]$ par scatter karo. $\mathbf{K}\leftarrow \mathbf{K}+k_2\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]=\left[\begin{array}{cc}{k}_1+k_2& -(k_1+k_2)\\ -(k_1+k_2)& k_1+k_2\end{array}\right].$ Yeh step kyun? Dono elements dono nodes share karte hain, isliye har entry overlap karti hai aur hum entry-by-entry sum karte hain. Yeh parallel-spring rule seedha scatter-add se nikalta hai.

Verify: combined stiffness $k_1+k_2$ hai — parallel mein springs add hote hain (average nahi), bilkul elementary physics kehti hai. Row sums abhi bhi $0$. Agar $k_2=0$ (koi doubler nahi) toh single bar recover hota hai. ✓

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Cell H — Real-World Word Problem

Forecast: kya tip same load ka ek single bar se zyaada move karega ya kam?

Step 1 — spring constant compute karo. $k=\frac{EA}{L}=\frac{(70\times 10^9)(2\times 10^{-4})}{0.5}=2.8\times 10^7\text{N/m}$. Yeh step kyun? Material + geometry ko ek number mein convert karta hai; dono bars isse share karte hain.

Step 2 — assemble karo aur BC apply karo (Ex 1 + Ex 4 reuse karo). Reduced system (DOF 1 drop karo): $k\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}{u}_2\\ u_3\end{array}\right]=\left[\begin{array}{c}0\\ F\end{array}\right].$ Yeh step kyun? Node 1 fixed hai, isliye solve karne se pehle uski row/column remove ki jaati hai.

Step 3 — solve karo. Row 1 se: $2u_2-u_3=0\Rightarrow u_3=2u_2$. Row 2: $-u_2+u_3=F/k$. Substitute karo: $u_2=F/k$, $u_3=2F/k$. $u_3=\frac{2F}{k}=\frac{2(1400)}{2.8\times 10^7}=1.0\times 10^{-4}\text{m}=0.1\text{mm}.$ Yeh step kyun? Series mein do bars softer hote hain (compliances add hoti hain), isliye tip ek single bar ke stretch se double move karta hai.

Verify: units $\text{N}/(\text{N/m})=\text{m}$ ✓. Series check: har bar $F/k=1400/2.8\times 10^7=5\times 10^{-5}\text{m}$ stretch karta hai, total $2\times 5\times 10^{-5}=1\times 10^{-4}\text{m}$ ✓. Forecast confirmed: ek single bar se zyaada. ✓

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Cell I — Exam Twist (Non-Adjacent DOF Numbers)

Forecast: kya $\mathbf{K}$ abhi bhi tridiagonal rahega, ya nonzeros diagonal band ke bahar appear honge?

Step 1 — Bar I ko rows/cols $[1,3]$ mein scatter karo. $K_{11}+=k,K_{13}+=-k,K_{31}+=-k,K_{33}+=k.$ Yeh step kyun? $\mathbf{k}$ ki block entries $(1,1),(1,2),(2,1),(2,2)$ DOF list $[1,3]$ use karke global $(1,1),(1,3),(3,1),(3,3)$ par map hoti hain.

Step 2 — Bar II ko rows/cols $[3,2]$ mein scatter karo. $K_{33}+=k,K_{32}+=-k,K_{23}+=-k,K_{22}+=k.$ Yeh step kyun? Node 3 (shared wala) globally index 3 hai, isliye $K_{33}$ dono bars se accumulate karta hai.

Step 3 — assembled matrix likho. $\mathbf{K}=k\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& -1\\ -1& -1& 2\end{array}\right].$ Yeh step kyun? Rows/cols $(1,3,2)\to (1,2,3)$ reorder karo aur Ex 1 ka tridiagonal matrix recover hoga — same physics, permuted labels.

Verify: $K_{33}=2k$ (shared node ✓); row sums zero ✓; nonzero $K_{13},K_{23}$ tight band se bahar baithe hain — buri numbering bandwidth badhaa deti hai, Sparse Matrix Storage aur solver cost ko hurt karta hai. Achha node ordering matter karta hai — ise Mesh Refinement and Convergence se connect karo. ✓

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Recall Rapid self-test

Series chain mein $K_{22}=2k$ kyun hai? ::: Node 2 dono bars se shared hai; har ek $k$ add karta hai. $K_{ij}=0$ kab guaranteed hai? ::: Jab DOFs $i,j$ koi common element share na karen. Same node pair ke beech do bars — kya stiffnesses add hoti hain ya average? ::: Add hoti hain: $k_1+k_2$ (parallel springs). Unconstrained $\mathbf{K}$ singular kyun hai? ::: Rigid-body motion ek zero-energy mode hai $\Rightarrow \det \mathbf{K}=0$. $\theta =0^{\circ },45^{\circ },90^{\circ }$ par bar ki $x$-stiffness kya hai? ::: $k$, phir $k/2$, phir $0$.

Har ${\mathbf{k}}^{(e)}$ ki mechanics ke liye dekho Element Stiffness Matrices; big-picture pipeline ke liye dekho Finite Element Method Overview.