Exercises — FEM for structures — assembling global stiffness

Bar element stiffness yeh hai: ${\mathbf{k}}^{(e)}=\frac{EA}{L}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right].$ Yeh kahan se aaya: bar force hai $F=\frac{EA}{L}(u_j-u_i)$. Node $i$ par force hai $\frac{EA}{L}(u_i-u_j)$, node $j$ par force hai $\frac{EA}{L}(u_j-u_i)$. Un do equations ko matrix ke roop mein likhne par upar wala pattern milta hai.

Row 1, column 2 = $-\frac{EA}{L}$. Matlab: DOF $i$ par woh force jo tab aata hai jab hum DOF $j$ ko ek unit move karein aur DOF $i$ ko fixed rakhen. Yeh negative hai kyunki node $j$ ko door kheenchne par node $i$ bhi usi direction mein khicha jaata hai — bar use drag karta hai. Yahi pura matlab hai stiffness entry $K_{ij}$ ka: DOF $j$ ke unit displacement par DOF $i$ par force.

Sirf $K_{22}$. Shared DOF dekho: dono elements DOF 2 list karte hain. Entry $K_{ij}$ ek element se contribution tabhi leti hai jab dono $i$ aur $j$ us element ke DOFs hon. Element 1 ke paas DOFs $\{1,2\}$ hain, element 2 ke paas $\{2,3\}$. Sirf pair $(i,j)$ jo dono sets ke andar hai woh hai $(2,2)$. Toh $K_{22}$ do baar sum hota hai; baaki har entry zyada se zyada ek element se aati hai. (Yahi reason hai ki parent ke two-bar truss mein $K_{22}=2$ aata hai.)

Maano $k=\frac{EA}{L}$. Har bar apne do DOF rows/columns mein $k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ contribute karta hai.

• Bar 1 → DOFs $[1,2]$
• Bar 2 → DOFs $[2,3]$
• Bar 3 → DOFs $[3,4]$

Scatter-add karo. Interior nodes 2 aur 3 dono do bars ke beech shared hain, isliye unke diagonals $1+1=2$ ho jaate hain: $\mathbf{K}=k\left[\begin{array}{cccc}1& -1& 0& 0\\ -1& 2& -1& 0\\ 0& -1& 2& -1\\ 0& 0& -1& 1\end{array}\right].$ Pattern check karo: har row ka sum zero hai. Yeh rigid-body signature hai — poori chain ko equally slide karo aur koi bar stretch nahi hoga, toh koi force nahi. Accha sign hai ki assembly sahi hai.

Row/column 1 delete karo (fixed DOF). Reduced system, $k=\frac{EA}{L}$ ke saath: $k\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right]\left[\begin{array}{c}{u}_2\\ u_3\\ u_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ F\end{array}\right].$ Row 1: $2u_2-u_3=0\Rightarrow u_3=2u_2$. Row 2: $-u_2+2u_3-u_4=0\Rightarrow u_4=-u_2+2u_3=-u_2+4u_2=3u_2$. Row 3: $-u_3+u_4=\frac{F}{k}\Rightarrow -2u_2+3u_2=u_2=\frac{F}{k}=\frac{FL}{EA}$. Toh $u_2=\frac{FL}{EA},u_3=\frac{2FL}{EA},u_4=\frac{3FL}{EA}.$ Physical check: series mein teen bars mein se har ek full force $F$ carry karta hai, har ek $\frac{FL}{EA}$ stretch hota hai, toh displacements chain mein $1,2,3$ times badhte hain. ✓

Har bar apne do DOFs par $k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ hai:

• A–C → DOFs $[2,1]$
• B–C → DOFs $[3,1]$
• D–C → DOFs $[4,1]$

C (DOF 1) teeno elements mein hai, isliye uska diagonal $1+1+1=3$ collect karta hai. Har outer node ek bar mein hai (diagonal 1) aur sirf C se couple hota hai ($-1$): $\mathbf{K}=k\left[\begin{array}{cccc}3& -1& -1& -1\\ -1& 1& 0& 0\\ -1& 0& 1& 0\\ -1& 0& 0& 1\end{array}\right],K_{11}=3k.$ Insight: ek node par diagonal entry wahan milne wale stiffness contributions ki count ke barabar hoti hai. Ek hub node jo kai elements se connected hai uska diagonal bada hoga — jitne zyada springs aapko kheenchein, utna mushkil aapko move karna. Note karo $K_{23}=K_{24}=K_{34}=0$: outer nodes ek hi element mein nahi hain, isliye woh kabhi directly "baat" nahi karte, sirf C ke through.

Symmetry: har element matrix symmetric hai ($k_{ij}=k_{ji}$, Maxwell–Betti reciprocity). Symmetric blocks ka scatter-add same $(i,j)$ aur $(j,i)$ slots mein symmetry preserve karta hai, isliye $\mathbf{K}={\mathbf{K}}^T$.

$K_{23}=0$ kyun: DOF 2 (node A) aur DOF 3 (node B) kabhi kisi single element mein saath nahi aate. Koi element nahi = koi coupling nahi = zero entry. Dekho Sparse Matrix Storage — isi wajah se $\mathbf{K}$ mostly zeros hoti hai.

General rule: $K_{ij}\ne 0$ tabhi jab DOFs $i$ aur $j$ kam se kam ek common element mein hon. Warna $K_{ij}=0$.

DOF ordering (entries padhne se pehle state karo): element matrix ki chaar rows/columns hain, order mein, $[u_{1x},u_{1y},u_{2x},u_{2y}],$ yaani node 1 ka $x$-displacement, node 1 ka $y$-displacement, phir node 2 ka $x$ aur $y$. Neeche har entry $k_{ab}$ "DOF $a$ mein force per unit displacement of DOF $b$" hai bilkul isi list ka use karke.

Maano $c=\cos \theta$, $s=\sin \theta$ jahan $\theta$ bar ka $x$-axis se angle hai. 45° bar ke liye, $c=s=\frac{1}{\sqrt{2}}$, toh $c^2=s^2=cs=\frac{1}{2}$.

$c$ aur $s$ kyun chahiye: bar sirf apni khud ki axis ke saath stretching ko resist karta hai. Ek global displacement $(u_x,u_y)$ axial stretch mein sirf apne projection ke through contribute karta hai bar direction par, aur woh projection $\cos$ aur $\sin$ use karta hai (yeh hai Coordinate Transformations in FEM). Standard result, rows/columns ordered $[u_{1x},u_{1y},u_{2x},u_{2y}]$ ke saath: ${\mathbf{k}}_{\text{global}}^{(e)}=\frac{EA}{L}\left[\begin{array}{cccc}{c}^2& cs& -c^2& -cs\\ cs& s^2& -cs& -s^2\\ -c^2& -cs& c^2& cs\\ -cs& -s^2& cs& s^2\end{array}\right].$ $c^2=s^2=cs=\frac{1}{2}$ plug in karo: ${\mathbf{k}}_{\text{global}}^{(e)}=\frac{EA}{2L}\left[\begin{array}{cccc}1& 1& -1& -1\\ 1& 1& -1& -1\\ -1& -1& 1& 1\\ -1& -1& 1& 1\end{array}\right].$ Sanity check: har row abhi bhi zero sum hai (rigid translation → koi force nahi), aur $x$–$y$ entries ab coupled hain ($cs\ne 0$) bilkul isliye kyunki bar tilted hai.

Maano $g=\frac{EA}{2L}$.

Bar A (45°, node 1→2), DOFs $[1,2,3,4]$ ordered $[u_{1x},u_{1y},u_{2x},u_{2y}]$: last $2\times 2$ block (node-2 self-block, rows/cols $[u_{2x},u_{2y}]$) hai $g\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$.

Bar B (−45°, node 2→3): yahan $c=\frac{1}{\sqrt{2}}$, $s=-\frac{1}{\sqrt{2}}$, toh $c^2=s^2=\frac{1}{2}$ lekin $cs=-\frac{1}{2}$. Iska element matrix, rows/cols ordered $[u_{2x},u_{2y},u_{3x},u_{3y}]$ → DOFs $[3,4,5,6]$: $g\left[\begin{array}{cccc}1& -1& -1& 1\\ -1& 1& 1& -1\\ -1& 1& 1& -1\\ 1& -1& -1& 1\end{array}\right].$ Iska node-2 self-block (pehla $2\times 2$, rows/cols $[u_{2x},u_{2y}]$) hai $g\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$.

Node 2 shared hai → do self-blocks add karo: $\mathbf{K}[3:4,3:4]=g\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]+g\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]=g\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\frac{EA}{L}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].$ Toh chaar entries hain $K_{33}=K_{44}=\frac{EA}{L}$ aur $K_{34}=K_{43}=0$. Khoobsurat check: cross-coupling cancel ho jaati hai ($+1$ bar A se, $-1$ bar B se), node 2 ko $x$ aur $y$ mein equally stiff chhod ke. Do symmetric diagonals use evenly brace karte hain — figure dekhte hi physically obvious lagta hai.

(a) Un-reduced $\mathbf{K}$ ka 1-dimensional null space hai, toh rank 3 (4 mein se). Null vector hai $\mathbf{u}=[1,1,1,1{]}^T$: har node ko same amount slide karo toh koi bar stretch nahi hoga, zero strain energy milegi, isliye $\mathbf{Ku}=0$. Yeh 1D chain ka single rigid-body translation mode hai — dekho Finite Element Method Overview.

(b) Energy: ${\mathbf{u}}^T\mathbf{Ku}=2U={\sum }_e\frac{EA}{L}(\Delta L_e{)}^2\ge 0$ hamesha. Jab node 1 fixed ho, toh ek hi tarika hai har bar mein zero strain paane ka: $u_2=u_3=u_4=0$ (har bar ki stretch wall tak fixed neighbour chain par depend karti hai). Toh ${\mathbf{u}}^T{\mathbf{K}}_r\mathbf{u}>0$ har nonzero $\mathbf{u}$ ke liye → positive definite → invertible → unique displacement solution. Isi wajah se solvers (Direct vs Iterative Solvers) sirf boundary conditions ke baad kaam karte hain.

Determinant check: $k=\frac{EA}{L}$ ke saath, reduced matrix hai $k\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right]$. Iska determinant hai $k^3\cdot 1=k^3>0$, aur saare leading minors ($2k>0$, $3k^2>0$, $k^3>0$) positive hain → Sylvester's criterion se positive definite. ✓

Half-bandwidth $b$ = saare nonzero $K_{ij}$ par sabse bada $|i-j|$. Ek bar apne dono ends ke do DOFs couple karta hai, isliye $b$ do physically-adjacent nodes ke beech sabse bada label-gap hai.

(a) Consecutive numbering $1,2,3,4,5$. Har bar physically-adjacent nodes ko join karta hai jinke global labels exactly 1 se differ karte hain. Toh worst nonzero off-diagonal ka $|i-j|=1$ → $b=1$ (ek tridiagonal matrix). Minimal storage.

(b) Scrambled labelling $1,3,5,4,2$ (physical node $1\to$ label 1, node $2\to$ label 3, node $3\to$ label 5, node $4\to$ label 4, node $5\to$ label 2). Ab har bar (physical neighbours) ke do labels check karo:

• bar 1–2: labels $1,3$ → gap $2$
• bar 2–3: labels $3,5$ → gap $2$
• bar 3–4: labels $5,4$ → gap $1$
• bar 4–5: labels $4,2$ → gap $2$

Maximum gap $2$ hai, toh $b=2$: band consecutive case se do guna wide hai, even though physics (same 4 bars, same nonzero count) unchanged hai.

Lesson: same structure worse numbering se sirf wider band deta hai, aur wider band mein zyada memory aur zyada solve time lagta hai (direct factorization band ke andar fill in hoti hai). Accha node ordering — bandwidth-minimizing schemes jaise Cuthill–McKee — isliye free performance hai, ek real concern jab meshes refined hote hain (Mesh Refinement and Convergence, Sparse Matrix Storage, Direct vs Iterative Solvers).