FEM for structures — assembling global stiffness: Question bank

3.6.19 · D5 · Physics › Spacecraft Structures & Systems Engineering › FEM for structures — assembling global stiffness

Yeh global stiffness assembly step ke liye ek rapid-fire misconception hunt hai. Neeche har item ek line ka hai: prompt padho, answer cover karo, phir reveal karo. Answers reasons hain, verdicts nahi — agar tum kyun nahi bata sakte, toh idea abhi tumhara nahi hua.

Traps se pehle, ek figure aur ek chhota symbol dictionary, taaki neeche kuch bhi surprise na kare.

Definition Is page par use kiye gaye symbols (ek baar, yahaan define kiye gaye)

$ℓ$ = ek element ki length (toh ek bar ki axial stiffness $E A / ℓ$ hai). Hum $ℓ$ use karte hain, $L$ nahi, precisely isliye taaki length ko neeche wali localization matrix se kabhi confuse na kiya jaaye.
$L^{(e)}$ = element $e$ ki localization (Boolean gather) matrix. Iske ek row per element DOF hoti hai; har row mein sirf ek $1$ hota hai matching global DOF ke column mein, baaki sab zeros. Ise global displacement vector se multiply karne par element ke DOFs pick out hote hain: $u^{(e)} = L^{(e)} u$ .
$N$ = poore mesh mein degrees of freedom (DOFs) ki total number. $K$ ka size $N \times N$ hota hai.
$T$ = rotation (transformation) matrix jo bar ke apne axes mein likhe vector ko shared global $x, y$ axes mein re-express karta hai. Yeh bar ke direction cosines ( $cos$ aur $sin$ of the angle jo bar global $x$ se banata hai) se banta hai. Yeh orthogonal hai, isliye $T^{- 1} = T^{T}$ ; yahi reason hai ki transform $k_{global} = T^{T} k_{local} T$ padhta hai.
Local bar axes: $ξ$ bar ke saath (along), $η$ us ke perpendicular. Ek bar mein $ξ$ ke along stiffness hoti hai aur $η$ ke along zero stiffness.

Figure — FEM for structures — assembling global stiffness

Upar ki picture teeno ideas ek saath dikhati hai: left mein, ek element ka $2 \times 2$ block bade $K$ ke overlapping slot mein scatter-add ho raha hai (note karo woh shared entry jo ek existing value ke upar aa ke add hoti hai); middle mein, ek tilted bar ke $ξ$ – $η$ local axes aur global $x, y$ mein rotation $T$ ; right mein, bandwidth kyun small hai — har node sirf apne immediate neighbours ko touch karta hai.

True ya false — justify karo

Assembly ka matlab hai har element matrix ko

K

ke ek empty slot mein rakhna, ek block per element.

False. Blocks shared DOFs par overlap karte hain; un entries par hum add (superimpose) karte hain contributions. Replace karne se neighbour element ki stiffness delete ho jaayegi.

Do-bar truss mein

K_{22} = 2 \frac{E A}{ℓ}

isliye hai kyunki humne arithmetic slip ki.

False — yeh exactly sahi hai. Node 2 dono bars se belong karta hai, isliye dono

\frac{E A}{ℓ}

K_{22}

mein scatter karte hain; correct action hai

1 + 1 = 2

Global stiffness

K

symmetric hoti hai sirf tab agar tum ise symmetric order mein banao.

False. Symmetry aati hai kyunki har

k^{(e)}

symmetric hoti hai (Maxwell–Betti reciprocity); scatter-add ise preserve karta hai element ordering ki parwah kiye bina.

Boundary conditions se pehle,

K

invertible hai, isliye hum seedha

Ku = F

solve kar sakte hain.

False. Un-constrained

K

sirf positive semi-definite hoti hai — rigid-body modes zero eigenvalues dete hain, isliye

K

singular hai. Pehle DOFs fix karne padte hain (dekho Boundary Conditions in FEM).

Ek bada element stiffness matrix hamesha ek bada global matrix deta hai.

False.

K

ka size total DOF count

N

se set hota hai (saare mesh DOFs), kisi ek element ke block size se nahi. Do elements jo apne saare nodes share karte hain woh

N

ko unchanged chhod sakte hain.

Scatter-add algorithm ek alag

K

deta hai formula

\sum_{e} L^{(e) T} k^{(e)} L^{(e)}

se.

False. Dono identical hain — scatter-add sirf us sum ko evaluate karne ka sasta tarika hai, yeh exploit karte hue ki

L^{(e)}

Boolean gather matrix hai (ek row mein ek

1

, baaki zeros).

K

ki sparsity ek numerical accident hai jis par rely nahi karna chahiye.

False.

K_{ij} \neq = 0

sirf tab jab DOFs

i

aur

j

ek element share karein, jo mesh ki ek topological fact hai. Yeh guaranteed hai aur Sparse Matrix Storage dwara exploit ki jaati hai.

Error dhundo

"Element 1 DOFs [1,2] use karta hai aur element 2 DOFs [3,4] use karta hai, isliye do-bar truss ka

K

4 \times 4

ka hai."

Error: node 2 shared hai, isliye uska DOF dono elements mein appear karta hai. Sahi numbering mein DOFs [1,2] aur [2,3] hain →

K

3 \times 3

ka hai,

4 \times 4

ka nahi.

"Ek bar transverse pushing ko resist karta hai, isliye uske local

4 \times 4

stiffness mein koi zero row nahi."

Error: ek axial bar mein apne axis ke perpendicular (

η

direction) mein zero stiffness hoti hai — transverse motion koi strain cause nahi karta. Yahi reason hai ki local matrix mein

η

DOFs ke liye zero rows/columns hoti hain.

"Hum har element ko global coordinates mein

K

assemble karne ke baad transform karte hain."

Error: rotation

k_{global} = T^{T} k_{local} T

(jahan

T

bar ke direction cosines se bana ho) har element ke liye, scatter se pehle hona chahiye — warna tum incompatible coordinate systems add kar rahe ho. Dekho Coordinate Transformations in FEM.

"Maine

K

assemble ki, phir fixed support apply karne ke liye sirf row 1 delete karke

u_{1} = 0

set kiya."

Error: tumhe row 1 aur column 1 dono delete karne chahiye (ya penalty/partition scheme use karo). Sirf row drop karne se

K

non-square ho jaati hai aur symmetry destroy ho jaati hai.

"Do-bar truss mein

K_{13} = 0

matlab main ek contribution bhool gaya."

Error: yeh correct hai. DOFs 1 aur 3 alag elements se belong karte hain aur kabhi ek share nahi karte, isliye koi bhi element kabhi

K_{13}

mein nahi likhta — yeh ek genuine structural zero hai.

"Kyunki har

k^{(e)}

singular hai, isliye assembled

K

bhi hamesha singular rehni chahiye."

Error: individual element matrices singular hoti hain (rigid-body modes), lekin bahut saare elements assemble karne ke baad plus boundary conditions apply karne par woh modes remove ho jaate hain aur

K_{reduced}

positive definite ban jaati hai.

"Ek node par milne wale do elements ka wahan alag displacement hai, isliye main unhe average karta hoon."

Error: compatibility ek single shared displacement ko node par force karti hai — isliye hi woh ek DOF share karte hain. Averaging karne se no-gap/no-overlap condition violate hogi.

Why questions

Hum shared DOF par element contributions ko add kyun karte hain, overwrite kyun nahi karte?

Kyunki total strain energy additive hoti hai:

U = \sum_{e} U^{(e)}

, aur har element independently contribute karta hai. Stiffness ka superposition directly energy expression ke sum se aata hai.

Localization matrix

L^{(e)}

Boolean (sirf 0s aur 1s) kyun hoti hai?

Yeh sirf select karti hai ki kaun se global DOFs element

e

ko feed karte hain — koi scaling nahi, sirf picking. Har row mein ek single 1 hota hai jo matching global DOF ki taraf point karta hai, yahi scatter-add ko exact banata hai.

2D mesh mein

K

ki bandwidth roughly

O (N)

kyun hoti hai?

Bandwidth wo largest gap hai jo kisi bhi do nodes ke global DOF numbers ke beech hoti hai jo ek element share karte hain. Agar tum ek

n \times n

grid (

N \approx n^{2}

DOFs) ko row-by-row number karo, neighbouring nodes mein zyada se zyada ek poori row

\approx n = N

ka difference hoga — toh nonzeros width

N

ki ek band ke paas rehti hain. Narrow band direct solvers ko sasta rakhta hai.

Ek DOF fix karne se singular

K

invertible kyun ban jaati hai?

DOFs fix karne se rigid-body modes (free translations/rotations) eliminate ho jaate hain jo zero eigenvalues ka source the, aur

u^{T} Ku > 0

reh jaata hai saare nonzero admissible

u

ke liye.

Quadratic form

\frac{1}{2} u^{T} Ku

strain energy ke equal kyun hai, aur yeh

K

ke symmetric hone ki guarantee kyun deta hai?

Kyunki energy usi quadratic form ke roop mein build ki gayi thi; aur koi bhi quadratic form matrix ke sirf symmetric part ko "dekhta" hai, isliye physically meaningful

K

symmetric hoti hai.

Element stiffness ko assembly se pehle

T

se global coordinates mein transform kyun karte hain, baad mein

K

ko kyun nahi?

Har element ke apne alag local

ξ, η

axes hote hain; sirf har ek ko common global

x, y

frame mein rotate karne ke baad hi unke shared-DOF entries same coordinate directions mein hoti hain aur add ho sakti hain.

Mesh refine karne par

N

kyun badhta hai lekin

K

sparse kyun rehti hai?

Zyada elements/nodes matlab zyada DOFs, lekin har naya node phir bhi sirf apne thode se immediate neighbours se connect hota hai — connectivity local rehti hai, isliye nonzeros per row bounded rehte hain. Dekho Mesh Refinement and Convergence.

Edge cases

Agar ek node dangling chhod diya jaaye (kisi bhi element se belong na kare) toh

K

ka kya hoga?

Uske rows aur columns poori tarah zero hote hain, isliye

K

supports fix karne ke baad bhi singular rehti hai — free node mein koi stiffness nahi hoti aur ek spurious rigid-body mode produce hoti hai.

Do identical elements same do nodes par stack kiye gaye hain (duplicated element). Assembly kya deta hai?

Unke contributions add ho jaate hain, shared block double ho jaata hai — physically do parallel springs, isliye combined stiffness ek single element ki double hoti hai. Correct hai, lekin agar unintended ho toh warning sign hai.

Ek bar element ki zero length hai (

ℓ \to 0

). Assembly mein kya toot-ta hai?

E A / ℓ

factor infinity tak blow up karta hai,

K

mein ek ill-conditioned near-rigid link inject karta hai aur solve bigaad deta hai — degenerate elements ko meshing ke dauran pakdna zaroori hai.

Agar ek bar load ke bilkul transverse ho (uska axis

ξ

force se

90°

angle par aligned ho), toh yeh

K

mein kaise enter karta hai?

T

transform ke baad uski stiffness poori tarah ek global direction mein aa jaati hai; perpendicular direction mein yeh zero contribute karta hai, isliye woh load component bilkul carry nahi kar sakta.

Agar do elements ek node share karte hain lekin tumne galti se unhe wahan alag global DOF numbers de diye?

Tum compatibility tod dete ho — shared point do independent points mein split ho jaata hai.

K

bada ho jaata hai,

K_{22} = 2

-style summation kabhi nahi hoti, aur structure aisa behave karta hai jaise woh us node par disconnected ho.

Assembled, un-reduced

K

mein ek all-zero row solve karne se pehle hi tumhe kya bata deti hai?

Woh DOF kisi bhi element dwara unrestrained hai — ya toh genuine free/rigid mode hai ya modelling error (orphan node). Solvers singularity report karte rahenge jab tak use constrain ya connect na kiya jaaye.

Recall Ek-line survival summary

Assembly = har $k^{(e)}$ ko $T$ se global mein transform karo, phir uske entries ko $K$ mein shared global DOFs par scatter-add karo jo $L^{(e)}$ dwara pick out hote hain; symmetry aur sparsity physics aur topology se aate hain; singularity sirf boundary conditions se cure hoti hai.

Parent par wapas jaao: FEM for structures — assembling global stiffness · overview in Finite Element Method Overview · element blocks in Element Stiffness Matrices.