Finite element method — nodes, elements, stiffness matrix: Question bank

3.6.18 · D5 · Physics › Spacecraft Structures & Systems Engineering › Finite element method — nodes, elements, stiffness matrix

Symbol dictionary — neeche jo bhi use hua hai, pehle yahan define hai

Har quiz item ek hi chote set of symbols pe rely karta hai. Taaki kuch bhi use hone se pehle samjha ja sake, yahan plain words mein diya gaya hai — ek seedhe bar element ki picture ke saath jiske do ends hain.

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

Definition Bar element aur uske symbols

Ek seedhi rod (element) ki picture karo jisme ek end node 1 (left) aur doosra node 2 (right) label kiya gaya hai. Uske saath ek axis $x$ hai jo node 1 par $0$ se shuru hokar node 2 par $L$ tak jaata hai.

$L$ = element ki length (metres), dono nodes ke beech ki doori.
$A$ = cross-sectional area (m²), rod ko cross-wise kaatne par kitna mota hai.
$E$ = Young's modulus (Pa), material ki stiffness — ek diye hue fraction tak stretch karne ke liye kitna stress lagta hai. Bada $E$ = stretch karna mushkil (steel); chhota $E$ = aasaan (rubber).
$u (x)$ = continuous displacement field: position $x$ par jo material hai woh kitna move hua hai. Yeh $x$ ka ek smooth function hai.
$d_{1}, d_{2}$ = do nodal displacements — $u$ ki values sirf nodes par, $d_{1} = u (0)$ aur $d_{2} = u (L)$ . Vector $d = [d_{1}, d_{2}]^{⊤}$ woh hai jo FEM actually solve karta hai; $u (x)$ ko interpolation se unse rebuild kiya jaata hai.
$ϵ$ = strain (dimensionless): stretch per unit length, $ϵ = \frac{d u}{d x}$ .
$σ$ = stress (Pa): internal force per unit area, Hooke's law $σ = E ϵ$ se strain se juda hua.

Do visual anchors hain jinhe hum baar baar use karte hain — inhe ek baar study karo, phir quiz fast lagega.

True or false — justify karo

FEM exact analytical solution deta hai agar mesh kaafi fine ho.

False. FEM ek approximation hai; refine karne se answer sirf truth ki taraf converge hota hai. Infinite linear elements bhi sirf converge karte hain — woh exact answer tab hi reproduce karte hain jab true field khud piecewise-linear ho.

Ek single bar element ki stiffness matrix invertible hoti hai.

False.

det (K^{e}) = 0

kyunki ek free bar rigid body ki tarah slide kar sakta hai (

d_{1} = d_{2}

) zero strain aur zero force ke saath — woh motion ek eigenvector hai eigenvalue

0

ke saath.

Global stiffness matrix symmetric hoti hai.

True. Har

K^{e}

symmetric hai (reciprocity: node

j

par unit push se node

i

par force, node

i

par unit push se node

j

par force ke barabar hoti hai), aur assembly ke dauran symmetric matrices ko add karne se symmetry bani rehti hai.

K^{e}

ki har row ka sum zero hota hai.

True. Row 1 ka sum:

\frac{E A}{L} (1 + (- 1)) = 0

; Row 2 ka sum:

\frac{E A}{L} ((- 1) + 1) = 0

. Physically yahi statement hai ki rigid translation (

d_{1} = d_{2}

) force

\frac{E A}{L} (1 \cdot d_{1} - 1 \cdot d_{2}) = 0

produce karta hai.

Zyada nodes hamesha zyada accurate result dete hain.

False. Element quality dominant hoti hai: ek sliver triangle ya 1:1000 aspect ratio accuracy ko kharab kar deta hai dense mesh mein bhi, aur galat loads/materials kisi bhi density par galat hi rehte hain.

Shape functions har element ke andar har point par

1

sum honi chahiye.

True. Yeh partition of unity guarantee karta hai ki rigid-body displacement (sab node values barabar) exactly reproduce hoti hai:

u = N_{1} d + N_{2} d = (N_{1} + N_{2}) d = d

. Agar

N_{1} + N_{2} \neq = 1

to element wahan strain invent kar leta jahan koi nahi hai — upar wali crossing-lines figure dekho.

Linear bar elements har element ke andar constant strain produce karte hain.

True. Strain hai

ϵ = d u / d x

, aur ek seedhi line (

u

x

mein linear hai) ki derivative constant hoti hai, isliye ek linear element varying strain represent nahi kar sakta.

Boundary conditions apply karne se

K

bada hota hai.

Yeh depend karta hai method par. Elimination (fixed DOF ki row aur column delete karna)

K

ko chhota karta hai; penalty method har row ko rakhte hue ek bada diagonal term add karta hai,

K

ko same size rakhta hai (kabhi bada nahi). Dono tarike rigid-body modes remove karte hain aur

K

non-singular ho jaata hai.

Tum kisi element par kahin bhi load apply kar sakte ho, sirf node par nahi.

False (directly). Is formulation mein forces nodes par rehti hain; distributed ya mid-span load ko pehle equivalent nodal forces mein convert karna padta hai taaki woh

F

mein enter ho sake.

Galti dhundho

"Node 2 ke diagonal par

2

hai kyunki woh double load carry karta hai."

Galat cause.

2

isliye aata hai kyunki node 2 do elements ke beech shared hai jinki stiffnesses assembly ke dauran add hoti hain — yeh extra stiffness ko reflect karta hai, extra load ko nahi (assembly figure dekho).

"Kyunki hum

Kd = F

solve karte hain,

K

shuru se invertible hona chahiye."

Error. Boundary conditions se pehle

K

singular hota hai; sirf enough DOFs constrain karne ke baad (2D mein ≥3, 3D mein ≥6) woh invertible hota hai.

"Main 2D mein rigid-body motion remove karne ke liye ek seedhe edge par teen DOFs fix karunga."

Error. Teen constraints non-colinear hone chahiye — ek line par teen points phir bhi body ko us line ke axis ke around rotate karne ya translate karne dete hain, ek rigid-body mode chhor ke.

"Error near a hole ko reduce karne ka sabse safe tarika poori mesh ko fine karna hai."

Wasteful hai, galat nahi. Steep gradient sirf cutout edge par hai; wahan refine karna (adaptive/h-refinement) poore panel ko blanket karne se kahin zyada error cut karta hai per element.

N_{1} = 1 - x / L

aur

N_{2} = x / L

, isliye

N_{1}

node 2 par

1

ke barabar hai."

Error. Node 2 par,

x = L

, isliye

N_{1} = 1 - L / L = 0

milta hai.

N_{1}

node 1 par

1

ke barabar hai (

x = 0

); writer ne endpoints swap kar diye.

"Stress solver se directly nodal unknown ki tarah nikalta hai jaise displacement."

Error. Sirf displacements

d

directly solve ki jaane wali unknowns hain; stress post-processed hota hai strain

ϵ = d u / d x

ke zariye phir Hooke's law

σ = E ϵ

se.

Why questions

Bar ke nodes 1 aur 2 par opposite sign ki forces kyun aati hain?

Newton's third law ki wajah se: element har node par exactly utna hi pull back karta hai jitna har node element par karta hai, isliye pure axial element ke liye

F_{1} = - F_{2}

hota hai.

FEM ko spacecraft panel ke liye directly elasticity PDE solve karne ki jagah kyun use kiya jaata hai?

Kyunki exact PDE real geometry (curved skins, cutouts, ribs) ke liye analytically unsolvable hai; discretization infinite DOFs ko simple elements ke finite set se replace karta hai jo hum compute kar sakte hain. 2.4.8-Numerical-methods-for-differential-equations dekho.

Middle diagonal entry physically "stiffer" kyun lagti hai?

Ek hi node par do springs kheenchti hain, dono uski motion ko resist karti hain, isliye unki resistances add hoti hain — us node ko move karne ke liye dono ko overcome karna padta hai, isliye stiffness value badi hoti hai.

Distributed load ko nodes par lump kyun karna padta hai?

Kyunki FEM mein equilibrium sirf nodes par likha jaata hai; energy-equivalent nodal forces true distributed load ke sahi kaam ko preserve karti hain.

High stress gradient wale areas mein chhote elements kyun choose karte hain?

Stress concentration (hole, fillet, joint ke paas) bahut chhoti distance mein fast change karti hai; chhote elements us curvature ko resolve karte hain, jabki bade use average kar dete hain aur peak ko underestimate karte hain — 3.6.5-Buckling-of-columns-and-shells ke hotspots se relevant.

Badi structure ke liye stiffness matrix sparse kyun hoti hai?

Zyaatar nodes sirf kuch neighbours se connected hote hain, isliye ek node ka DOF door ke nodes ke saath zero stiffness coupling rakhta hai — woh entries

0

hoti hain, sparse solvers unhe skip karne deta hai.

Edge cases

Agar aapne koi bhi boundary condition apply nahi ki to kya hoga?

K

singular rehta hai, isliye structure freely float karta hai — solver ya to fail ho jaata hai ya infinitely many solutions return karta hai (koi bhi rigid-body drift "free" hai).

Zero-length element (

L \to 0

)

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

ke saath kya karta hai?

Stiffness blow up hoti hai (

\to \infty

), severe ill-conditioning hoti hai — matrix badi aur normal numbers ko mix karti hai aur rounding accuracy destroy kar deta hai.

Agar do adjacent elements ek common node share nahi karte?

Woh mechanically disconnected hain — assembly kabhi unhe link nahi karti, isliye "structure" independent pieces mein split ho jaata hai, har ek apni rigid-body freedom ke saath.

Agar mesh fine hai lekin element aspect ratios 1:1000 tak pahunch jaayein?

Matrix ill-conditioned ho jaata hai; tiny numerical rounding badi displacement errors produce karta hai, isliye quality checks (aspect ratio

< 10

) raw count se zyada matter karti hain.

Natural-frequency (modal) study ke liye,

K

ka singular hona acceptable kyun ho sakta hai?

Ek free-floating structure ke legitimate zero-frequency rigid-body modes hote hain; woh

K

ke null space ke corresponding hote hain aur expected hain, error nahi — 3.6.12-Modal-analysis-and-natural-frequencies dekho.

Thermal expansion is framework mein kaise enter karni chahiye?

Equivalent nodal load ki tarah: temperature-induced strain ek force vector ban jaata hai jo

F

mein add hota hai,

K

unchanged rehta hai — coupling 3.7.4-Thermal-analysis-of-spacecraft-components mein handle ki gayi.

Agar mesh refine karo aur solution

2%

se zyada change hota rahe?

Tumne converge nahi kiya — ya to mesh abhi bhi gradients ke liye coarse hai ya kuch (element shape, BCs, material) galat hai; results trust karne se pehle convergence study jaari rakho jo 3.6.1-Stress-and-strain-in-structural-members par built hai.

Recall Fast self-test

Diagonal "2" isliye aata hai kyunki... ::: do elements woh node share karte hain aur unki stiffnesses add hoti hain. $K$ BCs se pehle singular hoti hai kyunki... ::: rigid-body motion mein zero strain energy hoti hai. Element mein constant strain aata hai... ::: linear shape functions se (line ki derivative constant hoti hai). Non-colinear constraints ki zaroorat hoti hai... ::: saare rigid-body modes kill karne ke liye (2D mein ≥3 DOF, 3D mein ≥6). $K^{e}$ ki row sum hoti hai... ::: zero, kyunki har row mein $1 + (- 1) = 0$ .