FEM software — NASTRAN, ABAQUS (concepts and use): Exercises

3.6.20 · D4 · Physics › Spacecraft Structures & Systems Engineering › FEM software — NASTRAN, ABAQUS (concepts and use)

Intuition Is page ko kaise use karein

Answers ko cover kar lo. Ek problem padho, paper par solve karo, PHIR collapsible solution kholo. Neeche har symbol parent note mein banaya gaya tha — agar koi symbol unfamiliar lage, toh aage badhne se pehle wahan jaao. Problems paanch levels climb karti hain: "is card ko naam do" se lekar "poora analysis campaign design karo" tak.

Hamare alphabet ki ek chhoti si reminder, simple words mein:

$K$ = stiffness matrix — numbers ka ek bada grid jo kehta hai "structure ke har point ko push karna kitna mushkil hai". Bold matlab "bahut saare numbers ek saath", koi ek number nahi.
$u$ = displacement vector — ek list ki har point kitni door move hui. Jab hum ek hi point wala model solve karte hain, is list mein sirf ek entry hoti hai, aur hum bold hata ke plain $u$ likh sakte hain us ek number ke liye.
$F$ = force vector — hamare apply kiye gaye pushes ki list (plain $F$ jab yeh sirf ek number ho).
$M$ = mass matrix — har point par kitni inertia baithi hai.
$ϕ$ = mode shape vector — ek vibration mode ki movement ka pattern: yeh kehta hai "jab structure is frequency par ring karta hai, node A itna swing karta hai, node B utna". Yeh shape hai, size nahi.
$ω$ = angular frequency radians per second mein; $f = ω /2 π$ everyday frequency hai hertz mein (cycles per second).
DOF = degree of freedom = ek point ko move karne ka ek independent tarika (3D mein ek free node ke 6 hote hain: teen slides $x, y, z$ aur teen spins).

Level 1 — Recognition

Exercise 1.1 (L1)

Neeche diye NASTRAN input deck mein, har card ko uske kaam se match karo.

SOL 103
EIGRL,1,,,10
CQUAD4,1,1,2,3,4
MAT1,1,70.E9,,2700
SPC1,10,123456,1

Recall Solution

SOL 103 ::: solution sequence select karta hai; 103 modal (natural-frequency) analysis hai.
EIGRL,1,,,10 ::: Lanczos eigenvalue solver, method set 1, 10 modes maangta hai.
CQUAD4 ::: ek 4-node shell element, thin plate-jaise structures ke liye ideal.
MAT1,1,70.E9,,2700 ::: ek isotropic material: Young's modulus $E = 70 \times 1 0^{9}$ Pa, density $ρ = 2700$ kg/m³.
SPC1,10,123456,1 ::: ek single-point constraint jo node 1 par saare 6 DOF (digits 1-2-3-4-5-6) fix karta hai — ek clamp.

Exercise 1.2 (L1)

Har scenario ke liye, analysis type batao (Linear Static / Modal / Frequency Response / Random Vibration) aur yeh bhi ki NASTRAN ya ABAQUS pehli natural choice hai.

(a) Pata karo ki ek solar panel steady 2 g acceleration mein zyada door deflect to nahi ho raha. (b) Ek bracket ki pehli bending frequency nikalo taaki launcher ke tone se bacha ja sake. (c) Compression ke under ek 8-ply carbon panel mein delamination predict karo. (d) Broadband launch shaker environment se RMS stress predict karo.

Recall Solution

(a) ::: Linear Static ( $Ku = F$ ), NASTRAN. Steady load, koi material nonlinearity nahi. (b) ::: Modal, NASTRAN. Governing equation hai $(K - ω^{2} M) ϕ = 0$ , jahan $ϕ$ (upar alphabet mein defined) mode shape vector hai — woh pattern jismein structure natural frequency $ω$ par sway karta hai. Dekho Vibration and Modal Analysis. (c) ::: Nonlinear progressive-failure, ABAQUS. Hashin damage material nonlinearity hai — dekho Composite Materials. (d) ::: Random Vibration (PSD), NASTRAN. Dekho Launch Vehicle Loads.

Level 2 — Application

Exercise 2.1 (L2)

Ek single spring ki stiffness $k = 5000$ N/m hai. 1-DOF finite-element model ke roop mein treat karne par, $K = [5000]$ . $F = [250]$ N apply karo. Displacement ke liye $Ku = F$ solve karo.

Recall Solution

Kya hai: global system ek number tak collapse ho jaata hai. Kyun: ek DOF ke saath displacement vector $u$ mein sirf ek entry hai, toh hum ise plain $u$ likhte hain, aur $Ku = F$ bas $5000 u = 250$ hai. $u = \frac{F}{k} = \frac{250}{5000} = 0.05 m = 50 mm .$

Exercise 2.2 (L2)

Ek bolt ko $T = 10$ N·m torque diya gaya hai. Diameter $d = 6$ mm $= 0.006$ m, nut factor $k = 0.2$ . $T = k F d$ use karke preload force $F$ nikalo.

Recall Solution

Yeh formula kyun: wrench par jo torque apply karte ho woh teen kamon mein bant jaata hai — bolt ko stretch karna, thread friction, aur under-head friction. Empirical nut factor $k$ teeno ko bundle karta hai taaki $T = k F d$ jis twist ko feel karo use bolt ke andar ke tension se connect kare. $F = \frac{T}{k d} = \frac{10}{0.2 \times 0.006} = \frac{10}{0.0012} \approx 8333 N .$ Yeh parent ke bolted-joint example mein *CLOAD preload hai.

Exercise 2.3 (L2)

Ek NASTRAN modal run pehla mode $ω_{1} = 5341$ rad/s par report karta hai. Ise hertz mein ordinary frequency $f_{1}$ mein convert karo.

Recall Solution

$2 π$ se kyun divide karo: ek poora cycle $2 π$ radians hota hai, toh cycles-per-second $=$ radians-per-second $\div 2 π$ . $f_{1} = \frac{ω _{1}}{2 π} = \frac{5341}{2 π} \approx 850 Hz .$ Parent ke L-bracket first bending mode se match karta hai.

Level 3 — Analysis

Exercise 3.1 (L3)

Do identical springs (har ek $k = 5000$ N/m) series mein ek wall aur ek free tip ke beech join hain. Node 1 wall par hai (fixed), node 2 junction hai, node 3 free tip hai jahan force $F = 250$ N apply hai. $3 \times 3$ stiffness matrix assemble karo, node 1 par boundary condition apply karo, aur tip displacement $u_{3}$ solve karo.

Figure — FEM software — NASTRAN, ABAQUS (concepts and use)

Recall Solution

Step 1 — element matrices (KYA aur KYUN). DOFs $i, j$ ke beech ek single spring contribute karta hai $k_{e} = k [1 - 1 - 1 1] .$ Diagonal par $+ k$ matlab "kisi node ko push karna use stiffen karta hai"; off-diagonal $- k$ matlab "ek end ko move karna doosre ko drag karta hai".

Step 2 — global grid mein assemble karo. Spring A nodes 1–2 ko link karta hai, spring B nodes 2–3 ko. Inhe shared node 2 par overlap karo (yahi "assembly" hai — figure mein middle diagonal dekho jahan do matrices pile up karti hain): $K = k 1 - 1 0 - 1 2 - 1 0 - 1 1 .$

Step 3 — wall (boundary condition) apply karo. Node 1 clamped hai, toh $u_{1} = 0$ . Row/column 1 delete karo, $u_{2}, u_{3}$ mein ek $2 \times 2$ system bachta hai: $k [2 - 1 - 1 1] [u_{2} u_{3}] = [0250] .$

Step 4 — solve karo. Do equations $5000 (2 u_{2} - u_{3}) = 0$ aur $5000 (- u_{2} + u_{3}) = 250$ se: pehli equation deti hai $u_{3} = 2 u_{2}$ ; substitute karo: $5000 (u_{2}) = 250 \Rightarrow u_{2} = 0.05$ m, toh $u_{3} = 0.10$ m $= 100$ mm.

Sanity check (ISKA MATLAB KYA HAI): series mein do springs softer hoti hain — effective stiffness $k /2 = 2500$ N/m, toh $u_{3} = 250/2500 = 0.10$ m. Wahi answer. Series mein springs zyada stretch karti hain, bilkul figure mein tip ki tarah jo single-spring case ko overshoot karta hai.

Exercise 3.2 (L3)

Parent note ke composite panel mein, ek fiber-tension check Hashin form use karta hai $(\frac{σ _{11}}{X _{T}})^{2} + α (\frac{τ _{12}}{S})^{2} \geq 1.$ Ek $0^{\circ}$ ply $σ_{11} = 1400$ MPa aur $τ_{12} = 60$ MPa carry karta hai. Strengths: $X_{T} = 2000$ MPa, $S = 100$ MPa, aur $α = 1$ . Failure index compute karo aur batao ki ply fail hoti hai ya nahi.

Recall Solution

Yeh form kyun: har squared term hai "stress apni limit ke kitne paas hai". Inhe sum karna poochta hai "combined, kya humne 100% capacity reach kar li?" $\geq 1$ value matlab failure initiate ho gayi. $(\frac{1400}{2000})^{2} + 1 \cdot (\frac{60}{100})^{2} = (0.7)^{2} + (0.6)^{2} = 0.49 + 0.36 = 0.85.$ $0.85 < 1$ , toh ply abhi fail nahi hui — yeh Hashin criterion ke 85% par hai. Ise margin mein convert karne ke liye dekho Stress Analysis and Margins.

Level 4 — Synthesis

Exercise 4.1 (L4)

Ek bracket par pehla modal run $f_{1} = 850$ Hz deta hai. Launcher ka dominant tonal excitation $820$ Hz par baitha hai — khataraanak close. Tumhe stiffening se $f_{1}$ ko $1000$ Hz se upar uthana hai. Ek simple mode ke liye, $f_{1} \propto k / m$ jahan $k$ effective stiffness hai aur $m$ effective mass. Agar tum ek rib add kar sako jo $k$ badhaye lekin $m$ essentially unchanged rahe, toh $k$ ko kis factor se badhana hoga?

Recall Solution

$f \propto k / m$ kyun: ek mode mass-spring oscillator ki tarah behave karta hai; uski frequency $f = \frac{1}{2 π} k / m$ hai. Stiffness double karne se frequency $2$ se badhti hai, 2 se nahi — square root hi poori kahani hai.

Ratio set up karo. $m$ fixed ke saath: $\frac{f _{new}}{f _{old}} = \frac{k _{new}}{k _{old}} \Rightarrow \frac{k _{new}}{k _{old}} = (\frac{f _{new}}{f _{old}})^{2} .$ Plug in karo target $f_{new} = 1000$ , $f_{old} = 850$ : $\frac{k _{new}}{k _{old}} = (\frac{1000}{850})^{2} \approx 1.384.$ Tumhe effective stiffness lagbhag 38% badhani hogi. Yeh ek classic Vibration and Modal Analysis design decision hai Launch Vehicle Loads mein identify ki gayi resonance se bachne ke liye.

Exercise 4.2 (L4)

Tumhe ek deployable boom ke liye analysis campaign choose karni hai jo 90° rotation ke through fold hoti hai, carbon-fiber tape springs use karta hai, aur frictional contact ke saath latch hota hai. Order mein modelling decisions list karo (analysis type, geometric setting, material model, contact) aur har ek ko ek line mein justify karo.

Recall Solution

Analysis type ::: nonlinear static (Riks/arc-length) — boom unstable configurations ke through snap karta hai; arc-length equilibrium path ko peak load ke baad follow karta hai jahan ordinary load-stepping diverge ho jaata hai.
Geometric setting ::: NLGEOM=YES (large deformation) — 90° rotation "small" ke kahin paas nahi hai, toh linear geometry stiffness ko bahut galat predict karega. ABAQUS chuno.
Material model ::: orthotropic composite (Hashin damage) — carbon tape ki stiffness direction-dependent hai aur uske apne failure modes hain; dekho Carbon Fiber Structures aur Composite Materials.
Contact ::: friction pair at the latch — friction ke bina latch hold nahi kar sakta; ek friction coefficient model karo taaki mechanism lock ho. Parent mein bolted-joint example dekho.

Level 5 — Mastery

Definition Internal vs external force (subscript convention)

Nonlinear solve mein hum forces ko do lists mein split karte hain:

$F_{ext}$ = external force jo tum structure par apply karte ho (load jo tum command karte ho — gravity, bolt preload, shaker push). Yeh structure ke deform hone par change nahi hota.
$F_{int} (u)$ = internal force jo structure push back karta hai jab uska material displacement $u$ tak stretch hota hai. Kyunki ek stiffening material jitna aage push karo utna zyada resist karta hai, $F_{int}$ $u$ par depend karta hai — wahi dependence problem ke nonlinear hone ki poori wajah hai. Equilibrium matlab "apply equals push-back", yani $F_{ext} = F_{int} (u)$ . Bacha hua imbalance $R = F_{ext} - F_{int}$ residual hai jise hum zero tak drive karte hain. Neeche hum ek scalar (1-DOF) case work karte hain, toh bold vectors plain numbers ban jaate hain.

Exercise 5.1 (L5)

Ek nonlinear ABAQUS step Newton–Raphson use karta hai. Iteration $i$ par internal force $F_{int} (u) = 3 u^{3}$ hai (ek stiffening spring, scalar 1-DOF), aur external load $F_{ext} = 24$ hai. $u^{(0)} = 1$ se start karke, ek Newton–Raphson update perform karo aur $u^{(1)}$ report karo. (Tangent stiffness $K_{T} = d F_{int} / d u = 9 u^{2}$ .)

Recall Solution

Iterate kyun karte hain: $3 u^{3} = 24$ nonlinear hai — koi one-shot matrix inverse nahi hai. Newton–Raphson locally linearize karta hai, ek step leta hai, aur repeat karta hai. Exact root $u = 2$ hai (kyunki $3 \cdot 2^{3} = 24$ ); dekho iteration uski taraf badhti hai.

Step 1 — residual (woh imbalance jise hum khatam karna chahte hain), external minus internal: $R^{(0)} = F_{ext} - F_{int} (u^{(0)}) = 24 - 3 (1)^{3} = 21.$ Step 2 — guess par tangent stiffness: $K_{T} = 9 u^{2} = 9 (1)^{2} = 9.$ Step 3 — correction solve karo $K_{T} Δ u = R$ : $Δ u = \frac{21}{9} \approx 2.3333.$ Step 4 — update: $u^{(1)} = u^{(0)} + Δ u = 1 + 2.3333 = 3.3333.$ Iska matlab: ek step overshoot karta hai ( $3.33$ vs true $2$ ) kyunki cubic sharply curve karta hai, lekin agli iterations ise wapis kheenchegi — convergence fast hai (quadratic) ek baar close ho jaane ke baad.

Exercise 5.2 (L5)

Exercise 5.1 continue karo: $u^{(1)} = 3.3333$ se doosri Newton–Raphson iteration perform karo aur $u^{(2)}$ report karo. True root $u = 2$ ki taraf convergence par comment karo.

Recall Solution

Residual (external minus internal): $R^{(1)} = F_{ext} - F_{int} (u^{(1)}) = 24 - 3 (3.3333)^{3} = 24 - 3 (37.037) = 24 - 111.11 = - 87.11$ . Tangent: $K_{T} = 9 (3.3333)^{2} = 9 (11.111) = 100.0$ . Correction: $Δ u = R / K_{T} = - 87.11/100.0 = - 0.8711$ . Update: $u^{(2)} = 3.3333 - 0.8711 = 2.4622$ . Comment: hum $3.33 \to 2.46$ move hue, true root $2$ ke paas aa rahe hain. Residual ka sign flip hua (overshoot phir correct), jo stiffening curve par Newton ke liye normal hai; ek real ABAQUS run tab tak iterate karta rehta hai jab tak $∥ R ∥ < tol$ . Dekho Model Correlation ki aise solver output ko phir Ground Test Procedures data se kaise match kiya jaata hai.

Exercise 5.3 (L5)

Ek linear-static stress result ek joint mein peak von-Mises stress $σ = 180$ MPa deta hai jiska material allowable $σ_{allow} = 250$ MPa hai. Margin of safety $MoS = \frac{σ _{allow}}{σ} - 1$ compute karo aur batao ki joint pass karta hai ya nahi. Phir required factor of safety $F o S = 1.25$ load par apply karke (yani design stress $= 1.25 σ$ ) re-check karo.

Recall Solution

Reminder — FoS ka matlab: factor of safety (FoS) ek multiplier hai jo tum deliberately load par apply karte ho taaki structure un loads se zyada strong design ho jo woh actually dekhega; $F o S = 1.25$ matlab "expected se 25% zyada load ke liye design karo".

MoS is tarah kyun define hai: margin of safety (MoS) jawaab deta hai "jo use ho raha hai usse pare kitna extra capacity hai?" $MoS = 0$ matlab "bilkul limit par"; positive matlab room to spare; negative matlab failure.

FoS ke bina: $MoS = \frac{σ _{allow}}{σ} - 1 = \frac{250}{180} - 1 = 1.3889 - 1 = 0.3889 (\approx 39%) . Pass.$ FoS = 1.25 load par apply karne ke saath, design stress $1.25 \times 180 = 225$ MPa ho jaata hai, aur hum usi ko unchanged material allowable se compare karte hain: $MoS = \frac{σ _{allow}}{1.25 σ} - 1 = \frac{250}{225} - 1 = 1.1111 - 1 = 0.1111 (\approx 11%) . Abhi bhi pass (positive).$ Joint safety factor ke saath bhi survive karta hai, parent ke "joint is safe" conclusion se match karta hai. Dekho Stress Analysis and Margins.

Recall One-line self-test (answers cover karo)

NASTRAN mein modal solution number? ::: SOL 103. Woh card jo ek node par saare 6 DOF clamp karta hai? ::: SPC1 with digits 123456. Fully-free stiffness matrix singular kyun hoti hai? ::: rigid-body modes — structure float kar sakta hai, toh koi unique solution nahi. Frequency stiffness ki kis power ke saath scale karti hai? ::: square root, $f \propto k$ . Newton–Raphson correction equation? ::: $K_{T} Δ u = R$ , with $R = F_{ext} - F_{int}$ . Margin of safety formula? ::: $MoS = σ_{allow} / σ - 1$ . Modal analysis mein $ϕ$ kya stand karta hai? ::: mode shape vector — natural frequency par motion ka pattern.