Worked examples — FEM software — NASTRAN, ABAQUS (concepts and use)

3.6.20 · D3 · Physics › Spacecraft Structures & Systems Engineering › FEM software — NASTRAN, ABAQUS (concepts and use)

Shuru karne se pehle, woh symbols jo hum baar baar use karenge. Jab bhi dekho:

• $\mathbf{K}$ — stiffness matrix. Socho isko "structure unit squeeze pe kitna push back karta hai." Bada $\mathbf{K}$ = stiff = hilana mushkil.
• $\mathbf{M}$ — mass matrix. "Kitni inertia shake kiye jaane ka virodh karti hai."
• $\mathbf{u}$ — displacement (koi point kitna door move karta hai), $\mathbf{F}$ — lagaya gaya force.
• $\omega$ — angular frequency (circular frequency bhi kehte hain): ek oscillation kitni tezi se angle sweep karta hai, radians per second (rad/s) mein measure hota hai. Ek puri cycle $2\pi$ radians ki hoti hai, isliye Hz mein ordinary frequency (cycles per second) hai $f=\omega /(2\pi )$.
• $\mathbf{\varphi }$ (Greek "phi") — mode shape vector: movement ka pattern jab structure apni natural frequencies mein se kisi ek pe freely vibrate karta hai (kaunsa point upar jaata hai jab doosra neeche jaata hai).
• DOF — degree of freedom: ek independent tarika jisme koi point move kar sakta hai (jaise $x$ ke saath slide karna, ya rotate karna). Model ka "size" uske total DOF ki number hai.

Neeche sab kuch $\mathbf{Ku}=\mathbf{F}$ (statics) aur modal eigenvalue/eigenvector equation $(\mathbf{K}-{\omega }^2\mathbf{M})\mathbf{\varphi }=0$ se bana hai (yeh poochta hai: kin frequencies $\omega$ aur shapes $\mathbf{\varphi }$ ke liye structure apne aap vibrate karta hai?), dono parent note se hain.

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The scenario matrix

Har FEM problem jo tumhare haath aaye woh inn cells mein se kisi ek mein fit hoti hai. Neeche ke worked examples mein har ek [Cell n] tag carry karta hai taaki tum dekh sako ki poori grid cover ho rahi hai. ("Sign case" neeche ka matlab sirf yeh hai: dekho kaun si quantities positive, negative, ya zero ja sakti hain aur har choice ka physically kya matlab hai.)

#	Case class	Isme chupi trap	Example
1	Linear static, single DOF — sabse simple positive-force case	$\mathbf{Ku}=\mathbf{F}$ mein units bhool jaana	Ex 1
2	Stiffness scaling / limiting value — kya hota hai jab stiffness $\to \infty$ ya $\to 0$	zero se divide karna (rigid vs free body)	Ex 2
3	Modal, natural frequency — eigenvalue positive root	kaun sa root ($\pm$) physical hai	Ex 3
4	Resonance margin — real-world "kya yeh launch survive karega?"	Hz vs rad/s ka confusion	Ex 4
5	Nonlinear Newton–Raphson — iteration converging	tangent stiffness har step pe change hoti hai	Ex 5
6	Contact / bolt preload — word problem, torque→force	factor $0.2$ (nut factor)	Ex 6
7	Composite failure index — degenerate case: index exactly = 1	index $\ge 1$ boundary	Ex 7
8	Exam twist — combined modal + mass change, shortcut dhundho	scaling law $f\propto \sqrt{K/M}$	Ex 8

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Ex 1 — Linear static, single spring [Cell 1]

Forecast: aage padhne se pehle guess karo — kya answer millimetres mein hoga ya metres mein? Bada $K$ matlab…?

Figure mein wahi $500\text{N}$ pull (pink arrow) do springs pe dikhaya gaya hai: thick blue spring stiff hai aur barely stretches karta hai; thin yellow spring soft hai aur bahut stretch karta hai. Lesson padho — fixed force ke liye, displacement stiffness badhne pe ghatti hai. Yahi kaam step 2 mein $K$ se divide karna karta hai.

1. Equilibrium likho. $Ku=F$. Yeh step kyun? Yeh $\mathbf{Ku}=\mathbf{F}$ hai jisme har cheez ek number tak shrink ho gayi — sabse chhoti possible FEM. Yeh kehta hai "spring force = balance pe applied force."
2. Unknown ke liye solve karo. $u=F/K=500/(2.0\times 10^5)$. Yeh step kyun? Hum displacement chahte hain, isliye driving force ko resistance se divide karte hain. Figure dekho: ek stiff spring (thick coils) usi pull ke liye barely stretches karta hai.
3. Calculate karo. $u=2.5\times 10^{-3}\text{m}=2.5\text{mm}$. Yeh step kyun? Hum arithmetic ko ek single number with units tak le jaate hain taaki result ek actual engineering quantity ho jise hum tolerance se compare kar sakein — sirf formula nahi.

Verify: Units: $\frac{\text{N}}{\text{N/m}}=\text{m}$ ✓. Plug back karo: $Ku=2\times 10^5\times 2.5\times 10^{-3}=500\text{N}=F$ ✓.

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Ex 2 — Limiting stiffness: rigid aur free bodies [Cell 2]

Forecast: inme se ek $u=0$ deta hai, doosra blow up ho jaata hai. Kaun sa kaun sa hai?

1. Rigid limit. $u=F/K$. Jab $K\to \infty$, $u\to 0$. Yeh step kyun? Ek truly rigid body deform nahi ho sakti — infinite pushback matlab zero movement. Isliye ek fully-fixed stress model ko rigid points pe koi displacement solution ki zaroorat nahi.
2. Free limit. Jab $K\to 0$, $u=F/K\to \infty$. Yeh step kyun? Zero stiffness = kuch bhi part ko hold nahi kar raha; ek force usse koi equilibrium ke bagair accelerate karke le jaata hai. FEM mein yahi dreaded singular stiffness matrix hai — solver "zero pivot" report karta hai.

Verify: Sanity — space mein free spacecraft panel actually drift karta hai; statics ka koi answer nahi, isliye tum ek modal analysis run karoge, jahan free-free modes legal hain.

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Ex 3 — Ek mode ki natural frequency [Cell 3]

Forecast: natural frequency tab badhti hai jab structure stiffer ya lighter ho — guess karo yeh 100 Hz se upar hogi ya neeche.

Figure mein frequency $f$ ko mass $M$ ke against plot kiya gaya hai hamare fixed $K$ ke liye. Dekho curve mass badhne ke saath girta hai (bhaari = dhheema) aur humara chosen point $M=0.5\text{kg}$ (yellow dot) pink dashed line par $100.7\text{Hz}$ par land karta hai. Har modal answer ko sanity-check karne ke liye iska use karo: ek bhaari structure is curve par lower-left land karni chahiye.

1. Modal equation yaad karo. $(\mathbf{K}-{\omega }^2\mathbf{M})\mathbf{\varphi }=0$. Ek DOF ke liye: $K-{\omega }^{2}M=0$. Yeh step kyun? Ek non-zero shape $\mathbf{\varphi }$ tabhi survive kar sakta hai jab bracket $(K-{\omega }^{2}M)$ zero ho — warna ek hi solution hai "koi motion nahi." Woh zero condition hi natural frequency hai.
2. $\omega$ ke liye solve karo. ${\omega }^2=K/M\Rightarrow \omega =\sqrt{K/M}$. Yeh step kyun? Hum positive square root lete hain: $\omega$ (angular frequency) angle-sweep ki ek rate hai, physically positive. Negative root usi oscillation ko reverse mein describe karta hai — same frequency.
3. $\omega$ calculate karo. $\omega =\sqrt{2.0\times 10^5/0.5}=\sqrt{4.0\times 10^5}=632.5\text{rad/s}$. Yeh step kyun? Symbolic $\sqrt{K/M}$ ko ek concrete rad/s number mein badate hain taaki isse Hz value mein convert kiya ja sake jo launch specifications actually quote karti hain.
4. Hz mein convert karo. $f=\omega /(2\pi )=632.5/6.2832=100.7\text{Hz}$. Yeh step kyun? Hz full cycles per second count karta hai; $2\pi$ radians ek cycle banate hain. Engineers Hz quote karte hain kyunki launch specs Hz mein hoti hain.

Verify: ${\omega }^{2}M=(632.5{)}^2\times 0.5=2.0\times 10^5=K$ ✓. $\sqrt{K/M}$ ke units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{1/{\text{s}}^2}=1/\text{s}$ ✓.

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Ex 4 — Resonance margin (real-world) [Cell 4]

Forecast: required minimum $1.4\times 80$ hai. Number guess karo, phir check karo ki 100.7 clear karta hai ya nahi.

1. Required frequency floor calculate karo. $f_{\text{min}}=1.4\times 80=112\text{Hz}$. Yeh step kyun? $1.4$ ek frequency separation factor of safety hai — mode ko forcing band se door rakho taaki response kabhi resonance pe blow up na ho.
2. Compare karo. $f_1=100.7\text{Hz}<112\text{Hz}$. Fails. Yeh step kyun? 80 Hz se upar hona kaafi nahi; margin gap ko absorb kar leta hai.
3. Fix karo — kitna stiffer chahiye? Chahiye $f_1\ge 112$. Kyunki $f\propto \sqrt{K}$, required stiffness ratio $=(112/100.7{)}^2=1.237$ hai. Toh $K$ ko $\approx 24\%$ badhao. Yeh step kyun? Frequency stiffness ke square root ke saath scale karti hai, isliye thoda sa frequency bump proportionally bada stiffness bump maangta hai.

Verify: Naya $f=100.7\sqrt{1.237}=112.0\text{Hz}$ ✓ — exactly the floor. Yahi logic hai ek stiffener add karne ke peeche, jaisa parent bracket example ne kiya.

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Ex 5 — Nonlinear spring, Newton–Raphson [Cell 5]

Forecast: sahi answer wahan hai jahan $F_{\text{int}}=200$. Kya ek step overshoot karega ya undershoot?

Figure mein curved blue force law $F_{\text{int}}(u)$ dikhaya gaya hai, target $200\text{N}$ par flat yellow line, aur hamare guess $u^{(0)}=0.1$ par curve ko touch karti pink straight tangent line. Newton–Raphson curve pe nahi balki us tangent pe slide karta hai next guess dhundhne ke liye — dekho tangent ka slope (tangent stiffness $K_T$) set karta hai ki hum kitna bada correction lete hain.

1. Guess par residual. $R=F-F_{\text{int}}(u^{(0)})=200-(1000(0.1)+5\times 10^6(0.1{)}^3)$. Yeh step kyun? $R$ measure karta hai "hum kitne out of balance hain." Yahan $F_{\text{int}}(0.1)=100+5000=5100\text{N}$, toh $R=200-5100=-4900\text{N}$ — bada negative residual matlab internal force badly overshoot kar rahi hai, toh guess $0.1$ bahut bada hai.
2. Tangent stiffness. $K_T=\frac{d{F}_{\text{int}}}{du}=K_0+3\beta u^2=1000+3(5\times 10^6)(0.1{)}^2=1000+150000=151000\text{N/m}$. Yeh step kyun? Tangent force curve ka local slope hai (figure mein pink line). Linear FEM ek fixed slope use karta; nonlinear ko har iteration mein recompute karna padta hai kyunki curve bend karti hai.
3. Correction. $\Delta u=R/K_T=-4900/151000=-0.03245\text{m}$. Yeh step kyun? $K_T\Delta u=R$ solve karo — ${\mathbf{K}}_T\Delta \mathbf{u}=\mathbf{R}$ ka 1-D version. Negative guess ko wapas neeche pull karta hai, jaise expected tha.
4. Update karo. $u^{(1)}=0.1+(-0.03245)=0.06755\text{m}$. Yeh step kyun? Hum improved guess produce karne ke liye correction ko purane guess mein add karte hain — yahi actual "step" hai jo Newton–Raphson repeat karta hai, $u$ ko balance point ki taraf le jaata hai.

Verify: $F_{\text{int}}(0.06755)=1000(0.06755)+5\times 10^6(0.06755{)}^3=67.55+1541\approx 1609\text{N}$. Abhi bhi 200 se upar hai — ek step kaafi nahi; method iterate karta rehta hai (isliye parent note "repeat until $\Vert R\Vert <\text{tol}$" loop karta hai). Residual $4900$ se $\approx 1409$ tak gira ✓ — sahi direction mein ja raha hai.

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Ex 6 — Torque se bolt preload (word problem) [Cell 6]

Forecast: usi torque ke liye chhota diameter — zyada force ya kam?

1. Torque–tension law yaad karo. $T=kFd\Rightarrow F=\frac{T}{kd}$. Yeh step kyun? Torque ko axial clamp force mein thread geometry ke through convert kiya jaata hai, empirical nut factor $k$ mein lump karke.
2. M8 case. $F=\frac{20}{0.2\times 0.008}=\frac{20}{0.0016}=12500\text{N}$. Yeh step kyun? Units: $\frac{\text{N⋅m}}{\text{m}}=\text{N}$ ✓.
3. M6 case (parent value). $F=\frac{10}{0.2\times 0.006}=\frac{10}{0.0012}=8333\text{N}$. Yeh step kyun? Parent note ke "$\approx 8333\text{N}$" se match karta hai — chhota $d$ usi torque ke liye zyada force deta hai, kyunki $F\propto 1/d$.

Verify: M8 back-substitute karo: $kFd=0.2\times 12500\times 0.008=20\text{N⋅m}=T$ ✓. Yah clamp force woh *CLOAD preload hai jo ABAQUS contact model ko chahiye.

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Ex 7 — Composite failure index, boundary case [Cell 7]

Forecast: na stress akela apni strength se zyada hai — lekin combined? Guess karo: index 1 se upar ya neeche?

1. Hashin fiber-tension index likho. $\text{FI}={\left(\frac{{\sigma }_{11}}{X_T}\right)}^2+\alpha {\left(\frac{{\tau }_{12}}{S}\right)}^2$. Yeh step kyun? Failure ek combined effect hai. Squaring ka matlab hai tension aur shear dono positively pile on karte hain; ratios har term ko dimensionless banate hain.
2. Fiber term. $(1400/1500{)}^2=(0.9333{)}^2=0.8711$.
3. Shear term. $(200/70{)}^2=(2.857{)}^2=8.163$. Yeh step kyun? Ghaur se dekho — shear akela strength se zyada hai ($200>70$), toh uska term fiber term ko dwarf kar deta hai.
4. Sum karo. $\text{FI}=0.8711+8.163=9.03$. Yeh step kyun? Rule: $\text{FI}\ge 1\Rightarrow$ failure. Yahan $9.03\gg 1$. Failed — aur degenerate boundary $\text{FI}=1$ ka matlab hoga "exactly edge par, damage bas start ho raha hai."

Verify: Agar hum ${\tau }_{12}=0$ set karen, $\text{FI}=0.871<1$ (safe) — confirm hota hai ki shear is failure ko drive karta hai. Test coupons se aise predictions tie karne ke liye Model Correlation dekho.

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Ex 8 — Exam twist: re-solve kiye bagair mass change [Cell 8]

Forecast: bhaari structure — frequency upar jaayegi ya neeche? Kis factor se?

1. Scaling law spot karo. $f=\frac{1}{2\pi }\sqrt{K/M}$, toh $f\propto 1/\sqrt{M}$ (K fixed). Yeh step kyun? Yahi shortcut hai. Kyunki sirf $M$ badla, poora eigenvalue solve ek single factor se rescale ho jaata hai — koi naya run nahi chahiye.
2. Ratio apply karo. $\frac{f_{\text{new}}}{f_{\text{old}}}=\sqrt{\frac{M_{\text{old}}}{M_{\text{new}}}}=\sqrt{\frac{1}{2}}=0.7071$. Yeh step kyun? Mass double karna frequency ko $1/\sqrt{2}$ se multiply karta hai.
3. Calculate karo. $f_{\text{new}}=850\times 0.7071=601\text{Hz}$. Yeh step kyun? Bhaari = dhheema vibration = lower frequency — physically sensible.

Verify: $850^2/601^2=2.00$ — mass ratio recover hua ✓. Warning: yeh trick assume karti hai ki mode shape $\mathbf{\varphi }$ same rehta hai; agar box shape shift kare, toh re-solve karna padega. Yahi judgement hai jiske liye Thermal-Structural Coupling aur full Vibration and Modal Analysis runs exist karte hain.

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Recall Self-test

K=2e5, F=500 ke liye Ex1 displacement ::: 2.5 mm K→0 hone par static displacement ::: infinity ho jaata hai (singular, koi equilibrium nahi) K=2e5, M=0.5 kg ke liye pehli frequency ::: 100.7 Hz 80 Hz drive, factor 1.4 ke liye frequency separation floor ::: 112 Hz M8, T=20 N·m, k=0.2 ke liye clamp force ::: 12500 N σ11=1400/X_T=1500, τ12=200/S=70 ke saath Hashin index ::: 9.03 (failed) 850 Hz se modal mass double hone par naya f1 ::: 601 Hz