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

Before anything, the symbols we will lean on. When you see:

• $\mathbf{K}$ — the stiffness matrix. Think of it as "how hard the structure pushes back per unit of squeeze." Big $\mathbf{K}$ = stiff = hard to move.
• $\mathbf{M}$ — the mass matrix. "How much inertia resists being shaken."
• $\mathbf{u}$ — the displacement (how far a point moves), $\mathbf{F}$ — the force applied.
• $\omega$ — the angular frequency (also called circular frequency): how fast an oscillation sweeps through angle, measured in radians per second (rad/s). One full cycle is $2\pi$ radians, so ordinary frequency in Hz (cycles per second) is $f=\omega /(2\pi )$.
• $\mathbf{\varphi }$ (Greek "phi") — the mode shape vector: the pattern of movement when the structure vibrates freely at one of its natural frequencies (which point goes up while another goes down).
• DOF — degree of freedom: one independent way a point can move (e.g. slide along $x$, or rotate). A model's "size" is its total number of DOF.

Everything below is built from $\mathbf{Ku}=\mathbf{F}$ (statics) and the modal eigenvalue/eigenvector equation $(\mathbf{K}-{\omega }^2\mathbf{M})\mathbf{\varphi }=0$ (it asks: for which frequencies $\omega$ and shapes $\mathbf{\varphi }$ does the structure vibrate on its own?), both from the parent note.

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The scenario matrix

Every FEM problem you can be handed falls into one of these cells. The worked examples below each carry a tag [Cell n] so you can see the whole grid gets covered. ("Sign case" below just means: watch which quantities can go positive, negative, or zero and what each choice means physically.)

#	Case class	The trap it hides	Example
1	Linear static, single DOF — simplest positive-force case	forgetting units in $\mathbf{Ku}=\mathbf{F}$	Ex 1
2	Stiffness scaling / limiting value — what happens as stiffness $\to \infty$ or $\to 0$	dividing by zero (rigid vs free body)	Ex 2
3	Modal, natural frequency — eigenvalue positive root	which root ($\pm$) is physical	Ex 3
4	Resonance margin — real-world "will it survive launch?"	Hz vs rad/s confusion	Ex 4
5	Nonlinear Newton–Raphson — iteration converging	tangent stiffness changes each step	Ex 5
6	Contact / bolt preload — word problem, torque→force	the 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, spot the shortcut	scaling law $f\propto \sqrt{K/M}$	Ex 8

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

Forecast: guess before reading on — is the answer millimetres or metres? Bigger $K$ means…?

The figure draws the same $500\text{N}$ pull (pink arrow) acting on two springs: the thick blue spring is stiff and barely stretches; the thin yellow spring is soft and stretches far. Read off the lesson — for a fixed force, displacement shrinks as stiffness grows. That is exactly what dividing by $K$ in step 2 does.

1. Write the equilibrium. $Ku=F$. Why this step? This is $\mathbf{Ku}=\mathbf{F}$ shrunk to one number each — the smallest possible FEM. It says "spring force = applied force at balance."
2. Solve for the unknown. $u=F/K=500/(2.0\times 10^5)$. Why this step? We want displacement, so we divide the driving force by the resistance. Look at the figure: a stiff spring (thick coils) barely stretches for the same pull.
3. Compute. $u=2.5\times 10^{-3}\text{m}=2.5\text{mm}$. Why this step? We carry the arithmetic through to a single number with units so the result is an actual engineering quantity we can compare to a tolerance — not just a formula.

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

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

Forecast: one of these gives $u=0$, the other blows up. Which is which?

1. Rigid limit. $u=F/K$. As $K\to \infty$, $u\to 0$. Why this step? A truly rigid body cannot deform — infinite pushback means zero movement. This is why a fully-fixed stress model needs no displacement solution at rigid points.
2. Free limit. As $K\to 0$, $u=F/K\to \infty$. Why this step? Zero stiffness = nothing holds the part; a force accelerates it away with no equilibrium. In FEM this is the dreaded singular stiffness matrix — the solver reports "zero pivot."

Verify: Sanity — a spacecraft panel free in space does just drift; statics has no answer, so you'd run a modal analysis instead, where free-free modes are legal.

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Ex 3 — Natural frequency of one mode [Cell 3]

Forecast: natural frequency rises when the structure is stiffer or lighter — guess whether this is above or below 100 Hz.

The figure plots frequency $f$ against mass $M$ for our fixed $K$. Notice the curve falls as mass grows (heavier = slower) and our chosen point $M=0.5\text{kg}$ (yellow dot) lands on the pink dashed line at $100.7\text{Hz}$. Use it to sanity-check every modal answer: a heavier structure must land lower-left on this curve.

1. Recall the modal equation. $(\mathbf{K}-{\omega }^2\mathbf{M})\mathbf{\varphi }=0$. For one DOF: $K-{\omega }^{2}M=0$. Why this step? A non-zero shape $\mathbf{\varphi }$ can only survive if the bracket $(K-{\omega }^{2}M)$ is zero — otherwise the only solution is "no motion." That zero condition is the natural frequency.
2. Solve for $\omega$. ${\omega }^2=K/M\Rightarrow \omega =\sqrt{K/M}$. Why this step? We take the positive square root: $\omega$ (angular frequency) is a rate of angle-sweep, physically positive. The negative root describes the same oscillation running backwards — same frequency.
3. Compute $\omega$. $\omega =\sqrt{2.0\times 10^5/0.5}=\sqrt{4.0\times 10^5}=632.5\text{rad/s}$. Why this step? We turn the symbolic $\sqrt{K/M}$ into a concrete rad/s number so it can be converted to the Hz value that launch specifications actually quote.
4. Convert to Hz. $f=\omega /(2\pi )=632.5/6.2832=100.7\text{Hz}$. Why this step? Hz counts full cycles per second; $2\pi$ radians make one cycle. Engineers quote Hz because launch specs are in Hz.

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

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

Forecast: the required minimum is $1.4\times 80$. Guess the number, then check whether 100.7 clears it.

1. Compute the required frequency floor. $f_{\text{min}}=1.4\times 80=112\text{Hz}$. Why this step? The $1.4$ is a frequency separation factor of safety — keep the mode clear of the forcing band so response never blows up at resonance.
2. Compare. $f_1=100.7\text{Hz}<112\text{Hz}$. Fails. Why this step? Being above 80 Hz isn't enough; the margin swallows the gap.
3. Fix it — how much stiffer? Need $f_1\ge 112$. Since $f\propto \sqrt{K}$, required stiffness ratio $=(112/100.7{)}^2=1.237$. So raise $K$ by $\approx 24\%$. Why this step? Frequency scales with the square root of stiffness, so a small frequency bump needs a proportionally larger stiffness bump.

Verify: New $f=100.7\sqrt{1.237}=112.0\text{Hz}$ ✓ — exactly the floor. This is the logic behind adding a stiffener, as the parent bracket example did.

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

Forecast: the true answer lies where $F_{\text{int}}=200$. Will one step overshoot or undershoot?

The figure shows the curved blue force law $F_{\text{int}}(u)$, the flat yellow line at the target $200\text{N}$, and the pink straight tangent line touching the curve at our guess $u^{(0)}=0.1$. Newton–Raphson slides down that tangent, not the curve, to find the next guess — watch how the tangent's slope (the tangent stiffness $K_T$) sets how big a correction we take.

1. Residual at the guess. $R=F-F_{\text{int}}(u^{(0)})=200-(1000(0.1)+5\times 10^6(0.1{)}^3)$. Why this step? $R$ measures "how far out of balance we are." Here $F_{\text{int}}(0.1)=100+5000=5100\text{N}$, so $R=200-5100=-4900\text{N}$ — a big negative residual means the internal force badly overshoots, so the guess $0.1$ is too large.
2. Tangent stiffness. $K_T=\frac{d{F}_{\text{int}}}{du}=K_0+3\beta u^2=1000+3(5\times 10^6)(0.1{)}^2=1000+150000=151000\text{N/m}$. Why this step? The tangent is the local slope of the force curve (pink line in the figure). Linear FEM would use a fixed slope; nonlinear must recompute it each iteration because the curve bends.
3. Correction. $\Delta u=R/K_T=-4900/151000=-0.03245\text{m}$. Why this step? Solve $K_T\Delta u=R$ — the 1-D version of ${\mathbf{K}}_T\Delta \mathbf{u}=\mathbf{R}$. Negative pulls the guess back down, as expected.
4. Update. $u^{(1)}=0.1+(-0.03245)=0.06755\text{m}$. Why this step? We add the correction to the old guess to produce the improved guess — this is the actual "step" that Newton–Raphson repeats, marching $u$ toward the balance point.

Verify: $F_{\text{int}}(0.06755)=1000(0.06755)+5\times 10^6(0.06755{)}^3=67.55+1541\approx 1609\text{N}$. Still above 200 — one step isn't enough; the method keeps iterating (that's why the parent note loops "repeat until $\Vert R\Vert <\text{tol}$"). The residual dropped from $4900$ to $\approx 1409$ ✓ — moving the right way.

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Ex 6 — Bolt preload from torque (word problem) [Cell 6]

Forecast: smaller diameter for the same torque — more or less force?

1. Recall the torque–tension law. $T=kFd\Rightarrow F=\frac{T}{kd}$. Why this step? Torque is converted to axial clamp force through the thread geometry, lumped into the empirical nut factor $k$.
2. M8 case. $F=\frac{20}{0.2\times 0.008}=\frac{20}{0.0016}=12500\text{N}$. Why this step? Units: $\frac{\text{N⋅m}}{\text{m}}=\text{N}$ ✓.
3. M6 case (parent value). $F=\frac{10}{0.2\times 0.006}=\frac{10}{0.0012}=8333\text{N}$. Why this step? Matches the parent note's "$\approx 8333\text{N}$" — smaller $d$ gives more force for the same torque, since $F\propto 1/d$.

Verify: Back-substitute M8: $kFd=0.2\times 12500\times 0.008=20\text{N⋅m}=T$ ✓. This clamp force is the *CLOAD preload the ABAQUS contact model needs.

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

Forecast: neither stress alone exceeds its strength — but combined? Guess: index above or below 1?

1. Write the Hashin fiber-tension index. $\text{FI}={\left(\frac{{\sigma }_{11}}{X_T}\right)}^2+\alpha {\left(\frac{{\tau }_{12}}{S}\right)}^2$. Why this step? Failure is a combined effect. Squaring means both tension and shear pile on positively; the ratios make each term dimensionless.
2. Fiber term. $(1400/1500{)}^2=(0.9333{)}^2=0.8711$.
3. Shear term. $(200/70{)}^2=(2.857{)}^2=8.163$. Why this step? Look closely — shear alone already exceeds strength ($200>70$), so its term dwarfs the fiber term.
4. Sum. $\text{FI}=0.8711+8.163=9.03$. Why this step? The rule: $\text{FI}\ge 1\Rightarrow$ failure. Here $9.03\gg 1$. Failed — and the degenerate boundary $\text{FI}=1$ would mean "exactly on the edge, damage just initiating."

Verify: If we set ${\tau }_{12}=0$, $\text{FI}=0.871<1$ (safe) — confirming shear drives this failure. See Model Correlation for tying such predictions to test coupons.

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Ex 8 — Exam twist: mass change without re-solving [Cell 8]

Forecast: heavier structure — frequency goes up or down? By what factor?

1. Spot the scaling law. $f=\frac{1}{2\pi }\sqrt{K/M}$, so $f\propto 1/\sqrt{M}$ (K fixed). Why this step? This is the shortcut. Because only $M$ changed, the whole eigenvalue solve rescales by a single factor — no new run needed.
2. Apply the ratio. $\frac{f_{\text{new}}}{f_{\text{old}}}=\sqrt{\frac{M_{\text{old}}}{M_{\text{new}}}}=\sqrt{\frac{1}{2}}=0.7071$. Why this step? Doubling mass multiplies frequency by $1/\sqrt{2}$.
3. Compute. $f_{\text{new}}=850\times 0.7071=601\text{Hz}$. Why this step? Heavier = slower vibration = lower frequency — physically sensible.

Verify: $850^2/601^2=2.00$ — mass ratio recovered ✓. Warning: this trick assumes the mode shape $\mathbf{\varphi }$ stays the same; if the box shifts the shape, you must re-solve. This is exactly the judgement Thermal-Structural Coupling and full Vibration and Modal Analysis runs exist to check.

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Recall Self-test

Ex1 displacement for K=2e5, F=500 ::: 2.5 mm As K→0 the static displacement ::: goes to infinity (singular, no equilibrium) First frequency for K=2e5, M=0.5 kg ::: 100.7 Hz Frequency separation floor for 80 Hz drive, factor 1.4 ::: 112 Hz Clamp force for M8, T=20 N·m, k=0.2 ::: 12500 N Hashin index with σ11=1400/X_T=1500, τ12=200/S=70 ::: 9.03 (failed) New f1 when modal mass doubles from 850 Hz ::: 601 Hz