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

Intuition How to use this page

Each line below is a trap. Read the left side, commit to an answer out loud, then reveal the right side. If your reasoning was hollow ("I think... true?"), that's exactly the misconception this page is hunting. Every answer gives you the why, not just the verdict.

Before you start, make sure the vocabulary is solid. When we say DOF (degree of freedom) we mean one number the solver is allowed to change — for a single node in 3D that's 3 translations plus 3 rotations, so 6 numbers. When we say mesh we mean the structure chopped into small pieces (elements) joined at nodes. When we say $Ku = F$ , read it as "stiffness times displacement equals force" — the machine knows $K$ and $F$ , and solves for the unknown displacements $u$ . Keep those three pictures in your head.

Related vault topics: parent topic, Vibration and Modal Analysis, Stress Analysis and Margins, Model Correlation, Composite Materials.

True or false — justify

A finer mesh always gives a more accurate answer

False. Finer usually converges toward the exact answer, but a badly shaped element (a "sliver" with a tiny angle) or a mesh refined in the wrong place wastes cost and can worsen conditioning; accuracy comes from mesh convergence, not raw element count. Always check that the answer stops changing as you refine.

A modal analysis needs no applied force

F

True. Modes are a property of the structure itself — you solve the eigenvalue problem

(K - ω^{2} M) ϕ = 0

, which has zero on the right. Frequencies and mode shapes exist whether or not anyone pushes on the structure. See Vibration and Modal Analysis.

NASTRAN cannot do dynamics, only static

Ku = F

False. NASTRAN's heritage is precisely launch dynamics — modal, frequency response, and random vibration (PSD). Static is just its simplest solution sequence, not its only one.

A linear static solution scales exactly: double the load, double the stress

True — by definition of linear. Linear means

K

is constant, so

u = K^{- 1} F

is proportional to

F

. The moment this stops being true (contact opening, plasticity, buckling) you are no longer in the linear world and must switch to a nonlinear solver.

Newton-Raphson gives the exact nonlinear answer in one solve

False. It's iterative: each step solves a linearised system

K_{T} Δ u = R

and shrinks the residual

R

. You repeat until

∥ R ∥

drops below tolerance — one solve only lands the exact answer if the problem was linear all along.

ABAQUS is simply a "better NASTRAN" — always use it

False. For a 5-million-DOF linear launch model, NASTRAN's optimised linear solvers are far faster and carry decades of flight heritage. Reach for ABAQUS only when the physics is genuinely nonlinear (contact, plasticity, large rotation, composite damage).

Fixing all six DOF at one node is enough to make a static run solve

Usually false for a full 3D body. A single fully-fixed node still lets the rest of the structure hinge or spin about it if the load line allows; the solver needs enough constraints to remove all rigid-body motion, or

K

stays singular and the solve fails.

The stiffness matrix

K

depends on the applied load

False in linear analysis —

K

comes only from geometry and material (

\int B^{T} DB d V

). It becomes load/displacement-dependent (the tangent stiffness

K_{T}

) only once you go nonlinear.

Spot the error

"My model has no boundary conditions, but static solved fine and gave huge displacements."

The huge displacements are the tell. With no constraints

K

is singular (rigid-body modes have zero stiffness); a "solved" run that survives numerically is riding on round-off noise. Those displacements are meaningless — pin down the rigid-body motion first.

"I ran SOL 103 (modal) and asked why the stresses look wrong."

Modal analysis outputs mode shapes, which are unscaled patterns of motion (often normalised to unit mass). Their "stresses" are relative shapes, not real stress values. For real stress you need a static or response run with actual forces.

"I found a mode at 850 Hz but the launch environment has no energy there, so I ignored it."

That is actually the right call for that mode — but check the whole spectrum. Resonance only bites where the environment's energy overlaps a natural frequency, so compare every low mode against the launch PSD, not just the first.

"I modelled a bolted joint in NASTRAN linear static and trusted the contact pressure."

Contact (surfaces touching, separating, sliding) is inherently nonlinear — the contact area changes with load. Linear NASTRAN can't open or close a gap; it either glues the surfaces or leaves them free. Use ABAQUS contact for true pressure distribution.

"Failure index came out 0.8, so the part failed."

A failure index of 1.0 means onset of failure; 0.8 means 80% of the way there — the part has not failed but the margin is thin. See Stress Analysis and Margins; index below 1 with a factor of safety is what you want.

"My composite ply directions are all 0°, and I'm surprised the panel is weak in the 90° direction."

A single-direction layup is strong only along the fibres. Quasi-isotropic strength needs mixed orientations (0/45/-45/90); with all-0° plies the transverse (matrix-dominated) direction is far weaker by design. See Composite Materials.

"I switched the mesh from coarse to fine and the frequency dropped, so the coarse one was wrong and I'm done."

You've only shown two points. Coarse meshes tend to over-stiffen (frequencies too high), so a drop is expected — but you must refine again and confirm the frequency has stopped moving. Two points is not convergence.

"I applied the bolt torque directly as a load in the model."

Torque isn't a structural load the mesh understands; you convert it to an equivalent axial preload

F = T / (k d)

and apply that. Feeding raw torque in the wrong DOF gives nonsense.

Why questions

Why formulate FEM through virtual work instead of solving the PDE directly?

The stress-equilibrium PDE

\nabla \cdot σ + f = 0

demands exact solutions over odd geometries — impossible by hand. Virtual work turns it into an integral (weak) form that only needs approximate shape functions per element, which discretise cleanly into the matrix

Ku = F

Why does the tangent stiffness

K_{T}

appear specifically as a derivative

\partial F_{int} / \partial u

Because Newton-Raphson linearises the residual with a Taylor step:

R (u + Δ u) \approx R (u) + K_{T} Δ u

. The derivative is the local slope that predicts how the residual changes, so setting the prediction to zero gives the best correction

Δ u

Why is the Riks (arc-length) method needed for post-buckling instead of ordinary load stepping?

After a peak load the structure can snap through — displacement grows while load drops. Load-controlled stepping can't follow a decreasing-load path (it would jump). Riks advances along the equilibrium curve's arc length, tracking load and displacement together, so it survives the turnaround.

Why do spacecraft engineers prefer text input decks over pure GUI clicking?

Text decks (.bdf, .inp) are diffable and version-controllable, and they let you script parametric studies — sweep a plate thickness across 50 runs automatically. GUI clicks can't be replayed or audited the same way. This also underpins Model Correlation.

Why does the modal problem take the form

(K - ω^{2} M) ϕ = 0

with a mass matrix

M

Free vibration means inertia balances stiffness:

M \ddot{u} + Ku = 0

. Assuming sinusoidal motion

u = ϕ e^{iω t}

turns

\ddot{u}

into

- ω^{2} u

, so mass enters as

M

and the whole thing collapses to that eigenvalue equation.

Why check the mode shapes, not just the frequency numbers?

Two designs can share a frequency yet vibrate completely differently — one twisting a solar panel, one flexing a bracket. The shape tells you where to stiffen and whether the mode is even relevant to the load path. Frequency alone hides that.

Why does a bolt preload increase joint stiffness (geometric nonlinearity)?

Pulling the plates hard together closes gaps and puts the members in a stressed state that resists further motion — the stiffness now depends on the deformed, preloaded geometry, which is exactly what NLGEOM=YES captures. A linear model, blind to this, would understate the joint's rigidity.

Edge cases

What frequencies does a completely unconstrained (free-free) structure show first?

Six rigid-body modes at (near) zero Hz — three translations, three rotations that cost no strain energy. They're physically correct for a spacecraft floating in orbit; the first flexible mode is the one you compare against loads.

What happens to

Ku = F

if the structure is a mechanism (has an internal hinge with no stiffness)?

K

is singular along the mechanism direction, so no unique

u

exists — the solve either fails or returns unbounded motion. FEM assumes a load-carrying structure; a free mechanism needs constraints or a stiffness there.

A material's Young's modulus is entered as zero by mistake — what does the solve do?

That element contributes no stiffness, effectively a hole in the load path. Neighbouring elements over-deform, displacements blow up, and

K

may go singular. Always sanity-check material cards before trusting output.

At exactly the failure index = 1.0, has the part broken?

It's at the onset of damage — the criterion is just satisfied. For progressive failure (like Hashin in composites) this marks where cracking begins, not total collapse; the structure may still carry load as damage spreads until fibre breakage.

A load is applied but no motion results (

u = 0

) — is this a bug?

Not necessarily. If the load line runs straight into a fully constrained node with no free DOF in its direction, the reaction absorbs it and displacement is genuinely zero. Confirm by checking reaction forces balance the applied load.

The nonlinear solve won't converge — the residual

∥ R ∥

plateaus instead of dropping. What's the physical reading?

The structure has likely hit an instability (buckling, snap-through, or full contact separation) where the tangent stiffness

K_{T}

becomes singular or negative-definite. Newton-Raphson stalls there; you switch to an arc-length (Riks) scheme or add stabilisation.

Two identical meshes give slightly different frequencies on two machines — real error?

Usually not physical — tiny differences come from floating-point round-off and solver ordering, well below engineering significance. Worry only if the gap is large enough to change a resonance decision, which points to an ill-conditioned model, not a real design difference.

Recall Quick self-test

Modal analysis right-hand side is... ::: Zero — modes are a property of the structure, no external force. Linear analysis assumes $K$ is... ::: Constant (independent of load and displacement). Failure index of 1.0 means... ::: Onset of failure, not necessarily total collapse. A free-free structure's first six modes are at... ::: Near-zero Hz — the rigid-body modes. Contact and plasticity force you into... ::: Nonlinear analysis (e.g. ABAQUS, Newton-Raphson).