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

Intuition How to use this page

Cover the answers. Read a problem, work it on paper, THEN open the collapsible solution. Every symbol below was built in the parent note — if a symbol feels unfamiliar, jump back there before continuing. The problems climb five levels: from "name this card" up to "design a whole analysis campaign".

A tiny reminder of the alphabet we use, in plain words:

$K$ = the stiffness matrix — a big grid of numbers that says "how hard is it to push this structure at each point". Bold means "many numbers stacked together", not one number.
$u$ = the displacement vector — a list of how far every point moves. When we solve a model with just one point, this list has a single entry, and we may drop the bold and write plain $u$ for that one number.
$F$ = the force vector — a list of the pushes we apply (plain $F$ when it is a single number).
$M$ = the mass matrix — how much inertia sits at each point.
$ϕ$ = the mode shape vector — the pattern of movement of one vibration mode: it says "when the structure rings at this frequency, node A swings this much, node B that much". It is the shape, not the size.
$ω$ = angular frequency in radians per second; $f = ω /2 π$ is the everyday frequency in hertz (cycles per second).
DOF = degree of freedom = one independent way a point can move (a free node in 3D has 6: three slides $x, y, z$ and three spins).

Level 1 — Recognition

Exercise 1.1 (L1)

In the NASTRAN input deck below, match each card to its job.

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

Recall Solution

SOL 103 ::: selects the solution sequence; 103 is modal (natural-frequency) analysis.
EIGRL,1,,,10 ::: the Lanczos eigenvalue solver, method set 1, asking for 10 modes.
CQUAD4 ::: a 4-node shell element, ideal for thin plate-like structures.
MAT1,1,70.E9,,2700 ::: an isotropic material: Young's modulus $E = 70 \times 1 0^{9}$ Pa, density $ρ = 2700$ kg/m³.
SPC1,10,123456,1 ::: a single-point constraint fixing all 6 DOF (digits 1-2-3-4-5-6) at node 1 — a clamp.

Exercise 1.2 (L1)

For each scenario, name the analysis type (Linear Static / Modal / Frequency Response / Random Vibration) and whether NASTRAN or ABAQUS is the natural first choice.

(a) Find whether a solar-panel deflects too far under a steady 2 g acceleration. (b) Find the first bending frequency of a bracket to dodge a launcher's tone. (c) Predict delamination in an 8-ply carbon panel under compression. (d) Predict RMS stress from a broadband launch shaker environment.

Recall Solution

(a) ::: Linear Static ( $Ku = F$ ), NASTRAN. Steady load, no material nonlinearity. (b) ::: Modal, NASTRAN. The governing equation is $(K - ω^{2} M) ϕ = 0$ , where $ϕ$ (defined in the alphabet above) is the mode shape vector — the pattern in which the structure sways at natural frequency $ω$ . See Vibration and Modal Analysis. (c) ::: Nonlinear progressive-failure, ABAQUS. Hashin damage is material nonlinearity — see Composite Materials. (d) ::: Random Vibration (PSD), NASTRAN. See Launch Vehicle Loads.

Level 2 — Application

Exercise 2.1 (L2)

A single spring has stiffness $k = 5000$ N/m. Treated as a 1-DOF finite-element model, $K = [5000]$ . Apply $F = [250]$ N. Solve $Ku = F$ for the displacement.

Recall Solution

What: the global system collapses to one number. Why: with one DOF the displacement vector $u$ has a single entry, so we write it as plain $u$ , and $Ku = F$ is just $5000 u = 250$ . $u = \frac{F}{k} = \frac{250}{5000} = 0.05 m = 50 mm .$

Exercise 2.2 (L2)

A bolt is torqued to $T = 10$ N·m. Diameter $d = 6$ mm $= 0.006$ m, nut factor $k = 0.2$ . Using $T = k F d$ , find the preload force $F$ .

Recall Solution

Why this formula: torque you apply at the wrench spreads into three jobs — stretching the bolt, thread friction, under-head friction. The empirical nut factor $k$ bundles all three so that $T = k F d$ links the twist you feel to the tension inside the bolt. $F = \frac{T}{k d} = \frac{10}{0.2 \times 0.006} = \frac{10}{0.0012} \approx 8333 N .$ This is the *CLOAD preload in the parent's bolted-joint example.

Exercise 2.3 (L2)

A NASTRAN modal run reports the first mode at angular frequency $ω_{1} = 5341$ rad/s. Convert to ordinary frequency $f_{1}$ in hertz.

Recall Solution

Why divide by $2 π$ : one full cycle is $2 π$ radians, so cycles-per-second $=$ radians-per-second $\div 2 π$ . $f_{1} = \frac{ω _{1}}{2 π} = \frac{5341}{2 π} \approx 850 Hz .$ Matches the parent's L-bracket first bending mode.

Level 3 — Analysis

Exercise 3.1 (L3)

Two identical springs (each $k = 5000$ N/m) are joined in series between a wall and a free tip. Node 1 is at the wall (fixed), node 2 is the junction, node 3 is the free tip where a force $F = 250$ N is applied. Assemble the $3 \times 3$ stiffness matrix, apply the boundary condition at node 1, and solve for the tip displacement $u_{3}$ .

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

Recall Solution

Step 1 — element matrices (WHAT & WHY). A single spring between DOFs $i, j$ contributes $k_{e} = k [1 - 1 - 1 1] .$ The $+ k$ on the diagonal means "pushing a node stiffens it"; the $- k$ off-diagonal means "moving one end drags the other".

Step 2 — assemble into the global grid. Spring A links nodes 1–2, spring B links nodes 2–3. Overlap them at the shared node 2 (this is what "assembly" is — look at the middle diagonal in the figure where the two matrices pile up): $K = k 1 - 1 0 - 1 2 - 1 0 - 1 1 .$

Step 3 — apply the wall (boundary condition). Node 1 is clamped, so $u_{1} = 0$ . We delete row/column 1, leaving a $2 \times 2$ system in $u_{2}, u_{3}$ : $k [2 - 1 - 1 1] [u_{2} u_{3}] = [0250] .$

Step 4 — solve. From the two equations $5000 (2 u_{2} - u_{3}) = 0$ and $5000 (- u_{2} + u_{3}) = 250$ : the first gives $u_{3} = 2 u_{2}$ ; substitute: $5000 (u_{2}) = 250 \Rightarrow u_{2} = 0.05$ m, so $u_{3} = 0.10$ m $= 100$ mm.

Sanity check (WHAT IT MEANS): two springs in series are softer — effective stiffness $k /2 = 2500$ N/m, so $u_{3} = 250/2500 = 0.10$ m. Same answer. Springs in series stretch more, exactly like the tip in the figure overshooting the single-spring case.

Exercise 3.2 (L3)

In the composite panel of the parent note, a fiber-tension check uses the Hashin form $(\frac{σ _{11}}{X _{T}})^{2} + α (\frac{τ _{12}}{S})^{2} \geq 1.$ A $0^{\circ}$ ply carries $σ_{11} = 1400$ MPa and $τ_{12} = 60$ MPa. Strengths: $X_{T} = 2000$ MPa, $S = 100$ MPa, and $α = 1$ . Compute the failure index and state whether the ply fails.

Recall Solution

Why this form: each squared term is "how close a stress is to its own limit". Summing them asks "combined, have we reached 100% of capacity?" A value $\geq 1$ means failure has initiated. $(\frac{1400}{2000})^{2} + 1 \cdot (\frac{60}{100})^{2} = (0.7)^{2} + (0.6)^{2} = 0.49 + 0.36 = 0.85.$ $0.85 < 1$ , so the ply has not yet failed — it sits at 85% of the Hashin criterion. See Stress Analysis and Margins for turning this into a margin.

Level 4 — Synthesis

Exercise 4.1 (L4)

A first modal run on a bracket gives $f_{1} = 850$ Hz. The launcher's dominant tonal excitation sits at $820$ Hz — dangerously close. You must raise $f_{1}$ above $1000$ Hz by stiffening. For a simple mode, $f_{1} \propto k / m$ where $k$ is effective stiffness and $m$ effective mass. If you can add a rib that raises $k$ but leaves $m$ essentially unchanged, by what factor must $k$ increase?

Recall Solution

Why $f \propto k / m$ : a mode behaves like a mass-spring oscillator; its frequency is $f = \frac{1}{2 π} k / m$ . Doubling stiffness raises frequency by $2$ , not by 2 — the square root is the whole story.

Set up the ratio. With $m$ fixed: $\frac{f _{new}}{f _{old}} = \frac{k _{new}}{k _{old}} \Rightarrow \frac{k _{new}}{k _{old}} = (\frac{f _{new}}{f _{old}})^{2} .$ Plug in the target $f_{new} = 1000$ , $f_{old} = 850$ : $\frac{k _{new}}{k _{old}} = (\frac{1000}{850})^{2} \approx 1.384.$ You must raise effective stiffness by about 38%. This is a classic Vibration and Modal Analysis design decision to escape a resonance identified in Launch Vehicle Loads.

Exercise 4.2 (L4)

You must choose an analysis campaign for a deployable boom that folds through 90° of rotation, uses carbon-fiber tape springs, and latches with frictional contact. List, in order, the modelling decisions (analysis type, geometric setting, material model, contact) and justify each in one line.

Recall Solution

Analysis type ::: nonlinear static (Riks/arc-length) — the boom snaps through unstable configurations; arc-length follows the equilibrium path past the peak load where ordinary load-stepping diverges.
Geometric setting ::: NLGEOM=YES (large deformation) — a 90° rotation is nowhere near "small", so linear geometry would grossly mis-predict stiffness. Choose ABAQUS.
Material model ::: orthotropic composite (Hashin damage) — carbon tape has direction-dependent stiffness and its own failure modes; see Carbon Fiber Structures and Composite Materials.
Contact ::: friction pair at the latch — without friction the latch cannot hold; model a friction coefficient so the mechanism locks. See the bolted-joint example in the parent.

Level 5 — Mastery

Exercise 5.1 (L5)

A nonlinear ABAQUS step uses Newton–Raphson. At iteration $i$ the internal force is $F_{int} (u) = 3 u^{3}$ (a stiffening spring, scalar 1-DOF), and the external load is $F_{ext} = 24$ . Starting from $u^{(0)} = 1$ , perform one Newton–Raphson update and report $u^{(1)}$ . (Tangent stiffness $K_{T} = d F_{int} / d u = 9 u^{2}$ .)

Recall Solution

Why iterate at all: $3 u^{3} = 24$ is nonlinear — there's no one-shot matrix inverse. Newton–Raphson linearizes locally, takes a step, and repeats. The exact root is $u = 2$ (since $3 \cdot 2^{3} = 24$ ); watch the iteration march toward it.

Step 1 — residual (the imbalance we want to kill), external minus internal: $R^{(0)} = F_{ext} - F_{int} (u^{(0)}) = 24 - 3 (1)^{3} = 21.$ Step 2 — tangent stiffness at the guess: $K_{T} = 9 u^{2} = 9 (1)^{2} = 9.$ Step 3 — solve the correction $K_{T} Δ u = R$ : $Δ u = \frac{21}{9} \approx 2.3333.$ Step 4 — update: $u^{(1)} = u^{(0)} + Δ u = 1 + 2.3333 = 3.3333.$ What it means: one step overshoots ( $3.33$ vs true $2$ ) because the cubic curves sharply, but the next iterations will pull it back — convergence is fast (quadratic) once close.

Exercise 5.2 (L5)

Continue Exercise 5.1: perform a second Newton–Raphson iteration from $u^{(1)} = 3.3333$ and report $u^{(2)}$ . Comment on convergence toward the true root $u = 2$ .

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: we moved from $3.33 \to 2.46$ , closing on the true root $2$ . The residual sign flipped (overshoot then correct), which is normal for Newton on a stiffening curve; a real ABAQUS run keeps iterating until $∥ R ∥ < tol$ . See Model Correlation for how such solver output is then matched to Ground Test Procedures data.

Exercise 5.3 (L5)

A linear-static stress result gives a peak von-Mises stress $σ = 180$ MPa in a joint whose material allowable is $σ_{allow} = 250$ MPa. Compute the margin of safety $MoS = \frac{σ _{allow}}{σ} - 1$ and state whether the joint passes. Then re-check with a required factor of safety $F o S = 1.25$ applied to the load (i.e. design stress $= 1.25 σ$ ).

Recall Solution

Reminder — what FoS means: the factor of safety (FoS) is a multiplier you deliberately apply to the load so the structure is designed stronger than the loads it will actually see; $F o S = 1.25$ means "design for 25% more load than expected".

Why MoS is defined this way: the margin of safety (MoS) answers "how much extra capacity beyond what's used?" $MoS = 0$ means "exactly at the limit"; positive means room to spare; negative means failure.

Without FoS: $MoS = \frac{σ _{allow}}{σ} - 1 = \frac{250}{180} - 1 = 1.3889 - 1 = 0.3889 (\approx 39%) . Passes.$ With FoS = 1.25 applied to the load, the design stress becomes $1.25 \times 180 = 225$ MPa, and we compare that against the unchanged material allowable: $MoS = \frac{σ _{allow}}{1.25 σ} - 1 = \frac{250}{225} - 1 = 1.1111 - 1 = 0.1111 (\approx 11%) . Still passes (positive).$ The joint survives even with the safety factor, matching the parent's "joint is safe" conclusion. See Stress Analysis and Margins.

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

Modal solution number in NASTRAN? ::: SOL 103. Card that clamps all 6 DOF at a node? ::: SPC1 with digits 123456. Why is a fully-free stiffness matrix singular? ::: rigid-body modes — the structure can float, so no unique solution. Frequency scales with which power of stiffness? ::: the square root, $f \propto k$ . Newton–Raphson correction equation? ::: $K_{T} Δ u = R$ , with $R = F_{ext} - F_{int}$ . Margin of safety formula? ::: $MoS = σ_{allow} / σ - 1$ . What does $ϕ$ stand for in modal analysis? ::: the mode shape vector — the pattern of motion at a natural frequency.