Visual walkthrough — FEM software — NASTRAN, ABAQUS (concepts and use)

Everything below uses only these plain ideas: a force is a push (measured in newtons, N), a displacement is how far something moves (in metres, m), and stiffness is "how hard you must push to move it one metre" (N/m). That is the entire vocabulary we start with.

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Step 1 — One spring: force, stretch, stiffness

WHAT. Take the simplest structure imaginable: a single spring, fixed at the left wall, with a hand pulling the right end.

WHY. Every finite element model is, at its heart, a giant collection of springs. If we understand one spring completely, the rest is bookkeeping. We start here because you cannot understand a matrix of stiffnesses before you understand one number of stiffness.

PICTURE.

The spring obeys Hooke's law:

$F=ku$

• $F$ — the force you apply at the free end, in newtons (N). It is the cause.
• $u$ — the displacement, how far the end moves, in metres (m). It is the effect.
• $k$ — the stiffness, in N/m. It is the exchange rate between them: "push $k$ newtons to get one metre."

Rearranged, $k=F/u$ — stiffness is slope. Keep that mental image: stiffness is the steepness of the force–stretch line.

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Step 2 — Two springs: why one number becomes a matrix

WHAT. Now connect two springs in a line: wall → node 1 → node 2 → node 3. Each spring can push or pull the nodes it touches.

WHY. A real bracket has many connection points ("nodes"). Moving one node stretches the springs on both sides of it, so one node's motion creates forces at its neighbours too. A single number $k$ can no longer describe this cross-talk. We need a table — a matrix.

PICTURE.

Let $u_1,u_2,u_3$ be the displacements of the three nodes, and $F_1,F_2,F_3$ the forces applied there. Spring A (stiffness $k_A$) links nodes 1–2; spring B (stiffness $k_B$) links nodes 2–3.

Write the force balance at each node. Spring A pulls node 2 back by how much node 2 outran node 1, i.e. by $u_2-u_1$:

$$F_2=\underset{\text{spring A’s pull on node 2}}{\underbrace{k_A(u_2-u_1)}}+\underset{\text{spring B’s pull on node 2}}{\underbrace{k_B(u_2-u_3)}}$$

• $u_2-u_1$ — the stretch of spring A (how much longer it got). If node 2 moves and node 1 does not, spring A stretches by exactly $u_2$'s extra.
• $k_A(u_2-u_1)$ — force = stiffness × stretch, straight from Step 1.
• The two terms add because node 2 feels both springs at once.

Doing this at every node and collecting the coefficients of $u_1,u_2,u_3$ gives a table of stiffnesses:

$$\begin{array}{rl}{F}_1& =k_A{u}_1-k_A{u}_2\\ F_2& =-k_A{u}_1+(k_A+k_B)u_2-k_B{u}_3\\ F_3& =-k_B{u}_2+k_B{u}_3\end{array}$$

That table is the matrix. Read on.

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Step 3 — Packing the table: $\mathbf{Ku}=\mathbf{F}$ is born

WHAT. Stack the three equations from Step 2 into one block of numbers times a column of displacements.

WHY. A matrix equation is just a compact way to say all three balance equations at once. Nothing new physically — but now a computer can solve thousands of nodes with one command instead of chasing each spring by hand.

PICTURE.

$$\underset{\mathbf{K}}{\underbrace{\left[\begin{array}{ccc}{k}_A& -k_A& 0\\ -k_A& k_A+k_B& -k_B\\ 0& -k_B& k_B\end{array}\right]}}\underset{\mathbf{u}}{\underbrace{\left[\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right]}}=\underset{\mathbf{F}}{\underbrace{\left[\begin{array}{c}{F}_1\\ F_2\\ F_3\end{array}\right]}}$$

• $\mathbf{K}$ — the global stiffness matrix. Bold capital = a whole table of numbers. Each entry $K_{ij}$ answers "what force appears at node $i$ when node $j$ moves one metre and all others are held?"
• $\mathbf{u}$ — the displacement vector: the list of all node motions, our unknowns.
• $\mathbf{F}$ — the force vector: the list of applied loads, known.

This is exactly the $\mathbf{Ku}=\mathbf{F}$ of the parent note — for a real bracket, $\mathbf{K}$ is a million-by-million version of this same table.

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Step 4 — Where the springs come from: an element in a triangle

WHAT. A bracket is not made of literal springs. So how does a solid triangle of aluminium turn into stiffness numbers? We watch one triangular element deform.

WHY. The parent note's scary integral ${\mathbf{k}}_e=\int {\mathbf{B}}^T\mathbf{DB}dV$ is nothing more than "compute the effective spring stiffness of a chunk of material." We need to see that a solid patch resists stretching just like a spring, so its stiffness slots into $\mathbf{K}$ the same way.

PICTURE.

Pick a triangle with three corner nodes. Inside it we make one honest assumption: displacement varies linearly across the triangle. That lets us write the displacement of any interior point from the three corner values:

$\mathbf{u}(x,y)=\mathbf{N}{\mathbf{u}}_e$

• ${\mathbf{u}}_e$ — the three corner displacements of this element (a short list).
• $\mathbf{N}$ — the shape functions: recipes that blend the corner values to fill the interior. Picture a tent stretched over three poles — $\mathbf{N}$ is the tent surface, the poles are the corners.

From displacement we get strain (fractional stretch) by measuring how fast displacement changes across the triangle:

$\mathbf{ϵ}=\mathbf{B}{\mathbf{u}}_e$

• $\mathbf{ϵ}$ — strain: stretch per unit length, unitless. Our "$u_2-u_1$" from Step 2, but spread over an area.
• $\mathbf{B}$ — the strain–displacement table: it turns corner motions into stretch. It contains the slopes of the shape functions — literally how steeply the tent tilts.

Then stress (internal force per area) via Hooke's law again:

$\mathbf{\sigma }=\mathbf{D}\mathbf{ϵ}$

• $\mathbf{\sigma }$ — stress, force per area (Pa). The internal push the material sets up when stretched.
• $\mathbf{D}$ — the material stiffness: this element's version of $k$, built from Young's modulus $E$ (how stiff the metal is). Stiff metal → big $\mathbf{D}$.

Chain them and integrate the resistance over the triangle's volume:

${\mathbf{k}}_e={\int }_V{\mathbf{B}}^T\mathbf{D}\mathbf{B}dV$

Read it as stretch-recipe × material-stiffness × stretch-recipe, summed over the whole chunk = "the spring constant of this triangle." That is the ${\mathbf{k}}_e$ NASTRAN computes millions of times.

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Step 5 — Assembly: gluing element springs into the global table

WHAT. Each element produced its own tiny stiffness table ${\mathbf{k}}_e$. We now overlap them at shared nodes, exactly like Step 2 added two springs at node 2.

WHY. Neighbouring triangles share corners. A shared node feels every triangle touching it, so their stiffnesses must be added together at that node's row and column. This overlapping-and-adding is called assembly, and it is the whole reason $\mathbf{K}$ has that banded overlap pattern.

PICTURE.

$\mathbf{K}={\sum }_e{\mathbf{k}}_e,\mathbf{F}={\sum }_e{\mathbf{f}}_e$

• ${\sum }_e$ — sum over all elements $e$ in the mesh.
• Each ${\mathbf{k}}_e$ drops its numbers into the global rows/columns of its own nodes; where two elements share a node, their numbers land on the same slot and add — precisely the $k_A+k_B$ we saw in Step 3's middle diagonal.

Same for the loads: pushes on a shared node add up into one $\mathbf{F}$ entry. After assembly we again hold $\mathbf{Ku}=\mathbf{F}$ — now for the real structure.

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Step 6 — The degenerate case: why you MUST fix something first

WHAT. Try to solve $\mathbf{Ku}=\mathbf{F}$ for a bracket floating free in space, before bolting it down.

WHY. If nothing is held fixed, the whole structure can drift sideways with zero internal stretch — every spring keeps its length, so any rigid slide costs no force. The matrix cannot tell a floating bracket where "zero" is. This is the single most common beginner crash, and it has a clean geometric cause.

PICTURE.

Mathematically, a rigid drift ${\mathbf{u}}_{\text{rigid}}$ (all nodes move the same amount) produces no internal force:

$\mathbf{K}{\mathbf{u}}_{\text{rigid}}=0$

• The right side is the zero vector: no forces generated. A non-zero displacement giving zero force means $\mathbf{K}$ is singular — it has no unique inverse, so the solve fails.
• Geometrically: sliding the wall-and-spring picture bodily to the right stretches nothing.

The cure is a boundary condition — the parent note's SPC1 card. Bolt at least enough nodes to kill all rigid drifts (in 3D: 6 of them — 3 slides, 3 spins). This is the SPC = 10 line in the NASTRAN deck. Once pinned, $\mathbf{K}$ becomes invertible and $\mathbf{u}={\mathbf{K}}^{-1}\mathbf{F}$ has one answer.

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Step 7 — When the spring line bends: the nonlinear case

WHAT. So far one $\mathbf{K}$ solved everything, because our force–stretch line was straight. For big deflections, contact, or plasticity, that line curves — $\mathbf{K}$ is no longer a fixed number.

WHY. A carbon panel that buckles, or a bolt that begins to slip, changes its own stiffness as it deforms. The slope of the force–stretch curve is now different at every point. One straight-line solve can't chase a curved line, so ABAQUS takes small tangent steps — the Newton–Raphson idea.

PICTURE.

We hunt for the displacement where internal and external forces balance:

$\mathbf{R}(\mathbf{u})={\mathbf{F}}_{\text{ext}}-{\mathbf{F}}_{\text{int}}(\mathbf{u})=0$

• $\mathbf{R}$ — the residual, the leftover imbalance. When it hits zero, we are in equilibrium.
• ${\mathbf{F}}_{\text{int}}(\mathbf{u})$ — internal resisting force, now a curved function of $\mathbf{u}$.

Use the local slope (the tangent) to guess the correction:

${\mathbf{K}}_T\Delta \mathbf{u}=\mathbf{R},{\mathbf{K}}_T=\frac{\partial {\mathbf{F}}_{\text{int}}}{\partial \mathbf{u}}$

• ${\mathbf{K}}_T$ — the tangent stiffness: the slope of the curved line at your current guess. This is exactly the Step 1 idea "$k$ = slope," but measured fresh at each iteration.
• $\Delta \mathbf{u}$ — the step that slides your guess toward the balance point.

Repeat: aim along the tangent, land closer, re-measure the slope, aim again — like walking to the bottom of a valley by always stepping downhill. See Composite Materials and Stress Analysis and Margins for what these curved regions mean physically.

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The one-picture summary

One spring gives $F=ku$. Many springs sharing nodes force us to stack the balances into $\mathbf{Ku}=\mathbf{F}$. A solid triangle earns its own spring constant ${\mathbf{k}}_e=\int {\mathbf{B}}^T\mathbf{DB}dV$; assembly adds them at shared corners into the global $\mathbf{K}$. Pin the rigid drifts or the solve is singular. Straight line ⇒ solve once (NASTRAN); curved line ⇒ tangent-step to equilibrium (ABAQUS). Modal analysis is the same $\mathbf{K}$ asking a different question — see Vibration and Modal Analysis and Launch Vehicle Loads.

Recall Feynman retelling — say it like a story

Imagine one spring on a wall. Pull it: the force you feel is stiffness times stretch. Now chain springs — every node you push tugs its neighbours, so instead of one stiffness number you need a whole table telling you who tugs whom. Line the equations up and that table becomes $\mathbf{K}$, your displacements become $\mathbf{u}$, your pushes become $\mathbf{F}$: $\mathbf{Ku}=\mathbf{F}$. Real metal isn't springs, but a little triangle of it resists stretching in exactly the same way — we compute its own stiffness by adding up its resistance over its volume, and that's the ${\mathbf{k}}_e$ integral. Glue all the triangles together, adding stiffness wherever they share a corner, and you rebuild the big table. One warning: if you never bolt the thing down, it can drift with no resistance, the table has no unique answer, and the solver quits — so always pin it. If the material behaves like a straight line, solve once and you're done (that's NASTRAN's happy place). If it curves — buckling, plastic, contact — you can't leap to the answer; you feel the slope where you stand, step a little, feel again, and creep to balance (that's what ABAQUS does). Same equation, one straight, one curved.

Recall

What does each entry $K_{ij}$ of the stiffness matrix mean? ::: The force appearing at node $i$ when node $j$ is moved one unit and all other nodes are held fixed. Why does a free-floating (unconstrained) structure give a singular $\mathbf{K}$? ::: A rigid drift moves all nodes equally, stretching no spring, so $\mathbf{K}{\mathbf{u}}_{\text{rigid}}=0$ for non-zero $\mathbf{u}$ — no unique inverse exists. What is the tangent stiffness ${\mathbf{K}}_T$ physically? ::: The local slope of the curved force–displacement relation at the current guess, used to take one Newton–Raphson step. Where do the off-diagonal entries of $\mathbf{K}$ come from? ::: From springs (elements) that connect two nodes, coupling their motions; zeros appear between nodes that share no element.