Finite element method — nodes, elements, stiffness matrix: Visual walkthrough

Everything here assumes only that you know what a length, a force, and a ratio are. Each new idea — stretch, strain, stiffness — is built on the picture before it. If a stretched rubber band and a spring make sense to you, you are ready.

Step 1 — Draw the rod and name its two ends

WHAT. We take one straight rod. We mark its two ends with dots and call them node 1 (left) and node 2 (right). A node is just a tracked point — a place where we agree to measure how far the material has moved.

WHY. We cannot follow every atom in the rod; that is infinitely many numbers. The whole trick of the finite element method is: track only the ends, then fill in the middle with a rule. So the very first act is to pick those two ends.

PICTURE. The rod lies flat. Its natural (unloaded) length is $L$ . The left dot is node 1, the right dot is node 2.

Figure — Finite element method — nodes, elements, stiffness matrix

Three symbols now have meaning, anchored to the picture:

$L$ — the rod's rest length (the distance between the dots before anything happens).
$A$ — the rod's cross-section area (how thick it is, seen end-on).
$E$ — Young's modulus, a number for how stiff the material is (steel is big, rubber is small). We meet it properly in Step 4; see 3.6.1-Stress-and-strain-in-structural-members.

Step 2 — Describe every inside point with the two end values

WHAT. Pick any point inside the rod, a distance $x$ from node 1. Its displacement $u (x)$ is a weighted blend of the two end displacements: $u (x) = N_{1} (x) (1 - \frac{x}{L}) d_{1} + N_{2} (x) \frac{x}{L} d_{2}$

Reading it term by term:

$d_{1}$ — the left end's movement, the raw quantity we are blending.
$(1 - \frac{x}{L})$ — how much of node 1's movement reaches point $x$ . Full ( $= 1$ ) at the left end, fading to $0$ at the right.
$d_{2}$ — the right end's movement.
$\frac{x}{L}$ — how much of node 2's movement reaches $x$ . Zero at the left, full at the right.

These two weights $N_{1}, N_{2}$ are the shape functions.

WHY. We need some rule to guess the middle from only the ends. The simplest honest guess is a straight line between the two end movements. A straight line is exactly the linear blend above — and it has the one property we insist on: at each node, the blend gives back that node's own value ( $N_{1} = 1$ at node 1, and $= 0$ at node 2). Anything fancier is unnecessary for one small rod.

PICTURE. The two shape functions are straight ramps crossing like an X: $N_{1}$ slides down from $1$ to $0$ , $N_{2}$ climbs from $0$ to $1$ . At every $x$ they add to $1$ (nothing is lost) — that is called the partition of unity.

Recall Why must

$N_{1} + N_{2} = 1$ everywhere? If the whole rod slides right by the same amount $c$ (so $d_{1} = d_{2} = c$ ), no point should move differently: $u (x) = N_{1} c + N_{2} c = (N_{1} + N_{2}) c$ . For this to equal $c$ at every $x$ , we need $N_{1} + N_{2} = 1$ . ::: The weights must sum to one so that pure sliding produces pure sliding — no fake stretching.

Step 3 — Measure the stretch: strain

WHAT. Stretch is not "how far did it move" but "how much did it move per length". That per-length stretch is the strain, written $ϵ$ . We get it by asking how fast $u$ changes as we walk along $x$ — the slope of the $u$ -versus- $x$ line: $ϵ = \frac{d u}{d x} = - \frac{1}{L} d_{1} + \frac{1}{L} d_{2} = \frac{d _{2} - d _{1}}{L}$

Term by term:

$\frac{d u}{d x}$ — the derivative: the slope of the displacement line, i.e. change in movement divided by change in position.
$- \frac{1}{L} d_{1}$ — node 1 moving right un-stretches the rod (pushes the left end toward the right end), hence the minus.
$+ \frac{1}{L} d_{2}$ — node 2 moving right stretches the rod.
$\frac{d _{2} - d _{1}}{L}$ — the gap change $(d_{2} - d_{1})$ spread over the rest length $L$ .

WHY the derivative and not something else? Strain answers "how much longer, as a fraction of original length?" That is a rate of change of position along the rod — and the tool that measures rate of change of one quantity against another is the derivative. No other operation gives you "per unit length."

Notice: because $u$ is a straight line (Step 2), its slope is the same everywhere. So this element has one constant strain throughout — fine, as long as the element is small.

PICTURE. The straight $u (x)$ line has a single slope; that slope is $ϵ$ . Stretch $=$ (new length $-$ old length) $\div$ old length.

Step 4 — Turn stretch into internal push/pull: Hooke's law

WHAT. A stretched rod pulls back. The internal pull-per-area is the stress $σ$ , tied to strain by Hooke's law: $σ = E ϵ = \frac{E}{L} (d_{2} - d_{1})$

$σ$ — internal force spread over the cross-section (force ÷ area).
$E$ — Young's modulus: the material's stiffness constant. Bigger $E$ ⇒ more stress for the same stretch.
$ϵ$ — the strain from Step 3.

Then the total internal force is stress times how much area is pulling: $F_{internal} = σ A = \frac{E A}{L} (d_{2} - d_{1})$

$A$ — the cross-section area; a thicker rod carries more total force at the same stress.
The bundle $\frac{E A}{L}$ is the rod's spring constant — stiffness $=$ (material stiffness $\times$ thickness) ÷ length.

WHY. For small stretches, real materials behave like a linear spring: double the stretch, double the pull. Hooke's law is the mathematical statement of that straight-line behaviour, and it is what lets a whole continuous rod be replaced by a single number $E A / L$ — a plain spring. Compare with 3.6.1-Stress-and-strain-in-structural-members.

PICTURE. The rod is redrawn as a coil spring of stiffness $k = E A / L$ . Longer rod = softer spring; thicker or stiffer material = stronger spring.

Step 5 — Balance forces at each end (Newton's third law)

WHAT. The element pulls on its nodes; by reaction the nodes feel forces $F_{1}, F_{2}$ . Write what each end feels: $F_{1} = - F_{internal} = \frac{E A}{L} (d_{1} - d_{2}), F_{2} = + F_{internal} = \frac{E A}{L} (d_{2} - d_{1})$

$F_{2} = + k (d_{2} - d_{1})$ — if node 2 is ahead of node 1, the rod is stretched and pulls node 2 back left... wait: $F_{2}$ here is the force the node must supply to hold that stretched state, so it points outward. Sign convention: a stretched rod needs outward forces at both ends to keep it stretched.
$F_{1} = - F_{2}$ — equal and opposite. Whatever we push on one end, the other end feels the mirror image.

WHY the opposite signs? Newton's third law. The rod cannot invent force; every pull on node 1 is matched by an equal pull on node 2 in the opposite direction. That is why $F_{1} = - F_{2}$ always, and it is the deep reason the matrix rows will each sum to zero.

PICTURE. Two arrows on the ends of the spring: for a stretched rod they point away from each other; for a squashed rod they point toward each other. Same size, opposite directions.

Step 6 — Pack the two equations into one matrix

WHAT. Expand the two force equations: $F_{1} = \frac{E A}{L} d_{1} - \frac{E A}{L} d_{2}, F_{2} = - \frac{E A}{L} d_{1} + \frac{E A}{L} d_{2}$ Stack them and factor out $\frac{E A}{L}$ : $[F_{1} F_{2}] = \frac{E A}{L} [1 - 1 - 1 1] [d_{1} d_{2}]$

Reading the matrix as a grid of influences, $K_{ij}$ = force at node $i$ caused by a unit movement of node $j$ :

top-left $1$ : move node 1 right by one unit ⇒ force $+ k$ at node 1 (it resists).
top-right $- 1$ : move node 2 right by one unit ⇒ force $- k$ at node 1 (it gets pulled along).
bottom row: the mirror, by Newton's third law.

WHY a matrix? Because two forces depend on two displacements, and every force depends on both displacements. A matrix is exactly the bookkeeping device for "each output built from every input." The clean split into a constant number $\frac{E A}{L}$ times a pure pattern $[1 - 1 - 1 1]$ shows the geometry of connection (the pattern) separated from the material size (the number).

PICTURE. A 2×2 grid whose cells are coloured by sign — orange for $+$ , teal for $-$ — with an arrow from each column (a displacement) to each row (the force it produces).

Step 7 — The degenerate case: why $K^{e}$ cannot be inverted

WHAT. Slide the whole rod right by the same amount: $d_{1} = d_{2} = c$ . Plug in: $K^{e} [c c] = \frac{E A}{L} [c - c - c + c] = [00]$ Zero force from a nonzero motion. The determinant is $det K^{e} = (\frac{E A}{L})^{2} (1 \cdot 1 - (- 1) (- 1)) = 0.$

WHY this matters. A rigid slide stretches nothing, so it stores no energy and needs no force — this is a rigid body mode. Mathematically it means many displacement states give the same (zero) force, so we cannot undo $K^{e}$ to recover a unique $d$ . The cure, applied in the parent note's assembly, is a boundary condition: pin at least one node so sliding is forbidden. Only then does the reduced matrix become invertible.

PICTURE. The rod floating in space with both arrows equal — it drifts, no spring stretch, no restoring force. A ghost copy shows it pinned at node 1, now well-behaved.

The one-picture summary

WHAT. One diagram walks the whole chain: two dots → a straight blend → a slope (strain) → a spring pull (stress×area) → equal-and-opposite end forces → the 2×2 matrix, with the zero-determinant rigid slide hanging off the side.

This same recipe, with fancier shape functions and more DOFs, produces the beam, shell, and solid matrices used in 3.6.12-Modal-analysis-and-natural-frequencies, 3.6.5-Buckling-of-columns-and-shells, and 3.7.4-Thermal-analysis-of-spacecraft-components.

Recall Feynman retelling — say it back in plain words

We had a rod and we refused to track every point, so we tracked only its two ends and their movements $d_{1}, d_{2}$ . To fill in the middle we drew a straight line between the two end movements — that's the shape-function blend. The slope of that line tells us how much the rod stretched per unit length; that's the strain. Multiply the strain by the material stiffness $E$ to get the internal stress, and by the area $A$ to get the total pull; that's a plain spring with stiffness $k = E A / L$ . A stretched spring pushes its two ends outward with equal and opposite force, which is just Newton's third law. Writing "force at each end in terms of movement at each end" needs a little 2×2 grid — and out pops $\frac{E A}{L} [1 - 1 - 1 1]$ . The $+ 1$ 's are a node resisting its own motion; the $- 1$ 's are a node being dragged by its neighbour. Finally, if you slide the whole rod without stretching it, you feel no force at all — so the matrix has determinant zero and can't be inverted until you pin the rod down.

Prerequisites revisited: 3.6.1-Stress-and-strain-in-structural-members · 2.4.8-Numerical-methods-for-differential-equations Where this leads: 3.6.12-Modal-analysis-and-natural-frequencies · 3.6.5-Buckling-of-columns-and-shells · 3.7.4-Thermal-analysis-of-spacecraft-components