Finite element method — nodes, elements, stiffness matrix: Worked examples

This deep dive assumes you have read the parent note and know what a node, an element, a shape function, and the 1D bar stiffness matrix $K^{e} = \frac{E A}{L} [1 - 1 - 1 1]$ are. Here we do nothing new in theory — we simply use those tools on every kind of situation the topic can throw at you, so that no exam or real design task surprises you.

Before we start, three plain-word reminders so every symbol below is earned:

Recall What each symbol means (tap to open)

$E$ ::: Young's modulus — how hard the material resists being stretched (stiff = big $E$ ), units pascals ( $Pa = N / m^{2}$ ). $A$ ::: cross-section area of the bar — how thick it is, units $m^{2}$ . $L$ ::: length of the element, units metres. $k = E A / L$ ::: the "spring constant" of one bar — pull it with force $k$ and it stretches by 1 metre. Units $N / m$ . $d_{i}$ ::: the displacement (how far it moved) of node $i$ , units metres. $F_{i}$ ::: the force applied at node $i$ , units newtons.

The scenario matrix

Every problem this topic can pose is one of these case classes. The worked examples below are labelled with the cell(s) they cover, so together they fill the whole table.

#	Case class	What makes it distinct	Covered by
C1	Single element, tension	positive displacement, force pulls	Ex 1
C2	Single element, compression	negative displacement, sign flip	Ex 2
C3	Two elements in series, assembly	the "2 in the middle" of global $K$	Ex 3
C4	Different stiffnesses $k_{1} \neq = k_{2}$	non-symmetric assembly, weakest link	Ex 4
C5	Degenerate / singular $K$	no boundary condition → rigid body motion	Ex 5
C6	Zero / limiting input	$A \to 0$ , $L \to \infty$ , $E \to \infty$	Ex 6
C7	Real-world word problem	spacecraft strut sizing, units throughout	Ex 7
C8	Exam twist — load at a middle node, both ends fixed	force splits by stiffness ratio	Ex 8

Example 1 — Single bar in tension (C1)

Forecast: guess first — will node 2 move toward or away from node 1, and roughly how many microns?

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

Compute the spring constant $k = E A / L$ . Why this step? One bar behaves exactly like one spring; $k$ is all the geometry and material we need. $k = \frac{E A}{L} = \frac{( 70 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 4} )}{0.5} = 2.8 \times 1 0^{7} N/m$
Write $F = Kd$ and delete the fixed row. Why this step? Node 1 cannot move, so its row carries no unknown; we keep only node 2's equation. $k d_{2} = F_{2}$
Solve for the displacement. Why this step? This is the whole point — displacement is our primary unknown in FEM. $d_{2} = \frac{F _{2}}{k} = \frac{1400}{2.8 \times 1 0 ^{7}} = 5.0 \times 1 0^{- 5} m = 50 μ m$
Recover stress via strain then Hooke's law (the parent's Steps 2–3). Why this step? Displacement alone never breaks a structure — stress does. $ε = \frac{d _{2} - d _{1}}{L} = \frac{5.0 \times 1 0 ^{- 5}}{0.5} = 1 0^{- 4}, σ = E ε = 70 \times 1 0^{9} \times 1 0^{- 4} = 7 MPa$

Verify: $d_{2} > 0$ → node moves away from node 1 (outward pull → tension → elongation). ✔ Units: $N \div (N/m) = m$ . ✔ 7 MPa is far below aluminium yield (~270 MPa), so linear elastic FEM is valid. ✔ See 3.6.1-Stress-and-strain-in-structural-members.

Example 2 — Single bar in compression (C2)

Forecast: which way does node 2 move, and does anything about our equations change?

Nothing in $K$ changes — only the sign of $F_{2}$ . Why this step? Stiffness is a property of geometry/material, not of load direction. Only the right-hand side flips. $d_{2} = \frac{- 1400}{2.8 \times 1 0 ^{7}} = - 5.0 \times 1 0^{- 5} m$
Interpret the negative sign. Why this step? A minus sign in FEM is not an error — it is direction information. Negative $d_{2}$ means node 2 moves toward node 1: the bar shortens → compression.
Stress. $σ = E \frac{d _{2} - d _{1}}{L} = 70 \times 1 0^{9} \cdot \frac{- 5.0 \times 1 0 ^{- 5}}{0.5} = - 7 MPa$ The negative sign labels it as compressive.

Verify: magnitude identical to Ex 1, sign flipped — because our equation is linear ( $d_{2} \propto F_{2}$ ). ✔ A real strut in compression must also be checked against 3.6.5-Buckling-of-columns-and-shells; FEM axial stress alone does not catch buckling. ✔

Example 3 — Two bars in series, assembly (C3)

Forecast: which node moves more, and by what ratio?

Assemble each element into the global $3 \times 3$ matrix. Why this step? Node 2 is shared, so both elements deposit stiffness into position $(2, 2)$ — they add. $K_{global} = k 1 - 1 0 - 1 2 - 1 0 - 1 1$
Apply the boundary condition $d_{1} = 0$ : strike out row/column 1. Why this step? Removes the rigid-body freedom (see Ex 5) and shrinks to the solvable $2 \times 2$ . $k [2 - 1 - 1 1] [d_{2} d_{3}] = [0 P]$
Solve the $2 \times 2$ system. Row 1: $2 d_{2} - d_{3} = 0 \Rightarrow d_{3} = 2 d_{2}$ . Row 2: $- d_{2} + d_{3} = P / k$ . Substitute: $d_{2} = P / k$ . $d_{2} = \frac{P}{k}, d_{3} = \frac{2 P}{k}$

Verify: $d_{3} / d_{2} = 2$ — node 3 is two springs away from the wall, so it accumulates twice the stretch. ✔ Each bar carries the same internal force $P$ (series springs share load): bar 1 stretch $= d_{2} - 0 = P / k$ , bar 2 stretch $= d_{3} - d_{2} = P / k$ . Equal. ✔

Example 4 — Unequal stiffnesses, weakest link (C4)

Forecast: which bar stretches more, and does node 3's total motion look more like the soft bar or the stiff bar?

Assemble with the two different $k$ 's. Why this step? Position $(2, 2)$ now gets $k_{1} + k_{2}$ , not $2 k$ . $K_{global} = k_{1} - k_{1} 0 - k_{1} k_{1} + k_{2} - k_{2} 0 - k_{2} k_{2}$
Delete row/column 1 (node 1 fixed). $[k_{1} + k_{2} - k_{2} - k_{2} k_{2}] [d_{2} d_{3}] = [0 P]$
Solve. From row 1: $(k_{1} + k_{2}) d_{2} = k_{2} d_{3}$ . With $k_{1} = 3 k, k_{2} = k$ : $4 k d_{2} = k d_{3} \Rightarrow d_{3} = 4 d_{2}$ . Row 2: $- k d_{2} + k d_{3} = P \Rightarrow d_{2} (- 1 + 4) = P / k$ . $d_{2} = \frac{P}{3 k}, d_{3} = \frac{4 P}{3 k}$
Compare bar stretches. Stiff bar 1: $d_{2} - 0 = P /3 k$ . Soft bar 2: $d_{3} - d_{2} = 4 P /3 k - P /3 k = P / k$ .

Verify: each bar again carries force $P$ (series). Bar 1 stretch $= P / k_{1} = P /3 k$ ✔, bar 2 stretch $= P / k_{2} = P / k$ ✔ — the soft bar stretches 3× more. The structure's compliance is dominated by the weakest link. ✔

Example 5 — The degenerate, singular case (C5)

Forecast: guess whether a unique displacement exists.

Compute the determinant of the full $3 \times 3$ $K$ . Why this step? A unique solution needs $det \neq = 0$ . This is the litmus test for a solvable system. $det k 1 - 1 0 - 1 2 - 1 0 - 1 1 = 0$
Identify the rigid-body mode. Why this step? A zero determinant means a nonzero $d$ exists with $Kd = 0$ . Find it. $d = (1, 1, 1)^{⊤}$ gives $Kd = 0$ : every node slides the same distance → no bar stretches → zero internal force. This is the structure floating in space.
The fix. Why this step? We must remove the free-float freedom before the system is solvable. Fix at least one DOF (in 1D). In 2D fix 3 non-colinear DOFs; in 3D fix 6. That deletes the rigid-body columns and restores $det \neq = 0$ .

Verify: row sums of $K$ are $1 - 1 + 0 = 0$ , $- 1 + 2 - 1 = 0$ , $0 - 1 + 1 = 0$ — every row sums to zero, the algebraic fingerprint of a rigid-body (constant-displacement) null vector. ✔ This matches the parent's warning that unconstrained $K$ is singular.

Example 6 — Zero and limiting inputs (C6)

Forecast: which limits make the strut go floppy (huge displacement) and which make it rigid (zero displacement)?

(a) Vanishing area $A \to 0$ . Why this step? Tests a degenerate geometry — a bar with no material. $d_{2} = \frac{F _{2} L}{E A} A \to 0 + \infty$ No cross-section means no resistance: an infinitely thin strut stretches without bound. Physically, it would have snapped — a signal the model has left its valid range.
(b) Infinite length $L \to \infty$ . $d_{2} = \frac{F _{2} L}{E A} L \to \infty + \infty$ Longer springs are softer ( $k = E A / L \to 0$ ): more material in series accumulates more total stretch. Sensible.
(c) Infinite stiffness $E \to \infty$ . $d_{2} = \frac{F _{2} L}{E A} E \to \infty 0$ A perfectly rigid material does not deform at all — the FEM answer collapses to the rigid-body idealisation.

Verify: the three limits ( $\infty, \infty, 0$ ) match intuition: softness raises displacement, stiffness kills it. ✔ Note that as $k \to 0$ the element stiffness matrix tends to $0$ , which is again singular — a limiting form of Ex 5. ✔

Example 7 — Real-world word problem: sizing a spacecraft strut (C7)

Forecast: guess the area in $mm^{2}$ — a few, a few hundred, a few thousand?

Write the stiffness requirement from $d = F L / (E A)$ . Why this step? Deflection is the binding design limit here, so we invert the displacement formula for $A$ . $A \geq \frac{F L}{E d _{m a x}}$
Substitute numbers (SI throughout). Why this step? Mixed units are the #1 source of orders-of-magnitude errors — convert millimetres to metres first: $d_{m a x} = 0.2 \times 1 0^{- 3} m$ . $A \geq \frac{5000 \times 0.8}{110 \times 1 0 ^{9} \times 0.2 \times 1 0 ^{- 3}} = \frac{4000}{2.2 \times 1 0 ^{7}} = 1.818 \times 1 0^{- 4} m^{2} = 181.8 mm^{2}$
Check the resulting stress against titanium yield (~830 MPa). Why this step? Meeting stiffness does not guarantee strength — always check both. $σ = \frac{F}{A} = \frac{5000}{1.818 \times 1 0 ^{- 4}} = 2.75 \times 1 0^{7} Pa = 27.5 MPa$ Safety factor $= 830/27.5 \approx 30$ — stiffness, not strength, governs this design.

Verify: back-substitute $A = 1.818 \times 1 0^{- 4}$ : $d = \frac{5000 \times 0.8}{110 \times 1 0 ^{9} \times 1.818 \times 1 0 ^{- 4}} = 2.0 \times 1 0^{- 4} m = 0.2 mm$ — exactly the limit. ✔ Units of $A$ : $\frac{N \cdot m}{Pa \cdot m} = \frac{N \cdot m}{( N/m ^{2} ) \cdot m} = m^{2}$ . ✔ A thermal load would add to this — see 3.7.4-Thermal-analysis-of-spacecraft-components.

Example 8 — Exam twist: load at the middle node, both ends fixed (C8)

Forecast: if $k_{1} = k_{2}$ , does node 2 move at all? If bar 1 is much stiffer, which bar takes most of $P$ ?

Assemble, then apply the two boundary conditions. Why this step? With node 1 and node 3 fixed, only node 2's row survives — a single equation. Global $(2, 2)$ entry is $k_{1} + k_{2}$ ; the coupling terms to nodes 1 and 3 multiply zero displacements and vanish. $(k_{1} + k_{2}) d_{2} = P$
Solve. $d_{2} = \frac{P}{k _{1} + k _{2}}$ The two bars act like parallel springs here (both resist node 2's motion simultaneously), so their stiffnesses add.
Force in each bar. Why this step? The reaction each fixed end feels is what the exam usually asks for. Bar 1 stretches by $d_{2}$ : $F_{1} = k_{1} d_{2} = \frac{k _{1}}{k _{1} + k _{2}} P$ . Bar 2 compresses by $d_{2}$ : $F_{2} = k_{2} d_{2} = \frac{k _{2}}{k _{1} + k _{2}} P$ .

Verify: $F_{1} + F_{2} = \frac{( k _{1} + k _{2} )}{k _{1} + k _{2}} P = P$ ✔ (load fully reacted). With $k_{1} = k_{2} = k$ : $d_{2} = P /2 k$ and each bar takes $P /2$ — symmetric, as expected. ✔ The stiffer bar takes the larger share — the opposite of the series case (Ex 4), where the softer bar stretched more. Know which topology you are in!

Recall Self-test

In Ex 4, which bar stretched more and why? ::: The soft bar ( $k_{2} = k$ ), because both bars carry the same force $P$ in series and stretch $= P / k$ , so smaller $k$ → larger stretch. In Ex 8 with $k_{1} ≫ k_{2}$ , which bar carries most of $P$ ? ::: Bar 1 (the stiff one), fraction $k_{1} / (k_{1} + k_{2}) \to 1$ . Why is unconstrained $K$ singular? ::: A rigid-body mode $(1, 1, 1)$ gives zero force → non-unique solutions → $det = 0$ .

For dynamic loads and how these stiffness matrices feed natural-frequency calculations, continue to 3.6.12-Modal-analysis-and-natural-frequencies; for the underlying solver mathematics see 2.4.8-Numerical-methods-for-differential-equations.