Finite element method — nodes, elements, stiffness matrix: Question bank

Symbol dictionary — everything used below, defined here first

Every quiz item leans on the same small cast of symbols. So nothing is used before it is earned, here they are in plain words, anchored to the picture of one straight bar element with two ends.

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

Definition The bar element and its symbols

Picture a straight rod (the element) with an end labelled node 1 (left) and node 2 (right). Along it we place an axis $x$ running from $0$ at node 1 to $L$ at node 2.

$L$ = the length of the element (metres), the distance between the two nodes.
$A$ = the cross-sectional area (m²), how thick the rod is when you slice across it.
$E$ = Young's modulus (Pa), the material's stiffness — how much stress it takes to stretch it by a given fraction. Big $E$ = hard to stretch (steel); small $E$ = easy (rubber).
$u (x)$ = the continuous displacement field: how far the material at position $x$ has moved. It is a smooth function of $x$ .
$d_{1}, d_{2}$ = the two nodal displacements — the values of $u$ only at the nodes, $d_{1} = u (0)$ and $d_{2} = u (L)$ . The vector $d = [d_{1}, d_{2}]^{⊤}$ is what FEM actually solves for; $u (x)$ is rebuilt from them by interpolation.
$ϵ$ = strain (dimensionless): the stretch per unit length, $ϵ = \frac{d u}{d x}$ .
$σ$ = stress (Pa): the internal force per unit area, tied to strain by Hooke's law $σ = E ϵ$ .

Two visual anchors we lean on repeatedly — study them once, then the quiz reads faster.

True or false — justify

FEM gives the exact analytical solution if the mesh is fine enough.

False. FEM is an approximation; refining only makes the answer converge toward the truth. Even infinite linear elements only converge — they reproduce the exact answer only when the true field is itself piecewise-linear.

A single bar element's stiffness matrix is invertible.

False.

det (K^{e}) = 0

because a free bar can slide as a rigid body (

d_{1} = d_{2}

) with zero strain and zero force — that motion is an eigenvector with eigenvalue

0

The global stiffness matrix is symmetric.

True. Each

K^{e}

is symmetric (reciprocity: the force at node

i

from a unit push at

j

equals the force at

j

from a unit push at

i

), and adding symmetric matrices during assembly keeps symmetry.

Each row of

K^{e}

sums to zero.

True. Sum row 1:

\frac{E A}{L} (1 + (- 1)) = 0

; sum row 2:

\frac{E A}{L} ((- 1) + 1) = 0

. Physically this is the statement that a rigid translation (

d_{1} = d_{2}

) produces force

\frac{E A}{L} (1 \cdot d_{1} - 1 \cdot d_{2}) = 0

More nodes always means a more accurate result.

False. Element quality dominates: a sliver triangle or a 1:1000 aspect ratio ruins accuracy even in a dense mesh, and wrong loads/materials stay wrong at any density.

Shape functions must sum to

1

at every point inside an element.

True. This partition of unity guarantees a rigid-body displacement (all node values equal) is reproduced exactly:

u = N_{1} d + N_{2} d = (N_{1} + N_{2}) d = d

. If

N_{1} + N_{2} \neq = 1

the element would invent strain where there is none — see the crossing-lines figure above.

Linear bar elements produce constant strain within each element.

True. Strain is

ϵ = d u / d x

, and the derivative of a straight line (

u

linear in

x

) is a constant, so one linear element cannot represent a varying strain.

Applying boundary conditions makes

K

larger.

It depends on the method. Elimination (deleting the fixed DOF's row and column) shrinks

K

; the penalty method instead keeps every row and adds a huge diagonal term, leaving

K

the same size (never larger). Either way rigid-body modes are removed and

K

becomes non-singular.

You can apply a load anywhere on an element, not just at a node.

False (directly). In this formulation forces live on nodes; a distributed or mid-span load must first be converted into equivalent nodal forces before it enters

F

Spot the error

"Node 2 has a

2

on the diagonal because it carries double the load."

Wrong cause. The

2

appears because node 2 is shared by two elements whose stiffnesses add during assembly — it reflects extra stiffness, not extra load (see the assembly figure).

"Since we solve

Kd = F

K

must be invertible from the start."

Error. Before boundary conditions

K

is singular; only after constraining enough DOFs (≥3 in 2D, ≥6 in 3D) is it invertible.

"I'll fix three DOFs on one straight edge to remove rigid-body motion in 2D."

Error. The three constraints must be non-colinear — three points on one line still allow the body to rotate about that line's axis / translate, leaving a rigid-body mode.

"A finer mesh everywhere is the safest way to reduce stress error near a hole."

Wasteful, not wrong-headed. The steep gradient is only at the cutout edge; refining there (adaptive/h-refinement) cuts error far more per element than blanketing the whole panel.

N_{1} = 1 - x / L

and

N_{2} = x / L

, so

N_{1}

equals

1

at node 2."

Error. At node 2,

x = L

, giving

N_{1} = 1 - L / L = 0

N_{1}

equals

1

at node 1 (

x = 0

); the writer swapped the endpoints.

"Stress comes straight out of the solver as a nodal unknown like displacement."

Error. Only displacements

d

are unknowns solved directly; stress is post-processed via strain

ϵ = d u / d x

then Hooke's law

σ = E ϵ

Why questions

Why do nodes 1 and 2 of a bar feel forces of opposite sign?

Because of Newton's third law: the element pulls back on each node exactly as each node pulls on the element, so

F_{1} = - F_{2}

for a pure axial element.

Why is FEM used instead of solving the elasticity PDE directly for a spacecraft panel?

Because the exact PDE is analytically unsolvable for real geometry (curved skins, cutouts, ribs); discretization replaces infinite DOFs with a finite set of simple elements we can compute. See 2.4.8-Numerical-methods-for-differential-equations.

Why does the middle diagonal entry act "stiffer" physically?

Two springs pulling on the same node both resist its motion, so their resistances add — moving that node requires overcoming both, hence the larger stiffness value.

Why must a distributed load be lumped onto nodes?

Because equilibrium in FEM is written only at nodes; energy-equivalent nodal forces preserve the correct work done by the true distributed load.

Why choose smaller elements where stress gradients are high?

A stress concentration (near a hole, fillet, joint) changes fast over a tiny distance; small elements resolve that curvature, while large ones average it away and underestimate the peak — relevant to 3.6.5-Buckling-of-columns-and-shells hotspots.

Why is the stiffness matrix sparse for a large structure?

Most nodes connect to only a few neighbours, so a node's DOF has zero stiffness coupling to distant nodes — those entries are

0

, letting sparse solvers skip them.

Edge cases

What happens if you never apply any boundary condition?

K

stays singular, so the structure floats freely — the solver either fails or returns infinitely many solutions (any rigid-body drift is "free").

What does a zero-length element (

L \to 0

) do to

K^{e} = \frac{E A}{L} [1 - 1 - 1 1]

Stiffness blows up (

\to \infty

), causing severe ill-conditioning — the matrix mixes huge and normal numbers and rounding destroys accuracy.

What if two adjacent elements don't share a common node?

They are mechanically disconnected — assembly never links them, so the "structure" splits into independent pieces, each with its own rigid-body freedom.

What if the mesh is fine but element aspect ratios reach 1:1000?

The matrix becomes ill-conditioned; tiny numerical rounding produces large displacement errors, so quality checks (aspect ratio

< 10

) matter more than raw count.

For a natural-frequency (modal) study, why can

K

being singular still be acceptable?

A free-floating structure has legitimate zero-frequency rigid-body modes; those correspond to

K

's null space and are expected, not an error — see 3.6.12-Modal-analysis-and-natural-frequencies.

How should thermal expansion enter this framework?

As an equivalent nodal load: the temperature-induced strain becomes a force vector added to

F

, keeping

K

unchanged — the coupling handled in 3.7.4-Thermal-analysis-of-spacecraft-components.

What if you refine the mesh and the solution keeps changing more than

2%

You have not converged — either the mesh is still too coarse for the gradients or something (element shape, BCs, material) is wrong; continue the convergence study before trusting results built on 3.6.1-Stress-and-strain-in-structural-members.

Recall Fast self-test

Diagonal "2" appears because... ::: two elements share that node and their stiffnesses add. $K$ is singular before BCs because... ::: rigid-body motion has zero strain energy. Constant strain per element comes from... ::: linear shape functions (derivative of a line is constant). Non-colinear constraints are needed to... ::: kill all rigid-body modes (≥3 DOF in 2D, ≥6 in 3D). Row of $K^{e}$ sums to... ::: zero, since $1 + (- 1) = 0$ in each row.