FEM for structures — assembling global stiffness: Question bank

This is a rapid-fire misconception hunt for the global stiffness assembly step. Every item below is one line: read the prompt, cover the answer, then reveal. The answers are reasons, not verdicts — if you can't say why, you don't own the idea yet.

Before the traps, one figure and a tiny symbol dictionary so nothing below is a surprise.

Definition Symbols used on this page (defined once, here)

$ℓ$ = an element's length (so a bar's axial stiffness is $E A / ℓ$ ). We use $ℓ$ , not $L$ , precisely so length is never confused with the localization matrix below.
$L^{(e)}$ = the localization (Boolean gather) matrix of element $e$ . It has one row per element DOF; each row is all zeros except a single $1$ in the column of the matching global DOF. Multiplying it by the global displacement vector picks out that element's DOFs: $u^{(e)} = L^{(e)} u$ .
$N$ = the total number of degrees of freedom (DOFs) across the whole mesh. $K$ is $N \times N$ .
$T$ = the rotation (transformation) matrix that re-expresses a vector written in the bar's own axes into the shared global $x, y$ axes. It is built from the bar's direction cosines ( $cos$ and $sin$ of the angle the bar makes with global $x$ ). It is orthogonal, so $T^{- 1} = T^{T}$ ; that is why the transform reads $k_{global} = T^{T} k_{local} T$ .
Local bar axes: $ξ$ along the bar, $η$ perpendicular to it. A bar has stiffness along $ξ$ and zero stiffness along $η$ .

Figure — FEM for structures — assembling global stiffness

The picture above shows all three ideas at once: on the left, one element's $2 \times 2$ block being scatter-added into the overlapping slot of the big $K$ (note the shared entry that lands on top of an existing value and adds); in the middle, the $ξ$ – $η$ local axes of a tilted bar and the rotation $T$ into global $x, y$ ; on the right, why bandwidth is small — each node only touches its immediate neighbours.

True or false — justify

Assembly means placing each element matrix into an empty slot of

K

, one block per element.

False. Blocks overlap at shared DOFs; at those entries we add (superimpose) contributions. Replacing would delete a neighbouring element's stiffness.

K_{22} = 2 \frac{E A}{ℓ}

in the two-bar truss because we made an arithmetic slip.

False — it is exactly right. Node 2 belongs to both bars, so both scatter

\frac{E A}{ℓ}

into

K_{22}

; the correct action is

1 + 1 = 2

The global stiffness

K

is symmetric only if you build it in a symmetric order.

False. Symmetry comes from each

k^{(e)}

being symmetric (Maxwell–Betti reciprocity); scatter-add preserves it regardless of element ordering.

Before boundary conditions,

K

is invertible so we can just solve

Ku = F

False. Un-constrained

K

is only positive semi-definite — rigid-body modes give zero eigenvalues, so

K

is singular. You must fix DOFs first (see Boundary Conditions in FEM).

A larger element stiffness matrix always means a larger global matrix.

False. The size of

K

is set by the total DOF count

N

(all mesh DOFs), not by any single element's block size. Two elements sharing all their nodes could leave

N

unchanged.

The scatter-add algorithm gives a different

K

than the formula

\sum_{e} L^{(e) T} k^{(e)} L^{(e)}

False. They are identical — scatter-add is just the cheap way to evaluate that sum, exploiting that

L^{(e)}

is the Boolean gather matrix (one

1

per row, rest zeros).

Sparsity of

K

is a numerical accident you should not rely on.

False.

K_{ij} \neq = 0

only if DOFs

i

and

j

share an element, which is a topological fact of the mesh. It is guaranteed and exploited by Sparse Matrix Storage.

Spot the error

"Element 1 uses DOFs [1,2] and element 2 uses DOFs [3,4], so the two-bar truss has a

4 \times 4

K

Error: node 2 is shared, so its DOF appears in both elements. Correct numbering gives DOFs [1,2] and [2,3] →

K

3 \times 3

, not

4 \times 4

"A bar resists transverse pushing, so its local

4 \times 4

stiffness has no zero rows."

Error: an axial bar has zero stiffness perpendicular to its axis (the

η

direction) — transverse motion causes no strain. That is exactly why the local matrix has zero rows/columns for the

η

DOFs.

"We transform each element to global coordinates after assembling

K

Error: the rotation

k_{global} = T^{T} k_{local} T

(with

T

built from the bar's direction cosines) must happen per element, before scatter — otherwise you would add incompatible coordinate systems. See Coordinate Transformations in FEM.

"I assembled

K

, then set

u_{1} = 0

by deleting only row 1 to apply the fixed support."

Error: you must delete row 1 and column 1 (or use a penalty/partition scheme). Dropping only the row leaves

K

non-square and destroys symmetry.

K_{13} = 0

in the two-bar truss must mean I forgot a contribution."

Error: it is correct. DOFs 1 and 3 belong to different elements and never share one, so no element ever writes to

K_{13}

— a genuine structural zero.

"Since each

k^{(e)}

is singular, the assembled

K

must also be singular forever."

Error: individual element matrices are singular (rigid-body modes), but assembling many elements plus applying boundary conditions removes those modes and makes

K_{reduced}

positive definite.

"Two elements meeting at a node have different displacements there, so I average them."

Error: compatibility forces a single shared displacement at the node — that is why they share a DOF. Averaging would violate the no-gap/no-overlap condition.

Why questions

Why do we add element contributions rather than overwrite them at a shared DOF?

Because total strain energy is additive:

U = \sum_{e} U^{(e)}

, and each element contributes independently. Superposition of stiffness follows directly from the sum in the energy expression.

Why is the localization matrix

L^{(e)}

Boolean (only 0s and 1s)?

It merely selects which global DOFs feed element

e

— no scaling, just picking. Each row has a single 1 pointing at the matching global DOF, which is what makes scatter-add exact.

Why is the bandwidth of

K

roughly

O (N)

in a 2D mesh?

Bandwidth is the largest gap between the global DOF numbers of any two nodes that share an element. If you number an

n \times n

grid (

N \approx n^{2}

DOFs) row-by-row, neighbouring nodes differ by at most about one full row

\approx n = N

— so the nonzeros hug a band of width

N

. A narrow band keeps direct solvers cheap.

Why does fixing a DOF turn a singular

K

into an invertible one?

Fixing DOFs eliminates the rigid-body modes (free translations/rotations) that were the source of the zero eigenvalues, leaving

u^{T} Ku > 0

for all nonzero admissible

u

Why does the quadratic form

\frac{1}{2} u^{T} Ku

equal the strain energy, and why does that guarantee

K

is symmetric?

Because energy was built as that quadratic form; and any quadratic form only "sees" the symmetric part of a matrix, so the physically meaningful

K

is symmetric.

Why transform element stiffness with

T

into global coordinates before assembly rather than transforming

K

Each element has its own local

ξ, η

axes; only after rotating each into the common global

x, y

frame do their shared-DOF entries live in the same coordinate directions and become addable.

Why does refining the mesh increase

N

but keep

K

sparse?

More elements/nodes means more DOFs, but each new node still only connects to its few immediate neighbours — connectivity stays local, so nonzeros per row stay bounded. See Mesh Refinement and Convergence.

Edge cases

What happens to

K

if a node is left dangling (belongs to no element)?

Its rows and columns are entirely zero, so

K

is singular even after fixing supports — the free node has no stiffness and produces a spurious rigid-body mode.

Two identical elements are stacked on the same two nodes (duplicated element). What does assembly give?

Their contributions add, doubling the shared block — physically two parallel springs, so the combined stiffness is twice a single element's. Correct, but a warning sign if unintended.

A bar element has zero length (

ℓ \to 0

). What breaks in assembly?

The

E A / ℓ

factor blows up to infinity, injecting an ill-conditioned near-rigid link into

K

and wrecking the solve — degenerate elements must be caught during meshing.

If a bar is perfectly transverse to a load (aligned so its axis

ξ

90°

from the force), how does it enter

K

After the

T

transform its stiffness lands entirely in one global direction; in the perpendicular direction it contributes zero, so it cannot carry that load component at all.

What if two elements share a node but you accidentally give them different global DOF numbers there?

You break compatibility — the shared point splits into two independent points.

K

grows, the

K_{22} = 2

-style summation never happens, and the structure behaves as if disconnected at that node.

What does an all-zero row in the assembled, un-reduced

K

tell you before you even solve?

That DOF is unrestrained by any element — either a genuine free/rigid mode or a modelling error (orphan node). Solvers will report singularity until it is constrained or connected.

Recall One-line survival summary

Assembly = transform each $k^{(e)}$ to global with $T$ , then scatter-add its entries into $K$ at the shared global DOFs picked out by $L^{(e)}$ ; symmetry and sparsity come from physics and topology; singularity is cured only by boundary conditions.

Back to the parent: FEM for structures — assembling global stiffness · overview in Finite Element Method Overview · element blocks in Element Stiffness Matrices.