Module 10 · Lesson 10.1
Assembling the matrices
The two-storey pattern, extended — and the reason every entry of a shear-building stiffness matrix can still be checked by inspection at fifty storeys.
Why this matters
Module 9 built a two-by-two system by hand. Nothing conceptual changes at ten storeys or a hundred; the same pattern repeats and the same entries mean the same things.
What does change is that you can no longer solve it by hand, so you have to be able to check it instead. This lesson is about what a correctly assembled matrix looks like, so that an incorrect one announces itself.
By the end of this lesson you should be able to
- Assemble the stiffness matrix of an n-storey shear building
- State the properties every correct M and K must have
- Explain why the matrix is tridiagonal and when it would not be
- Use rigid-body and symmetry checks on an assembled matrix
What you should already know
- Two-degree-of-freedom matrices (Module 9)
- The shear-building idealisation (Module 2)
The pattern extends
For an n-storey shear building with storey stiffnesses k₁ … kₙ (k₁ at the bottom):
K[i][i] = kᵢ + kᵢ₊₁ (the last floor has no storey above, so K[n][n] = kₙ) K[i][i+1] = K[i+1][i] = −kᵢ₊₁ K[i][j] = 0 for |i − j| > 1
And the mass matrix is simply diag(m₁ … mₙ).
Why tridiagonal
Because floor 1 has no direct connection to floor 3. Moving floor 3 deforms only the storeys immediately above and below it, so it can only exert a force on floors 2 and 4.
That is a statement about the STRUCTURE, not about the method. Introduce an outrigger connecting floor 3 to floor 10, or a braced bay spanning several storeys, and the matrix stops being tridiagonal — entries appear wherever a direct load path exists.
The sparsity pattern of a stiffness matrix is a map of what is connected to what.
Four properties every correct M and K must have
Symmetry. K[i][j] = K[j][i], by Betti's reciprocal theorem. An unsymmetric stiffness matrix has an assembly error in it, without exception.
Positive definiteness. For a properly restrained structure, both M and K are positive definite. Practically: xᵀKx > 0 for any non-zero x, meaning every possible deformation stores strain energy. A K that fails this has a mechanism.
Positive diagonals. K[i][i] > 0 always. A zero diagonal means a coordinate the structure does not resist at all — usually an unconnected node.
Correct total mass. The sum of a lumped mass matrix's diagonal is the total mass. For a consistent mass matrix, ιᵀMι over the translational coordinates gives the total. Either way, it is checkable against a hand estimate, and it is the single most valuable check available.
The rigid-body test
For an UNRESTRAINED structure — an aircraft, a satellite, a bridge deck on isolators before they are engaged — a uniform translation stores no strain energy, so Kι = 0 exactly.
For a grounded building it does not, because the lowest storey deforms. Kι then returns k₁ at the ground-connected floor and zero everywhere else, which is itself a useful check: any OTHER non-zero entry means a storey is not connected to the one below it as intended.
Try it
Assembling the matrices, entry by entry
Every number in these matrices has a physical meaning. Change a storey and watch which entries move.
Uncheck to taper mass and stiffness with height.
Reduces that storey's stiffness to 40%.
Mass matrix M(tonne)
| 400 | 0 | 0 |
| 0 | 400 | 0 |
| 0 | 0 | 400 |
Stiffness matrix K(MN/m)
| 1200 | -600 | 0 |
| -600 | 1200 | -600 |
| 0 | -600 | 600 |
Reading these matrices
- M is diagonal because the mass is lumped at the floors: accelerating floor 2 produces no inertia force on floor 3. That is an assumption, not a fact, and it is the shear-building assumption.
- K[i][i] is the sum of the storey stiffnesses above and below floor i, because moving that floor deforms both. Here K[0][0] = 600 + 600 = 1200 MN/m.
- K[i][i+1] is minus the storey stiffness between them. The minus sign says that pushing floor i towards floor i+1 pulls floor i+1 along.
- Every entry more than one floor away is zero, because floor 1 has no direct connection to floor 3 — only through floor 2. That is why the matrix is tridiagonal, and why a real frame with outriggers or bracing is not.
- K is symmetric. It must be: Betti's theorem guarantees it, and a stiffness matrix that is not symmetric has an assembly error in it.
| Mode | T (s) | f (Hz) | Γ | Effective mass | Cumulative |
|---|---|---|---|---|---|
| 1 | 0.365 | 2.74 | 1047.33 | 91.4% | 91.4% |
| 2 | 0.130 | 7.69 | 299.75 | 7.5% | 98.9% |
| 3 | 0.090 | 11.11 | -115.12 | 1.1% | 100.0% |
Total mass 1200 tonnes, all of it accounted for across 3 modes. The first mode alone carries 91% of it.
What this shows: The stiffness matrix entry K[i][j] is the force needed at i to hold every other floor still while floor j moves one unit. That definition, not a formula, is what makes the tridiagonal pattern obvious.
Worked example
Assembling and checking a four-storey building
Given
- Four storeys, masses 400, 400, 380 and 300 tonnes from the ground up
- Storey stiffnesses 600, 600, 520 and 440 MN/m
Find
The matrices, and every check that can be applied to them.
Assumptions
- Shear building: rigid floors, axially rigid columns, mass at floor level
Predict first
A shear-building stiffness matrix is assembled and the third interior row sums to −80 MN/m instead of zero. What has most likely happened?
Practice
A five-storey shear building has all storey stiffnesses equal to 400 MN/m. What is K[2][2] — the second floor's diagonal entry — in MN/m?
Practice
For that same building, what is K[4][4] — the top floor's diagonal entry — in MN/m?
Worked example
Assembling a three-storey stiffness matrix by hand
Given
- A three-storey shear building
- Each storey has a lateral stiffness of 30 MN/m
Find
The stiffness matrix, and what its pattern means
Summary
- The two-storey pattern repeats unchanged at any number of storeys
- K is tridiagonal because each floor connects only to its immediate neighbours
- The sparsity pattern is a map of what is connected to what
- K must be symmetric, with positive diagonals and positive definite
- Interior rows sum to zero; the ground-connected row sums to k₁
- The top diagonal is kₙ, not kₙ + kₙ₊₁ — there is no storey above the roof
- Total mass against a hand estimate is the single most valuable check
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint