Matrix Stiffness & Flexibility Methods Notes

Last reviewed 16 Sept 2026 · 6 min read

Why matrices

Classical methods become unwieldy for large structures. Matrix methods write the same principles — equilibrium, compatibility and member force–displacement laws — in a systematic form that a computer can solve for thousands of unknowns. Every structural analysis program (STAAD, ETABS, SAP) uses the stiffness method.

Flexibility and stiffness

For a structure with coordinates (directions of forces/displacements) $1 \dots n$ :

Flexibility coefficient $f_{ij}$ = displacement at coordinate $i$ due to a unit force at coordinate $j$ (all other forces zero).
Stiffness coefficient $k_{ij}$ = force at coordinate $i$ required to produce a unit displacement at coordinate $j$ with all other displacements held at zero.

{Δ} = [F] {P}, {P} = [K] {Δ}, [K] = [F]^{- 1}

RememberProperties

Both $[F]$ and $[K]$ are square and symmetric (Maxwell's reciprocal theorem).
Diagonal elements are positive.
$[K]$ of a stable, supported structure is positive definite (non-singular); the stiffness matrix of an unsupported element or structure is singular (rigid-body motion is possible).
The flexibility matrix exists only for a stable structure with the chosen coordinates.

Force (flexibility) method in matrix form

Choose redundants ${X}$ . With the released structure:

[F] {X} = {Δ} - {Δ_{L}} \Rightarrow {X} = [F]^{- 1} ({Δ} - {Δ_{L}})

The unknowns equal the degree of static indeterminacy. Selection of redundants is not unique and affects convenience.

Stiffness (displacement) method

Unknowns are joint displacements — their number equals the degree of kinematic indeterminacy. The procedure is automatic and does not require choosing redundants, which is why it dominates software.

Element stiffness matrices (local coordinates)

Axial (truss) element, length $L$ , area $A$ :

[k] = \frac{A E}{L} [1 - 1 - 1 1]

Beam element (coordinates: transverse displacement and rotation at each end, $v_{1}, θ_{1}, v_{2}, θ_{2}$ ):

[k] = \frac{E I}{L ^{3}} 12 6 L - 12 6 L 6 L 4 L^{2} - 6 L 2 L^{2} - 12 - 6 L 12 - 6 L 6 L 2 L^{2} - 6 L 4 L^{2}

Reading the rotation terms: $4 E I / L$ at the near end and $2 E I / L$ at the far end (carry-over ½); translation gives $6 E I / L^{2}$ moments and $12 E I / L^{3}$ shears — the same numbers as in slope deflection.

Plane frame element — 6×6: combines the axial matrix (terms $A E / L$ ) and the beam matrix.

Beam element for rotations only (translations restrained, common in hand problems):

[k] = \frac{2 E I}{L} [2112]

Transformation to global coordinates

For an inclined truss member at angle $θ$ with $c = cos θ$ , $s = sin θ$ :

[k]_{g l o ba l} = \frac{A E}{L} c^{2} cs - c^{2} - cs cs s^{2} - cs - s^{2} - c^{2} - cs c^{2} cs - cs - s^{2} cs s^{2}

In general $[k]_{g l o ba l} = [T]^{T} [k]_{l oc a l} [T]$ .

Assembly, boundary conditions and solution

Number joints and degrees of freedom; list each element's global DOF numbers.
Assemble $[K]$ by adding each element's global stiffness terms into the rows and columns of its DOFs (direct stiffness method).
Form the load vector ${P}$ : joint loads plus equivalent joint loads from member loads (negative of fixed-end forces).
Apply boundary conditions: delete rows and columns of restrained DOFs (or use a large number on the diagonal).
Solve $[K] {Δ} = {P}$ .
Member forces: ${p} = [k] {δ} + {p_{F}}$ (fixed-end forces added back).
Reactions from the restrained rows.

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Matrix Stiffness & Flexibility Methods