Chapter 2 of 11

Stiffness & flexibility methods

In the TNPSC AE Civil syllabus under Structural Analysis · 2 parts

Matrix Stiffness & Flexibility Methods · Force (Flexibility) Method — Consistent Deformation & Three-Moment Equation

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📑 Contents (13 sections)

Part 1 of 2

Matrix Stiffness & Flexibility Methods

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.

Part 2 of 2

Force (Flexibility) Method — Consistent Deformation & Three-Moment Equation

Last reviewed 16 Sept 2026 · 6 min read

Idea of the force method

A statically indeterminate structure has more unknown forces than equilibrium equations. The force method (method of consistent deformation, flexibility method) treats selected unknown forces — the redundants — as the primary unknowns.

Choose redundants equal in number to the degree of static indeterminacy $D_{s}$ .
Remove them to obtain a stable, determinate released (primary) structure.
Find the displacements in the released structure at the redundants' locations due to the actual loads.
Find the displacements there due to unit values of each redundant (flexibility coefficients).
Write compatibility equations: total displacement at each redundant = its actual value (usually zero, or a known settlement).
Solve for the redundants; then complete the analysis by equilibrium.

FormulaCompatibility equations

For $n$ redundants $X_{1} \dots X_{n}$ :

Δ_{i L} + j \sum f_{ij} X_{j} = Δ_{i} (i = 1 \dots n)

$Δ_{i L}$ = displacement at $i$ in the released structure due to loads; $f_{ij}$ = displacement at $i$ due to a unit value of $X_{j}$ (flexibility coefficient); $Δ_{i}$ = actual displacement at $i$ (0 for a rigid support). By Maxwell's theorem $f_{ij} = f_{j i}$ .

The force method is attractive when $D_{s}$ is small (one or two redundants).

Propped cantilever

Fixed at A, propped at B, span $L$ . Take the prop reaction $R_{B}$ as the redundant; the released structure is a cantilever.

UDL $w$ : deflection at B of the cantilever $= w L^{4} /8 E I$ (down); due to $R_{B}$ $= R_{B} L^{3} /3 E I$ (up).

\frac{R _{B} L ^{3}}{3 E I} = \frac{w L ^{4}}{8 E I} \Rightarrow R_{B} = \frac{3 w L}{8}, R_{A} = \frac{5 w L}{8}, M_{A} = \frac{w L ^{2}}{8} (hogging)

Maximum sagging moment $= \frac{9 w L ^{2}}{128}$ at $3 L /8$ from the prop; point of contraflexure at $L /4$ from the fixed end.

Central point load $W$ : $R_{B} = \frac{5 W}{16}$ , $R_{A} = \frac{11 W}{16}$ , $M_{A} = \frac{3 W L}{16}$ , moment under load $= \frac{5 W L}{32}$ .

Fixed beams and fixed-end moments

A beam with both ends fixed has two redundant moments (for vertical loads). Using compatibility (zero slope and deflection at the ends) — or moment-area on the "free BMD" and "fixing moment" diagrams:

RememberMoment-area rules for fixed beams

Area of the free BMD = area of the fixing moment diagram.
The centroids of the two diagrams lie on the same vertical line.

Loading on a fixed beam of span $L$	Fixed-end moments ( $M_{A B}$ , $M_{B A}$ )	Mid-span moment	Max deflection
Central point load $W$	$- \frac{W L}{8}$ , $+ \frac{W L}{8}$	$+ \frac{W L}{8}$	$\frac{W L ^{3}}{192 E I}$
Point load $W$ at $a$ from A ( $b$ from B)	$- \frac{W a b ^{2}}{L ^{2}}$ , $+ \frac{W a ^{2} b}{L ^{2}}$	—	—
UDL $w$ over whole span	$- \frac{w L ^{2}}{12}$ , $+ \frac{w L ^{2}}{12}$	$+ \frac{w L ^{2}}{24}$	$\frac{w L ^{4}}{384 E I}$
Triangular load, zero at A to $w$ at B	$- \frac{w L ^{2}}{30}$ , $+ \frac{w L ^{2}}{20}$	—	—
Sinking of B by $δ$ relative to A	$- \frac{6 E I δ}{L ^{2}}$ at both ends	—	—
Rotation $θ$ at A only	$\frac{4 E I θ}{L}$ at A, $\frac{2 E I θ}{L}$ at B	—	—

(Sign convention: clockwise end moment on the member positive, as used in slope-deflection.)

Clapeyron's three-moment equation

For a continuous beam, take the support moments as the unknowns. For two adjacent spans AB ( $L_{1}$ , $I_{1}$ ) and BC ( $L_{2}$ , $I_{2}$ ) with support moments $M_{A}$ , $M_{B}$ , $M_{C}$ (hogging positive in the classical form):

FormulaThree-moment equation (uniform EI in each span)

M_{A} \frac{L _{1}}{I _{1}} + 2 M_{B} (\frac{L _{1}}{I _{1}} + \frac{L _{2}}{I _{2}}) + M_{C} \frac{L _{2}}{I _{2}} = \frac{6 A _{1} x ˉ _{1}}{L _{1} I _{1}} + \frac{6 A _{2} x ˉ _{2}}{L _{2} I _{2}}

$A_{1}$ , $A_{2}$ = areas of the free (simply supported) BMDs of spans 1 and 2; $\overset{x}{ˉ}_{1}$ = distance of the centroid of $A_{1}$ from A (outer end); $\overset{x}{ˉ}_{2}$ = distance of the centroid of $A_{2}$ from C (outer end).

For constant $E I$ and UDL on both spans: $M_{A} L_{1} + 2 M_{B} (L_{1} + L_{2}) + M_{C} L_{2} = \frac{w _{1} L _{1}^{3}}{4} + \frac{w _{2} L _{2}^{3}}{4}$ .

Support settlement (B lower than A by $δ_{1}$ and lower than C by $δ_{2}$ ): add $6 E I (\frac{δ _{1}}{L _{1}} + \frac{δ _{2}}{L _{2}})$ to the right-hand side (constant $E I$ ).

Boundary conditions: a simply supported end has zero moment; a fixed end is handled by adding an imaginary zero-length span beyond it.

Exam TipFree BMD area terms

UDL $w$ on span $L$ : $A = \frac{2}{3} \cdot L \cdot \frac{w L ^{2}}{8} = \frac{w L ^{3}}{12}$ , $\overset{x}{ˉ} = L /2$ → $\frac{6 A x ˉ}{L} = \frac{w L ^{3}}{4}$ .
Central point load $W$ : $A = \frac{W L ^{2}}{8}$ , $\overset{x}{ˉ} = L /2$ → $\frac{6 A x ˉ}{L} = \frac{3 W L ^{2}}{8}$ .

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