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

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

Idea of the force method

Propped cantilever

Fixed beams and fixed-end moments

Clapeyron's three-moment equation

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