Chapter 3 of 6

Statically Indeterminate Structures (Force & Displacement Methods)

In the GATE Civil syllabus under Structural Analysis · 3 parts

Force (Flexibility) Method — Consistent Deformation & Three-Moment Equation · Slope Deflection Method · Moment Distribution Method

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (22 sections)

Part 1 of 3

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 $W$ : $A = \frac{W L ^{2}}{8}$ , $\overset{x}{ˉ} = L /2$ → $\frac{6 A x ˉ}{L} = \frac{3 W L ^{2}}{8}$ .

Part 2 of 3

Slope Deflection Method

Last reviewed 16 Sept 2026 · 5 min read

The displacement approach

In the slope deflection method (G. A. Maney, 1915) the unknowns are joint displacements — rotations of rigid joints and, in sway frames, lateral translations. Member end moments are written in terms of these displacements, and joint equilibrium equations are solved for them. It is the basis of moment distribution and of the stiffness matrix method.

It is convenient when the kinematic indeterminacy $D_{k}$ is small, even if the static indeterminacy is large (a fixed beam with many spans, a multi-bay frame).

Assumptions: members are prismatic and linearly elastic; axial and shear deformations are neglected; joints are rigid (all members at a joint rotate by the same angle).

Sign convention

Moments at member ends: clockwise positive (acting on the member).
Rotations $θ$ of joints: clockwise positive.
Chord rotation $ψ = Δ/ L$ : clockwise positive (the far end moving so that the chord rotates clockwise).

Slope-deflection equations

For member AB of length $L$ and flexural rigidity $E I$ :

FormulaSlope-deflection equations

M_{A B} = M_{F A B} + \frac{2 E I}{L} (2 θ_{A} + θ_{B} - 3 ψ)

M_{B A} = M_{F B A} + \frac{2 E I}{L} (2 θ_{B} + θ_{A} - 3 ψ)

$M_{F A B}$ , $M_{F B A}$ = fixed-end moments due to loads on the span (clockwise positive: e.g. UDL gives $- w L^{2} /12$ at A and $+ w L^{2} /12$ at B); $ψ = Δ/ L$ with $Δ$ the relative transverse displacement of B with respect to A.

Where the equations come from

Superpose on a fixed-ended member: (1) the loads with ends fixed → fixed-end moments; (2) rotation $θ_{A}$ with B fixed → $4 E I θ_{A} / L$ at A and $2 E I θ_{A} / L$ at B (carry-over factor ½); (3) rotation $θ_{B}$ similarly; (4) relative translation $Δ$ with no rotation → $- 6 E I Δ/ L^{2}$ at both ends.

Modified equation for a pinned far end

If end B is a pin or roller (so $M_{B A} = 0$ ), eliminate $θ_{B}$ :

M_{A B} = (M_{F A B} - \frac{M _{F B A}}{2}) + \frac{3 E I}{L} (θ_{A} - ψ)

This saves one unknown per pinned end.

Procedure

Identify unknown joint rotations (and sway $Δ$ if the frame can sway). Fixed supports have $θ = 0$ .
Compute fixed-end moments for every loaded member.
Write slope-deflection equations for every member end.
Joint equilibrium: at each rigid joint with unknown rotation, $\sum M_{m e mb er e n d s} =$ external moment at the joint (often 0).
Shear (sway) equation for each sway degree: horizontal equilibrium of the storey, using column shears from their end moments.
Solve the simultaneous equations; substitute back for end moments.
Find reactions, shear and moment diagrams by statics.

Frames without and with sway

A frame does not sway when it is symmetric in geometry and loading, or when lateral movement is prevented (a brace or a support).

For a portal with columns AB and DC (feet A, D) and beam BC, with horizontal load $P$ at beam level, the shear equation is:

\frac{M _{A B} + M _{B A}}{h _{1}} + \frac{M _{D C} + M _{C D}}{h _{2}} + P = 0

(sign depends on the chosen sway direction; column shear $= - (M_{t o p} + M_{b o tt o m}) / h$ for clockwise-positive moments, plus any load on the column).

Worked examples

Worked ExampleExample 1 — two-span continuous beam

A beam ABC is fixed at A and simply supported at B and C. AB = 6 m carries a UDL of 20 kN/m; BC = 4 m carries a central load of 40 kN. $E I$ constant. Find the support moments.

Solution. Unknowns: $θ_{B}$ , $θ_{C}$ ( $θ_{A} = 0$ ).

Fixed-end moments: $M_{F A B} = - \frac{20 \times 36}{12} = - 60$ , $M_{F B A} = + 60$ ; $M_{F B C} = - \frac{40 \times 4}{8} = - 20$ , $M_{F C B} = + 20$ (kN·m).

Use the modified equation for BC (C is a simple support, $M_{C B} = 0$ ):

$M_{B C} = (- 20 - \frac{20}{2}) + \frac{3 E I}{4} θ_{B} = - 30 + 0.75 E I θ_{B}$

$M_{B A} = 60 + \frac{2 E I}{6} (2 θ_{B}) = 60 + 0.6667 E I θ_{B}$

$M_{A B} = - 60 + \frac{2 E I}{6} θ_{B} = - 60 + 0.3333 E I θ_{B}$

Joint B: $M_{B A} + M_{B C} = 0$ → $30 + 1.4167 E I θ_{B} = 0$ → $E I θ_{B} = - 21.18$

$M_{A B} = - 60 - 7.06 = - 67.06 kN \cdot m$ $M_{B A} = 60 - 14.12 = + 45.88 kN \cdot m$ , $M_{B C} = - 30 - 15.88 = - 45.88 kN \cdot m$ ✔ (equal and opposite)

So hogging moments: 67.06 kN·m at A and 45.88 kN·m at B.

Worked ExampleExample 2 — non-sway portal frame

A symmetric portal frame ABCD has fixed feet A and D, columns 4 m high and beam BC 6 m long carrying a UDL of 24 kN/m. All members have the same $E I$ . Find the moments.

Solution. Symmetry → no sway, and $θ_{C} = - θ_{B}$ .

FEM for BC: $M_{F B C} = - \frac{24 \times 36}{12} = - 72$ , $M_{F C B} = + 72$ .

$M_{A B} = \frac{2 E I}{4} θ_{B} = 0.5 E I θ_{B}$ ; $M_{B A} = E I θ_{B}$

$M_{B C} = - 72 + \frac{2 E I}{6} (2 θ_{B} + θ_{C}) = - 72 + \frac{E I}{3} θ_{B}$ (using $θ_{C} = - θ_{B}$ )

Joint B: $E I θ_{B} - 72 + \frac{E I θ _{B}}{3} = 0$ → $E I θ_{B} = 54$

$M_{B A} = 54 kN \cdot m$ , $M_{B C} = - 54 kN \cdot m$ , $M_{A B} = 27 kN \cdot m$

Mid-span moment of beam $= \frac{w L ^{2}}{8} - 54 = 108 - 54 = 54 kN \cdot m$ (sagging).

Worked ExampleExample 3 — effect of support settlement

A two-span continuous beam ABC (AB = BC = $L$ , constant $E I$ , simply supported at A and C and at B) has no load, but support B settles by $δ$ . Find the moment at B.

Solution. Chord rotations: AB: B moves down relative to A → $ψ_{A B} = + δ / L$ (clockwise); BC: B lower than C → $ψ_{B C} = - δ / L$ .

With pinned ends at A and C (modified equations), by symmetry $θ_{B} = 0$ :

$M_{B A} = \frac{3 E I}{L} (0 - ψ_{A B}) = - \frac{3 E I δ}{L ^{2}}$ ; $M_{B C} = \frac{3 E I}{L} (0 - ψ_{B C}) = + \frac{3 E I δ}{L ^{2}}$

Joint B balances ✔. Support moment at B $= \frac{3 E I δ}{L ^{2}}$ (sagging at B — the settling support "hangs" from the beam).

Part 3 of 3

Moment Distribution Method

Last reviewed 16 Sept 2026 · 6 min read

The idea

The moment distribution method (Hardy Cross, 1930) solves indeterminate beams and frames by successive corrections instead of simultaneous equations.

Imagine every joint locked against rotation; loaded members develop fixed-end moments.
Release one joint at a time. The unbalanced moment at the joint is shared among the members meeting there in proportion to their stiffnesses (distribution).
Each distributed moment induces half of itself (for a fixed far end) at the far end of the member (carry-over).
Repeat until the carried-over moments are negligible.

Each step is an exact equilibrium statement, and the process converges quickly. It is a displacement method, equivalent to solving slope-deflection equations iteratively.

Definitions

DefinitionStiffness, distribution and carry-over

Stiffness $K$ — moment needed at the near end to rotate it by one radian.
- Far end fixed: $K = \frac{4 E I}{L}$
- Far end pinned/roller (simple end): $K = \frac{3 E I}{L}$
- Symmetric member in a symmetric frame (far end rotates equal and opposite): $K = \frac{2 E I}{L}$
- Antisymmetric case (far end rotates equally in the same sense): $K = \frac{6 E I}{L}$
Relative stiffness — use $I / L$ (fixed far end) and $\frac{3}{4} I / L$ (pinned far end) when $E$ is common.
Distribution factor at a joint: $D F_{i} = \frac{K _{i}}{\sum K}$ ; the factors at a joint add to 1. At a fixed support $D F = 0$ ; at a free pinned end $D F = 1$ .
Carry-over factor — ratio of the moment induced at the far end to the moment applied at the near end: $+ \frac{1}{2}$ when the far end is fixed; $0$ when the far end is pinned.

Sign convention and fixed-end moments

Use clockwise end moments positive (same as slope deflection). Fixed-end moments for a span loaded with a UDL $w$ : $- w L^{2} /12$ at the left end, $+ w L^{2} /12$ at the right end; central point load: $\mp W L /8$ ; eccentric load: $- W a b^{2} / L^{2}$ , $+ W a^{2} b / L^{2}$ .

Procedure

Compute stiffness and distribution factors at every joint.
Compute fixed-end moments.
Balance each joint: unbalanced moment $U$ = algebraic sum of end moments at the joint (plus any external moment); distribute $- U \times D F$ to each member end.
Carry over half of each distributed moment to the far end (if that end is fixed or continuous).
Repeat balancing and carrying over; stop when changes are small.
Sum each column to get final end moments; check that the moments at each internal joint balance.

Exam TipPinned outer ends

Two ways: (a) use stiffness $3 E I / L$ for the member and do not carry over to the pinned end after releasing it once; or (b) use $4 E I / L$ and balance the pinned end to zero in every cycle. Method (a) converges faster.

Overhangs

The moment at the support due to the overhang is statically known ( $=$ load × arm). Treat that support like a pinned end carrying a known external moment; the overhang itself has zero stiffness.

Frames with sway

If a frame can sway (unsymmetrical geometry or loading, lateral loads):

Non-sway analysis: prevent sway with an imaginary horizontal restraint; distribute as usual; find the restraint force $R$ from horizontal equilibrium of column shears.
Sway analysis: allow an arbitrary sway $Δ$ with joints locked. Fixed-end moments due to sway in each column $= - \frac{6 E I Δ}{L ^{2}}$ (with far end fixed) or $- \frac{3 E I Δ}{L ^{2}}$ (far end pinned). Choose convenient round values in the correct ratio; distribute; find the corresponding lateral force $S$ .
Combine: final moments $= M_{n o n - s w a y} + \frac{R}{S} M_{s w a y}$ (with signs so the restraint force is cancelled).

For columns of equal height and fixed feet, sway FEMs are in the ratio of $I / L^{2}$ .

Support settlement

A settlement $δ$ of one support produces fixed-end moments $- 6 E I δ / L^{2}$ at both ends of each affected span (sign by chord rotation). Distribute them like any other fixed-end moments.

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