Module 8: Exact Analysis of Indeterminate Structures - Displacement Methods - Theory & Concepts

Exact Analysis of Indeterminate Structures: Displacement Methods

$Learning Objectives$

Identify joint translations and rotations used as displacement-method unknowns.
Apply slope-deflection equations to prismatic members with joint rotation, chord rotation, and fixed-end effects.
Calculate member rotational stiffness and joint distribution factors.
Execute moment distribution using balancing and carry-over operations.
Distinguish non-sway and sway frame analysis.
Verify displacement-method results using joint equilibrium, member-end actions, boundary conditions, and symmetry.

$Displacement Method$

A displacement method uses independent joint translations and rotations as the primary unknowns, relates member-end forces to those displacements through stiffness relationships, and enforces equilibrium to solve the unknown generalized displacements.

$Kinematic Indeterminacy$

Kinematic indeterminacy is the number of independent generalized displacement coordinates required to describe the structural deformation after supports, releases, symmetry, and other constraints are applied.

$Force Methods versus Displacement Methods$

Force methods choose redundant forces and add compatibility equations. Displacement methods choose independent displacements and add equilibrium equations after expressing member actions in terms of those displacements. The preferable method depends strongly on the number of redundants versus active displacement degrees of freedom.

Slope-Deflection Equation at End A

Prismatic member AB using a consistent member-end moment and signed chord-rotation convention.

M_{AB}=M_{AB}^{F}+\frac{2EI}{L}\left(2\theta_A+\theta_B-3\psi\right)

Variables

Symbol	Description	Unit
$M_{AB}$	Final member-end moment at end A	-
$M_{AB}^{F}$	Fixed-end moment at A caused by member loading	-
$\theta_A,\theta_B$	Joint rotations under the adopted sign convention	-
$\psi$	Signed chord rotation associated with relative transverse joint translation	-
$E,I,L$	Member elastic modulus, second moment of area, and length	-

Slope-Deflection Equation at End B

Companion equation for the opposite end using the same sign convention.

M_{BA}=M_{BA}^{F}+\frac{2EI}{L}\left(2\theta_B+\theta_A-3\psi\right)

Variables

Symbol	Description	Unit
$M_{BA}$	Final member-end moment at end B	-
$M_{BA}^{F}$	Fixed-end moment at B caused by member loading	-

$Slope-Deflection Signs Depend on the Adopted Convention$

The equations above require one internally consistent convention for positive end moments, joint rotations, and chord rotation. Fixed-end-moment tables from another convention may use opposite signs. Establish the convention before substituting values.

$Fixed-End Moments$

Member loads first create fixed-end actions for the hypothetical condition in which both member ends are rotationally fixed. For a prismatic member carrying a full-span uniform load, the fixed-end moment magnitudes are $wL^2/12$ at both ends with opposite rotational senses. For a centered point load, the corresponding magnitudes are $PL/8$ .

$Member Rotational Stiffness$

Rotational stiffness at a joint is the end moment required to produce a unit rotation at that joint under the specified restraint condition at the far end.

$Common Prismatic Member Stiffnesses$

For bending of a prismatic member:

far end fixed: $K=4EI/L$ at the near joint;
far end hinged: $K=3EI/L$ at the near joint. These values are local rotational stiffnesses used in classical moment distribution and are not interchangeable with complete matrix stiffness terms without context.

Moment-Distribution Factor

Fraction of an unbalanced joint moment assigned to a connected member.

DF_i=\frac{K_i}{\sum_j K_j}

Variables

Symbol	Description	Unit
$DF_i$	Distribution factor for member i at the joint	-
$K_i$	Rotational stiffness of member i at the joint	-

$Carry-Over Factor$

For a prismatic member whose far end is fixed, an applied near-end joint moment carries one-half of that moment to the far end in the corresponding classical sign sense. If the far end is an ideal hinge, the far-end moment must remain zero and the classical carry-over factor to that hinge is zero.

$Moment Distribution Procedure$

Start with fixed-end moments. At each unrestrained joint, sum the member-end moments and any applied joint moment to determine the unbalance. Apply an equal and opposite balancing moment and distribute it according to the member distribution factors. Carry appropriate portions to far ends, then repeat until the remaining corrections are negligible for the required precision.

$Interactive Exploration$

Use the moment-distribution simulator to inspect fixed-end moments, stiffnesses, distribution factors, balancing moments, carry-over, and convergence. Match the simulator's sign convention before comparing a hand table.

Moment Distribution Method Simulator

A (Fixed)

B (Roller)

C (Fixed)

Uniform Load: 20 kN/m on spans AB and BC (L = 10m)

Uniform Load (w)20 kN/m

Joint	A	B		C
Member	AB	BA	BC	CB
DF	0	0.5	0.5	0
Initial FEMs	-166.67	166.67	-166.67	166.67
Cycle 1 (Dist)		0.00	0.00
Cycle 1 (CO)	0.00			0.00
Cycle 2 (Dist)		0.00	0.00
Cycle 2 (CO)	0.00			0.00
Cycle 3 (Dist)		0.00	0.00
Cycle 3 (CO)	0.00			0.00
Cycle 4 (Dist)		0.00	0.00
Cycle 4 (CO)	0.00			0.00
Cycle 5 (Dist)		0.00	0.00
Cycle 5 (CO)	0.00			0.00
Final Moments	-166.67	166.67	-166.67	166.67

Note how the unbalanced moment at joint B is distributed (multiplied by DF and reversed in sign), and then half of that distributed moment is carried over to the fixed ends A and C.

$Non-Sway Frame$

A non-sway idealization constrains the relevant lateral joint translation, so the corresponding chord-rotation term is zero for the frame-level lateral degree of freedom being neglected.

$Sway Frame$

A sway frame has one or more relevant joint translations that must be included as displacement unknowns or solved through an equivalent equilibrium condition.

$Sway Analysis$

When lateral translation is possible, member chord rotation contributes to end moments. The frame must satisfy both joint moment equilibrium and horizontal force/story-shear equilibrium. Classical hand approaches may separate no-sway and sway components by superposition when the system remains linear.

Displacement-Method Analysis Workflow

Start by identifying the unknown joint rotations and translations. Then choose slope-deflection equations or moment distribution based on the structure and desired hand-solution format. A sway assumption must be decided before omitting translation terms.

Stable indeterminate beam or frame → Identify independent joint rotations and translations; Identify independent joint rotations and translations → Relevant joint translation/sway present?; Relevant joint translation/sway present? — No → Set applicable chord-rotation terms to zero; Relevant joint translation/sway present? — Yes → Include translation/chord rotation and lateral equilibrium equation; Set applicable chord-rotation terms to zero → Calculate fixed-end actions and member rotational stiffnesses; Include translation/chord rotation and lateral equilibrium equation → Calculate fixed-end actions and member rotational stiffnesses; Calculate fixed-end actions and member rotational stiffnesses → Preferred hand solution?; Preferred hand solution? — Slope-defl. → Write slope-deflection member equations; Preferred hand solution? — Moment dist. → Build moment-distribution table with DF and carry-over; Write slope-deflection member equations → Enforce joint equilibrium and solve rotations/translations or balance moments; Build moment-distribution table with DF and carry-over → Enforce joint equilibrium and solve rotations/translations or balance moments; Enforce joint equilibrium and solve rotations/translations or balance moments → Recover final member-end moments, shears, and reactions; Recover final member-end moments, shears, and reactions → Joint and global equilibrium satisfied?; Joint and global equilibrium satisfied? — Yes → Verified displacement-method solution; Joint and global equilibrium satisfied? — No → Review signs, stiffnesses, restraints, sway assumption, and fixed-end moments; Review signs, stiffnesses, restraints, sway assumption, and fixed-end moments → Identify independent joint rotations and translations

Stable indeterminate beam or frame: terminator
Identify independent joint rotations and translations: process
Relevant joint translation/sway present?: decision
Set applicable chord-rotation terms to zero: process
Include translation/chord rotation and lateral equilibrium equation: process
Calculate fixed-end actions and member rotational stiffnesses: process
Preferred hand solution?: decision
Write slope-deflection member equations: process
Build moment-distribution table with DF and carry-over: process
Enforce joint equilibrium and solve rotations/translations or balance moments: process
Recover final member-end moments, shears, and reactions: process
Joint and global equilibrium satisfied?: decision
Review signs, stiffnesses, restraints, sway assumption, and fixed-end moments: process
Verified displacement-method solution: terminator

Key Takeaways

Displacement methods use joint rotations and translations as primary unknowns.
Slope-deflection expresses member-end moments as fixed-end effects plus stiffness times generalized displacements.
Moment distribution enforces joint equilibrium iteratively using member stiffness, distribution factors, and carry-over.
A far-fixed prismatic member has rotational stiffness $4EI/L$ ; a far-hinged member has $3EI/L$ .
Sway frames require lateral translation effects and an additional equilibrium condition.
Sign convention, restraint condition, joint equilibrium, and final reactions must be checked together.

$References$

R. C. Hibbeler, Structural Analysis, 11th ed., Pearson, 2024

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