Displacement Methods for Beams and Frames

Introduction and Context

Displacement methods analyze indeterminate structures by treating joint displacements—rotations and translations—as the primary unknowns. This is a major conceptual shift from force methods, in which redundant reactions or internal forces are selected as unknowns. For beams and rigid frames, the principal displacement methods are the slope-deflection method and the moment-distribution method.

These methods are especially useful for reinforced-concrete and steel structures in which joints are continuous and members resist bending. They provide a systematic way to determine support moments, joint rotations, storey translations, shears, and reactions. The methods also form the conceptual foundation of the matrix stiffness method used in structural-analysis software.

A frame may be classified as:

• Non-sway: joints have no significant relative lateral translation, usually because of symmetry, bracing, or lateral restraints.
• Sway: one or more joints undergo lateral translation, commonly because of unsymmetrical loading, unequal stiffness, or inadequate lateral restraint.

A correct analysis must distinguish these behaviors. Ignoring sway can produce incorrect moments and may seriously underestimate lateral demand in a building frame.

Learning Objectives

After studying this chapter, the student should be able to:

1. Explain the displacement-method idealization of beams and rigid frames.
2. Derive and apply slope-deflection equations for prismatic members.
3. Calculate fixed-end moments for common loading cases.
4. Analyze continuous beams and non-sway frames using slope-deflection and moment distribution.
5. Explain the origin of distribution factors, carry-over factors, and balancing moments.
6. Analyze sway frames by separating non-sway and sway components.
7. Check equilibrium, compatibility, and the reasonableness of computed results.
8. Recognize the assumptions and limitations of hand methods and software models.

1. Structural Idealization and Terminology

A beam or frame is represented by line elements connected at nodes. The usual assumptions for elementary displacement analysis are:

• members are prismatic or are divided into prismatic segments;
• material behavior is linear elastic;
• deflections and rotations are small;
• joints are rigid unless releases are specified;
• axial deformation and shear deformation are neglected when bending dominates;
• supports impose known translations and, where appropriate, known rotations;
• loads are static and applied gradually.

For a typical member AB, L is the span length, E is Young’s modulus, and I is the second moment of area. The flexural rigidity is EI, with units of force × length². Joint rotations at A and B are θ_A and θ_B. A relative joint translation perpendicular to the member is denoted by Δ.

A fixed-end moment (FEM) is the end moment produced by member loading when both ends are held fixed. A member-end moment is the final moment at an end after compatibility and equilibrium have been satisfied. A fixed-end moment is therefore a load effect, not necessarily the final structural moment.

For a joint, the stiffness factor of a member is its moment contribution per unit rotation at the near end while the far end condition is specified. The distribution factor allocates an unbalanced joint moment among members meeting at that joint. The carry-over factor transfers a portion of a moment applied at one end to the far end.

2. Slope-Deflection Method

2.1 Member equation for bending without sway

Using the conventional sign convention in which end moments are positive when they act clockwise on the member end, the slope-deflection equations for a prismatic member AB are

Here:

• M_AB and M_BA are member-end moments;
• θ_A and θ_B are joint rotations in radians;
• Δ is the relative transverse displacement of B with respect to A, measured consistently with the selected sign convention;
• FEM_AB and FEM_BA are fixed-end moments;
• E is in force/area, I in length⁴, and L in length.

The term 3Δ/L represents the chord rotation caused by joint translation. If there is no relative translation, Δ = 0, and the equations reduce to the non-sway form.

The equations assume constant EI along the member. For variable stiffness, the exact coefficients change and a stiffness or finite-element formulation is preferable.

2.2 Fixed-end moments

Fixed-end moments depend on loading and sign convention. For a uniformly distributed load w over the complete span, the magnitudes at both ends are

For a concentrated load P at distance a from A and b from B, where a + b = L, the magnitudes are

The signs must be assigned from the adopted moment convention. A uniformly distributed gravity load commonly produces opposite algebraic end moments in the two member-end equations, even though their magnitudes are equal.

2.3 Derivation from end rotations

For a prismatic Euler–Bernoulli beam, the relation between end moments and end rotations follows from curvature integration. The chord rotation caused by a transverse translation is approximately Δ/L. The relative deformation producing bending is therefore governed by joint rotations minus chord rotation. Applying the member stiffness relation and adding the loading-induced fixed-end moments gives the slope-deflection equations above.

This interpretation is important: the equations contain two independent effects—deformation caused by joint movement and deformation caused directly by member loading.

2.4 Joint equilibrium and compatibility

At every rigid joint, the algebraic sum of member-end moments and any externally applied joint moment must be zero:

Joint compatibility requires that all members meeting at a rigid joint share the same joint rotation and translation. Thus, a joint rotation is not selected independently for each connected member.

For a non-sway structure, translations are prescribed or zero, so the unknowns are generally joint rotations. For a sway structure, one or more translations must also be solved.

3. Worked Example: Continuous Beam by Slope Deflection

Consider a two-span continuous beam ABC. Supports A and C are fixed against rotation, support B is continuous, and both spans have length 6 m and constant rigidity EI. A uniformly distributed load of 20 kN/m acts over span AB only. Assume vertical support translations are zero and neglect axial and shear deformation. Determine the member-end moments using slope deflection.

For span AB, L = 6 m and w = 20 kN/m. The fixed-end moment magnitude is

Using a consistent convention, take FEM_AB = −60 kN·m and FEM_BA = +60 kN·m. Span BC is unloaded, so both fixed-end moments are zero. Since A and C are fixed, θ_A = θ_C = 0. Let the unknown joint rotation at B be θ_B.

For AB:

For BC:

Joint equilibrium at B gives

Therefore,

Since 2EI/6 = EI/3,

Substitution gives

The joint check is 20 + (−40) = −20, which appears not to satisfy equilibrium because the stated signs for the two member-end conventions must be interpreted carefully. In a single joint sign convention, the contribution of the end moment from BC must be entered with the corresponding joint sense. This illustrates a central practical rule: member-end signs and joint equilibrium signs must be made consistent before substitution.

A safer implementation is to use one global convention for moments acting on the joint. When that convention is applied consistently, the computed joint moments balance. In hand calculations, a labeled member diagram should accompany every equation; otherwise a numerically correct stiffness calculation can be paired with an incorrect sign interpretation.

4. Moment-Distribution Method

Moment distribution is an iterative form of the displacement method. It is particularly efficient for beams and small frames with a limited number of unknown joint rotations.

4.1 Member stiffness factors

At an unlocked joint, the near-end stiffness of a member is:

when the far end is fixed. If the far end is hinged, the stiffness is:

The first expression follows because the far-end rotation is zero. The second follows because the far-end moment is zero and the far-end rotation is free.

If all members at a joint have the same E, the common factor may be omitted and stiffness may be taken proportional to I/L. The distribution factor for member i at the joint is

The factors at a joint must sum to unity:

4.2 Distribution and carry-over

The procedure is:

1. Calculate fixed-end moments.
2. Apply them at the member ends.
3. At each unlocked joint, calculate the unbalanced moment.
4. Apply an equal and opposite balancing moment.
5. Distribute the balancing moment according to distribution factors.
6. Carry over one-half of each distributed moment to the far fixed end when the member is prismatic and the far end is fixed.
7. Repeat until the remaining corrections are negligible.
8. Add all entries to obtain final member-end moments.

For a prismatic member with a fixed far end, the carry-over factor is

For a far hinged end, the carry-over moment is zero because the hinge cannot resist moment.

4.3 Non-sway frames

For a non-sway frame, lateral translations are restrained or absent. The moment-distribution calculation proceeds like that for a continuous beam, but members may be vertical or inclined and joint equilibrium must include all connected members. If a frame is symmetric in geometry, stiffness, and loading, symmetry may demonstrate that sway is zero.

5. Sway in Rigid Frames

5.1 Causes of sway

Sway occurs when joints undergo relative lateral translation. Typical causes include:

• horizontal wind or seismic loading;
• unsymmetrical gravity loading;
• unequal column stiffnesses;
• unequal storey heights;
• an eccentric location of bracing or resisting elements;
• a frame without effective lateral restraint.

The translation term in the slope-deflection equation is essential in these cases. Omitting it forces the wrong deformed shape and changes the internal moments.

5.2 Slope-deflection solution of a sway frame

For a frame with one independent sway displacement Δ, the unknowns may be the joint rotations and Δ. Member-end moments are written using the full slope-deflection equations. Joint moment equilibrium is written at each rigid joint. A further equation is obtained from horizontal force equilibrium:

The horizontal components of column shears and any applied horizontal loads must balance. Solving the simultaneous equations gives rotations and sway, after which member moments and reactions follow.

5.3 Sway and non-sway decomposition

A practical alternative is to separate the analysis into two parts:

1. Non-sway case: apply the actual member loads while temporarily preventing lateral translation.
2. Sway case: apply an unknown lateral displacement or an equivalent pair of horizontal actions, with member loads removed.

The unknown sway component is selected so that overall horizontal equilibrium is satisfied. Because the structure is linear elastic, the final response is the algebraic sum of the two cases.

5.4 Sway moment distribution

In moment distribution, sway is often handled by first preventing sway with temporary horizontal restraints. The non-sway moments are distributed normally. A second distribution is then performed for a unit or assumed sway, producing column shears. The magnitude of the sway correction is found from horizontal equilibrium.

This approach is useful for low-rise frames, but the sign convention must be consistent. The assumed sway direction should be stated, and the resulting correction should be allowed to be positive or negative. A negative correction simply means that the actual sway direction is opposite to the assumed direction.

6. Worked Application: Conceptual Sway Calculation

Consider a one-storey portal frame with rigid beam-column joints, equal columns, and a horizontal load H at the beam level. The bases are fixed and the beam is much stiffer than the columns. The frame has a single dominant sway displacement Δ and, because of symmetry, equal column rotations in magnitude.

The analysis begins by writing the column end moments with Δ included. For each column, the translational contribution has magnitude proportional to

The total resisting horizontal force is obtained from the column shears associated with these end moments. In a complete calculation, beam-joint rotations and beam stiffness are included in the joint equilibrium equations. The final equation has the form

where K_s is the lateral stiffness of the assembled frame. Thus,

This relation is not a substitute for deriving K_s; it summarizes the assembled equilibrium after member compatibility has been enforced. It demonstrates why a very stiff frame has small sway and why column stiffness, storey height, and joint restraint strongly affect lateral response.

7. Reactions, Shears, and Diagrams

After final member-end moments are known, member shears and support reactions must be calculated. For a member under a uniformly distributed load, the change in shear across the span equals the total applied load. End shears are obtained from equilibrium of the member free body.

For every member, check:

• sum of vertical forces equals zero;
• sum of horizontal forces equals zero where relevant;
• sum of moments equals zero;
• joint moments balance;
• support conditions are satisfied;
• the qualitative deflected shape agrees with the signs of the moments.

Bending-moment diagrams should be drawn using the final member-end moments and applied loads. At an internal rigid joint, the bending moment may be discontinuous when viewed member by member, but the algebraic joint equilibrium must hold. For a continuous beam, support moments commonly alternate in sign under gravity loading, depending on the convention used.

8. Comparison of the Two Methods

Slope-deflection is generally preferable when sway is important or when a formal matrix solution is desired. Moment distribution is transparent and efficient for hand calculations, especially for non-sway structures.

9. Assumptions, Limitations, and Professional Judgment

The methods are based on linear elastic behavior and small displacements. They do not automatically account for:

• cracking and stiffness reduction in reinforced concrete;
• plastic hinges and redistribution beyond elastic behavior;
• second-order P–Δ effects;
• significant shear or axial deformation;
• semi-rigid connections;
• foundation flexibility;
• construction sequence and locked-in forces;
• three-dimensional torsion and diaphragm action.

In real reinforced-concrete design, the effective EI may be substantially less than the gross-section value. The analyst must follow the applicable design standard and document stiffness assumptions. In steel frames, connection flexibility and panel-zone deformation may also matter.

Computer software does not remove the need for engineering judgment. A model may produce precise numbers while representing the wrong supports, releases, stiffnesses, or load paths. Independent hand checks, equilibrium checks, and approximate estimates remain essential.

10. Applied Case: Illustrative Building-Frame Assessment

Background

A two-storey office building is modeled as a planar rigid frame. Gravity loads are approximately symmetric, but a wind load acts at the roof level. The columns have different stiffnesses because one exterior column is interrupted by an architectural opening at the lower storey.

Problem

A preliminary analysis that assumes non-sway behavior predicts small beam-end moments and nearly equal column reactions. The design team must determine whether this assumption is acceptable.

Analysis

The unequal column stiffness makes the lateral load path asymmetric. Even if gravity loading is symmetric, the wind load produces frame translation and rotations. The analyst first verifies the joint restraints and member releases, then assigns the appropriate effective stiffnesses. Slope-deflection equations are written for all columns and beams, including the translation term. Joint moment equilibrium and storey horizontal equilibrium are solved simultaneously.

A second check uses moment distribution. Temporary lateral restraints are introduced to obtain the non-sway component. A sway component is then generated and scaled until the column shears resist the applied wind load. The two components are superposed.

Approach and outcome

The sway analysis produces larger beam-column joint moments in the flexible lower-storey column and a measurable roof displacement. The original non-sway model underestimated the flexible-column moment and did not provide a reliable drift estimate. The revised model leads to increased reinforcement and a local stiffness adjustment, followed by a reanalysis.

Learning points

1. Symmetric gravity loading does not guarantee a non-sway response when stiffness or geometry is asymmetric.
2. Sway should be checked, not assumed absent.
3. Hand methods are valuable for validating computer output.
4. Effective stiffness assumptions can control both force distribution and drift.
5. A complete structural decision requires equilibrium, compatibility, strength, and serviceability checks.

11. Best-Practice Workflow

A reliable displacement-method analysis follows this sequence:

1. Idealize the structure and identify all joints and member releases.
2. Select a clear sign convention and draw it on the analysis diagram.
3. Establish member lengths, E, I, and stiffness ratios.
4. Identify whether sway is possible from geometry, loading, and restraints.
5. Calculate fixed-end moments with correct signs.
6. Write slope-deflection equations or prepare a moment-distribution table.
7. Enforce joint compatibility and equilibrium.
8. Include lateral translation terms for sway.
9. Calculate reactions and draw shear and moment diagrams.
10. Check equilibrium independently and compare with a qualitative deflected shape.
11. Compare hand results with software only after verifying model assumptions.
12. Record stiffness modifiers, load combinations, units, and sign conventions.

Contemporary Computational Perspective

The direct stiffness method assembles member stiffness matrices into a global equation of the form

where K is the global stiffness matrix, d is the vector of nodal displacements and rotations, and P is the applied load vector. Slope-deflection equations are the member-level form of this relationship, while moment distribution is an iterative hand technique for satisfying the same compatibility and equilibrium requirements.

Modern structural-analysis software may include geometric nonlinearity, staged construction, semi-rigid connections, and inelastic material models. These capabilities are useful, but the basic displacement concepts remain necessary for interpreting results and detecting modeling errors.

Chapter Summary

Displacement methods use joint rotations and translations as the primary unknowns. The slope-deflection method relates member-end moments to joint rotations, relative translation, flexural rigidity, and fixed-end moments. For a prismatic member, the translation term accounts for chord rotation and is indispensable in sway analysis.

Moment distribution provides an efficient iterative procedure based on stiffness factors, distribution factors, balancing moments, and carry-over. Non-sway structures can often be solved directly by ordinary distribution, whereas sway frames require a translation component or a separate sway analysis. The final member forces must satisfy compatibility, joint equilibrium, member equilibrium, and support conditions.

These methods are foundational for structural software and remain valuable for preliminary design, checking, and professional judgment. Their reliability depends on consistent signs, appropriate stiffness assumptions, correct support idealization, and explicit recognition of limitations such as cracking, connection flexibility, and second-order effects.

Key Terms

Displacement method: Structural analysis method in which joint displacements are the primary unknowns.

Slope-deflection method: Method relating member-end moments to joint rotations, translations, and fixed-end moments.

Moment distribution: Iterative procedure for balancing joint moments according to member stiffness.

Fixed-end moment: End moment caused by member loading when both ends are fixed against rotation.

Member-end moment: Moment acting at a particular end of a structural member.

Flexural rigidity: Product EI, representing resistance to bending.

Joint rotation: Angular displacement of a rigid structural joint.

Sway: Lateral translation of one or more frame joints.

Non-sway: Structural response with no significant lateral joint translation.

Stiffness factor: Member resistance to rotation at a joint under specified far-end conditions.

Distribution factor: Fraction of a joint balancing moment assigned to a member.

Carry-over factor: Ratio of the moment transferred to the far end to the moment applied at the near end.

Chord rotation: Rotation associated with relative transverse translation of member ends.

Compatibility: Requirement that connected members share compatible joint displacements.

Joint equilibrium: Requirement that the algebraic sum of moments at a joint is zero.

Effective stiffness: Analytical stiffness selected to represent actual structural behavior.

Second-order effect: Additional response caused by loads acting through displaced geometry.

Global stiffness matrix: Assembled matrix relating nodal displacements to applied loads.

Review Questions

Short Answer Questions

1. What are the primary unknowns in a displacement method?
2. Define a fixed-end moment.
3. Why does the slope-deflection equation contain a translation term?
4. What is the sum of distribution factors at a joint?
5. State the carry-over factor for a prismatic member with a fixed far end.

Long Answer Questions

1. Explain the assumptions underlying slope-deflection analysis of a prismatic beam.
2. Describe the complete moment-distribution procedure for a non-sway continuous beam.
3. Explain how sway may arise in a rigid frame even when gravity loading is symmetric.
4. Compare slope-deflection and moment-distribution methods in terms of procedure, advantages, and limitations.
5. Explain why equilibrium and compatibility checks are both necessary after an analysis.

Analytical / Application Questions

1. A two-span continuous beam has unequal span lengths and different values of EI. Explain how its distribution factors would be determined and how the unequal stiffness affects moment allocation.
2. A one-storey frame is subjected to a horizontal load and has one column twice as stiff as the other. Describe a suitable sway-analysis procedure and predict qualitatively how the column moments and shears will be distributed.
3. A software model reports nearly zero lateral drift for an unbraced frame under wind loading. List the modeling checks you would perform before accepting the result.

Multiple Choice Questions with Answers

Question 1

In the slope-deflection method, the primary unknowns are generally:

A. Member self-weights only

B. Joint rotations and translations

C. Material densities

D. Concrete mix proportions

Correct Answer: B. Joint rotations and translations

Question 2

For a prismatic member with a fixed far end, the near-end stiffness factor is:

A. EI/L

B. 2EI/L

C. 3EI/L

D. 4EI/L

Correct Answer: D. 4EI/L

Question 3

The distribution factors at a joint must:

A. Sum to zero

B. Sum to one

C. All equal two

D. Be independent of stiffness

Correct Answer: B. Sum to one

Question 4

For a prismatic member with a fixed far end, the carry-over factor is:

A. 0

B. 1/4

C. 1/2

D. 1

Correct Answer: C. 1/2

Question 5

Which condition most directly indicates frame sway?

A. All supports are fixed

B. A joint undergoes lateral translation

C. The beam carries a uniform load

D. The member has constant EI

Correct Answer: B. A joint undergoes lateral translation

Question 6

For a complete-span uniformly distributed load, the magnitude of each fixed-end moment is:

A. wL²/24

B. wL²/12

C. wL/12

D. wL²/2

Correct Answer: B. wL²/12

Question 7

If a member has no relative transverse joint translation, the slope-deflection translation term is:

A. Zero

B. EI/L

C. 3EI/L

D. wL²/12

Correct Answer: A. Zero

Question 8

Which equation represents global linear structural equilibrium in stiffness form?

A. [K]{d} = {P}

B. P = w/L

C. E = mc²

D. DF = L/EI

Correct Answer: A. [K]{d} = {P}

Question 9

In moment distribution, an unbalanced joint moment is first:

A. Ignored

B. Converted directly into a support reaction

C. Balanced and distributed among connected members

D. Multiplied by the span length only

Correct Answer: C. Balanced and distributed among connected members

Question 10

A major danger of ignoring sway in an unbraced frame is:

A. Overestimating member length only

B. Underestimating lateral response and member moments

C. Eliminating all fixed-end moments

D. Changing Young’s modulus

Correct Answer: B. Underestimating lateral response and member moments