Displacement Methods of Analysis - Theory & Concepts

Learning Objectives

  • Understand the principles of the Displacement Method and Kinematic Indeterminacy (Degrees of Freedom).
  • Calculate common Fixed-End Moments (FEM) for various loading conditions.
  • Apply the Slope-Deflection Method to solve indeterminate beams and frames.
  • Apply the Moment Distribution Method (Hardy Cross Method) for iterative structural analysis.
  • Understand the analysis of frames with sidesway and Kani's Method.

In contrast to the Force Method (where redundant forces are the primary unknowns), the Displacement Method treats the joint displacements (rotations θ\theta and translations Δ\Delta) as the primary unknowns. Once the displacements are found, the internal moments and forces are calculated back-substituting into load-displacement relationships.

The Displacement Method is generally preferred for highly indeterminate structures and is the foundation for computer-based Matrix Structural Analysis because the procedure is systematic and identical regardless of the degree of indeterminacy.

Degrees of Freedom (Kinematic Indeterminacy)

Before applying displacement methods, you must identify the Degrees of Freedom (DOF) or Kinematic Indeterminacy. This is the number of independent joint displacements (translations and rotations) that are free to occur under load.

Degrees of Freedom by Support Type

Common Fixed-End Moment (FEM) Formulas

A fundamental step in all displacement methods is calculating the Fixed-End Moments (FEM) generated by external loads on individual spans, assuming the ends are rigidly locked.

Standard FEM Values

Assuming a clockwise positive convention (standard for Slope-Deflection and Moment Distribution):

  • Uniformly Distributed Load (ww) on Span LL:
    • FEMleft=−wL212FEM_{left} = -\frac{wL^2}{12} (Counter-Clockwise)
    • FEMright=+wL212FEM_{right} = +\frac{wL^2}{12} (Clockwise)
  • Concentrated Point Load (PP) at Midspan:
    • FEMleft=−PL8FEM_{left} = -\frac{PL}{8}
    • FEMright=+PL8FEM_{right} = +\frac{PL}{8}
  • Concentrated Point Load (PP) at distances aa and bb from ends:
    • FEMleft=−Pab2L2FEM_{left} = -\frac{Pab^2}{L^2}
    • FEMright=+Pa2bL2FEM_{right} = +\frac{Pa^2b}{L^2}

The Slope-Deflection Method

Introduced by G.A. Maney in 1915, this method relates the internal moments at the ends of a member to the member's end rotations (slopes) and its relative end displacements (deflections).

Slope-Deflection Equation

For any prismatic member AB with length LL and flexural rigidity EIEI, the internal moments at ends A (MABM_{AB}) and B (MBAM_{BA}) are determined by the joint rotations, chord rotation due to sidesway, and the fixed-end moments.

Slope-Deflection Equation

Relates the internal moments at the ends of a member to end rotations and displacements.

MAB=2EIL(2θA+θB−3ψ)+FEMABM_{AB} = \frac{2EI}{L} \left( 2\theta_A + \theta_B - 3\psi \right) + FEM_{AB}MBA=2EIL(2θB+θA−3ψ)+FEMBAM_{BA} = \frac{2EI}{L} \left( 2\theta_B + \theta_A - 3\psi \right) + FEM_{BA}

Variables

SymbolDescriptionUnit
MAB,MBAM_{AB}, M_{BA}Internal moments at ends A and B-
EEModulus of elasticity-
IIMoment of inertia-
LLLength of member-
θA,θB\theta_A, \theta_BJoint rotations at ends A and B (positive is clockwise)-
ψ\psiThe chord rotation due to relative lateral displacement Δ\Delta between ends A and B (ψ=Δ/L\psi = \Delta / L)-
FEMAB,FEMBAFEM_{AB}, FEM_{BA}Fixed-End Moments caused by external loads on the span, assuming both ends are perfectly fixed-

Slope-Deflection Sign Convention

Sign Convention: The Slope-Deflection Method strictly uses a clockwise-positive sign convention for moments and rotations.

Slope-deflection rotation and moment signs

Clockwise joint rotations and member-end moments are positive. With A held transversely and B moving downward relative to A, the chord rotation is positive and equals psi = Delta/L. The displacement is enlarged for visibility.

Member AB with clockwise-positive joint rotations and end moments; downward relative movement at B gives positive chord rotation psi equal to Delta over L.

Procedure

Slope-Deflection Procedure

  • Identify Unknown DOFs: Determine the unknown joint rotations (θ\theta) and chord rotations (ψ\psi).
  • Calculate FEMs: Use standard tables to find Fixed-End Moments for each loaded span.
  • Write Slope-Deflection Equations: Write the equations for the moment at each end of every member in terms of the unknown DOFs.
  • Write Equilibrium Equations: For every unknown rotation (θi\theta_i), write a joint moment equilibrium equation (∑Mjoint_i=0\sum M_{joint\_i} = 0). For every unknown translation (Δ\Delta), write a global shear equation (∑F=0\sum F = 0).
  • Solve for DOFs: Substitute the Slope-Deflection equations into the Equilibrium equations and solve the resulting system of simultaneous linear equations for the unknown displacements.
  • Calculate Final Moments: Plug the found displacements back into the original Slope-Deflection equations.
Slope-Deflection Analysis

Identify displacement unknowns, form member-end and equilibrium equations, include sway equilibrium when needed, then solve and recover moments.

Use SI consistently: E in N/m², I in m⁴, member lengths in m, loads in N or N/m, and moments in N·m. The lesson uses clockwise-positive member-end moments and rotations. Define the positive global lateral displacement and shear directions once and keep them consistent. For each member, ψ is the relative transverse end displacement divided by its length. Use member-end equations for moments and global-axis equilibrium (ΣF_x=0) for frame shear. Check structural stability and the coefficient-system rank before solving; correct any mechanism, restraint, release, or degree-of-freedom mismatch first.

Slope-Deflection AnalysisIdentify displacement unknowns, form member-end and equilibrium equations, include sway equilibrium when needed, then solve and recover moments.. Use SI consistently: E in N/m², I in m⁴, member lengths in m, loads in N or N/m, and moments in N·m. The lesson uses clockwise-positive member-end moments and rotations. Define the positive global lateral displacement and shear directions once and keep them consistent. For each member, ψ is the relative transverse end displacement divided by its length. Use member-end equations for moments and global-axis equilibrium (ΣF_x=0) for frame shear. Check structural stability and the coefficient-system rank before solving; correct any mechanism, restraint, release, or degree-of-freedom mismatch first.. Start slope-deflection analysis → Define members, restraints, loads, EI, and sign convention; Define members, restraints, loads, EI, and sign convention → Is the idealized structure stable?; Is the idealized structure stable? — Yes → Identify unknown joint rotations θ and translations Δ; Is the idealized structure stable? — No → Revise the model, then restart analysis; Identify unknown joint rotations θ and translations Δ → Compute fixed-end moments for loaded members; Compute fixed-end moments for loaded members → Write both end-moment equations per member, including ψ=Δ/L; Write both end-moment equations per member, including ψ=Δ/L → Write moment equilibrium at every unknown rotation; Write moment equilibrium at every unknown rotation → Is an independent sidesway translation unknown?; Is an independent sidesway translation unknown? — Yes → Add global shear equation (ΣF_x = 0) for sway; Is an independent sidesway translation unknown? — No → No sway DOF; omit global shear equation; Add global shear equation (ΣF_x = 0) for sway → Substitute end moments and assemble the required equations; No sway DOF; omit global shear equation → Substitute end moments and assemble the required equations; Substitute end moments and assemble the required equations → Is the displacement coefficient system full-rank?; Is the displacement coefficient system full-rank? — Yes → Solve for the unknown rotations and translations; Is the displacement coefficient system full-rank? — No → Stop: correct the singular DOF system; Solve for the unknown rotations and translations → Substitute solved DOFs to recover member-end moments; Substitute solved DOFs to recover member-end moments → Check joint moments and global force equilibrium; Check joint moments and global force equilibrium → Report displacements and member-end moments

Start slope-deflection analysis → Define members, restraints, loads, EI, and sign convention; Define members, restraints, loads, EI, and sign convention → Is the idealized structure stable?; Is the idealized structure stable? — Yes → Identify unknown joint rotations θ and translations Δ; Is the idealized structure stable? — No → Revise the model, then restart analysis; Identify unknown joint rotations θ and translations Δ → Compute fixed-end moments for loaded members; Compute fixed-end moments for loaded members → Write both end-moment equations per member, including ψ=Δ/L; Write both end-moment equations per member, including ψ=Δ/L → Write moment equilibrium at every unknown rotation; Write moment equilibrium at every unknown rotation → Is an independent sidesway translation unknown?; Is an independent sidesway translation unknown? — Yes → Add global shear equation (ΣF_x = 0) for sway; Is an independent sidesway translation unknown? — No → No sway DOF; omit global shear equation; Add global shear equation (ΣF_x = 0) for sway → Substitute end moments and assemble the required equations; No sway DOF; omit global shear equation → Substitute end moments and assemble the required equations; Substitute end moments and assemble the required equations → Is the displacement coefficient system full-rank?; Is the displacement coefficient system full-rank? — Yes → Solve for the unknown rotations and translations; Is the displacement coefficient system full-rank? — No → Stop: correct the singular DOF system; Solve for the unknown rotations and translations → Substitute solved DOFs to recover member-end moments; Substitute solved DOFs to recover member-end moments → Check joint moments and global force equilibrium; Check joint moments and global force equilibrium → Report displacements and member-end moments

  • Start slope-deflection analysis: terminator
  • Define members, restraints, loads, EI, and sign convention: process
  • Is the idealized structure stable?: decision
  • Revise the model, then restart analysis: terminator
  • Identify unknown joint rotations θ and translations Δ: process
  • Compute fixed-end moments for loaded members: process
  • Write both end-moment equations per member, including ψ=Δ/L: process
  • Write moment equilibrium at every unknown rotation: process
  • Is an independent sidesway translation unknown?: decision
  • Add global shear equation (ΣF_x = 0) for sway: process
  • No sway DOF; omit global shear equation: process
  • Substitute end moments and assemble the required equations: process
  • Is the displacement coefficient system full-rank?: decision
  • Stop: correct the singular DOF system: terminator
  • Solve for the unknown rotations and translations: process
  • Substitute solved DOFs to recover member-end moments: process
  • Check joint moments and global force equilibrium: process
  • Report displacements and member-end moments: terminator

Moment Distribution Method (Hardy Cross Method)

Before computers, solving large systems of simultaneous equations for slope-deflection was practically impossible by hand. The Moment Distribution Method is a brilliant iterative technique that avoids simultaneous equations entirely. It works by mathematically "locking" all joints, calculating the artificial moments required to hold them locked, and then "unlocking" them one by one to allow the moments to distribute proportionally based on the relative stiffness of the connected members.

Distribution Factor (DF)

When a joint is unlocked, the unbalanced moment acting on the joint is distributed to the connecting members. The DF is the fraction of that moment taken by a specific member ii.

Distribution Factor Equation

Calculates the fraction of unbalanced moment taken by a specific member.

DFi=Ki∑KDF_i = \frac{K_i}{\sum K}

Variables

SymbolDescriptionUnit
DFiDF_iDistribution Factor for member i-
KiK_iStiffness Factor of member i-
∑K\sum KSum of Stiffness Factors of all members connected to the joint-

Where KK is the Stiffness Factor of the member.

  • If the far end of the member is fixed (or continuous): K=4EILK = \frac{4EI}{L}
  • If the far end of the member is a simple pin/roller: K=3EILK = \frac{3EI}{L}

Carry-Over Factor (COF)

When you apply a moment to one end of a member to rotate it, a moment is naturally induced at the far "locked" end to maintain equilibrium. The COF is the ratio of this induced moment to the applied moment.

  • If the far end is Fixed: COF=0.5COF = 0.5 (half the moment carries over, with the same sign).
  • If the far end is Pinned: COF=0COF = 0 (a pin cannot resist moment, so nothing carries over).
Moment distribution joint balance and carry-over

The signed balancing moment is the negative of the joint's unbalanced moment. Distribution factors divide that correction in proportion to member stiffness and sum to one. Carry-over is one half with the same sign at a fixed far end, and zero at a pinned far end.

At a joint, distribute the balancing moment opposite the unbalanced sum by stiffness factors; corrections carry over by one half with the same sign to a fixed far end and by zero to a pinned far end.

Modified Stiffness (Pinned End Trick)

To significantly speed up manual Moment Distribution calculations, you can drastically reduce the number of iterations required if a member's far end is a simple pin or roller support (like an exterior overhanging beam or an exterior column resting on a pin foundation).

Instead of treating the pinned end as a normal joint to be repeatedly locked and unlocked (which bounces moments back and forth unnecessarily), you can simply use the modified stiffness K=3EILK = \frac{3EI}{L} for that member from the very beginning. By using this reduced stiffness, the pinned end is allowed to rotate freely to its final position immediately. You do not carry over any moments to the pinned end (COF=0COF = 0), and you never have to balance it again.

Exploiting Symmetry and Anti-Symmetry

For frames that are geometrically symmetric, the amount of computation in Moment Distribution can be halved if the loading is either perfectly symmetric or perfectly anti-symmetric. We modify the stiffness factor (KK) of the central member to reflect how it rotates.

Modified Stiffness Factors

  • Symmetric Loading: The structure rotates symmetrically. The centerline axis acts perfectly vertical (no rotation). The stiffness of the central member crossing the axis of symmetry is modified to K=2EILK = \frac{2EI}{L} (half stiffness), and its carry-over factor becomes zero.
  • Anti-Symmetric Loading: The structure rotates anti-symmetrically. The centerline axis acts perfectly horizontal. The stiffness of the central member crossing the axis is modified to K=6EILK = \frac{6EI}{L} (1.5x stiffness), and its carry-over factor becomes zero.

General Procedure

Moment Distribution Procedure

  • Preparation: Calculate the Fixed End Moments (FEM) for all loaded spans. Calculate the Stiffness Factors (KK) and Distribution Factors (DF) for every member at every joint.
  • Lock Joints: Assume all joints are artificially locked against rotation. The initial moments at the ends of members are simply the FEMs.
  • Distribute: Identify a joint with an unbalanced moment. "Unlock" the joint. To re-establish equilibrium, a balancing moment (equal and opposite to the unbalanced sum) is distributed to the connecting members according to their DFs.
  • Carry-Over: For every moment distributed in step 3, immediately multiply it by the COF (0.5 or 0) and "carry it over" to the opposite end of that specific member.
  • Iterate: Re-lock the joint. Move to the next joint and repeat the distribute and carry-over process. The carried-over moments create new (but smaller) unbalanced moments at adjacent joints. Continue iterating until the unbalanced moments become negligibly small.
  • Summation: Sum the initial FEMs, all distributed moments, and all carried-over moments at each end of every member to find the final internal moments.
Moment Distribution Analysis

Calculate member stiffness and distribution factors, balance and carry over moments to convergence, apply the separate sway correction when needed, and verify the final moments before reporting.

Use SI consistently: E in N/m², I in m⁴, member lengths in m, loads in N or N/m, and moments in N·m. The lesson uses clockwise-positive member-end moments. For each freely rotating joint, use K=4EI/L when the far end is fixed or continuous and K=3EI/L when it is pinned; DF=Kᵢ/ΣK. Distribute the moment opposite to the joint's unbalanced sum, carry each correction with its sign, and use COF=0.5 at a fixed far end and 0 at a pinned far end. Continue until the remaining unbalanced moments meet the stated tolerance. For sidesway superposition, obtain signed R from the actual-load no-sway case with a fictitious lateral restraint, then obtain signed R′ and sway-only moments from a compatible trial-sway case with external loads removed and the fictitious restraint released; keep all physical supports unchanged. Scale the sway moments by R/R′ and add them to the no-sway moments. Do not scale if the sway case is unstable or R′ is zero. If final equilibrium fails, review the model, data, signs, and R/R′ arithmetic, correct the issue, and recalculate before reporting.

Moment Distribution AnalysisCalculate member stiffness and distribution factors, balance and carry over moments to convergence, apply the separate sway correction when needed, and verify the final moments before reporting.. Use SI consistently: E in N/m², I in m⁴, member lengths in m, loads in N or N/m, and moments in N·m. The lesson uses clockwise-positive member-end moments. For each freely rotating joint, use K=4EI/L when the far end is fixed or continuous and K=3EI/L when it is pinned; DF=Kᵢ/ΣK. Distribute the moment opposite to the joint's unbalanced sum, carry each correction with its sign, and use COF=0.5 at a fixed far end and 0 at a pinned far end. Continue until the remaining unbalanced moments meet the stated tolerance. For sidesway superposition, obtain signed R from the actual-load no-sway case with a fictitious lateral restraint, then obtain signed R′ and sway-only moments from a compatible trial-sway case with external loads removed and the fictitious restraint released; keep all physical supports unchanged. Scale the sway moments by R/R′ and add them to the no-sway moments. Do not scale if the sway case is unstable or R′ is zero. If final equilibrium fails, review the model, data, signs, and R/R′ arithmetic, correct the issue, and recalculate before reporting.. Start moment-distribution analysis → Define stable structure, member data, loads, and sign convention; Define stable structure, member data, loads, and sign convention → Is the idealized structure stable?; Is the idealized structure stable? — Yes → Compute fixed-end moments for loaded members; Is the idealized structure stable? — No → Correct restraints, releases, or connectivity; Correct restraints, releases, or connectivity → Define stable structure, member data, loads, and sign convention; Compute fixed-end moments for loaded members → Compute end stiffnesses K and joint distribution factors DF; Compute end stiffnesses K and joint distribution factors DF → Does this case require an independent lateral translation?; Does this case require an independent lateral translation? — No → Lock joints and initialize end moments with FEMs; Does this case require an independent lateral translation? — Yes → Add fictitious lateral restraint; retain actual loads; Lock joints and initialize end moments with FEMs → Balance joint moments by DF; carry over by COF; Balance joint moments by DF; carry over by COF → Converged to tolerance with joint and global equilibrium satisfied?; Converged to tolerance with joint and global equilibrium satisfied? — No → Review and correct the calculation, then repeat balancing and carry-over; Review and correct the calculation, then repeat balancing and carry-over → Balance joint moments by DF; carry over by COF; Converged to tolerance with joint and global equilibrium satisfied? — Yes → Use converged standard-case moments as final member-end moments; Use converged standard-case moments as final member-end moments → Report final member-end moments; Add fictitious lateral restraint; retain actual loads → Balance and carry over the no-sway moments; Balance and carry over the no-sway moments → Converged to tolerance?; Converged to tolerance? — No → Balance and carry over the no-sway moments; Converged to tolerance? — Yes → Record holding force R at the fictitious restraint; Record holding force R at the fictitious restraint → Remove actual loads and fictitious restraint; keep physical supports; impose trial sway; Remove actual loads and fictitious restraint; keep physical supports; impose trial sway → Compute sway FEMs; balance and carry over to convergence; Compute sway FEMs; balance and carry over to convergence → Converged to tolerance?; Converged to tolerance? — No → Compute sway FEMs; balance and carry over to convergence; Converged to tolerance? — Yes → Compute signed lateral force R′ for the trial sway; Compute signed lateral force R′ for the trial sway → Is the sway case stable with R′ nonzero?; Is the sway case stable with R′ nonzero? — Yes → Scale sway moments by R/R′ and add to no-sway moments; Is the sway case stable with R′ nonzero? — No → Stop and inspect the sway model before scaling; Scale sway moments by R/R′ and add to no-sway moments → Form the final member-end moments for the analysis; Form the final member-end moments for the analysis → Do final member-end moments satisfy joint moment and global force equilibrium?; Do final member-end moments satisfy joint moment and global force equilibrium? — Yes → Report final member-end moments; Do final member-end moments satisfy joint moment and global force equilibrium? — No → Review model, data, signs, and R/R′ arithmetic; correct and recalculate; Review model, data, signs, and R/R′ arithmetic; correct and recalculate → Define stable structure, member data, loads, and sign convention

Start moment-distribution analysis → Define stable structure, member data, loads, and sign convention; Define stable structure, member data, loads, and sign convention → Is the idealized structure stable?; Is the idealized structure stable? — Yes → Compute fixed-end moments for loaded members; Is the idealized structure stable? — No → Correct restraints, releases, or connectivity; Correct restraints, releases, or connectivity → Define stable structure, member data, loads, and sign convention; Compute fixed-end moments for loaded members → Compute end stiffnesses K and joint distribution factors DF; Compute end stiffnesses K and joint distribution factors DF → Does this case require an independent lateral translation?; Does this case require an independent lateral translation? — No → Lock joints and initialize end moments with FEMs; Does this case require an independent lateral translation? — Yes → Add fictitious lateral restraint; retain actual loads; Lock joints and initialize end moments with FEMs → Balance joint moments by DF; carry over by COF; Balance joint moments by DF; carry over by COF → Converged to tolerance with joint and global equilibrium satisfied?; Converged to tolerance with joint and global equilibrium satisfied? — No → Review and correct the calculation, then repeat balancing and carry-over; Review and correct the calculation, then repeat balancing and carry-over → Balance joint moments by DF; carry over by COF; Converged to tolerance with joint and global equilibrium satisfied? — Yes → Use converged standard-case moments as final member-end moments; Use converged standard-case moments as final member-end moments → Report final member-end moments; Add fictitious lateral restraint; retain actual loads → Balance and carry over the no-sway moments; Balance and carry over the no-sway moments → Converged to tolerance?; Converged to tolerance? — No → Balance and carry over the no-sway moments; Converged to tolerance? — Yes → Record holding force R at the fictitious restraint; Record holding force R at the fictitious restraint → Remove actual loads and fictitious restraint; keep physical supports; impose trial sway; Remove actual loads and fictitious restraint; keep physical supports; impose trial sway → Compute sway FEMs; balance and carry over to convergence; Compute sway FEMs; balance and carry over to convergence → Converged to tolerance?; Converged to tolerance? — No → Compute sway FEMs; balance and carry over to convergence; Converged to tolerance? — Yes → Compute signed lateral force R′ for the trial sway; Compute signed lateral force R′ for the trial sway → Is the sway case stable with R′ nonzero?; Is the sway case stable with R′ nonzero? — Yes → Scale sway moments by R/R′ and add to no-sway moments; Is the sway case stable with R′ nonzero? — No → Stop and inspect the sway model before scaling; Scale sway moments by R/R′ and add to no-sway moments → Form the final member-end moments for the analysis; Form the final member-end moments for the analysis → Do final member-end moments satisfy joint moment and global force equilibrium?; Do final member-end moments satisfy joint moment and global force equilibrium? — Yes → Report final member-end moments; Do final member-end moments satisfy joint moment and global force equilibrium? — No → Review model, data, signs, and R/R′ arithmetic; correct and recalculate; Review model, data, signs, and R/R′ arithmetic; correct and recalculate → Define stable structure, member data, loads, and sign convention

  • Start moment-distribution analysis: terminator
  • Define stable structure, member data, loads, and sign convention: process
  • Is the idealized structure stable?: decision
  • Correct restraints, releases, or connectivity: process
  • Compute fixed-end moments for loaded members: process
  • Compute end stiffnesses K and joint distribution factors DF: process
  • Does this case require an independent lateral translation?: decision. Answer Yes when the frame model and load case require a lateral translation to be solved by sway analysis; use the no-sway path when they do not.
  • Lock joints and initialize end moments with FEMs: process
  • Balance joint moments by DF; carry over by COF: process
  • Converged to tolerance with joint and global equilibrium satisfied?: decision
  • Review and correct the calculation, then repeat balancing and carry-over: process
  • Use converged standard-case moments as final member-end moments: process
  • Report final member-end moments: terminator
  • Add fictitious lateral restraint; retain actual loads: process
  • Balance and carry over the no-sway moments: process
  • Converged to tolerance?: decision
  • Record holding force R at the fictitious restraint: process
  • Remove actual loads and fictitious restraint; keep physical supports; impose trial sway: process
  • Compute sway FEMs; balance and carry over to convergence: process
  • Converged to tolerance?: decision
  • Compute signed lateral force R′ for the trial sway: process
  • Is the sway case stable with R′ nonzero?: decision
  • Stop and inspect the sway model before scaling: terminator
  • Scale sway moments by R/R′ and add to no-sway moments: process
  • Form the final member-end moments for the analysis: process
  • Do final member-end moments satisfy joint moment and global force equilibrium?: decision
  • Review model, data, signs, and R/R′ arithmetic; correct and recalculate: process
  • Report final member-end moments: terminator

Interactive Simulation

The interactive simulation below demonstrates the Moment Distribution Method:

Moment Distribution Interactive Lab

Adjust the initial Fixed End Moments (FEM) and the Distribution Factor at Joint B for a two-span continuous beam (fixed at A and C). Watch how the unbalanced moment at B is iteratively distributed and carried over.

JointABC
MemberABBABCCB
DF00.500.500
* Check Equilibrium at B: + = NaN

Analysis of Frames with Sidesway

Frames will undergo lateral translation, or sidesway (Δ\Delta), if they lack geometric symmetry, loading symmetry, or both. Sidesway fundamentally changes the analysis because it introduces an unknown chord rotation (ψ=Δ/L\psi = \Delta/L) into the equations.

If using Slope-Deflection, you simply include ψ\psi as an unknown and write an additional global shear equilibrium equation (∑Fx=0\sum F_x = 0 for the entire frame) to solve for it. This requires analyzing the entire frame as a free body to relate the shear forces in the columns to the externally applied lateral loads.

If using Moment Distribution, the process requires superposition of two separate analyses:

Sidesway Moment Distribution

  • Non-Sway Analysis: Apply a fictitious lateral support to physically prevent the frame from swaying. Run the standard moment distribution. Calculate the reaction force (RR) generated at this fictitious support by summing horizontal forces.
  • Sway Analysis: Remove all external loads. Apply an arbitrary lateral displacement to the frame, calculate the resulting fixed-end moments due purely to this sway (FEM=6EIΔL2FEM = \frac{6EI\Delta}{L^2}), and distribute them. Calculate the lateral force (R′R') required to cause this arbitrary sway by summing the horizontal column shears.
  • Superposition: The true sway moments are a scaled version of the arbitrary sway moments. The scaling factor is R/R′R / R'. The final results are the Non-Sway moments plus the scaled Sway moments.

Kani's Method

Kani's method (developed by Gaspar Kani) is an iterative procedure similar to the Moment Distribution Method but offers advantages in self-correcting calculation errors and handling frames with sidesway more efficiently. It distributes fixing moments based on rotation factors and continuously updates support moments until convergence is achieved.

Global Shear Equation (Sway Frames) Context

When using the Slope-Deflection method for a single-bay, single-story frame with columns AB and CD, you need an extra equation to solve for the unknown sway Δ\Delta. Take a horizontal section through both columns:

Global Shear Equation (Sway Frames)

Calculates shear for sway frames.

∑Fx=0⇒VAB+VCD−Plat=0\sum F_x = 0 \Rightarrow V_{AB} + V_{CD} - P_{lat} = 0

Variables

SymbolDescriptionUnit
VAB,VCDV_{AB}, V_{CD}Shear forces in columns AB and CD-
PlatP_{lat}Externally applied lateral load-

Where VAB=MAB+MBALABV_{AB} = \frac{M_{AB} + M_{BA}}{L_{AB}} and VCD=MCD+MDCLCDV_{CD} = \frac{M_{CD} + M_{DC}}{L_{CD}}. Substitute the Slope-Deflection equations for the moments into this shear equation to solve for ψ\psi.

Key Takeaways
  • Displacement Methods solve for unknown geometric displacements (rotations θ\theta and translations Δ\Delta) as the primary step, rather than solving for redundant forces.
  • Slope-Deflection Method: Formulates an exact system of simultaneous linear equations by writing the internal end moments of members as functions of their joint displacements.
  • Fixed-End Moments (FEM): Crucial starting points representing moments when joints are completely locked against rotation.
  • Moment Distribution Method: An ingenious iterative, mathematical relaxation technique that solves the slope-deflection equations without requiring matrix algebra. It relies on alternating steps of locking joints, distributing unbalanced moments based on relative stiffness (DFDF), and carrying over moments (COFCOF) to adjacent joints.
  • Stiffness (KK): A measure of a member's resistance to rotation. A member fixed at the far end is stiffer (4EI/L4EI/L) than a member pinned at the far end (3EI/L3EI/L).
  • Symmetry: Utilizing symmetric (2EI/L2EI/L) and anti-symmetric (6EI/L6EI/L) modified stiffnesses drastically speeds up the manual calculation for applicable structures.
  • Sidesway: Frames that are asymmetric in geometry or loading will sway laterally. This requires analyzing the effects of chord rotation (ψ\psi), significantly complicating manual analysis but easily handled by computer matrix methods.