Displacement Methods of Analysis - Theory & Concepts

Displacement Methods of Analysis - Theory & Concepts

In contrast to the Force Method (where redundant forces are the primary unknowns), the Displacement Method treats the joint displacements (rotations

\theta

and translations

\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.

Fixed Support: 0 DOF ( $\theta=0$ , $\Delta=0$ )
Pinned Support: 1 DOF (rotation $\theta$ )
Roller Support: 2 DOF (rotation $\theta$ , translation $\Delta$ parallel to the surface)
Internal Free Joint (Frame): 3 DOF (rotation $\theta$ , translation $\Delta_x$ , translation $\Delta_y$ )

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 ( $w$ ) on Span $L$ :
- $FEM_{left} = -\frac{wL^2}{12}$ (Counter-Clockwise)
- $FEM_{right} = +\frac{wL^2}{12}$ (Clockwise)
Concentrated Point Load ( $P$ ) at Midspan:
- $FEM_{left} = -\frac{PL}{8}$
- $FEM_{right} = +\frac{PL}{8}$
Concentrated Point Load ( $P$ ) at distances $a$ and $b$ from ends:
- $FEM_{left} = -\frac{Pab^2}{L^2}$
- $FEM_{right} = +\frac{Pa^2b}{L^2}$

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).

The Slope-Deflection Equation

For a typical member

AB

with uniform

EI

and length

L

, the internal moment at end

A

(

M_{AB}

) is expressed as:

M_{AB} = \frac{2EI}{L} \left( 2\theta_A + \theta_B - 3\frac{\Delta}{L} \right) + \text{FEM}_{AB}

Where:

$E$ is the modulus of elasticity, $I$ is the moment of inertia.
$\theta_A$ and $\theta_B$ are the unknown rotations at nodes $A$ and $B$ .
$\Delta$ is the relative lateral displacement (sidesway) between ends $A$ and $B$ .
$\text{FEM}_{AB}$ is the Fixed-End Moment at $A$ due to external loads on the span.

Sign Convention: Clockwise moments and rotations are typically considered positive.

Slope-Deflection Equation

For any prismatic member AB with length

L

and flexural rigidity

EI

, the internal moment at end A (

M_{AB}

) is given by:

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

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

Where:

$\theta_A, \theta_B$ : Joint rotations at ends A and B (positive is clockwise).
$\psi = \Delta / L$ : The chord rotation due to relative lateral displacement $\Delta$ between ends A and B.
$FEM_{AB}, FEM_{BA}$ : Fixed-End Moments caused by external loads on the span, assuming both ends are perfectly fixed.

Important

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

Procedure

Procedure

STEP-BY-STEP

Identify Unknown DOFs: Determine the unknown joint rotations ( $\theta$ ) and chord rotations ( $\psi$ ).
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 ( $\theta_i$ ), write a joint moment equilibrium equation ( $\sum M_{joint\_i} = 0$ ). For every unknown translation ( $\Delta$ ), write a global shear equation ( $\sum F = 0$ ).
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.

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

i

DF_i = \frac{K_i}{\sum K}

Where

K

is the Stiffness Factor of the member.

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

Important

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: $COF = 0.5$ (half the moment carries over, with the same sign).
If the far end is Pinned: $COF = 0$ (a pin cannot resist moment, so nothing carries over).

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 = \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 = 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 (

K

) 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 $K = \frac{2EI}{L}$ (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 $K = \frac{6EI}{L}$ (1.5x stiffness), and its carry-over factor becomes zero.

General Procedure

Procedure

STEP-BY-STEP

Preparation: Calculate the Fixed End Moments (FEM) for all loaded spans. Calculate the Stiffness Factors ( $K$ ) 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 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.

Distribution Factor DF_BA: 0.50(DF_BC = 0.50)

FEM_AB

FEM_BA

FEM_BC

FEM_CB

Joint	A	B		C
Member	AB	BA	BC	CB
DF	0	0.50	0.50	0

* Check Equilibrium at B: + = NaN

Analysis of Frames with Sidesway

Frames will undergo lateral translation, or sidesway (

\Delta

), if they lack geometric symmetry, loading symmetry, or both. Sidesway fundamentally changes the analysis because it introduces an unknown chord rotation (

\psi = \Delta/L

) into the equations.

If using Slope-Deflection, you simply include

\psi

as an unknown and write an additional global shear equilibrium equation (

\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:

Procedure

STEP-BY-STEP

Non-Sway Analysis: Apply a fictitious lateral support to physically prevent the frame from swaying. Run the standard moment distribution. Calculate the reaction force ( $R$ ) 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 ( $FEM = \frac{6EI\Delta}{L^2}$ ), and distribute them. Calculate the lateral force ( $R'$ ) 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 $R / R'$ . 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)

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

\Delta

. Take a horizontal section through both columns:

\sum F_x = 0 \Rightarrow V_{AB} + V_{CD} - P_{lat} = 0

Where

V_{AB} = \frac{M_{AB} + M_{BA}}{L_{AB}}

and

V_{CD} = \frac{M_{CD} + M_{DC}}{L_{CD}}

. Substitute the Slope-Deflection equations for the moments into this shear equation to solve for

\psi

Key Takeaways

Displacement Methods solve for unknown geometric displacements (rotations $\theta$ and translations $\Delta$ ) 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 ( $DF$ ), and carrying over moments ( $COF$ ) to adjacent joints.
Stiffness ( $K$ ): A measure of a member's resistance to rotation. A member fixed at the far end is stiffer ( $4EI/L$ ) than a member pinned at the far end ( $3EI/L$ ).
Symmetry: Utilizing symmetric ( $2EI/L$ ) and anti-symmetric ( $6EI/L$ ) 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 ( $\psi$ ), significantly complicating manual analysis but easily handled by computer matrix methods.

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