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

Exact Analysis of Indeterminate Structures: Displacement Methods

Using joint rotations and translations as primary unknowns.

In contrast to Force Methods (which use redundant forces as unknowns), Displacement Methods use the displacements (rotations and translations) of the structure's joints as the primary unknown variables. Once the joint displacements are found, the internal member forces can be calculated using equilibrium equations.

Displacement methods are generally preferred for highly indeterminate structures because the number of unknown joint displacements is often much smaller than the number of redundant forces. The Matrix Stiffness Method, the foundation of all modern structural analysis software, is a generalized displacement method.

Core Principle of Displacement Methods

Identify Degrees of Freedom (DOF): Determine the possible independent displacements (rotations $\theta$ and translations $\Delta$ ) at the joints of the structure.
Lock the Joints: Imagine all joints are clamped or "locked" to prevent any displacement.
Calculate Fixed-End Moments (FEM): Apply the actual external loads to these locked members and calculate the moments required at the ends to keep them locked (Fixed-End Moments).
Write Equilibrium Equations: For each degree of freedom, write an equilibrium equation stating that the sum of the moments acting on that joint must be zero. These equations relate the internal member end moments to the unknown joint displacements.
Solve for Displacements: Solve the system of equilibrium equations simultaneously for the unknown joint displacements ( $\theta, \Delta$ ).
Calculate Member Forces: Substitute the known displacements back into the member slope-deflection equations to find the final internal moments and shears.

Key Takeaways

Displacement Methods take joint rotations ( $\theta$ ) and translations ( $\Delta$ ) as the primary unknown variables.
They are highly systematic and often require fewer equations to solve for highly indeterminate structures compared to Force Methods.

The Slope-Deflection Method

Relating end moments to joint rotations and translations.

The fundamental equation of the displacement method relates the moment at the end of a member to its fixed-end moment and the displacements (rotations and translation) of its ends. Before using the method, one must calculate the Fixed-End Moments.

Common Fixed-End Moment (FEM) Formulas

Assuming a member is perfectly fixed at both ends (A and B) with length

L

Uniformly Distributed Load ( $w$ ): $\text{FEM}_{AB} = \frac{wL^2}{12}$ , $\text{FEM}_{BA} = -\frac{wL^2}{12}$
Concentrated Load ( $P$ ) at midspan: $\text{FEM}_{AB} = \frac{PL}{8}$ , $\text{FEM}_{BA} = -\frac{PL}{8}$
Concentrated Load ( $P$ ) at distance $a$ from A and $b$ from B: $\text{FEM}_{AB} = \frac{Pab^2}{L^2}$ , $\text{FEM}_{BA} = -\frac{Pa^2b}{L^2}$

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

Sign Convention

Strict rules for writing slope-deflection equations.

Unlike the standard shear and moment diagram convention, the Slope-Deflection method uses a joint-oriented sign convention:

Moments ( $M, \text{FEM}$ ): Clockwise moments acting on the member ends are positive. Counter-clockwise moments are negative.
Rotations ( $\theta$ ): Clockwise rotations of the joints are positive.
Translation ( $\Delta$ ): A relative displacement that rotates the chord of the member clockwise is positive.

Key Takeaways

The Slope-Deflection method relates internal moments to joint displacements ( $\theta, \Delta$ ) and Fixed-End Moments (FEM).
A strict joint-oriented sign convention (clockwise is positive) must be followed for all moments and rotations.

The Moment Distribution Method

An iterative relaxation technique for solving slope-deflection equations without solving simultaneous equations.

Developed by Hardy Cross in 1930, the Moment Distribution Method is a highly efficient, iterative technique for analyzing continuous beams and non-sway frames. It essentially solves the slope-deflection equations by successive approximations, avoiding the need to solve large sets of simultaneous equations manually.

Key Concepts in Moment Distribution

Fixed-End Moments (FEM): Initial moments assuming all joints are locked.
Stiffness Factor ( $K$ ): The moment required to rotate a specific joint by one radian while the other end is fixed. $K = \frac{4EI}{L}$ for a member far-end fixed.
Modified Stiffness Factor ( $K_{mod}$ ): If the far end of a member is known to be a real pin or roller, its stiffness is reduced, allowing calculation without needing to distribute moments to that pinned end. $K_{mod} = \frac{3EI}{L}$ .
Distribution Factor (DF): When a moment is applied to a joint, it is distributed among the connecting members in proportion to their relative stiffnesses. $\text{DF} = \frac{K}{\sum K_{joint}}$ . Note: The sum of DFs at any joint must equal 1.0. A fixed support has $\text{DF} = 0$ ; a pin/roller end has $\text{DF} = 1.0$ .
Carry-Over Factor (COF): When a moment is applied at one end of a member, a portion of that moment is "carried over" to the other end. If the far end is fixed, the COF is $1/2$ . If the far end is pinned, the COF is $0$ .

Procedure

STEP-BY-STEP

Calculate FEMs: Determine fixed-end moments for all loaded spans.
Calculate Stiffness ( $K$ ) and Distribution Factors (DF): Compute the relative stiffness ( $I/L$ ) for each member at every joint and calculate the DF for each member end. Use $K_{mod}$ for members with pinned far-ends.
Lock Joints: Assume all joints are initially locked against rotation. The moments at the ends are the FEMs.
Release a Joint and Distribute: Select a joint that is unbalanced (sum of moments $\neq 0$ ). "Release" it by applying a balancing moment equal and opposite to the unbalanced moment. Distribute this balancing moment to the connecting members according to their Distribution Factors (DF).
Carry Over: Multiply the distributed moments at that joint by the Carry-Over Factor (usually 1/2) and apply them to the far ends of those members.
Re-lock the Joint: The joint is now balanced. Lock it again.
Repeat: Move to the next unbalanced joint and repeat steps 4-6. Continue this iterative process of releasing, distributing, carrying over, and locking until the unbalanced moments at all joints become negligibly small.
Sum Final Moments: Sum the initial FEMs and all distributed and carry-over moments in each column to find the final member end moments.

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.

Key Takeaways

The Moment Distribution Method is an iterative technique that distributes unbalanced moments at joints based on member stiffnesses until equilibrium is reached.
Key factors include Stiffness ( $K$ ), Distribution Factor (DF), and Carry-Over Factor (COF = 1/2 for fixed far-ends). Using $K_{mod}$ simplifies calculations for pinned ends.

Stiffness, Distribution, and Carry-Over Factors

Stiffness Factor ( $K$ ): The amount of moment required to rotate a joint by one radian. For a far-end fixed member, $K = \frac{4EI}{L}$ . For a far-end pinned member, $K = \frac{3EI}{L}$ .
Distribution Factor ( $DF$ ): The proportion of unbalanced moment distributed to a specific member meeting at a joint. $DF = \frac{K}{\sum K}$ . At a fixed support, $DF = 0$ . At a pin support, $DF = 1$ .
Carry-Over Factor ( $COF$ ): The fraction of distributed moment that "carries over" to the far end of the member. For a straight, prismatic beam, the $COF$ is always $0.5$ towards a fixed end, and $0$ towards a pinned end.

Sway vs. Non-Sway Frames

Handling lateral translation in unbraced frames.

The classification of a frame depends on whether its joints can undergo significant lateral translation (sway).

Non-Sway Frames

Frames in which lateral translation is prevented or severely restricted. This classification requires careful assessment:

Physical Bracing: The frame is tied to a rigid element like a shear wall, elevator core, or diagonal cross-bracing that explicitly restricts lateral movement.
Perfect Symmetry: The frame geometry, material properties, column stiffnesses, AND the applied loading are perfectly symmetric. If the load is even slightly asymmetric, the frame will sway.

In true non-sway frames, the $\Delta$ term in the slope-deflection equations is considered zero. Moment Distribution can be applied directly without needing a separate shear equation.

Sway Frames

Frames where lateral joints can translate. Sway occurs when a frame is unbraced and subjected to:

Lateral loads (wind, earthquake).
Asymmetrical geometry or supports.
Asymmetrical gravity loading.

For sway frames, an unknown translation $\Delta$ is introduced. Analyzing joint rotations alone is insufficient. You must write an additional equilibrium equation for the entire story based on lateral shear ( $\sum F_x = 0$ ) to solve for the unknown sway displacement $\Delta$ .

Key Takeaways

Frames that can translate laterally (sway frames) due to asymmetry or lateral loads require an additional story shear equilibrium equation.
This extra equation is necessary to solve for the additional unknown sway displacement ( $\Delta$ ).

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