CELE Structural Theory & Analysis — Indeterminate Structures: Displacement MethodsSummary

In force (flexibility) methods, unknown internal forces and reactions are found first by satisfying compatibility of deformations; member forces are then determined from equilibrium. In displacement (stiffness) methods, unknown joint displacements and rotations are found first by satisfying equilibrium at each joint; member-end moments are then expressed in terms of these displacements using the member stiffness relationships. The slope-deflection equation M = (2EI/L)(2θ_near + θ_far − 3ψ) + FEM exemplifies this: moment is written directly as a linear function of rotations. Displacement methods are superior for computer implementation because the stiffness matrix is symmetric, positive-definite, and banded—properties that make large sparse systems computationally efficient.

The moments that develop at the ends of a fully restrained (fixed-fixed) beam when external loads are applied, assuming no rotation or translation occurs. FEMs are always computed first, then incorporated into the slope-deflection or moment distribution equations. The convention used throughout this course is clockwise moments on the member-free body diagram taken as positive. For a UDL w over span L: FEM_AB = −wL²/12 (negative at the left, opposing the load's tendency to rotate the member clockwise), and FEM_BA = +wL²/12 (positive at the right). For a central point load P: FEM_AB = −PL/8 and FEM_BA = +PL/8. For an off-center load P at distance a from A and b from B: FEM_AB = −Pab²/L² and FEM_BA = +Pa²b/L². Memorizing this table (or deriving values quickly) is non-negotiable for board success.

The foundational relationship relating member-end moment to end rotations and chord rotation: M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB, where θ_A and θ_B are the rotations at ends A and B respectively (positive counterclockwise), ψ = Δ/L is the chord rotation due to relative displacement Δ perpendicular to the member, and FEM_AB is the fixed-end moment. The coefficients 2 and 1 on the near and far rotations reflect the relative stiffness contributions—the near end controls the moment more strongly. The −3ψ term accounts for chord rotation (common in frames with sidesway). To solve a problem: (1) write this equation for every member end, (2) use support boundary conditions (e.g., M_pin = 0 at a pinned support, or M = 0 at a cantilever tip), (3) impose joint equilibrium ΣM = 0 at each rigid joint where members meet, (4) solve the resulting system of linear equations for unknown rotations, (5) back-substitute to find all member moments.

An elegant hand-calculation procedure that avoids solving simultaneous equations by iteratively balancing joint moments. The method works as follows: (1) Compute FEMs for all members. (2) At each interior joint, calculate the distribution factor DF = k/(Σk) for each member, where k is the relative stiffness of that member. Standard stiffness k = 4EI/L for a member with both ends free to rotate (or one end fixed, one continuous); use k = 3EI/L when the far end is pinned (it cannot receive carry-over moments). (3) At the first joint with unbalanced moment M_unbal = −(ΣM of all member ends meeting at that joint), distribute it to each member in proportion to its DF. Each member receives a distributed moment ΔM = DF × M_unbal. (4) Carry half of each distributed moment to the opposite end of that member (carry-over factor = 0.5), unless the far end is a pin (in which case no carry-over). (5) Move to the next joint and repeat. Continue iterating until unbalanced moments are negligible (usually 3–5 cycles for hand work). (6) Sum the column under each member end to find the final moment. The method is self-correcting and converges monotonically—a major advantage over setting up and solving large systems by hand.

At any joint where multiple members meet, the moment carried by each member is proportional to its relative stiffness. Stiffness k = (moment required to produce unit rotation). For a member with length L and bending stiffness EI: if both ends are free to rotate or one is fixed and the other continuous, k = 4EI/L; if the far end is pinned (rotation occurs but far end moment must be zero), k = 3EI/L; if the far end is fixed, the member cannot contribute to rotation at the joint (it is a support reaction, not a frame member, so it is not included in ΣK). The distribution factor is DF_i = k_i / (Σk) for each member i meeting at that joint. A member with larger stiffness (shorter L, larger EI, or both ends continuous) receives a larger share of the unbalanced moment and thus exerts greater control over the joint equilibrium. A pinned end typically has DF approaching 1 at its near end because the reduced stiffness of 3EI/L compared to 4EI/L reflects that less moment is needed to achieve the same rotation.

When moment ΔM is distributed to a member at one end, half of it appears at the opposite end of that member. This is the carry-over coefficient = 0.5. For example, if 20 kN·m is distributed to member AB at joint A, then 10 kN·m is carried over to the opposite end B of that same member. The carry-over represents the fact that a rotation at A induces a moment at B proportional to the slope-deflection equation (θ_B term has coefficient 1, half of the coefficient 2 on θ_A). Carry-over does NOT occur to a pinned end (since a pin cannot develop a moment). After carry-over, the far joint becomes unbalanced, so it is balanced in the next iteration. This iterative process continues until the unbalanced moments fall below a threshold (often 1–5% of the largest moment in the structure). The order of balancing does not matter (the method converges regardless), and convergence is usually fast—typically 3–5 iterations for hand calculation.

Chord rotation ψ = Δ/L occurs when the two ends of a member displace perpendicular to the original axis, changing the slope of the member centerline. In a continuous beam with no settlement, ψ = 0 for all members. However, if support B settles by Δ relative to support A, the chord of member AB rotates by ψ = Δ/L, and this affects the moment distribution: each moment is modified by the term −3ψ(2EI/L) in the slope-deflection equation. In frames with sidesway (lateral translation), each column develops a chord rotation equal to the lateral displacement divided by its height. For example, if the top of a frame sways 50 mm and the column height is 4000 mm, then ψ = 0.050/4 = 0.0125 rad. Including this in the slope-deflection equation ensures that lateral deflections and the induced bending moments are correctly calculated. A common pitfall: forgetting to include ψ when the problem states a support settlement or when analyzing a frame under lateral load.

At any rigid joint where members meet, the sum of all member-end moments at that joint must equal zero (no external moment applied at the joint): ΣM = 0. This is the condition that allows you to set up equations when using slope-deflection. At a pinned support, the moment must be zero (M_pin = 0), which provides a constraint equation. At a fixed support, the moment is unknown but the rotation is zero (θ_fixed = 0). At a free (cantilever) end, both moment and rotation are zero. At a joint where an external moment is applied, the equilibrium condition is ΣM = M_applied. These boundary conditions are applied to the slope-deflection equations to reduce the number of unknowns and yield a system solvable for the joint rotations. In moment distribution, boundary conditions are incorporated implicitly: a fixed end receives no carry-over (its DF = 0, so it does not balance), and a pinned end has DF = 1 and receives the full unbalanced moment.

Reinforced concrete floor beams in multi-story buildings are typically continuous over multiple supports. Using moment distribution, an engineer quickly determines negative moments over interior supports and positive moments in spans. These moments then guide the placement of reinforcing steel (top bars over supports, bottom bars in spans) and ensure compliance with NSCP 2015 Section 6 (reinforced concrete) and Section 4 (loads and load combinations). The method is much faster than analyzing each span in isolation.

Rigid-jointed steel or concrete portal frames in industrial buildings or bridge piers experience lateral forces (wind per NSCP 2015 Chapter 4, or seismic per Chapter 5). The frame sways, inducing chord rotations ψ in the columns. Using slope-deflection with non-zero ψ, or moment distribution with explicit accounting for sidesway, the engineer determines column moments (which govern bending stresses and required section capacity per AISC 360) and base shear forces. Displacement methods handle this systematically without guessing the form of deformations.

A cantilever beam with a prop (simple support) at its free end is a common bridge or building element. If the prop support settles (common in soil-structure problems or when support plates compress), the settlement Δ creates a chord rotation ψ = Δ/L in the cantilever span, which modifies all bending moments. Using slope-deflection with ψ non-zero, the engineer calculates the true moment distribution and verifies that the structure remains safe. This application highlights the importance of including the −3ψ term.

ACI 318-19 allows moment redistribution in continuous reinforced concrete members when ductility is assured (section curvature capacity exceeds yield curvature by a prescribed amount). Displacement methods provide the elastic moments; ACI limits allow up to 10–20% redistribution to reduce peak negative moments and improve practical design. The engineer uses moment distribution to find elastic moments, then applies redistribution factors to optimize reinforcement layout and control cracking.

Professional engineers often use commercial software to analyze complex structures. To verify that the software is set up correctly (supports, loads, material properties) or to catch data entry errors, a hand calculation using moment distribution on a simplified section of the frame provides a quick sanity check. If the software results and hand calculation agree closely, confidence in the model is high. This is a critical professional practice step.

For steel moment-resisting frames, displacement methods determine the moments and shear forces at each member end. These are then used to select member sizes (AISC Section G for shear strength, Section H for bending capacity, Section E for compression) and verify strength limit states. Moment distribution is preferred because it is fast and transparent—the engineer sees where moments are highest and why, aiding in member selection strategy.

When a column-foundation connection experiences an eccentric load (e.g., due to frame moment), the foundation and cap beam behave as an indeterminate structure. Moment distribution on the cap beam and foundation ring determines the distribution of bearing pressure and the induced moments, which inform the foundation design. This is particularly important in Philippine bridge and building design where soil bearing capacity and foundation geometry must be carefully coordinated.