Indeterminate Structures: Displacement Methods - Revision Notes
Displacement methods (also called stiffness methods) are the workhorse of modern structural analysis. Unlike force methods that treat redundant forces as unknowns, displacement methods treat joint rotations and displacements (translations) as the primary unknowns. Once these are found, all member-end moments follow by substitution. The two hand-calculation techniques tested in the PRC Civil Engineer Licensure Examination are the Slope-Deflection Method and the Moment Distribution Method. Both begin from the same foundation: Fixed-End Moments (FEM). Mastering FEMs, the slope-deflection equation, and the moment distribution procedure is non-negotiable for exam success.
Fixed-End Moments (FEM) — The Essential Starting Point
Formulas
FEM_AB = −wL²/12 ; FEM_BA = +wL²/12
Example
w = 10 kN/m, L = 6 m → FEM_AB = −10(6)²/12 = −30 kN·m; FEM_BA = +30 kN·m
Variables
w = uniformly distributed load (kN/m), L = span length (m)
Application
Full-span UDL on a fixed-fixed beam
FEM_AB = −PL/8 ; FEM_BA = +PL/8
Example
P = 48 kN, L = 6 m → FEM_AB = −48(6)/8 = −36 kN·m; FEM_BA = +36 kN·m
Variables
P = concentrated midspan load (kN), L = span length (m)
Application
Central point load on a fixed-fixed beam
FEM_AB = −Pab²/L² ; FEM_BA = +Pa²b/L²
Example
P = 40 kN, a = 2 m, b = 4 m, L = 6 m → FEM_AB = −40(2)(4²)/36 = −35.6 kN·m; FEM_BA = +40(4)(2²)/36 = +17.8 kN·m
Variables
P = concentrated load (kN), a = distance from A to load (m), b = distance from load to B (m), L = span (m), a + b = L
Application
Off-center point load on a fixed-fixed beam
FEM_AB = −wL²/30 (zero-load end A) ; FEM_BA = +wL²/20 (peak-load end B)
Example
w = 12 kN/m, L = 5 m → FEM_AB = −12(25)/30 = −10 kN·m; FEM_BA = +12(25)/20 = +15 kN·m
Variables
w = peak intensity of triangular load (kN/m), L = span (m)
Application
Triangular load increasing from zero at A to peak w at B
Exam tips
- Memorize the four FEM formulas in the table — they appear directly or as sub-steps in nearly every displacement-method board problem.
- For a UDL: FEM = wL²/12. For a central P: FEM = PL/8. These two are the most frequently tested.
- Always write down the FEM table as your very first step before touching any slope-deflection or distribution equation.
- Quick check: the sum of |FEM_AB| + |FEM_BA| for an off-center P equals P·a·b/L (the product PL/4 only for midspan). Use this to catch arithmetic errors.
Key points
- FEMs are the moments at the ends of a beam that is assumed to be FULLY FIXED (both ends restrained against rotation and translation) under a given applied load.
- Sign convention (clockwise-on-member = positive) must be adopted consistently throughout a problem — mixing conventions is the #1 source of errors.
- For a UDL w over full span L: FEM_AB = −wL²/12 (near end, counterclockwise = negative) and FEM_BA = +wL²/12 (far end, clockwise = positive).
- For a central point load P: FEM_AB = −PL/8 and FEM_BA = +PL/8.
- For an off-center point load P at distance a from A and b from B (a + b = L): FEM_AB = −Pab²/L² and FEM_BA = +Pa²b/L².
- For a triangular load (zero at A, peak w at B): FEM_AB = −wL²/30 and FEM_BA = +wL²/20.
- FEM magnitudes are LARGER at the end closer to the load for off-center loads (the far-end FEM has the a² term while the near-end has the b² term — and a < b when load is near A).
- When no load acts on a span, all FEMs for that span are zero.
Definitions
Fixed-End Moment (FEM)
The moment reaction at the end of a fully restrained (fixed-fixed) beam member due to applied loads, before any joint rotation is permitted. Denoted FEM_AB (moment at near end A) and FEM_BA (moment at far end B).
Importance
FEMs are the mandatory first step in both slope-deflection and moment distribution methods. Incorrect FEMs invalidate all subsequent calculations.
Clockwise-Positive Sign Convention
A moment acting clockwise on the cut face of the member end is taken as positive. This is the most common convention in slope-deflection/moment distribution literature and Philippine review materials.
Importance
Consistency in sign convention is critical. A single sign error at the FEM stage will propagate through every subsequent equation.
Common mistakes
- Swapping the formulas for FEM_AB and FEM_BA on an off-center load — remember: FEM at the near end (A) carries b², FEM at the far end (B) carries a².
- Forgetting to assign negative sign to FEM_AB and positive to FEM_BA for standard downward loads with clockwise-positive convention.
- Using the wrong formula when only part of the span is loaded — always identify the load type and the exact distances a and b.
- Not zeroing out FEMs for unloaded spans — FEM = 0 for every member end that carries no load on that span.
- Confusing the triangular-load FEMs: the LARGER magnitude (wL²/20) goes at the PEAK-LOAD end.
Slope-Deflection Method
Formulas
M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB
Example
Span AB, L = 6 m, uniform EI, UDL w = 10 kN/m, B is an interior joint, A is a pin (θ_A free, M_AB = 0). Setting M_AB = 0: (2EI/6)(2θ_A + θ_B) − wL²/12 = 0. If by symmetry θ_B = 0, then θ_A = wL³/(48EI).
Variables
M_AB = moment at near end A of member AB (kN·m); E = modulus of elasticity (MPa or kN/m²); I = moment of inertia (m⁴); L = member length (m); θ_A, θ_B = joint rotations at A and B (rad), positive counterclockwise; ψ = chord rotation = Δ/L (rad); FEM_AB = fixed-end moment at A (kN·m)
Application
Write one equation per member end. Two equations per member (near and far). Then apply boundary conditions and joint equilibrium to solve.
ψ = Δ/L
Example
Support B settles 12 mm relative to A, L = 6 m → ψ = 0.012/6 = 0.002 rad
Variables
ψ = chord rotation (rad); Δ = relative transverse displacement of far end relative to near end (m); L = member length (m)
Application
Include ψ whenever a support settles or the frame sways. For settlement at B of amount δ relative to A: ψ = δ/L (positive if B settles downward relative to A with the common sign convention).
M_near = (3EI/L)(θ_near − ψ) + FEM_modified
Example
Propped cantilever AB (A = fixed, B = pin), UDL w: M_AB = (3EI/L)θ_A + (−wL²/12 + wL²/24) = (3EI/L)θ_A − wL²/8. Setting joint A moment equation gives M_A.
Variables
Modified slope-deflection equation used when the FAR end of the member is pinned (M_far = 0). FEM_modified = FEM_near − (1/2)FEM_far for the standard case.
Application
Use this to eliminate the θ of the far-pinned end directly, reducing the number of equations.
Exam tips
- Systematic procedure: (1) Identify unknowns (θ at each rigid joint; ψ if settlement/sway exists). (2) Write slope-deflection equations for ALL member ends. (3) Apply BCs (fixed → θ = 0; pin → M = 0). (4) Write ΣM = 0 at each unknown-θ joint. (5) Solve the system. (6) Back-substitute for moments.
- For a two-span symmetric beam under symmetric load, exploit symmetry: θ at the center support = 0 by symmetry, reducing to a single equation.
- Board exams often ask for ONE specific moment (usually the interior support moment). Practice solving only for the unknowns needed to find that moment.
- Always verify your answer using an alternative method (e.g., moment distribution or three-moment equation) if time permits — a quick check saves points.
- In a propped cantilever (fixed one end, pin the other), use the modified 3EI/L equation to solve in a single step without introducing θ at the pinned end.
Key points
- The slope-deflection method expresses every member-end moment as a function of the joint rotations (θ) and chord rotation (ψ = Δ/L), plus the FEM.
- The fundamental slope-deflection equation is: M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB
- A corresponding equation is written for the far end: M_BA = (2EI/L)(2θ_B + θ_A − 3ψ) + FEM_BA
- Unknowns are the joint rotations θ and any chord rotation ψ due to settlement or sidesway.
- Boundary conditions at supports constrain the unknowns: a fixed support has θ = 0; a simple (pin/roller) support has M = 0 (use this to eliminate θ at that end).
- Joint equilibrium: at every rigid interior joint, the algebraic sum of all member-end moments meeting at that joint equals zero (ΣM_joint = 0).
- The equilibrium equations form a system of simultaneous equations in the unknown θ values — solve by substitution or matrix methods.
- After solving for θ, back-substitute into each slope-deflection equation to get the final member-end moments.
- Chord rotation ψ = Δ/L is included when a support settles by Δ or when a frame experiences sidesway Δ. For no settlement and no sidesway, ψ = 0.
- For a near end that is a pin (simple support): use the modified equation M_AB = (3EI/L)(θ_A − ψ) + FEM_AB_modified, where the modified FEM accounts for the released far end.
Definitions
Joint Rotation (θ)
The angle through which a rigid joint rotates from its original (unloaded) position. In the slope-deflection method, θ is positive counterclockwise (consistent with beam theory sign convention in most Philippine review books).
Importance
θ values are the PRIMARY unknowns in slope-deflection analysis. Every member-end moment is expressed through θ.
Chord Rotation (ψ)
The angle that the straight line connecting the two ends of a deformed member makes with its original (undeformed) axis. ψ = Δ/L where Δ is the relative settlement or sway displacement.
Importance
Neglecting ψ when settlement or sidesway exists leads to completely wrong moments. Always check if the problem mentions support settlement or lateral loads on frames.
Joint Equilibrium
The condition that the algebraic sum of all member-end moments at a rigid joint equals zero (plus any applied external moment at that joint): ΣM = 0.
Importance
This is the equation that allows you to solve for the unknown θ values after writing all slope-deflection equations.
Common mistakes
- Writing the slope-deflection equation with θ_A and θ_B in the wrong positions — the near-end rotation is always multiplied by 2, the far-end by 1.
- Forgetting the factor of 2 outside: the equation is (2EI/L)(2θ_near + θ_far − 3ψ), not (EI/L)(2θ_near + θ_far − 3ψ).
- Setting θ = 0 at a pin/roller support — this is WRONG. A pin support has M = 0 (zero moment), not zero rotation. θ is generally non-zero at a simple support.
- Omitting ψ in settlement problems — if a support settles, ψ must appear in all equations for that member.
- Forgetting to apply joint equilibrium after writing all member-end moment equations — the equilibrium equations ARE the equations you solve for θ.
- Not including applied external moments at joints in the equilibrium equation — ΣM_joint = 0 includes both member-end moments AND any external moment at the joint.
Moment Distribution Method
Formulas
DF_i = k_i / Σk where k = 4EI/L (far end fixed) or k = 3EI/L (far end pinned)
Example
Joint B with two equal spans of equal EI, L = 6 m, both far ends fixed: k_BA = k_BC = 4EI/6. DF_BA = DF_BC = (4EI/6)/[(4EI/6)+(4EI/6)] = 0.5 each. Sum = 1.0 ✓
Variables
DF_i = distribution factor for member i at the joint; k_i = stiffness of member i; Σk = sum of stiffnesses of all members meeting at that joint
Application
Compute one DF per member at every interior joint. The DFs at any joint must sum to 1.0 (use this as a check).
ΔM_i = −DF_i × M_unbalanced
Example
Unbalanced moment at B = +36 kN·m, DF_BA = DF_BC = 0.5 → ΔM_BA = −0.5(36) = −18 kN·m; ΔM_BC = −0.5(36) = −18 kN·m
Variables
ΔM_i = distributed moment to member i; DF_i = distribution factor of member i; M_unbalanced = algebraic sum of all moments currently at the joint (the imbalance to be corrected)
Application
Apply at every cycle of iteration. The negative sign ensures the distributed moment opposes the unbalanced moment, bringing the joint into equilibrium.
Carry-over = (1/2) × ΔM_distributed
Example
ΔM_BA = −18 kN·m → Carry-over to A: CO_A = (1/2)(−18) = −9 kN·m (added to M_AB column)
Variables
Carry-over factor = 1/2 for standard prismatic members with far end fixed or continuous; carry-over = 0 when far end is pinned and modified stiffness 3EI/L is used
Application
After distributing at joint B, carry half of each distributed moment to the respective far end of each member.
M_final_end = FEM + Σ(all distributed and carried-over moments at that end)
Example
M_AB_final = −36 (FEM) + (−9) (carry-over from first cycle) = −45 kN·m
Variables
M_final_end = final member-end moment (kN·m); FEM = fixed-end moment; Σ = sum of all distributed and carried-over increments accumulated at that end across all cycles
Application
Summing each column of the moment distribution table gives the final moment at each member end.
Exam tips
- Set up a neat tabular format: rows for FEM, Distribution Cycle 1, Carry-Over 1, Distribution Cycle 2, Carry-Over 2, etc.; columns for each member end. This prevents losing track of values.
- For board exam problems with fixed ends at the terminals and a single interior joint, one complete cycle (distribute + carry-over) followed by a check of the joint often gives the exact answer.
- If a terminal support is a pin, use modified stiffness (3EI/L) from the start — this converts an infinite iteration to a single-pass calculation for simple beams.
- The DF check: sum of all DFs at every joint MUST equal 1.0. If not, recompute before proceeding.
- For symmetric beams under symmetric loads, only analyze HALF the structure. The interior joint moment can often be found without any iteration by recognizing that the unbalanced moment distributes equally.
- Board problems commonly ask: find the moment at support B. Set up the MD table, perform 1–2 cycles, and read M_B from the column sums.
- After obtaining final moments, use free-body diagrams of individual spans to compute reactions and shear — this is often the follow-up question.
Key points
- Moment distribution is an iterative relaxation method — it reaches the exact slope-deflection solution as the number of cycles approaches infinity, but converges rapidly in practice.
- The method imagines all joints initially LOCKED (fixed), computes the FEMs, then UNLOCKS one joint at a time and distributes the unbalanced moment.
- The key quantities are: Stiffness (k), Distribution Factor (DF), Carry-Over Factor (COF = 1/2 for prismatic members with far end fixed), and the FEMs.
- Stiffness k = 4EI/L when the far end is fixed (or is a continuous interior joint).
- Stiffness k = 3EI/L when the far end is a pin or roller (free to rotate). Using this modified stiffness eliminates carry-overs to the pinned end, speeding up convergence.
- Distribution Factor at a joint: DF_member = k_member / Σk_all members at that joint.
- At a FIXED end: DF = 0 (the fixed support absorbs moments without distributing — it can NEVER rotate).
- At a FREE end (pin/roller): DF = 1 if using modified stiffness, OR the joint is balanced to zero in a single step.
- Carry-Over: After balancing a joint, half of the distributed moment is carried over (sent) to the far end of each member at that joint. The carry-over factor is 1/2 for standard prismatic members.
- When the far end is a pin and you use k = 3EI/L, carry-over to that pinned end is ZERO — this is the advantage of the modified stiffness approach.
- The process: Compute FEMs → Lock all joints → At each joint, sum all FEMs (unbalanced moment) → Distribute negative of unbalanced moment using DFs → Carry over 1/2 to far ends → Repeat at every joint until moments are negligibly small → Sum all values in each column for the final end moment.
- Iteration typically converges in 2–4 cycles for most board exam problems. The board exam may ask you to perform only one or two cycles.
Definitions
Stiffness (k)
A measure of a member's resistance to rotation at its near end when the far end is fixed. k = 4EI/L for a fixed far end; k = 3EI/L for a pinned far end. Higher EI or shorter L = stiffer member.
Importance
Stiffness determines how much moment each member attracts when a joint is released. Stiffer members attract more moment.
Distribution Factor (DF)
The fraction of the unbalanced moment at a joint that is distributed to a particular member. DF_member = k_member / Σk_joint. Always between 0 and 1. Sum of all DFs at a joint = 1.0.
Importance
DFs are computed once at the start of the problem and used throughout all iterations. An error in DF propagates through every cycle.
Carry-Over Factor (COF)
The fraction of the distributed moment at the near end that is transmitted to the far end. For a standard prismatic member with the far end fixed, COF = +1/2. The carry-over is always to the FAR end of the member whose near end was just balanced.
Importance
Carry-over is what causes the iterative nature of moment distribution — balancing one joint disturbs adjacent joints, requiring further iterations.
Unbalanced Moment
The net moment at a joint that violates equilibrium. In the first cycle, the unbalanced moment at a joint equals the algebraic sum of all FEMs of members meeting there. In subsequent cycles, it includes all accumulated carry-overs.
Importance
The unbalanced moment is what is distributed (with a negative sign) in each balancing step. Convergence is complete when all unbalanced moments are negligibly small.
Modified Stiffness
When the far end of a member is a pin (simple support), the stiffness is reduced to k = 3EI/L instead of 4EI/L, and the carry-over factor to that pinned end becomes zero. This is a one-time adjustment that eliminates the need to carry over moments to the pinned end in all subsequent cycles.
Importance
Using modified stiffness for pin-ended spans dramatically speeds up the moment distribution process and is standard practice for board-level problems.
Common mistakes
- Forgetting that DF at a FIXED support = 0 (the fixed support cannot rotate, so it attracts moment but distributes none).
- Distributing POSITIVE unbalanced moment instead of NEGATIVE — the distribution OPPOSES the unbalance: ΔM = −DF × M_unbalanced.
- Carrying over to the NEAR end instead of the FAR end — carry-over always goes to the OTHER end of the member.
- Applying carry-over factor of 1/2 to a pinned far end when using modified stiffness 3EI/L — with modified stiffness, carry-over to the pin = 0.
- Not carrying over to fixed far ends — fixed ends DO receive carry-overs (they are locked and cannot balance themselves, so carry-overs accumulate there).
- Checking joint equilibrium by summing FEMs only, ignoring subsequent carry-overs — always check the FINAL moments at each joint sum to zero.
- Using k = 4EI/L for a pinned far end when this modification would simplify the problem significantly.
- Forgetting to carry over in BOTH directions from a balanced joint — if joint B is balanced, carry-overs go to A (via member BA) AND to C (via member BC).
Comparison of Methods and Choice of Approach
Formulas
DKI = r_j + r_s
Example
A two-span continuous beam (fixed at both ends, one interior pin joint): r_j = 1 (interior joint B can rotate), r_s = 0 (no sway). DKI = 1.
Variables
DKI = degree of kinematic indeterminacy; r_j = number of unknown joint rotations (one per rigid joint that is not fully fixed); r_s = number of independent sway (translation) DOFs
Application
Used to count how many slope-deflection equations (and equilibrium equations) you need to set up before solving.
Exam tips
- On the board exam, if the structure has only 1 unknown rotation (one interior joint, no sway), slope-deflection gives the answer in 2–3 lines. Use it for speed.
- If the problem says 'by moment distribution,' set up the table carefully — partial credit may be awarded for correct FEMs and DFs even if arithmetic errors occur later.
- Know when a frame HAS sidesway vs. does NOT: symmetric frame + symmetric load = NO sidesway; everything else = assume sidesway unless proven otherwise.
- The PRC board exam typically tests: (a) a two-span beam by moment distribution, (b) a propped cantilever or single-span beam by slope-deflection, or (c) a fixed-end moment calculation. All three should be practiced thoroughly.
Key points
- Slope-deflection gives exact closed-form results in a single solution step but requires solving simultaneous equations — practical for structures with 1–3 unknowns.
- Moment distribution is iterative but avoids solving simultaneous equations — preferred by hand for structures with many joints.
- Both methods yield IDENTICAL final moments if executed correctly — they are simply different mathematical paths to the same answer.
- For a two-span beam with one interior joint: either method works equally well. Slope-deflection needs 1 equation; moment distribution needs 1 balancing cycle.
- For a multi-story frame with sidesway: slope-deflection requires setting up shear equations (additional equilibrium) alongside the rotation equations. Moment distribution requires the Hardy Cross sway correction procedure.
- Both methods are precursors to the matrix stiffness method (finite element analysis) — understanding them gives insight into how computer structural software works.
- The degree of kinematic indeterminacy (DKI) determines the number of unknowns in the slope-deflection method. DKI = number of independent joint DOFs (rotations + translations).
- For frames with sidesway: identify the number of independent sway mechanisms (story mechanism count) — each adds one ψ unknown to the system.
Definitions
Degree of Kinematic Indeterminacy (DKI)
The number of independent displacement unknowns (joint rotations and/or translations) that must be solved in a displacement method analysis. Equal to the number of equations needed.
Importance
Determines the size of the system of equations in the slope-deflection method. High DKI favors iterative moment distribution over direct solution.
Sidesway
The lateral translation (horizontal displacement) of the joints in a frame, occurring when the frame or its loading is not symmetric. Sidesway introduces chord rotations ψ in all columns and requires an additional equilibrium equation (shear equation) to solve.
Importance
Board exam portal frame problems often include sidesway. Recognize sidesway by: unsymmetric frame, unsymmetric loading, or presence of horizontal loads.
Common mistakes
- Assuming no sidesway in a portal frame simply because there is no horizontal load — unsymmetric geometry or loading causes sidesway even under vertical loads.
- Applying moment distribution to a frame with sidesway without the sway correction — the standard (no-sway) distribution must be followed by a separate sway analysis.
- Not recognizing that both methods give the same answer — students sometimes doubt their slope-deflection result and redo in moment distribution, getting the same answer, and thinking they made an error both times.
Connections
- Fixed-End Moments → Slope-Deflection Method: FEMs are the load terms in the slope-deflection equation. Without correct FEMs, no displacement method analysis can be started.
- Slope-Deflection Method → Moment Distribution Method: Moment distribution is a numerical iterative technique that converges to the exact slope-deflection solution. The two methods are mathematically equivalent.
- Displacement Methods → Three-Moment Equation (Clapeyron): The three-moment equation (a force method) gives the same support moments for continuous beams as slope-deflection/moment distribution — useful as a cross-check.
- Moment Distribution → Matrix Stiffness Method (FEM): The distribution factor concept generalizes to the stiffness matrix. Understanding moment distribution builds intuition for reading structural software output.
- Slope-Deflection Equation → Beam Deflection Theory: The equation derives from the elastic curve equation (EI d²y/dx² = M) integrated with fixed boundary conditions — connects to conjugate beam and moment-area methods.
- Chord Rotation (ψ) → Support Settlement Problems: Any settlement problem from the previous chapter (using the three-moment equation) can also be solved by slope-deflection with the ψ = Δ/L term — same physics, different method.
- Stiffness (k) → Relative Stiffness and Load Distribution: The concept of stiffness-proportional load distribution appears in grid analysis, pile group analysis, and seismic base shear distribution — all tested in advanced structures problems.
- Member-End Moments → Shear Force and Bending Moment Diagrams: Once member-end moments are found by either displacement method, free-body diagrams of individual spans give reactions, shears, and the complete BMD — frequently the follow-up question on the board exam.
- Degree of Kinematic Indeterminacy → Degree of Static Indeterminacy: DKI and DSI are related by DKI = m·(n+1) − r for specific frame types; knowing both helps choose the more efficient method for a given structure.
- Moment Distribution → Building Frame Analysis (NSCP 2015 Sec. 406): NSCP 2015 Section 406 (ACI 318-14 Ch. 6) permits approximate analysis of regular frames using simplified moment coefficients — these coefficients are derived from moment distribution results for idealized conditions.
Quick review questions
A fixed-fixed beam of span L = 8 m carries a full-span UDL of w = 15 kN/m. What are FEM_AB and FEM_BA?
Answer
FEM_AB = −80 kN·m; FEM_BA = +80 kN·m
Using FEM = wL²/12: FEM = 15(8²)/12 = 15(64)/12 = 80 kN·m. FEM_AB is negative (counterclockwise at A) and FEM_BA is positive (clockwise at B) by the clockwise-positive convention.
A concentrated load P = 60 kN is applied at a = 3 m from A on a fixed-fixed beam of span L = 9 m (b = 6 m). Calculate FEM_AB and FEM_BA.
Answer
FEM_AB = −80 kN·m; FEM_BA = +40 kN·m
FEM_AB = −Pab²/L² = −60(3)(6²)/81 = −60(3)(36)/81 = −6480/81 = −80 kN·m. FEM_BA = +Pa²b/L² = +60(3²)(6)/81 = +60(9)(6)/81 = +3240/81 = +40 kN·m. Note: |FEM_AB| > |FEM_BA| because the load is closer to A (a < b).
In the moment distribution method, what is the distribution factor at a fixed end support, and why?
Answer
DF = 0 at a fixed end because the fixed support has infinite stiffness compared to the members — it cannot rotate and therefore does not distribute moments.
Although the fixed support provides a reaction moment (and receives carry-overs from adjacent joints), it is never 'balanced' — it simply holds whatever moment is required. Mathematically, if we consider the fixed support as a member with k → ∞, its DF → 1 and adjacent members' DFs → 0; but from the joint's perspective, the fixed support IS the joint boundary with DF = 0 for distribution OUT from the support.
For a two-span continuous beam — fixed at A, continuous at B, fixed at C — with equal spans L = 6 m, equal EI, and UDL w = 12 kN/m on span AB only. Using moment distribution (one full cycle), what is M_B (the moment at the interior joint)?
Answer
M_B = M_BA + M_BC = 18 + (−18) = 0 at the balanced joint; the moments in the MEMBERS are M_BA = +18 kN·m and M_BC = −18 kN·m.
FEMs: FEM_AB = −36, FEM_BA = +36, FEM_BC = FEM_CB = 0 kN·m. At joint B: k_BA = k_BC = 4EI/6. DF_BA = DF_BC = 0.5. Unbalanced = FEM_BA + FEM_BC = 36 + 0 = +36. Distribute −36: ΔM_BA = −18, ΔM_BC = −18. Final: M_BA = 36 − 18 = +18 kN·m; M_BC = 0 − 18 = −18 kN·m. Check: M_BA + M_BC = 18 − 18 = 0 ✓ (joint B is in equilibrium).
State the slope-deflection equation for M_AB and identify what each term represents physically.
Answer
M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB. The term (2EI/L)(2θ_A) represents the moment due to rotation of joint A (near end, double contribution); (2EI/L)(θ_B) is the moment induced by rotation of joint B (far end, single contribution); (2EI/L)(−3ψ) is the moment correction for chord rotation (settlement or sway); FEM_AB is the moment due to loads on the fixed-fixed member.
The factors of 2 and 1 for near and far end rotations arise from the fixed-end stiffness derivation using the unit displacement method. The chord rotation term −3ψ reflects that relative settlement of one end introduces a hogging (negative) effect. This equation is the cornerstone of the slope-deflection method.
Why is k = 3EI/L used instead of 4EI/L when the far end is pinned in moment distribution?
Answer
Because when the far end is a pin (free to rotate, M = 0), the member is less stiff in resisting near-end rotation. The reduced stiffness 3EI/L accounts for the released far-end condition and automatically makes the carry-over factor to that end zero.
Derivation: For a beam with near end restrained and far end pinned, applying unit rotation at the near end and enforcing M = 0 at the far end gives a near-end moment of 3EI/L (not 4EI/L). Using this modified stiffness means no carry-over is sent to the pinned end, which eliminates one source of iteration and speeds up convergence. This is a standard board-exam simplification.
A propped cantilever (fixed at A, pin at B) has span L = 5 m, EI = constant, and carries a UDL w = 8 kN/m. Using the slope-deflection method, find the fixed-end moment M_A.
Answer
M_A = −3wL²/8 = −3(8)(25)/8 = −75 kN·m (75 kN·m hogging at A)
FEMs: FEM_AB = −wL²/12 = −8(25)/12 = −16.67 kN·m; FEM_BA = +wL²/12 = +16.67 kN·m. Slope-deflection equations (ψ = 0, θ_A = 0 since A is fixed): M_AB = (2EI/5)(2(0) + θ_B) − 16.67 = (2EI/5)θ_B − 16.67; M_BA = (2EI/5)(2θ_B + 0) + 16.67 = (4EI/5)θ_B + 16.67. BC at B (pin): M_BA = 0 → (4EI/5)θ_B = −16.67 → θ_B = −16.67(5)/(4EI) = −20.83/EI. Back-substitute into M_AB: M_AB = (2EI/5)(−20.83/EI) − 16.67 = −8.33 − 16.67 = −25 kN·m. [Alternative direct formula: M_A = −3wL²/8 = −3(8)(25)/8 = −75 kN·m — Note: this formula applies for the fixed-end moment with pin at B, not the slope-deflection result. The slope-deflection result M_AB = −25 kN·m is the final fixed-end moment at A for this propped cantilever, not the FEM of the fully-fixed beam. Verification: for a propped cantilever with UDL, M_A = wL²/8 = 8(25)/8 = 25 kN·m ✓]
What condition determines whether a frame undergoes sidesway? Give two examples.
Answer
A frame undergoes sidesway when the frame geometry or loading is UNSYMMETRIC, or when horizontal loads are applied. Example 1: A portal frame with columns of unequal height or stiffness under vertical load will sway. Example 2: A symmetric portal frame subjected to a horizontal wind load will sway in the direction of the load.
Sidesway occurs because the lateral resistance of the columns is unequal or because an external horizontal force directly induces lateral displacement. In slope-deflection analysis, sidesway introduces an additional unknown ψ for each independent story, requiring an additional equation from the shear condition: the sum of column shears at a story must equal the total horizontal force at that level.
Diagrams
Chapter Overview: Displacement Methods for Indeterminate Structures
A comprehensive mind map showing how Fixed-End Moments, Slope-Deflection Method, and Moment Distribution Method are interrelated, with all key sub-concepts and formulas organized hierarchically.
Slope-Deflection Method: Step-by-Step Procedure
A decision flowchart showing the complete systematic procedure for the slope-deflection method, from identifying unknowns through applying boundary conditions and joint equilibrium to solving and verifying the final moments.
Moment Distribution Method: Step-by-Step Procedure
A decision flowchart showing the complete iterative procedure for the moment distribution method, including the choice of stiffness (4EI/L vs 3EI/L), the balancing and carry-over steps, the convergence check, and the final summation.
Fixed-End Moment Selection Guide
A decision flowchart that guides the student to the correct fixed-end moment formula based on the type of load applied to the span, reinforcing the sign convention (near end negative, far end positive for downward loads).
Moment Distribution Iteration States
A state diagram illustrating the iterative cycle of moment distribution: from the initially locked state through repeated balancing and carry-over cycles until convergence, followed by the final summation step.
Development of Displacement Methods in Structural Analysis
A timeline showing the historical development from the Three-Moment Equation through the Slope-Deflection Method and Moment Distribution Method, leading to modern matrix and finite element methods, grounding the classical methods in their engineering context.
Exam strategy
For the PRC Civil Engineer Licensure Examination, Structural Theory & Analysis problems on displacement methods almost always fall into one of three patterns: (1) A two-span or three-span continuous beam asking for the support moment, solved most efficiently by moment distribution in 1–2 cycles; (2) A propped cantilever or single-span fixed beam with an off-center load, requiring the FEM formulas and possibly the slope-deflection equation; (3) A portal frame with or without sidesway. Your attack plan: ALWAYS start by writing down all FEMs — this earns partial credit and prevents downstream errors. For beams, use moment distribution (faster and less error-prone than simultaneous equations). For frames with one unknown rotation and no sway, use slope-deflection (1 equation, 1 unknown). Check your final moments using joint equilibrium (all moments at a joint must sum to zero) and global equilibrium (sum of all support reactions must equal total applied load). Time allocation: FEM computation 2 minutes, method application 5–8 minutes, verification 2 minutes. Do not spend more than 12 minutes on any single displacement-method problem — if you are stuck, move on and return. Practice at least 10 complete worked examples of each method before the exam.