Indeterminate Structures: Displacement Methods - Cheat Sheet
Rapid-fire reference for slope-deflection and moment distribution methods. Master FEM tables, stiffness, and distribution factors in the final 30 minutes before the exam.
Fixed-End Moments (FEM) — The Foundation
Formulas
FEM_AB = −wL²/12 (UDL full span)
Meaning
w = UDL (kN/m), L = span length (m). Negative at end A (clockwise by convention).
Watch out
Sign convention: clockwise-on-member = positive. FEM_BA is opposite sign and often half the moment (asymmetric loading). Always confirm your sign convention with the problem statement or diagram.
When to use
Any uniformly distributed load across entire beam length.
FEM_AB = −PL/8 (central point load)
Meaning
P = point load (kN), L = span. Equal and opposite at both ends.
Watch out
If load is NOT central, use the general formula below. This shortcut only works at midspan.
When to use
Symmetric point load at L/2.
FEM_AB = −Pab²/L² ; FEM_BA = +Pa²b/L²
Meaning
a = distance from A to load, b = distance from load to B, L = a + b.
Watch out
Always check that a + b = L. FEM_AB is NEGATIVE (near end), FEM_BA is POSITIVE (far end). Memorize this asymmetry—it's a common PRC trap.
When to use
Any point load at arbitrary distance on a fixed-fixed span.
FEM_AB = −wL²/30 (triangular: zero at A, peak w at B)
Meaning
w = maximum intensity at far end (kN/m), L = span.
Watch out
This is LESS common; memorize the zero-end FEM value. If peak is at A, reverse the sign and swap indices.
When to use
Triangular load with zero at starting end.
Important facts
- FEM sign convention: clockwise-positive on the member element itself.
- The sum of both FEMs (FEM_AB + FEM_BA) equals the total external moment on the span; they are NOT always equal in magnitude.
- FEM tables assume FIXED-FIXED ends; modify for pinned/free ends by clearing moments at those ends.
- Multiple loads: superpose FEMs from each load individually, then sum.
- Temperature and pre-stress also create FEMs; use material handbooks (not usually on PRC exam).
Key definitions
Fixed-End Moment (FEM)
Example
A 6 m beam with 12 kN/m UDL has FEM_AB = −12(6)²/12 = −36 kN·m (clockwise at A).
Moment at member end of a fully fixed beam under external load, with both ends artificially locked (θ = 0).
Chord Rotation (ψ)
Example
Support B settles 50 mm relative to A on a 5 m span: ψ = −0.05/5 = −0.01 rad (negative = clockwise rotation of chord).
Rotation of the member chord due to relative settlement of ends: ψ = Δ/L (rad), where Δ = vertical displacement difference.
Diagrams to know
- Diagrams showing load position (a, b) on a span; label each end clearly.
- Sign convention diagram: clockwise moment on member face = positive (draw arrows).
- Deflected shape vs. fixed-end shape to identify chord rotation direction.
Slope-Deflection Method
Formulas
M_AB = (2EI/L)[2θ_A + θ_B − 3ψ] + FEM_AB
Meaning
M_AB = moment at end A of member AB; E = elastic modulus, I = second moment of inertia, L = span; θ_A, θ_B = end rotations; ψ = chord rotation; FEM_AB = fixed-end moment.
Watch out
The coefficient 2EI/L is the STIFFNESS of the member. Do NOT forget the (−3ψ) term if there is any settlement or sidesway. Include FEM_AB directly into the formula; do not add separately.
When to use
ALWAYS in slope-deflection. One equation per member end (write both M_AB and M_BA for each span).
Stiffness k = 4EI/L (fully fixed far end)
Meaning
Rotational stiffness of a member; used in moment distribution.
Watch out
If the far end is pinned/free, use k = 3EI/L instead. Using the wrong k will make the whole distribution collapse.
When to use
When the far end is rigidly connected (fixed or continuous).
Stiffness k = 3EI/L (pinned or free far end)
Meaning
Reduced stiffness when the far end cannot develop full rotational resistance.
Watch out
Many students forget this modification; check support conditions FIRST before picking k.
When to use
End cantilever, pinned terminal, or frictionless roller.
Important facts
- The slope-deflection method solves directly for ALL rotations by setting up a system of linear equations: (# unknowns) = (# unknown rotations) + (# unknown displacements).
- For a continuous beam with no settlement, the only unknowns are the interior joint rotations.
- Every member contributes TWO equations (one for each end: M_AB and M_BA), but they share the same θ values.
- Once you solve for θ values, substitute back to find all moments. Reactions follow from shear equilibrium.
- Sidesway (lateral displacement of a storey) adds unknown chord rotations ψ (usually one per storey).
Key definitions
Slope-Deflection Equation
Example
For a continuous beam span with no settlement, set ψ = 0 and write M_AB and M_BA for each span.
Linear relationship between member-end moment and end rotations (θ_A, θ_B) and chord rotation (ψ), derived from the curvature-moment relation.
Joint Equilibrium (Compatibility)
Example
At interior support B of a two-span beam: M_BA (from span AB) + M_BC (from span BC) = 0 (no applied moment at B).
At any rigid joint, the sum of moments from all members meeting at that joint MUST equal zero: ΣM = 0.
Support Conditions (Boundary Conditions)
Example
At a simply supported end: M = 0 (write this equation to solve for one unknown rotation).
Constraints that reduce unknowns: pinned end has θ free but M = 0; fixed end has θ = 0 and M free; free end has both M = 0 and θ free.
Diagrams to know
- FBD of each member showing internal moments M_AB, M_BA and external loads/FEMs.
- Joint equilibrium diagram: all moments at an interior joint drawn in a circle, summing to zero.
- Chord rotation: draw settled support position vs. original position to visualize ψ direction.
Reactions or equations
ΣM_joint = 0
Note
Write this for each interior joint and each pin/roller support (where M = 0 anyway).
Conditions
At every rigid joint with no applied external moment.
θ_fixed = 0
Note
Eliminates unknown rotation at that end.
Conditions
At a fixed support (built-in end).
M_pinned = 0 or M_free = 0
Note
The S-D equation evaluated at that end gives you a relationship among the other end's rotation and ψ.
Conditions
At a simply supported or free end.
Moment Distribution Method
Formulas
DF = k / Σk (at a joint)
Meaning
Distribution Factor; k = stiffness of member; Σk = sum of all stiffness values at the joint.
Watch out
DF is dimensionless (0 to 1). At a fixed end, DF = 0 (no distribution of moment there). At a joint with only one member, DF = 1. Always check: sum of all DF at a joint must equal 1.
When to use
FIRST step in moment distribution. Calculate once per member at each joint.
Distributed moment = −(unbalanced moment) × DF
Meaning
Unbalanced moment = sum of FEMs at that joint. Distribute the negative of this sum in proportion to DF.
Watch out
The sign is CRITICAL: distribute the NEGATIVE of the unbalanced moment. If FEM_BA + FEM_BC = +30 kN·m at B, then distribute −30 in proportion to DF values.
When to use
In the 'balance' step of moment distribution. Repeat at each joint until negligible.
Carry-over = (distributed moment) × 1/2
Meaning
Half of the distributed moment at one end carries over to the far end of the same member.
Watch out
Carry-over happens ONLY from the balanced joint to the far end (one direction only). Do NOT carry back. At a fixed end, the carry-over is absorbed (no further balance). At a free end, do not balance it; the moment stays as is.
When to use
After balancing a joint, carry over 1/2 of each distributed moment along each member to its far end.
k = 4EI/L
Meaning
Stiffness for a member with both ends continuous or fixed.
Watch out
See below for modification when far end is pinned.
When to use
Standard case in moment distribution.
k = 3EI/L (with pinned/free far end)
Meaning
Reduced stiffness when the far end cannot resist rotation fully (e.g., pin support, cantilever end).
Watch out
Forgetting this is a frequent mistake. It directly changes DF and thus the distribution pattern.
When to use
If moment distribution involves a pinned terminal, use this stiffness (and skip carry-over to that end).
Important facts
- Moment distribution avoids solving a system of equations; it is an ITERATIVE hand-calculation method.
- Always compute FEMs FIRST; errors here corrupt the entire solution.
- The method converges rapidly (usually 3–5 iterations for hand work) if done by hand; use a table or spreadsheet for accuracy.
- The final moment at each member end is the SUM of the FEM column plus all distributed and carry-over moments in that column.
- Pinned or free ends: set DF = 0 (no distribution to a pinned support) or use k = 3EI/L (pinned terminal).
- For a cantilever, the fixed end is never balanced; just apply FEM and carry-over from the continuous joint.
Key definitions
Distribution Factor (DF)
Example
At joint B with two equal spans, k_BA = k_BC = 4EI/6 → DF_BA = DF_BC = 0.5 each.
Fraction of unbalanced moment distributed to each member at a joint, proportional to its stiffness.
Unbalanced Moment
Example
At B: FEM_BA = +36 kN·m, FEM_BC = −18 kN·m → unbalanced = +18 kN·m.
Algebraic sum of all FEMs converging at a joint (both directions); if nonzero, the joint is not in equilibrium and must be balanced.
Carry-Over
Example
If −18 kN·m is distributed at B to the BA direction, then +9 kN·m (i.e., −18 × 1/2 = −9, but with opposite sign) carries to A.
Transmission of half the distributed moment from a balanced joint to the far end of the same member.
Iteration (Successive Balancing)
Example
Multi-span continuous beam: balance B, then C, then B again, etc., until all unbalanced moments are < 1 kN·m (or negligible).
Process of balancing one joint, carrying over, then balancing the next, and repeating until convergence.
Diagrams to know
- Moment distribution table: rows = iterations, columns = member ends (AB, BA, BC, CB, etc.); entries = FEM, distributions, carry-overs, final moments.
- Sketch of structure with stiffness and DF labels at each joint.
- Bending moment diagram (BMD) from final moments + reactions to verify equilibrium.
Reactions or equations
Σ(distributed moments + carry-overs) = 0 at each joint (final check)
Note
A quick sanity check: if not zero (within rounding), you made an arithmetic error.
Conditions
After all iterations converge.
Comparison: Slope-Deflection vs. Moment Distribution
Special Cases & Support Conditions
Formulas
θ_fixed = 0
Meaning
Rotation at a fixed (built-in) end is zero.
Watch out
Apply this early to reduce unknowns. Do not write a slope-deflection equation at a fixed end; instead, use θ = 0 as a boundary condition.
When to use
Any fixed support (cantilever, fixed–pinned, fixed–fixed).
M_pinned = 0
Meaning
Moment at a simply supported or pinned end is zero.
Watch out
Evaluate the slope-deflection equation at a pinned end with M = 0; this gives a relation among the other end's rotation and ψ.
When to use
Pin or roller support, or free end.
ψ = Δ/L (chord rotation from settlement)
Meaning
Δ = relative vertical displacement of supports (positive downward); L = span.
Watch out
Sign: if support B settles DOWN relative to A, then Δ > 0 and ψ > 0 (chord rotates clockwise). The −3ψ term in slope-deflection is CRITICAL for this.
When to use
When one support settles relative to another.
Important facts
- For a cantilever with end load, the slope-deflection equation at the fixed end is automatically satisfied by θ_fixed = 0; use the free-end condition (M_free = 0) to solve the other unknowns.
- Portal frames and multi-storey buildings introduce MULTIPLE unknown chord rotations (one per storey if all joints move together).
- Asymmetric frames may require writing equilibrium of shear forces and moments together to find ψ.
- In moment distribution with sidesway, the stiffness and DF calculations remain the same, but carry-over to a pinned end goes nowhere (use k = 3EI/L to eliminate it).
Key definitions
Sidesway (Lateral Displacement)
Example
Wind load on a building frame causes all joints in a storey to move right by Δ = 30 mm; this is sidesway, and ψ = Δ/h (h = storey height).
Horizontal displacement of a frame storey due to lateral load or asymmetric vertical load; introduces unknown chord rotation ψ.
Propped Cantilever
Example
A 5 m beam fixed at A, roller at B; load 20 kN/m → M_A = (20)(5)²/8 − reaction × moment arm.
Beam fixed at one end, freely supported (pinned or roller) at the other; used to model simple end conditions.
Cantilever (Fixed at One End Only)
Example
In moment distribution, a cantilever is never balanced; the fixed end absorbs all moment.
Beam fixed at one end, free (unsupported) at the other; useful as a sub-system.
Diagrams to know
- Fixed-end beam vs. pinned-end beam: show the difference in moment distribution.
- Cantilever sketch: show that the free end has M = 0 and V ≠ 0, while the fixed end has M ≠ 0 and V ≠ 0.
- Sidesway diagram: draw a frame in original and deflected configuration; label ψ and Δ.
Reactions or equations
ΣH = 0 (horizontal equilibrium of frame storey)
Note
This gives a second equation to solve for ψ (in addition to moment equilibrium at joints).
Conditions
Sidesway case: external horizontal load + shear reactions must sum to zero.
Worked Example Patterns (Exam Classics)
Important facts
- Example 1: Two-span continuous beam, equal spans, UDL on both → by symmetry, θ_B = 0; find M_B by slope-deflection.
- Example 2: Two-span beam, UDL on one span only → asymmetric; requires slope-deflection or moment distribution with asymmetric carry-overs.
- Example 3: Fixed–pinned beam with point load → use k = 3EI/L at the pinned end; solve for interior rotation.
- Example 4: Propped cantilever → write two equations (FEM + boundary conditions); solve for the prop reaction.
- Example 5: Portal frame with horizontal load → introduce sidesway ψ; use shear equilibrium ΣH = 0 plus moment equilibrium to solve for ψ and all θ values.
- Example 6: Continuous beam with support settlement → ψ ≠ 0; include −3ψ term in slope-deflection; compare final moments to settlement case.
Diagrams to know
- FBD for each member: show FEM, internal moments M_AB & M_BA, external load, and shear reactions.
- Shear and moment diagrams: draw from the final moments + equilibrium to verify closure.
Quick Reference Tables
Common values
Value
wL²/12
Symbol
FEM_UDL
Quantity
FEM for UDL
Value
PL/8
Symbol
FEM_P
Quantity
FEM for central point load
Value
Pab²/L²
Symbol
FEM_AB
Quantity
FEM for arbitrary point load (near end A)
Value
4EI/L
Symbol
k_fixed
Quantity
Stiffness (both ends fixed)
Value
3EI/L
Symbol
k_pinned
Quantity
Stiffness (pinned far end)
Value
0.5 (or 1/2)
Symbol
CO
Quantity
Carry-over factor
Must remember
- FEM values are SIGNED (clockwise-positive by convention); FEM_AB is NEGATIVE for most loads (moment acts clockwise on element A-end); FEM_BA is OPPOSITE sign.
- Slope-deflection: M = (2EI/L)[2θ_near + θ_far − 3ψ] + FEM. The −3ψ term is critical for settlement/sidesway problems; omitting it causes wrong answers.
- In moment distribution, Distribution Factor DF = k/Σk; always check that DF values sum to 1.0 at each joint. If they don't, you've miscounted the members or mislabeled stiffness.
- Use k = 4EI/L for fully continuous or fixed-end members; use k = 3EI/L ONLY when the far end is a pin or a free cantilever. Mixing these is the #1 exam killer.
- Carry-over always = (distributed moment) × 1/2, and it flows ONLY from the balanced joint to the far end of each member (one direction). Carry-over to a pinned end is discarded (or use k = 3EI/L to avoid it).
- At a fixed support, θ = 0 and DF = 0. Never balance or distribute moment at a fixed end; the moment is whatever equilibrium requires.
- At a simply supported end, M = 0 (boundary condition). In slope-deflection, this gives a relation among the other end's rotation and ψ; in moment distribution, the FEM itself may be asymmetric if the support isn't symmetrically loaded.
- Multiple loads on one span: superpose individual FEMs (e.g., FEM_total = FEM_UDL + FEM_point_load at that end). Do this BEFORE starting moment distribution or slope-deflection.
- Convergence in moment distribution: iterate until unbalanced moments at all joints are < ~1 kN·m (or < 0.5% of the largest FEM). Usually 3–5 iterations by hand suffice.
- After solving for moments by either method, always check by drawing a free-body diagram and verifying joint equilibrium (ΣM = 0) and span equilibrium (ΣV correct, no gaps). A failed check means you have an error; do not move on.
Last minute tips
- Memorize the FEM table: UDL → wL²/12, central load → PL/8, arbitrary load → Pab²/L² (near end), Pa²b/L² (far end). Triangular is rare but know −wL²/30 and +wL²/20. Test yourself on these 5 formulas the night before the exam.
- Before you start any calculation, write down your sign convention (e.g., 'clockwise-positive on member' or 'tension-positive') and STICK TO IT. Exam invigilators often ask you to state your convention; if yours is non-standard but consistent, you'll get credit.
- In moment distribution, if you make an arithmetic error early on (e.g., wrong FEM or DF), it propagates through every iteration. Always double-check FEMs and DFs in the first row of your table BEFORE doing any balancing.
- For slope-deflection, set up the system of equations FIRST (write out all unknowns and all equations, including support conditions and equilibrium). THEN solve (by substitution or matrix). This discipline prevents missing an equation or swapping signs.
- If the problem mentions 'support settlement,' 'relative displacement,' 'sidesway,' or 'frame movement'—IMMEDIATELY introduce ψ = Δ/L and include the −3ψ term. This is a reflex that saves you from a 20-point deduction.
Comparison tables
Slope-Deflection vs. Moment Distribution
| Method | Unknowns | Equation per Member | Solution Process | Hand Calculation Speed | Convergence | |
|---|---|---|---|---|---|---|
| Slope-Deflection | Joint rotations θ and displacements Δ | M_AB = (2EI/L)[2θ_A + θ_B − 3ψ] + FEM_AB | Solve system of linear equations (joint equilibrium + boundary conditions) | Direct (matrix or Gaussian elimination) | Medium (depends on system size) | Exact after solving linear system |
| Moment Distribution | Member-end moments (not rotations) | DF = k/Σk; distribute unbalanced FEM and carry over | Iterative relaxation of slope-deflection equations (implicit) | Successive approximation by balancing and carry-over | Fast (usually 3–5 hand iterations) | Asymptotic (converges rapidly if DF ≤ 0.5) |
Fixed-End Moments (FEM) Cheat Table
| Load Type | Load Diagram | FEM_AB (near end A) | FEM_BA (far end B) | Applicable Span |
|---|---|---|---|---|
| UDL full span | w [kN/m] across entire L | −wL²/12 | +wL²/12 | Full length L |
| Central point load | P at L/2 | −PL/8 | +PL/8 | L (symmetric) |
| Arbitrary point load | P at distance a from A (b from B) | −Pab²/L² | +Pa²b/L² | Full length L |
| Triangular (0 at A, peak w at B) | w [kN/m] at B only | −wL²/30 | +wL²/20 | Full length L |
| Triangular (peak w at A, 0 at B) | w [kN/m] at A only | +wL²/20 | −wL²/30 | Full length L |
Support Conditions & Their Effect on Analysis
| Support Type | Reaction(s) | Moment at End | Rotation at End | Stiffness k to Use | Distribution Factor |
|---|---|---|---|---|---|
| Fixed (Built-in) | V, H, M | M ≠ 0 (free to balance) | θ = 0 (locked) | 4EI/L | 0 (no distribution) |
| Simply Supported (Pin/Roller) | V (and H if pin) | M = 0 (boundary condition) | θ ≠ 0 (free to rotate) | 3EI/L or 4EI/L (if continuous) | Limited by M = 0 constraint |
| Free (Cantilever end) | None (overhanging end) | M = 0 | θ ≠ 0 | Not applicable | 0 (never balanced) |
| Continuous (Interior) | V only (reactions from adjacent spans) | M ≠ 0 (equilibrium ΣM = 0 at joint) | θ ≠ 0 (unknown, solve for) | 4EI/L | k/Σk (shared among members) |
Common PRC Exam Mistakes & How to Avoid Them
| Mistake | Why It Happens | How to Catch It | Correct Approach | |
|---|---|---|---|---|
| Wrong FEM sign | Flipping clockwise-positive convention mid-problem | Inconsistent diagram or reference | Check FEM table and draw arrows on your work | Always confirm sign convention at problem start; draw FEM moments consistently |
| Using k = 4EI/L at a pinned end | Not modifying stiffness for pinned supports | Forgetting RA 544 design notes and textbook warnings | DF values are abnormally large at pinned joint | Use k = 3EI/L (or skip that end entirely in MD) |
| Forgetting −3ψ in slope-deflection | Skipping chord rotation when there is support settlement or sidesway | Assuming all supports are at the same level | Final moments don't match check calculations | Always include −3ψ; set ψ = 0 only if explicitly no settlement |
| Distributing moment at a fixed end | Treating fixed ends as if they can unbalance and distribute | Confusion about DF = 0 at fixed ends | Moment balance equation fails; iterations don't converge | Never balance a fixed end; it absorbs carry-over without distribution |
| Forgetting to carry over | Omitting the × 1/2 carry-over step after distribution | Rushing or misunderstanding the iterative process | Moments at far ends are too small; equilibrium fails | After each balance, always carry over 1/2 of distributed moment to far end |
| Carry-over in the wrong direction | Carrying over BACK to the joint just balanced (or to pinned ends) | Misunderstanding one-way flow of carry-over | Joint moments don't converge; arithmetic loop | Carry-over flows ONLY from balanced joint → far end (one way). Stop at fixed ends; skip pinned ends |
| Arithmetic in moment distribution table | Accumulating rounding errors across iterations | Mental math errors with many iterations | Final moments are off by > 2% or equilibrium check fails | Use a spreadsheet or table with neat columns; carry at least 1 decimal; do one more iteration if unsure |
Diagrams
Displacement Methods: Concept Map
Hierarchical concept map showing the relationship between FEM, slope-deflection, and moment distribution methods, with key formulas and support conditions branching from the central theme.
Moment Distribution: Step-by-Step Algorithm
Complete flowchart of the moment distribution algorithm, showing decision points for support types, stiffness selection, and iterative balancing until convergence. Includes verification steps.
Moment Distribution: Iteration State Machine
State diagram showing the iterative nature of moment distribution, where the system cycles through joint balancing and carry-over until the convergence tolerance is met, then verification occurs.
Choosing Between Slope-Deflection and Moment Distribution
Decision flowchart to help students select the most appropriate displacement method based on problem complexity, number of unknowns, and availability of computational aids.
Typical Exam Problem Workflow
Timeline showing the typical sequence of steps in solving an indeterminate structure exam problem, from problem interpretation to final verification and answer reporting.