CELE Structural Theory & Analysis — Indeterminate Structures: Force MethodsRevision Notes
Revision notes for CELE Structural Theory & Analysis — Indeterminate Structures: Force Methods. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Structural Theory & Analysis subtest is marked as "Core" in the official pattern, and Indeterminate Structures: Force Methods appears in position 3rd of 6 in the CELE Structural Theory & Analysis review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Indeterminate Structures: Force Methods - Revision Notes
Statically indeterminate structures cannot be solved by equilibrium equations alone — the number of unknowns exceeds the number of available static equilibrium equations. The force (flexibility) method — also called the method of consistent deformation — resolves this by treating the extra unknowns (redundants) as the primary variables. Once redundants are determined from compatibility conditions, the entire structure is solved by statics. This chapter covers two cornerstone techniques for the PRC CE board exam: (1) the Method of Consistent Deformation and (2) the Three-Moment Equation for continuous beams. Mastery of these methods is essential, as indeterminate beam and frame problems appear in virtually every CE licensure examination.
Sections
Formulas
Example
A propped cantilever has 4 reactions (Ax, Ay, MA, RB) → DI = 4 − 3 = 1 (once-indeterminate). A fixed-fixed beam has 6 reactions → DI = 6 − 3 = 3.
Formula
DI = r − 3 (for 2D beam/frame, no internal hinges)
Variables
r = total number of external reaction components; 3 = number of 2D equilibrium equations (ΣFx=0, ΣFy=0, ΣM=0)
Application
Quickly classify structures as determinate (DI=0), indeterminate (DI>0), or unstable (DI<0).
Exam Tips
- On board exams, always state DI before starting any solution — it signals to the examiner you understand the problem type.
- For quick DI checks: propped cantilever = 1, two-span continuous beam (simple ends) = 1, fixed-fixed beam = 3, fixed-propped beam = 2.
- If a problem gives EI, it likely requires the flexibility method — this is your cue to apply force methods.
Key Points
- The Degree of Indeterminacy (DI) equals the number of unknown reactions/internal forces minus the number of independent equilibrium equations available.
- For a beam: DI = total reactions − 3 (for 2D structures with no internal hinges). Add 1 for each internal hinge.
- A redundant is any reaction or internal force that can be removed WITHOUT making the structure geometrically unstable — it is the 'extra' unknown.
- Removing all redundants from the original structure produces the PRIMARY (released) structure, which must be statically determinate and stable.
- The number of compatibility (deformation) equations required equals the degree of indeterminacy.
- Common redundants: prop reaction of a propped cantilever, interior support reaction of a continuous beam, one fixed-end moment of a fixed-fixed beam.
- Choice of redundant is not unique — different choices lead to the same final answer but different solution complexity. Choose for easiest deflection calculations.
Definitions
Term
Redundant
Definition
A reaction or internal force in excess of what is needed for static equilibrium; its removal leaves a stable, determinate primary structure.
Importance
Identifying the redundant correctly is the first and most critical step in any force-method solution.
Term
Primary (Released) Structure
Definition
The statically determinate structure obtained after removing the redundants from the original indeterminate structure.
Importance
All deflection calculations (δ₀ and δ₁₁) are performed on this simpler structure using standard beam formulas.
Term
Flexibility Coefficient (δ₁₁)
Definition
The deflection at the location of redundant R₁ due to a UNIT value of R₁ applied to the primary structure.
Importance
It forms the 'stiffness per unit' of the redundant's effect; used to set up the compatibility equation.
Section Title
1. Degree of Static Indeterminacy and Redundants
Common Mistakes
- Counting reactions incorrectly: a pin gives 2 reactions, a roller gives 1, a fixed support gives 3.
- Releasing a reaction that makes the structure unstable (e.g., releasing the only horizontal support when horizontal loads are present).
- Confusing degree of indeterminacy with degree of kinematic indeterminacy (which is used in the stiffness/displacement method).
- Forgetting that an internal hinge reduces the DI by 1 (each hinge adds one equation: moment = 0 at hinge).
Formulas
Example
Propped cantilever, UDL w: δ₀ = wL⁴/(8EI) ↓, δ₁₁ = L³/(3EI) ↑ per unit upward R. Compatibility: wL⁴/(8EI) = R·L³/(3EI) → R = 3wL/8.
Formula
δ₀ + R · δ₁₁ = Δ
Variables
δ₀ = deflection at the redundant location due to applied loads on the primary structure (without redundant); R = redundant force/moment; δ₁₁ = deflection at the redundant location due to unit value of the redundant on the primary structure; Δ = prescribed displacement at that support (0 if rigid, +Δ if settlement)
Application
The fundamental compatibility equation for a once-indeterminate structure. Directly gives the redundant R.
Example
If δ₀ = −wL⁴/(8EI) (downward, negative in chosen sign convention) and δ₁₁ = +L³/(3EI) (upward per unit R), then R = −(−wL⁴/8EI)/(L³/3EI) = 3wL/8 (upward). ✓
Formula
R = (Δ − δ₀) / δ₁₁
Variables
Δ = known displacement at support; δ₀ = load-induced displacement on primary structure; δ₁₁ = flexibility coefficient
Application
Direct formula for one redundant. When support is rigid (no settlement), Δ = 0, so R = −δ₀/δ₁₁.
Example
w = 12 kN/m, L = 4 m: δ₀ = 12(4)⁴/(8EI) = 384/EI m (downward).
Formula
δ₀ (UDL on cantilever, free end) = wL⁴ / (8EI)
Variables
w = uniform distributed load (kN/m); L = span length (m); E = modulus of elasticity (GPa); I = moment of inertia (m⁴)
Application
Used as δ₀ when primary structure is a cantilever with UDL and the redundant is the prop at the free end.
Example
L = 4 m: δ₁₁ = (4)³/(3EI) = 64/(3EI) m per kN. Then R = [384/EI] / [64/(3EI)] = 384 × 3/64 = 18 kN ✓ (= 3wL/8 = 3×12×4/8).
Formula
δ₁₁ (point load at free end of cantilever) = L³ / (3EI)
Variables
L = cantilever span length; E = modulus of elasticity; I = moment of inertia
Application
Used as δ₁₁ when unit redundant is a concentrated upward force at the free end of a cantilever primary structure.
Example
P = 16 kN, L = 4 m, a = 2 m: δ₀ = 16(2)²[3(4)−2]/(6EI) = 16(4)(10)/(6EI) = 640/(6EI) = 5(16)(4)³/(48EI) = 5PL³/48EI ✓
Formula
δ₀ (point load P at distance a from fixed end, deflection at free end of cantilever) = Pa²(3L − a) / (6EI)
Variables
P = applied point load; a = distance from fixed support to load; L = total cantilever length
Application
Used for propped cantilever with point load NOT at free end. When a = L/2 (midspan load): δ₀ = 5PL³/(48EI).
Example
w = 12 kN/m, L = 4 m: R_prop = 3(12)(4)/8 = 18 kN; M_fixed = 12(4)²/8 = 24 kN·m (hogging).
Formula
R_prop (propped cantilever, UDL) = 3wL/8; M_fixed = wL²/8
Variables
w = UDL (kN/m); L = span (m); R_prop = prop (roller) reaction; M_fixed = fixed-end moment (hogging)
Application
Standard result — MEMORIZE for board exams. Appears repeatedly in problems.
Exam Tips
- Memorize the five most common beam deflection formulas: cantilever (UDL and point load at free end), simply supported beam (midspan point load and UDL midspan deflection).
- R_prop = 3wL/8 for propped cantilever under UDL is a direct board-exam answer — recognize the problem type instantly.
- When EI is not given numerically, it cancels in the compatibility equation — no need to substitute actual EI values.
- For settlement problems: δ₀ is downward (positive), δ₁₁ is upward per unit R (positive), Δ is the settlement value — write carefully.
- Draw the primary structure and label all forces clearly before calculating any deflections.
Key Points
- Core principle: the deflection at the point of the redundant in the original structure equals a known value (usually zero for a rigid support, or Δ for a settled support).
- Step 1 — Release: Remove the redundant to obtain the primary structure. Calculate deflection δ₀ at the released point due to APPLIED LOADS ONLY.
- Step 2 — Unit load: Apply unit redundant (R = 1) to the primary structure. Calculate deflection δ₁₁ at the released point.
- Step 3 — Compatibility: Set up the equation δ₀ + R·δ₁₁ = Δ (where Δ = 0 for no settlement, or Δ = support settlement value).
- Step 4 — Solve for R, then use equilibrium to find all remaining reactions and draw shear/moment diagrams.
- For n redundants: write n compatibility equations simultaneously — δᵢ₀ + Σⱼ Rⱼ δᵢⱼ = Δᵢ for each released coordinate i.
- All deflection calculations use the PRIMARY (released) determinate structure — apply standard beam deflection formulas or virtual work.
- Sign convention: define positive direction consistently. Typically, downward deflection is positive. If loads deflect the point downward and redundant pushes it upward, the compatibility equation correctly yields a positive R.
Definitions
Term
Compatibility Condition
Definition
A geometric requirement that the actual deformation of the structure must be consistent with the support conditions — e.g., zero deflection at a rigid support.
Importance
This is the additional equation that supplements equilibrium to solve indeterminate structures in the force method.
Term
Support Settlement (Δ)
Definition
A known prescribed downward displacement of a support due to soil compaction or differential settlement.
Importance
When settlement Δ occurs at the redundant support, the compatibility equation becomes δ₀ + R·δ₁₁ = Δ (NOT zero). Failure to account for this is a major exam pitfall.
Section Title
2. Method of Consistent Deformation
Common Mistakes
- Using the wrong deflection formula — always verify whether the load is on the primary structure (cantilever, simply supported, etc.).
- Sign error: if δ₀ and δ₁₁ act in the same direction, the compatibility equation may give a negative R — this is valid, it means the redundant acts opposite to the assumed direction.
- Forgetting the compatibility equation changes to δ₀ + R·δ₁₁ = Δ (not zero) when support settlement is given.
- Using the original indeterminate structure's deflection formulas instead of the primary structure's formulas.
- Not verifying the answer by checking equilibrium after finding R.
Formulas
Example
Two equal spans L, UDL w, simple ends: M_A=0, M_C=0, 6A₁x̄₁/L₁=wL³/4, 6A₂x̄₂/L₂=wL³/4. Equation: 0 + 2M_B(2L) + 0 = −(wL³/4 + wL³/4) → M_B = −wL²/8.
Formula
M_A·L₁ + 2M_B(L₁ + L₂) + M_C·L₂ = −6(A₁x̄₁/L₁ + A₂x̄₂/L₂)
Variables
M_A, M_B, M_C = bending moments at supports A, B, C (negative for hogging per beam sign convention); L₁ = span AB length; L₂ = span BC length; A₁, A₂ = areas of the simple-beam M-diagram for span AB and BC respectively; x̄₁ = centroid distance of A₁ from support A; x̄₂ = centroid distance of A₂ from support C
Application
Write one equation per interior support. For n-span beam with n+1 supports, there are n−1 interior supports and n−1 equations.
Example
Span L = 6 m, w = 10 kN/m: 6Ax̄/L = 10(6)³/4 = 10(216)/4 = 540 kN·m².
Formula
6Ax̄/L (UDL w, span L) = wL³/4
Variables
w = uniform distributed load (kN/m); L = span length; this is the per-span RHS load term for a UDL
Application
Substitute directly into the RHS of the three-moment equation for any span with a full UDL.
Example
P = 20 kN, L = 5 m (midspan): 6Ax̄/L = 3(20)(5)²/8 = 3(20)(25)/8 = 187.5 kN·m².
Formula
6Ax̄/L (central point load P, span L) from left support = 3PL²/8; from right support = 3PL²/8 (symmetric)
Variables
P = concentrated point load at midspan; L = span length
Application
For a concentrated load at midspan, both x̄ terms equal 3PL²/8 (symmetric case). For a load NOT at midspan, use 6A₁x̄₁/L = Pb(L²−b²)/L and 6A₂x̄₂/L = Pa(L²−a²)/L where a+b=L.
Example
If only support B settles by Δ_B = 10 mm = 0.01 m: settlement term = 6EI·Δ_B·(1/L₁ + 1/L₂) added to RHS with negative sign.
Formula
Three-moment equation with settlement: M_A·L₁ + 2M_B(L₁+L₂) + M_C·L₂ = −6(A₁x̄₁/L₁ + A₂x̄₂/L₂) + 6EI(Δ_A/L₁ + Δ_C/L₂ − Δ_B(1/L₁ + 1/L₂))
Variables
Δ_A, Δ_B, Δ_C = settlements at supports A, B, C respectively (positive downward); EI = flexural rigidity of beam
Application
Used when differential settlement occurs at supports. The settlement term modifies the RHS.
Example
w = 10 kN/m, L = 6 m: M_B = −10(6)²/8 = −45 kN·m (hogging). Reactions: R_A = R_C = wL/2 − |M_B|/L = 30 − 7.5 = 22.5 kN; R_B = 2wL − 2(22.5) = 75 kN.
Formula
M_B (two equal spans, UDL, simple ends) = −wL²/8
Variables
w = UDL; L = span length (equal spans)
Application
Direct result for the most common exam scenario — two equal spans, UDL, simply supported ends. MEMORIZE.
Exam Tips
- The two key RHS terms to memorize: UDL → wL³/4; central point load → 3PL²/8. These appear on almost every board exam problem involving the three-moment equation.
- For two-span equal beams with equal UDL and simple ends: M_B = −wL²/8. Write this down immediately when you see this configuration.
- Always draw the continuous beam, label spans L₁ and L₂, identify interior supports, and write the three-moment equation systematically.
- After solving for M_B, compute reactions span by span: treat each span as a simply supported beam with the computed end moments as additional applied couples.
- Always verify reactions: ΣR = total load applied. If the beam has symmetry, exploit it to reduce unknowns.
Key Points
- The three-moment equation is the most efficient method for continuous beams on multiple supports — it is the standard board-exam tool for multi-span beams.
- It relates bending moments at THREE consecutive supports (A, B, C) across spans L₁ (AB) and L₂ (BC).
- For each interior support, write one three-moment equation → n interior supports = n equations for n unknown interior moments.
- Boundary conditions: for simple (pin/roller) end supports, M = 0; for fixed end supports, M ≠ 0 (add a virtual span of zero length).
- The right-hand side (RHS) of the equation uses the 6Ax̄/L terms, computed from the SIMPLE-BEAM moment diagrams of each span loaded independently.
- For equal spans with UDL on all spans, the three-moment equation gives M_interior = −wL²/8 for a two-span beam (classical result).
- After finding all support moments, reactions are determined span-by-span using free body diagrams.
- The method can handle different span lengths, different loads per span, and support settlements.
Definitions
Term
Simple-Beam Moment Diagram Area (A) and Centroid (x̄)
Definition
For each span, imagine the span as an isolated simply supported beam carrying its own load. Compute the area A of the resulting M-diagram and its centroid distance x̄ from the reference support.
Importance
These values populate the RHS of the three-moment equation. Incorrect computation of 6Ax̄/L is the most common source of error on board exams.
Term
Interior Support Moment
Definition
The bending moment at a continuous intermediate support; always hogging (negative by the standard beam sign convention) for downward loads.
Importance
This is typically the main unknown in three-moment equation problems. Once found, all reactions and shear/moment diagrams follow.
Section Title
3. Three-Moment Equation (Clapeyron's Theorem)
Common Mistakes
- Using 6Ax̄/L = wL³/4 but forgetting it represents the term for ONE span only — add both spans' terms on the RHS.
- Applying the three-moment equation at end supports instead of interior supports only.
- Setting M ≠ 0 at simply supported ends — simple supports have M = 0, always.
- For a fixed end, failing to add a fictitious zero-length span to correctly handle the boundary condition.
- Computing reactions directly from total load without accounting for the effect of support moments on each span.
- Not checking: the sum of all reactions must equal the total applied load on the entire beam.
Formulas
Example
w = 8 kN/m, L = 5 m: δ₀ = 8(5)⁴/(8EI) = 8(625)/(8EI) = 625/EI m
Formula
δ_max (cantilever, UDL w, free end) = wL⁴/(8EI)
Variables
w = UDL; L = cantilever length; E = elastic modulus; I = moment of inertia
Application
δ₀ for propped cantilever under UDL when redundant is the prop force at free end.
Example
L = 5 m: δ₁₁ = (5)³/(3EI) = 125/(3EI) m per kN
Formula
δ (cantilever, point load P at free end) = PL³/(3EI)
Variables
P = concentrated load at free end; L = cantilever length
Application
δ₁₁ for propped cantilever (prop at free end) — apply unit load P=1 to get δ₁₁ = L³/(3EI).
Example
P = 16 kN, a = L/2 = 2 m, L = 4 m: δ₀ = 16(2)²[3(4)−2]/(6EI) = 16(4)(10)/(6EI) = 640/(6EI) = 5PL³/(48EI) ✓
Formula
δ (cantilever free end, load P at distance a from fixed end) = Pa²(3L−a)/(6EI)
Variables
a = distance from fixed support to load P; L = total cantilever length; P = point load
Application
δ₀ for propped cantilever with interior point load, measuring free-end deflection.
Example
w = 10 kN/m, L = 6 m: δ = 5(10)(6)⁴/(384EI) = 5(10)(1296)/(384EI) = 168.75/EI mm (if EI in kN·m²)
Formula
δ_midspan (simply supported, UDL w) = 5wL⁴/(384EI)
Variables
w = UDL; L = span; E, I = section properties
Application
Used when primary structure is a simply supported beam and the redundant is a midspan support or moment.
Example
P = 20 kN, L = 4 m: δ = 20(4)³/(48EI) = 20(64)/(48EI) = 26.67/EI
Formula
δ_midspan (simply supported, central load P) = PL³/(48EI)
Variables
P = central concentrated load; L = span
Application
Used when primary is simply supported and load is at center.
Exam Tips
- Create a quick reference card with all five standard deflection formulas — board exams often require direct substitution.
- The formula δ = Pa²(3L−a)/(6EI) for an interior load on a cantilever is frequently tested; derive it mentally from integration if you forget.
- When EI cancels in the compatibility equation, you do not need the actual EI value — this is intentional in most board exam problems.
Key Points
- In the force method, you MUST know deflection formulas for determinate beams — these are used for both δ₀ and δ₁₁.
- All formulas apply to the PRIMARY (released, determinate) structure, NOT the original indeterminate structure.
- The five most critical formulas: (1) cantilever free-end deflection under UDL, (2) cantilever free-end deflection under point load at free end, (3) cantilever free-end deflection under point load at interior point, (4) simply supported midspan deflection under UDL, (5) simply supported midspan deflection under central point load.
- EI typically cancels in the compatibility equation — keep it symbolic.
- For fixed-fixed or other complex primary structures, use superposition or virtual work (unit load method) to compute deflections.
Definitions
Term
Flexibility Coefficient δᵢⱼ
Definition
The displacement at coordinate i due to a unit force at coordinate j on the primary structure. δᵢⱼ = δⱼᵢ by Maxwell's reciprocal theorem.
Importance
Forms the coefficient matrix in multi-redundant problems. Symmetry (δᵢⱼ = δⱼᵢ) provides a useful check.
Section Title
4. Standard Beam Deflection Formulas for Force Methods
Common Mistakes
- Using midspan deflection formulas when the actual point of interest is not at midspan.
- Confusing δ₀ (due to applied loads) with δ₁₁ (due to unit redundant) — these must be computed separately.
- Applying cantilever formulas to the primary structure when the primary is actually a simply supported beam (wrong primary structure choice).
Formulas
Example
w = 12 kN/m, L = 4 m: R_B = 3(12)(4)/8 = 18 kN; R_A = 5(12)(4)/8 = 30 kN; M_A = 12(16)/8 = 24 kN·m (hogging). Check: R_A + R_B = 48 = wL ✓
Formula
Example 1 Summary: R_prop = 3wL/8; R_fixed = 5wL/8; M_fixed = wL²/8
Variables
For propped cantilever under UDL w, span L. R_prop = prop reaction, M_fixed = fixed-end moment (hogging).
Application
Standard propped cantilever result — given directly on board exams asking for any of these three values.
Example
w = 10 kN/m, L = 6 m: M_B = −10(36)/8 = −45 kN·m; R_A = R_C = 10(6)/2 − 45/6 = 30 − 7.5 = 22.5 kN; R_B = 2(10)(6) − 45 = 120 − 45 = 75 kN. Check: 22.5+75+22.5 = 120 = 2wL ✓
Formula
Example 2 Summary: M_B = −wL²/8; R_A = R_C = wL/2 − |M_B|/L; R_B = 2wL − R_A − R_C
Variables
Two-span continuous beam, equal spans L, full UDL w, simple ends. M_B = interior support moment (hogging).
Application
Most common multi-span beam type on board exams. Results R_A = R_C = 3wL/8; R_B = 10wL/8 = 5wL/4.
Example
P = 16 kN, L = 4 m: R_B = 5(16)/16 = 5 kN; R_A = 16−5 = 11 kN; M_A = 16(2) − 5(4) = 32−20 = 12 kN·m. Check: ΣM_A = 0: 16(2) − 5(4) − 12 = 0 ✓
Formula
Example 3 Summary: R_B = 5P/16; R_A = 11P/16; M_A = PL/4 − R_B·L = 3PL/16 (when load at midspan)
Variables
Propped cantilever, point load P at midspan a = L/2. R_B = prop reaction; M_A = fixed-end moment.
Application
Propped cantilever with concentrated midspan load — tests knowledge of the δ₀ = 5PL³/(48EI) formula.
Exam Tips
- For any propped cantilever exam problem: immediately write R_prop = 3wL/8 (UDL) or R_prop = 5P/16 (midspan point load) if you recognize the configuration.
- For any two-span equal beam with UDL: write M_B = −wL²/8 immediately, then compute reactions.
- Time management: these standard results should be recalled in under 30 seconds on the board exam.
Key Points
- Example 1: Propped cantilever, UDL — direct application of consistent deformation.
- Example 2: Two-span continuous beam, equal spans UDL — direct three-moment equation application.
- Example 3: Propped cantilever, interior point load — demonstrates non-standard δ₀ formula.
- Understanding the full solution process is critical — board exams may ask for reaction, moment, shear, or deflection at specific points.
- Always verify answers using equilibrium checks.
Section Title
5. Worked Board-Exam Problems (Fully Solved)
Common Mistakes
- In Example 2, forgetting that R_B gets contributions from BOTH spans — use R_B = 2wL − R_A − R_C.
- In Example 3, using δ₀ = wL⁴/(8EI) for a point load instead of the correct Pa²(3L−a)/(6EI) formula.
- After finding support moments, not redrawing free body diagrams span by span to correctly compute reactions.
Connections
- Deflection formulas from Chapter on Beam Deflections (double integration, moment-area, conjugate beam) are prerequisites — consistent deformation uses these formulas directly.
- Virtual Work / Unit Load Method provides an alternative way to compute δ₀ and δ₁₁ for any structure, extending force methods beyond beams to frames and trusses.
- The results of force methods (support reactions and moments) feed directly into Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) construction.
- Stiffness (displacement) methods — slope-deflection equation and moment distribution — solve the same indeterminate structures from the opposite direction (treating displacements as unknowns); force and stiffness methods must yield identical results.
- Influence lines for indeterminate beams can be constructed using Müller-Breslau's principle, which extends the concept of unit redundant loads from consistent deformation.
- NSCP 2015 Section 406 (structural analysis requirements) recognizes that indeterminate structures require consideration of relative stiffness (EI) ratios — the force method quantifies this.
- The fixed-end moments derived by consistent deformation are the same values used as 'carry-over' and 'distribution' starting values in the moment distribution method (Hardy Cross method).
- Settlement analysis: differential settlement at supports of indeterminate beams generates secondary moments — the compatibility equation with Δ ≠ 0 directly gives these moments, which are critical in foundation engineering.
Exam Strategy
On the PRC CE board exam, force method problems typically appear in the Structural Theory and Analysis subject (15–20% of questions). Prioritize these strategies: (1) INSTANT RECOGNITION — identify the structure type (propped cantilever, two-span beam, fixed-fixed beam) and write the standard result immediately (R_prop=3wL/8, M_B=−wL²/8). This saves 2–3 minutes per problem. (2) THREE-MOMENT EQUATION — practice setting up and solving this equation systematically. The RHS load terms (wL³/4 for UDL; 3PL²/8 for midspan load) must be recalled instantly. (3) DEFLECTION FORMULAS — memorize the five standard formulas (cantilever UDL, cantilever point load at free end, cantilever interior load, SS-UDL, SS-central load). EI always cancels. (4) SIGN CONSISTENCY — choose a sign convention at the start and stick with it. Most errors are sign errors. (5) VERIFICATION — always check: ΣR = total load (1 min verification saves 5 min of rechecking). (6) TIME ALLOCATION — with 100 questions in 5 hours, spend no more than 3 minutes per force-method problem. If you recognize the standard configuration, answer in under 1 minute. (7) SETTLEMENT PROBLEMS — if EI is given numerically, suspect a settlement problem is involved; substitute carefully into the modified compatibility equation.
Quick Review Questions
A propped cantilever of span L = 5 m carries a UDL of w = 8 kN/m. What is the prop reaction R_B?
For a propped cantilever under UDL, the compatibility equation δ₀ = R_B·δ₁₁ gives: wL⁴/(8EI) = R_B·L³/(3EI), solving: R_B = 3wL/8. With w = 8 kN/m and L = 5 m: R_B = 3(8)(5)/8 = 15 kN. The fixed-end reaction R_A = wL − R_B = 40 − 15 = 25 kN, and M_fixed = wL²/8 = 8(25)/8 = 25 kN·m (hogging).
A two-span continuous beam has equal spans of L = 6 m. Each span carries a UDL of 10 kN/m. The ends are simply supported. What is the bending moment at the interior support B?
Apply the three-moment equation at support B with M_A = M_C = 0 (simple ends). The RHS term for each span with UDL is 6Ax̄/L = wL³/4. Equation: 0 + 2M_B(6+6) + 0 = −(10·216/4 + 10·216/4) = −(540+540) = −1080. Thus 24M_B = −1080 → M_B = −45 kN·m (hogging).
What is the compatibility equation for a propped cantilever if the prop support settles by Δ = 5 mm downward?
When the support settles by Δ downward, the compatibility condition is no longer zero deflection but Δ deflection. The equation becomes: δ₀ − R_B·δ₁₁ = Δ (when loads push down and redundant pushes up). Substituting: wL⁴/(8EI) − R_B·L³/(3EI) = Δ, solving for R_B = [wL⁴/(8EI) − Δ]·3EI/L³ = 3wL/8 − 3EIΔ/L³. Settlement reduces the prop reaction.
A propped cantilever of span 4 m with fixed end at A and roller at B carries a 16 kN point load at midspan (2 m from A). What is the prop reaction R_B?
Primary structure is a cantilever (release prop). δ₀ = Pa²(3L−a)/(6EI) = 16(2²)(3×4−2)/(6EI) = 16(4)(10)/(6EI) = 640/(6EI). δ₁₁ = L³/(3EI) = 64/(3EI). Compatibility: δ₀ = R_B·δ₁₁ → 640/(6EI) = R_B×64/(3EI) → R_B = [640/(6EI)]×[3EI/64] = 640×3/(6×64) = 1920/384 = 5 kN. This equals 5P/16 = 5(16)/16 = 5 kN ✓
For the three-moment equation applied to a two-span beam, what is the 6Ax̄/L term for a span of length L = 5 m carrying a central point load P = 20 kN?
For a concentrated load P at midspan of a simply supported beam, the simple-beam M-diagram is a triangle with peak M = PL/4 at midspan. The area A = (1/2)(L)(PL/4) = PL²/8 and centroid x̄ = L/2 from either support (symmetric). Thus 6Ax̄/L = 6·(PL²/8)·(L/2)/L = 6·PL²/16 = 3PL²/8. With P=20, L=5: = 3(20)(25)/8 = 1500/8 = 187.5 kN·m².
What is the degree of static indeterminacy of a fixed-fixed beam?
A fixed-fixed beam has reactions: 2 vertical reactions + 2 horizontal reactions + 2 fixed-end moments = 6 total reaction components. Equilibrium equations available = 3 (ΣFx=0, ΣFy=0, ΣM=0). DI = 6 − 3 = 3. This means 3 compatibility equations are needed. If the beam has no horizontal load, one equation is trivial (horizontal equilibrium), so effectively 2 non-trivial compatibility equations govern the 2 end moments.
For a two-span continuous beam (spans L₁ = 5 m and L₂ = 7 m) with UDL w = 10 kN/m on both spans and simple supports at both ends, set up the three-moment equation at interior support B.
Apply three-moment equation: M_A(L₁) + 2M_B(L₁+L₂) + M_C(L₂) = −6(A₁x̄₁/L₁ + A₂x̄₂/L₂). Simple ends: M_A = M_C = 0. UDL terms: 6Ax̄/L = wL³/4. Span AB: 10(5)³/4 = 10(125)/4 = 312.5. Span BC: 10(7)³/4 = 10(343)/4 = 857.5. Equation: 2M_B(12) = −(312.5 + 857.5) = −1170. 24M_B = −1170 → M_B = −48.75 kN·m (hogging).
What is the relationship between the number of compatibility equations required and the degree of indeterminacy?
Each redundant introduces one unknown that cannot be found by statics. For each such redundant, you need one additional equation — the compatibility (deformation) equation at the point of the removed redundant. For a DI=1 structure: 1 equation. For DI=2: 2 simultaneous equations with 2 unknowns. For DI=n: n equations. This is the mathematical core of the force method.
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