CELE Strength of Materials — Beam DeflectionsRevision Notes
Condensed revision notes for Beam Deflections, built for the final weeks before the CELE 2026. These are the distilled key points you need when there is no time left for full study notes — just the concepts, formulas, and traps 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 Strength of Materials subtest is marked as "Core" in the official pattern, and Beam Deflections appears in position 5th of 8 in the CELE Strength of Materials 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.
Beam Deflections - Revision Notes
Beam deflection is a critical serviceability limit state in structural design. A beam may be adequately strong in bending and shear yet fail in service if it deflects excessively — causing cracked finishes, misaligned doors, or an uncomfortable bouncing floor. The NSCP 2015 (Section 506) prescribes maximum allowable deflections, commonly L/360 for live load on members supporting brittle finishes. This chapter covers four principal methods for computing deflections: (1) Double Integration, (2) Area-Moment Theorems, (3) Conjugate-Beam Method, and (4) Superposition using standard formulas. Mastery of these methods is essential for both the Strength of Materials and Structural Theory portions of the PRC Civil Engineer Licensure Examination.
Sections
Formulas
Example
For a simply supported beam of span L under central load P, M(x) = (P/2)x for 0 ≤ x ≤ L/2. Substituting into EI y'' = M(x) and integrating twice (with BCs y=0 at x=0 and x=L) yields the elastic curve.
Formula
EI (d²y/dx²) = M(x)
Variables
E = modulus of elasticity (MPa or N/mm²), I = moment of inertia (mm⁴), y = deflection (mm), x = position along beam (mm), M(x) = bending moment at x (N·mm)
Application
Master differential equation relating curvature to bending moment. Foundation of the double-integration method.
Example
For a cantilever with a free-end point load P: integrating EI y'' = P(L - x) gives EI y' = PLx - x²/2 + C₁. Applying y'(0) = 0 (fixed end) gives C₁ = 0.
Formula
EI (dy/dx) = ∫M(x) dx + C₁
Variables
dy/dx = slope of elastic curve, C₁ = constant of integration determined from boundary/symmetry conditions
Application
First integration of the governing equation gives the slope function.
Example
Continuing the cantilever example: EI y = PLx²/2 - x³/6 + C₂. Applying y(0) = 0 gives C₂ = 0. At x = L (free end): EI δ = PL³/2 - PL³/6 = PL³/3, so δ = PL³/(3EI).
Formula
EI y = ∬M(x) dx + C₁x + C₂
Variables
y = deflection, C₁ and C₂ = constants of integration determined from two boundary conditions
Application
Second integration gives the deflection function y(x). Two boundary conditions are always needed to find C₁ and C₂.
Exam Tips
- In board exams, the double-integration method is rarely the fastest — use standard formulas and superposition whenever the load case matches a tabulated formula.
- For simply supported beams with symmetric loading, use symmetry: slope at midspan = 0, so set dy/dx = 0 at x = L/2 to find C₁ directly.
- Macaulay brackets integrate like polynomials but are NOT evaluated until x > a — practice this technique separately.
- Always state units explicitly in your solution and do a rough magnitude check — deflections for typical beam spans are usually in the range of 1–30 mm.
Key Points
- The elastic curve is the deflected shape of the beam's neutral axis under load.
- For small deflections of a linearly elastic beam, the exact curvature κ = M/EI reduces to d²y/dx² = M(x)/EI.
- EI is the flexural rigidity — E is Young's modulus (MPa) and I is the second moment of area (mm⁴).
- Slope θ ≈ dy/dx (in radians) for small deflections.
- The sign convention must be established before integration: downward deflection is typically taken as positive in most textbooks, but be consistent throughout a problem.
- Boundary conditions physically constrain the solution: a simple support gives y = 0; a fixed support gives y = 0 AND dy/dx = 0; a free end has no geometric constraints.
- A larger EI means a stiffer beam and smaller deflections for the same loading.
Definitions
Term
Elastic Curve
Definition
The deflected shape of the beam's neutral axis, described mathematically by y(x). It must satisfy both the differential equation EI y'' = M(x) and the boundary conditions of the support system.
Importance
Understanding the elastic curve shape helps predict where maximum deflection occurs and aids in checking computed results.
Term
Flexural Rigidity (EI)
Definition
The product of the modulus of elasticity E and the moment of inertia I of the cross-section. It measures the resistance of a beam to bending deformation.
Importance
EI appears in every deflection formula. For steel beams, E = 200 GPa = 200,000 MPa. Increasing I (by using a deeper section) is the most effective way to reduce deflection.
Term
Boundary Conditions
Definition
Geometric constraints at beam supports used to evaluate constants of integration. A pin or roller support: y = 0. A fixed (clamped) support: y = 0 AND y' = 0. A free end: no geometric constraint (but shear and moment are zero).
Importance
Incorrectly applying boundary conditions is the most common source of error in double-integration problems. Always identify the number and type of supports before integrating.
Term
Macaulay Bracket ⟨x − a⟩ⁿ
Definition
A singularity function defined as: ⟨x−a⟩ⁿ = 0 if x < a; (x−a)ⁿ if x ≥ a. Used to write a single M(x) expression valid over the entire beam span, avoiding piecewise integration.
Importance
Essential for beams with multiple point loads, concentrated moments, or partial UDLs. Saves significant algebraic work compared to writing separate equations for each segment.
Section Title
The Elastic Curve and Governing Differential Equation
Common Mistakes
- Using mixed units — always convert everything to N and mm (or kN and m) before substituting into formulas. The L⁴ or L³ terms magnify unit errors enormously.
- Applying only one boundary condition at a fixed end — a fixed support requires BOTH y = 0 AND y' = 0.
- Forgetting that M(x) must be written as a function of x measured from the chosen origin — not as a constant.
- Not identifying the correct origin for x — always state where x = 0 is measured from.
- Assuming maximum deflection always occurs at midspan — this is only true for symmetric loading on a simply supported beam.
Formulas
Example
A 6 m beam with R_A = 20 kN, a 15 kN load at 2 m, and a UDL of 5 kN/m from 3 m to 6 m: M(x) = 20x − 15⟨x−2⟩ − (5/2)⟨x−3⟩² [kN·m, x in m]
Formula
M(x) = R_A · x − P₁⟨x − a₁⟩¹ − P₂⟨x − a₂⟩¹ − (w/2)⟨x − a₃⟩²
Variables
R_A = left reaction, P₁, P₂ = point loads at positions a₁, a₂ from left, w = UDL intensity, a₃ = start of UDL
Application
General Macaulay moment equation for a simply supported beam with multiple loads. Brackets ⟨ ⟩ activate only when x exceeds the subscript value.
Example
For a simply supported beam with an off-center point load, set the slope expression (first integral) to zero and solve for x. Then substitute into the second integral to get δ_max.
Formula
δ_max occurs where dy/dx = 0
Variables
δ_max = maximum deflection; for symmetric simply supported beams this is always at x = L/2
Application
To find the location and magnitude of maximum deflection when the load is not symmetric.
Exam Tips
- In the PRC exam, double integration problems usually involve a single load on a simple beam or cantilever — check if a standard formula applies first.
- When using Macaulay brackets, integrate and differentiate them just like regular polynomials: ∫⟨x−a⟩ⁿ dx = ⟨x−a⟩ⁿ⁺¹/(n+1).
- Check your answer dimensionally: δ must have units of length (mm or m). EI has units of N·mm², so PL³/(3EI) is [N·mm³]/[N·mm²] = mm. ✓
Key Points
- Integrate EI y'' = M(x) twice to obtain y(x); apply boundary conditions to find integration constants C₁ and C₂.
- Write M(x) as a function of position x along the beam, using Macaulay brackets for multi-segment loading.
- For a beam with n segments, double integration without Macaulay brackets requires 2n constants of integration, all found by matching continuity conditions (y and y' must be equal at internal boundaries) plus end boundary conditions.
- Macaulay's method collapses this to just C₁ and C₂ regardless of the number of loads, making it far more efficient.
- The method gives the complete elastic curve y(x) — from this, the slope at any point and the deflection at any point can be found.
- Useful rule: at a point of maximum deflection, dy/dx = 0. Set the slope equation to zero and solve for x, then substitute back into y(x).
Definitions
Term
Macaulay's Method (Singularity Functions)
Definition
A systematic way to write a single bending-moment equation valid along the entire beam using singularity functions ⟨x−a⟩ⁿ. These brackets behave like ordinary polynomial terms (x−a)ⁿ only when x > a; they equal zero when x < a.
Importance
Eliminates the need to write and integrate separate M(x) expressions for each beam segment. Essential for multi-load problems in board exams.
Term
Integration Constants (C₁, C₂)
Definition
Two constants arising from the indefinite double integration of EI y'' = M(x). They represent the slope at x = 0 (C₁/EI) and the deflection at x = 0 (C₂/EI) before boundary conditions are applied.
Importance
These constants are physically meaningful — C₂ = 0 whenever the deflection at the origin is zero (e.g., simple support at x = 0).
Section Title
Method 1 — Double Integration Method
Common Mistakes
- Evaluating Macaulay brackets at the wrong x — the bracket ⟨x−a⟩ equals zero when x ≤ a, NOT when x < a only.
- Forgetting to carry the negative sign for point loads downward (gravity loads produce negative M contribution relative to positive reaction).
- Integrating a UDL term as w⟨x−a⟩ instead of the correct (w/2)⟨x−a⟩² — the factor of 1/2 is part of the integration.
- Applying boundary conditions before fully integrating — always complete both integrations first, then substitute BCs.
Formulas
Example
Cantilever, length L, free-end point load P: M/EI diagram is triangular, area = (1/2)(L)(PL/EI) = PL²/(2EI). This equals the slope at the free end (since slope at fixed end = 0).
Formula
θ_{B/A} = Area of (M/EI) diagram from A to B
Variables
θ_{B/A} = change in slope from A to B (rad), Area = net area of M/EI diagram between the two points
Application
Theorem 1. Use when you need the slope at a point relative to another known tangent.
Example
Cantilever, UDL w, length L: M/EI diagram is parabolic. Area = (1/3)(L)(wL²/2EI) = wL³/(6EI). Centroid of parabolic area is at 3L/4 from the free end. t = (wL³/6EI)(3L/4) = wL⁴/(8EI) = free-end deflection.
Formula
t_{B/A} = (Area of M/EI from A to B) × x̄_B
Variables
t_{B/A} = tangential deviation of B from the tangent at A (mm), x̄_B = distance from the centroid of the M/EI area to the vertical line through B
Application
Theorem 2. Compute deflections by finding the vertical distance from the elastic curve to a reference tangent line.
Example
For midspan deflection of a simply supported beam with symmetric loading, use symmetry directly: slope at midspan = 0, so t_{midspan/A} = δ_midspan (the deviation from the horizontal tangent at midspan equals the midspan deflection).
Formula
δ_C = t_{C/A} − (x_C/L) × t_{B/A} [for simply supported beam A–B, point C between A and B]
Variables
δ_C = actual deflection at C, t_{C/A} = deviation of C from tangent at A, t_{B/A} = deviation of B from tangent at A, x_C = distance from A to C
Application
Geometric relationship for simply supported beams where neither end tangent is horizontal. Relates actual deflection to computed deviations.
Exam Tips
- For board exam problems using area-moment: sketch the M/EI diagram first, identify its shape, recall the centroid location, then apply the theorems.
- Standard M/EI area centroids to memorize: Rectangle → centroid at L/2; Triangle → centroid at L/3 from the larger end; Parabola (apex at one end) → centroid at L/4 from the base end.
- Area-moment is fastest for cantilevers and propped cantilevers. For simple beams with a single midspan load, the standard formula is faster.
Key Points
- This method uses the M/EI diagram (the bending moment diagram divided by EI) to compute slopes and deflections geometrically.
- Theorem 1 (Slope): The change in slope between two points A and B on the elastic curve equals the area of the M/EI diagram between A and B.
- Theorem 2 (Deflection): The vertical deviation of point B from the tangent drawn at A equals the first moment of the M/EI area between A and B, taken about a vertical line through B.
- The tangential deviation t_{B/A} is NOT the same as the actual deflection at B — you must use geometry to relate deviations to actual deflections.
- Most efficient for cantilevers: the tangent at the fixed end is horizontal, so the tangential deviation from the fixed-end tangent equals the actual deflection directly.
- For simply supported beams: use symmetry (tangent at midspan is horizontal for symmetric loading) or compute deviations from both ends and apply similar triangles.
- The M/EI diagram for common load cases: a UDL gives a parabolic M diagram; a point load gives a triangular M diagram.
Definitions
Term
M/EI Diagram
Definition
The bending moment diagram scaled by dividing each ordinate by the corresponding EI value. If EI is constant, the M/EI diagram has the same shape as the M diagram but with ordinates reduced by factor 1/EI.
Importance
The M/EI diagram is the 'load' for both the area-moment method and the conjugate-beam method. Recognizing common shapes (triangles, parabolas) and their centroids is essential.
Term
Tangential Deviation t_{B/A}
Definition
The vertical distance from point B on the elastic curve to the tangent line drawn to the elastic curve at point A. It is positive when B is above the tangent at A (depends on sign of moment). Crucially, t_{B/A} ≠ t_{A/B} in general.
Importance
Often confused with actual deflection. For cantilevers, t_{B/A} from the fixed end equals the actual deflection at B. For other beams, geometry must be used to extract actual deflection.
Section Title
Method 2 — Area-Moment (Moment-Area) Theorems
Common Mistakes
- Confusing t_{B/A} with t_{A/B} — the subscript order matters; t_{B/A} is the deviation of B from the tangent at A.
- Forgetting to locate the centroid of the M/EI area correctly — triangular areas have centroid at 1/3 from the larger base; parabolic areas have centroid at 3/4 from the vertex (or 1/4 from the base).
- Using Theorem 2 directly as the deflection for a simply supported beam — you must apply the geometric correction relating deviations to actual deflections.
- Getting the sign of t_{B/A} wrong — for a positive bending moment (sagging), the elastic curve is concave up, and the tangential deviation sign depends on the direction of computation.
Formulas
Example
For a cantilever with free-end load P: the conjugate beam is a fixed-end beam at right, free at left, loaded with a triangular M/EI load (apex at free end). The 'reaction' (fixed-end moment) of the conjugate beam = free-end deflection of the real beam = PL³/(3EI).
Formula
V'(x) = θ(x) and M'(x) = y(x)
Variables
V'(x) = shear in conjugate beam at x, θ(x) = slope of real beam at x, M'(x) = bending moment in conjugate beam at x, y(x) = deflection of real beam at x
Application
The fundamental conjugate-beam analogy. Use it to read off slopes and deflections as if solving a statics problem.
Exam Tips
- The conjugate-beam method is fastest when you only need the deflection at one specific point — find the conjugate beam's bending moment at that point using equilibrium.
- For board exam conjugate-beam problems: (1) Draw M/EI diagram; (2) Transform supports; (3) Apply M/EI as distributed load; (4) Find reactions of conjugate beam; (5) Compute shear (slope) or moment (deflection) at desired point.
- For a simply supported beam (real), the conjugate beam is also simply supported — both have pin and roller supports, making the conjugate-beam reactions easy to find.
Key Points
- The conjugate-beam method converts the deflection problem into an equivalent statics problem (shear and moment of a loaded beam).
- The conjugate beam has the same span as the real beam but different support conditions, and is loaded with the real beam's M/EI diagram as a distributed load.
- Key analogy: Shear in conjugate beam = Slope in real beam; Moment in conjugate beam = Deflection in real beam.
- Support transformation rules: Real pin/roller → Conjugate pin/roller (y=0 becomes V=0... wait — actually: Real pin: y=0 and M'=0 → Conjugate pin: M_conj=0 which matches. Real fixed end: y=0 and y'=0 → Conjugate free end. Real free end: y≠0 and y'≠0 → Conjugate fixed end.
- Correct support conversion: Pin/Roller → Pin/Roller (unchanged); Fixed End → Free End; Free End → Fixed End; Internal Hinge → Internal Support.
- After setting up the conjugate beam, find the shear and moment at any point using standard statics (sum of forces and moments).
- The method is particularly efficient when the deflection/slope is needed at only one or two points.
Definitions
Term
Conjugate Beam
Definition
An imaginary beam of the same length as the real beam, loaded with the real beam's M/EI diagram as a distributed load, but with supports transformed according to specific rules (fixed ↔ free, pins stay as pins). Its shear and moment diagrams give the slope and deflection of the real beam.
Importance
Converts a calculus problem (integration) into a statics problem (equilibrium), which many engineers find more intuitive.
Term
Support Transformation Rule
Definition
The rule for converting real beam supports to conjugate beam supports: (1) Real pin or roller → Conjugate pin or roller (same); (2) Real fixed end → Conjugate free end; (3) Real free end → Conjugate fixed end. This ensures the boundary conditions of the real beam are satisfied automatically.
Importance
Applying the wrong support conditions to the conjugate beam gives completely wrong answers. This is a frequent error.
Section Title
Method 3 — Conjugate-Beam Method
Common Mistakes
- Applying the wrong support conversions — the most critical step. A fixed end MUST become a free end in the conjugate beam.
- Not recognizing that the 'load' on the conjugate beam is M/EI (not just M) — always divide by EI.
- Forgetting to use the sign of the shear and moment in the conjugate beam correctly — positive V' means upward slope (beam slopes upward from left to right).
- Using conjugate-beam for a beam with an internal hinge without correctly applying the modified internal support condition.
Formulas
Example
Simply supported steel beam, L = 6 m, P = 20 kN, I = 60×10⁶ mm⁴, E = 200,000 MPa: δ = (20,000)(6,000)³ / (48 × 200,000 × 60×10⁶) = 4.32×10¹⁵ / 5.76×10¹⁴ = 7.5 mm
Formula
δ_max = PL³ / (48EI)
Variables
P = central point load (N), L = span (mm), E = modulus of elasticity (MPa), I = moment of inertia (mm⁴). Deflection occurs at midspan.
Application
Simply supported beam, single central concentrated load. Most common formula in board exams.
Example
L = 6,000 mm, w = 10 N/mm, E = 200,000 MPa, I = 80×10⁶ mm⁴: δ = 5(10)(6000)⁴ / (384 × 200,000 × 80×10⁶) = 6.48×10¹⁶ / 6.144×10¹⁵ = 10.5 mm. Check: L/360 = 16.7 mm. ∴ OK.
Formula
δ_max = 5wL⁴ / (384EI)
Variables
w = UDL intensity (N/mm), L = span (mm), E (MPa), I (mm⁴). Deflection at midspan.
Application
Simply supported beam, full-span uniformly distributed load. Second most common formula.
Example
Cantilever L = 2,500 mm, P = 8,000 N, EI = 1.2×10¹³ N·mm²: δ = 8,000 × (2,500)³ / (3 × 1.2×10¹³) = 1.25×10¹⁴ / 3.6×10¹³ = 3.47 mm
Formula
δ_max = PL³ / (3EI)
Variables
P = point load at free end (N), L = cantilever length (mm), E (MPa), I (mm⁴). Deflection at free end.
Application
Cantilever beam, concentrated load at free end. EIGHT times more flexible than an equivalent simply supported beam with central load.
Example
Cantilever L = 3,000 mm, w = 3 N/mm, EI = 2.0×10¹³ N·mm²: δ = 3(3,000)⁴ / (8 × 2.0×10¹³) = 2.43×10¹⁴ / 1.6×10¹⁴ = 1.52 mm
Formula
δ_max = wL⁴ / (8EI)
Variables
w = UDL intensity (N/mm), L = cantilever length (mm). Deflection at free end.
Application
Cantilever beam, full uniformly distributed load.
Example
Cantilever L = 2,000 mm, M = 5×10⁶ N·mm, EI = 1.5×10¹³ N·mm²: δ = 5×10⁶ × (2,000)² / (2 × 1.5×10¹³) = 2.0×10¹³ / 3.0×10¹³ = 0.67 mm
Formula
δ_max = ML² / (2EI)
Variables
M = applied moment at free end (N·mm), L = cantilever length (mm). Deflection at free end.
Application
Cantilever beam, concentrated moment applied at the free end.
Example
6 m floor beam: δ_allow = 6,000/360 = 16.7 mm. A computed live-load deflection of 12 mm satisfies this limit; a deflection of 20 mm does NOT.
Formula
δ_allowable = L / 360 (live load on brittle finishes)
Variables
L = beam span (mm), δ_allowable = maximum permissible live-load deflection (mm). Per NSCP 2015 Table 506-1.
Application
Serviceability check for floor beams supporting plaster ceilings or tile finishes. Other limits: L/240 for total load, L/180 for roof members not supporting plaster.
Exam Tips
- Memorize the 'big five' formulas as a numbered list and their denominators: 48, 384 (with 5 in numerator), 3, 8, 2. Drill these until automatic.
- For the serviceability check: compute δ_computed, then compute L/360. If δ_computed ≤ L/360, write 'SATISFACTORY' or 'OK'. If not, you need a larger I.
- To find required I for a deflection limit: rearrange the standard formula. Example: I_req = 5wL⁴/(384E × δ_allow). Substitute δ_allow = L/360.
- Superposition of two cantilever loads: δ_total = δ_P + δ_w = PL³/(3EI) + wL⁴/(8EI). Both are in the same direction (downward), so add them.
Key Points
- Superposition: for linear elastic beams, the deflection under combined loads equals the sum of deflections from each load acting alone.
- Validity requires: material is linearly elastic (Hooke's Law applies), small deflections (so geometry is not significantly changed by loading), no interaction between load effects.
- Superposition is the FASTEST method for board exam problems when the loading can be decomposed into tabulated standard cases.
- Five formulas to memorize absolutely: (1) Simply supported + central P; (2) Simply supported + full UDL w; (3) Cantilever + free-end P; (4) Cantilever + full UDL w; (5) Cantilever + free-end moment M.
- For partial UDL, point load off-center, or overhanging beams: use superposition of the standard cases creatively (e.g., add a full UDL and subtract the portion not loaded).
- Deflection compatibility: Setting deflections from multiple load effects equal to zero (or to some prescribed value) at a support is the key to solving statically indeterminate beams.
Definitions
Term
Principle of Superposition
Definition
For a linear elastic structure under multiple loads, the total response (deflection, slope, stress) equals the algebraic sum of the individual responses to each load acting separately. Valid only when deflections are small and material remains elastic.
Importance
Enables rapid computation of deflections for complex loading by breaking it into simple tabulated cases. Essential for board exam efficiency.
Term
Serviceability Limit State
Definition
A design condition related to the in-service performance of a structure — deflection, vibration, cracking — rather than its ultimate load-carrying capacity. NSCP 2015 Section 506 governs deflection limits for beams.
Importance
A beam can pass all strength checks (bending, shear) but still fail the deflection serviceability check and require a larger section.
Term
Deflection Compatibility
Definition
The condition that geometric constraints at supports must be satisfied. For a propped cantilever, the deflection at the prop must be zero. Setting the deflection from the applied load equal to the deflection from the unknown prop reaction gives the compatibility equation used to solve for indeterminate reactions.
Importance
This concept is the bridge from determinate to indeterminate structural analysis.
Section Title
Method 4 — Superposition with Standard Formulas
Common Mistakes
- Mixing up the denominators: 48 (simple + central P) vs 384/5 (simple + UDL) vs 3 (cantilever + P) vs 8 (cantilever + UDL). A memory trick: cantilever formulas have smaller denominators because cantilevers are much more flexible.
- Forgetting the factor of 5 in the numerator for the UDL formula for simple beams: 5wL⁴/(384EI), not wL⁴/(384EI).
- Applying the superposition principle when the beam is not linearly elastic or deflections are large (geometric nonlinearity) — invalid in those cases.
- Using L in meters in the formula but E and I in MPa and mm⁴ — all lengths must be in the same unit system. Recommended: use N and mm throughout.
- Checking L/360 against the TOTAL load deflection instead of just the LIVE load deflection — the L/360 limit in NSCP applies to live load deflection only.
Formulas
Example
8 m beam: δ_allow = 8,000/360 = 22.2 mm. A computed LL deflection of 18 mm passes; 25 mm fails.
Formula
δ_allow = L / 360 (LL only, brittle finishes)
Variables
L = beam span in mm, δ_allow = allowable live-load deflection in mm. NSCP 2015 Table 506-1.
Application
Most common serviceability check. Must be applied after computing the live-load-only deflection (not dead + live).
Example
Propped cantilever, L = 4 m, w = 15 kN/m: R_prop = 3(15)(4)/8 = 22.5 kN. Fixed-end reaction = 15×4 − 22.5 = 37.5 kN.
Formula
R_prop = 3wL / 8 (propped cantilever, full UDL)
Variables
R_prop = prop reaction (N), w = UDL intensity (N/mm), L = span (mm)
Application
Classic indeterminate result from deflection compatibility. The prop takes 3/8 of the total UDL, and the fixed end takes 5/8.
Example
Simply supported, L = 5,000 mm, w = 12 N/mm, E = 200,000 MPa, δ_allow = 5,000/360 = 13.89 mm: I_req = 5(12)(5,000)⁴ / (384 × 200,000 × 13.89) = 9.375×10¹⁶ / 1.068×10¹² = 87.8×10⁶ mm⁴. Select a steel section with I ≥ 87.8×10⁶ mm⁴.
Formula
I_req = 5wL⁴ / (384 × E × δ_allow) [for simply supported + full UDL]
Variables
I_req = required moment of inertia (mm⁴), δ_allow = L/360 or other limit (mm)
Application
Design formula: find the minimum I needed to satisfy a deflection limit under a given UDL on a simply supported beam.
Exam Tips
- In board exam problems, the deflection check is always: compute δ, compute L/360 (or the stated limit), compare. Three numbers, one comparison.
- When asked to find the required I for a deflection limit: isolate I from the appropriate standard formula and substitute δ_allow = L/360.
- Memorize that the prop reaction in a propped cantilever under UDL is 3wL/8 — this appears frequently in structural theory problems.
Key Points
- NSCP 2015 Section 506 (formerly NSCP 2001 Section 409) specifies maximum computed deflections for beams and one-way slabs.
- The most commonly tested limit: L/360 for live load on floors supporting brittle finishes (plaster, tile).
- Other NSCP limits: L/240 for live load + long-term effects (total deflection after creep); L/180 for roof members not supporting plastered ceilings.
- In steel design (AISC 360 / NSCP Steel), L/360 for live load and L/240 for total load are standard serviceability benchmarks.
- For timber members (NSCP Timber), deflection limits also apply: L/360 for floors, L/240 for roofs with plaster.
- Camber is a precamber (upward curve) built into the beam to compensate for expected dead-load deflection, so the beam appears level under its own weight.
- The propped cantilever example of deflection compatibility: For a cantilever of length L under UDL w, with a prop at the free end providing reaction R: δ_free_end_from_w = wL⁴/(8EI) (downward); δ_free_end_from_R = RL³/(3EI) (upward). Setting them equal: R = 3wL/8.
Definitions
Term
Camber
Definition
An intentional upward curvature built into a beam during fabrication to offset the anticipated downward deflection under dead load (and sometimes a portion of live load). After loading, the cambered beam appears straight.
Importance
Commonly specified for long-span steel beams (AISC recommends cambering for spans > 9 m or deflections > 20 mm under dead load).
Term
Long-term Deflection (Creep)
Definition
For concrete and timber members, deflection increases over time due to creep and shrinkage. ACI 318-19 Section 24.2.4 requires multiplying the immediate deflection by a multiplier λ_Δ = ξ/(1 + 50ρ') to obtain long-term additional deflection, where ξ depends on duration (2.0 for 5 years or more).
Importance
Total deflection (immediate + long-term) must be checked against L/240. This is tested in both Strength of Materials and Reinforced Concrete Design portions.
Section Title
NSCP 2015 Deflection Limits and Design Application
Common Mistakes
- Checking L/360 against the combined dead + live load deflection — the code limit applies to live load deflection only for the L/360 criterion.
- Forgetting to apply the correct NSCP limit — using L/360 for a roof member not supporting plaster when L/180 actually applies (less restrictive).
- Not checking deflection at all when the problem asks to 'design' a beam — serviceability is a separate and mandatory check from strength.
Connections
- Beam Deflections → Statically Indeterminate Beams: Deflection compatibility equations (e.g., δ = 0 at a prop) are the fundamental tool for analyzing propped cantilevers, fixed-fixed beams, and continuous beams by the force method (consistent deformations).
- Beam Deflections → Serviceability Design (NSCP 2015 Section 506): The computed maximum deflection is checked against code-prescribed limits (L/360, L/240, L/180) as a mandatory serviceability verification separate from strength design.
- Beam Deflections → Flexural Stress (Bending Theory): Both bending stress (σ = Mc/I) and deflection depend on EI and M(x) — the same beam properties govern both. Understanding bending stress is prerequisite to understanding the elastic curve.
- Beam Deflections → Reinforced Concrete Design (ACI 318 / NSCP): ACI 318-19 Section 24.2 uses moment-of-inertia formulas (including the effective moment of inertia Ie for cracked sections) in deflection calculations. Long-term deflection multipliers (creep factor λ_Δ) extend the basic elastic formulas.
- Beam Deflections → Steel Design (AISC 360 / NSCP Steel): Serviceability in steel design requires checking live-load deflection (typically L/360) and total deflection (L/240). Camber is specified based on computed dead-load deflection.
- Beam Deflections → Shear and Bending Moment Diagrams: Drawing accurate M(x) diagrams (or using Macaulay brackets) is the essential first step for both the double-integration method and the area-moment/conjugate-beam methods.
- Beam Deflections → Structural Theory (PRC Exam): The deflection compatibility method for indeterminate beams (propped cantilever, fixed-ended beams, two-span continuous beams) is a core topic in Structural Theory, which builds directly on the formulas derived in this chapter.
- Beam Deflections → Moment Distribution Method: Fixed-end moments used in the moment distribution method are derived from beam deflection formulas (carry-over factor = 0.5 comes from the elastic curve of a fixed-pinned beam).
Exam Strategy
For PRC board exam Beam Deflection problems, follow this decision tree: (1) IDENTIFY the beam type (simply supported, cantilever, propped, fixed-fixed) and loading (point load, UDL, moment, or combination). (2) CHECK if the loading matches one of the five standard formula cases — if yes, apply the formula directly (90% of board exam problems fit standard cases). (3) For COMBINED loads, use superposition: apply each standard formula and add. (4) For the SERVICEABILITY CHECK, always compute both δ_computed and δ_allow = L/360, then explicitly state the comparison. (5) For REQUIRED I problems, rearrange the standard formula algebraically to isolate I. (6) For INDETERMINATE beams (propped cantilever), use deflection compatibility: (a) remove the redundant reaction, (b) compute deflection at the redundant's location from applied loads, (c) compute deflection there from the unknown redundant, (d) equate and solve. UNIT DISCIPLINE is critical — always convert to N and mm before substituting into any formula. The L³ or L⁴ terms will amplify even small unit errors by three or four orders of magnitude. Practice the five standard formulas until you can write them from memory in under 10 seconds. In the actual exam, allocate about 3–4 minutes per deflection item.
Quick Review Questions
A simply supported steel beam spans 8 m and carries a central point load of 30 kN. Given E = 200 GPa and I = 120×10⁶ mm⁴, compute the midspan deflection and check against the L/360 serviceability limit.
Use the standard formula δ = PL³/(48EI) for a central point load on a simply supported beam. Work in N and mm throughout: P = 30,000 N, L = 8,000 mm, E = 200,000 N/mm², I = 120×10⁶ mm⁴. The denominator is 48 (not 384). Always compute L/360 and compare explicitly.
A cantilever beam, 3 m long, carries both a free-end point load P = 5 kN and a full UDL w = 3 kN/m. If EI = 2.0×10¹³ N·mm², find the total free-end deflection using superposition.
Superposition: apply each load's formula separately and add results (both act in the same direction — downward). Convert w = 3 kN/m = 3 N/mm before substituting. The denominators are 3 (point load) and 8 (UDL) for cantilevers.
What are the correct support transformations for the conjugate-beam method? List the conversions for: (a) pin/roller, (b) fixed end, (c) free end.
The transformation is based on matching boundary conditions: A fixed end in the real beam has zero deflection AND zero slope (two constraints), which in the conjugate beam corresponds to a free end where shear = 0 and moment = 0 have no such constraints — wait, actually the conjugate free end allows non-zero shear and moment, which represent non-zero slope and deflection at the real fixed end... but we KNOW they're zero from the real beam. The transformation is applied so that the conjugate beam's equilibrium equations automatically reproduce the correct geometric boundary conditions of the real beam.
State the two Area-Moment Theorems precisely.
These theorems convert the double integration into geometry of the M/EI diagram. Theorem 1 gives slopes; Theorem 2 gives deviations (which must then be related to actual deflections using geometry). The most common error is applying Theorem 2 deviation directly as the deflection without the geometric correction.
A simply supported beam of span L = 5 m carries a full UDL of w = 12 kN/m. E = 200 GPa. What minimum moment of inertia I is required to ensure the beam does not exceed the L/360 deflection limit?
Rearrange the standard UDL formula to isolate I. Use N and mm: w = 12 N/mm, L = 5,000 mm, E = 200,000 MPa. This is the design approach: find the required I, then select a steel section (from AISC or local tables) with I ≥ 87.8×10⁶ mm⁴.
A propped cantilever of span L carries a full UDL w. Using deflection compatibility, determine the prop reaction R.
This is deflection compatibility applied to an indeterminate beam. Remove the prop (redundant reaction), compute the deflection at the prop location from the applied load, then find the force R that would produce equal and opposite deflection. The two expressions are equated and solved for R. The factor 3wL/8 is a fundamental result in structural analysis.
For the governing differential equation EI y'' = M(x), what are the appropriate boundary conditions for: (a) a simply supported beam at its supports, (b) a fixed (clamped) end?
Boundary conditions translate physical support conditions into mathematical constraints used to determine the integration constants C₁ and C₂. Each supported beam has exactly two boundary conditions (one for each integration constant). A fixed end provides both a geometric (y=0) and a kinematic (y'=0) constraint. Forgetting y'=0 at a fixed end is the most common error in double-integration problems.
Which deflection formula applies to a simply supported beam with a full UDL, and why does the maximum deflection occur at midspan?
Symmetry is a powerful tool. For symmetric loading on a simply supported beam, set dy/dx = 0 at x = L/2 to find C₁ without evaluating the second boundary condition. The factor 5/384 (approximately 0.01302) comes from the integration of a parabolic M(x) diagram twice. Note that 5/384 ≈ 0.013 is about 2.5 times the factor 1/48 ≈ 0.0208 for a central point load (with the same total load wL = P, UDL gives less deflection than a central point load because the load is more uniformly distributed).
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