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CELE Strength of Materials — Beam DeflectionsDetailed Explanation

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Detailed explanation of Beam Deflections for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Strength of Materials subtest.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Strength of Materials section sits under a "Core" weighting, and Beam Deflections is the 5th chapter in the 8-chapter CELE Strength of Materials rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Strength of Materials.

Beam Deflections - Detailed Explanation

Beam deflection is one of the most frequently tested topics in the Strength of Materials portion of the PRC Civil Engineer Licensure Examination. A beam may be perfectly safe against bending stress failure yet still be unacceptable in service if it deflects excessively — causing cracked plaster, jammed doors, vibrating floors, or misaligned machinery. The National Structural Code of the Philippines (NSCP 2015, Section 406) imposes deflection limits as a serviceability requirement, the most familiar being L/360 for live-load deflection of members supporting brittle finishes. This chapter systematically develops four methods for computing deflections — the Double-Integration Method, the Area-Moment (Moment-Area) Method, the Conjugate-Beam Method, and Superposition using standard formulas — and shows how each method is applied in board-style problems. Mastery of this chapter also unlocks the analysis of statically indeterminate beams, which appears in Structural Theory, because the compatibility equations that solve those problems are nothing more than deflection equations set equal to zero or to each other.

Concepts

The Elastic Curve and the Governing Differential Equation

When a beam bends under load within the elastic range, its longitudinal axis deforms into a smooth curve called the elastic curve. At any cross-section located at distance x from the chosen origin, the beam has a deflection y (transverse displacement) and a slope θ = dy/dx. For small deflections — the standard assumption in structural engineering — the curvature of the elastic curve can be approximated as d²y/dx², and the fundamental relationship between curvature and bending moment gives the governing equation: EI · d²y/dx² = M(x) Here, E is the modulus of elasticity of the beam material (e.g., 200 GPa for structural steel per AISC 360 and NSCP Table 702.3.1), I is the second moment of area of the cross-section about the neutral axis (mm⁴), and M(x) is the internal bending moment at position x (N·mm). The product EI is called the flexural rigidity — it is the beam's resistance to bending deformation. Sign convention: In the most common convention used in Philippine textbooks and board exams, x is measured from the left support, y is positive upward, and M is positive when it causes sagging (tension on bottom fiber). Downward deflection is therefore negative y, but many solutions simply compute the magnitude and state the direction. Differentiating the moment equation links all four beam functions: EI · y'''' = w(x) [distributed load intensity, positive upward] EI · y''' = V(x) [shear force] EI · y'' = M(x) [bending moment] EI · y' = ∫M dx + C₁ [slope θ] EI · y = ∬M dx + C₁x + C₂ [deflection] The constants C₁ and C₂ are evaluated from boundary conditions (BCs): at a simple (pin or roller) support, y = 0; at a fixed (clamped) support, both y = 0 AND y' = 0; at a free end, M = 0 AND V = 0.

Examples

This example demonstrates the complete double-integration procedure: write M(x), integrate twice, apply BCs. The symmetry condition (y' = 0 at midspan) replaces the second support BC and avoids writing a second-segment equation. Always verify with the standard formula.

Scenario

A simply supported steel beam of span L = 4 m carries a point load P = 20 kN at midspan. E = 200 GPa, I = 50×10⁶ mm⁴. Write the elastic curve equation for the left half (0 ≤ x ≤ L/2) and identify the boundary conditions.

Solution

Step 1 — Reactions: By symmetry, RA = RB = P/2 = 10 kN. Step 2 — M(x) for 0 ≤ x ≤ L/2 (before the load): M(x) = RA·x = 10x (kN·m) = 10,000x (N·mm, with x in mm). Step 3 — Differential equation: EI·y'' = M(x) = 10,000x. Step 4 — Integrate: EI·y' = 5,000x² + C₁. Step 5 — Integrate: EI·y = (5,000/3)x³ + C₁x + C₂. Step 6 — Boundary conditions: (a) At x = 0 (left support): y = 0 → C₂ = 0. (b) At x = L/2 = 2,000 mm (midspan): by symmetry, slope y' = 0 (the elastic curve is horizontal at midspan for a symmetric load). ∴ EI·(0) = 5,000(2,000)² + C₁ → C₁ = −5,000(4×10⁶) = −2×10¹⁰. Step 7 — Elastic curve: EI·y = (5,000/3)x³ − (2×10¹⁰)x. Step 8 — Max deflection at x = 2,000 mm: EI·y = (5,000/3)(2,000)³ − (2×10¹⁰)(2,000) = 1.333×10¹³ − 4×10¹³ = −2.667×10¹³ N·mm³. EI = (200,000)(50×10⁶) = 10¹³ N·mm². y = −2.667×10¹³ / 10¹³ = −2.667 mm (negative = downward). |δ_max| = 2.667 mm ✓ [Agrees with PL³/48EI = 20,000×(4,000)³/(48×10¹³) = 2.667 mm]

Applications

Setting up the correct bending-moment equation M(x) before integrating — critical for accuracy.
Checking that the elastic curve shape (concave up where M > 0, concave down where M < 0) is physically reasonable.
Expressing E and I in consistent units (N and mm, or kN and m) before computing — the most common source of numerical errors on board exams.

Misconceptions

MISCONCEPTION: 'The maximum deflection always occurs at midspan.' TRUTH: It occurs at midspan only for symmetric loads on simply supported beams. For asymmetric loads, the maximum deflection occurs where dy/dx = 0, which must be found by differentiation.
MISCONCEPTION: 'A larger moment of inertia I always means a smaller deflection, so any I increase helps equally.' TRUTH: Deflection is inversely proportional to I, so doubling I halves the deflection — but using a deeper section (larger d) raises I by d³/12 for rectangular sections, making depth far more effective than width.
MISCONCEPTION: 'EI·y'' = M(x) applies to any loading.' TRUTH: It is valid only for linearly elastic, isotropic, prismatic beams with small deflections. Non-prismatic beams require variable EI in the equation.

Related Concepts

Bending moment diagrams (prerequisite)
Second moment of area (moment of inertia)
Modulus of elasticity
Double-Integration Method
Serviceability limit states (NSCP 2015 Section 406)

Common Exam Questions

Example

A cantilever of length L carries a concentrated load P at mid-length. Find the deflection at the free end.

Approach

Identify support conditions, draw FBD, write M(x), integrate twice, apply boundary conditions, solve for constants, then evaluate y at the required point.

Question Type

Write and solve the elastic curve equation

Example

State all boundary conditions for a propped cantilever (fixed at A, roller at B).

Approach

Simple support: y = 0 only. Fixed support: y = 0 AND y' = 0. Free end: no geometric BC (natural BCs are M = 0, V = 0). Internal hinge: M = 0 at the hinge location.

Question Type

Identify boundary conditions for a given support configuration

Key Points To Remember

Master equation: EI·y'' = M(x). Memorize this — it is the starting point of both double-integration and conjugate-beam methods.
Flexural rigidity EI combines material stiffness (E) and geometric stiffness (I); increasing either reduces deflection.
Small-deflection assumption (θ ≈ tan θ) is valid for most structural beams; it is the basis of all four methods.
Sign convention must be consistent throughout the solution; establish it at the start and never switch.
Boundary conditions: simple support → y = 0; fixed end → y = 0 AND y' = 0; free end → M = 0 AND V = 0.
The elastic curve is continuous and smooth (no kinks) for beams with no internal hinges.
For steel: E = 200 GPa (200,000 MPa = 200,000 N/mm²). For concrete: E = 4700√f'c (ACI 318-19 §19.2.2).

Double-Integration Method (with Macaulay's Bracket Method)

The Double-Integration Method integrates the equation EI·y'' = M(x) twice to obtain the slope and deflection equations. It is systematic and works for any loading, but for beams with multiple loads at different positions, the bending moment has a different expression in each segment. Writing separate equations for each segment — and enforcing continuity of slope and deflection at every segment boundary — becomes algebraically heavy. Macaulay's Bracket Method (also called the singularity function method) eliminates this by writing a single M(x) expression valid for the entire span using angle-bracket notation: ⟨x − a⟩ⁿ = 0 if x < a = (x − a)ⁿ if x ≥ a For integration: ∫⟨x−a⟩ⁿ dx = ⟨x−a⟩ⁿ⁺¹ / (n+1), treating the bracket as a unit. Common singularity function terms used in board problems: • Point load P at x = a: contribution to M(x) = +P·⟨x−a⟩¹ • UDL of intensity w starting at x = a: contribution = +(w/2)·⟨x−a⟩² • UDL ending at x = b (add a canceling UDL from x = b onward): +(w/2)·⟨x−a⟩² − (w/2)·⟨x−b⟩² • Applied moment M₀ at x = a: contribution = +M₀·⟨x−a⟩⁰ Procedure: 1. Draw FBD; solve reactions (include unknown reactions as symbols if indeterminate). 2. Write M(x) for the full span using Macaulay brackets. 3. Integrate twice; keep Macaulay terms intact — only integrate the bracket as a whole. 4. Apply two BCs to find C₁ and C₂. 5. Substitute x values to find desired slopes and deflections. With Macaulay, there is only ONE set of integration constants C₁ and C₂ for the entire beam, regardless of how many loads are applied — a major computational saving.

Examples

This is a typical board problem combining UDL and point load. Key moves: write a single Macaulay M(x), integrate once (slope) and twice (deflection), apply the two simple-support BCs (y=0 at x=0 and x=L), and convert units of EI consistently. Working in kN and m avoids the L⁴ unit explosion that occurs in mm.

Scenario

BOARD-TYPE PROBLEM: A simply supported beam AB of span L = 6 m carries a UDL of w = 15 kN/m over the entire span and a point load P = 40 kN at x = 2 m from A. E = 200 GPa, I = 100×10⁶ mm⁴. Using the double-integration/Macaulay method, find (a) the midspan deflection and (b) the slope at support A.

Solution

Step 1 — Reactions (taking moments about B): RA × 6 = 40(4) + 15(6)(3) → RA = (160 + 270)/6 = 71.67 kN RB = 40 + 15(6) − 71.67 = 58.33 kN Step 2 — M(x) using Macaulay (x from A, in meters for now, convert at end): M(x) = 71.67x − (15/2)x² − 40⟨x−2⟩¹ [kN·m, x in m] Step 3 — EI·y'' = M(x) [keep in kN·m, convert EI later] EI·y' = 71.67x²/2 − (15/6)x³ − 40⟨x−2⟩²/2 + C₁ = 35.835x² − 2.5x³ − 20⟨x−2⟩² + C₁ EI·y = 35.835x³/3 − 2.5x⁴/4 − 20⟨x−2⟩³/3 + C₁x + C₂ = 11.945x³ − 0.625x⁴ − (20/3)⟨x−2⟩³ + C₁x + C₂ Step 4 — Boundary conditions: At x=0: y=0 → C₂ = 0. At x=6: y=0: 0 = 11.945(216) − 0.625(1296) − (20/3)(64) + 6C₁ 0 = 2580.12 − 810 − 426.67 + 6C₁ 6C₁ = −1343.45 → C₁ = −223.91 kN·m² Step 5 — Midspan deflection at x = 3 m: EI·y = 11.945(27) − 0.625(81) − (20/3)(1) + (−223.91)(3) = 322.52 − 50.625 − 6.667 − 671.73 = −406.50 kN·m³ Convert EI to kN·m²: EI = 200×10⁶ kN/m² × 100×10⁶ mm⁴ × (1 m/1000 mm)⁴ = 200×10⁶ × 100×10⁻⁶ = 20,000 kN·m² y_midspan = −406.50 / 20,000 = −0.02033 m = −20.33 mm (downward) Step 6 — Slope at A (x = 0): EI·y'|ₓ₌₀ = C₁ = −223.91 kN·m² θ_A = −223.91 / 20,000 = −0.01120 rad = −11.20 × 10⁻³ rad |θ_A| = 11.20 mrad (clockwise, i.e., sloping downward to the right) Answers: δ_midspan ≈ 20.3 mm; θ_A ≈ 11.2 × 10⁻³ rad

Applications

Computing deflection and slope at any point on a beam with multiple loads at arbitrary locations.
Finding the location of maximum deflection (set dy/dx = 0 and solve for x).
Setting up compatibility equations for propped cantilevers and other once-indeterminate beams.

Misconceptions

MISCONCEPTION: 'I need separate integration constants for each beam segment.' TRUTH: Macaulay's method uses ONE pair of constants (C₁, C₂) for the ENTIRE beam, regardless of the number of loads.
MISCONCEPTION: 'I can drop the Macaulay bracket terms that are zero at the evaluation point.' TRUTH: Never drop a Macaulay term mid-solution; only evaluate it to zero when substituting a specific x value, and the bracket is zero only when x < a.
MISCONCEPTION: 'The maximum deflection occurs where the bending moment is maximum.' TRUTH: Maximum deflection occurs where dy/dx = 0 (zero slope). For symmetric loads this coincides with maximum moment, but in general they are different locations.

Related Concepts

Elastic curve and governing differential equation
Macaulay singularity functions
Boundary conditions for beams
Bending moment diagrams
Statically indeterminate beam analysis (compatibility method)

Common Exam Questions

Example

A 5-m simply supported beam carries a 25 kN point load at 2 m from the left. Find the deflection at midspan. EI = 15,000 kN·m².

Approach

Write M(x) with Macaulay brackets, integrate twice, solve BCs, substitute x. Convert EI to consistent units first.

Question Type

Find midspan or specific-point deflection by double integration

Example

For the beam above, find the slope at the left support.

Approach

Use the slope equation (first integral), evaluate at x = 0 or x = L after solving C₁.

Question Type

Find the slope at a support

Key Points To Remember

In Macaulay notation, write the reaction forces first, then point loads, then UDL contributions — in order of increasing x.
When a UDL does not extend to the right end, close it off by adding an equal and opposite UDL starting at the right end of the load.
Integrate the entire Macaulay bracket as one term: ∫⟨x−a⟩ⁿ dx = ⟨x−a⟩ⁿ⁺¹/(n+1).
Boundary conditions at simple supports give y = 0 at both ends. At a fixed-free (cantilever): y = 0 AND y' = 0 at fixed end.
After solving, VERIFY the answer with a standard formula if the loading matches a tabulated case.
Deflection is proportional to L³ for point loads and to L⁴ for UDLs — doubling the span multiplies deflection by 8 or 16 respectively!

Area-Moment (Moment-Area) Method

The Moment-Area Method, developed by Mohr, is a powerful geometric approach that extracts slope changes and deflections directly from the M/EI diagram without integrating algebraically. It is especially efficient for cantilevers (where the tangent at the fixed end is known to be horizontal) and for finding deflections or slopes at specific points on simple beams. The two theorems: THEOREM 1 — Change in Slope: The change in slope (in radians) between two points A and B on the elastic curve equals the area of the M/EI diagram between those two points. θ_B/A = ∫[A to B] (M/EI) dx = Area of M/EI diagram from A to B The sign: θ_B/A is positive (B has rotated counterclockwise relative to A) when the M/EI area is positive (sagging). THEOREM 2 — Tangential Deviation: The vertical distance from point B on the elastic curve to the tangent drawn at point A (measured at the vertical line through B) equals the first moment of the M/EI area between A and B, taken about B. t_B/A = ∫[A to B] (M/EI) · x̄_B · dx = (Area of M/EI from A to B) × x̄_B where x̄_B is the horizontal distance from the centroid of the M/EI area to the point B. Note carefully: t_B/A ≠ t_A/B in general (different reference tangents). PROCEDURE: 1. Draw the M diagram. Divide by EI to get the M/EI diagram. 2. Break the M/EI diagram into simple geometric shapes (rectangles, triangles, parabolas) whose areas and centroids are known. 3. Compute the areas and locate centroids. 4. Apply Theorem 1 (for slope changes) or Theorem 2 (for deviations). 5. Use geometry of the elastic curve to find actual deflections from deviations. For common shapes of M/EI diagrams: Rectangle (base b, height h): Area = bh; centroid at b/2 from either end. Triangle (base b, height h): Area = bh/2; centroid at b/3 from the tall end, 2b/3 from the short end. Parabola (apex at one end, base b, height h): Area = bh/3; centroid at 3b/4 from apex.

Examples

For cantilevers, the moment-area method is the fastest available — there is no need to write and integrate algebraic equations. The key insight is that the fixed-end tangent is horizontal, so the tangential deviation at the free end IS the deflection directly. The centroid location of the parabola (3/4 from the apex end) is a critical formula to memorize.

Scenario

BOARD-TYPE PROBLEM: A cantilever beam of length L = 3 m carries a UDL w = 10 kN/m over its full length. E = 200 GPa, I = 60×10⁶ mm⁴. Using the Moment-Area Method, find the slope and deflection at the free end B.

Solution

Step 1 — Bending moment at fixed end A (x = 0 at B): M_A = wL²/2 = 10(3)²/2 = 45 kN·m (negative, hogging; treat magnitude). The M diagram is a parabola opening from 0 at B (free end) to 45 kN·m at A (fixed end). M/EI at A = 45 / EI. Step 2 — EI value: EI = (200×10⁶ kN/m²)(60×10⁶ mm⁴)(1 m/1000 mm)⁴ = 200×10⁶ × 60×10⁻⁶ = 12,000 kN·m². M/EI at A = 45 / 12,000 = 0.00375 rad/m. Step 3 — M/EI area (parabola from B to A): For a parabola with apex at B (height = 0) and height h = M_A/EI at A: Area = (1/3)(base)(height) = (1/3)(3)(0.00375) = 0.00375 rad. Step 4 — Theorem 1 (slope at B): The fixed end A has zero slope (tangent is horizontal). θ_B/A = area of M/EI from B to A = 0.00375 rad. Since A has zero slope, the slope at B = θ_B = θ_B/A = 0.00375 rad = 3.75 × 10⁻³ rad. ✓ [Check: θ_B = wL³/6EI = 10(3)³/6/12,000 = 270/72,000 = 0.00375 rad ✓] Step 5 — Theorem 2 (deflection at B = t_B/A for a cantilever): The tangent at A is horizontal. Deviation of B from that horizontal tangent = deflection at B. Centroid of parabolic area is at 3L/4 from the apex (B) = 3(3)/4 = 2.25 m from B. t_B/A = Area × x̄_B = 0.00375 × 2.25 = 0.008438 m = 8.44 mm (downward). [Check: δ_B = wL⁴/8EI = 10(3)⁴/8/12,000 = 810/96,000 = 0.00844 m = 8.44 mm ✓]

Applications

Cantilever beams: fastest method since fixed-end tangent is horizontal and t_free/fixed = deflection directly.
Simple beams with complex multi-point loading: decompose the M/EI diagram into triangles and rectangles.
Finding the point of maximum deflection in asymmetrically loaded simple beams (find x where the M/EI area from left support gives a tangential deviation equal to the linear interpolation point).
Checking double-integration results quickly by computing the same deflection geometrically.

Misconceptions

MISCONCEPTION: 'Theorem 2 gives the deflection directly.' TRUTH: Theorem 2 gives the TANGENTIAL DEVIATION — the vertical distance between the elastic curve and a TANGENT LINE at the reference point. Only for a cantilever (horizontal fixed-end tangent) does the tangential deviation equal the deflection.
MISCONCEPTION: 'The centroid of a triangular M/EI area is at the middle.' TRUTH: For a right triangle, the centroid is at L/3 from the tall side and 2L/3 from the zero end. This asymmetry causes major errors in board exams.
MISCONCEPTION: 'I can use the total M/EI area in Theorem 2 without locating its centroid from the correct reference point (B).' TRUTH: In Theorem 2, the centroid distance x̄ must be measured from the point B where the deviation is being computed, not from A.

Related Concepts

M/EI diagram construction
Centroids of common geometric shapes
Double-Integration Method (both give the same numerical results)
Conjugate-Beam Method (reframes moment-area as a statics problem)
Principle of superposition

Common Exam Questions

Example

A 2-m cantilever has P = 12 kN at the free end and w = 6 kN/m along the full length. EI = 8,000 kN·m². Find the free-end deflection.

Approach

Draw M/EI diagram. Find area and centroid. t_free/fixed = deflection because fixed-end tangent is horizontal.

Question Type

Cantilever free-end deflection under various loads

Example

A 6-m simple beam carries a 30 kN point load at 2 m from the left. Find the midspan deflection using moment-area. EI = 18,000 kN·m².

Approach

Draw M/EI diagram. Compute t_A/B (deviation of A from tangent at B) and t_C/B (deviation of midpoint C from tangent at B). Use geometry: δ_C = (t_A/B)/2 − t_C/B [for C at midspan, with tangent at B as reference].

Question Type

Midspan deflection of a simply supported beam under non-symmetric load

Key Points To Remember

Theorem 1 gives the CHANGE IN SLOPE between two points, not the absolute slope. You need a known slope (e.g., horizontal tangent at fixed end, or at midspan of symmetric beam) as a reference.
Theorem 2 gives a TANGENTIAL DEVIATION (a vertical distance between the elastic curve and a tangent line), NOT the deflection directly — except for cantilevers where the fixed-end tangent is horizontal.
For a simply supported beam under symmetric loading, the tangent at midspan is horizontal; use that as the reference tangent.
For an asymmetrically loaded simple beam, find the tangent at one support using the geometry: t_B/A = RA×L (no, use compatibility: t_B/A / L = vertical rise of tangent from A to B, then subtract to get deflection).
Standard M/EI area shapes: triangle (UDL on cantilever gives parabola M, hence parabolic M/EI), rectangle (constant moment), parabola (UDL on simple beam gives parabolic M).
Units check: M/EI has units of 1/length (rad/mm or rad/m). Area of M/EI diagram has units of angle (radians). Moment of M/EI area has units of length (mm or m) — which is the tangential deviation.

Conjugate-Beam Method

The Conjugate-Beam Method is an elegant restatement of the Moment-Area Method: it converts the slope and deflection problem into a standard shear-force and bending-moment problem on a fictitious 'conjugate beam.' This makes it accessible to engineers who are very comfortable with statics and SFD/BMD, which is virtually every civil engineering student. Key correspondences: Real beam — Slope (θ) ↔ Conjugate beam — Shear force (V*) Real beam — Deflection (y) ↔ Conjugate beam — Bending moment (M*) The conjugate beam is loaded with the M/EI diagram of the real beam as a distributed load (the 'M/EI load'). Regions of positive (sagging) moment produce downward M/EI load; regions of negative (hogging) moment produce upward M/EI load. Support transformation rules (critical to get right): Real support → Conjugate support Pin or roller (y=0, θ≠0) → Pin or roller (M*=0, V*≠0) Fixed end (y=0, θ=0) → Free end (M*=0, V*=0) Free end (y≠0, θ≠0) → Fixed end (M*≠0, V*≠0) Internal hinge (y continuous, M=0) → Internal fixed support PROCEDURE: 1. Draw the real beam and compute M(x). 2. Draw the M/EI diagram (this becomes the conjugate load). 3. Apply support-transformation rules to create the conjugate beam. 4. Find conjugate-beam reactions by statics (ΣF = 0, ΣM = 0). 5. Cut the conjugate beam at the desired point: — The conjugate shear V*(x) = real slope θ(x). — The conjugate moment M*(x) = real deflection y(x). For a simply supported real beam, the conjugate beam is also simply supported — same structure, just loaded with M/EI. This makes the method especially intuitive for simple beams: the maximum deflection occurs at the point of zero conjugate shear (just as maximum real moment occurs where real shear = 0).

Examples

The conjugate-beam method converts deflection finding into a familiar statics problem. The conjugate reactions equal the real end slopes (a bonus result!), and cutting the conjugate beam at any section gives both slope (conjugate V*) and deflection (conjugate M*) at that point on the real beam. The support-transformation rules are the most critical thing to memorize.

Scenario

BOARD-TYPE PROBLEM: A simply supported beam AB of span L = 5 m carries a central point load P = 20 kN. E = 200 GPa, I = 40×10⁶ mm⁴. Using the conjugate-beam method, find the midspan deflection and the slope at support A.

Solution

Step 1 — Real beam M diagram: RA = RB = 10 kN (by symmetry). M at midspan = 10 × 2.5 = 25 kN·m (triangle peaking at center). Step 2 — M/EI diagram: EI = (200×10⁶)(40×10⁶)(10⁻⁶)² = 200×10⁶ × 0.04 = 8,000 kN·m². M/EI at midspan = 25/8,000 = 3.125×10⁻³ rad/m. Shape: two triangles meeting at center, each with base L/2 = 2.5 m, height 3.125×10⁻³ rad/m. Step 3 — Conjugate beam: Real simple beam → Conjugate simple beam (same support layout). Load: the M/EI diagram (two symmetric triangles) applied as distributed load. Step 4 — Conjugate reactions: By symmetry, RA* = RB* = (total area of M/EI load)/2. Total M/EI area = 2 × (1/2)(2.5)(3.125×10⁻³) = 2 × 3.906×10⁻³ = 7.8125×10⁻³ rad. RA* = RB* = 7.8125×10⁻³/2 = 3.906×10⁻³ rad. ∴ Slope at A = θ_A = V*_A = RA* = 3.906×10⁻³ rad ✓ [Check: θ_A = PL²/16EI = 20(5)²/(16×8,000) = 500/128,000 = 3.906×10⁻³ rad ✓] Step 5 — Conjugate moment at midspan (x = 2.5 m): M* at midspan = RA* × 2.5 − (moment of left M/EI area about midspan) Left M/EI area = (1/2)(2.5)(3.125×10⁻³) = 3.906×10⁻³ rad. Centroid of left triangle from A = 2×2.5/3 = 5/3 m from A, so (2.5 − 5/3) = 5/6 m from midspan. M* = 3.906×10⁻³ × 2.5 − 3.906×10⁻³ × (5/6) = 9.766×10⁻³ − 3.255×10⁻³ = 6.510×10⁻³ kN·m³... Wait — units: V* (= slope) is dimensionless (rad); M/EI has units rad/m; conjugate load is in rad/m; conjugate reaction RA* is in rad (load × distance); conjugate M* (= deflection) is in rad × m = rad·m. Since EI has been divided out already, M* is in m (deflection). M*_midspan = 3.906×10⁻³(2.5) − 3.906×10⁻³(5/6) = 9.766×10⁻³ − 3.255×10⁻³ = 6.510×10⁻³ m ... Let us redo cleanly. Centroid of left-half triangle from left support A: A triangle with the peak at the far end (midspan): centroid is at 2/3 × 2.5 = 1.667 m from A. M*_C = RA* × 2.5 − (area of M/EI to left of C) × (distance from centroid of that area to C) Distance from centroid to C = 2.5 − 1.667 = 0.833 m. M*_C = (3.906×10⁻³)(2.5) − (3.906×10⁻³)(0.833) = 9.766×10⁻³ − 3.255×10⁻³ = 6.510×10⁻³ m = 6.51 mm. [Check: δ = PL³/48EI = 20(5)³/(48×8,000) = 2,500/384,000 = 6.51×10⁻³ m = 6.51 mm ✓]

Applications

Any beam for which you can quickly draw the M/EI diagram and find conjugate-beam reactions by statics.
Particularly powerful for continuous beams and those with multiple loads, where area-moment geometry becomes complex.
Finding maximum deflection location by locating zero conjugate shear (same technique as finding maximum moment in SFD analysis).

Misconceptions

MISCONCEPTION: 'The conjugate beam has the same supports as the real beam.' TRUTH: Only for simply supported beams. Fixed ends become free, and free ends become fixed in the conjugate beam. Getting these wrong leads to completely incorrect results.
MISCONCEPTION: 'The M/EI load on the conjugate beam acts downward everywhere.' TRUTH: Positive (sagging) M/EI loads act downward; negative (hogging) M/EI loads act upward. For a cantilever with a downward point load, the moment is hogging throughout, so the entire conjugate load is upward.
MISCONCEPTION: 'Conjugate M* directly gives deflection in mm without conversion.' TRUTH: M* gives deflection in whatever units EI is divided through — if you work in kN and m, M* is in meters. Convert to mm only at the final step.

Related Concepts

Moment-Area Method (conjugate-beam is equivalent, just reframed)
Shear force and bending moment diagrams (prerequisite)
Statics — finding beam reactions and internal forces by free-body diagrams
Elastic curve equation
Indeterminate beam analysis by compatibility

Common Exam Questions

Example

A cantilever 4 m long carries P = 15 kN at free end. EI = 10,000 kN·m². Find free-end slope and deflection using conjugate-beam method.

Approach

Draw M/EI diagram as conjugate load. Apply support transformations. Find conjugate reactions by statics. Cut conjugate beam and compute V* (slope) and M* (deflection).

Question Type

Find deflection and slope using conjugate-beam method

Example

Sketch the conjugate beam for a propped cantilever (fixed at A, pin at B).

Approach

Identify real support type, apply transformation table: fixed→free, free→fixed, simple→simple.

Question Type

Conjugate beam support identification

Key Points To Remember

The conjugate beam is loaded with M/EI as distributed load; the load intensity at any x equals M(x)/EI.
Conjugate shear V* = real slope θ. Conjugate moment M* = real deflection y.
A fixed real end becomes a FREE conjugate end (both M* = 0 and V* = 0 there, matching the zero slope and zero deflection).
A free real end (cantilever tip) becomes a FIXED conjugate end — the large conjugate moment and shear at that 'support' represent the large deflection and slope of the real free end.
For a simply supported real beam: conjugate is also simply supported. Conjugate reactions = real end slopes.
Maximum real deflection occurs where conjugate shear V* = 0 (same logic as finding maximum BMD from SFD = 0).

Superposition Method and Standard Deflection Formulas

The Principle of Superposition states that for a linearly elastic structure, the response (deflection, slope, reaction) due to a combination of loads is the algebraic sum of the responses due to each load acting individually. This is valid provided: (a) The material is linearly elastic (stress ∝ strain). (b) Deflections are small (geometry does not change significantly). (c) The supports do not change with loading. For board exams, this is the FASTEST method when you can decompose the actual loading into tabulated standard cases. Memorizing the four or five most common formulas is essential. STANDARD FORMULAS (for simply supported beam, span L, constant EI): 1. Central point load P: δ_max = PL³ / (48EI) at x = L/2 θ_max = PL² / (16EI) at supports 2. Uniform distributed load w (full span): δ_max = 5wL⁴ / (384EI) at x = L/2 θ_max = wL³ / (24EI) at supports 3. Point load P at distance a from left (a ≤ b, b = L − a): δ_max = Pab(a+2b)√(3a(a+2b)) / (27EIL) at x = √(a(a+2b)/3) from left δ_at_midspan ≈ Pab(3L²−4a²−4b²+...)/(48EI) [use tables for exact] STANDARD FORMULAS (for cantilever of length L, fixed at left, constant EI): 4. Point load P at free end: δ_free = PL³ / (3EI) θ_free = PL² / (2EI) 5. Uniform distributed load w (full length): δ_free = wL⁴ / (8EI) θ_free = wL³ / (6EI) 6. Moment M₀ at free end: δ_free = M₀L² / (2EI) θ_free = M₀L / (EI) 7. Point load P at distance a from fixed end (load at intermediate point): δ_free = Pa³/(3EI) + Pa²(L−a)/(2EI) [free end to the right of load] MNEMONIC for denominators — 'simply three, four, eight: 48, 384, 3, 8': SS central P → 48 SS full UDL → 384 (numerator has 5) Cantilever P → 3 Cantilever UDL → 8 For complex loading (e.g., cantilever with both P and w), simply ADD the individual deflections at the free end. This is the fastest board-exam approach. SERVICEABILITY CHECK (NSCP 2015, Table 406.2.1): Members supporting plaster or brittle finishes: δ_L ≤ L/360 Members supporting masonry: δ_L ≤ L/480 Roofs not supporting brittle finishes: δ_L ≤ L/240 Total deflection (DL + LL): often ≤ L/240 or L/180

Examples

This is the prototypical board serviceability problem. Note the factor '5' in the numerator of the UDL formula — a classic source of errors. Always compute L/360 explicitly; never estimate it mentally. The NSCP citation (Table 406.2.1) demonstrates code awareness expected at licensure level.

Scenario

NSCP SERVICEABILITY CHECK — BOARD TYPE: A simply supported steel floor beam spans L = 8 m. It supports a live load of w_L = 12 kN/m (UDL). E = 200 GPa, I = 250×10⁶ mm⁴. (a) Compute the live-load midspan deflection. (b) Check whether it satisfies NSCP 2015 Table 406.2.1 for a member supporting a plaster ceiling (limit L/360).

Solution

Step 1 — Identify formula: Simply supported, full UDL: δ = 5wL⁴ / (384EI) Step 2 — Convert units to N and mm: w = 12 kN/m = 12 N/mm L = 8 m = 8,000 mm E = 200 GPa = 200,000 N/mm² = 200,000 MPa I = 250×10⁶ mm⁴ EI = 200,000 × 250×10⁶ = 5×10¹³ N·mm² Step 3 — Compute deflection: δ = 5(12)(8,000)⁴ / [384(5×10¹³)] Numerator: 5 × 12 × (8,000)⁴ = 60 × 4.096×10¹⁵ = 2.458×10¹⁷ N·mm³ Denominator: 384 × 5×10¹³ = 1.920×10¹⁶ N·mm² δ = 2.458×10¹⁷ / 1.920×10¹⁶ = 12.8 mm Step 4 — NSCP serviceability check: Allowable δ = L/360 = 8,000/360 = 22.2 mm Computed δ = 12.8 mm < 22.2 mm → SATISFACTORY ✓ Answer: The live-load deflection is 12.8 mm, which is within the NSCP 2015 limit of 22.2 mm for members supporting brittle finishes.

Superposition is the fastest method here — two standard cantilever formulas added together. This is the most common board-exam format for combined loading. The critical discipline is to NOT mix the formulas (don't use PL³/48 for a cantilever or PL³/3 for a simple beam).

Scenario

SUPERPOSITION PROBLEM: A cantilever AB (L = 4 m, fixed at A) carries BOTH a point load P = 10 kN at the free end B AND a UDL w = 5 kN/m over the entire length. E = 200 GPa, I = 80×10⁶ mm⁴. Find the free-end deflection.

Solution

Step 1 — EI: EI = 200,000 × 80×10⁶ = 1.6×10¹³ N·mm² Step 2 — Deflection due to point load alone: δ_P = PL³/(3EI) = (10,000)(4,000)³ / [3(1.6×10¹³)] = (10,000)(6.4×10¹⁰) / (4.8×10¹³) = 6.4×10¹⁴ / 4.8×10¹³ = 13.33 mm Step 3 — Deflection due to UDL alone: δ_w = wL⁴/(8EI) = (5)(4,000)⁴ / [8(1.6×10¹³)] = (5)(2.56×10¹⁴) / (1.28×10¹⁴) = 1.28×10¹⁵ / 1.28×10¹⁴ = 10.00 mm Step 4 — Total deflection (superposition): δ_total = δ_P + δ_w = 13.33 + 10.00 = 23.33 mm (downward) Note: w = 5 kN/m = 5 N/mm for consistent N-mm units.

Applications

Fastest approach on board exams when the loading matches or can be decomposed into standard cases.
NSCP serviceability checks for floor beams, roof beams, and members supporting brittle finishes.
Preliminary design: determine the required moment of inertia I to satisfy a deflection limit (solve the formula for I).
Compatibility equations for statically indeterminate beams (set sum of deflections at redundant support to zero).

Misconceptions

MISCONCEPTION: 'The formula 5wL⁴/384EI applies to a cantilever.' TRUTH: It applies ONLY to a simply supported beam with full-span UDL. For a cantilever, the formula is wL⁴/8EI (no factor of 5).
MISCONCEPTION: 'I can use L in meters in the formula as long as I convert the final answer.' TRUTH: You must convert ALL quantities to the same unit system BEFORE substituting. L⁴ in m⁴ but I in mm⁴ makes EI unit-inconsistent.
MISCONCEPTION: 'Superposition can always be used.' TRUTH: Superposition is valid only for linear-elastic, small-deflection behavior. It must not be used in plasticity problems or large-deformation situations.

Related Concepts

Principle of superposition (prerequisite)
NSCP 2015 Table 406.2.1 — Deflection limits
Serviceability versus strength limit states
Compatibility method for indeterminate beams
Required moment of inertia for deflection control (design application)

Common Exam Questions

Example

A 6-m simply supported beam carries w_L = 20 kN/m. EI = 25,000 kN·m². Check for L/360 compliance.

Approach

Apply appropriate standard formula. Compute L/360 (or other applicable NSCP limit). Compare and state PASS or FAIL.

Question Type

Compute deflection and check NSCP serviceability limit

Example

Determine the minimum I for a 5-m SS beam under 15 kN/m so that δ ≤ L/360. E = 200 GPa.

Approach

Set deflection formula equal to L/360 (or given limit). Solve for I algebraically. Round up to next available standard section.

Question Type

Find required I for deflection limit

Example

A cantilever 3.5 m long carries P = 8 kN at free end and w = 4 kN/m full span. Find total free-end deflection. EI = 12,000 kN·m².

Approach

Apply each load formula separately, then add (for same direction deflections) or subtract (for opposing directions).

Question Type

Superposition with two or more loads

Key Points To Remember

Denominators: SS beam, central P → 48; SS beam, full UDL → 384 (with factor 5); Cantilever, free-end P → 3; Cantilever, full UDL → 8.
For SS beam under full UDL: δ = 5wL⁴/384EI. The factor 5 in the numerator is frequently forgotten — a common board exam trap.
Convert ALL units to N and mm (or kN and m) BEFORE substituting into formulas. Mixing units gives wildly wrong answers.
L/360 is for live-load deflection of members supporting brittle finishes (NSCP 2015 Table 406.2.1). Different limits apply for total load or for different finish types.
Superposition is valid only for LINEAR-ELASTIC structures under small deflections. Do not use it for large-deflection or plastic analysis.
When a UDL acts only over part of the span, decompose it into a full-span UDL minus a UDL over the unloaded portion — then superpose.

Application to Statically Indeterminate Beams — Compatibility (Consistent Deformation) Method

The deflection methods learned in this chapter are the direct tool for analyzing statically indeterminate (hyperstatic) beams — beams that have more unknowns than equilibrium equations alone can solve. The Compatibility Method (also called the Method of Consistent Deformations or the Force Method) proceeds as follows: 1. IDENTIFY the REDUNDANT reaction(s) — the reaction(s) in excess of what statics needs. For a propped cantilever (fixed + roller), statics gives 3 unknowns (M_A, RA, RB) but only 2 equations (ΣF = 0, ΣM = 0), so it is once-indeterminate — one redundant, say R_B. 2. RELEASE the redundant by removing the corresponding support (if the redundant is a roller reaction, remove the roller). This creates a statically determinate 'released' or 'primary' structure. 3. COMPUTE the deflection at the released support point in the primary structure due to the ACTUAL LOADS (call it δ_₀, the 'load deflection'). 4. COMPUTE the deflection at that same point due to the REDUNDANT acting as a unit load (call it δ_₁₁, the 'flexibility coefficient' or 'unit redundant deflection'). 5. APPLY the COMPATIBILITY CONDITION: The actual deflection at the support must be zero (or whatever the actual support settlement is, usually zero): δ₀ + R_B · δ₁₁ = 0 ∴ R_B = −δ₀ / δ₁₁ 6. With R_B now known, treat the beam as statically determinate and compute all other reactions and internal forces. EXAMPLE SETUP — Propped Cantilever under UDL: Real beam: Fixed at A (left), roller at B (right), span L, UDL w. Redundant: R_B (roller reaction at B). Primary structure: cantilever fixed at A, free at B. δ₀ = wL⁴/(8EI) (downward, at B, due to UDL on cantilever). δ₁₁ = L³/(3EI) (upward at B, due to unit upward load at B on cantilever). Compatibility: δ₀ − R_B · δ₁₁ = 0 (note: R_B acts upward, opposing δ₀). R_B = δ₀/δ₁₁ = [wL⁴/(8EI)] / [L³/(3EI)] = 3wL/8. This result (R_B = 3wL/8 for a propped cantilever under full UDL) is a CLASSIC board-exam result that should be memorized.

Examples

This is a cornerstone problem in Structural Theory that originates entirely from beam deflection formulas. The two classic results — R_B = 3wL/8 and M_A = wL²/8 for a propped cantilever under full UDL — must be committed to memory for board exams, as they are used as 'given' results in more complex problems.

Scenario

CLASSIC BOARD PROBLEM: A propped cantilever of span L = 5 m is fixed at A and has a roller at B. It carries a UDL w = 20 kN/m. E = 200 GPa, I = 150×10⁶ mm⁴. Find the prop reaction R_B and the fixed-end moment M_A.

Solution

Step 1 — Identify redundant: R_B at B. Primary structure: cantilever fixed at A. Step 2 — δ₀ (free-end deflection of cantilever under UDL): δ₀ = wL⁴/(8EI) [downward] EI = 200,000 × 150×10⁶ = 3×10¹³ N·mm² = 3×10¹³ / (10⁶) = 30,000 kN·m² δ₀ = 20(5)⁴/(8×30,000) = 20×625/240,000 = 12,500/240,000 = 0.05208 m = 52.08 mm Step 3 — δ₁₁ (free-end deflection of cantilever under unit upward load at B): δ₁₁ = (1)L³/(3EI) = (5)³/(3×30,000) = 125/90,000 = 1.389×10⁻³ m/kN Step 4 — Compatibility: R_B = δ₀/δ₁₁ = 0.05208 / 1.389×10⁻³ = 37.5 kN [Check: R_B = 3wL/8 = 3(20)(5)/8 = 37.5 kN ✓] Step 5 — Fixed-end moment M_A: ΣM_A = 0: M_A + R_B(L) − w(L)(L/2) = 0 M_A = wL²/2 − R_B × L = 20(25)/2 − 37.5(5) = 250 − 187.5 = 62.5 kN·m [Check: M_A = wL²/8 = 20(25)/8 = 62.5 kN·m — standard result for propped cantilever under UDL ✓] Step 6 — Reaction at A: RA = wL − R_B = 20(5) − 37.5 = 62.5 kN

Applications

Analysis of propped cantilevers (once-indeterminate) in floors and decks.
Continuous beam analysis (multi-span, multi-degree indeterminate).
Foundation design: differential settlement compatibility.
Checking of beam-to-column connections in structural steel (AISC 360).

Misconceptions

MISCONCEPTION: 'For a propped cantilever, R_B = wL/2 (same as a simple beam).' TRUTH: R_B = 3wL/8 for a propped cantilever under full UDL. The fixed end attracts more reaction; the prop (roller) takes only 3/8 of total load, not half.
MISCONCEPTION: 'I can use superposition on the indeterminate beam directly.' TRUTH: You can use superposition on the primary (determinate) structure to find δ₀ and δ₁₁, but the original indeterminate beam cannot be superposed without first finding the redundants.
MISCONCEPTION: 'Any reaction can be chosen as the redundant.' TRUTH: While theoretically true, choosing the wrong redundant creates a primary structure that is either unstable or still indeterminate. Always choose the redundant that, when removed, leaves a stable determinate primary structure.

Related Concepts

Degree of static indeterminacy
Superposition method for deflection
Standard cantilever and simple-beam deflection formulas
Three-Moment Equation (Clapeyron) — alternative method for continuous beams
Slope-Deflection Method and Moment Distribution (advanced structural analysis)

Common Exam Questions

Example

A propped cantilever L = 4 m with central point load P = 24 kN. Find prop reaction. EI = constant.

Approach

Remove roller, compute δ₀ under applied loads on cantilever, compute δ₁₁ for unit load at roller, set δ₀ = R × δ₁₁, solve for R.

Question Type

Find redundant reaction of a propped cantilever

Example

For the propped cantilever above, find the fixed-end moment and maximum bending moment.

Approach

After finding R_B by compatibility, use statics (ΣM_A = 0) to find M_A.

Question Type

Find fixed-end moment after determining redundant

Key Points To Remember

The compatibility condition states that the actual deflection at the removed support = 0 (or the given settlement).
The primary structure must be statically determinate and stable after removing the redundant.
R_B = 3wL/8 for a propped cantilever under full UDL — memorize this classic result.
For a propped cantilever under central point load P: R_B = 5P/16 — another classic result.
After finding the redundant, treat the beam as statically determinate and draw SFD, BMD, and compute stresses normally.
This method extends to multi-degree indeterminate beams by setting up simultaneous compatibility equations.

Practice Problems

This problem uses the two central point-load formulas for a simply supported beam. Note that the denominator 48 (deflection) and 16 (slope) are easy to confuse; the deflection denominator is always 3× the slope denominator for this case. The NSCP check is a one-line computation but is mandatory in design-oriented board problems.

Problem

PROBLEM 1 (Double Integration — Simply Supported Beam): A simply supported beam of span L = 8 m carries a central point load P = 30 kN. E = 200 GPa, I = 120×10⁶ mm⁴. (a) Find the midspan deflection. (b) Find the slope at the left support. (c) Check the L/360 serviceability limit (NSCP 2015).

Solution

Given: P = 30 kN = 30,000 N; L = 8 m = 8,000 mm; E = 200,000 N/mm²; I = 120×10⁶ mm⁴. EI = 200,000 × 120×10⁶ = 2.4×10¹³ N·mm². (a) Midspan deflection: δ = PL³/(48EI) = (30,000)(8,000)³/(48 × 2.4×10¹³) Numerator: 30,000 × 5.12×10¹¹ = 1.536×10¹⁶ Denominator: 48 × 2.4×10¹³ = 1.152×10¹⁵ δ = 1.536×10¹⁶ / 1.152×10¹⁵ = 13.33 mm (downward) (b) Slope at left support A: θ_A = PL²/(16EI) = (30,000)(8,000)²/(16 × 2.4×10¹³) = (30,000)(6.4×10⁷)/(3.84×10¹⁴) = 1.92×10¹² / 3.84×10¹⁴ = 5.0×10⁻³ rad = 5.0 mrad (c) Serviceability check (NSCP 2015 Table 406.2.1): L/360 = 8,000/360 = 22.22 mm δ = 13.33 mm < 22.22 mm → PASS ✓

Superposition of two cantilever standard cases. Note that both the point load and UDL create downward deflection and clockwise slope (both additive). Unit conversions are critical: w in N/mm (not kN/m) when L is in mm. Checking units: (N/mm)(mm⁴)/N·mm² = mm ✓.

Problem

PROBLEM 2 (Superposition — Combined Loading, Cantilever): A cantilever beam AB is 3.5 m long, fixed at A and free at B. It supports a point load P = 12 kN at the free end B and a UDL w = 8 kN/m over its full length. E = 200 GPa, I = 95×10⁶ mm⁴. Compute: (a) the free-end deflection; (b) the free-end slope.

Solution

Given: L = 3.5 m = 3,500 mm; P = 12,000 N; w = 8 N/mm; EI = 200,000 × 95×10⁶ = 1.9×10¹³ N·mm². (a) Free-end deflection: δ_P = PL³/(3EI) = (12,000)(3,500)³/[3(1.9×10¹³)] = (12,000)(4.2875×10¹⁰)/(5.7×10¹³) = 5.145×10¹⁴/5.7×10¹³ = 9.026 mm δ_w = wL⁴/(8EI) = (8)(3,500)⁴/[8(1.9×10¹³)] = (8)(1.50063×10¹⁴)/(1.52×10¹⁴) = 1.20050×10¹⁵/1.52×10¹⁴ = 7.900 mm δ_total = 9.026 + 7.900 = 16.93 mm (downward) ✓ (b) Free-end slope: θ_P = PL²/(2EI) = (12,000)(3,500)²/[2(1.9×10¹³)] = (12,000)(1.225×10⁷)/(3.8×10¹³) = 1.47×10¹¹/3.8×10¹³ = 3.868×10⁻³ rad θ_w = wL³/(6EI) = (8)(3,500)³/[6(1.9×10¹³)] = (8)(4.2875×10¹⁰)/(1.14×10¹⁴) = 3.43×10¹¹/1.14×10¹⁴ = 3.009×10⁻³ rad θ_total = 3.868×10⁻³ + 3.009×10⁻³ = 6.877×10⁻³ rad ≈ 6.88 mrad

The moment-area method for an asymmetric load requires careful centroid tracking and the use of the geometry relationship δ_D = θ_A × x_D − t_D/A. The sign convention here: the upward intercept of the tangent at A above point D minus the downward deviation of D from that tangent gives the elastic curve at D. This problem illustrates why area-moment can be more laborious than superposition for asymmetric loading — on the board exam, use superposition if the loading matches standard cases.

Problem

PROBLEM 3 (Area-Moment Method — Simple Beam): A simply supported beam AB spans 6 m. A point load P = 24 kN acts at x = 2 m from A. E = 200 GPa, I = 75×10⁶ mm⁴. Using the moment-area method, find (a) the slope at A; (b) the deflection at x = 3 m (midspan).

Solution

Given: L = 6 m; a = 2 m (load from A); b = 4 m (load from B); P = 24 kN. RA = Pb/L = 24(4)/6 = 16 kN; RB = Pa/L = 24(2)/6 = 8 kN. M at load point C (x=2 m): M_C = RA × a = 16 × 2 = 32 kN·m. EI = 200×10⁶ kN/m² × 75×10⁶ mm⁴ × (10⁻³)⁴ m⁴/mm⁴ ... Easier: EI = 200,000 MPa × 75×10⁶ mm⁴ = 1.5×10¹³ N·mm² = 15,000 kN·m². M_C/EI = 32/15,000 = 2.133×10⁻³ rad/m. M/EI diagram: triangle from A (zero) to C (peak = 2.133×10⁻³) to B (zero). Left sub-triangle (A to C): base = 2 m, height = 2.133×10⁻³; Area_AC = ½(2)(2.133×10⁻³) = 2.133×10⁻³ rad. Right sub-triangle (C to B): base = 4 m, height = 2.133×10⁻³; Area_CB = ½(4)(2.133×10⁻³) = 4.267×10⁻³ rad. Total area = 6.400×10⁻³ rad. (a) Slope at A — use Theorem 2 and geometry: t_B/A (deviation of B from tangent at A) = total M/EI area × x̄_A x̄_A = centroid of total M/EI diagram from A. Centroid of left triangle from A = 2×2/3 = 1.333 m. Centroid of right triangle from A = 2 + (1/3)(4) = 3.333 m. [centroid at 1/3 from the tall end = 1/3 from C] x̄_A = [2.133×10⁻³(1.333) + 4.267×10⁻³(3.333)] / 6.400×10⁻³ = [2.844×10⁻³ + 14.223×10⁻³] / 6.400×10⁻³ = 17.067×10⁻³ / 6.400×10⁻³ = 2.667 m from A. t_B/A = (total area) × (x̄_A from B) = 6.400×10⁻³ × (6 − 2.667) = 6.400×10⁻³ × 3.333 = 0.02133 m. Since t_B/A = distance B is above the tangent at A (for a simple beam loaded downward), and geometrically: θ_A × L = t_B/A (for small angles, tangent at A intercepts B at height θ_A × L above elastic curve at B, which is at y=0) Wait — more precisely: t_B/A is the vertical intercept at B of the tangent drawn at A. Since B is a support (y_B = 0) and the tangent at A is inclined at θ_A: t_B/A = θ_A × L (positive: B is above the tangent at A for a downward-loaded beam? No — B is BELOW the tangent at A.) Actually for a downward loaded beam: the tangent at A slopes downward to the right, so the tangent line at A projected to x = L is ABOVE point B (which is at y = 0). So t_B/A (measured downward from tangent to elastic curve at B) is negative in some conventions. Use magnitudes: |t_B/A| = θ_A × L → θ_A = t_B/A / L = 0.02133/6 = 3.556×10⁻³ rad. (b) Midspan deflection at x = 3 m: First, find t_D/A (deviation of midspan D from tangent at A): M/EI area from A to D (x=0 to x=3): Left triangle (A to C, 0 to 2 m): Area = 2.133×10⁻³, centroid at 1.333 m from A, so distance from D = 3 − 1.333 = 1.667 m. Partial right triangle (C to D, 2 to 3 m): a triangle with base 1 m, height = M/EI at C going down toward zero at B; at x=3: M(3) = RB×(6−3) = 8×3 = 24 kN·m... wait, M diagram from C to B is a triangle declining from 32 kN·m at C to 0 at B. At x=3: M = 32 − (32/4)(3−2) = 32 − 8 = 24 kN·m. So from C (x=2) to D (x=3): it's a trapezoid with height 32/EI at C and 24/EI at D. Area_CD = ½(2.133×10⁻³ + 1.600×10⁻³)(1) = ½(3.733×10⁻³) = 1.867×10⁻³ rad. [where 24/15,000 = 1.600×10⁻³] Centroid of trapezoid from D: x̄_D for trapezoid (h₁=2.133, h₂=1.600, width b=1) → x̄ from D = b(2h₁+h₂)/(3(h₁+h₂)) = 1(2×2.133+1.600)/(3×3.733) = (5.866)/(11.199) = 0.5238 m from D. t_D/A = Area_AC(dist of centroid of AC from D) + Area_CD(dist of centroid of CD from D) = 2.133×10⁻³(3 − 1.333) + 1.867×10⁻³(0.5238) = 2.133×10⁻³(1.667) + 1.867×10⁻³(0.5238) = 3.556×10⁻³ + 0.978×10⁻³ = 4.534×10⁻³ m. Midspan deflection: δ_D = θ_A × 3 − t_D/A = 3.556×10⁻³ × 3 − 4.534×10⁻³ = 10.668×10⁻³ − 4.534×10⁻³ = 6.134×10⁻³ m = 6.13 mm. Answer: θ_A = 3.56 mrad; δ_midspan = 6.13 mm.

This is a design-type board problem that reverses the computation: instead of checking δ, you solve for the required I. The NSCP code reference (L/360 for live load on members supporting plaster) is an expected citation at licensure level. Note the phrase 'minimum required' — select the next LARGER standard section, never interpolate downward.

Problem

PROBLEM 4 (Required I for Serviceability — Design-Type): A simply supported steel floor beam spans 7 m and carries a live load of w = 18 kN/m. The member supports a plaster ceiling. E = 200 GPa. Determine the minimum required moment of inertia I such that the beam satisfies the NSCP 2015 live-load deflection limit (Table 406.2.1).

Solution

Given: L = 7 m; w = 18 kN/m; E = 200 GPa; Limit: δ ≤ L/360 (plaster ceiling). Step 1 — Allowable deflection: δ_allow = L/360 = 7,000/360 = 19.44 mm Step 2 — Set deflection formula equal to allowable and solve for I: δ = 5wL⁴/(384EI) ≤ δ_allow I ≥ 5wL⁴/(384E × δ_allow) Step 3 — Substitute (N, mm): w = 18 N/mm; L = 7,000 mm; E = 200,000 N/mm²; δ_allow = 19.44 mm. I ≥ [5(18)(7,000)⁴] / [384(200,000)(19.44)] Numerator: 5 × 18 × (7,000)⁴ = 90 × 2.4010×10¹⁵ = 2.1609×10¹⁷ Denominator: 384 × 200,000 × 19.44 = 384 × 3.888×10⁶ = 1.4929×10⁹ I ≥ 2.1609×10¹⁷ / 1.4929×10⁹ = 1.4476×10⁸ mm⁴ ≈ 144.8×10⁶ mm⁴ Answer: The minimum required moment of inertia is I_min ≈ 145×10⁶ mm⁴. In design practice, select a standard wide-flange section with I ≥ 145×10⁶ mm⁴ (e.g., a W-section from AISC/PSSC tables).

For a cantilever, the conjugate beam has a fixed end at the FREE end of the real beam and a free end at the FIXED end of the real beam. The conjugate reactions at the fixed end of the conjugate beam directly give the slope (from V*) and deflection (from M*) at the free end of the real beam. The centroid of the triangular M/EI load is at 2/3 from the zero end (B) — i.e., 2L/3 from B, confirming the formula δ = PL³/3EI.

Problem

PROBLEM 5 (Conjugate-Beam — Cantilever): A cantilever of length L = 4 m (fixed at A, free at B) carries a point load P = 18 kN at the free end. E = 200 GPa, I = 60×10⁶ mm⁴. Using the conjugate-beam method, find the slope and deflection at the free end B.

Solution

Given: P = 18,000 N; L = 4,000 mm; EI = 200,000 × 60×10⁶ = 1.2×10¹³ N·mm². Step 1 — Real beam M diagram: M is maximum at A: M_A = PL = 18,000 × 4,000 = 7.2×10⁷ N·mm (hogging). M diagram: triangle from B (M=0) to A (M = −7.2×10⁷ N·mm); M/EI at A = −7.2×10⁷/1.2×10¹³ = −6×10⁻⁶ rad/mm. Sign: hogging moment → M/EI diagram is negative → conjugate load acts UPWARD. Step 2 — Conjugate beam: Real cantilever: fixed at A, free at B. Conjugate beam: fixed at B (free end → fixed), free at A (fixed end → free). Load: upward triangular load from 0 at B to 6×10⁻⁶ rad/mm at A. Step 3 — Conjugate beam reactions at B (fixed end): Total area of M/EI diagram = ½ × 4,000 × 6×10⁻⁶ = 0.012 rad. Centroid from B = (2/3) × 4,000 = 2,667 mm from B [centroid of triangle with peak at A, measured from zero-end B is 2L/3]. Conjugate reaction (upward force at fixed end B): V*_B = total M/EI area = 0.012 rad (upward, per statics) M*_B = total area × centroid distance from B = 0.012 × 2,667 = 32 mm = 32 mm Step 4 — Read results: Real slope at B = conjugate shear at B = V*_B = 0.012 rad = 12×10⁻³ rad. Real deflection at B = conjugate moment at B = M*_B = 32 mm (downward). Verification: θ_B = PL²/(2EI) = 18,000(4,000)²/(2 × 1.2×10¹³) = 18,000×1.6×10⁷/2.4×10¹³ = 2.88×10¹¹/2.4×10¹³ = 0.012 rad ✓ δ_B = PL³/(3EI) = 18,000(4,000)³/(3 × 1.2×10¹³) = 18,000×6.4×10¹⁰/3.6×10¹³ = 1.152×10¹⁵/3.6×10¹³ = 32 mm ✓

Exam Preparation Tips

MEMORIZE THE BIG FOUR formulas and their denominators: SS+central P → 48; SS+full UDL → 384 (with factor 5 in numerator); Cantilever+free P → 3; Cantilever+full UDL → 8. Write these on your scratch paper the moment the exam starts.
UNIT DISCIPLINE is non-negotiable: convert ALL quantities to N and mm (or kN and m) BEFORE substituting. L⁴ magnifies mm-vs-m errors by a factor of 10¹²— a two-order-of-magnitude mistake in I leads to a 12-order-of-magnitude error in the numerator if L units are mixed.
For NSCP SERVICEABILITY CHECKS, always cite Table 406.2.1. The most tested limit is L/360 for live load on members supporting brittle finishes (plaster, tile). For total deflection limits or roof members, the limit may differ — read the problem statement carefully.
CONJUGATE BEAM SUPPORT RULES: Fixed → Free, Free → Fixed, Simple → Simple. These three rules, wrongly applied, invalidate the entire solution. State them explicitly in your solution for partial credit.
AREA CENTROIDS of M/EI shapes must be memorized: Triangle — centroid at L/3 from the TALL side (not the middle); Parabola with apex at one end — centroid at 3L/4 from the apex. These are tested indirectly in every moment-area problem.
PROPPED CANTILEVER classics: R_B = 3wL/8 (full UDL); M_A = wL²/8 (full UDL); R_B = 5P/16 (central point load); M_A = 5PL/16. Memorizing these saves 5–8 minutes per problem.
IDENTIFY THE METHOD that is fastest for each problem type: Superposition if load matches standard cases (use this first on board exams); Moment-area for cantilevers or for finding slope/deflection at specific points; Conjugate beam if you can quickly read shear and moment from statics; Double integration only if no standard formula applies or if the full elastic curve equation is required.
SIGN CONVENTION: establish it once at the start and never switch. Most Philippine textbooks take downward deflection as positive (consistent with gravity loading). State your convention explicitly to avoid confusion in multi-part problems.
MAXIMUM DEFLECTION LOCATION: For asymmetric loading on a simple beam, max deflection is NOT at midspan. It occurs where dy/dx = 0 (double-integration) or where conjugate shear = 0. On the board exam, if asked for 'maximum deflection' under an off-center load, locate the zero-slope point first.
PRACTICE CONVERTING between methods: for every double-integration result you compute, verify it with the superposition formula (if available). This cross-checking habit catches arithmetic errors before they become wrong answers.
EI IN kN·m² vs N·mm²: both are valid but must be consistent. 1 kN·m² = 10⁶ N·mm². E in GPa × I in m⁴ gives EI in GN·m² which must be converted. Safest habit: always use N·mm² (E in MPa = N/mm², I in mm⁴).
FOR RA 544 RELEVANCE: Republic Act 544 (Civil Engineering Law of the Philippines) requires that a registered CE sign and seal all structural computations. Board-exam problems on deflection relate directly to the professional responsibility of ensuring serviceability of structures — frame your understanding in this professional context.

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In summary

Beam deflection is both a standalone examination topic and the gateway to structural analysis of indeterminate beams — making it one of the highest-leverage subjects to master for the PRC Civil Engineer Licensure Examination. The four methods covered in this chapter are complementary, not competing: superposition with standard formulas wins on speed for simple loading cases; the area-moment method excels for cantilevers and specific-point calculations on simple beams; the conjugate-beam method reframes the problem as statics, which plays to every civil engineer's core strength; and double integration, though the most algebraically intensive, is the universally applicable baseline. The two most important takeaways for the board exam are: 1. MEMORIZE the standard formulas with their correct denominators — PL³/48EI (SS, central P), 5wL⁴/384EI (SS, full UDL), PL³/3EI (cantilever, free-end P), and wL⁴/8EI (cantilever, full UDL) — and never confuse which formula belongs to which beam-load configuration. 2. ALWAYS APPLY the NSCP 2015 serviceability check. Structural design in the Philippines under RA 544 requires registered civil engineers to ensure that structures are safe (strength limit state) AND serviceable (serviceability limit state). A beam that does not yield or fracture but deflects so much that it damages finishes, creates excessive vibration, or causes misalignment has failed its professional duty. Connect deflection formulas to the indeterminate beam problems you will encounter in Structural Theory: the propped cantilever result R_B = 3wL/8 is derived directly from the cantilever deflection formula by compatibility — a perfect example of how Strength of Materials and Structural Analysis are one continuous chain of engineering reasoning. Practice the five worked problems in this chapter, verify every result against the standard formula, and you will be well-prepared for this topic in the licensure examination.

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