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CELE Structural Theory & Analysis — Deflections of StructuresDetailed Explanation

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Deflections of Structures has a reputation among CELE reviewers for being deceptively tricky in the Structural Theory & Analysis subtest. PRC likes to hide the hard part in the phrasing rather than the concept. This long-form explanation untangles the phrasing traps and takes you through the concept the way someone who scored at the top of the CELE papers would.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Structural Theory & Analysis section sits under a "Core" weighting, and Deflections of Structures is the 2nd chapter in the 6-chapter CELE Structural Theory & Analysis 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 Structural Theory & Analysis.

Deflections of Structures - Detailed Explanation

Structural deflection is not merely a serviceability concern — it is the mathematical bridge that connects statically determinate analysis to the compatibility equations governing indeterminate structures. Under NSCP 2015 Section 306, deflection limits protect attached non-structural elements and ensure user comfort: for example, live-load deflections of beams and girders supporting plaster ceilings must not exceed L/360. Beyond code compliance, every method for analyzing continuous beams, propped cantilevers, and portal frames (force method, slope-deflection, moment distribution) demands a reliable tool for computing deflections and rotations. This chapter develops two such tools — the Virtual Work (Unit-Load) Method and Castigliano's Second Theorem — and applies them uniformly to trusses, beams, and frames. Mastery of these methods is consistently tested in the PRC Civil Engineer Licensure Examination under Structural Engineering and Construction (SEC).

Concepts

Strain Energy in Elastic Structures

When external forces deform a linearly elastic structure, the work done by those forces is stored as internal strain energy U. This energy is recoverable — remove the loads and the structure springs back. The two most important strain energy expressions for board exam purposes are: 1. AXIAL (truss members): U = N²L / (2AE) where N = axial force (kN), L = member length (m), A = cross-sectional area (m²), E = modulus of elasticity (kN/m²). 2. BENDING (beams and frames): U = ∫[M²/(2EI)] dx where M = bending moment at section x, EI = flexural rigidity. Shear and torsional strain energy exist but are normally neglected in slender beams (shear strain energy is typically less than 1% of bending strain energy for L/d > 10). These strain energy expressions are the starting point for both virtual work and Castigliano's theorem.

Examples

This is the energy stored in a single member. For a full truss, repeat for each member and sum. Note that N is in kN, so the result is in kN·m.

Scenario

A steel truss member 4 m long with A = 1200 mm² = 1.2×10⁻³ m², E = 200 GPa = 2×10⁸ kN/m², carries N = 60 kN tension. Compute U.

Solution

U = N²L/(2AE) = (60)²(4) / [2 × (1.2×10⁻³) × (2×10⁸)] = 14400 / [4.8×10⁵] = 0.030 kN·m = 30 N·m = 30 J

Applications

Foundation of both virtual work and Castigliano methods
Computing stored energy for impact/dynamic loading estimates
Verifying deflection calculations by energy balance

Misconceptions

Students often forget to square the force before summing — strain energy is NOT Σ NL/(AE).
Mixing units (kN with mm, or N with m) is the most common arithmetic error.
Strain energy cannot be superimposed directly — it is quadratic, not linear.

Related Concepts

Virtual Work Method
Castigliano's Second Theorem
Elastic Behavior and Hooke's Law

Common Exam Questions

Example

A cantilever with EI = 4×10¹² N·mm² and M(x) = -Px carries P = 8 kN, L = 2 m. Find U. Answer: U = P²L³/(6EI) = (8000)²(2000)³/[6×(4×10¹²)] = 5.33×10¹⁸/2.4×10¹³ = 222 222 N·mm = 222.2 N·m

Approach

Identify member type (axial or bending), substitute directly into U formula, ensure consistent units.

Question Type

Strain energy computation

Key Points To Remember

Strain energy is always positive — it is a quadratic function of internal forces.
For trusses: U_total = Σ N²L/(2AE), summed over all members.
For beams: U = ∫ M²/(2EI) dx, integrated over the entire length.
Shear strain energy is neglected for slender beams (L/d > 10) — standard board-exam assumption.
Units check: (kN)²(m)/[(m²)(kN/m²)] = kN·m = J ✓

Virtual Work (Unit-Load) Method — Trusses

The virtual work method is the most versatile deflection tool in structural analysis. The key idea: impose a small, fictitious (virtual) unit load at the point and in the direction of the desired deflection, compute the internal forces it causes, then combine with the real internal forces using the principle of virtual work. For TRUSSES, the formula is: δ = Σ (n · N · L) / (AE) where: δ = deflection at the point of interest (positive = in direction of unit load) N = real member force (tension positive, compression negative) n = member force caused by the UNIT virtual load (same sign convention) L = member length A = cross-sectional area E = modulus of elasticity PROCEDURE: Step 1 — Analyze the real truss; find all N forces. Step 2 — Remove the real loads. Apply a unit load (1 kN) at the desired joint in the desired direction. Step 3 — Analyze the virtual truss; find all n forces. Step 4 — For each member, compute n·N·L/(AE). Step 5 — Sum algebraically. A positive result confirms deflection in the assumed unit-load direction. For ROTATION at a joint, apply a unit MOMENT (1 kN·m) instead of a unit force.

Examples

All three products are positive, meaning all members contribute deflection in the downward direction (consistent with physical expectation under downward load). The deflection is 0.693 mm downward at joint C. Note: The factor-of-2 for symmetrical members was handled by computing each member separately.

Scenario

A Pratt truss with apex C at (3, 2.5) m and supports at A(0,0) and B(6,0) carries 20 kN downward at C. Member properties: AE = 2×10⁵ kN for all members. Member forces: N_AC = N_BC = −15.63 kN (compression), N_AB = +12.00 kN (tension). Lengths: L_AC = L_BC = 3.905 m, L_AB = 6.00 m. Find the vertical deflection at C.

Solution

STEP 1 — Real forces N are given above. STEP 2 — Apply 1 kN downward at C. By geometry similarity (loads scale with magnitude), unit-load forces are n = N/20: n_AC = n_BC = −15.63/20 = −0.7815 kN n_AB = 12.00/20 = +0.600 kN STEP 3 — Compute nNL for each member: Member AC: (−0.7815)(−15.63)(3.905) = +47.70 kN²·m Member BC: (−0.7815)(−15.63)(3.905) = +47.70 kN²·m [by symmetry] Member AB: (+0.600)(+12.00)(6.000) = +43.20 kN²·m STEP 4 — Sum: Σ nNL = 47.70 + 47.70 + 43.20 = 138.60 kN²·m STEP 5 — Divide by AE: δ_C = 138.60 / (2×10⁵) = 6.93×10⁻⁴ m = 0.693 mm ↓

Applications

Checking serviceability of bridge trusses against NSCP 2015 deflection limits
Computing camber requirements for fabricated steel trusses
Finding joint displacements as compatibility conditions for indeterminate trusses

Misconceptions

Students sometimes re-apply the real loads during the virtual analysis — the virtual truss carries ONLY the unit load.
Forgetting that a horizontal unit load changes the support reactions — use equilibrium carefully.
Assuming AE is always the same for all members when the problem specifies different sections.

Related Concepts

Strain Energy in Elastic Structures
Method of Joints and Sections (for computing N and n)
Castigliano's Second Theorem

Common Exam Questions

Example

Given a 3-member truss, AE uniform = 150,000 kN, find δ at loaded joint. Typical board answer: 1.2 mm to 5.0 mm range.

Approach

Tabulate members with N, n, L, AE. Compute nNL/AE for each. Sum. Watch sign conventions.

Question Type

Vertical deflection at a loaded joint

Example

For the same truss above, apply 1 kN horizontal at C to get n forces, then compute Σ nNL/AE.

Approach

Apply 1 kN horizontal at that joint. Re-analyze for n forces. The real load system is unchanged.

Question Type

Horizontal deflection at an unloaded joint

Key Points To Remember

The unit load is 1 kN (dimensionless numerically, but direction-specific).
Tension is positive, compression is negative — be consistent for both N and n.
Like signs (both T or both C) contribute positive terms → deflection in unit-load direction.
Unlike signs contribute negative terms → deflection is opposite to unit-load direction.
If all members have the same AE, factor it out: δ = (1/AE) Σ nNL.
For horizontal deflection, apply unit horizontal load; for vertical, apply unit vertical load.

Virtual Work (Unit-Load) Method — Beams and Frames

For beams and frames, internal actions are primarily bending moments, and the virtual work formula integrates the product of real and virtual moment diagrams: δ = ∫ [m(x) · M(x) / EI] dx where: M(x) = real bending moment function m(x) = bending moment function from the unit virtual load EI = flexural rigidity (may vary along the member) For ROTATION at a section, apply a unit MOMENT (1 kN·m) at that section and use the same integral: θ = ∫ [m_θ(x) · M(x) / EI] dx MULTIPLICATION TABLE FOR COMMON M AND m DIAGRAMS: When M and m are both polynomial (constant, linear, parabolic), the integral ∫ mM dx can be evaluated using the standard diagram-multiplication table (Mohr's integral table). For board exams, these key results are essential: Rectangle × Rectangle: (b)(ab) = ab² [for uniform M × uniform m] Triangle × Triangle: (1/3)(base)(h₁)(h₂) Triangle × Parabola: (1/3 or 1/6) variations — memorize the table! PROCEDURE FOR BEAMS: Step 1 — Draw the real bending moment diagram M(x). Step 2 — Apply 1 kN (or 1 kN·m for rotation) at the target location; draw the virtual moment diagram m(x). Step 3 — Integrate ∫ mM/EI dx over the entire structure. Step 4 — Positive result = deflection in the direction of the unit load.

Examples

The unit load and real load both produce hogging moments (negative), so their product is positive throughout, giving a positive deflection — downward, consistent with physical expectation. This recovers the textbook formula PL³/(3EI).

Scenario

Cantilever beam of length L = 3 m, end load P = 10 kN, EI = 2×10¹³ N·mm². Find tip deflection using the unit-load method.

Solution

Set up coordinate x from the free end (tip). Real moment: M(x) = −Px = −10x kN·m (hogging = negative) Unit load (1 kN ↓ at tip): m(x) = −1·x = −x kN·m Integrate (both negative, product is positive): δ = ∫₀ᴸ [m·M / EI] dx = ∫₀ᴸ [(−x)(−10x) / EI] dx = (10/EI) ∫₀³ x² dx = (10/EI) · [x³/3]₀³ = (10/EI) · (27/3) = 90 / EI Convert: L = 3 m = 3000 mm, P = 10 kN = 10,000 N Using formula δ = PL³/(3EI): δ = (10,000)(3000)³ / [3 × (2×10¹³)] = (10,000)(2.7×10¹⁰) / (6×10¹³) = 2.7×10¹⁴ / 6×10¹³ = 4.50 mm ↓

The factor of 2 accounts for the symmetric right half. Integrating only over one half and doubling is computationally efficient. The result 7.5 mm matches the standard formula PL³/(48EI).

Scenario

Simply supported beam, span L = 6 m, central point load P = 20 kN, EI = 1.2×10¹³ N·mm². Find midspan deflection.

Solution

By symmetry, analyze left half only (0 ≤ x ≤ L/2). Real reactions: R_A = R_B = 10 kN Real M(x) = 10x for 0 ≤ x ≤ 3 m Unit load 1 kN at midspan: Virtual reactions: r_A = r_B = 0.5 kN Virtual m(x) = 0.5x for 0 ≤ x ≤ 3 m δ = 2 × ∫₀³ [m·M / EI] dx [factor of 2 for symmetry] = 2 × ∫₀³ [(0.5x)(10x) / EI] dx = 2 × (5/EI) ∫₀³ x² dx = 2 × (5/EI) × (9) = 90/EI EI in N·mm²: 1.2×10¹³ N·mm² Convert: P=20,000 N, L=6000 mm δ = PL³/(48EI) = (20,000)(6000)³/[48(1.2×10¹³)] = (20,000)(2.16×10¹¹) / (5.76×10¹⁴) = 4.32×10¹⁵ / 5.76×10¹⁴ = 7.50 mm ↓

Applications

Computing deflections of floor beams to verify NSCP 2015 serviceability limits (L/360 for LL, L/240 for LL+ΔDL)
Checking girder deflections under crane runway loads in industrial buildings
Computing compatibility deflections in the force method for indeterminate beams

Misconceptions

Students forget to include the column contribution for frames — every member that bends must be integrated.
Using the wrong sign convention (sagging vs. hogging) mid-problem leads to wrong answers.
Applying the unit load at the wrong location — it must go exactly where the deflection is desired.

Related Concepts

Bending Moment Diagrams
Castigliano's Second Theorem
Moment Area Method (alternative for beams)

Common Exam Questions

Example

A 5-m simply supported beam carries w = 15 kN/m. EI = 8×10¹² N·mm². Find δ_max. Answer: 5(15)(5000)⁴/[384(8×10¹²)] = 5(15)(6.25×10¹⁴)/3.072×10¹⁵ = 15.26 mm.

Approach

Use unit-load integration to derive δ = PL³/(48EI) for simply supported beam with central load, or δ = 5wL⁴/(384EI) for UDL.

Question Type

Standard formula recovery

Example

L-shaped frame: vertical column (height h) + horizontal beam (length L). Apply load at beam tip. Compute δ at tip by integrating column and beam contributions separately.

Approach

For each member of the frame, draw M and m diagrams. Integrate ∫mM/EI dx for each member. Sum all members.

Question Type

Frame deflection by virtual work

Key Points To Remember

The sign convention for M and m must be consistent (sagging positive throughout).
Break the beam into segments where M(x) or m(x) changes form.
For a simply supported beam, m(x) is always a triangle (linear).
For a cantilever, m(x) is also a triangle with maximum at the fixed end.
∫ mM/EI dx = (1/EI) ∫ mM dx when EI is constant — factor it out.
A positive result means deflection is in the direction of the assumed unit load (downward if unit load is downward).

Castigliano's Second Theorem

Castigliano's Second Theorem states that the deflection at the point of application of a load equals the partial derivative of the total strain energy with respect to that load: δᵢ = ∂U/∂Pᵢ (for linear deflection at load Pᵢ) θᵢ = ∂U/∂Mᵢ (for rotation at moment Mᵢ) For beams and frames with bending dominating: U = ∫ M²/(2EI) dx δ = ∂U/∂P = ∫ [M/(EI)] · (∂M/∂P) dx The key insight: ∂M/∂P is algebraically identical to the virtual moment m(x) in the unit-load method! Castigliano's theorem and the unit-load method are mathematically equivalent — they are two views of the same principle. DUMMY LOAD TECHNIQUE: If no real load exists at the point where deflection is desired: 1. Add a dummy (fictitious) load Q at that point in the desired direction. 2. Express M in terms of Q and the real loads. 3. Compute ∂M/∂Q. 4. Evaluate δ = ∫ [M/(EI)] · (∂M/∂Q) dx. 5. Set Q = 0 in the final expression. For TRUSSES: δ = ∂U/∂P = Σ [NL/(AE)] · (∂N/∂P) where ∂N/∂P = n (the force in each member per unit of P) — again identical to virtual work.

Examples

This derivation recovers the famous formula δ_max = 5wL⁴/(384EI) for a uniformly loaded simply supported beam. Note that Q = 0 is substituted into M(x) BEFORE computing the integral for efficiency, but ∂M/∂Q is computed BEFORE setting Q = 0.

Scenario

A simply supported beam of span L carries a UDL of intensity w (kN/m). Using Castigliano's theorem, find the midspan deflection.

Solution

Since no point load exists at midspan, introduce dummy load Q at midspan. Reactions (with Q at center): R_A = wL/2 + Q/2 R_B = wL/2 + Q/2 For 0 ≤ x ≤ L/2: M(x) = (wL/2 + Q/2)x − wx²/2 ∂M/∂Q = x/2 By symmetry, integrate left half and double: δ_mid = 2 ∫₀^(L/2) [M/(EI)] · (∂M/∂Q) dx |_(Q=0) With Q = 0: M(x) = wLx/2 − wx²/2 = wx(L−x)/2 ∂M/∂Q = x/2 δ_mid = 2/EI ∫₀^(L/2) [wx(L−x)/2] · (x/2) dx = 2/EI · w/4 ∫₀^(L/2) x²(L−x) dx = w/(2EI) ∫₀^(L/2) (Lx² − x³) dx = w/(2EI) [Lx³/3 − x⁴/4]₀^(L/2) = w/(2EI) [L(L/2)³/3 − (L/2)⁴/4] = w/(2EI) [L⁴/24 − L⁴/64] = w/(2EI) · L⁴(8−3)/192 = wL⁴·5 / (384EI) = 5wL⁴/(384EI) ✓

Applications

Systematic computation of deflections in indeterminate structures (combined with compatibility)
Finding rotations at beam ends for moment distribution setup
Deflection at any point using dummy loads without re-solving the entire structure

Misconceptions

Setting Q = 0 before differentiating — this destroys the dummy load effect and gives a wrong answer.
Confusing Castigliano's First Theorem (force = ∂U/∂δ) with the Second Theorem (δ = ∂U/∂P) — the board exam tests the Second Theorem.
Forgetting that the theorem requires linear elastic behavior — it fails for nonlinear or inelastic structures.

Related Concepts

Strain Energy in Elastic Structures
Virtual Work (Unit-Load) Method
Force Method for Indeterminate Structures

Common Exam Questions

Example

w = 12 kN/m, L = 4 m, EI = 6×10¹² N·mm². δ = 5(12)(4000)⁴/[384(6×10¹²)] = 5(12)(2.56×10¹⁴)/2.304×10¹⁵ = 8.33 mm

Approach

Place dummy Q at midspan. Write M(x) with Q. Differentiate. Set Q = 0. Integrate.

Question Type

Midspan deflection under UDL using Castigliano

Example

θ_tip = PL²/(2EI) for cantilever with end load P. Verify by Castigliano.

Approach

Place dummy moment M₀ at free end. Write M(x) with M₀. Differentiate with respect to M₀. Set M₀ = 0. Integrate.

Question Type

End rotation of cantilever under end load

Key Points To Remember

∂M/∂P acts as the virtual moment m(x) — the two methods are equivalent.
Always differentiate M with respect to the specific load before integrating, not after.
Dummy load Q must be introduced before writing M(x); set Q = 0 only at the very end.
Castigliano's theorem applies only to linear elastic structures (Hooke's Law must hold).
For rotation: apply a dummy moment, differentiate with respect to it, then set it to zero.

Serviceability and Deflection Limits (NSCP 2015 Context)

Computing deflections is only half the task; verifying them against code limits is the other half. NSCP 2015 Table 306-1 (equivalent to ASCE 7 and analogous to ACI 318-14 Table 24.2.2) prescribes maximum allowable deflections as fractions of span L: Beams and girders supporting plaster ceilings: L/360 (live load only) Beams and girders not supporting plaster ceilings: L/240 (live load only) Roof members not supporting ceilings: L/180 (live load only) Beams considering long-term effects (live + creep): L/240 For reinforced concrete members (ACI 318-19 Section 24.2): Immediate deflection under live load: L/360 (floors with partitions) Long-term deflection (including creep and shrinkage): not to exceed L/240 − (immediate DL deflection) In board exam problems, after computing the elastic deflection δ by virtual work or Castigliano, the follow-up question often asks: 'Does this beam satisfy NSCP requirements?' — simply compare δ_computed with L/360 or L/240 depending on the context.

Examples

The beam passes by a margin of only 0.22 mm. In practice, a deeper section would be selected for additional safety. This type of final check is common in board exam problems where computation leads to a pass/fail conclusion.

Scenario

A simply supported steel floor beam, L = 8 m, carries live load w_L = 15 kN/m. Computed midspan deflection δ_LL = 22 mm. Does it satisfy NSCP 2015 for a beam with a plaster ceiling?

Solution

Allowable δ = L/360 = 8000/360 = 22.22 mm Computed δ = 22 mm < 22.22 mm ✓ (SATISFIES — barely!)

Applications

Structural design verification for building permits under the National Building Code of the Philippines
Shop drawing review for pre-engineered steel buildings
Deflection checks for post-tensioned concrete slabs in high-rise buildings

Misconceptions

Using L/360 for total load when NSCP specifies L/360 only for live load — total load limit is L/240.
Using span in meters when computing L/360 — the result must also be in meters (or consistently in mm).
Forgetting that RC slabs have additional long-term multipliers for creep (ACI 318 Section 24.2.4, multiplier λ_Δ = ξ/(1 + 50ρ')).

Related Concepts

NSCP 2015 Section 306
ACI 318-19 Chapter 24
Long-term deflection in reinforced concrete

Common Exam Questions

Example

If δ_computed = 18 mm and L = 5 m: allowable = 5000/360 = 13.9 mm. Fails! Ratio = 18/13.9 = 1.29 > 1.

Approach

Compute δ using appropriate formula or virtual work. Compute L/360 or L/240. Compare.

Question Type

Pass/fail serviceability check

Key Points To Remember

NSCP 2015 Table 306-1: L/360 for LL on beams with plaster ceilings.
L/240 for total load deflection on beams without plaster ceilings.
ACI 318 controls for RC members — check both immediate and long-term deflections.
Steel beams (AISC 360 / NSCP Chapter 5): deflection ≤ L/360 for LL is the typical service criterion.
Always state the allowable deflection in mm before comparing.

Practice Problems

Both compression members contribute positively because n and N are both negative (product is positive). The tension bottom chord also contributes positively. The answer 1.602 mm downward is the vertical displacement at the apex joint C under the 30 kN load.

Problem

PROBLEM 1 (Truss — Virtual Work): A simple Pratt truss has span 8 m (joints A at origin, B at 8 m). The top chord joint C is at (4, 3) m. A vertical load of 30 kN acts at C. Member areas and E are: A_AC = A_BC = 900 mm², A_AB = 1200 mm², E = 200 GPa for all. Determine the vertical deflection at C.

Solution

STEP 1 — Geometry: L_AC = √(4² + 3²) = 5.00 m L_BC = √(4² + 3²) = 5.00 m L_AB = 8.00 m STEP 2 — Real member forces (by equilibrium at joint C): ΣFy = 0 at C: 2·N_AC·(3/5) = 30 → N_AC = N_BC = +25.00 kN (T) ΣFx = 0 at C: N_AB = −2·N_AC·(4/5) = −40.00 kN (C) [wait — let us re-check] Actually at joint A (pin): R_A = 15 kN↑, R_B = 15 kN↑ At joint C (free body): 2 top-chord members at angle θ where sinθ = 3/5, cosθ = 4/5 In the vertical: 2·F_chord·sin θ = 30 → F = 25 kN (C) [compression in top chord for a loaded apex] Wait: Apex carries load DOWN → top-chord members are in COMPRESSION. N_AC = N_BC = −25.00 kN (C) At joint A: F_AB = N_AC·cos θ = 25·(4/5) = +20.00 kN (T) STEP 3 — Unit virtual load (1 kN ↓ at C): By proportionality: n = N/30: n_AC = n_BC = −25/30 = −0.8333 n_AB = +20/30 = +0.6667 STEP 4 — Compute nNL/(AE) for each member: AE_AC = (900×10⁻⁶ m²)(200×10⁶ kN/m²) = 180,000 kN AE_BC = 180,000 kN AE_AB = (1200×10⁻⁶)(200×10⁶) = 240,000 kN Member AC: (−0.8333)(−25)(5.00)/180,000 = 104.17/180,000 = 5.787×10⁻⁴ m Member BC: same by symmetry = 5.787×10⁻⁴ m Member AB: (0.6667)(20)(8.00)/240,000 = 106.67/240,000 = 4.444×10⁻⁴ m STEP 5 — Sum: δ_C = 5.787×10⁻⁴ + 5.787×10⁻⁴ + 4.444×10⁻⁴ = 16.018×10⁻⁴ m = 1.602 mm ↓

For part (b), the unit virtual moment is a constant (−1 kN·m hogging) throughout because any section between the tip and a cut at x is subjected to the unit moment. The product m_θ·M = (−1)(−4x²) = 4x² is positive, giving a positive rotation — the tip rotates clockwise (consistent with downward UDL causing hogging).

Problem

PROBLEM 2 (Beam — Virtual Work, Rotation): A cantilever of length L = 4 m carries a UDL w = 8 kN/m over its full length. EI = 1.5×10¹³ N·mm². Using the unit-load method, find (a) the tip deflection and (b) the tip rotation.

Solution

Coordinate x from free end (tip), 0 ≤ x ≤ L = 4 m. Real moment (hogging — negative convention): M(x) = −wx²/2 = −8x²/2 = −4x² kN·m --- PART (a): Tip deflection --- Unit load: 1 kN ↓ at tip → m(x) = −x kN·m δ_tip = ∫₀⁴ [m·M/EI] dx = ∫₀⁴ [(−x)(−4x²)/EI] dx = (4/EI) ∫₀⁴ x³ dx = (4/EI) · [x⁴/4]₀⁴ = (4/EI) · 64 = 256/EI Convert: EI = 1.5×10¹³ N·mm², L = 4000 mm, w = 8 N/mm = 8×10⁻³ kN/mm Using formula: δ = wL⁴/(8EI) = (8)(4000)⁴/[8×(1.5×10¹³)] = (8)(2.56×10¹⁴) / (1.2×10¹⁴) = 2.048×10¹⁵ / 1.2×10¹⁴ = 17.07 mm ↓ Verification with calculus in kN, m: δ = 256 kN·m³ / EI EI in kN·m²: 1.5×10¹³ N·mm² = 1.5×10¹³ × 10⁻⁶ kN·m² = 1.5×10⁷ kN·m² δ = 256 / (1.5×10⁷) = 1.707×10⁻⁵ m = 0.01707 m = 17.07 mm ✓ --- PART (b): Tip rotation --- Unit moment: 1 kN·m ↺ at tip → m_θ(x) = −1 kN·m (constant, hogging) θ_tip = ∫₀⁴ [m_θ·M/EI] dx = ∫₀⁴ [(−1)(−4x²)/EI] dx = (4/EI) ∫₀⁴ x² dx = (4/EI) · [64/3] = 256/(3EI) = 256/(3 × 1.5×10⁷) = 5.689×10⁻⁶ rad = 0.326° Formula check: θ = wL³/(6EI) = (8)(4)³/(6 × 1.5×10⁷) = 512/(9×10⁷) = 5.689×10⁻⁶ rad ✓

The L-frame problem illustrates that horizontal deflection at C has TWO contributions: (1) the column sways like a cantilever under horizontal load H, giving Hh³/(3EI); and (2) the beam acts as a rigid arm of length L rotating with the column top, adding Hh²L/(2EI). The Castigliano method, when correctly set up with moments due to all effects of the unit load, captures both. Always sketch the frame deformation to confirm the contributions.

Problem

PROBLEM 3 (Castigliano — Frame): An L-shaped frame has a vertical member (column) of height h = 3 m (fixed at base A, free at top B) and a horizontal member (beam) of length L = 4 m from B to free end C. A horizontal load H = 6 kN acts at C. EI is constant = 2×10⁷ kN·m². Find the horizontal deflection at C.

Solution

Set up coordinates: Beam BC: x from C toward B, 0 ≤ x ≤ 4 m (horizontal) Column AB: y from B toward A, 0 ≤ y ≤ 3 m (vertical) --- Real Moment Diagrams --- Horizontal load H = 6 kN at C (rightward, say). Beam BC (x from C): M_beam(x) = −H·x = −6x kN·m (H causes moment in beam) Wait: H is horizontal, beam is horizontal → H causes no moment in the horizontal beam. Actually: H at C is horizontal. In the beam BC (horizontal), V = 0, M = 0, N = H = 6 kN (axial in beam). The moment in the column AB at height y from B: M_col(y) = H·(h − y)... let us restate carefully. PROPER FREE-BODY: At fixed base A: reactions are V_A (vertical), H_A (horizontal) = 6 kN, M_A. For column AB (y measured from A upward): At cut at height y: M_col(y) = M_A − H_A·y = −6(h−y)... Let y measured from B downward (0 at B, h at A): M_col(y) = H × (distance from C to the column section) = H × (L + 0) for the column? No. Re-do with y from A (base, y=0) to B (top, y=h=3m): The horizontal reaction at A = 6 kN (left, to balance H at C). At cut in column at height y (from A): Taking the lower free body: M = 6·y kN·m (clockwise = positive per our convention) Wait, H at C is to the right → H_A at A is to the LEFT = 6 kN. At cut at height y from A, lower free-body: M = H_A × y = 6y kN·m. In the beam BC (horizontal, from B to C): Cut in beam at distance x from B. The horizontal force in the beam is 6 kN (compression, axial only). No transverse load on beam → M_beam = 0 throughout (the load at C is in the axis of the beam). Wait — re-read: H is horizontal at C. Beam BC is horizontal. Column AB is vertical. H at C acts along the axis of beam BC → beam is in pure axial compression (no bending). All bending is in the column. DEFLECTION AT C (horizontal) = deflection at B (horizontal) + rotation at B × L Castigliano: δ_C = ∂U/∂H = ∫_col [M_col/EI · ∂M_col/∂H] dy M_col(y) = Hy (y from A) ∂M_col/∂H = y δ_C = ∫₀³ [Hy · y / EI] dy = (H/EI) ∫₀³ y² dy = (H/EI) · [y³/3]₀³ = (H/EI) · 9 = 6×9/(2×10⁷) = 54/(2×10⁷) = 2.7×10⁻⁶ m BUT: The deflection at C also includes the cantilever effect of the beam rotating with joint B: δ_C = δ_B + θ_B × L δ_B = Hh³/(3EI) = 6(3)³/[3×2×10⁷] = 162/(6×10⁷) = 2.7×10⁻⁶ m θ_B = Hh²/(2EI) = 6(9)/(2×2×10⁷) = 54/(4×10⁷) = 1.35×10⁻⁶ rad δ_C = 2.7×10⁻⁶ + 1.35×10⁻⁶ × 4 = 2.7×10⁻⁶ + 5.4×10⁻⁶ = 8.1×10⁻⁶ m = 0.0081 mm Correcting Castigliano (include beam rotation effect properly): Using Castigliano directly with virtual complementary energy in both members: δ_C = ∫_col [M/EI · ∂M/∂H] dy + ∫_beam [M/EI · ∂M/∂H] dx = ∫₀ʰ [Hy·y/EI] dy + 0 [beam has M = 0] But the deflection at C differs from B because the RIGID beam carries the rotation of B. Castigliano gives δ at the LOAD point C directly. The moment in the column is: At cut y from A: M = Hy (from horizontal reaction at A). The ROTATION at B also moves C horizontally by θ_B × L. In Castigliano, M_col = Hy is correct; the integration already accounts for δ_B. The beam contributes zero bending, so the frame horizontal deflection at C: δ_C = Hh³/(3EI) + Hh²L/(2EI) [sway + rigid-body rotation of beam] = H/EI × [h³/3 + h²L/2] = 6/(2×10⁷) × [(27/3) + (9×4/2)] = 3×10⁻⁷ × [9 + 18] = 3×10⁻⁷ × 27 = 8.1×10⁻⁶ m = 0.0081 mm Correct Castigliano setup (with x measured from C in beam, y from A in column): In beam (x from C, length L): M_beam(x) = 0 (H is axial) In column (y from A, height h): M_col(y) = H·y ∂M_col/∂H = y; ∂M_beam/∂H = 0 BUT: We missed that H at C causes moment in the column PLUS the beam transmits the rotation. For virtual work equivalence, consider the full geometry with ξ from C along beam, then up the column: At point C, apply 1 kN horizontal. At cut in column at distance y from A: Virtual moment m = 1·y + 1·0 = y (the load travels horizontally to B, then down column)? Actually at cut y from A in column: m(y) = 1×(h−y)... let η = h−y (distance from B): m(η) = η (the moment arm from the virtual load at C to the column section = η + 0 since beam is horizontal) FINAL CORRECT APPROACH (y from B downward, 0 at B, h at A): M_col(η) = H·η where η = distance from B downward ∂M/∂H = η δ_C = ∫₀ʰ [Hη·η/EI] dη + L·θ_B Using direct formula: δ_C = Hh³/(3EI) + Hh²L/(2EI) = (Hh²/EI)(h/3 + L/2) = [6×9/(2×10⁷)]×[3/3 + 4/2] = [54/(2×10⁷)]×[1 + 2] = [2.7×10⁻⁶]×3 = 8.1×10⁻⁶ m ≈ 0.0081 mm

The total load is 25 kN/m but only the live-load portion is checked against L/360. The dead-load deflection is typically pre-cambered. The maximum permissible live load intensity is 21.3 kN/m. If w_LL > 21.3 kN/m, the beam must be deepened (larger I) to comply with NSCP 2015.

Problem

PROBLEM 4 (Board-Exam Style — Multiple Choice Format): A simply supported steel beam of span L = 10 m carries a uniform load w = 25 kN/m (including self-weight). The beam has I = 5×10⁸ mm⁴ and E = 200 GPa. The beam supports a plastered ceiling. What is the maximum permissible live load intensity w_LL (in kN/m) if the beam must satisfy NSCP 2015 deflection requirements?

Solution

NSCP 2015 Table 306-1: For beams with plaster ceiling, δ_LL ≤ L/360. Allowable δ_LL = L/360 = 10,000/360 = 27.78 mm Formula: δ_LL = 5w_LL·L⁴ / (384EI) Solve for w_LL: w_LL = 384·EI·δ_all / (5·L⁴) Substitute (consistent units: N, mm): E = 200,000 N/mm² = 200 GPa I = 5×10⁸ mm⁴ EI = 200,000 × 5×10⁸ = 10¹⁴ N·mm² δ_all = 27.78 mm L = 10,000 mm L⁴ = (10⁴)⁴ = 10¹⁶ mm⁴ w_LL = 384 × 10¹⁴ × 27.78 / (5 × 10¹⁶) = 384 × 27.78 × 10¹⁴ / (5 × 10¹⁶) = 10,667.52 × 10¹⁴ / (5 × 10¹⁶) = 10,667.52 / 500 = 21.34 N/mm = 21.34 kN/m ANSWER: w_LL_max ≈ 21.3 kN/m

Exam Preparation Tips

MEMORIZE THE BIG FOUR deflection formulas (they appear in at least one problem per board exam): δ = PL³/(3EI) for cantilever with end load; δ = wL⁴/(8EI) for cantilever with UDL; δ = PL³/(48EI) for simply supported beam with center load; δ = 5wL⁴/(384EI) for simply supported beam with UDL.
UNIT CONSISTENCY is the #1 source of errors in deflection problems. Choose one consistent system at the start — either (kN, m, kN/m², m⁴) or (N, mm, N/mm², mm⁴) — and never mix. If EI is given in N·mm², convert all other quantities to N and mm.
For TRUSS virtual work problems, tabulate your work: make a table with columns [Member | N (kN) | n (kN) | L (m) | AE (kN) | nNL/AE]. Sum the last column. This organized approach prevents sign and arithmetic errors under exam pressure.
Remember the DUMMY LOAD RULE: If you need a deflection where NO real load acts, you MUST introduce a dummy load Q. Write M(x) with Q in it, differentiate ∂M/∂Q, THEN substitute Q = 0. Never substitute Q = 0 before differentiating.
For NSCP 2015 serviceability problems, always state the allowable deflection numerically in mm before comparing. Example: 'Allowable δ = L/360 = 6000/360 = 16.67 mm.' Then compare with computed δ.
SIGN CONVENTION TIP: In the unit-load method for beams, if you adopt 'sagging positive' consistently for both M and m, the integral ∫mM/EI dx is positive when both M and m are on the same side (both sagging or both hogging). A positive result means deflection in the direction of the unit load.
Use SYMMETRY aggressively. For symmetric structures with symmetric loading, integrate over half the span and double. This reduces computation time by 50% — critical in a timed board exam.
The CASTIGLIANO and UNIT-LOAD methods give identical answers — if your two methods disagree, there is an arithmetic error. Use this as a self-check: solve with one method, verify with the other on a key problem.
For FRAME problems, do not forget the column's bending contribution when computing horizontal deflection. A common error is integrating only the beam moment and ignoring the column.
In the PRC board exam, deflection problems often appear as part of a 3-part question: Part 1 asks for reaction/moment (from the determinate analysis); Part 2 asks for the virtual member forces; Part 3 asks for the deflection. Answer Part 1 carefully — an error there propagates through the entire problem.
Practice the DIAGRAM MULTIPLICATION TABLE (Vereshchagin's rule) for quickly evaluating ∫mM dx when both diagrams are polygons: Rectangle×Rectangle = bL, Triangle×Triangle = bL/3, Triangle×Rectangle = bL/2, Parabola×Triangle = bL/3. This saves significant integration time.
When a problem says 'AE = constant for all members,' immediately factor it out of the truss virtual work sum: δ = (1/AE) Σ nNL. This prevents the error of dividing by AE multiple times.

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In summary

Deflection analysis using the Virtual Work (Unit-Load) Method and Castigliano's Second Theorem is a cornerstone of structural analysis in the PRC Civil Engineer Licensure Examination. These two methods, though derived from different theoretical viewpoints (principle of virtual work vs. energy methods), produce identical results — because ∂M/∂P is mathematically the same as the virtual moment m(x). The key to exam success is systematic execution: tabulate for trusses, integrate for beams, always use the dummy load when no real load acts at the target point, and differentiate before substituting the dummy load as zero. Equally important is the connection to NSCP 2015 serviceability requirements — computing the deflection is only useful if you then verify it against the code limit (L/360 for beams with plaster ceilings, L/240 for others). Beyond the board exam, these deflection calculations are the compatibility conditions that unlock the force method for indeterminate structures — every continuous beam and portal frame analysis you will perform in practice rests on the skills developed here. Master the four standard formulas, practice the sign conventions rigorously, maintain unit discipline, and you will handle any deflection problem the PRC throws at you with confidence.

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