Deflections of Structures - Revision Notes
Deflection analysis is a cornerstone of structural engineering and a consistent topic in the PRC Civil Engineer Licensure Examination. Understanding how structures deform under load is essential for two reasons: (1) serviceability — codes such as NSCP 2015 Section 406 impose maximum allowable deflection limits (e.g., L/360 for live load on beams supporting brittle finishes) to protect occupants and finishes; and (2) compatibility — every method used to analyze statically indeterminate structures (force method, slope-deflection, moment distribution) uses deflection equations as compatibility conditions. This chapter focuses on the two most powerful energy-based deflection methods: the Virtual Work (Unit-Load) Method and Castigliano's Second Theorem. Mastery of these two methods allows you to find deflections and rotations in trusses, beams, and frames from a single, unified framework.
1. Why Deflections Matter — Serviceability and Compatibility
Formulas
U_axial = N²L / (2AE)
Example
A steel bar with N = 50 kN, L = 2 m, A = 10×10⁻⁴ m², E = 200×10⁶ kN/m²: U = (50²×2)/(2×200×10⁶×10⁻³) = 12.5 N·m = 12.5 J
Variables
N = axial force (kN), L = member length (m), A = cross-sectional area (m²), E = modulus of elasticity (kN/m²)
Application
Strain energy stored in a single axially loaded truss member.
U_bending = ∫[M²/(2EI)] dx
Example
Cantilever, length L, end load P: M(x) = Px (from fixed end). U = ∫₀ᴸ (Px)²/(2EI) dx = P²L³/(6EI)
Variables
M = bending moment at section x (kN·m), EI = flexural rigidity (kN·m²), integrated over the full member length
Application
Strain energy stored in a bent beam or frame member.
W_ext = Pδ/2
Example
If P = 20 kN produces δ = 5 mm = 0.005 m, then W = 20×0.005/2 = 0.05 kN·m = 50 J
Variables
P = applied load, δ = deflection at point of load application (same direction)
Application
External work by a single gradually applied point load on a linear elastic structure.
Exam tips
- Memorize the most tested NSCP 2015 deflection limits: L/360 (live load, floor members supporting brittle finishes), L/240 (total load), L/180 (roof, not supporting brittle finishes).
- When a problem asks 'does the beam satisfy deflection requirements?' always compute δ, then compare with L/limit.
- The elastic curve equation (EI d²y/dx² = M) is rarely tested directly; the board favors the energy methods.
Key points
- Deflections are checked against NSCP 2015 Table 406.2.1 allowable limits (e.g., L/240 for total load, L/360 for live load on floors supporting non-structural elements).
- ACI 318-19 Section 24.2 limits immediate and long-term deflections of concrete members to protect attached non-structural elements.
- Energy stored in a deformed elastic structure is called strain energy U; it equals the work done by the applied loads (W_ext = U for linear elastic systems).
- For a linear elastic structure, the external work done by a slowly applied load P moving through displacement δ is W = Pδ/2.
- Deflection methods based on energy are general — they apply equally to trusses, beams, and frames without re-deriving separate equations.
- The principle of superposition applies when material is linearly elastic and deformations are small (both are standard assumptions in licensure exam problems).
Definitions
Strain Energy (U)
The elastic potential energy stored in a deformed structure equal to the total internal work done by stresses on strains. For a linear elastic structure, U = W_ext.
Importance
Foundation of all energy-based deflection methods; understanding U leads directly to both the virtual work method and Castigliano's theorem.
Serviceability Limit State
A design condition (per NSCP 2015, ACI 318, AISC 360) where the structure remains functional and comfortable for occupants even though it has not collapsed. Excessive deflection, vibration, and cracking are serviceability failures.
Importance
NSCP 2015 Section 406 specifies maximum deflection limits; board exams test whether computed deflections are within these limits.
Principle of Superposition
For linear elastic structures with small deformations, the total deflection due to multiple loads equals the algebraic sum of deflections caused by each load acting alone.
Importance
Allows complex loading cases to be broken into simpler sub-problems; widely used in board exam solutions.
Common mistakes
- Applying deflection formulas to structures with large deformations or inelastic behavior where superposition is invalid.
- Confusing the deflection limit with the span: L/360 means deflection ≤ span/360, not 360 mm.
- Using NSCP live-load deflection limits for total-load checks — always identify which limit applies to which load combination.
2. Virtual Work Method (Unit-Load Method) — Trusses
Formulas
δ = Σ (n · N · L) / (AE)
Example
Triangular truss: apex C carries 20 kN↓. N_AC = N_BC = −15.63 kN (compression), N_AB = +12 kN (tension). Unit load 1 kN↓ at C: n_AC = n_BC = −0.781, n_AB = +0.60. AE = 2×10⁵ kN for all members. L_AC = L_BC = 3.905 m, L_AB = 6 m. δ_C = [2(−0.781)(−15.63)(3.905) + (0.60)(12)(6)] / (2×10⁵) = [2(47.7) + 43.2] / (2×10⁵) = 138.6 / (2×10⁵) = 6.93×10⁻⁴ m ≈ 0.69 mm ↓
Variables
n = member force due to the unit virtual load (kN/kN = dimensionless), N = member force due to real loads (kN), L = member length (m), AE = axial rigidity (kN), summation over all members
Application
Deflection at any joint of a statically determinate (or indeterminate, after forces are known) plane or space truss.
Exam tips
- Always set up a tabular solution for board exams: columns for Member | L | A | E | N | n | nNL/AE. This organizes work and prevents arithmetic errors.
- For symmetric trusses with symmetric loading, n and N will be equal for symmetric members — compute once and multiply by 2.
- If the answer is negative, the deflection is opposite to the assumed direction of the unit load. State this explicitly in your answer.
- The most commonly tested truss deflection in board exams is the vertical deflection at a loaded joint of a simple Pratt or Warren truss.
Key points
- The unit-load method is derived from the principle of virtual work: virtual external work = virtual internal work.
- Procedure: (a) Solve the real structure for member forces N under the actual loads. (b) Remove actual loads and apply a unit load (1 kN or 1 kN·m) at the point and direction of the desired deflection. (c) Solve for member forces n under the unit load. (d) Apply the truss deflection formula.
- The unit load must be in the same direction as the desired deflection (vertical unit load → vertical deflection; horizontal unit load → horizontal deflection; unit moment → rotation).
- A positive result means the deflection is in the same direction as the unit load; negative means opposite direction.
- For an L-shaped truss or any statically determinate truss, method of joints or sections gives both N and n.
- All members must use consistent units: if AE is in kN, then N in kN, L in m, δ in m.
Definitions
Virtual Load (Unit Load)
A fictitious load of magnitude 1 (unit) applied at the point and in the direction of the desired deflection, used only to create a stress state for the virtual work equation. It is NOT part of the real loading.
Importance
The internal forces n produced by this unit load are the 'weighting factors' that extract the desired deflection component from the total strain energy.
Axial Rigidity (AE)
The product of cross-sectional area A and Young's modulus E of a truss member. It measures resistance to axial deformation. Units: kN (when A in m² and E in kN/m²) or N (SI base).
Importance
Every term in the truss deflection sum contains AE in the denominator; a larger AE means smaller member deformation and smaller deflection contribution.
Common mistakes
- Applying the unit load in the wrong direction: a horizontal unit load computes horizontal deflection, not vertical.
- Using mixed units — N in kN with L in mm and AE in N produces a nonsensical answer. Always standardize (N in kN, L in m, AE in kN).
- Forgetting members with zero N or zero n — they contribute zero to the sum, but must not be omitted from the tabulation (to show awareness).
- Sign error: tension is (+), compression is (−). Like signs multiply to give a positive contribution (deflection in the direction of the unit load).
- Confusing n (unit-load forces) with N (real forces) in the summation table.
3. Virtual Work Method (Unit-Load Method) — Beams and Frames
Formulas
δ = ∫ (m · M / EI) dx
Example
Cantilever length L, end load P. Origin at free end. M(x) = −Px, m(x) = −x (unit load 1 kN at free end). δ = ∫₀ᴸ (−x)(−Px)/EI dx = P/EI × [x³/3]₀ᴸ = PL³/(3EI). Numerically: P=10 kN, L=3 m, EI=2×10¹³ N·mm² = 2×10⁴ kN·m²; δ = 10×27/(3×2×10⁴) = 270/60000 = 4.5×10⁻³ m = 4.5 mm ↓
Variables
m = bending moment diagram due to unit virtual load (kN·m / kN = m), M = bending moment diagram due to real loads (kN·m), EI = flexural rigidity (kN·m²), integration over member length
Application
Deflection at any point of a beam or frame; also applies to the rotation if a unit moment (1 kN·m) replaces the unit force.
θ = ∫ (m_θ · M / EI) dx
Example
Simply supported beam, span L, central point load P. To find slope at left support A: apply unit couple at A. m_θ at A = 1−x/L (varies linearly). Integrate with M over left half by symmetry: θ_A = PL²/(16EI)
Variables
m_θ = bending moment due to unit couple (1 kN·m) applied at the point where rotation is desired (dimensionless, m/m), M = real bending moment (kN·m), EI = flexural rigidity (kN·m²)
Application
Slope (rotation in radians) at any cross-section of a beam or frame.
δ_mid (SS beam, central load) = PL³ / (48EI)
Example
P = 20 kN, L = 6 m, EI = 1.2×10⁴ kN·m²: δ = 20×216/(48×1.2×10⁴) = 4320/576000 = 7.5×10⁻³ m = 7.5 mm ↓
Variables
P = central point load (kN), L = span (m), EI = flexural rigidity (kN·m²)
Application
Standard formula derived by unit-load method; most frequently tested beam deflection formula.
δ_mid (SS beam, UDL) = 5wL⁴ / (384EI)
Example
w = 15 kN/m, L = 8 m, EI = 4×10⁴ kN·m²: δ = 5×15×4096/(384×4×10⁴) = 307200/15360000 = 0.02 m = 20 mm
Variables
w = uniformly distributed load (kN/m), L = span (m), EI = flexural rigidity (kN·m²)
Application
Maximum midspan deflection of a simply supported beam under full-span UDL.
δ_tip (cantilever, UDL) = wL⁴ / (8EI)
Example
w = 10 kN/m, L = 2 m, EI = 5000 kN·m²: δ = 10×16/(8×5000) = 160/40000 = 4×10⁻³ m = 4 mm ↓
Variables
w = UDL (kN/m), L = cantilever length (m), EI = flexural rigidity (kN·m²)
Application
Maximum deflection at the free end of a cantilever beam under full-span UDL.
Exam tips
- Memorize the four standard formulas: cantilever-tip-point (PL³/3EI), cantilever-tip-UDL (wL⁴/8EI), SS-midspan-point (PL³/48EI), SS-midspan-UDL (5wL⁴/384EI). These appear in roughly 60% of beam deflection board exam problems.
- When integrating ∫mM/EI dx for linearly varying m and parabolic M, use the tabulated integral formulas (e.g., area of parabola = 2/3 × base × height) to avoid long integration.
- Check units at the end: the result must be in meters (if L is in m) or millimeters (if L in mm). A deflection of 0.007 m = 7 mm is reasonable; 700 m is not.
- For frames, sketch the m-diagram first (it is always simpler — usually triangular or trapezoidal) before drawing M.
Key points
- For beams and frames, bending is the dominant internal action; shear and axial contributions are neglected unless stated otherwise.
- The procedure mirrors the truss method: (a) Compute M(x) due to real loads. (b) Apply a unit load at the point/direction of interest; compute m(x). (c) Integrate mM/EI over each member.
- The integration ∫mM/EI dx must be done member by member, with x measured consistently (usually from a free end or left support).
- For rotation, apply a unit couple (1 kN·m) instead of a unit force; the result is in radians.
- Graphical (moment-area) integration can be used when m is linear and M is parabolic or linear — use the M·A table (area of M-diagram times ordinate of m at centroid).
- For frames, each member contributes a separate integral; sum all contributions.
- EI may vary between members of a frame — use the correct EI for each member's integral.
- Results match classical formulas: cantilever end deflection = PL³/(3EI); simply supported midspan = PL³/(48EI) for central load, 5wL⁴/(384EI) for UDL.
Definitions
Flexural Rigidity (EI)
The product of the modulus of elasticity E and the second moment of area I of a cross-section. It quantifies the beam's resistance to bending deformation. For concrete, E_c = 4700√f'c (MPa) per ACI 318-19 Section 19.2.2.
Importance
Appears in every beam deflection formula. Knowing how to compute EI for steel (AISC tables), concrete (ACI), and composite sections is essential for board exams.
Elastic Curve
The deformed shape (y vs. x plot) of the centroidal axis of a beam after loading. The governing differential equation is EI d²y/dx² = M(x) (with appropriate sign convention).
Importance
The virtual work integral ∫mM/EI dx can be visualized as the weighted area product of the m and M diagrams along the elastic curve.
Common mistakes
- Setting up M(x) with inconsistent sign conventions between the real and virtual systems — use the same sign convention throughout (positive sagging or positive hogging) for both M and m.
- Forgetting to integrate over ALL members of a frame, not just the loaded member.
- Using mm for L and kN for loads without adjusting E and I units — this is the single most common arithmetic error; use a consistent SI set (kN, m, kN·m²) or (N, mm, N·mm²).
- Applying the midspan formula PL³/48EI to a beam that is not simply supported or where the load is not at midspan.
- Confusing deflection (mm) with rotation (radians) when reading the question — 'find the slope' means compute θ, not δ.
4. Castigliano's Second Theorem
Formulas
δ_i = ∂U / ∂P_i
Example
Cantilever, end load P: U = P²L³/(6EI). ∂U/∂P = 2PL³/(6EI) = PL³/(3EI). This recovers the known tip deflection formula.
Variables
U = total strain energy of the structure (kN·m or J), P_i = the load at the point where deflection δ_i is desired (kN), partial derivative with respect to P_i only
Application
General statement of Castigliano's Second Theorem for a concentrated load.
δ = ∫ [M/EI × ∂M/∂P] dx
Example
SS beam, span L, central load P at midspan. Left half (0 ≤ x ≤ L/2): M = Px/2, ∂M/∂P = x/2. By symmetry: δ_mid = 2∫₀^(L/2) (x/2)(Px/2)/(EI) dx = 2∫₀^(L/2) Px²/(4EI) dx = P/(2EI)[x³/3]₀^(L/2) = PL³/(48EI) ✓
Variables
M = bending moment as a function of x and P, ∂M/∂P = rate of change of moment with respect to load P (dimensionless or m), EI = flexural rigidity
Application
Working form of Castigliano for beams/frames — differentiate inside the integral (valid by Leibniz rule) to avoid first computing U.
θ_i = ∂U / ∂M_i
Example
Fixed-end beam with applied moment M₀ at free end: θ = ∂U/∂M₀ = M₀L/(EI) — the tip rotation of a cantilever under end moment.
Variables
U = total strain energy, M_i = applied couple at the point where rotation θ_i is desired
Application
Finding rotation at a point where a moment is applied; if no moment exists there, introduce dummy couple.
δ_truss = Σ [NL/(AE) × ∂N/∂P]
Example
In the triangular truss example: ∂N/∂P for each member equals n (unit-load forces). Result is identical: δ_C = 0.69 mm ↓
Variables
N = real member forces (kN), L = member length (m), AE = axial rigidity (kN), ∂N/∂P = change in member force per unit increase in P (equivalent to n in unit-load method)
Application
Castigliano form for truss deflections — ∂N/∂P is numerically the same as n from the unit-load method.
Exam tips
- Whenever you see a beam/frame deflection problem with no load at the target point, think Castigliano + dummy load immediately.
- The fastest approach for standard beams (SS, cantilever) is to use the memorized formulas. Use Castigliano only when the problem geometry or loading is non-standard.
- In multi-span frames, the dummy load Q must be placed correctly — wrong placement changes ALL the M expressions.
- Castigliano and unit-load always give identical answers. If you solve by both methods and get different numbers, there is an arithmetic error.
- Board exams often ask: 'Using Castigliano's theorem, find the deflection at the free end.' This is a signal to show the ∫[M/EI × ∂M/∂P]dx procedure explicitly.
Key points
- Castigliano's Second Theorem states: the partial derivative of the total strain energy U with respect to a load P_i equals the deflection δ_i at the point and in the direction of P_i.
- Similarly, the partial derivative of U with respect to an applied moment M_i equals the rotation θ_i at that point.
- For bending-dominated structures: δ = ∫[M/EI × ∂M/∂P] dx — this is algebraically identical to the unit-load method because ∂M/∂P plays the exact role of m.
- Dummy Load Technique: If no real load acts where the deflection is needed, introduce a fictitious dummy load Q at that point, compute ∂M/∂Q, integrate, then set Q = 0. This is the practical advantage of Castigliano over unit-load.
- Castigliano applies to any elastic structure — trusses, beams, frames, curved members.
- For trusses: δ = Σ[NL/(AE) × ∂N/∂P].
- The theorem is only valid for linear elastic structures (Hooke's Law obeyed, small deformations).
Definitions
Castigliano's Second Theorem
For a linearly elastic structure, the partial derivative of the total strain energy U with respect to any load P_i gives the displacement (deflection or rotation) at the point of application of P_i in the direction of P_i. Stated by Italian engineer Alberto Castigliano in 1879.
Importance
Provides a systematic algebraic route to deflections, especially useful when a dummy load is needed — avoids re-drawing the unit-load diagram from scratch.
Dummy Load
A fictitious load Q (force or moment) of zero magnitude introduced at the point and in the direction of the desired deflection when no real load acts there. After differentiating, Q is set to zero in the final integral.
Importance
Without the dummy load, Castigliano cannot find deflections at unloaded points. This is the most tested application of Castigliano in board exams.
Leibniz Integration Rule
Allows the partial derivative to be moved inside the integral sign: ∂/∂P ∫f(x,P)dx = ∫[∂f/∂P]dx, valid when limits do not depend on P. This converts ∂U/∂P into ∫[M/EI × ∂M/∂P]dx.
Importance
Justifies the working form of Castigliano's theorem and shows that it is equivalent to the unit-load method.
Common mistakes
- Differentiating U with respect to the wrong variable — you must differentiate with respect to the load AT the point of interest, not with respect to any load.
- Forgetting to set Q = 0 after differentiating when using the dummy load technique — leaving Q in the answer makes it a function of Q, not a number.
- Confusing Castigliano's First Theorem (∂U/∂δ = P, force from energy) with the Second Theorem (∂U/∂P = δ, deflection from energy).
- Not recognizing that ∂M/∂P is simply the m-diagram — students who see this connection solve problems much faster.
- Forgetting the factor of 2 when differentiating U = N²L/(2AE): ∂U/∂N = NL/(AE), then apply chain rule ∂N/∂P.
5. Key Standard Deflection Formulas Reference
Formulas
δ_max (cantilever, tip point load P) = PL³ / (3EI)
Example
P=15 kN, L=2 m, EI=6000 kN·m²: δ = 15×8/(3×6000) = 120/18000 = 6.67×10⁻³ m = 6.67 mm
Variables
P = tip load (kN), L = cantilever span (m), EI = flexural rigidity (kN·m²)
Application
Free-end deflection of a cantilever with single point load at tip.
δ_max (cantilever, UDL w) = wL⁴ / (8EI)
Example
w=12 kN/m, L=3 m, EI=8000 kN·m²: δ = 12×81/(8×8000) = 972/64000 = 15.2×10⁻³ m = 15.2 mm
Variables
w = UDL over full span (kN/m), L = cantilever span (m), EI = flexural rigidity (kN·m²)
Application
Free-end deflection of a cantilever under uniformly distributed load over full length.
δ_max (SS, central point load P) = PL³ / (48EI)
Example
P=30 kN, L=6 m, EI=15000 kN·m²: δ = 30×216/(48×15000) = 6480/720000 = 9×10⁻³ m = 9 mm
Variables
P = central load (kN), L = span (m), EI = flexural rigidity (kN·m²)
Application
Midspan deflection of a simply supported beam with single central point load.
δ_max (SS, full UDL w) = 5wL⁴ / (384EI)
Example
w=20 kN/m, L=5 m, EI=10000 kN·m²: δ = 5×20×625/(384×10000) = 62500/3840000 = 16.3×10⁻³ m = 16.3 mm
Variables
w = UDL (kN/m), L = span (m), EI = flexural rigidity (kN·m²)
Application
Midspan deflection of a simply supported beam under full-span UDL.
δ_max (SS, end moments M₀) = M₀L² / (8EI)
Example
M₀=40 kN·m, L=4 m, EI=8000 kN·m²: δ = 40×16/(8×8000) = 640/64000 = 0.01 m = 10 mm
Variables
M₀ = equal end moments (kN·m), L = span (m), EI = flexural rigidity (kN·m²)
Application
Midspan deflection of a simply supported beam loaded by equal and opposite end moments.
θ_A (SS beam, central load P) = PL² / (16EI)
Example
P=20 kN, L=6 m, EI=12000 kN·m²: θ = 20×36/(16×12000) = 720/192000 = 3.75×10⁻³ rad
Variables
P = central load (kN), L = span (m), EI = flexural rigidity (kN·m²), θ_A = slope at left support (radians)
Application
End slope of a simply supported beam under central point load.
Exam tips
- Write all four fundamental formulas at the top of your scratch paper at the start of any board exam session — saves time during the test.
- Ratio trick: a fixed-fixed beam is 4× stiffer than SS (PL³/192 vs PL³/48), and a propped cantilever is intermediate. Useful for quick sanity checks.
- If EI is given in N·mm², convert to kN·m² by dividing by 10¹²: 2×10¹³ N·mm² = 2×10¹³/10¹² = 20 kN·m².
Key points
- These formulas are derived by integration or virtual work; they must be memorized for board exam efficiency.
- All formulas assume homogeneous, prismatic beams (constant EI) with small, linear elastic deformations.
- For beams with varying EI, the formulas do not apply directly — use virtual work integration.
- The 'maximum deflection' in a simply supported beam under a non-central load occurs near but not exactly at midspan.
- For composite loading, use superposition: δ_total = δ_from_load1 + δ_from_load2 + ...
- NSCP 2015 requires that deflections be checked after construction and under service loads (not factored loads).
Definitions
Maximum Deflection Point
The location along a beam where slope dy/dx = 0 (elastic curve has horizontal tangent). For symmetric loading on a symmetric span, this is at midspan. For asymmetric loading, it shifts toward the load.
Importance
Board exam problems sometimes ask for 'maximum deflection' which occurs at a different point than the load; use the virtual work integral or appropriate formula.
Common mistakes
- Using PL³/48EI when the load is NOT at midspan — this formula is only for a central load on a simply supported beam.
- Applying the SS beam formula to a fixed-end beam (δ_max for fixed-fixed, central load is PL³/192EI — four times stiffer).
- Superposing deflections for loads in different problems without verifying same span, same E, same I.
6. Step-by-Step Procedures and Board Exam Strategy
Exam tips
- In 4-choice MCQ format (typical board exam), use process of elimination: wrong units, negative magnitudes, or values outside the 1–50 mm range for typical structural spans are immediately suspect.
- If the problem specifies E and I separately (e.g., E = 200 GPa, I = 450×10⁶ mm⁴), compute EI = 200,000 MPa × 450×10⁶ mm⁴ = 9×10¹³ N·mm² = 90,000 kN·m² before substituting.
- Time management: standard deflection formula problems (types 5.1–5.6 above) should take under 2 minutes each. Unit-load integration problems may take 5–8 minutes — budget accordingly.
- For frame problems with right-angle joints: treat each member separately; choose x-origin at the free end or joint to simplify m(x) expressions.
Key points
- For unit-load (virtual work) method on beams: draw FBDs, write M(x), apply unit load, write m(x), integrate ∫mM/EI dx.
- For unit-load on trusses: find reactions and member forces for real loads (N), then for unit load (n), tabulate, sum nNL/AE.
- For Castigliano on beams: write M as a function of x AND P, differentiate ∂M/∂P, integrate ∫(M/EI)(∂M/∂P)dx.
- For Castigliano with dummy load: insert Q at target point, differentiate ∂M/∂Q, integrate, set Q=0.
- Always state units and direction of deflection in the final answer.
- Check against known formulas or order of magnitude estimates (mm range for typical spans is reasonable).
Common mistakes
- Skipping the free body diagram — without a proper FBD, moment equations M(x) will be wrong.
- Not checking whether EI is constant — if it varies, you cannot factor it out of the integral.
- Solving for deflection in kN·m (energy units) instead of m (deflection units) — unit check catches this: [kN·m × kN·m] / [kN·m²] = m ✓
- In Castigliano with dummy Q: differentiating M with respect to real load P instead of Q.
Connections
- Deflection compatibility equations are the foundation of the Force (Flexibility) Method for indeterminate structures: the redundant forces are found by requiring that deflections at the release points equal zero (or specified values). The virtual work integrals in this chapter are directly reused.
- Slope-deflection method (Chapter on Indeterminate Beams) requires knowing end rotations in terms of member stiffness EI/L — these are derived from the same differential equation and energy principles.
- NSCP 2015 Section 406 serviceability checks connect computed deflections to code limits; every deflection computation in practice ends with a code comparison.
- ACI 318-19 Section 24.2 long-term deflection multiplier (λ_Δ = ξ/(1+50ρ')) accounts for creep and shrinkage in concrete members — the initial deflection from this chapter is multiplied by (1 + λ_Δ) for long-term checks.
- AISC 360-16 Commentary Chapter L addresses deflection and drift limits for steel structures; the allowable story drift (H/400 to H/500) uses the same virtual work principle applied to lateral loads.
- Influence lines for deflection use the Mueller-Breslau principle, which is a direct application of virtual work — a unit displacement at the point of interest creates the influence line for reaction or internal force.
- Moment distribution and stiffness methods (Chapters on Indeterminate Analysis) use carry-over factors and fixed-end moments derived from beam deflection equations of this chapter.
- RA 544 (Civil Engineering Law of the Philippines, as amended by RA 1582) requires civil engineers to ensure structural safety and serviceability — deflection control is a direct professional responsibility under this law.
Quick review questions
A simply supported steel beam has span L = 8 m and carries a uniformly distributed load w = 25 kN/m over the full span. Given EI = 5×10⁴ kN·m², compute the maximum midspan deflection in millimeters.
Answer
δ_max = 5wL⁴/(384EI) = 5×25×8⁴/(384×5×10⁴) = 5×25×4096/19,200,000 = 512,000/19,200,000 = 0.02667 m = 26.7 mm
Apply the standard UDL formula for a simply supported beam: δ = 5wL⁴/(384EI). Convert all to kN and m: w=25 kN/m, L=8 m, EI=5×10⁴ kN·m². Numerator: 5×25×(8⁴) = 125×4096 = 512,000 kN·m³. Denominator: 384×50,000 = 19,200,000 kN·m². δ = 512,000/19,200,000 = 0.02667 m = 26.67 mm. If span L = 8 m and NSCP 2015 live-load limit is L/360 = 8000/360 = 22.2 mm, this beam would fail serviceability — a practical interpretation.
State the Virtual Work equation for finding the vertical deflection at joint C of a plane truss. Define each symbol.
Answer
δ_C = Σ(n·N·L)/(AE), where N = axial force in each member due to real loads, n = axial force in each member due to a unit vertical load at C, L = member length, A = cross-sectional area, E = modulus of elasticity. Summation is over all members.
The unit-load method equates virtual external work (1×δ_C) to virtual internal work (Σ n×N×L/AE). The unit virtual load at C creates the virtual force state (n-forces), and the real loads create the real deformation state (N-forces). Their product summed over all members gives the desired deflection component. Sign convention: tension is positive; like signs multiply to give positive (deflection in direction of unit load).
A cantilever beam of length L = 4 m carries an end point load P = 8 kN. EI = 1.6×10⁴ kN·m². (a) Find the tip deflection δ. (b) Find the tip slope θ.
Answer
(a) δ = PL³/(3EI) = 8×64/(3×16,000) = 512/48,000 = 0.01067 m = 10.67 mm ↓. (b) θ = PL²/(2EI) = 8×16/(2×16,000) = 128/32,000 = 0.004 rad
(a) Use the standard cantilever tip deflection formula. EI = 1.6×10⁴ kN·m² = 16,000 kN·m². δ = 8×(4³)/(3×16,000) = 8×64/48,000 = 512/48,000 = 0.01067 m = 10.67 mm downward. (b) Tip slope: θ = PL²/(2EI) = 8×16/(2×16,000) = 128/32,000 = 0.004 rad. Both formulas are derivable from EI d²y/dx² = M(x) = P(L−x) with boundary conditions y(0)=0, y'(0)=0, or by virtual work (unit load for δ, unit moment for θ).
In Castigliano's Second Theorem, what is the 'dummy load' and when is it used?
Answer
A dummy load Q is a fictitious concentrated force (or moment) of zero magnitude introduced at a point where no real load acts but where a deflection (or rotation) is desired. After computing ∂M/∂Q and integrating ∫[M/EI × ∂M/∂Q]dx, Q is set to zero in the final expression to recover the answer.
Castigliano requires a real load at the point of interest so the partial derivative ∂U/∂P is meaningful. When the problem asks for deflection at an unloaded point (e.g., midspan deflection when loads are at quarter points), there is no load P to differentiate with respect to. The dummy load Q provides that variable. Setting Q=0 after differentiation ensures that Q does not affect the actual structural response — it was only a mathematical device.
A 2-member plane truss has member AB (length 3 m, AE = 9×10⁴ kN) with N_AB = +45 kN and member BC (length 4 m, AE = 12×10⁴ kN) with N_BC = −60 kN. Under a unit vertical load at B, n_AB = +0.6 and n_BC = −0.8. Find the vertical deflection at B.
Answer
δ_B = Σ nNL/AE = [(0.6)(45)(3)/(9×10⁴)] + [(−0.8)(−60)(4)/(12×10⁴)] = [81/90,000] + [192/120,000] = 9×10⁻⁴ + 1.6×10⁻³ = 9×10⁻⁴ + 16×10⁻⁴ = 25×10⁻⁴ m = 2.5 mm ↓
Tabulate: Member AB: n=+0.6, N=+45 kN, L=3 m, AE=90,000 kN → nNL/AE = 0.6×45×3/90,000 = 81/90,000 = 9×10⁻⁴ m. Member BC: n=−0.8, N=−60 kN, L=4 m, AE=120,000 kN → nNL/AE = (−0.8)(−60)(4)/120,000 = 192/120,000 = 1.6×10⁻³ m. Sum = 0.0009 + 0.0016 = 0.0025 m = 2.5 mm. Both contributions are positive (like signs) → deflection is in the direction of the unit load (downward).
How does Castigliano's theorem for beams (δ = ∫[M/EI × ∂M/∂P]dx) relate to the virtual work method (δ = ∫mM/EI dx)?
Answer
They are mathematically identical. The term ∂M/∂P in Castigliano's method is exactly equal to m(x) in the virtual work method. Both expressions represent the bending moment produced per unit of load P at the point of interest.
In virtual work, m(x) is the moment diagram produced by a unit load (1 kN) at the point of interest. In Castigliano, ∂M/∂P is the rate of change of the real moment with respect to load P — but since M is linear in P for a linear elastic structure, ∂M/∂P equals the moment produced per unit of P, which is exactly m(x) when P=1 kN. The two methods are therefore two different derivations of the same equation. Recognizing this identity helps in checking work and in choosing the faster method for a given problem.
A simply supported beam of span 6 m carries a central load P = 24 kN and EI = 9,000 kN·m². Check whether the midspan deflection satisfies NSCP 2015 live-load limit for a floor beam supporting brittle finishes.
Answer
δ = PL³/(48EI) = 24×216/(48×9,000) = 5,184/432,000 = 0.012 m = 12 mm. Allowable = L/360 = 6,000/360 = 16.67 mm. Since 12 mm < 16.67 mm, the beam SATISFIES the serviceability requirement.
Per NSCP 2015 Table 406.2.1, the maximum deflection due to live load for a floor member supporting brittle finishes is L/360. Here L = 6 m = 6,000 mm, so allowable = 6,000/360 = 16.67 mm. Computed δ = 12 mm < 16.67 mm — the beam passes. Had δ exceeded the limit, the designer would need to increase EI (deeper section, higher-strength steel) or reduce the span.
Write the Castigliano expression for the deflection at the free end of a cantilever of length L under a UDL w, using the dummy load approach if necessary.
Answer
No dummy load is needed because the UDL produces a resultant effect along the beam, but to find the free-end deflection δ, introduce dummy load Q at the free end. M(x) = wx²/2 + Qx (x from fixed end, but more conveniently from free end: M(x) = −wx²/2 − Qx). ∂M/∂Q = x. δ = ∫₀ᴸ (−wx²/2)(x)/EI dx |_{Q=0} = w/(2EI) × ∫₀ᴸ x³ dx = w/(2EI) × L⁴/4 = wL⁴/(8EI).
Measuring x from the free end: M(x) = −wx²/2 − Qx (both UDL and Q cause hogging at the section). ∂M/∂Q = −x. With signs: δ = ∫₀ᴸ [(−wx²/2)(−x)]/(EI) dx = ∫₀ᴸ wx³/(2EI) dx = w/(2EI) × L⁴/4 = wL⁴/(8EI). Note: the product of two negatives gives positive integrand. After setting Q=0, we recover the standard cantilever UDL tip deflection formula.
Diagrams
Chapter Overview: Deflections of Structures
A comprehensive mind map showing all major topics of the Deflections of Structures chapter: the motivation (serviceability and compatibility), the two energy quantities (strain energy), the two main methods (virtual work and Castigliano), the standard formulas to memorize, and board exam strategy. Use this as a quick visual orientation before studying each section in detail.
Step-by-Step: Virtual Work (Unit-Load) Method for Beams and Frames
A detailed decision-tree flowchart for applying the unit-load (virtual work) method to beams and frames. Shows the complete procedure from drawing the FBD to interpreting the sign of the result. Key decision points are whether EI is constant and whether multiple members need separate integration. This flowchart serves as a step-by-step checklist during problem-solving.
Step-by-Step: Castigliano's Second Theorem with Dummy Load
Flowchart for applying Castigliano's Second Theorem, with a key branch for the dummy load technique. The left path handles the case where a real load exists at the target point; the right path shows the dummy load procedure when no real load acts there. Both paths converge at the unit check and optional NSCP serviceability comparison.
Choosing the Right Deflection Method
A decision-tree flowchart to help examinees quickly select the most efficient deflection method for any given problem. Starting from the structure type (truss vs. beam/frame), it branches into standard formula lookup, virtual work integration, or Castigliano with dummy load. This is particularly useful for time management during the board examination.
Distribution of Deflection Method Applications in Board Exam Problems
Approximate distribution of deflection-related problem types in PRC Civil Engineer Licensure Examinations based on past board exam analysis. Standard beam formulas dominate at ~45%, confirming that memorizing the four key formulas is the highest-ROI study activity. Virtual work (truss and beam) accounts for ~40% combined, and Castigliano with dummy load appears in ~10% of problems.
Exam strategy
In the PRC Civil Engineer Licensure Examination (Structural Theory & Analysis component), deflection problems appear in approximately 10–15% of questions. The most efficient strategy is: (1) Immediately classify the problem as truss-deflection, beam-deflection (standard case), or beam-deflection (non-standard/frame). (2) For standard beams, apply the memorized formula directly — PL³/48EI, 5wL⁴/384EI, PL³/3EI, or wL⁴/8EI. These 4 formulas alone solve roughly half of all deflection board exam items. (3) For non-standard problems (frames, partial loads, deflection at unloaded points), set up the virtual work integral or Castigliano + dummy load systematically. (4) Always perform a unit check: [kN × m³] / [kN·m²] = m ✓. (5) Compare the computed deflection against NSCP 2015 limits when the problem asks about serviceability — know L/360 (LL, brittle finishes), L/240 (total load), L/180 (roof). (6) For multiple-choice questions, use backward substitution: plug in the answer choices and check which satisfies the equation — faster than deriving from scratch when stuck. (7) Budget 2 minutes for formula-type problems and 6–8 minutes for integration-type problems. If integration takes longer than 8 minutes, switch strategy or use approximation.