Energy Methods and Virtual WorkIntroductionStructural analysis commonly begins with equilibrium equations, compatibility conditions, and constitutive relationships. These methods are powerful, but they can become cumbersome when the required result is a displacement at one particular point or when a structure is statically indeterminate. Energy methods provide an alternative approach. Instead of determining every internal force first, they relate external work, internal strain energy, and displacement directly. The central idea is that deformation stores energy in a structure. For a linearly elastic member, the energy stored by axial force, bending moment, shear force, or torsion can be integrated along the member. Castigliano’s theorems obtain displacements and rotations by differentiating strain energy with respect to loads. The unit-load or virtual-work method obtains a selected displacement through an integral involving the actual internal force system and a virtual internal force system caused by a unit load. These methods are especially useful in construction engineering for beams, trusses, frames, bridges, temporary works, and indeterminate structures. They also form the conceptual basis of the stiffness method and finite-element analysis. Learning ObjectivesBy the end of this chapter, the reader should be able to: - Explain strain energy, complementary energy, virtual work, and reciprocity.
- Derive and apply Castigliano’s theorems for elastic structures.
- Calculate beam and truss deflections using the unit-load method.
- Select appropriate deformation components and identify negligible effects.
- Use flexibility equations to analyse simple statically indeterminate structures.
- Check results using symmetry, dimensions, signs, and reciprocal relationships.
- Recognize assumptions, limitations, and professional applications of energy methods.
1. Fundamental Concepts1.1 External work and internal strain energyWhen a load is gradually applied to an elastic structure, the load moves through a displacement. The external work is stored internally as strain energy. For a gradually applied load P producing displacement Δ, the work is U = ½PΔ where U is strain energy, in joules (N·m), P is load in newtons, and Δ is displacement in metres. This expression applies to a linear load–displacement relationship. For a nonlinear relationship, the general expression is U = ∫₀^Δ P(δ) dδ where δ is an intermediate displacement and P(δ) is the corresponding load. For a member subjected to several deformation actions, the total strain energy is the sum of the energies associated with axial deformation, bending, shear, and torsion. For a prismatic, linearly elastic member: U = ∫₀ᴸ [N²/(2EA) + M²/(2EI) + V²/(2κGA) + T²/(2GJ)] dx Here N, M, V, and T are internal axial force, bending moment, shear force, and torque; E is Young’s modulus; G is shear modulus; A is cross-sectional area; I is the second moment of area; J is the torsional constant; κ is the shear correction factor; and L is member length. The coordinate x follows the member axis. For slender beams, shear deformation is often small compared with bending deformation, so the bending-only approximation is U ≈ ∫₀ᴸ M²/(2EI) dx For a pin-jointed truss with axial deformation only, U = ∑ᵢ Nᵢ²Lᵢ/(2AᵢEᵢ) where Nᵢ, Lᵢ, Aᵢ, and Eᵢ refer to member i. These equations assume small deformation, linear elasticity, known member properties, and compatible deformation. 1.2 Complementary energyComplementary energy is expressed in terms of force rather than displacement. For a linearly elastic member, strain energy and complementary energy have equal numerical values. In nonlinear materials they are generally different. Complementary-energy concepts are particularly useful in force-based formulations and in deriving flexibility relationships. 1.3 Work-conjugate quantitiesA force is work-conjugate to a translation, and a moment is work-conjugate to a rotation. Thus: | Generalized load | Corresponding displacement | Typical unit |
|---|
| Force P | Translation Δ | N, m | | Moment M | Rotation θ | N·m, radians | | Axial force N | Axial extension δ | N, m | | Torque T | Twist φ | N·m, radians |
Correct pairing is essential. Differentiating energy with respect to a force gives the displacement in the direction of that force; differentiating with respect to a moment gives the rotation in the direction of that moment. 2. Castigliano’s Theorems2.1 First theoremFor a linearly elastic structure, the partial derivative of strain energy with respect to a generalized displacement gives the corresponding generalized load: Qᵢ = ∂U/∂qᵢ where Qᵢ is a generalized force and qᵢ is its conjugate displacement. This form is useful when displacements are prescribed and reactions are required. 2.2 Second theoremThe second theorem is more frequently used for displacement calculations. The displacement corresponding to an applied force P is Δ = ∂U/∂P The rotation corresponding to an applied moment M₀ is θ = ∂U/∂M₀ The differentiation must be performed while treating all other independent loads as constant. If internal actions depend on P, substitute those expressions into U before differentiating. For bending of a beam, Δ = ∫₀ᴸ [M(x)/(EI)] [∂M(x)/∂P] dx and for a rotation caused by moment M₀, θ = ∫₀ᴸ [M(x)/(EI)] [∂M(x)/∂M₀] dx If axial, shear, and torsional effects are significant, their corresponding terms must be added. For example, the displacement in the direction of P is Δ = ∫₀ᴸ [N ∂N/∂P /(EA) + M ∂M/∂P /(EI) + V ∂V/∂P /(κGA) + T ∂T/∂P /(GJ)] dx 2.3 Worked example: cantilever tip deflectionA cantilever of length L carries a downward point load P at its free end. EI is constant. Determine the vertical deflection at the free end, neglecting shear deformation. At a section located x from the fixed end, the bending moment magnitude is M(x) = P(L − x) The strain energy is U = ∫₀ᴸ [P²(L − x)²/(2EI)] dx Since ∫₀ᴸ(L − x)² dx = L³/3, U = P²L³/(6EI) Applying Castigliano’s second theorem, Δ = ∂U/∂P = PL³/(3EI) The result has units N·m³/(N·m²) = m, as required. The negative sign may be included if upward displacement is defined as positive; the magnitude is PL³/(3EI). The corresponding free-end rotation can be obtained by applying a moment at the tip or by using the appropriate energy expression: θ = PL²/(2EI) 2.4 Practical cautions in applying Castigliano’s theoremThe method requires a differentiable force–displacement relationship and elastic behavior. For piecewise loading, the member must be divided into intervals. Internal-force signs should be selected consistently, although the final displacement sign depends on the chosen coordinate convention. If EI varies along the member, retain EI(x) inside the integral. Concentrated loads do not require a separate energy integral at the point of application; their influence appears through the internal-force functions. 3. Unit-Load Method and Virtual Work3.1 PrincipleSuppose an actual structure carries a real load system and has internal actions N, M, V, and T. To find a displacement at a selected location, apply a virtual unit force in the desired displacement direction. To find a rotation, apply a virtual unit moment in the desired rotational direction. Let the resulting virtual internal actions be n, m, v, and t. The displacement is Δ = ∫₀ᴸ [Nn/(EA) + Mm/(EI) + Vv/(κGA) + Tt/(GJ)] dx For a slender beam where bending dominates, Δ = ∫₀ᴸ Mm/(EI) dx For a truss, Δ = ∑ᵢ Nᵢ nᵢ Lᵢ/(AᵢEᵢ) The real system supplies N and M; the unit-load system supplies n and m. Because the virtual load has magnitude one, the resulting integral has the units of displacement. 3.2 Derivation from Castigliano’s theoremFor bending-only behavior, strain energy is U = ∫₀ᴸ M²/(2EI) dx If P is the selected external force and M depends on P, differentiation gives ∂U/∂P = ∫₀ᴸ [M/(EI)] [∂M/∂P] dx The derivative ∂M/∂P is precisely the bending moment diagram produced by a unit load placed in the direction of P. Therefore n = ∂M/∂P, yielding the unit-load equation. This establishes that the unit-load method and Castigliano’s second theorem are equivalent for the stated assumptions. 3.3 Worked example: simply supported beamA simply supported beam of span L carries a central point load P. Determine the midspan deflection using a unit load. EI is constant and shear deformation is neglected. By symmetry, the real reactions are P/2. On the left half, with x measured from the left support, M = Px/2, 0 ≤ x ≤ L/2 Apply a unit load at midspan. The virtual reactions are 1/2, and the virtual moment on the left half is m = x/2 Using symmetry, Δ = 2∫₀ᴸ/² [(Px/2)(x/2)/(EI)] dx Therefore, Δ = P/(2EI) ∫₀ᴸ/² x² dx = P/(2EI) × (L/2)³/3 Thus, Δ = PL³/(48EI) The deflection is downward in the direction of the real central load. 3.4 Truss displacement exampleConsider a truss in which member forces under the actual load are Nᵢ and a unit horizontal load at the joint of interest produces forces nᵢ. If all members have the same E and A, Δ = 1/(EA) ∑ᵢ NᵢnᵢLᵢ Members for which Nᵢnᵢ is positive contribute displacement in the unit-load direction; members for which it is negative reduce the displacement. A zero product means that the member makes no contribution to that selected displacement, even though it may carry force in one of the two load systems. 4. Reciprocal Theorems4.1 Maxwell–Betti reciprocityFor a stable, linearly elastic structure, the displacement at point A in direction 1 caused by a load at point B in direction 2 equals the displacement at B in direction 2 caused by the same load applied at A in direction 1: δAB = δBA More generally, for two load systems P and Q, ∑ Pᵢ δᵢ^(Q) = ∑ Qᵢ δᵢ^(P) This theorem follows from equality of mutual work and is valuable for checking calculations and establishing symmetry in flexibility matrices. 4.2 Flexibility coefficientsFor an indeterminate structure, select redundants X₁, X₂, …, Xₙ. The compatibility equations can be written δᵢ₀ + ∑ⱼ fᵢⱼXⱼ = Δᵢ where δᵢ₀ is the displacement at redundant coordinate i due to the primary determinate loading, fᵢⱼ is the flexibility coefficient at coordinate i caused by a unit load at coordinate j, and Δᵢ is the prescribed displacement, often zero. For a structure with one redundant, X = (Δ − δ₀)/f The coefficient f is obtained by a unit-load integral. Reciprocity gives fᵢⱼ = fⱼᵢ, so the flexibility matrix is symmetric for a conventional elastic structure. 4.3 Simple indeterminate applicationConsider a propped cantilever of length L carrying a downward load P at its free end, with the prop reaction R at the free end. Select R as the redundant and remove the prop, creating a cantilever. The downward displacement caused by P is δ₀ = PL³/(3EI) A unit upward load at the prop location produces a flexibility coefficient f = L³/(3EI) Compatibility requires the total free-end displacement to be zero. Taking downward displacement as positive, the upward reaction contributes −Rf: δ₀ − Rf = 0 Thus, R = δ₀/f = P The fixed-end vertical reaction is therefore P − R = 0, while the fixed-end moment and horizontal reaction depend on the complete equilibrium system and load arrangement. This result illustrates an important interpretation: the prop reaction is the force required to cancel the displacement that would occur in the released structure. 5. Energy Methods in Structural Practice5.1 Choosing the appropriate method| Need | Recommended method |
|---|
| One beam displacement or rotation | Castigliano or unit load | | Several selected displacements | Unit-load method | | Truss joint displacement | Truss unit-load summation | | One redundant reaction | Flexibility method | | Several redundants | Matrix flexibility method | | Complex geometry and many degrees of freedom | Stiffness method or finite elements |
5.2 Sign conventions and diagramsA reliable workflow is: - Define positive displacement and rotation directions.
- Draw the actual free-body diagram and calculate reactions.
- Divide the structure into intervals where internal-force equations are continuous.
- Draw the unit-load system separately.
- Obtain real and virtual internal-force functions using the same coordinate convention.
- Integrate products such as Mm/EI or NnL/AE.
- Interpret the sign relative to the selected direction.
- Check dimensions, limiting behavior, and an independent result when possible.
5.3 When shear and axial deformation matterFor deep beams, short spans, sandwich panels, and members with low shear stiffness, the shear term Vv/(κGA) may be important. In trusses and columns, axial deformation is usually central. In frames, bending is often dominant, but axial deformation can affect joint translations and indeterminate reactions. Neglecting a deformation component is an engineering approximation and should be justified rather than assumed automatically. 5.4 Variable stiffness and composite constructionIf stiffness varies, use E(x)I(x), A(x)E(x), or the appropriate member property inside the integral. For a beam with two segments, Δ = ∑ₖ ∫ₖ M m/[EₖIₖ] dx This is important in construction where members may be stepped, strengthened, cracked, composite, or assembled from materials with different moduli. The effective stiffness must reflect the adopted structural model and the relevant construction stage. 6. Illustrative Case: Deflection Assessment of a Temporary Access BeamBackgroundA temporary steel access beam spans between two supports while construction materials are transported across it. A concentrated service load acts at midspan. The engineer must estimate the deflection before deciding whether a deeper beam is required. ProblemThe beam is simply supported, has span 6 m, carries a central load of 20 kN, and has constant flexural rigidity EI = 1.20 × 10⁶ kN·m². Ignore shear deformation and beam self-weight for this illustrative calculation. AnalysisFor a central point load on a simply supported beam, Δmax = PL³/(48EI) Substituting consistent kN–m units, Δmax = (20)(6³)/[48(1.20 × 10⁶)] m Since 6³ = 216, Δmax = 4320/57,600 m = 0.075 m Thus the predicted maximum deflection is 75 mm downward at midspan. Approach and decisionThe engineer compares 75 mm with the project’s specified serviceability limit. If the allowable limit were 25 mm, the beam would not be acceptable in its present form. Since deflection varies inversely with EI, the required flexural rigidity for a 25 mm limit would be EIrequired = PL³/(48Δallow) EIrequired = 4320/(48 × 0.025) = 3.60 × 10⁶ kN·m² The beam would require three times its assumed flexural rigidity, or the load/span arrangement would need to be changed. Outcome and learning pointsThe unit-load and standard-formula approaches provide a rapid serviceability assessment. However, the final construction decision must also consider self-weight, impact, load placement, lateral stability, connection flexibility, construction stage, material variability, and the governing code. An energy calculation is an analytical model, not a substitute for engineering judgment or site controls. 7. Limitations, Safety, and Contemporary PracticeEnergy methods in their basic form assume small displacement, linear elastic material behavior, stable supports, and sufficiently idealized member behavior. They may be inaccurate when yielding, cracking, geometric nonlinearity, large rotations, contact, buckling, or plastic hinges dominate. In such cases, nonlinear analysis, experimental verification, or code-based advanced analysis may be required. Professional responsibility includes documenting assumptions, load combinations, stiffness values, boundary conditions, and sign conventions. Deflection predictions affect safety, accessibility, finishes, drainage, vibration, and public confidence. Engineers must not manipulate stiffness assumptions to obtain a preferred result. Independent review is especially important for temporary works and structures supporting workers or the public. Computer algebra, spreadsheets, structural-analysis software, and finite-element programs automate integrations and matrix operations. They do not eliminate the need to understand virtual systems, mesh or model adequacy, units, boundary conditions, and result verification. A useful quality-control practice is to compare software output with a hand-derived energy estimate for a simplified load case. Modern finite-element stiffness matrices are closely related to energy minimization and virtual-work principles, so mastery of this chapter supports later study of computational structural analysis. Chapter SummaryStrain energy is the work stored in an elastic structure as it deforms. It is calculated from axial, bending, shear, and torsional actions, with bending energy often governing slender beams. Castigliano’s second theorem obtains a displacement or rotation by differentiating strain energy with respect to its conjugate load. The unit-load method is an equivalent virtual-work procedure: real internal actions are multiplied by virtual internal actions produced by a unit force or unit moment. Truss displacements are found by summing NᵢnᵢLᵢ/(AᵢEᵢ), while beam displacements commonly use ∫Mm/(EI) dx. Maxwell–Betti reciprocity provides a powerful check and establishes symmetry of flexibility coefficients. By releasing selected redundants and enforcing compatibility, the flexibility method determines reactions in statically indeterminate structures. Correct results depend on consistent signs, appropriate stiffness assumptions, complete deformation terms, dimensional checks, and awareness of model limitations. Key TermsStrain energy: Elastic energy stored by deformation. Complementary energy: Force-based energy quantity used in elastic analysis. Castigliano’s first theorem: The derivative of energy with respect to displacement gives the corresponding load. Castigliano’s second theorem: The derivative of strain energy with respect to a load gives its conjugate displacement. Virtual work: Work equation involving a real state and a compatible virtual state. Unit-load method: Virtual-work method using a unit force or unit moment at the required response location. Work-conjugate quantities: A load and displacement pair whose product represents work. Flexibility coefficient: Displacement at one coordinate caused by a unit load at another coordinate. Redundant: An unknown force or reaction that makes a structure statically indeterminate. Compatibility: A condition requiring connected parts to deform consistently. Maxwell–Betti reciprocity: Equality of mutual displacements in a linear elastic structure. Axial strain energy: Energy associated with N²/(2EA). Bending strain energy: Energy associated with M²/(2EI). Shear correction factor: Factor κ accounting for nonuniform shear stress distribution. Flexibility method: Indeterminate analysis based on released structures and compatibility. Generalized force: Force or moment paired with a generalized displacement. Generalized displacement: Translation, rotation, extension, or twist paired with a load. Flexural rigidity: Product EI, representing resistance to beam bending. Virtual internal action: Internal force or moment produced by a virtual load system. Serviceability: Performance relating to deflection, vibration, appearance, and usability. Review QuestionsShort Answer Questions- What physical quantity does strain energy represent in an elastic structure?
- State Castigliano’s second theorem for a force and for a moment.
- Why is a unit load applied in the direction of the desired displacement?
- What is the meaning of a flexibility coefficient?
- State one assumption required for the basic unit-load method.
Long Answer Questions- Derive the bending-only unit-load equation from Castigliano’s second theorem.
- Explain the differences between actual and virtual internal-force systems.
- Describe how the flexibility method analyses a structure with one redundant.
- Discuss when shear deformation should be included in an energy calculation.
- Explain Maxwell–Betti reciprocity and its value as a quality-control check.
Analytical / Application Questions- A cantilever of length 4 m carries a 10 kN end load. Derive its tip deflection using strain energy, retaining EI symbolically.
- A truss designer needs the vertical displacement of a loaded joint. Explain the complete sequence for using member-force summation and identify how compression members enter the result.
- A temporary beam calculation predicts excessive deflection. Propose and justify three engineering changes, and explain which assumptions should be reviewed before final approval.
Multiple Choice Questions with AnswersQuestion 1Which expression represents bending strain energy for a beam with constant EI? A. ∫ M²/(2EI) dx B. ∫ 2EI/M² dx C. ∫ M/(EI) dx D. ∫ EI M² dx Correct Answer: A. ∫ M²/(2EI) dx Question 2According to Castigliano’s second theorem, the displacement corresponding to load P is obtained from: A. ∂P/∂U B. ∂U/∂P C. P/U D. U/P Correct Answer: B. ∂U/∂P Question 3In the unit-load method for a beam displacement, m represents: A. The actual bending moment due to service loads B. The beam’s mass per unit length C. The bending moment caused by a unit load D. The maximum allowable bending moment Correct Answer: C. The bending moment caused by a unit load Question 4A simply supported beam with a central point load P has maximum bending deflection equal to: A. PL³/(3EI) B. PL³/(16EI) C. PL³/(24EI) D. PL³/(48EI) Correct Answer: D. PL³/(48EI) Question 5For a truss, the unit-load displacement expression is: A. ∑NᵢnᵢLᵢ/(AᵢEᵢ) B. ∑Nᵢ²AᵢEᵢ/Lᵢ C. ∑Mᵢmᵢ/(EI) D. ∑AᵢEᵢ/(NᵢnᵢLᵢ) Correct Answer: A. ∑NᵢnᵢLᵢ/(AᵢEᵢ) Question 6Maxwell–Betti reciprocity states that, under suitable elastic conditions: A. All support reactions are equal B. δAB equals δBA C. Bending energy is always zero D. Virtual loads must equal actual loads Correct Answer: B. δAB equals δBA Question 7Which quantity is work-conjugate to a rotation? A. Axial force B. Shear force C. Moment D. Area Correct Answer: C. Moment Question 8If a member has Nᵢnᵢ < 0 in a truss unit-load calculation, its contribution generally: A. Increases displacement in the selected direction B. Reduces displacement in the selected direction C. Is always zero D. Must be omitted because the member is in compression Correct Answer: B. Reduces displacement in the selected direction Question 9For a slender beam, shear deformation is commonly neglected when: A. Shear stiffness is very low B. The beam is deep and short C. Bending deformation dominates the response D. The beam carries axial force Correct Answer: C. Bending deformation dominates the response Question 10In a one-redundant flexibility analysis with zero prescribed displacement, compatibility requires: A. δ₀ + Rf = 1 B. δ₀ − Rf = 0, with signs defined consistently C. δ₀f = R D. R = δ₀ + f Correct Answer: B. δ₀ − Rf = 0, with signs defined consistently
|