Energy methods: virtual work and unit load method

Strain energy, the principle of virtual work, the unit load method for trusses, beams, torsion and thin-walled shear, Castigliano's theorems and Maxwell's reciprocal theorem.

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Why it matters

Aircraft structures are designed against stiffness as well as strength: wing tip deflection, control-surface twist and flutter margins all depend on displacements. Energy methods give those displacements for trusses, beams, frames and thin-walled boxes with one systematic recipe, and the same tools are the basis of the force method for redundant structures and of the finite element method.

Key ideas

Strain energy. A linear elastic body loaded slowly stores strain energy U equal to the work done by the loads. For a bar it is P²L/(2AE); for a beam in bending ∫M²/(2EI) dx; in torsion ∫T²/(2GJ) dx. Strain energy is always positive and is quadratic in load, so superposition does not apply to U itself.

Principle of virtual work. Two forms are used.

  • Virtual displacements: a body is in equilibrium if, for any small compatible virtual displacement, the virtual work of the external forces equals the virtual work of the internal stresses. It is a statement of equilibrium.
  • Virtual forces (complementary virtual work): a set of real displacements is compatible if, for any virtual force system in equilibrium, the external complementary virtual work equals the internal one. It is a statement of compatibility, and it is what gives deflections.

Unit load method. To find the real displacement Δ at a point in a chosen direction, apply a virtual unit load there in that direction, find the internal actions it produces (m, t, p) in the same structure, and integrate them against the real strains produced by the actual loads (M/EI, T/GJ, P/AE). The result is a displacement in the direction of the unit load; a negative answer means it is opposite. For a rotation, use a unit couple. Required: linear elastic material, small displacements, and the unit-load system must be in equilibrium (it need not be the real system).

Castigliano's theorems. First theorem: ∂U/∂Δᵢ = Pᵢ. Second theorem (linear structures): ∂U/∂Pᵢ = Δᵢ, the displacement of the load point in the load's direction. Differentiating under the integral sign gives exactly the unit-load integrals, because ∂M/∂P = m. If no real load acts where the displacement is wanted, add a dummy load Q, differentiate, then set Q = 0.

Maxwell's reciprocal theorem. The displacement at point i due to a unit load at j equals the displacement at j due to a unit load at i: δᵢⱼ = δⱼᵢ. Flexibility matrices are therefore symmetric.

Which terms to keep. In slender beams the bending term dominates and the shear term is neglected. In trusses only axial terms exist. In thin-walled wing boxes the shear-flow term ∮q·q₁/(G·t) ds matters, because skins and webs are thin.

Formulas

Δ = Σ (P·p·L) / (A·E)

  • Pin-jointed truss. P: member force from real loads (N, tension +), p: member force from the unit load (N/N), L: member length (m), A: area (m²), E: Young's modulus (Pa), Δ: displacement (m).

Δ = ∫ M·m / (E·I) dx

  • Beam bending. M: real bending moment (N·m), m: moment due to unit load (m per N), EI: flexural rigidity (N·m²). Integrate over every segment; keep the same sign convention for M and m.

θ = ∫ M·m_θ / (E·I) dx

  • Rotation (rad); m_θ is the moment from a unit couple (dimensionless).

Δ = ∫ T·t / (G·J) dx

  • Torsion. T: real torque (N·m), t: torque from unit load, GJ: torsional rigidity (N·m²).

Δ = ∫∫ q·q₁ / (G·t) ds dz

  • Thin-walled sections in shear. q: real shear flow (N/m), q₁: shear flow from unit load, t: wall thickness (m), G: shear modulus (Pa); s round the section, z along the span.

U = P²·L / (2·A·E), U = ∫ M² / (2·E·I) dx, U = ∫ T² / (2·G·J) dx

  • Strain energy (J) for axial, bending and torsion.

Δᵢ = ∂U / ∂Pᵢ

  • Castigliano's second theorem, linear elastic structures.

Worked examples

Example 1 (standard): tip deflection of a cantilever. Given: cantilever, L = 2.0 m, tip load P = 5 kN, E = 200 GPa, I = 8 × 10⁻⁶ m⁴. Measure x from the free end.

  1. Flexural rigidity: EI = 200 × 10⁹ × 8 × 10⁻⁶ = 1.6 × 10⁶ N·m².
  2. Real moment (hogging taken positive): M = P·x. Unit load at the tip: m = 1·x.
  3. Deflection: Δ = ∫₀ᴸ P·x·x / (EI) dx = P·L³ / (3·EI).
  4. Numbers: Δ = 5000 × 2.0³ / (3 × 1.6 × 10⁶) = 40 000 / 4.8 × 10⁶ = 8.33 × 10⁻³ m.
  5. Check by Castigliano: U = P²·L³ / (6·EI), ∂U/∂P = P·L³ / (3·EI), the same. Answer: Δ = 8.33 mm, downward (in the direction of the load).

Example 2 (GATE level): two-bar truss. Given: joint B carries a vertical load P = 30 kN downward. Bar AB is horizontal, length 3 m, fixed to a wall at A. Bar CB runs from C, 4 m directly below A, so its length is 5 m. Both bars have A = 500 mm² and E = 70 GPa, so AE = 3.5 × 10⁷ N.

  1. Equilibrium at B (tension +). Vertical: 0.8·F_CB = −P, so F_CB = −1.25·P = −37.5 kN (compression). Horizontal: F_AB = −0.6·F_CB = 0.75·P = 22.5 kN (tension).
  2. Unit vertical load at B: by the same equations p_AB = 0.75, p_CB = −1.25.
  3. Vertical deflection: Δ_v = Σ P·p·L / (AE) = [22 500 × 0.75 × 3 + (−37 500) × (−1.25) × 5] / 3.5 × 10⁷.
  4. Terms: AB gives 1.446 mm, CB gives 6.696 mm, so Δ_v = 8.14 × 10⁻³ m.
  5. Horizontal deflection: a unit horizontal load at B (pointing away from the wall) gives p_AB = 1, p_CB = 0, so Δ_h = 22 500 × 1 × 3 / 3.5 × 10⁷ = 1.93 × 10⁻³ m, away from the wall. Answer: Δ_v = 8.14 mm downward, Δ_h = 1.93 mm away from the wall.

Common mistakes

  • Using different sign conventions (or different origins for x) for M and m, which flips terms.
  • Forgetting that a compression member with a compressive unit-load force contributes positively: (−)(−) = +.
  • Applying the unit load in the wrong direction or at the wrong point; the answer is the displacement along the unit load only.
  • Using Castigliano's second theorem on a nonlinear structure, or forgetting to set the dummy load to zero after differentiating.
  • Adding strain energies of two load cases separately; U is quadratic, so U(P₁ + P₂) ≠ U(P₁) + U(P₂).
  • Dropping the shear-flow term in thin-walled boxes, where it can be the largest.

For GATE AE

Common items are cantilever and simply supported beam deflections by the unit load method, truss joint deflections from a member force table, strain energy expressions, Castigliano's theorem and the reciprocal theorem. Practise building the P, p, L, PpL table for small trusses and writing M and m with one consistent origin.

Quick check

  1. What does the unit load method give if the result is negative?
  2. Write the strain energy of a bar of length L, area A, modulus E carrying axial force P.
  3. State Maxwell's reciprocal theorem.
  4. A simply supported beam (span L) with central load W: what is the central deflection by the unit load method? Answers: 1. a displacement opposite to the assumed unit load direction; 2. U = P²L/(2AE); 3. δᵢⱼ = δⱼᵢ, the displacement at i due to unit load at j equals that at j due to unit load at i; 4. W·L³/(48·EI).

Interview questions

All Aircraft Structures interview questions

Try answering each one aloud before you open it.

  1. 1.What is the principle of virtual work in the context of aircraft structures?Concept

    For a deformable body in equilibrium, the virtual work done by the external forces during any small compatible virtual displacement equals the virtual work done by the internal stresses on the corresponding virtual strains. The complementary form, with virtual forces in equilibrium acting through the real displacements, is what gives deflections through the unit load method. It holds for any material for equilibrium, and in aircraft structures it is used to find wing and fuselage deflections and to set up the force method for redundant structures.

  2. 2.Explain the unit load method and its application in aircraft structures.Concept

    To find a displacement at a point, apply a virtual unit load there in the required direction, find the internal actions it causes (moment m, axial force p, torque t, shear flow q₁), and integrate them against the real strains: Δ = ∫M·m/(EI) dx + Σ P·p·L/(AE) + ∫T·t/(GJ) dx + ∫q·q₁/(Gt) ds. A unit couple gives a rotation. It is used for wing tip deflection and twist, truss and frame joint displacements, and the flexibility coefficients of the force method.

  3. 3.How does the principle of virtual work help in analyzing complex aircraft structures?Application

    It replaces the solution of differential equations by an integration of internal actions, and it works the same way for trusses, beams, frames, torsion boxes and combinations of them. Each member's contribution is just added, so a structure with many load paths reduces to a table or a sum of integrals. It also supplies the compatibility equations for statically indeterminate structures and is the theoretical basis of the finite element method.

  4. 4.Why is the unit load method preferred for calculating deflections in aircraft structures?Application

    It gives the displacement at any chosen point and direction directly, without finding the whole deflected shape, and it handles varying cross-sections, several load types and combined bending, torsion, axial and shear deformation in one sum. The same unit-load systems are reused to get flexibility coefficients for redundant structures.

  5. 5.What does it mean if the virtual work equation is not satisfied for an assumed force or displacement field?Application

    With virtual displacements, a mismatch between external and internal virtual work means the assumed force or stress field is not in equilibrium. With virtual forces, a mismatch means the assumed displacements are not compatible. In practice it signals an error in the analysis (wrong member forces, sign or support condition) rather than a physical state of the structure.

  6. 6.Describe a scenario where the unit load method might not be applicable in aircraft structures.Application

    The unit load method might not be applicable in scenarios where the structure exhibits non-linear behavior or where material properties change significantly under load. In such cases, more advanced methods like finite element analysis may be required.

  7. 7.How would you use the unit load method to find the tip deflection of a wing idealised as a cantilever beam?Application

    Write the real bending moment M(x) along the span from the actual spanwise load, apply a unit load at the tip and write its moment m(x) = (distance from the tip), then evaluate Δ = ∫ M·m/(EI) dx from root to tip, splitting the integral wherever EI or the load changes. For a tip load P on uniform EI this gives P·L³/(3·EI). For a thin-walled wing the shear and torsion terms can be added in the same way.

  8. 8.Calculate the deflection at the free end of a cantilever beam with length 2 m, subjected to a point load of 500 N at the free end using the unit load method. Assume EI = 2000 N·m².Numerical

    Measure x from the free end: M = 500x and the unit-load moment m = x. Δ = ∫₀² 500x·x dx / EI = 500 × (2³/3) / 2000 = 500 × 2.667 / 2000 = 0.667 m, i.e. P·L³/(3·EI). The deflection is downward, in the direction of the load.

  9. 9.Using the unit load method, determine the deflection at the midpoint of a simply supported beam of length 4 m with a uniform load of 1000 N/m. Assume EI = 5000 N·m².Numerical

    Real moment M = 500x(4 − x) N·m; a unit load at mid-span gives m = x/2 for 0 ≤ x ≤ 2, symmetric on the other half. Δ = 2∫₀² (x/2)·500x(4 − x) dx / EI = 500 × ∫₀² (4x² − x³) dx / 5000 = 500 × (32/3 − 4)/5000 = 0.667 m, which matches 5wL⁴/(384EI) = 5 × 1000 × 256/(384 × 5000) = 0.667 m.

  10. 10.What are the limitations of using energy methods like virtual work in aircraft structural analysis?Application

    Energy methods like virtual work are limited to linear elastic systems and small deformations. They may not accurately predict behavior in structures with large deformations, non-linear material properties, or complex boundary conditions. In such cases, numerical methods like finite element analysis are more appropriate.

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