Strain energy and Castigliano's theorems

Why it matters

Energy methods are the backbone of aircraft structural analysis. They give deflections of frames, rings, curved members and trusses where integrating the elastic curve is clumsy, they solve statically indeterminate structures, and they lead directly to the finite-element method. Strain energy also tells you how well a part absorbs impact, as a landing gear or a crash structure must.

Key ideas

Strain energy. When a body is loaded gradually within the elastic range, the work done by the loads is stored as recoverable strain energy U. For a linear spring-like member under a load P that rises from zero to its final value with deflection δ, U = ½·P·δ (the area under a straight load–deflection line).

Strain energy density. Per unit volume, u = ½·σ·ε = σ²/(2E) for uniaxial stress and τ²/(2G) for shear. The value at the elastic limit, σy²/(2E), is the modulus of resilience; the total energy a body can store up to the elastic limit is its proof resilience.

Energy in each type of loading (linear elastic, integrate over the length):

• Axial: U = ∫ N²/(2AE) dx; for a uniform bar, U = P²L/(2AE).
• Bending: U = ∫ M²/(2EI) dx.
• Torsion: U = ∫ T²/(2GJ) dx.
• Transverse shear: U = ∫ k·V²/(2GA) dx (k = 1.2 for a rectangle); usually negligible in slender beams.

Gradual, sudden and impact loads. A load applied suddenly (full magnitude at once, no velocity) produces twice the stress of the same load applied gradually, because the work P·δ must equal the stored energy ½·(σ²/E)·A·L. A weight W falling through height h before striking a collar on a bar produces σ = (W/A)·[1 + √(1 + 2·A·E·h/(W·L))]. Increasing the volume or lowering E reduces impact stress, which is why long, slender bolts absorb shock better.

Castigliano's theorem for deflections. For a linear elastic structure, the deflection at the point and in the direction of a load Pᵢ is δᵢ = ∂U/∂Pᵢ; the rotation at a couple Mᵢ is θᵢ = ∂U/∂Mᵢ. (Naming: many Indian textbooks call this Castigliano's first theorem. In the original and international usage it is the second theorem, and the first theorem is Pᵢ = ∂U/∂δᵢ, which holds even for non-linear elastic materials. Learn the formulas, not just the names.)

Dummy load. To find the deflection where no load acts, add a fictitious load Q there in the required direction, compute ∂U/∂Q, then set Q = 0. In practice, differentiate under the integral: δ = ∫ (M/EI)·(∂M/∂Q) dx. For a truss, δ = Σ N·(∂N/∂Q)·L/(AE).

Least work. For a statically indeterminate structure with redundant reaction R and an unyielding support, ∂U/∂R = 0: the redundants take values that minimise strain energy. This is the compatibility condition in energy form.

Assumptions: linear elastic material, small deflections (geometry does not change significantly), loads applied gradually, supports unyielding unless stated, temperature effects excluded.

Formulas

U = ½·P·δ Strain energy (J) for load P (N) and deflection δ (m), linear elastic, gradual loading.

u = σ² / (2·E), u = τ² / (2·G) Strain energy per unit volume (J/m³).

U = P²·L / (2·A·E), U = ∫ M² / (2·E·I) dx, U = T²·L / (2·G·J) Axial, bending and torsion; N, N·m, m, m², m⁴, Pa.

σ = 2·P / A Suddenly applied load.

σ = (W/A)·[1 + √(1 + 2·A·E·h / (W·L))] Weight W (N) falling height h (m) onto a bar of length L, area A.

δᵢ = ∂U / ∂Pᵢ, θᵢ = ∂U / ∂Mᵢ Castigliano's theorem for deflections and rotations.

δ = ∫ (M / (E·I))·(∂M / ∂Q) dx (Q = 0 afterwards) Dummy-load form for beams and frames.

∂U / ∂R = 0 Least work for a redundant R.

Worked examples

Example 1 (standard). A steel bar 2 m long with A = 500 mm² and E = 200 GPa. Find (a) the stress, elongation and strain energy under a gradually applied 50 kN load; (b) the stress if 50 kN is applied suddenly; (c) the stress if a 5 kN weight falls 20 mm onto a collar at its end.

1. (a) σ = P/A = 50 000/500 = 100 MPa; δ = PL/(AE) = 50 000 × 2000/(500 × 200 000) = 1.0 mm.
2. U = ½·P·δ = 0.5 × 50 000 × 1.0 = 25 000 N·mm = 25 J. Check: P²L/(2AE) gives the same.
3. (b) σ = 2P/A = 200 MPa.
4. (c) W/A = 5000/500 = 10 MPa. 2AEh/(WL) = 2 × 500 × 200 000 × 20/(5000 × 2000) = 400.
5. σ = 10 × (1 + √401) = 10 × (1 + 20.025) = 210.2 MPa.
6. Check: δ = σL/E = 2.102 mm; work W(h + δ) = 5000 × 22.102 = 110.5 J; stored energy σ²AL/(2E) = 210.25² × 500 × 2000/(400 000) = 110.5 J.

Answer: (a) 100 MPa, 1.0 mm, 25 J; (b) 200 MPa; (c) about 210 MPa. A 5 kN weight dropped 20 mm is more severe than a 50 kN static load.

Example 2 (GATE level, Castigliano). An L-shaped frame has a vertical leg AB of height a = 1 m, fixed at A (bottom), and a horizontal arm BC of length b = 0.8 m. A downward load P = 2 kN acts at C. E·I = 2 × 10⁵ N·m² for both members; neglect axial and shear energy. Find the vertical and horizontal deflections of C.

1. Arm BC, x from C: M = P·x, so ∂M/∂P = x.
2. Leg AB, y measured down from B: M = P·b (constant), so ∂M/∂P = b. Add a horizontal dummy load Q at C: in AB, M = P·b + Q·y, so ∂M/∂Q = y; in BC, Q causes no moment.
3. Vertical: δv = (1/EI)·[∫₀ᵇ P·x·x dx + ∫₀ᵃ P·b·b dy] = P·b³/(3EI) + P·b²·a/(EI).
4. = 2000 × 0.512/(6 × 10⁵) + 2000 × 0.64 × 1/(2 × 10⁵) = 1.707 × 10⁻³ + 6.40 × 10⁻³ = 8.11 × 10⁻³ m.
5. Horizontal (set Q = 0): δh = (1/EI)·∫₀ᵃ P·b·y dy = P·b·a²/(2EI) = 2000 × 0.8 × 1/(4 × 10⁵) = 4.0 × 10⁻³ m.

Answer: δv ≈ 8.11 mm downward, δh = 4.0 mm, with C moving horizontally away from the leg, in the direction from B to C. Most of the vertical deflection comes from the leg AB rotating, not the arm bending.

Common mistakes

• Writing U = P·δ instead of ½·P·δ for a gradually applied load.
• Forgetting to set the dummy load to zero after differentiating.
• Squaring before differentiating when ∫ M·(∂M/∂Q) dx is simpler and less error-prone.
• Dropping the sign of ∂M/∂Q; a negative result means the deflection is opposite to the assumed dummy load.
• Using the static formula for impact loads, or forgetting that a suddenly applied load doubles the stress.
• Mixing up the names of Castigliano's theorems between textbooks; state the formula you are using.

For GATE AE

Expect strain energy of bars, beams and shafts, ratios of strain energy for different sections or loadings, sudden and impact loading, deflection of cantilevers, simply supported beams, bent frames and rings by Castigliano's theorem with a dummy load, and simple redundant problems by least work. Practise writing M and ∂M/∂Q for each member of a frame cleanly before integrating.

Quick check

1. What is the strain energy of a bar of length L and area A under axial load P?
2. By what factor does the stress increase if a load is applied suddenly instead of gradually?
3. Write the modulus of resilience in terms of the yield stress.
4. What is ∂U/∂R for a redundant reaction at an unyielding support?

Answers: 1. P²L/(2AE). 2. 2. 3. σy²/(2E). 4. Zero.