Strain energy and Castigliano's theorem

Strain energy in axial, bending and torsional members, resilience, sudden and impact loading, and Castigliano's theorem with dummy loads and least work for frames and indeterminate beams.

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Why it matters

Springs, bumpers, bolts in shock-loaded joints and suspension components are designed to absorb energy, not just to carry a static load. Strain energy tells you how much energy a part stores elastically and how high the stress rises when a load is dropped or applied suddenly. Castigliano's theorem turns the same energy into a fast way of finding deflections of bent frames, curved bars and indeterminate structures where geometry-based methods get messy.

Key ideas

Strain energy. When a load is applied gradually to an elastic body, the work done by the load is stored as strain energy U and is recovered on unloading. For a linear elastic member the load–deflection graph is a straight line, so U = ½·P·δ.

Energy density. Per unit volume, u = σ²/(2E) for normal stress and τ²/(2G) for shear. For a given stress, a material with lower E stores more energy, which is why springs use high-strength steel with large allowable stress rather than stiffness.

Resilience terms.

  • Resilience: total strain energy stored.
  • Proof resilience: the maximum strain energy stored up to the elastic limit.
  • Modulus of resilience: proof resilience per unit volume, σ_y²/(2E).
  • Toughness: energy absorbed up to fracture, including plastic work (area under the whole stress–strain curve).

Types of loading on a bar.

  • Gradual: σ = P/A.
  • Sudden (full load applied at once, no drop): energy balance P·δ = σ²·A·L/(2E) gives σ = 2·P/A, twice the static value.
  • Impact (load W falling through h): W·(h + δ) = σ²·A·L/(2E), giving σ = (W/A)·[1 + √(1 + 2·h·A·E/(W·L))]. When h is large compared with δ, σ ≈ √(2·E·W·h/(A·L)). These assume all the falling energy goes into the bar (no losses, no inertia of the bar).

Strain energy in members.

  • Axial: U = P²·L/(2·A·E).
  • Bending: U = ∫ M²/(2·E·I) dx.
  • Torsion: U = T²·L/(2·G·J).
  • Shear in beams: usually small for slender beams and neglected. A complete structure's U is the sum over all members and actions.

Castigliano's first theorem (displacement form). For a linear elastic structure, the displacement of the point of application of a load P, in the direction of P, is δ = ∂U/∂P. Similarly, the rotation at an applied couple M₀ is θ = ∂U/∂M₀. In practice, differentiate under the integral: δ = ∫ (M/EI)·(∂M/∂P) dx.

Dummy (fictitious) load. To find a displacement where no load acts, or in a direction with no load, apply a fictitious load Q there, write M including Q, differentiate with respect to Q and then set Q = 0.

Least work (Castigliano's second theorem, for redundants). For a statically indeterminate structure, the redundant reaction R adjusts so that ∂U/∂R = 0 when its support does not move. This gives the extra equation needed.

Validity. Linear elastic material, small displacements, and loads applied gradually with temperature constant. Castigliano's theorem does not apply to non-linear behaviour in this form.

Formulas

U = ½·P·δ

  • U: strain energy (J); P: load (N); δ: deflection in the direction of P (m). Linear elastic, gradually applied.

u = σ² / (2E), u = τ² / (2G)

  • u: strain energy per unit volume (J/m³).

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

σ_sudden = 2·P/A σ_impact = (W/A)·[1 + √(1 + 2·h·A·E / (W·L))]

  • W: falling weight (N); h: height of drop (m); A, L: bar area and length.

δ = ∂U/∂P = ∫ (M/(E·I))·(∂M/∂P) dx, θ = ∂U/∂M₀

  • Castigliano's first theorem.

∂U/∂R = 0

  • Least work for a redundant reaction R at an unyielding support.

Standard results: cantilever, end load: U = P²·L³ / (6EI); simply supported, central load: U = P²·L³ / (96EI).

Worked examples

Example 1 (standard): impact stress in a bar. Given: a vertical steel bar 2 m long, A = 500 mm², E = 200 GPa, has a collar at its lower end. A 1 kN weight falls 50 mm onto the collar. Find the maximum stress and extension, and compare with static and sudden loading.

  1. Static stress: W/A = 1000/500 = 2 MPa.
  2. Parameter: 2·h·A·E/(W·L) = 2 × 50 × 500 × 200 000 / (1000 × 2000) = 5000.
  3. σ = (W/A)·[1 + √(1 + 5000)] = 2 × (1 + 70.72) = 143.4 MPa.
  4. Extension: δ = σ·L/E = 143.4 × 2000 / 200 000 = 1.43 mm.
  5. Sudden application (h = 0) would give 4 MPa. Answer: σ_max ≈ 143 MPa, δ ≈ 1.43 mm, about 72 times the static stress. A 50 mm drop is far more damaging than the same weight resting on the bar.

Example 2 (GATE level): bent cantilever by Castigliano. Given: an L-shaped frame: a vertical post of height h = 1.5 m fixed at its base, and a horizontal arm of length a = 1 m rigidly joined at the top. A vertical load P = 2 kN acts at the free end of the arm. Both parts have EI = 500 kN·m². Neglect axial and shear energy. Find the vertical and horizontal deflections of the free end.

  1. Arm (x from the free end, 0 to a): M = P·x, ∂M/∂P = x.
  2. Post (y from the top, 0 to h): M = P·a (constant), ∂M/∂P = a.
  3. Vertical: δ_v = (1/EI)·[∫₀ᵃ P·x² dx + ∫₀ʰ P·a² dy] = P·a³/(3EI) + P·a²·h/(EI) = 2/1500 + 3/500 = 0.00133 + 0.00600 = 0.00733 m.
  4. Horizontal: add a dummy horizontal load Q at the free end. In the post, M = P·a + Q·y, so ∂M/∂Q = y; in the arm Q causes no moment. Set Q = 0: δ_h = (1/EI)·∫₀ʰ P·a·y dy = P·a·h²/(2EI) = 2 × 1 × 2.25 / 1000 = 0.0045 m. Answer: δ_v ≈ 7.33 mm downward, δ_h = 4.5 mm (away from the arm side, since the post bends that way).

Common mistakes

  • Writing U = P·δ instead of ½·P·δ for gradual loading.
  • Using the strain-energy formula for a sudden or impact load without the energy balance.
  • Forgetting the post (or other member) contribution in a frame.
  • Setting the dummy load to zero before differentiating.
  • Using U = P²L³/(48EI) for a simply supported beam with a central load; the correct value is P²L³/(96EI).
  • Applying Castigliano to non-linear or plastically deforming systems.

For GATE ME

Expect strain energy in bars of varying section, stress ratios for gradual, sudden and impact loading, proof resilience and modulus of resilience, deflection of cantilevers, frames and curved bars by Castigliano, and propped-cantilever reactions by least work. Practise writing M and ∂M/∂P member by member.

Quick check

  1. What is the stress under a suddenly applied load compared with the same load applied gradually?
  2. Steel with σ_y = 250 MPa and E = 200 GPa: find the modulus of resilience.
  3. Cantilever, L = 3 m, end load 8 kN, EI = 1500 kN·m². Find U.
  4. How do you find a displacement at a point with no applied load?

Answers: 1. Twice as large. 2. 250²/(2 × 200 000) = 0.156 MJ/m³. 3. 8² × 27/(6 × 1500) = 0.192 kN·m = 192 J. 4. Apply a dummy load there, differentiate with respect to it and then set it to zero.

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is strain energy in the context of materials?Concept

    Strain energy is the energy stored in a material due to deformation. When a material is subjected to stress, it deforms, and the work done on the material is stored as strain energy. This energy is recoverable when the material returns to its original shape, assuming the deformation is elastic.

  2. 2.Explain Castigliano's theorem and its significance in engineering mechanics.Concept

    Castigliano's theorem states that the partial derivative of the total strain energy of a structure with respect to an applied force gives the displacement in the direction of that force. It is significant because it allows engineers to calculate displacements in complex structures using energy methods, which can be simpler than using direct methods.

  3. 3.How is strain energy related to the modulus of elasticity?Concept

    For a linear elastic material the strain energy per unit volume is u = σ²/(2E) = E·ε²/2. So for the same strain a stiffer material (higher E) stores more energy, but for the same stress it stores less. That is why energy-absorbing parts such as springs are judged by σ_allow²/E: a high-strength spring steel stores far more energy than mild steel even though their E values are almost the same. The modulus of resilience, σ_y²/(2E), is the maximum elastic energy per unit volume.

  4. 4.Why is Castigliano's theorem particularly useful for statically indeterminate structures?Application

    In an indeterminate structure the equilibrium equations alone cannot give all the reactions. Using Castigliano, you treat a redundant reaction R as an unknown load, express the total strain energy in terms of the applied loads and R, and use the fact that the displacement at an unyielding support is zero: ∂U/∂R = 0 (the theorem of least work). This gives the missing compatibility equation directly, and it extends naturally to frames, curved members and several redundants.

  5. 5.What happens to the strain energy in a material if it is loaded beyond its elastic limit?Application

    If a material is loaded beyond its elastic limit, it undergoes plastic deformation, and the strain energy is no longer fully recoverable. The energy used to deform the material beyond the elastic limit is dissipated as heat or used in creating permanent deformations, and the material will not return to its original shape.

  6. 6.How can Castigliano's theorem be applied to calculate the deflection of a beam under a point load?Application

    To apply Castigliano's theorem to calculate the deflection of a beam under a point load, first express the strain energy in terms of the load and the geometry of the beam. Then, take the partial derivative of the strain energy with respect to the load to find the deflection at the point of application of the load.

  7. 7.A steel rod of length 2 m and cross-sectional area 0.01 m² is subjected to a tensile force of 50 kN. Calculate the strain energy stored in the rod. Assume Young's modulus for steel is 200 GPa.Numerical
    1. Calculate the stress: σ = F / A = 50,000 N / 0.01 m² = 5,000,000 N/m².
    2. Calculate the strain: ε = σ / E = 5,000,000 N/m² / 200,000,000,000 N/m² = 0.000025.
    3. Calculate the strain energy: U = 0.5 × σ × ε × Volume = 0.5 × 5,000,000 N/m² × 0.000025 × (2 m × 0.01 m²) = 1.25 J.
  8. 8.Using Castigliano's theorem, determine the deflection at the end of a cantilever beam of length L, modulus of elasticity E, and moment of inertia I, subjected to an end load P.Numerical
    1. The strain energy U for a cantilever beam under an end load P is U = (P²L³) / (6EI).
    2. According to Castigliano's theorem, the deflection δ at the point of load application is δ = ∂U/∂P = (PL³) / (3EI).
  9. 9.What are the limitations of using Castigliano's theorem in practical applications?Application

    Castigliano's theorem assumes that the material is linearly elastic and that the deformations are small. It may not be accurate for materials that exhibit significant plastic deformation or for structures with large deformations. Additionally, it requires that the structure's strain energy can be expressed as a function of the applied loads, which may not be straightforward for complex geometries.

  10. 10.Explain how strain energy can be used to assess the safety of a structure.Application

    Strain energy matters most when loads are dynamic. Equating the energy delivered by a falling or suddenly applied load to the strain energy stored gives the peak stress: a suddenly applied load doubles the static stress, and a dropped weight can raise it many times. Comparing that stress with yield, or the energy demand with the part's proof resilience, shows whether the part survives elastically. Designers increase energy capacity with longer or more flexible members, or by reducing the cross-section of a bolt shank to the root diameter so it stretches more under shock.

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