Strain energy methods

Strain energy and resilience, gradual, sudden and impact loading, strain energy in axial, torsional and bending members, and Castigliano's theorem for deflections and redundants.

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Why it matters

When a load deforms an elastic member, the work done is stored as strain energy and given back on unloading. Energy ideas explain why suddenly applied and impact loads are so damaging, how springs, bolts and crash barriers are sized to absorb shocks, and they give the quickest route to deflections of frames, curved bars and trusses, and to the redundant forces of statically indeterminate structures.

Key ideas

Strain energy. For a linear elastic member loaded gradually from zero, the work done by the load is the area under the load–deflection line, U = ½ P δ. Per unit volume (strain energy density) u = σ²/(2E) for normal stress and τ²/(2G) for shear.

Resilience terms.

  • Resilience: total strain energy stored in a body.
  • Proof resilience: maximum strain energy that can be stored without permanent set (at the elastic limit), σe² V/(2E).
  • Modulus of resilience: proof resilience per unit volume, σe²/(2E).
  • Modulus of toughness: energy per unit volume to fracture (area under the whole stress–strain curve).

Types of axial loading.

  • Gradual load P: σ = P/A.
  • Sudden load P (applied all at once with no drop): P δ = σ² A L/(2E) gives σ = 2P/A, twice the gradual value.
  • Impact load (weight W falling height h onto a collar): W (h + δ) = σ² A L/(2E), giving the formula below. If h ≫ δ, σ ≈ √(2 E W h/(A L)). The energy balance assumes the bar's own mass is negligible and no energy is lost in the impact.

Strain energy in members.

  • Axial: U = P² L/(2AE).
  • Torsion: U = T² L/(2GJ).
  • Bending: U = ∫ M² dx/(2EI).
  • Shear: usually small for slender beams and neglected. Strain energy is a quadratic function of load, so energies from different loads cannot simply be added; superpose the loads first, then compute U.

Castigliano's theorem. For a linear elastic structure, the displacement at a load point in the direction of that load is δi = ∂U/∂Pi; the rotation at a couple is θi = ∂U/∂Mi. To find a displacement where no load acts, add a dummy load Q there, differentiate, then set Q = 0. Differentiating under the integral sign is usually easier: δ = ∫ (M/EI)(∂M/∂P) dx.

Indeterminate structures. Choose a redundant reaction R; the corresponding displacement is known (often zero), so ∂U/∂R = 0 (theorem of least work). This gives the extra compatibility equation.

Maxwell's reciprocal theorem. The deflection at point 1 due to a unit load at 2 equals the deflection at 2 due to a unit load at 1.

Limits. All this assumes linear elastic behaviour and small deflections. Beyond the elastic limit, part of the energy is dissipated and not recoverable.

Formulas

U = ½ P δ = P² L/(2AE) (axial bar)

  • U in N·mm (= mJ) or J; P in N; L in mm; A in mm²; E in N/mm².

u = σ²/(2E) (energy per unit volume, normal stress); u = τ²/(2G) (shear)

U = T² L/(2GJ) (torsion); U = ∫ M² dx/(2EI) (bending)

σ_sudden = 2P/A

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

δi = ∂U/∂Pi, θi = ∂U/∂Mi (Castigliano)

U = P² L³/(6EI) (cantilever, end load); U = P² L³/(96EI) (simply supported, central load); U = w² L⁵/(240EI) (simply supported, UDL)

Worked examples

Example 1 (standard, impact). A vertical steel bar 2 m long with an area of 500 mm² (E = 200 GPa) has a collar at its lower end. A weight of 1 kN falls 100 mm onto the collar. Find the maximum stress, the extension and the energy absorbed. Compare with the same weight applied gradually.

  1. Static stress: W/A = 1000/500 = 2 MPa.
  2. Impact factor: 2 h A E/(W L) = 2 × 100 × 500 × 200 000 / (1000 × 2000) = 10 000; 1 + √(1 + 10 000) = 101.005.
  3. σ = 2 × 101.005 = 202.0 MPa.
  4. Extension: δ = σ L/E = 202.0 × 2000 / 200 000 = 2.02 mm.
  5. Energy check: W (h + δ) = 1000 × 102.02 = 102 020 N·mm; σ² A L/(2E) = 202.0² × 500 × 2000 / 400 000 = 102 020 N·mm. ✓

Answer: σ ≈ 202 MPa, δ ≈ 2.02 mm, energy ≈ 102 J; about 101 times the gradual-load stress of 2 MPa.

Example 2 (GATE level, Castigliano). An L-shaped bracket has a vertical column of height h = 1.5 m fixed at its base and a horizontal arm of length a = 1 m. A vertical load P = 2 kN acts downward at the free end of the arm. EI = 8 × 10¹¹ N·mm² for both parts (E = 200 GPa, I = 4 × 10⁶ mm⁴). Neglecting axial and shear energy, find the vertical deflection of the free end.

  1. Arm (x from the free end): M = P x, ∂M/∂P = x. Contribution: ∫₀ᵃ P x² dx /EI = P a³/(3EI).
  2. Column: the moment is constant, M = P a, ∂M/∂P = a. Contribution: ∫₀ʰ P a² dy /EI = P a² h/EI.
  3. δ = P a³/(3EI) + P a² h/EI.
  4. Numbers (N, mm): 2000 × 1000³/(3 × 8 × 10¹¹) = 0.833 mm; 2000 × 1000² × 1500/(8 × 10¹¹) = 3.75 mm.
  5. δ = 0.833 + 3.75 = 4.58 mm.

Answer: vertical deflection ≈ 4.58 mm downward (most of it from rotation of the column).

Common mistakes

  • Adding strain energies of separate loads; U is quadratic, so superpose loads first.
  • Using σ = P/A for a suddenly applied load; it is 2P/A.
  • Forgetting δ in W(h + δ) when h is not much larger than δ.
  • Leaving the dummy load in the final expression, or setting it to zero before differentiating.
  • Mixing J with N·mm (1 J = 1000 N·mm).
  • Applying Castigliano's theorem to non-linear or plastic behaviour.

For GATE PI

Expect strain energy of bars, shafts and beams, stress under sudden and impact loads, proof resilience and modulus of resilience, and deflections of cantilevers, bent frames and simple trusses by Castigliano's theorem. Practise writing M(x) for each member of a frame with a convenient origin.

Quick check

  1. How does the stress from a suddenly applied load compare with the same load applied gradually?
  2. Strain energy in a steel bar, P = 100 kN, L = 2 m, A = 0.01 m², E = 200 GPa?
  3. What is the modulus of resilience of a steel with σe = 250 MPa and E = 200 GPa?
  4. Strain energy of a simply supported beam, span 6 m, UDL 5 kN/m, EI = 21 MN·m²?

Answers: 1. Twice as large. 2. 5 J. 3. 0.156 N·mm/mm³ (156 kJ/m³). 4. w²L⁵/(240EI) = 38.6 J.

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 engineering mechanics?Concept

    Strain energy is the energy stored in a body 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.

  2. 2.Explain the principle of virtual work in the context of strain energy methods.Concept

    The principle of virtual work states that for a system in equilibrium, the virtual work done by external forces is equal to the virtual work done by internal forces. In strain energy methods, this principle helps in analyzing structures by equating the work done by applied loads to the strain energy stored in the structure.

  3. 3.How is Castigliano's theorem used in determining deflections in structures?Concept

    Castigliano's theorem states that the partial derivative of the total strain energy with respect to an applied force gives the displacement in the direction of that force. It is used to calculate deflections in structures by differentiating the strain energy expression with respect to the applied load.

  4. 4.Why are strain energy methods useful for analysing statically indeterminate structures?Application

    Equilibrium alone gives too few equations in an indeterminate structure; the missing ones come from compatibility of displacements. Energy methods supply them systematically: taking a redundant reaction R as an unknown load, Castigliano's theorem gives the displacement there as ∂U/∂R, which is set equal to its known value (usually zero, the theorem of least work). This works the same way for beams, frames, rings and trusses, without having to integrate the deflection equation segment by segment.

  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 is dissipated as heat and permanent deformation occurs, meaning the material will not return to its original shape when the load is removed.

  6. 6.How does the concept of strain energy apply to the design of safety devices like crash barriers?Application

    Strain energy is crucial in the design of safety devices like crash barriers because these devices are designed to absorb energy during an impact. By storing strain energy, they reduce the force transmitted to passengers or structures, thereby minimizing damage and injury.

  7. 7.Calculate the strain energy stored in a steel rod of length 2 m and cross-sectional area 0.01 m² when subjected to a tensile force of 100 kN. Assume Young's modulus for steel is 200 GPa.Numerical
    1. Calculate the stress: σ = Force / Area = 100,000 N / 0.01 m² = 10,000,000 N/m².
    2. Calculate the strain: ε = σ / E = 10,000,000 N/m² / 200,000,000,000 N/m² = 0.00005.
    3. Calculate the strain energy: U = 0.5 × σ × ε × Volume = 0.5 × 10,000,000 N/m² × 0.00005 × (2 m × 0.01 m²) = 5 Joules.
  8. 8.A simply supported beam of span 4 m carries a uniform load of 5 kN/m with EI = 5000 kN·m². Find the mid-span deflection, and explain how an energy method gets it.Numerical

    With Castigliano's theorem, place a dummy point load Q at mid-span, write M(x) = (wL/2 + Q/2)x − wx²/2 for the left half, and evaluate δ = 2∫₀^(L/2) (M/EI)(∂M/∂Q) dx with Q = 0. This gives the standard result δ = 5wL⁴/(384EI). Here δ = 5 × 5 × 4⁴ / (384 × 5000) = 6400/1 920 000 = 3.33 × 10⁻³ m = 3.33 mm. Note the dummy load must be set to zero only after differentiating.

  9. 9.What are the limitations of using strain energy methods in structural analysis?Concept

    Strain energy methods assume linear elastic behavior, which may not be valid for materials that exhibit significant plastic deformation. They also require accurate material properties and may not account for complex loading conditions or geometric non-linearities. These limitations can lead to inaccuracies in predicting real-world behavior.

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