Strain energy and Castigliano's theorem

Why it matters

Energy methods answer two practical questions in mechatronic design. First, how hard does a part get hit: a carriage striking an end stop, a load dropped onto a load cell, a gripper closing fast all produce stresses far above the static value, and strain energy gives them quickly. Second, how much does an awkward part deflect: an L-shaped sensor bracket, a curved flexure or a frame with bends is tedious by integration, but Castigliano's theorem gives any deflection from one energy expression.

Key ideas

Strain energy. When a load is applied gradually to an elastic body, the work it does is stored as strain energy U, recoverable on unloading. For a linear-elastic member, U = ½·P·δ (the area under the load-deflection line). Per unit volume, the strain energy density is 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 stored up to the elastic limit, (σ_e²/2E) × volume.
• Modulus of resilience: proof resilience per unit volume, σ_e²/(2E); the area under the elastic part of the stress-strain curve.
• Modulus of toughness: total area under the stress-strain curve up to fracture. Springs and energy-absorbing parts want high resilience: high strength, low modulus.

Gradual, sudden and impact loading of a bar.

• Gradual: σ = P/A.
• Suddenly applied (full load placed at once, no drop): work P·δ = stored ½·σ²/E·A·L gives σ = 2P/A, twice the static value.
• Impact (load P falling through height h onto a collar): P·(h + δ) = σ²·A·L/(2E). Solving gives σ = (P/A)·[1 + √(1 + 2h·A·E/(P·L))]. If h is large compared with δ, σ ≈ √(2E·P·h/(A·L)). A bar with larger volume (longer or thicker) stores the same energy at lower stress, so for impact a long, slender, uniform bolt is better than a short stubby one, and turning the shank down to the root diameter of the thread helps.

Strain energy in members.

• Axial: U = ∫P²/(2AE) dx.
• Bending: U = ∫M²/(2EI) dx.
• Torsion: U = ∫T²/(2GJ) dx.
• Shear in beams is usually neglected for slender members. For a structure, add the contributions of all members and all segments.

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

Dummy load. If you need the deflection at a point where no load acts, or in a direction with no load, apply a fictitious load Q there, carry it through, differentiate with respect to Q, and then set Q = 0.

Statically indeterminate structures. Treat a redundant reaction R as an unknown load. Since the support does not move, ∂U/∂R = 0. This compatibility equation gives R (theorem of least work).

Reciprocity. Maxwell's reciprocal theorem follows: the deflection at A due to a unit load at B equals the deflection at B due to a unit load at A. Useful for checking influence coefficients and stiffness matrices.

Formulas

U = ½·P·δ

• U: strain energy (J); P: gradually applied load (N); δ: corresponding deflection (m).

u = σ² / (2E), u = τ² / (2G)

• u: strain energy per unit volume (J/m³); σ, τ in Pa; E, G in Pa.

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

• Axial bar, beam in bending, shaft in torsion. A (m²), L (m), I and J (m⁴).

σ_sudden = 2P/A

• Suddenly applied load (no drop height).

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

• P: falling weight (N); h: drop height before contact (m); L: bar length (m).

δ_i = ∂U/∂P_i = ∫(M/EI)·(∂M/∂P_i) dx

• Castigliano's theorem, bending; add axial and torsion terms when they matter.

∂U/∂R = 0

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

Worked examples

Example 1 (standard): gradual, sudden and impact loading of a tie rod. Given: a steel rod, diameter 20 mm, length 2 m, E = 200 GPa. (a) 30 kN applied gradually: find σ and U. (b) A weight of 500 N falls 10 mm onto a collar at the end: find the maximum stress, extension and energy absorbed.

1. A = π/4 × 20² = 314.2 mm²; volume = 314.2 × 2000 = 628 300 mm³.
2. (a) σ = 30 000/314.2 = 95.5 MPa. U = σ²/(2E) × volume = 95.5²/(2 × 200 000) × 628 300 = 14 320 N·mm = 14.3 J.
3. (b) Static stress P/A = 500/314.2 = 1.592 MPa.
4. 2h·A·E/(P·L) = 2 × 10 × 314.2 × 200 000/(500 × 2000) = 1257.
5. σ_impact = (P/A)·[1 + √(1 + 1257)] = 1.592 × (1 + 35.47) = 58.0 MPa.
6. Extension: δ = σ·L/E = 58.0 × 2000/200 000 = 0.580 mm.
7. Energy check: P·(h + δ) = 500 × 10.58 = 5290 N·mm = 5.29 J; σ²/(2E) × volume = 58.0²/400 000 × 628 300 = 5290 N·mm. Answer: (a) 95.5 MPa, 14.3 J; (b) 58.0 MPa, 0.58 mm, 5.29 J. A 500 N weight dropped only 10 mm causes 36 times the static stress.

Example 2 (GATE level): L-shaped sensor bracket by Castigliano. Given: an L-bracket of 20 mm square steel bar (E = 200 GPa, I = 20⁴/12 = 13 333 mm⁴, EI = 2.667 × 10⁹ N·mm²). The vertical leg (a = 300 mm) is fixed at its base; the horizontal arm (b = 200 mm) carries P = 200 N downward at its tip. Neglect axial and shear energy. Find the vertical and horizontal deflections of the tip.

1. Arm, s measured from the tip (0 to b): M = P·s. Leg, y measured from the corner downward (0 to a): M = P·b (constant).
2. Vertical: ∂M/∂P = s in the arm and b in the leg.
3. δ_v = (1/EI)·[∫P·s² ds + ∫P·b² dy] = P·b³/(3EI) + P·b²·a/(EI).
4. δ_v = 200 × 200³/(3 × 2.667 × 10⁹) + 200 × 200² × 300/(2.667 × 10⁹) = 0.20 + 0.90 = 1.10 mm.
5. Horizontal: add a dummy horizontal load Q at the tip. It is axial in the arm (no moment) and adds Q·y to the moment in the leg: M = P·b + Q·y, so ∂M/∂Q = y.
6. δ_h = (1/EI)·∫(P·b)·y dy (with Q = 0) = P·b·a²/(2EI) = 200 × 200 × 300²/(2 × 2.667 × 10⁹) = 0.675 mm. Answer: δ_v = 1.10 mm downward, δ_h = 0.675 mm. The leg contributes 82 % of the vertical deflection, so stiffening the leg is the effective fix.

Common mistakes

• Writing U = P·δ instead of ½·P·δ for a gradually applied load.
• Using the static stress for a suddenly applied or impact load.
• Differentiating with respect to a load after substituting its numerical value; keep it symbolic until after differentiation.
• Forgetting to set the dummy load to zero after differentiating.
• Leaving out a segment of the structure, or using one coordinate for a member that bends in two segments.
• Applying Castigliano's theorem to a nonlinear (yielding or large-deflection) problem.

For GATE ME

Expect strain energy in bars, beams and shafts, comparison of gradual, sudden and impact stresses, proof resilience and modulus of resilience, deflection of cantilevers, simply supported beams, bent bars and rings by Castigliano, and redundant reactions by least work. Practise writing M and ∂M/∂P cleanly for each segment.

Quick check

1. A load is suddenly applied to a bar. How does the stress compare with the gradually applied value?
2. Write the strain energy of a cantilever with end load P.
3. State Castigliano's theorem for deflection.
4. A rod (L = 2 m, A = 0.01 m², E = 200 GPa) carries 1000 N gradually. What is U?
5. What is the modulus of resilience of steel with σ_e = 250 MPa and E = 200 GPa?

Answers: 1. Twice as large. 2. U = P²·L³/(6EI). 3. δ = ∂U/∂P at the load, in its direction. 4. 0.0005 J. 5. 250²/(2 × 200 000) = 0.156 N·mm/mm³ = 156 kJ/m³.