Strain energy and impact loading

In short

Strain energy is the work stored in a body when it is deformed elastically; released fully on unloading, it is the energy basis of deflection and impact calculations. Impact loading is a load applied with velocity — a falling weight, a hammer blow, a vehicle hitting a kerb — and it is analysed by equating the kinetic or potential energy lost by the striking body to the strain energy stored in the struck member. The topic carries guaranteed marks in SOM university papers and GATE (energy methods, Castigliano's theorem, impact factor) and is used daily in spring, bolt, crane, press and automotive crash design at Tata Motors, Mahindra, L&T, BHEL, Godrej, ISRO and DRDO labs. Interviewers also ask the materials-side question — Charpy and Izod toughness, and why a brittle part fails under a shock it would survive statically.

Explained

Strain energy, resilience and load types

• Strain energy U: work done by the external load in deforming the body, stored as internal elastic energy. Unit: N·mm or joule.
• For a gradually applied axial load the load-extension graph is a straight line, so U = area under it = ½Pδ.
• Strain energy density (u): energy per unit volume = σ²/2E for uniaxial stress; equals the area under the stress-strain curve.
• Resilience: total strain energy stored. Proof resilience: strain energy at the elastic limit. Modulus of resilience: proof resilience per unit volume = σy²/2E — the ability to absorb energy *elastically*.
• Toughness: total area under the whole stress-strain curve up to fracture (includes plastic work) — the energy measured in an impact test. Resilience is elastic, toughness is total.
• Three load types:
• Gradual: load rises 0 → P; σ = P/A, U = ½Pδ.
• Sudden: full load P in place from the start, no velocity; work done = Pδ, so U = Pδ → stress and deflection are double the gradual values.
• Impact: load arrives with velocity (falling height h or speed v) — stress is many times the static value.

Strain energy under the four basic actions

• Axial: U = P²L/2AE = σ²AL/2E = σ²V/2E. For varying P or A, integrate: U = ∫P²dx/2AE.
• Shear: u = τ²/2G, so U = τ²V/2G (G = shear modulus).
• Torsion: U = T²L/2GJ = ½Tθ. In a solid circular shaft the stress varies linearly, so U = (τmax²/4G)·V — only half the energy a uniform shear stress τmax would store. A hollow shaft stores more energy per kg because more material sits near τmax.
• Bending: U = ∫M²dx/2EI (flexural energy). Energy from transverse shear is usually neglected in slender beams.
• General elastic body: U = ∫(σ²/2E)dV summed over all stress components; for principal stresses it splits into volumetric (dilatational) and shear (distortion) parts — the distortion part is the basis of the von Mises theory of failure.
• Section shape matters for energy: a bar of uniform section stores the most energy for a given maximum stress, because every fibre reaches σmax. A stepped or notched bar reaches σmax only at the small section, so its energy capacity drops sharply — the reason long, slender "resilient" bolts are used where shock is expected, and why a notch or a sharp fillet makes a part fail under impact.

Impact loading of a bar (falling weight)

Set-up: a weight W falls through height h onto a collar at the end of a vertical bar of length L, area A, modulus E; the bar extends δ as it arrests the weight.

1. Energy lost by weight = W(h + δ); energy stored by bar = σ²AL/2E, with δ = σL/E.
2. Equate: W(h + σL/E) = σ²AL/2E → quadratic in σ.
3. Solution: σ = (W/A)[1 + √(1 + 2hAE/WL)].
4. Then δ = σL/E, and the equivalent static (gradually applied) load is P_eq = σA.
5. Impact factor n = σ/σst = δ/δst, where σst = W/A. Equivalently n = 1 + √(1 + 2h/δst).
6. Checks: put h = 0 and you get σ = 2W/A — the suddenly applied load case, factor 2. For large h, 2hAE/WL ≫ 1 and σ ≈ √(2WhE/AL) — stress varies as √h, and increasing L or A reduces it.
7. Horizontal impact / known velocity: replace potential energy by kinetic energy — ½mv² = σ²AL/2E (ignoring the small work of the load moving through δ).
8. Any elastic member: the same result holds with δst = static deflection under W, so δ = δst[1 + √(1 + 2h/δst)]. For a weight dropped at mid-span of a simply supported beam δst = WL³/48EI; for a cantilever tip δst = WL³/3EI; for a spring δst = W/k.
9. Assumptions and limits: material stays elastic, the whole kinetic energy goes into the member (no loss in the collar, support or striking body), the mass of the struck member is neglected, and stress is assumed uniform at every instant — real impact travels as a stress wave and local yielding absorbs energy. The method is therefore conservative (over-predicts stress) for short, stiff members and must not be used beyond the elastic limit.

Energy methods for deflection

• Clapeyron / strain energy equality: external work = strain energy, so U = ½Pδ gives the deflection at the point of a single load: δ = 2U/P.
• Castigliano's first theorem: δi = ∂U/∂Pi — partial derivative of total strain energy with respect to a load gives the deflection in the direction of that load; ∂U/∂M gives the slope/rotation under a couple.
• Procedure: write M (or T, P) as a function of x including the load Pi, use δ = ∫(M/EI)(∂M/∂Pi)dx, then substitute numbers. If no load acts where deflection is wanted, apply a dummy load Q, differentiate, then set Q = 0.
• Castigliano's second theorem (least work): in a redundant structure the redundant forces take values that minimise the total strain energy — ∂U/∂R = 0 (used for fixed beams, closed rings, redundant trusses).
• Maxwell-Betti reciprocal theorem: deflection at A due to a load at B equals deflection at B due to the same load at A — used to check influence lines.
• Unit load / virtual work method: δ = Σ(PuL/AE) for trusses, δ = ∫(Mm/EI)dx for beams, where u and m are forces/moments from a unit load at the point.

Impact testing, toughness and Indian practice

• Charpy test: notched bar (standard 10 × 10 × 55 mm, 2 mm deep V-notch, 45°) supported as a simple beam, struck behind the notch by a pendulum; energy absorbed in joules read off the dial. Indian standard IS 1757; equivalent ASTM E23.
• Izod test: 10 × 10 mm specimen held as a vertical cantilever, notch level with the clamp, struck 22 mm above it — IS 1598.
• The test gives a comparative energy, not a design stress — it measures notch toughness under a high strain rate.
• Ductile-brittle transition: absorbed energy falls steeply below a transition temperature; BCC steels show it, FCC metals (austenitic stainless, aluminium) largely do not. Structural steel grades in IS 2062 are supplied in subgrades (BR, B0 etc.) with a specified minimum Charpy energy at a stated temperature — check the current edition for the exact values.
• Design practice: codes handle impact by a factor on the static load — IRC:6 adds an impact fraction to bridge live loads, crane and hoist codes (IS 807, IS 3177) add a percentage to vertical wheel loads, and IS 875 (Part 2) requires an allowance for machinery and moving loads; IS 800:2007 governs fatigue and brittle-fracture checks in steel members. Values are revised periodically — refer to the current edition rather than memorised numbers.
• Energy absorbers in practice: leaf and coil springs, rubber and hydraulic buffers, crumple zones, shear pins, buffer stops on railway sidings — all sized by equating kinetic energy to strain energy or to plastic work.

Key points

• Strain energy - work stored elastically in a strained body, fully recovered on unloading; U = ½Pδ for a gradual load.
• Strain energy density - σ²/2E (direct), τ²/2G (shear); area under the stress-strain curve per unit volume.
• Modulus of resilience - σy²/2E; energy absorbed per unit volume up to the elastic limit (high for spring steel).
• Toughness - total area under the full stress-strain curve; includes plastic work, measured by Charpy/Izod.
• Axial - U = P²L/2AE; torsion - U = T²L/2GJ; bending - U = ∫M²dx/2EI; shear - U = τ²V/2G.
• Solid shaft - stores τmax²V/4G, half of a uniformly stressed body; hollow shafts are better energy stores per unit mass.
• Suddenly applied load - stress and deflection are exactly twice the gradual values (impact factor 2).
• Impact factor - n = 1 + √(1 + 2h/δst); for large h, σ ≈ √(2WhE/AL).
• Equivalent static load - P_eq = σA, the gradual load that would produce the same stress as the impact.
• Longer, softer member absorbs impact better - raising L or lowering stiffness raises δst and cuts the impact factor.
• Notches and steps slash energy capacity - only the small section reaches σmax; hence slender resilient bolts.
• Castigliano I - δ = ∂U/∂P, θ = ∂U/∂M; use a dummy load where no load exists.
• Castigliano II - ∂U/∂R = 0 for redundant forces (least work).
• Maxwell-Betti - reciprocal deflections are equal in a linear elastic system.
• Distortion energy - the shear part of strain energy density; basis of the von Mises yield criterion.
• Charpy = simple beam, Izod = cantilever - IS 1757 and IS 1598; energy in joules at a stated temperature.
• Energy method is conservative - it assumes no loss and ignores the struck member's inertia and stress waves.

Formulas and examples

1. Gradual, sudden and impact on the same bar. Steel bar L = 2 m, d = 25 mm, E = 200 GPa. A = (π/4)(25²) = 490.9 mm².

• Gradual 50 kN: σ = 50 000/490.9 = 101.9 MPa; δ = σL/E = 101.9 × 2000/200 000 = 1.019 mm; U = ½ × 50 000 × 1.019 = 25 470 N·mm = 25.5 J.
• Sudden 50 kN: σ = 203.7 MPa, δ = 2.037 mm (exactly double).
• Impact, W = 500 N falling h = 100 mm: 2hAE/WL = (2 × 100 × 490.9 × 200 000)/(500 × 2000) = 19 635. σ = (500/490.9)[1 + √19 636] = 1.019 × (1 + 140.1) = 143.7 MPa; δ = 143.7 × 2000/200 000 = 1.437 mm; impact factor = 143.7/1.019 = 141. Energy check: W(h + δ) = 500 × 101.44 = 50 720 N·mm ≈ σ²AL/2E = 50 680 N·mm ✓

2. Bending energy and Castigliano. Cantilever L = 2 m, P = 10 kN at free end, E = 200 GPa, I = 1 × 10⁷ mm⁴.

• U = P²L³/6EI = (10 000² × 2000³)/(6 × 200 000 × 10⁷) = 8 × 10¹⁷/1.2 × 10¹³ = 66 667 N·mm = 66.7 J.
• δ = ∂U/∂P = PL³/3EI = (10 000 × 8 × 10⁹)/(6 × 10¹²) = 13.33 mm. Check ½Pδ = ½ × 10 000 × 13.33 = 66 667 N·mm ✓

3. Torsional strain energy. Solid shaft d = 60 mm, L = 1.5 m, T = 2 kN·m, G = 80 GPa.

• J = πd⁴/32 = π × 1.296 × 10⁷/32 = 1.272 × 10⁶ mm⁴.
• U = T²L/2GJ = (2 × 10⁶)² × 1500/(2 × 80 000 × 1.272 × 10⁶) = 6 × 10¹⁵/2.036 × 10¹¹ = 29 470 N·mm = 29.5 J.
• θ = TL/GJ = 3 × 10⁹/1.018 × 10¹¹ = 0.0295 rad = 1.69°; ½Tθ = 29 500 N·mm ✓

4. Strain energy density, split into volumetric and distortion. σ1 = 100 MPa, σ2 = 50 MPa, σ3 = 0, E = 200 GPa, ν = 0.3 → G = E/2(1+ν) = 76 923 MPa.

• Total u = (1/400 000)[100² + 50² − 2(0.3)(100 × 50)] = (12 500 − 3000)/400 000 = 0.02375 MPa = 23.75 kJ/m³.
• Distortion u_d = (1/(12 × 76 923))[50² + 50² + 100²] = 15 000/923 076 = 0.01625 MPa.
• Volumetric u_v = (1−2ν)(σ1+σ2+σ3)²/6E = 0.4 × 150²/(1.2 × 10⁶) = 0.0075 MPa; 0.01625 + 0.0075 = 0.02375 ✓
• Modulus of resilience of mild steel (σy = 250 MPa): 250²/400 000 = 0.156 MPa = 156 kJ/m³.

Where it is used

• Spring and suspension design: leaf springs, coil springs and torsion bars sized by energy stored per unit mass — automotive (Tata Motors, Mahindra, Ashok Leyland) and railway coach suspensions (ICF/RCF, LHB bogies).
• Fastener and pressure-part design: long resilient bolts in cylinder heads, flanges and turbine casings (BHEL, L&T) so shock energy is absorbed without yielding the thread root.
• Impact and forming machinery: pile hammers, drop hammers, forging presses, punching tools, rock breakers; crane buffers and railway buffer stops sized by kinetic energy.
• Crashworthiness and safety: crumple zones, bumper beams, helmet liners and seat-belt anchorages validated to AIS/BIS test standards; energy absorption per unit mass is the design metric.
• Materials QC and failure analysis: Charpy/Izod toughness testing to IS 1757 / IS 1598 in steel plants (SAIL, Tata Steel), boiler and pipeline fabrication, and weld-procedure qualification.
• Job roles: design engineer (springs, fasteners, structures), CAE/structural analyst (explicit dynamics for drop and crash), NDT and materials testing engineer, bridge and crane design engineer.

Interview questions

• Q: What is strain energy and how is it different from toughness? A: Strain energy is the recoverable elastic work stored in a strained body; toughness is the total energy absorbed up to fracture, including plastic work.
• Q: Why does a suddenly applied load produce twice the stress of a gradually applied one? A: The full load acts over the whole deflection, so work done is Pδ instead of ½Pδ; equating that to strain energy doubles the stress.
• Q: How do you handle an impact load in a hand calculation? A: Equate the energy lost by the striking body, W(h + δ) or ½mv², to the strain energy stored, then solve for stress — giving an equivalent static load.
• Q: Will a longer bar or a shorter bar survive a falling weight better? A: The longer bar, because it stores more energy at the same stress (U = σ²AL/2E), so the impact stress is lower.
• Q: Why are high-tensile bolts in shock-loaded joints made long and slender? A: More elastic volume at the same maximum stress means more energy absorbed; a short, stiff bolt reaches yield at a fraction of that energy.
• Q: Why is a notched or stepped member so weak under impact? A: Only the small or notched section reaches peak stress, so the rest of the body stores little energy, and the notch adds stress concentration.
• Q: State Castigliano's first theorem and one use. A: The deflection at a load point equals the partial derivative of total strain energy with respect to that load; used to find beam, frame and curved-bar deflections, with a dummy load where no load acts.
• Q: Why is the strain energy of a solid shaft only τmax²V/4G? A: Shear stress varies linearly from zero at the axis to τmax at the surface, so the average of τ² over the section is half of τmax².
• Q: Difference between Charpy and Izod tests? A: Charpy holds the notched specimen as a simple beam struck behind the notch (IS 1757); Izod holds it as a vertical cantilever struck above the notch (IS 1598).
• Q: Why does a steel bracket that is safe statically fail in a cold-weather impact? A: Below the ductile-brittle transition temperature the steel's absorbed energy collapses, so a notch can trigger brittle fracture at a load it carries easily when static or warm.
• Q: Is the energy method conservative or unsafe? A: Conservative — it assumes the entire kinetic energy enters the member with no loss and ignores member inertia and stress-wave effects, so real stresses are usually lower.
• Q: Where does distortion strain energy appear in design? A: It is the quantity limited by the von Mises (maximum distortion energy) theory of failure, the standard yield criterion for ductile metals.

Common mistakes

• Using U = ½Pδ for a suddenly applied load — the correct work done is Pδ, giving double the stress.
• Dropping the "1 +" in σ = (W/A)[1 + √(1 + 2hAE/WL)], or forgetting the bar's own extension δ in the energy balance W(h + δ).
• Taking strain energy of a shaft as τmax²V/2G instead of τmax²V/4G (forgetting the linear stress variation).
• Mixing units: h in mm with L in m, E in GPa with σ in MPa; keep everything in N and mm so U comes out in N·mm.
• Applying elastic strain-energy formulas past yield, or treating the Charpy energy in joules as a usable design stress.
• Differentiating a numerical value of U in Castigliano's method instead of keeping the load symbolic until after differentiation.
• Assuming a stiffer, thicker member is always better against impact — extra stiffness lowers δst and raises the impact factor.

Related topics

Strength of materials: stress and strain · Bending, torsion and columns · Deflection of beams · Principal stresses and theories of failure · Springs and spring design · Fatigue and fracture mechanics